Bukhovtsev-et-al-Problems-in-Elementary-Physics
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B. B. 6yX08I4e8. B. JI. KpU8IleHIC08.<br />
r. fl. MRKuwe8, B. n. llJaAbH08<br />
CBOPHHK 3A)lAq<br />
no 3JIEMEHTAPHOA
PROBLEMS<br />
IN ELEMENTARY<br />
PHYSICS<br />
B.BUKHOVTSEV. v. KRIVCHENKOV,<br />
G. MYAKISHEV, V. SHALNOV<br />
Translated from the Russian<br />
bll A~ T roitsky<br />
Translatton edited<br />
by<br />
G. Leib<br />
MIR<br />
PUBLISHERS • MOSCOW
First published 1971<br />
Second pr<strong>in</strong>t<strong>in</strong>g 1978<br />
This collection of 816 problems is based on the textbook<br />
"<strong>Elementary</strong> <strong>Physics</strong>" edited by Academician G. S. Landsberg.<br />
For this reason the content and nature of the problems and their<br />
arrangement ma<strong>in</strong>ly conform with this textbook. There is no section<br />
devoted to "Atomic <strong>Physics</strong>", however, s<strong>in</strong>ce the exercises <strong>in</strong><br />
Landsberg's book illustrate the relevant materi<strong>al</strong> <strong>in</strong> sufficient<br />
d<strong>et</strong>ail. Some problems on this subject have been <strong>in</strong>cluded <strong>in</strong> other<br />
chapters.<br />
The problems, most of which are unique, require a fundament<strong>al</strong><br />
knowledge of the basic laws of physics, and the ability<br />
to apply them <strong>in</strong> the most diverse conditions. A number of problems<br />
In the book have been revised from those used at the annu<strong>al</strong><br />
contests organized by the <strong>Physics</strong> faculty of the Moscow University.<br />
The solutions of <strong>al</strong>l the difficult problems are given <strong>in</strong> great<br />
d<strong>et</strong>ail. Solutions are <strong>al</strong>so given for some of the simpler ones.<br />
The book is recommended for self-education of senior pupils<br />
of gener<strong>al</strong> and speci<strong>al</strong> secondary and technic<strong>al</strong> schools. Many<br />
problems will be useful for first and second year students of<br />
higher schools.
CONTENTS<br />
Chapter 1. Mechanics. • • • • • . • . . • • . • • • . • •<br />
1-1. K<strong>in</strong>ematics of Uniform Rectil<strong>in</strong>ear Motion ..•.<br />
1-2. K<strong>in</strong>ematics of Non..Uniform and Uniformly Variable<br />
Rectil<strong>in</strong>ear Motion ••.•..•...••.••<br />
1-3. Dynamics of Rectil<strong>in</strong>ear Motion .<br />
1-4. The Law of Conservation of Momentum<br />
1-5. Statics . . . . • . . . . . . . . •<br />
1-6. Work and Energy •.•...•••<br />
1-7. K<strong>in</strong>ematics of Curvil<strong>in</strong>ear Motion .•<br />
1-8. Dynamics of Curvil<strong>in</strong>ear Motion<br />
1-9. The Law of Gravitation . • • • • • • .<br />
1-10. Hydro- and Aerostatics .•<br />
1-11. Hydro- and Aerodynamics<br />
<strong>Problems</strong><br />
Answers<br />
and<br />
solutions<br />
7 160<br />
7 160<br />
11 169<br />
15 174<br />
23 185<br />
25 188<br />
31 200<br />
38 214<br />
44 230<br />
53 248<br />
55 253<br />
62 261<br />
Chapter 2. Heat. Molecular <strong>Physics</strong> . 68 268<br />
2-1. Therm<strong>al</strong> Expansion of Solids and Liquids . . . .. 68 268<br />
2-2. The Law of Conservation of Energy. Therm<strong>al</strong> Con..<br />
ductivity. . . . . . . . • • • • . • . 69 270<br />
2-3. Properties of Gases . • • • • • • • • . 71 273<br />
2-4. Properties of Liquids . • . • . • . . . • . 76 283<br />
2-5. Mutu<strong>al</strong> Conversion of Liquids.and Solids 80 287<br />
2-6. Elasticity and Strength . • • 81 289<br />
2-7. Properties of Vapours . . • • • 84 291<br />
Chapter 3. Electricity and Magn<strong>et</strong>ism 87 294<br />
3-1. Electrostatics. • • • • • • • 87 294<br />
3-2. Direct Current • • • •• ..•... 97 317<br />
3-3. Electric Current <strong>in</strong> Gases and a Vacuum • . • • •• 107 336<br />
3-4. Magn<strong>et</strong>ic Field of a Current. Action of a Magn<strong>et</strong>ic<br />
Field on a Current and Mov<strong>in</strong>g Charges . . . . •• 112 341<br />
3-5. Electromagn<strong>et</strong>ic Induction. Alternat<strong>in</strong>g Current . •. 116 347<br />
3-6. Electric<strong>al</strong> Mach<strong>in</strong>es . • • • . • • • • • • . • . •. 123 359<br />
Chapter 4. Oscillatlons and Waves<br />
4-1. Mechanic<strong>al</strong> Oscillations<br />
4-2. Electric<strong>al</strong> Oscillations •<br />
4-3. Waves ••••••••<br />
127<br />
127<br />
131<br />
133<br />
365<br />
365<br />
373<br />
376
6<br />
Chapter 5. Geom<strong>et</strong>ric<strong>al</strong> Optics<br />
••••<br />
5-1. Photom<strong>et</strong>ry ....•••.<br />
5-2. Fundament<strong>al</strong> Laws of Optics • • • • •<br />
5.-3. Lenses and Spheric<strong>al</strong> Mirrors<br />
5-4. Optic<strong>al</strong> Systems and Devices<br />
Chapter 6. Physic<strong>al</strong> Optics • •<br />
~l. Interference of Light . • •.•••<br />
~2. Diffraction of Light. . . . • . . . . . . . . • • •<br />
6-3. Dispersion of Light and Colours of Bodies .<br />
<strong>Problems</strong><br />
CONTENTS<br />
Answen<br />
and<br />
solutions<br />
135 380<br />
135 380<br />
136 382<br />
142 395<br />
146 402<br />
151 421<br />
151 421<br />
156 429<br />
158 434
PROBLEMS<br />
CHAPTER 1<br />
MECHANICS<br />
1·1. K<strong>in</strong>ematics of Uniform; Rectil<strong>in</strong>ear Motion<br />
1. A motor-boat travell<strong>in</strong>g upstream m<strong>et</strong> rafts tioat<strong>in</strong>g downstream.<br />
One hour after this the eng<strong>in</strong>e of the. boat st<strong>al</strong>led.<br />
It took 30 m<strong>in</strong>utes to repair it, and dur<strong>in</strong>g this time the boat<br />
freely ftoated downstream. When the eng<strong>in</strong>e was repaired, the<br />
boat travelled downstream with the same speed relative to the<br />
current as before and overtook the rafts at a distance of s = 7.5 km<br />
from the po<strong>in</strong>t where they had m<strong>et</strong> the first time. D<strong>et</strong>erm<strong>in</strong>e<br />
the velocity of the river current, consider<strong>in</strong>g it constant.<br />
2. A man w<strong>al</strong>k<strong>in</strong>g with a speed v constant <strong>in</strong> magnitude arid<br />
direction passes under a lantern hang<strong>in</strong>g at a height H above<br />
the ground. F<strong>in</strong>d the velocity which the edge of the shadow of<br />
the man's head moves over the ground with if his height is h.<br />
3. The distance b<strong>et</strong>ween a town and a mill is 30 km. A man<br />
started to w<strong>al</strong>k from the mill to the town at 6:30 8. m., while<br />
a cyclist left the town for the mill at 6:40 a.m., rid<strong>in</strong>g at a<br />
speed of 18 krn/h. The man m<strong>et</strong> the cyclist after w<strong>al</strong>k<strong>in</strong>g 6 km,<br />
D<strong>et</strong>erm<strong>in</strong>e at what time they m<strong>et</strong> and the man's speed.<br />
Also f<strong>in</strong>d the place where the man was when he m<strong>et</strong> the<br />
twelfth bus com<strong>in</strong>g from the town and the number of -buses<br />
which overtook the cyclist if bus traffic beg<strong>in</strong>s at 6 8. m. The<br />
buses leave their term<strong>in</strong><strong>al</strong>s every 15 m<strong>in</strong>utes and their speed<br />
is 45 km/h, .<br />
• 4. Two tra<strong>in</strong>s left MosCow for Pushk<strong>in</strong>o at an <strong>in</strong>terv<strong>al</strong> of<br />
t= 10 m<strong>in</strong>utes and with 8 speed of V= 30 krn/h, What was the<br />
speed u of a tra<strong>in</strong> bound for Moscow if it m<strong>et</strong> these two tra<strong>in</strong>s<br />
at an <strong>in</strong>terv<strong>al</strong> of 't = 4 m<strong>in</strong>utes?<br />
5. An eng<strong>in</strong>eer works at a plant out-of-town. A car is sent<br />
for him from the plant every day that arrives at the railway<br />
station at the same time as the tra<strong>in</strong> he takes. One day the
8 PROBLEMS<br />
eng<strong>in</strong>eer arrived at the station one hour before his usu<strong>al</strong> time<br />
and, without wait<strong>in</strong>g for the car, started w<strong>al</strong>k<strong>in</strong>g to work"<br />
On his way. he m<strong>et</strong> the car and reached his plant 10 m<strong>in</strong>utes<br />
before the usu<strong>al</strong> time. How long did the eng<strong>in</strong>eer w<strong>al</strong>k before<br />
he m<strong>et</strong> the car? Give a graphic<strong>al</strong> solution.<br />
6. Two land<strong>in</strong>g-stages M and K are served by launches that<br />
<strong>al</strong>l travel at the same speed relative to the water. The distance<br />
b<strong>et</strong>ween the land<strong>in</strong>g-stages is 20 km. It is covered by each<br />
launch from M to K <strong>in</strong> one hour and from K to M <strong>in</strong> two hours.<br />
The .launches leave the two land<strong>in</strong>g-stages at the same time at<br />
an <strong>in</strong>terv<strong>al</strong> of 20 m<strong>in</strong>utes and stop at each of them <strong>al</strong>so for<br />
20 m<strong>in</strong>utes.<br />
D<strong>et</strong>erm<strong>in</strong>e: (1) the number of launches <strong>in</strong> service, (2) the number<br />
of launches m<strong>et</strong> by a launch travell<strong>in</strong>g from M to K, (3)<br />
the number of launches m<strong>et</strong> by a launch travell<strong>in</strong>g from K to M.<br />
7. Two tourists who are at a distance of 40 km from their<br />
camp must reach it tog<strong>et</strong>her <strong>in</strong> the shortest possible time. They<br />
have one bicycle which they decide to use iJ1. turn. One of them<br />
started w<strong>al</strong>k<strong>in</strong>g at a speed of VI = 5 km/h and the other rode<br />
off on the bicycle at a speed of v. = 15 krn/h. The tourists agreed<br />
to leave the bicycle at <strong>in</strong>termediate po<strong>in</strong>ts b<strong>et</strong>ween which one<br />
w<strong>al</strong>ks and the other rides. What will the mean speed of the<br />
tourists be? How long will the bicycle rema<strong>in</strong> unused?<br />
8. Two candles of equ<strong>al</strong> height h at the <strong>in</strong>iti<strong>al</strong> moment are<br />
at a distance of a from each other. The distance b<strong>et</strong>ween each<br />
candle and the nearest w<strong>al</strong>l is <strong>al</strong>so a (Fig. 1). With what speed<br />
will the shadows of the candles move <strong>al</strong>ong the w<strong>al</strong>ls if one<br />
candle burns down dur<strong>in</strong>g the time t, and the other dur<strong>in</strong>g the<br />
time tl?<br />
9. A bus is runn<strong>in</strong>g <strong>al</strong>ong a highway at a speed of VI = 16 m/s.<br />
A man is at a distance of a= 60 m<strong>et</strong>res from the highway and<br />
b= 400 m<strong>et</strong>res from the bus. In what direction should the man<br />
run to reach any po<strong>in</strong>t of the highway at the same time as<br />
the bus or before it? The man can run at a speed of VI = 4 <strong>in</strong>t».<br />
to. At what m<strong>in</strong>imum speed should the man run (see Problem 9)<br />
to be able to me<strong>et</strong> the bus, and <strong>in</strong> what direction?<br />
It. A man is on the shore of a lake at po<strong>in</strong>t A, and has<br />
to g<strong>et</strong> to po<strong>in</strong>t B on the lake <strong>in</strong> the shortest possible time<br />
(Fig. 2). The distance from po<strong>in</strong>t B to the shore Be = d and<br />
the distance AC= s. The man can swim <strong>in</strong> the water with<br />
a speed of VI and run <strong>al</strong>ong the shore with a speed of VI' greater<br />
than VI- Which way should he use-swim from po<strong>in</strong>t A straight
~ ,<br />
B"<br />
,,,,<br />
, ,<br />
I<br />
, I<br />
" d<br />
, I<br />
I<br />
I<br />
I<br />
I<br />
I<br />
I<br />
A....--8~C<br />
Fig. 1<br />
Fig. 2<br />
to B or run a certa<strong>in</strong> distance <strong>al</strong>ong the shore and then swim<br />
to po<strong>in</strong>t B?<br />
12. A motor-ship travell<strong>in</strong>g upstream me<strong>et</strong>s rafts float<strong>in</strong>g<br />
downstream. Forty-five m<strong>in</strong>utes (t 1<br />
) later the ship stops at a<br />
land<strong>in</strong>g-stage for I, = 1 hour. Then the ship travels downstream<br />
and overtakes the rafts <strong>in</strong> t a = 1 hour. The speed of the ship<br />
relative to the water is constant and equ<strong>al</strong> to VI = 10 krn/h.<br />
D<strong>et</strong>erm<strong>in</strong>e the river current velocity v" consider<strong>in</strong>g it constant.<br />
COnsider both graphic<strong>al</strong> and an<strong>al</strong>ytic<strong>al</strong> m<strong>et</strong>hods of solution.<br />
13. Mail is carried b<strong>et</strong>ween land<strong>in</strong>g-stages M and K by two<br />
launches. At the appo<strong>in</strong>ted time the launches leave their land<strong>in</strong>gstages,<br />
me<strong>et</strong> each other, exchange the mail and r<strong>et</strong>urn. If the<br />
launches leave their respective land<strong>in</strong>g-stages at the same time,<br />
the launch depart<strong>in</strong>g from M spends 3 hours both ways and that<br />
from K-1.5 hours. The speeds of both launches are the same<br />
relative to the water. D<strong>et</strong>erm<strong>in</strong>e graphic<strong>al</strong>ly how much later<br />
should the launch depart from M after the other one leaves K<br />
for the two to travel the same time.<br />
14. Use the conditions of the previous problem to d<strong>et</strong>erm<strong>in</strong>e<br />
the speed of the launches with respect to the water, the velocity<br />
of the river current and the po<strong>in</strong>t where the launches will<br />
me<strong>et</strong> if they leave their respective land<strong>in</strong>g-stages simultaneously.<br />
The distance b<strong>et</strong>ween the stages is 30 km.<br />
15. A rowboat travels from land<strong>in</strong>g-stage C to T at a speed<br />
of VI = 3 krn/h with respect to the water. At the same time a launch<br />
leaves T for C at a speed of VI = 10 km/h. While the boat moves
10<br />
PROBLEMS<br />
c<br />
rIII<br />
a<br />
Fig. 3<br />
01<br />
IIII<br />
~<br />
A<br />
B<br />
a<br />
B<br />
• j<br />
Fig. 4 Fig. 6
MEGHANICS 11<br />
the square, (2) the direction of the w<strong>in</strong>d co<strong>in</strong>cides with the diagon<strong>al</strong><br />
of the square? With no w<strong>in</strong>d the speed of the plane is v,<br />
greater than u.<br />
20. Two motor vehicles run at constant speeds Vi and V, <strong>al</strong>ong<br />
highways <strong>in</strong>tersect<strong>in</strong>g at an angle a. F<strong>in</strong>d the magnitude and<br />
direction of the speed of one vehicle with respect to the other.<br />
In what time after they me<strong>et</strong> at the <strong>in</strong>tersection will the distance<br />
b<strong>et</strong>ween the vehicles be s?<br />
21. Two <strong>in</strong>tersect<strong>in</strong>g straight l<strong>in</strong>es move translation<strong>al</strong>ly <strong>in</strong> opposite<br />
directions with velocities VI and V 2<br />
perpendicular to the<br />
correspond<strong>in</strong>g l<strong>in</strong>es. The angle b<strong>et</strong>ween the l<strong>in</strong>es is a. F<strong>in</strong>d the<br />
speed of the po<strong>in</strong>t of <strong>in</strong>tersection of these l<strong>in</strong>es .<br />
. 1-2. K<strong>in</strong>ematics of Non-Uniform and Uniformly<br />
Variable Rectil<strong>in</strong>ear Motion<br />
22. A motor vehicle travelled the first third of a distance s at<br />
a speed of V 1<br />
= 10 krn/h, the second third at a speed of V 2<br />
= 20 krn/h<br />
and the last third at a speed of Va = 60 krn/h. D<strong>et</strong>erm<strong>in</strong>e the<br />
mean speed of the vehicle over the entire distance s.<br />
23. D<strong>et</strong>erm<strong>in</strong>e the mean velocity and the mean acceleration<br />
of a po<strong>in</strong>t <strong>in</strong> 5 and 10 seconds if it moves as shown <strong>in</strong> Fig. 6.<br />
24. A man stands on a steep shore of a lake and pulls a boat<br />
<strong>in</strong> the water by a rope which he takes up at a constant speed of v.<br />
What will the speed of the boat be at the moment 'when the<br />
angle b<strong>et</strong>ween the rope and the surface of the water is a?<br />
25. A po<strong>in</strong>t source of light S is at a distance 1 from a vertic<strong>al</strong><br />
screen AB. An opaque object with a height h moves translation<strong>al</strong>ly<br />
at a constant speed v from the source to the screen <strong>al</strong>ong<br />
4"CIII!'<br />
5<br />
4<br />
~.~..6<br />
,~:.:Z<br />
. 'l,· ....<br />
,·~:::··t<br />
-<br />
tl IJ /<br />
d e<br />
c/ "r -, .f<br />
J<br />
'" ~..""....,,6' "<br />
·Ll -<br />
;,~.~<br />
t=5<br />
7<br />
tz=10<br />
~6ICIIIdI Fig. 6
12<br />
PROBLEMS<br />
....------1----....<br />
straight l<strong>in</strong>e SA. D<strong>et</strong>erm<strong>in</strong>e the <strong>in</strong>stantaneous velocity with which<br />
the upper edge of the shadow of the object moves <strong>al</strong>ong the<br />
screen (Fig. 7).<br />
26. The coord<strong>in</strong>ate of a po<strong>in</strong>t mov<strong>in</strong>g rectil<strong>in</strong>early <strong>al</strong>ong the<br />
x-axis changes <strong>in</strong> time accord<strong>in</strong>g to the law x = 1I +35t+41 t'<br />
(x is <strong>in</strong> centim<strong>et</strong>res and t is <strong>in</strong> seconds). F<strong>in</strong>d the velocity and<br />
acceleration of the po<strong>in</strong>t.<br />
27. Figures 8 and 9 show the velocity of a body and the change<br />
<strong>in</strong> its coord<strong>in</strong>ate (parabola) with time. The orig<strong>in</strong> of time read<strong>in</strong>g<br />
co<strong>in</strong>cides on both charts. Are the motions shown <strong>in</strong> the charts<br />
the same?<br />
28. Two motor vehicles left po<strong>in</strong>t A simultaneously and reached<br />
po<strong>in</strong>t B <strong>in</strong> to = 2 hours. The first vehicle travelled h<strong>al</strong>f of the<br />
distance at a speed of VI = 30 km/h and the other h<strong>al</strong>f at a speed<br />
of V 2<br />
= 45 krn/h, The second vehicle covered the entire distance<br />
with a constant acceleration. At what moment of time were the<br />
speeds of both the vehicles the same? Will one of them overtake<br />
the other en route?<br />
Fig. 8
4cm<br />
10<br />
A<br />
5<br />
10<br />
~ IICOfId8 Fig. 9<br />
29. A b<strong>al</strong>l freely drops from a height H 'onto an elastic,horizont<strong>al</strong><br />
support. Plot charts show<strong>in</strong>g the change <strong>in</strong> the coord<strong>in</strong>ate<br />
and velocity of the b<strong>al</strong>l versus the time neglect<strong>in</strong>g the duration<br />
Qf the collision. The impact is absolutely elastic.<br />
30.. Two steel b<strong>al</strong>ls freely drop onto an elastic plate, the first<br />
one from a height of hi = 44 em and the second <strong>in</strong> T > 0 seconds<br />
after the first from a height of h 2 = 11 em. After a certa<strong>in</strong> time<br />
the velocities of the b<strong>al</strong>ls co<strong>in</strong>cide <strong>in</strong> magnitude and direction.<br />
F<strong>in</strong>d the time 't and the <strong>in</strong>terv<strong>al</strong> dur<strong>in</strong>g which the velocities of<br />
the two b<strong>al</strong>ls rema<strong>in</strong> the same. The b<strong>al</strong>ls do not collide.<br />
31. How long will a body freely f<strong>al</strong>l<strong>in</strong>g without any <strong>in</strong>iti<strong>al</strong><br />
velocity pass the n-th centim<strong>et</strong>re of its path?<br />
32. Two bodies are thrown one after the other with the same<br />
velocities V o from a high tower. The first body is thrown vertic<strong>al</strong>ly<br />
upward, and the second one vertic<strong>al</strong>ly downward after the<br />
time 't'. D<strong>et</strong>erm<strong>in</strong>e the velocities of the bodies with respect to<br />
each other and the distance b<strong>et</strong>ween them at the moment of<br />
time t >T.<br />
33. At the <strong>in</strong>iti<strong>al</strong> moment three po<strong>in</strong>ts A, Band C are on a<br />
horizont<strong>al</strong> straight l<strong>in</strong>e at equ<strong>al</strong> distances from one another.<br />
Po<strong>in</strong>t A beg<strong>in</strong>s to move vertic<strong>al</strong>ly upward with a constant vee<br />
locity v and po<strong>in</strong>t C vertic<strong>al</strong>ly downward without any <strong>in</strong>iti<strong>al</strong>
14 PROBLEMS<br />
velocity at a constant acceleration a. How should po<strong>in</strong>t B move<br />
vertic<strong>al</strong>ly for <strong>al</strong>l the three po<strong>in</strong>ts to constantly be on one straight<br />
l<strong>in</strong>e? The po<strong>in</strong>ts beg<strong>in</strong> to move simultaneously.<br />
34. Two trucks tow a third one by means of a pulley tied<br />
to it (Fig. 10). The accelerations of the trucks are a 1<br />
and a 2<br />
•<br />
F<strong>in</strong>d the acceleration as of the truck be<strong>in</strong>g towed.<br />
35. A lift moves with an acceleration a. A passenger <strong>in</strong> the<br />
lift drops a book. What is the acceleration of the book with<br />
respect to the lift floor if: (1) the lift is go<strong>in</strong>g up, (2) the lift<br />
is go<strong>in</strong>g down?<br />
36. A railway carriage moves over a straight level track with<br />
an acceleration a. A passenger <strong>in</strong> the carriage drops a stone.<br />
What is the acceleration of the stone with respect to the carriage<br />
and the Earth?<br />
37. A man <strong>in</strong> a lift mov<strong>in</strong>g with an acceleration a drops<br />
a b<strong>al</strong>l from a height H above the floor. In t seconds the acceleration<br />
of the lift is reversed, and <strong>in</strong> 2t seconds becomes equ<strong>al</strong><br />
to zero. Directly after this the b<strong>al</strong>l touches the floor. What<br />
height from the floor of the lift will the b<strong>al</strong>l jump to after the<br />
impact? Consider the impact to be absolutely elastic.<br />
38. An overhead travell<strong>in</strong>g crane lifts a load from the ground<br />
with an upward acceleration of a l<br />
• At the same time the hook<br />
of the crane carry<strong>in</strong>g the load moves <strong>in</strong> a horizont<strong>al</strong> direction<br />
with an acceleration a, relative to the crane. Besides,<br />
the crane runs on its rails with a constant speed<br />
Vo (Fig. 11). The <strong>in</strong>iti<strong>al</strong> speed of the hook relative<br />
to the crane is zero. F<strong>in</strong>d the speed of the load<br />
with respect to the ground when it reaches the<br />
height h.<br />
39•. Body A is placed on a wedge form<strong>in</strong>g an<br />
angle a with the horizon (Fig. 12). What accele-<br />
Fig. 10<br />
Fig. 1J
MECHANICS<br />
15<br />
Fig. 12 Fig. 13<br />
ration should be imparted to the wedge <strong>in</strong> a horizont<strong>al</strong> direction<br />
for body A to freely f<strong>al</strong>l vertic<strong>al</strong>ly?<br />
1-3. Dynamics of Rectil<strong>in</strong>ear Motion<br />
40. A force F is applied to the centre of a homogeneous<br />
sphere (Fig. 13). In what direction will the sphere move?<br />
41. Six forces ly<strong>in</strong>g <strong>in</strong> one plane and form<strong>in</strong>g angles of 600<br />
relative to one another are applied to the centre of a homogeneous<br />
sphere with a mass of m = 4 kg. These forces are consecutively<br />
1, 2, 3, 4, 5 and 6 kgf (Fig. 14). In what direction and<br />
with what acceleration will the sphere move?<br />
42. How much does a body with a mass of one kilogram<br />
weigh?<br />
43. The resistance of the air causes a body thrown at a certa<strong>in</strong><br />
angle to the horizon to fly <strong>al</strong>ong a b<strong>al</strong>listic curve. At what<br />
angle to the horizon is the acceleration of the body directed at<br />
Bkgr<br />
A<br />
Flg. 14<br />
Fig. 15<br />
B
16 PROBLEMS<br />
the highest po<strong>in</strong>t of the trajectory A, if the mass of the body<br />
is m and the resistance of the air at this po<strong>in</strong>t is F?<br />
44. A disk arranged <strong>in</strong> a vertic<strong>al</strong> plane has sever<strong>al</strong> grooves<br />
directed <strong>al</strong>ong chords drawn from po<strong>in</strong>t A (Fig. 15). Sever<strong>al</strong> bodies<br />
beg<strong>in</strong> to slide down the respective grooves from po<strong>in</strong>t A<br />
simultaneously. In what time will each body reach the edge of<br />
the disk? Disregard friction and the resistance of the air (G<strong>al</strong>ileo's<br />
problem).<br />
45. What is the m<strong>in</strong>imum force of air resistance act<strong>in</strong>g on a<br />
parachutist and his parachute when the latter is compl<strong>et</strong>ely opened?<br />
The two weigh 75 kgf.<br />
46. What is the pressure force N exerted by a load weigh<strong>in</strong>g<br />
a kgf on the floor of a lift if the acceleration of the lift is a?<br />
What is this force equ<strong>al</strong> to upon free f<strong>al</strong>l<strong>in</strong>g of the lift?<br />
47. A puck with an <strong>in</strong>iti<strong>al</strong> velocity of 5 mls travels over<br />
a distance of 10 m<strong>et</strong>res and strikes the boards. After the impact<br />
which may be considered as absolutely elastic, the puck travelled<br />
another 2.5 m<strong>et</strong>res and stopped. F<strong>in</strong>d the coefficient of friction<br />
b<strong>et</strong>ween the puck and the ice. Consider the force of slid<strong>in</strong>g<br />
friction <strong>in</strong> this and subsequent problems to be equ<strong>al</strong> to the<br />
maximum force of friction of rest.<br />
48. The path travelled by a motor vehicle from the moment<br />
the brakes are applied until it stops is known as the brak<strong>in</strong>g<br />
distance. For the types of tyres <strong>in</strong> common use <strong>in</strong> the USSR<br />
and a norm<strong>al</strong> air pressure <strong>in</strong> the tubes, the dependence of the<br />
brak<strong>in</strong>g distance on the speed of the vehicle at the beg<strong>in</strong>n<strong>in</strong>g<br />
of brak<strong>in</strong>g and on the state and type of the pavement can be<br />
tabulated (see Table 1).<br />
F<strong>in</strong>d the coefficient of friction for the various k<strong>in</strong>ds of pavement<br />
surface to an accuracy of the first digit after the decim<strong>al</strong><br />
po<strong>in</strong>t, us<strong>in</strong>g the data <strong>in</strong> this table.<br />
49. D<strong>et</strong>erm<strong>in</strong>e the difference <strong>in</strong> the pressure of p<strong>et</strong>rol on the<br />
opposite w<strong>al</strong>ls of a fuel tank when a motor vehicle travels over<br />
a horizont<strong>al</strong> road if its speed <strong>in</strong>creases uniformly from Vo= 0<br />
to v dur<strong>in</strong>g t seconds. The distance b<strong>et</strong>ween the tank w<strong>al</strong>ls is I.<br />
The tank has the form of a par<strong>al</strong>lelepiped and its side w<strong>al</strong>ls are<br />
vertic<strong>al</strong>. It is compl<strong>et</strong>ely filled with p<strong>et</strong>rol. The density of p<strong>et</strong><br />
'rol is p.<br />
50. A homogeneous rod with a length L: is acted upon by<br />
two forces F 1 and F" applied to its ends and directed oppositely<br />
(Fig. 16). With what force F will the rod be str<strong>et</strong>ched <strong>in</strong> the<br />
cross section at a distance I from one of the ends?
MECHANICS<br />
17<br />
Table 1. Brak<strong>in</strong>g Distance In M<strong>et</strong>res<br />
Pavement<br />
I<br />
Speed. krn/h<br />
surface 10 I 20 I 30 I 40 I 50 I 60 I 70 I 80 I 90 I 100<br />
Ice 13.9115.6135.3162.9198.11141.41192.61251.61318.2' 393.0<br />
Dry snow II. 91 7·81 17· 61 3 1. 41 49. 0 I 70·71 96. 31 125 . 81 159 . I 196 . 6<br />
W<strong>et</strong> wood<br />
block 1.3 5.2 11.7 20.9 32.7 47.1 64.2 83.8 106.0 131.0<br />
Dry wood<br />
block 0.78 3.1 7.0 12.5 19.6 28.2 38.5 50.3 63.6 78.6<br />
W<strong>et</strong><br />
asph<strong>al</strong>t 0:97 3.9 8.8 15.7 24.5 35.3 48.1 62.9 79.5 98.2<br />
Dry asph<strong>al</strong>t 1 0 . 65 1 2.61 5.81 10.4 1 16. 3 1 23.51 32.1\ 4~ .91 53.0I 65.5<br />
D~ concr<strong>et</strong>e1 0 . 561 2.21 5.0I9.01 14. 0 I 20.21 27. 5 1 35.91 45.41 56.1<br />
51. A light b<strong>al</strong>l is dropped <strong>in</strong> air and photographed after it<br />
covers a distance of 20 m<strong>et</strong>res. A camera with a foc<strong>al</strong> length<br />
of 10 em is placed 15 m<strong>et</strong>res away from the plane which the<br />
b<strong>al</strong>l is f<strong>al</strong>l<strong>in</strong>g <strong>in</strong>. A disk with eight equ<strong>al</strong>ly spaced holes arranged<br />
<strong>al</strong>ong its circumference rotates with a speed of 3 test» <strong>in</strong><br />
front of the open lens of the camera. As a result a number of<br />
images of the b<strong>al</strong>l spaced 3 mm apart are produced on the film.<br />
Describe the motion of the b<strong>al</strong>l. What is the f<strong>in</strong><strong>al</strong> velocity of<br />
another b<strong>al</strong>l of the same radius, but with a mass four times<br />
greater than that of the first one? D<strong>et</strong>erm<strong>in</strong>e the coefficient of<br />
I~ l ~I<br />
I<br />
-t:<br />
.. ..<br />
F;<br />
I i I ~<br />
I<br />
... ~I<br />
t<br />
• Fig. 16<br />
2-2042
18<br />
PROBLEM~<br />
friction if the weight of the first b<strong>al</strong>l is 4.5 gf. At high velocities<br />
of f<strong>al</strong>l<strong>in</strong>g, the resistance of the air is proportion<strong>al</strong> to the<br />
square of the velocity.<br />
52. Two weights with masses m 1 = 100 g and m 2<br />
= 200 g are<br />
suspended from the ends of a str<strong>in</strong>g passed over a stationary<br />
pulley at a height of H = 2 m<strong>et</strong>res from the floor (Fig. 17).<br />
At the <strong>in</strong>iti<strong>al</strong> moment the weights are at rest. F<strong>in</strong>d the tension<br />
of the str<strong>in</strong>g when the weights move and the time dur<strong>in</strong>g which<br />
the weight m, reaches the floor. Disregard the mass of the pulley<br />
and· the str<strong>in</strong>g.<br />
53. A weight G is attached' to the axis of a mov<strong>in</strong>g pulley<br />
(Fig. 18). What force F should be applied to the end of the<br />
rope passed around the second pulley for the weight G to move<br />
upwards with an acceleration of a? For the weight G to be at<br />
rest? Disregard the mass of the pulleys and the rope.<br />
54. Two weights are suspended from a str<strong>in</strong>g thrown over<br />
a stationary pulley. The mass of one weight is 200 g. The<br />
str<strong>in</strong>g will not break if a very heavy weight is attached to its<br />
other end. What tension is the str<strong>in</strong>g designed for? Disregard<br />
the mass of the pulley and the str<strong>in</strong>g.<br />
55. Two pans with weights each equ<strong>al</strong> to a= 3 kgf are suspended<br />
from the ends of a str<strong>in</strong>g passed over two stationary<br />
pulleys. The str<strong>in</strong>g b<strong>et</strong>ween the pulleys is cut and the free ends<br />
are connected to a dynamom<strong>et</strong>er (Fig. 19). What will the dynamom<strong>et</strong>er<br />
show? What weight a 1 should be added to one of the<br />
pans for the read<strong>in</strong>g of the dynamom<strong>et</strong>er not to change after<br />
a weight O 2<br />
= 1 kgf is taken off the other pan? Disregard the<br />
'masses of the pans, pulleys, thread and dynamom<strong>et</strong>er.<br />
F<br />
Fig. 17 Fig. 18<br />
Fig. 19
MECHANICS 19<br />
Fig. 20<br />
56. A heavy sphere with a mass m is suspended on a th<strong>in</strong><br />
rope. Another rope as strong as the first one is attached to the<br />
bottom of the sphere. When the lower rope is sharply pulled<br />
it breaks. What accelera tion will be imparted to the sphere?<br />
57. Two weights with masses m 1 and m, are connected by<br />
a str<strong>in</strong>g pass<strong>in</strong>g over a pulley. The surfaces on which they rest<br />
form angles a and ~ with the horizont<strong>al</strong> (Fig. 20). The righthand<br />
weight is h m<strong>et</strong>res below the left-hand one. Both weights<br />
will be at the same height <strong>in</strong> 't seconds after motion beg<strong>in</strong>s.<br />
The coefficients of friction b<strong>et</strong>ween the weights and the surfaces<br />
are k. D<strong>et</strong>erm<strong>in</strong>e the relation b<strong>et</strong>ween the masses of the<br />
weights.<br />
58. A slide forms an angle of a = 30° with the horizon.<br />
A stone is thrown upward <strong>al</strong>ong it and covers a di-t.iuco of<br />
1= 16 m<strong>et</strong>res <strong>in</strong> t, == 2 seconds, after which it slides down. Wha t<br />
time t 2<br />
is required for the r<strong>et</strong>urn motion? What is the coefficient<br />
of friction b<strong>et</strong>ween the slide and the stone?<br />
59. A cart with a mass of M = 500 g is connected by a str<strong>in</strong>g<br />
to a weight hav<strong>in</strong>g a mass m = 200 g. At the <strong>in</strong>iti<strong>al</strong> moment<br />
the cart moves to the left <strong>al</strong>ong a horizont<strong>al</strong> plane at a speed<br />
of V o = 7 m/s (Fig. 21). F<strong>in</strong>d the magni tude and direction of<br />
I{,<br />
~<br />
Fig. 21<br />
Fig. 22<br />
.) .a-
20<br />
PROBLEMS<br />
Fig. 28<br />
H<br />
11/<br />
r<br />
~--I----'"<br />
F_<br />
m,<br />
Fig. 24<br />
the speed of the cart, the place it will be at and the distance<br />
it covers <strong>in</strong> t = 5 seconds.<br />
60. Can an ice-boat travel over a level surface faster than<br />
the w<strong>in</strong>d which it is propelled by?<br />
61. The fuel supply of a rock<strong>et</strong> is m=8 tons and its mass<br />
(<strong>in</strong>clud<strong>in</strong>g the fuel) is M = 15 tons. The fuel burns <strong>in</strong> 40 seconds.<br />
The consumption of fuel and the thrust F=20,OOO kgf are constant.<br />
(1) The rock<strong>et</strong> is placed horizont<strong>al</strong>ly on a trolley. F<strong>in</strong>d its<br />
acceleration at the moment of launch<strong>in</strong>g. F<strong>in</strong>d how the acceleration<br />
of the rock<strong>et</strong> depends on the duration of its motion<br />
and show the relation graphic<strong>al</strong>ly. Use the graph to f<strong>in</strong>d the<br />
velocity acquired by the rock<strong>et</strong> <strong>in</strong> 20 seconds after it beg<strong>in</strong>s<br />
to move. Disregard friction.<br />
(2) The rock<strong>et</strong> is launched vertic<strong>al</strong>ly upward. Measurements<br />
show that <strong>in</strong> 20 seconds the acceleration of the rock<strong>et</strong> is 0.8 g.<br />
C<strong>al</strong>culate the force of air resistance which acts on the rock<strong>et</strong><br />
at this moment. Consider the acceleration g to be constant.<br />
(3) The acceleration of the rock<strong>et</strong> is measured by an <strong>in</strong>strument<br />
hav<strong>in</strong>g the form of a spr<strong>in</strong>g secured <strong>in</strong> a vertic<strong>al</strong> tube.<br />
When at rest, the spr<strong>in</strong>g is str<strong>et</strong>ched a distance of 1 0<br />
= 1 em<br />
by a weight secured to its end. D<strong>et</strong>erm<strong>in</strong>e the relation b<strong>et</strong>ween<br />
the str<strong>et</strong>ch<strong>in</strong>g of the spr<strong>in</strong>g and the acceleration of the rock<strong>et</strong>.<br />
Draw the sc<strong>al</strong>e of the <strong>in</strong>strument.<br />
62. A bead of rnass In is fitted onto a rod with a length of<br />
21, and can move on it without friction. At the <strong>in</strong>iti<strong>al</strong> mo-
. M.ECHANICS 21<br />
ment the bead is <strong>in</strong> the middle of the rod. The rod moves<br />
translation<strong>al</strong>ly <strong>in</strong> a horizont<strong>al</strong> plane with an acceleration a <strong>in</strong><br />
a direction form<strong>in</strong>g an angle ex with the rod (Fig. 22). F<strong>in</strong>d<br />
the acceleration of the bead relative to the rod, the reaction<br />
force exerted by the rod on the bead, and the time when the<br />
bead will leave the rod.<br />
63. Solve the previous problem, assum<strong>in</strong>g that the mov<strong>in</strong>g<br />
bead is acted upon by a friction force. The coefficient of friction<br />
b<strong>et</strong>ween the bead and the rod is k. Disregard the force of<br />
gravity.<br />
64. A block with the mass M rests on a smooth horizont<strong>al</strong><br />
surface over which it can move without friction. A body with<br />
the mass m lies on the block (Fig. 23). The coefficient of friction<br />
b<strong>et</strong>ween the body and the block is k. At what force F<br />
applied to the block <strong>in</strong> a horizont<strong>al</strong> direction will the body<br />
beg<strong>in</strong> to slide over the block? In what time will the body f<strong>al</strong>l<br />
from the block if the length of the latter is I?<br />
65. A wagon with the mass M moves without friction over<br />
horizont<strong>al</strong> rails at a speed of V O • A body with the mass m is<br />
placed on the fran t edge of the wagon. The <strong>in</strong>iti<strong>al</strong> speed of<br />
the body is zero. At what length of the wagon will the body<br />
not slip off it? Disregard the dimensions of the body as compared<br />
with the length 1 of the wagon. The coefficient of friction<br />
b<strong>et</strong>ween the body and the wagon is k.<br />
66. A weightless str<strong>in</strong>g thrown over a stationary pulley is<br />
passed through a slit (Fig. 24). As the str<strong>in</strong>g moves it is acted<br />
upon by a constant friction force F on the side of the slit. The<br />
ends of the str<strong>in</strong>g carry two weights with masses m 1<br />
and m.,<br />
respectively. F<strong>in</strong>d the acceleration a of the weights.<br />
67. A stationary pulley is secured- to the end of a light bar.<br />
The bar is placed onto a b<strong>al</strong>ance pan and secured <strong>in</strong> a vertic<strong>al</strong><br />
direction. Different weights are attached to the ends of a str<strong>in</strong>g<br />
passed over the pulley. One of the weights slides over the bar<br />
with friction and therefore both weights move uniformly<br />
(Fig. 25). D<strong>et</strong>erm<strong>in</strong>e the force 'which the pulley acts on the bar<br />
with and .the read<strong>in</strong>gs of the b<strong>al</strong>ance when the weights move.<br />
Disregard the masses of the pulley, bar, str<strong>in</strong>g and the friction<br />
<strong>in</strong> the pulley axis. Consider two cases: (1) m 1<br />
= 1 kg, m 2<br />
= 3 kg,<br />
and (2) m 1 = 3 kg, m 2<br />
= 1 kg.<br />
68. A system consists of two stationary and one movable<br />
. pulleys (Fig. 26). A str<strong>in</strong>g thrown over them carries at its ends<br />
weights with masses m 1<br />
and m s , while a weight with a mass
22 PROBLEMS<br />
Fig. 25<br />
Fig. 26<br />
B<br />
m; is attached to the axis of the movable<br />
pulley. The parts of the str<strong>in</strong>g that are<br />
not on the pulleys are vertic<strong>al</strong>. F<strong>in</strong>d the<br />
acceleration of each weight, neglect<strong>in</strong>g<br />
the masses of the pulleys and the str<strong>in</strong>g,<br />
and <strong>al</strong>so friction.<br />
69. Two monkeys of the same weight<br />
are hang<strong>in</strong>g at the ends of a rope thrown<br />
over a stationary pulley. One monkey<br />
beg<strong>in</strong>s to climb the rope and the other<br />
stays where it is. Where will the second<br />
15kg,<br />
Fig. Z7<br />
Fig. 28
MECHANICS 23<br />
monkey be when the first one reaches the pulley? At the <strong>in</strong>iti<strong>al</strong><br />
moment both monkeys were at the same height from the floor.<br />
J)isregard the mass of the pulley and rope, and <strong>al</strong>so friction.<br />
70. D<strong>et</strong>erm<strong>in</strong>e the accelerations of the weights <strong>in</strong> the pulley<br />
system depicted <strong>in</strong> Fig. 27. Disregard the masses of the pulleys<br />
and str<strong>in</strong>g, and <strong>al</strong>so friction. In what direction will the pulleys<br />
rotate when the weights move?<br />
71. A table with a weight of at = 15 kgf can move without<br />
friction over a level floor. A weight of O 2<br />
= 10 kgf is placed<br />
on the table, and a rope passed over two pulleys fastened<br />
to the table is attached to it (Fig. 28). The coefficient of friction<br />
b<strong>et</strong>ween the weight and the table k = 0.6. What acceleration<br />
will the table move with if a constant force of 8 kgf is applied<br />
to the free end of the rope? Consider two cases: (1) the force<br />
is directed horizont<strong>al</strong>ly, (2) the force is directed vertic<strong>al</strong>ly upward.<br />
72. An old cannon without a counter-recoil device rests on<br />
a horizont<strong>al</strong> platform. A b<strong>al</strong>l with a mass m and an <strong>in</strong>iti<strong>al</strong><br />
velocity V o is fired at an angle of a to the horizon. What velocity<br />
V 1<br />
will be imparted to the cannon directly after the shot<br />
if the mass of the cannon is M and the acceleration of the b<strong>al</strong>l<br />
<strong>in</strong> the barrel is much greater than that of free f<strong>al</strong>l? The coefficient<br />
of friction b<strong>et</strong>ween the cannon and the platform is k.<br />
1..4. The Law of Conservation of Momentum<br />
73. A m<strong>et</strong>eorite burns <strong>in</strong> the atmosphere before it reaches the<br />
Earth's surface. What happens to its momentum?<br />
74. Does a homogeneous disk revolv<strong>in</strong>g about its axis have<br />
any momentum? The axis of the disk is stationary.<br />
75. The horizont<strong>al</strong> propeller of a helicopter can be driven by<br />
an eng<strong>in</strong>e mounted <strong>in</strong>side its fuselage or by the reactive forces<br />
of the gases ejected from speci<strong>al</strong> nozzles at the ends of the<br />
propeller blades. Why does a propeller-eng<strong>in</strong>e helicopter need<br />
a tail. rotor while a j<strong>et</strong> helicopter does not need it?<br />
76. A hunter discharges his gun from a light <strong>in</strong>flated boat.<br />
What velocity will be imparted to the boat when the gun is<br />
fired if the mass of the hunter and the boat is M = 70 kg, the<br />
mass of the shot m = 35 g and the mean <strong>in</strong>iti<strong>al</strong> velocity of the<br />
shot Vo= 320 m/s? The barrel of the gun is directed at an angle<br />
of '(x = 600 to the horizon.<br />
77. A rock<strong>et</strong> launched vertic<strong>al</strong>ly upward explodes at the highest<br />
po<strong>in</strong>t it reaches. The explosion produces three fr~ents. Prove
24 PROBLEMS<br />
~<br />
1 2<br />
•<br />
~<br />
Fig. 29<br />
that the vectors of the <strong>in</strong>iti<strong>al</strong> velocities of <strong>al</strong>l three fragments<br />
are <strong>in</strong> one plane.<br />
. 78. A man <strong>in</strong> a boat fac<strong>in</strong>g the bank with its stern w<strong>al</strong>ks to<br />
the bow. How will the distance b<strong>et</strong>ween the man and the bank<br />
change?<br />
79. A boat on a lake is perpendicular to the shore and faces<br />
it with its bow. The distance b<strong>et</strong>ween the bow and the shore is<br />
0.75 m<strong>et</strong>re. At the <strong>in</strong>iti<strong>al</strong> moment the boat was stationary. A man<br />
<strong>in</strong> the boat steps from its bow to its stern. Will the boat reach<br />
the shore if it is 2 m<strong>et</strong>res long? The mass of the boat M = 140 kg<br />
and that of the man m = 60 kg.<br />
80. Two identic<strong>al</strong> weights are connected by a spr<strong>in</strong>g. At the<br />
<strong>in</strong>iti<strong>al</strong> moment the spr<strong>in</strong>g is so compressed that the first weight<br />
is tightly pressed aga<strong>in</strong>st a w<strong>al</strong>l (Fig. 29) and the second weight<br />
is r<strong>et</strong>a<strong>in</strong>ed by a stop. How will the weights move if the stop is<br />
removed?<br />
81. A massive homogeneous cyl<strong>in</strong>der that can revolve without<br />
friction around a horizont<strong>al</strong> axis is secured on a cart stand<strong>in</strong>g on<br />
a smooth level surface (Fig. 30). A bull<strong>et</strong> fly<strong>in</strong>g horizont<strong>al</strong>ly<br />
with a velocity v strikes the cyl<strong>in</strong>der and drops onto the cart.<br />
Does the speed of the cart that it acquires after the impact<br />
depend on the po<strong>in</strong>t where the bull<strong>et</strong> strikes the cyl<strong>in</strong>der?<br />
82. At the <strong>in</strong>iti<strong>al</strong> moment a rock<strong>et</strong> with a mass M had a<br />
velocity VO' At the end of each second the rock<strong>et</strong> ejects a portion<br />
Fig. 30 Fig. 31
MECHANICS 25<br />
of gas with a mass m. The velocity of a portion of gas differs<br />
from that of the rock<strong>et</strong> before the given portion of gas burns by<br />
a constant v<strong>al</strong>ue u, i. e., the velocity of gas outflow is constant.<br />
D<strong>et</strong>erm<strong>in</strong>e the velocity of the rock<strong>et</strong> <strong>in</strong> n seconds, disregard<strong>in</strong>g<br />
the force of gravity.<br />
83. Will the velocity of a rock<strong>et</strong> <strong>in</strong>crease if the outflow velocity<br />
of the gases with respect to the rock<strong>et</strong> is sm<strong>al</strong>ler than the<br />
velocity of the rock<strong>et</strong> itself, so that the gases ejected from the<br />
nozzle follow the rock<strong>et</strong>?<br />
84. Two boats move towards each other <strong>al</strong>ong par<strong>al</strong>lel paths<br />
with the same velocities. When they me<strong>et</strong>, a sack is thrown from<br />
one boat onto the other and then an identic<strong>al</strong> sack is thrown from<br />
the second onto the first. The next time this is done simultaneously.<br />
When will the velocity of the boats be greater after the<br />
. sacks are thrown?<br />
85. A hoop is placed. on an absolutely smooth level surface.<br />
A be<strong>et</strong>le <strong>al</strong>ights on the hoop. What trajectory will be described.<br />
by the be<strong>et</strong>le and the centre of the hoop if the be<strong>et</strong>le beg<strong>in</strong>s to<br />
move <strong>al</strong>ong the hoop? The radius of the hoop is R, its mass -Is<br />
M and the mass of the be<strong>et</strong>le m.<br />
86. A wedge with an angle a at the base can move without<br />
friction over a smooth leveI surface (Fig. 31). At what ratio<br />
b<strong>et</strong>ween the masses m 1 and m, of the weights, that are connected<br />
by a str<strong>in</strong>g passed over a pulley, will the wedge rema<strong>in</strong> stationary,<br />
and at what ratio will it beg<strong>in</strong> to move to the right or<br />
left? The coefficient of friction b<strong>et</strong>ween the weight of mass m s<br />
and the wedge is k,<br />
1-5. Statics<br />
87. A homogeneous cha<strong>in</strong> with a length 1 lies ·on a table. What<br />
is the maximum length l1 of the part of the cha<strong>in</strong> hang<strong>in</strong>g over<br />
the table if the coefficient of friction b<strong>et</strong>ween the cha<strong>in</strong> and the<br />
table is k?<br />
88. Two identic<strong>al</strong> weights are suspended from the ends of a<br />
str<strong>in</strong>g thrown over two pulleys (Fig. 32). Over what distance<br />
will a third weight of the same mass lower if it is attached to<br />
the middle of the str<strong>in</strong>g? The distance b<strong>et</strong>ween the axes of the<br />
. pulleys is 21. The friction <strong>in</strong> the axes of the pulleys is negligible.<br />
89. An isosceles wedge with an acute angle a, is driven <strong>in</strong>to<br />
a slit. At what angle a will the wedge not be forced out of the<br />
slit if the coefficient of friction b<strong>et</strong>weenthe wedge and the slit is k?
2(;<br />
PROBLEMS<br />
m<br />
m<br />
Fig. 32<br />
Fig. 33<br />
90. What is the ratio b<strong>et</strong>ween the weights 0 1<br />
and G z if the<br />
system shown <strong>in</strong> Fig. 33 is <strong>in</strong> equilibrium. Bars AD, BC, CH,<br />
DI and arm 00. of the lever are twice as long as bars AE, EB,<br />
IJ, JH and arm FO, respectively. Disregard the weight of the<br />
bars and the arm.<br />
91. A horizont<strong>al</strong> force F is applied perpendicularly to an upper<br />
edge of a rectangular box with a length 1 and a height h to<br />
move it. What should the coefficient of friction k b<strong>et</strong>ween the<br />
box and the floor be so that the box moves without overturn<strong>in</strong>g?<br />
92. A homogeneous beam whose weight is G lies on a floor.<br />
The coefficient of friction b<strong>et</strong>ween the beam and the floor is k.<br />
What is easier for two men to do-turn the beam about its<br />
centre or move it translation<strong>al</strong>ly?<br />
93. An overhead travell<strong>in</strong>g crane (see Fig. 11) weigh<strong>in</strong>g a= 2<br />
tonf has a span of L = 26 m<strong>et</strong>res. The wire rope carry<strong>in</strong>g a load<br />
is at a distance of 1= 10 m<strong>et</strong>res from one of the rails. D<strong>et</strong>erm<strong>in</strong>e<br />
the force of pressure of the crane on the rails if it lifts a<br />
load of Go = 1 tonf with an acceleration of a = 9.8 m/s".<br />
94. A lever is so bent that its sides AR, BC and CD are<br />
equ<strong>al</strong> and form right angles with one another (Fig. 34). The axis<br />
of the lever is at po<strong>in</strong>t B. A force of 0= 1 kgf is applied at<br />
po<strong>in</strong>t A at right angles to arm AB. F<strong>in</strong>d the m<strong>in</strong>imum force that<br />
should be applied at D for the lever to be <strong>in</strong> equilibrium.<br />
Disregard the weight of the lever.<br />
95. A rod not reach<strong>in</strong>g the tloor is <strong>in</strong>serted b<strong>et</strong>ween two identic<strong>al</strong><br />
boxes (Fig. 35). A horizont<strong>al</strong> force F is applied to the<br />
upper end of the rod. Which of the boxes will move first?
MECHANICS 27<br />
A<br />
B<br />
C'- D Fig. 34<br />
96. A heavy homogeneous sphere is suspended from a str<strong>in</strong>g<br />
whose end is attached to a vertic<strong>al</strong> w<strong>al</strong>l. The po<strong>in</strong>t at which the<br />
str<strong>in</strong>g is fastened to the sphere lies on the same vertic<strong>al</strong> as the<br />
centre of the sphere. What should the coefficient of friction<br />
b<strong>et</strong>ween the sphere and the w<strong>al</strong>l be for the sphere to rema<strong>in</strong> <strong>in</strong><br />
equilibrium?<br />
97. A homogeneous rectangular brick lies on an <strong>in</strong>cl<strong>in</strong>ed plane<br />
(Fig. 36). What h<strong>al</strong>f of the brick (left or right) exerts a greater<br />
pressure on it?<br />
98. A horizont<strong>al</strong>ly directed force equ<strong>al</strong> to the weight of a heavy<br />
cyl<strong>in</strong>dric<strong>al</strong> roller with a radius R is applied to its axis to lift it<br />
onto a rectangular step. F<strong>in</strong>d the maximum height of the step.<br />
99. A sphere weigh<strong>in</strong>g a= 3 kgf lies on two <strong>in</strong>cl<strong>in</strong>ed planes<br />
form<strong>in</strong>g angles at = 30° and a, = 60° with the horizon. D<strong>et</strong>erm<strong>in</strong>e<br />
the pressure exerted by the sphere on each plane if there is no<br />
friction b<strong>et</strong>ween the sphere and one of the planes.<br />
100. The front w<strong>al</strong>l of a drawer <strong>in</strong> a cab<strong>in</strong><strong>et</strong> is provided with<br />
two symm<strong>et</strong>ric<strong>al</strong> handles. The distance b<strong>et</strong>ween the handles is 1<br />
and the length of the drawer a. The coefficient of friction b<strong>et</strong>ween<br />
the drawer and the cab<strong>in</strong><strong>et</strong> is k. Can the drawer <strong>al</strong>ways be<br />
pulled out by apply<strong>in</strong>g a force perpendicular to the w<strong>al</strong>l of the<br />
drawer only to one handle?<br />
101. A homogeneous board is b<strong>al</strong>anced on a rough horizont<strong>al</strong><br />
log (Fig. 37). After a weight has been added to one of the ends<br />
~<br />
F<br />
-,-<br />
----<br />
m<br />
m<br />
FI,. 86 Fig. 36<br />
~
28<br />
PROBLEMS<br />
o<br />
;...-. ..... 8<br />
D<br />
Fig. 37<br />
Fig. 38<br />
of the boasd, equilibrium can be obta<strong>in</strong>ed when the board forms<br />
an angle ex with the horizon. What is the coefficient of friction<br />
b<strong>et</strong>ween the board and the log?<br />
102. The upper end of a ladder rests aga<strong>in</strong>st a smooth" vertic<strong>al</strong><br />
w<strong>al</strong>l and its bottom end stands on a rough floor. The coefficient<br />
of friction b<strong>et</strong>ween the ladder and the floor is k. D<strong>et</strong>erm<strong>in</strong>e the<br />
angle ex b<strong>et</strong>ween the ladder and the w<strong>al</strong>l at which the former<br />
will be <strong>in</strong> equilibrium.<br />
103. Solve the previous problem, assum<strong>in</strong>g that the w<strong>al</strong>l is<br />
rough and the coefficient of friction b<strong>et</strong>ween the ladder and the<br />
w<strong>al</strong>l is <strong>al</strong>so equ<strong>al</strong> to k.<br />
104. A homogeneous th<strong>in</strong> rod AB with a length 1 is placed<br />
onto the horizont<strong>al</strong> surface of a table. A str<strong>in</strong>g with a length of<br />
21 is attached to end B of the rod (Fig. 38). How will the rod<br />
move if the other end C of the str<strong>in</strong>g is slowly lifted up a stationary<br />
vertic<strong>al</strong> straight l<strong>in</strong>e DO pass<strong>in</strong>g through end A of the<br />
rod. Disregard the weight of the str<strong>in</strong>g.<br />
·105. At what coefficient of friction will a man not slip when<br />
he runs <strong>al</strong>ong a straight hard path? The maximum angle b<strong>et</strong>ween<br />
a vertic<strong>al</strong> l<strong>in</strong>e and the l<strong>in</strong>e connect<strong>in</strong>g the man's centre of gravity<br />
wtth the po<strong>in</strong>t of support is cx.<br />
106. A ladder leans aga<strong>in</strong>st a smooth vertic<strong>al</strong> w<strong>al</strong>l of a house.<br />
The angle b<strong>et</strong>ween the ladder and the horizont<strong>al</strong> surface of the<br />
Earth is a = 60°. The length of the ladder is I, and its centre of<br />
gravity is at its middle. How is the force act<strong>in</strong>g on the ladder<br />
from the Earth directed?<br />
107. A ladder with its centre of gravity at the middle stands<br />
on an absolutely smooth floor and leans aga<strong>in</strong>st a smooth w<strong>al</strong>l
MECHANICS 29<br />
(Fig. 39). What should the tension of a rope tied to the middle<br />
of the ladder be to prevent its f<strong>al</strong>l<strong>in</strong>g down?<br />
108. A man climbs up a ladder lean<strong>in</strong>g aga<strong>in</strong>st 8 smooth<br />
vertic<strong>al</strong> w<strong>al</strong>l. The ladder beg<strong>in</strong>s to slip only when the man<br />
reaches a certa<strong>in</strong> height. Why? .<br />
109. A picture is attached to a vertic<strong>al</strong> w<strong>al</strong>l by means of str<strong>in</strong>g<br />
AC with a length I form<strong>in</strong>g an angle a with the w<strong>al</strong>l. The height<br />
of the picture BC = d (Fig. 40). The bottom of the picture is<br />
not fastened. At what coefficient of friction b<strong>et</strong>ween the picture<br />
and the w<strong>al</strong>l will the picture be <strong>in</strong> equilibrium?<br />
110. Four homogeneous rods are p<strong>in</strong>-connected to one another<br />
at po<strong>in</strong>ts B, C and D (Fig. 41). The two extreme rods AB and<br />
DE can freely revolve with respect to stationary po<strong>in</strong>ts A and<br />
E on a horizont<strong>al</strong> straight l<strong>in</strong>e. The lengths of the rod AB = ED<br />
and BC= CD. The masses of the rods are the same. Show that<br />
the angles a and ~ are related by the ratio tan ex, = 3 tan ~ when<br />
<strong>in</strong> equilibrium.<br />
111. What is the coefficient of friction b<strong>et</strong>ween a floor and<br />
a box weigh<strong>in</strong>g one ton-force if a m<strong>in</strong>imum force of 600 kgf is<br />
required to start the box mov<strong>in</strong>g? .<br />
112. A weightless unstr<strong>et</strong>chable str<strong>in</strong>g is wound around a cyl<strong>in</strong>der<br />
with a mass m (Fig. 42). With what m<strong>in</strong>imum force F m<strong>in</strong><br />
and at what angle
a<br />
Fig. 41 Fig. 42<br />
for the rotat<strong>in</strong>g cyl<strong>in</strong>der to rema<strong>in</strong> <strong>in</strong> place? The coefficient of<br />
friction b<strong>et</strong>ween the cyl<strong>in</strong>der and the floor is k.<br />
113. Figure 43 shows a simplified diagram of the steam eng<strong>in</strong>e<br />
and crank gear of a steam locomotive. Fig. 43a and b correspond<br />
to the moments when steam has been admitted <strong>in</strong>to the leftand<br />
right-hand parts of the cyl<strong>in</strong>der, respectively. C<strong>al</strong>culate the<br />
tractive effort for these cases when po<strong>in</strong>t C is on one vertic<strong>al</strong><br />
l<strong>in</strong>e with the axis of the driv<strong>in</strong>g wheel. The pressure of the<br />
steam <strong>in</strong> the cyl<strong>in</strong>der is p, the area of the piston is A, the<br />
radius of the driv<strong>in</strong>g wheel is R and the distance DC = r. Disregard<br />
the masses of the crank gear, the piston and the driv<strong>in</strong>g<br />
wheel.<br />
114. Bricks are so laid without a b<strong>in</strong>der that a part of each<br />
follow<strong>in</strong>g brick extends over the one below (Fig. 44). Over what<br />
maximum distance can the right-hand edge of the upper brick<br />
extend over the right-hand edge of the lowermost brick that<br />
serves as the base for the entire brickwork? The length of each<br />
brick is l,<br />
115. F<strong>in</strong>d the centre of gravity of a th<strong>in</strong> homogeneous wire<br />
bent to a semicircle with a radius 'f.<br />
(a)<br />
{Il}
MECHANICS 31<br />
Fig. 44 Fig. 45<br />
a<br />
r<br />
~~~-,A<br />
I<br />
I<br />
rl<br />
I<br />
C :<br />
- ..-----10<br />
I<br />
I<br />
I<br />
rl<br />
II<br />
Fig. 46 Fi,. ,g<br />
~~~";'6<br />
t 16. D<strong>et</strong>erm<strong>in</strong>e the centre of gravity of a homogeneous th<strong>in</strong><br />
semicircle with a radius r,<br />
t 17. D<strong>et</strong>erm<strong>in</strong>e the centre of gravity of a th<strong>in</strong> homogeneous<br />
wire bent over an arc with a radius r (Fig. 45).<br />
t 18. D<strong>et</strong>erm<strong>in</strong>e the centre of gravity of a th<strong>in</strong> homogeneous<br />
plate cut <strong>in</strong> the form of a sector with a radius r and a centr<strong>al</strong><br />
angle a (Fig. 46).<br />
119. D<strong>et</strong>erm<strong>in</strong>e the centre of gravity of a th<strong>in</strong> homogeneous<br />
plate hav<strong>in</strong>g the form of a rectangle with sides' and 2r which<br />
a semicircle with a radius, is cut out of (Fig. 47).<br />
1-6, Work and Energy<br />
120. What work will be performed if a force of 3 kgf is used<br />
to lift a load of 1 kgf to a height of 5 m<strong>et</strong>res?<br />
121. In the formula for work W=kFs, the coefficient k=l<br />
if <strong>al</strong>l the quantities are given <strong>in</strong> the same system of units.<br />
What is the coefficient k if the work is measured <strong>in</strong> joules, the<br />
force <strong>in</strong> kgf and the distance <strong>in</strong> centim<strong>et</strong>res?
32 PROBLEMS<br />
Fig. 48<br />
122. In Guericke's experiment two copper hemispheres were<br />
tightlv fitted to 'each other to form a hollow sphere from which<br />
the air was pumped out. The hemispheres were held so tightly<br />
tog<strong>et</strong>her by the atmospheric pressure that they could be pulled<br />
apart only with the aid of horses. How many horses are required<br />
to do this if each horse pulled with a force F? The radius<br />
of each hemisphere was R and the atmospheric pressure p.<br />
123. How do you expla<strong>in</strong> the fact that when a stone is<br />
dropped onto the ground, the change <strong>in</strong> the momentum of the<br />
Earth is equ<strong>al</strong> to that of the stone, while the change <strong>in</strong> the<br />
k<strong>in</strong><strong>et</strong>ic energy of the Earth is neglected?<br />
124. A pile with a weight of 100 kgf is forced <strong>in</strong>to the ground<br />
by a pile driver weigh<strong>in</strong>g 400 kgf. The latter freely drops from<br />
a height of 5 m<strong>et</strong>res and drives the pile to a depth of 5 em<br />
upon each impact. D<strong>et</strong>erm<strong>in</strong>e the resistance of the soil, assum<strong>in</strong>g<br />
it to be constant.<br />
125. A box with sand hav<strong>in</strong>g the mass M is suspended from<br />
a rope with a length L. The length of the rope is much greater<br />
than the l<strong>in</strong>ear dimensions of the box. A bull<strong>et</strong> with a mass m<br />
flies <strong>in</strong> a horizont<strong>al</strong> direction, strikes the box and g<strong>et</strong>s stuck <strong>in</strong><br />
it. After this the rope is deflected by angle ex, from the vertic<strong>al</strong>.<br />
D<strong>et</strong>erm<strong>in</strong>e the velocity of the bull<strong>et</strong>.<br />
126. Two carts are pushed apart by an explosion of a powder<br />
charge Q placed b<strong>et</strong>ween them (Fig. 48). The cart weigh<strong>in</strong>g<br />
100 gf travels a distance of 18 m<strong>et</strong>res and stops. What distance<br />
will be covered by the other cart weigh<strong>in</strong>g 300 gf? The coefficients<br />
of friction k b<strong>et</strong>ween the ground and the carts are the<br />
same.<br />
127. Solve Problem 65 us<strong>in</strong>g the law of conservation of momentum<br />
and tak<strong>in</strong>g <strong>in</strong>to account the change <strong>in</strong> the k<strong>in</strong><strong>et</strong>ic<br />
energy of the wagon and the body.<br />
12-8. A rock<strong>et</strong> launched vertic<strong>al</strong>ly upward ejects hot gases<br />
consecutively <strong>in</strong> two equ<strong>al</strong> portions. The outflow velocity of the<br />
gases relative to the rock<strong>et</strong> is constant and equ<strong>al</strong> to u. What<br />
should the time <strong>in</strong>terv<strong>al</strong> b<strong>et</strong>ween combustion of the portions be
MECHANICS<br />
c<br />
33<br />
~t:------;--------<br />
A<br />
Fig. 49 Fig. 50<br />
for the rock<strong>et</strong> to reach a maximum <strong>al</strong>ti tude? The fuel burns<br />
<strong>in</strong>stantaneously. Disregard the resistance of the air.<br />
129. The fuel <strong>in</strong> a rock<strong>et</strong> burns <strong>in</strong> equ<strong>al</strong> portions with a<br />
mass m. Combustion takes place <strong>in</strong>stantaneously. Will the outftow<br />
velocity of the gases be constant with respect to the rock<strong>et</strong> if<br />
the mechanic<strong>al</strong> energy of the system changes the same amount<br />
upon the combustion of each portion?<br />
130. A body is first lifted to the top of a mounta<strong>in</strong> over the<br />
path ADC, and aga<strong>in</strong> over the path ABC (Fig. 49). Prove that<br />
the work done will be the same provided the body is lifted<br />
slowly if the coefficient of friction is the same on both slopes.<br />
13t. What force should be applied to the handle of a screw<br />
jack to hold <strong>in</strong> equilibrium the load G lifted by the jack? The<br />
pitch of the screw is h and the length of the handle is R. There<br />
is no friction. ' '<br />
132. F<strong>in</strong>d the maximum efficiency of a screw jack hav<strong>in</strong>g no<br />
speci<strong>al</strong> device to prevent back travel.<br />
133. A rope ladder with a length I carry<strong>in</strong>g a man with a<br />
mass m at its end is attached to the bask<strong>et</strong> of a b<strong>al</strong>loon with<br />
a mass M. The entire system is <strong>in</strong> equilibrium <strong>in</strong> the air. F<strong>in</strong>d<br />
the work the man should do to climb <strong>in</strong>to the bask<strong>et</strong>. What is<br />
the velocity of the b<strong>al</strong>loon if the man climbs the ladder with a<br />
velocity v with respect to it?<br />
134. How should the power of a pump motor change for the<br />
pump to deliver twice as much water <strong>in</strong> a unit of time through<br />
a narrow orifice? Disregard friction.<br />
135. A rectangular pit with an area at its base A and a depth<br />
H is h<strong>al</strong>f filled with water. A pump forces the water up onto<br />
the surface through a cyl<strong>in</strong>dric<strong>al</strong> pipe with a radius R.<br />
'3-2042
34 PROBLEMS<br />
(I) What work is done by the pump if it pumps out <strong>al</strong>l the<br />
water dur<strong>in</strong>g the time 't?<br />
(2) What work is done by the pump dur<strong>in</strong>g the same time if a<br />
rectangular stone slab with an area at its base At and a height<br />
h lies at the bottom of the pit? The depth of the water <strong>in</strong> the<br />
pit is H/2 as before.<br />
136. What work should a man do to w<strong>al</strong>k up a subway esc<strong>al</strong>ator<br />
mov<strong>in</strong>g down? The height of the esc<strong>al</strong>ator is h, its speed is constant<br />
and equ<strong>al</strong> to v, and it is <strong>in</strong>cl<strong>in</strong>ed at an angle ex, to the horizon.<br />
137. C<strong>al</strong>culate the potenti<strong>al</strong> energy of a deformed spr<strong>in</strong>g if<br />
its elastic force F = kx, where k is the coefficient of spr<strong>in</strong>g elasticity,<br />
and x is the deformation.<br />
138. A man act<strong>in</strong>g with a force F upon a str<strong>et</strong>ched spr<strong>in</strong>g<br />
stands <strong>in</strong> a railway carriage of a uniformly mov<strong>in</strong>g tra<strong>in</strong> (Fig. 50).<br />
The tra<strong>in</strong> covered a distance L. What work will be done by the<br />
man <strong>in</strong> a coord<strong>in</strong>ate system related to the Earth?<br />
139. A man str<strong>et</strong>ches a spr<strong>in</strong>g attached to the front w<strong>al</strong>l<br />
of a railway carriage over a distance I <strong>in</strong> a uniformly mov<strong>in</strong>g<br />
tra<strong>in</strong>. Dur<strong>in</strong>g this time the tra<strong>in</strong> covered the distance L. What<br />
work will be performed by the man <strong>in</strong> a coord<strong>in</strong>ate system related<br />
to the Earth? What will this work be <strong>in</strong> a system related<br />
to the tra<strong>in</strong>? When the man str<strong>et</strong>ches the spr<strong>in</strong>g he moves <strong>in</strong><br />
a direction opposite to that of the tra<strong>in</strong>.<br />
140. Two absolutely elastic spheres with masses m, and m;<br />
collide. Their <strong>in</strong>iti<strong>al</strong> velocities are V t<br />
and V t • F<strong>in</strong>d the velocities<br />
of the spheres [-lfter collision.<br />
Consider the collision as centr<strong>al</strong>, the velocities of the spheres<br />
be<strong>in</strong>g directed <strong>al</strong>ong the l<strong>in</strong>e connect<strong>in</strong>g their centres. An<strong>al</strong>yse<br />
two cases: (1) the velocity of the second sphere before collision<br />
is zero, (2) the masses of the spheres are the same.<br />
141. Two elastic blocks of equ<strong>al</strong> mass m and connected by a<br />
spr<strong>in</strong>g with a length l rest on an absolutely smooth horizont<strong>al</strong><br />
surface (Fig. 51). The coefficient of elasticity of the spr<strong>in</strong>g is k.<br />
A third block of the same mass m strikes the left-hand block<br />
with a velocity o. Prove that the blocks connected by the<br />
spr<strong>in</strong>g will <strong>al</strong>ways move <strong>in</strong> the same direction.<br />
D<strong>et</strong>erm<strong>in</strong>e the velocities of the blocks when the spr<strong>in</strong>g is<br />
str<strong>et</strong>ched as much as possible.<br />
142. Two plates whose masses are m, and m l<br />
, respectively,<br />
are connected by a spr<strong>in</strong>g (Fig. 52). What force should be applied<br />
to the upper plate for it to raise the lower one after the<br />
pressure is removed? Disregard the mass of the spr<strong>in</strong>g.
MECH ..\NICS<br />
m m m<br />
J;//////;/C/////ffl////;=ffmv~ Fig. 51<br />
143. A b<strong>al</strong>l mov<strong>in</strong>g with a velocity v strikes a w<strong>al</strong>l mov<strong>in</strong>g<br />
toward the b<strong>al</strong>l with a velocity u (Fig. 53). An elastic impact<br />
occurs.<br />
D<strong>et</strong>erm<strong>in</strong>e the velocity of the b<strong>al</strong>l after the impact. What is<br />
the cause of the change <strong>in</strong> the k<strong>in</strong><strong>et</strong>ic energy of the b<strong>al</strong>l?<br />
Consider the mass of the w<strong>al</strong>l to be <strong>in</strong>f<strong>in</strong>itely great.<br />
144. Two stones identic<strong>al</strong> <strong>in</strong> mass and connected by a rope<br />
with a length of l = 39.2 m<strong>et</strong>res are dropped from a height of<br />
h = 73.5 m<strong>et</strong>res with an <strong>in</strong>iti<strong>al</strong> velocity of zero. The first stone<br />
beg<strong>in</strong>s to f<strong>al</strong>l 1: = 2 seconds before the second one. In what<br />
time will the stones reach the ground?<br />
(1) Consider the rope to be absolutely elastic.<br />
(2) Consider the rope to be absolutely <strong>in</strong>elastic.<br />
145. Sever<strong>al</strong> identic<strong>al</strong> elastic b<strong>al</strong>ls are so suspended <strong>in</strong> a row<br />
on str<strong>in</strong>gs of equ<strong>al</strong> length (Fig. 54) that the distances b<strong>et</strong>ween<br />
adjacent b<strong>al</strong>ls are very sm<strong>al</strong>l. How will the b<strong>al</strong>ls behave if an<br />
extreme b<strong>al</strong>l is moved aside and then released; two, three, <strong>et</strong>c.,<br />
b<strong>al</strong>ls are moved aside and released at the same time?<br />
146. A row of b<strong>al</strong>ls of identic<strong>al</strong> size are placed at sm<strong>al</strong>l<br />
<strong>in</strong>terv<strong>al</strong>s on a level surface (Fig. 55). The middle b<strong>al</strong>l is made<br />
of steel and the others of ivory (the mass of the steel b<strong>al</strong>l is<br />
greater). An ivory b<strong>al</strong>l (of the same mass) strikes the b<strong>al</strong>ls from<br />
the right <strong>al</strong>ong the l<strong>in</strong>e of centres. How will the b<strong>al</strong>ls move<br />
after the impact?<br />
147. Equ<strong>al</strong> weights of mass m are suspended from the ends<br />
of a very long str<strong>in</strong>g thrown over two sm<strong>al</strong>l stationary pulleys<br />
at a distance of 2l from each other (Fig. 56). F<strong>in</strong>d the vela-<br />
35<br />
1J<br />
Fig. 52 Fig. 53
36 PROBLEMS<br />
~ city of the weights after<br />
a sufficiently long time if<br />
a weight with a mass 2m<br />
is attached to the middle<br />
of the str<strong>in</strong>g.<br />
148. A weight of mass<br />
m, = 536 g first kept near<br />
the ceil<strong>in</strong>g b<strong>et</strong>ween po<strong>in</strong>ts<br />
A and B beg<strong>in</strong>s to lower<br />
(Fig. 57). At what angle<br />
ANB will Its velocity be<br />
equ<strong>al</strong> <strong>in</strong> absolute v<strong>al</strong>ue to<br />
Fig. 54 the velocity of a weight<br />
with a mass m; = 1,000 g?<br />
How will the weights move afterward?<br />
149. A heavy board form<strong>in</strong>g an angle ~ with the horizon<br />
rests on two rollers of different radii. D<strong>et</strong>erm<strong>in</strong>e how the board<br />
will move if it does not slip. Disregard the mass of the rollers.<br />
150. A homogeneous cha<strong>in</strong> with a length 21 and mass M<br />
lies on an absolutely smooth table. A sm<strong>al</strong>l part of the cha<strong>in</strong><br />
hangs from the table. At the <strong>in</strong>iti<strong>al</strong> moment of time the part<br />
of the cha<strong>in</strong> ly<strong>in</strong>g on the table is held and then released, after<br />
which the cha<strong>in</strong> beg<strong>in</strong>s to slide off the table under the weight<br />
of the hang<strong>in</strong>g end. F<strong>in</strong>d the velocity of the cha<strong>in</strong> when the<br />
length of the hang<strong>in</strong>g part is equ<strong>al</strong> to x (x < 1).<br />
D<strong>et</strong>erm<strong>in</strong>e (for the same moment of time) the acceleration of<br />
the cha<strong>in</strong> and the force with which it acts on the edge of the<br />
table.<br />
151. A wagon with a mass M can move without friction <strong>al</strong>ong<br />
horizont<strong>al</strong> raj-Is. A mathematic<strong>al</strong> pendulum (a sphere with a<br />
mass m suspended from a long str<strong>in</strong>g 1) is fastened on the wagon<br />
(Fig. 58). At the <strong>in</strong>iti<strong>al</strong> moment the wagon and the pendulum<br />
were at rest and the str<strong>in</strong>g was deflected through an angle<br />
a from the vertic<strong>al</strong>. What will the velocity of the wagon be<br />
when the pendulum str<strong>in</strong>g forms an angle p (P < a) with the<br />
vertic<strong>al</strong>?<br />
152. A wedge with a mass M rests on an absolutely smooth<br />
horizont<strong>al</strong> surface. A block with a mass m is placed on the
MECHANICS<br />
37<br />
21·<br />
2m<br />
m<br />
Fig. 56<br />
m<br />
Fig. 57<br />
m<br />
• •
38 PROBLEMS<br />
wedge. The block can slide <strong>al</strong>ong the wedge without friction<br />
under the force of gravity. Assum<strong>in</strong>g that the system was at<br />
rest at the <strong>in</strong>iti<strong>al</strong> moment of time, f<strong>in</strong>d the velocity of the<br />
wedge when the block lowers vertic<strong>al</strong>ly through a height h.<br />
153. A rod secured b<strong>et</strong>ween two coupl<strong>in</strong>gs can travel freely<br />
<strong>in</strong> a vertic<strong>al</strong> direction (Fig. 59). The lower tip of the rod<br />
bears aga<strong>in</strong>st a smooth wedge ly<strong>in</strong>g on a horizont<strong>al</strong> surface.<br />
The mass of the rod is m and that of the wedge is M. There<br />
is no friction. At the <strong>in</strong>iti<strong>al</strong> moment the rod and the wedge<br />
were at rest.<br />
F<strong>in</strong>d: the velocity v of the wedge at the moment the rod<br />
lowers through the height h, the velocity u r el of the rod relative<br />
to the mov<strong>in</strong>g wedge and the acceleration a of the rod.<br />
1-7. K<strong>in</strong>ematics of Curvil<strong>in</strong>ear Motion<br />
154. Two shafts A and B are jo<strong>in</strong>ed by an endless belt that<br />
transmits rotation from A to B. The speed of the driv<strong>in</strong>g shaft<br />
is n t = 3,000 rpm .. A pulley with a diam<strong>et</strong>er of D, = 500 mm<br />
is fitted onto the driven shaft which should rotate at n, = 600 rpm.<br />
What is the diam<strong>et</strong>er of the pulley to be fitted onto the driv<strong>in</strong>g<br />
shaft?<br />
155. The crawler of a tractor consists of n l<strong>in</strong>ks, each with a<br />
length of a. The radii of the wheels which the crawler is placed<br />
on are R. The tractor moves at a speed v. It is assumed that<br />
the crawler does not sag.<br />
(1) What number of l<strong>in</strong>ks move at this moment translation<strong>al</strong>ly t<br />
how many of them are at rest (relative to the Earth) and how<br />
many rotate on the wheels?<br />
(2) The tractor covered a distance s ~ na. How much time<br />
did each l<strong>in</strong>k of the crawler move translation<strong>al</strong>ly, rema<strong>in</strong> at rest<br />
and participate <strong>in</strong> rotation<strong>al</strong> motion?<br />
156. The follow<strong>in</strong>g device is used to d<strong>et</strong>erm<strong>in</strong>e the velocities<br />
of molecules. A silver-coated wire heated by current is arranged<br />
on a common axis of two cyl<strong>in</strong>ders fastened to each other<br />
and rotat<strong>in</strong>g with an angular velocity Q) (Fig. 60). The <strong>in</strong>tern<strong>al</strong><br />
cyl<strong>in</strong>der is provided with a slit which the molecules evaporat<strong>in</strong>g<br />
from the wire fly <strong>in</strong>to. The entire device is placed <strong>in</strong> a<br />
vacuum. If the cyl<strong>in</strong>ders are at rest. the trace of the deposited<br />
silver molecules appears at po<strong>in</strong>t A. If the cyl<strong>in</strong>ders rotate,<br />
the trace appears at po<strong>in</strong>t B at a distance l from A. F<strong>in</strong>d
MECHANICS<br />
39<br />
Fig. 60<br />
the velocity of the molecules. The<br />
radii of the cyl<strong>in</strong>ders are, and R.<br />
157. To turn a tractor mov<strong>in</strong>g<br />
at a speed of V o = 18 krn/h the<br />
8 driver so brakes one of the crawlers<br />
that the axle of its driv<strong>in</strong>g<br />
wheel beg<strong>in</strong>s to move forward at a<br />
speed of VI = 14 km/h. The distance<br />
b<strong>et</strong>ween the crawlers d = 1.5 m<strong>et</strong>res.<br />
An arc of what radius will be<br />
described by the centre of the trac-<br />
tor?<br />
158. In mounta<strong>in</strong>s the follow<strong>in</strong>g<br />
phenomenon can be observed: a star be<strong>in</strong>g watched quickly disappears<br />
beh<strong>in</strong>d a remote summit. (The same phenomenon can <strong>al</strong>so<br />
be observed on a pla<strong>in</strong> if there is a sufficiently remote t<strong>al</strong>l structure.)<br />
At what speed should the observer run to constantly<br />
see the star at the same angular distance from the mounta<strong>in</strong>?<br />
The distance b<strong>et</strong>ween the observer and the summit is<br />
10 km, Assume that observations are be<strong>in</strong>g made at the<br />
pole.<br />
159. The current velocity of a river grows <strong>in</strong> proportion to<br />
the distance from its bank and reaches its maximum V o <strong>in</strong> the<br />
middle. Near the banks the velocity is zero. A boat is so mov<strong>in</strong>g<br />
on the river that its velocity u relative to the water is<br />
constant and perpendicular to the current. F<strong>in</strong>d the distance<br />
through which the boat cross<strong>in</strong>g the river will be carried away<br />
by the current if the width of the river is c. Also d<strong>et</strong>erm<strong>in</strong>e<br />
the trajectory of the boat.<br />
160. Four tortoises are at the four corners of a square with<br />
a side a. They beg<strong>in</strong> to move at the same time with a constant<br />
speed v, the first one <strong>al</strong>ways mov<strong>in</strong>g <strong>in</strong> a direction toward the<br />
second, the second toward the third, the third toward the fourth<br />
and the fourth toward the first. Will the tortoises me<strong>et</strong>, and if<br />
they do, <strong>in</strong> what time?<br />
161. Two ships A and B orig<strong>in</strong><strong>al</strong>ly at a distance of a = 3 km<br />
from each other depart at the same time from a straight coastl<strong>in</strong>e.<br />
Ship A moves <strong>al</strong>ong a straight l<strong>in</strong>e perpendicular to the<br />
shore while ship B constantly heads for ship A, hav<strong>in</strong>g at each<br />
moment the same speed as the latter. After a sufficiently great<br />
<strong>in</strong>terv<strong>al</strong> of time the second ship will obviously follow the first<br />
one at a certa<strong>in</strong> distance. F<strong>in</strong>d this distance.
40<br />
PROBLEMS<br />
s<br />
II<br />
ne. 61<br />
Fl,. 62<br />
182. A body is thrown with an <strong>in</strong>iti<strong>al</strong> velocity Vo at an angle<br />
CI to the horizon. How long will the body fly? At what<br />
distance from the spot where it was thrown will the body f<strong>al</strong>l<br />
to the ground? At what angle ~ will the distance of the flight<br />
be maximum? At what height will the body be <strong>in</strong> the time ~<br />
after the motion beg<strong>in</strong>s? What will the velocity of the body be<br />
<strong>in</strong> magnitude and direction at the given moment of time?<br />
Consider 't to be greater than the time dur<strong>in</strong>g which the body<br />
reaches the maximum height. Disregard the resistance of the air.<br />
183. F<strong>in</strong>d the trajectory of the body thrown at an angle to<br />
the horizon (see Problem 162).<br />
184. A rubber b<strong>al</strong>l is to be. thrown from the ground over<br />
a vertic<strong>al</strong> w<strong>al</strong>l with 8 height h from a distance s (Fig. 61). At<br />
what m<strong>in</strong>imum <strong>in</strong>iti<strong>al</strong> velocity is this possible? At what angle ~<br />
to the horizon should the velocity be directed <strong>in</strong> this case?<br />
Fl,.63 Fi,. 64
MECHANICS 41<br />
165. A body is thrown <strong>in</strong>to a river from a steep bank with<br />
a height H. The <strong>in</strong>iti<strong>al</strong> velocity of the body forms an angle (X<br />
with the horizon and is equ<strong>al</strong> to v•. At what distance from the<br />
bank will the body f<strong>al</strong>l? In how many seconds will the body<br />
be at a height h above the water after motion beg<strong>in</strong>s? What is<br />
the velocity of the body when it f<strong>al</strong>ls <strong>in</strong>to the water?<br />
166. At what angle to the horizon should a stone be thrown<br />
from a steep bank for it to f<strong>al</strong>l <strong>in</strong>to the water as far as p0ssible<br />
from the bank? The height of the bank h o = 20 m<strong>et</strong>res and<br />
the <strong>in</strong>iti<strong>al</strong> velocity of the stone V o= 14 mt«.<br />
187. Two bodies are thrown at the same time and with an<br />
equ<strong>al</strong> <strong>in</strong>iti<strong>al</strong> velocity V o from po<strong>in</strong>t x = y = 0 at various angles ~<br />
and (x, to the horizon (Fig. 62). What is the velocity with which<br />
the bodies move relative to each other? What will the distance<br />
b<strong>et</strong>ween the bodies be after the time '( elapses?<br />
188. A dive-bomber drops a bomb from a height h at a distance<br />
I from the targ<strong>et</strong>. The speed of the bomber is e. At what<br />
angle to the horizon should it dive?<br />
169. Alassenger car is driv<strong>in</strong>g over 8 level highway beh<strong>in</strong>d<br />
a truck. stone got stuck b<strong>et</strong>ween double tyres of the rear<br />
wheels of the truck. At what distance should the car follow the<br />
truck so that the stone will not strike it if it fties out from<br />
b<strong>et</strong>ween the tyres~ Both vehicles have a speed of 50 km/h.<br />
170. A b<strong>al</strong>l freely f<strong>al</strong>ls from a height h onto an <strong>in</strong>cl<strong>in</strong>ed plane<br />
form<strong>in</strong>g an angle cz with the horizon (Fig. 63). F<strong>in</strong>d the ratio<br />
of the distances b<strong>et</strong>ween the po<strong>in</strong>ts at which the jump<strong>in</strong>g b<strong>al</strong>l<br />
strikes the <strong>in</strong>cl<strong>in</strong>ed plane. Consider the impacts b<strong>et</strong>ween the b<strong>al</strong>l<br />
and the plane to be absolutely elastic.<br />
171. F<strong>in</strong>d the acceleration of body A which slides without<br />
<strong>in</strong>iti<strong>al</strong> velocity down a helic<strong>al</strong> groove with a pitch h and a radius<br />
R at the end of the n-th turn (Fig. 64). Disregard friction.<br />
172. Two steel slabs M and N with a height h are placed on<br />
sand (Fig. 65). The distance b<strong>et</strong>ween the slabs I = 20 em. A b<strong>al</strong>l<br />
whose velocity has not been d<strong>et</strong>erm<strong>in</strong>ed exactly moves over<br />
slab M. It is only known that this velocity ranges b<strong>et</strong>ween<br />
200 cmls and 267 em/s.<br />
(1) At what height h is it impossible to predict the horizont<strong>al</strong><br />
direction of the velocity of the b<strong>al</strong>l at the moment when it<br />
f<strong>al</strong>ls onto the sand? (Before it touches the sand, the b<strong>al</strong>l strikes<br />
slab N at least once.)<br />
(2) At what m<strong>in</strong>imum height of the slabs is it impossible to<br />
predict the spot on section I which the b<strong>al</strong>l will strike? Dlsre-
42<br />
PROBLEMS<br />
A<br />
A<br />
N<br />
o<br />
B<br />
"<br />
Fig. 65 Fig. 66<br />
gard the duration of the impact b<strong>et</strong>ween the b<strong>al</strong>l and the slab,<br />
and consider the impact to be absolutely elastic.<br />
173. A solid homogeneous disk rolls without slipp<strong>in</strong>g over<br />
a horizont<strong>al</strong> path with a constant velocity v (Fig. 66).<br />
(1) Prove that the l<strong>in</strong>ear velocity of rotation of any po<strong>in</strong>t of<br />
the disk on its rim with respect to the centre 0 is equ<strong>al</strong> to the<br />
translation<strong>al</strong> velocity of the disk.<br />
(2) F<strong>in</strong>d the magnitude and direction of the velocities of<br />
po<strong>in</strong>ts A, B, C and D on the rim of the disk with respect to<br />
a stand<strong>in</strong>g observer.<br />
(3) What po<strong>in</strong>ts of the disk have the same absolute velocity<br />
as the centre of the disk relative to a stand<strong>in</strong>g observer?<br />
174. A mov<strong>in</strong>g cart is shown on a c<strong>in</strong>ema screen. The radius<br />
of the front wheels of the cart r= 0.35 m<strong>et</strong>re and that of the<br />
rear ones R = 1.5 r. The front wheels have N 1 = 6 spokes. The<br />
A<br />
A<br />
Fig. 61<br />
Fig. 68
MECHANICS 43<br />
film <strong>in</strong> the c<strong>in</strong>ema camera moves with<br />
a speed of 24 frames per second.<br />
Assum<strong>in</strong>g that the wheels of the cart<br />
do not slip, f<strong>in</strong>d the m<strong>in</strong>imum speed<br />
with which the cart should move for<br />
its front wheels to seem stationary on<br />
the screen. What m<strong>in</strong>imum number of<br />
spokes Nt should the rear wheels have<br />
for them <strong>al</strong>so to seem stationary?<br />
175. At what speeds of the cart mov<strong>in</strong>g<br />
from the left to the right (see Prob-<br />
Fig. 6' tern 174) will it seem to the audience<br />
that: (1) the spokes of the wheels rotate<br />
counterclockwise? (2) the spokes of the front and rear wheels<br />
rotate <strong>in</strong> opposite directions? There are six spokes on each front<br />
and rear wheel.<br />
176. A spool consists of a cyl<strong>in</strong>dric<strong>al</strong> core and two identic<strong>al</strong><br />
solid heads. The core rolls without slipp<strong>in</strong>g <strong>al</strong>ong a rough horizont<strong>al</strong><br />
block with a constant velocity v (Fig. 67). The radius<br />
of the core is r and that of the heads is R.<br />
What <strong>in</strong>stantaneous velocity will po<strong>in</strong>ts A and B on the rim<br />
of one of the heads have? What po<strong>in</strong>ts on the heads have an<br />
<strong>in</strong>stantaneous velocity equ<strong>al</strong> <strong>in</strong> magnitude to the velocity of the<br />
spool core?<br />
177. Draw the trajectories of po<strong>in</strong>ts A, Band C of a spool<br />
(Fig. 68) roll<strong>in</strong>g with its cyl<strong>in</strong>dric<strong>al</strong> core without slipp<strong>in</strong>g <strong>al</strong>ong<br />
a block (see Problem 176).<br />
Fig. 70
44 PROBLEMS<br />
178. A b<strong>al</strong>l bear<strong>in</strong>g supports the end of a shaft rotat<strong>in</strong>g with<br />
an angular velocity (I). The diam<strong>et</strong>er of the shaft axis is d and<br />
that of the race is D (Fig. 69). F<strong>in</strong>d the l<strong>in</strong>ear velocity of<br />
motion of the centre of one of the b<strong>al</strong>ls if the race is stationary,<br />
or rotates with an angular velocity of Q. Assume that <strong>in</strong><br />
both cases the b<strong>al</strong>ls do not slip when they rollover the shaft.<br />
179. A cone rolls without slipp<strong>in</strong>g <strong>al</strong>ong a plane. The axis of<br />
the cone rotates with a velocity (I) around a vertic<strong>al</strong> l<strong>in</strong>e pass<strong>in</strong>g<br />
through its apex. The height of the cone is h and the angle<br />
b<strong>et</strong>ween the axis and the generatrix of the cone is Cl. What is<br />
the angular velocity with which the cone rotates around its axis?<br />
Also d<strong>et</strong>erm<strong>in</strong>e the l<strong>in</strong>ear velocity of an arbitrary po<strong>in</strong>t on the<br />
diam<strong>et</strong>er of the cone base ly<strong>in</strong>g <strong>in</strong>" a vertic<strong>al</strong> plane.<br />
180. Figure 70 shows schematic<strong>al</strong>ly a differenti<strong>al</strong> transmission<br />
of a motor vehicle that does not <strong>al</strong>low the driv<strong>in</strong>g wheels of the<br />
vehicle to slip when negotiat<strong>in</strong>g a curve. (In this case the wheels<br />
should revolve at different speeds.)<br />
The eng<strong>in</strong>e rotates wheel B rigidly mounted on axle A around<br />
which a pair of bevel gears E can freely rotate. These gears<br />
mesh with another pair of bevel gears <strong>al</strong>ong which they run when<br />
axle A rotates. The axle of the driv<strong>in</strong>g wheels (usu<strong>al</strong>ly the rear<br />
ones) is divided <strong>in</strong>to two h<strong>al</strong>ves that carryon their ends gears C<br />
and D. These h<strong>al</strong>ves can rotate with various angular velocities,<br />
rema<strong>in</strong><strong>in</strong>g connected by the differenti<strong>al</strong> transmission.<br />
F<strong>in</strong>d the relationship b<strong>et</strong>ween the angular velocities '1, co,<br />
(1)1 and (I), of the differenti<strong>al</strong> transmission if the radii of gears E<br />
are equ<strong>al</strong> to r and the radii of gears D and 9 are '1.<br />
1·8. Dynamics of Curvil<strong>in</strong>ear Motion<br />
181. D<strong>et</strong>erm<strong>in</strong>e the tension of the rope of a b<strong>al</strong>listic pendulum<br />
(see Problem 125) at the moment when it is struck by a bull<strong>et</strong>.<br />
182. · Four identic<strong>al</strong> weights are secured on a ftexible unstr<strong>et</strong>chable<br />
str<strong>in</strong>g whose weight may be neglected (Fig. 71). The<br />
entire system rotates with an angular velocity co around a vertic<strong>al</strong><br />
axis pass<strong>in</strong>g through po<strong>in</strong>t O. The weights move over<br />
a smooth level surface. D<strong>et</strong>erm<strong>in</strong>e the tension of the str<strong>in</strong>g <strong>in</strong><br />
various sections.<br />
183. Masses nlt and m. are fastened to the ends of a weightless<br />
rod with a length I. The velocities of these masses are <strong>in</strong><br />
one plane and are equ<strong>al</strong> to VI and V" respectively (Fig. 72).<br />
F<strong>in</strong>d the velocity which the centre of gravity of the system
MECHANICS<br />
45<br />
mImI mtm<br />
• • • •<br />
?i ~ Jf Fig. 7/<br />
moves with and the angular velocity which the rod rotates with<br />
relative to an axis pass<strong>in</strong>g through the centre of gravity.<br />
184. A cannon is at the centre of a platform freely rotat<strong>in</strong>g<br />
around a vertic<strong>al</strong> axis. The axis of rotation passes through the<br />
breech of the cannon. A shell is fired <strong>in</strong> a horizont<strong>al</strong> direction<br />
<strong>al</strong>ong the radius of the platform. Will the velocity of rotation<br />
change <strong>in</strong> this case?<br />
185. A sm<strong>al</strong>l body beg<strong>in</strong>s to slide without <strong>in</strong>iti<strong>al</strong> velocity<br />
down an <strong>in</strong>cl<strong>in</strong>ed plane with a height H (Fig. 73). Assum<strong>in</strong>g<br />
that there is no friction and the impact of the body aga<strong>in</strong>st<br />
horizont<strong>al</strong> plane AB is absolutely elastic. d<strong>et</strong>erm<strong>in</strong>e the nature<br />
of motion of the body after it leaves the <strong>in</strong>cl<strong>in</strong>ed plane. Answer<br />
the same question if the impact is absolutely <strong>in</strong>elastic.<br />
186. What is the m<strong>in</strong>imum radius of an arc that can be negotiated<br />
by a motor-cyclist if his speed is v = 21 mls and the<br />
coefficient of friction b<strong>et</strong>ween the tyres and the ground k = O.3?<br />
To what angle a to the horizon should the motor-cycle be<br />
<strong>in</strong>cl<strong>in</strong>ed <strong>in</strong> this case?<br />
187. A massive sphere is fitted onto a light rod (Fig. 74).<br />
When will the rod f<strong>al</strong>l faster: if it is placed vertic<strong>al</strong>ly on<br />
end A or on end B? The end of the rod on the ground does<br />
not slip.<br />
188. A massive sphere is secured on the end of a light rod<br />
placed vertic<strong>al</strong>ly on a floor. The rod beg<strong>in</strong>s to f<strong>al</strong>l without any<br />
<strong>in</strong>iti<strong>al</strong> velocity. At what angle a b<strong>et</strong>ween the rod and a vertic<strong>al</strong><br />
will the end of the rod no longer press on the. floor?<br />
At what coefficient of friction will the end of the rod not<br />
slip up to this moment?<br />
t<br />
90<br />
Fig. 72
46 PROBLEMS<br />
Fig. 73 Fig. 74<br />
189. At, what distance from the bottom of the rod will the<br />
sphere (see Problem 188) f<strong>al</strong>l if the coefficient of friction<br />
k:» ~5.<br />
190. A wire is bent <strong>al</strong>ong an arc with a radius R (Fig. 75).<br />
A bead is placed on the wire that can move <strong>al</strong>ong it without<br />
friction. At the <strong>in</strong>iti<strong>al</strong> moment the bead was at po<strong>in</strong>t O. What<br />
horizont<strong>al</strong> velocity should be imparted to the bead for it to<br />
g<strong>et</strong> onto the wire aga<strong>in</strong> at po<strong>in</strong>t B after fiy<strong>in</strong>g a certa<strong>in</strong> distance<br />
(AB) through the air?<br />
191. A sm<strong>al</strong>l body slides down an <strong>in</strong>cl<strong>in</strong>ed surface pass<strong>in</strong>g<br />
<strong>in</strong>to a loop from the m<strong>in</strong>imum height ensur<strong>in</strong>g that the body<br />
does not leave the surface of the loop (Fig. 76). What symm<strong>et</strong>ric<strong>al</strong><br />
segment with an angle ex < 90° can be cut out of the loop<br />
for the body to reach po<strong>in</strong>t B after travell<strong>in</strong>g a certa<strong>in</strong> distance<br />
<strong>in</strong> the air?<br />
A I 8<br />
I<br />
\ I I<br />
\«,'a,IH<br />
~ , I I<br />
", YII<br />
,<br />
A<br />
I<br />
t<br />
o<br />
Fig. 75 Fig. 76
MECHANICS<br />
47<br />
I<br />
I<br />
I<br />
I<br />
I<br />
I<br />
I<br />
I<br />
I<br />
(.--. I _---- -..<br />
Fig. 77<br />
..........--------<br />
How will the body move if the angle (X is 'greater or sm<strong>al</strong>ler<br />
than the found one? Disregard friction and the resistance of<br />
the air.<br />
192-: Three weights mov<strong>in</strong>g circularly are attached {f) the ends<br />
of a str<strong>in</strong>g pass<strong>in</strong>g over two nails (Fig. 77). Two weights with<br />
a mass m each hang on the left and one weight with a mass<br />
2m on the right. Will this system be <strong>in</strong> equilibrium?<br />
193. A b<strong>al</strong>l is suspended from a very th<strong>in</strong> str<strong>in</strong>g. The latter<br />
is brought <strong>in</strong>to a horizont<strong>al</strong> position and is then released. At<br />
what po<strong>in</strong>ts of the trajectory is the acceleration of the b<strong>al</strong>l<br />
directed vertic<strong>al</strong>ly downward, vertic<strong>al</strong>ly upward and horizont<strong>al</strong>ly?<br />
The str<strong>in</strong>g is not tensioned at the <strong>in</strong>iti<strong>al</strong> moment.<br />
194. A weightless rod can rotate <strong>in</strong> a vertic<strong>al</strong> plane with<br />
respect to po<strong>in</strong>t O. The rod carries masses m, and m, at distan..<br />
a<br />
"tIJ<br />
Fig. 79<br />
0'
a<br />
0'<br />
Fig. 80<br />
Fig- 8/<br />
ces '1 and " from 0 (Fig. 78). The rod is released without any<br />
<strong>in</strong>iti<strong>al</strong> velocity from a position form<strong>in</strong>g an angle ex with the<br />
vertic<strong>al</strong>. D<strong>et</strong>erm<strong>in</strong>e the l<strong>in</strong>ear velocities of masses m, and m,<br />
when the rod reaches a vertic<strong>al</strong> position.<br />
195. A plumb to which a sm<strong>al</strong>l b<strong>al</strong>l is attached by means<br />
of a str<strong>in</strong>g with a length of I = 12.5 cm is secured on the axis<br />
of a centrifug<strong>al</strong> mach<strong>in</strong>e. F<strong>in</strong>d the angle ex through which the<br />
str<strong>in</strong>g deflects from the vertic<strong>al</strong> if.. the mach<strong>in</strong>e makes one revolution<br />
per second, two revolutions per second.<br />
196. A rigid rod bent as shown <strong>in</strong> Fig. 79 rotates with an<br />
angular velocity
MECHANICS 49<br />
the rod. F<strong>in</strong>d the distance I from po<strong>in</strong>t 0 at which the bead<br />
will be <strong>in</strong> equilibrium if the coefficient of friction b<strong>et</strong>ween the<br />
bead and the rod is k.<br />
198. A weight with a mass m is attached to the end of<br />
a str<strong>in</strong>g with a length I fastened to a vertic<strong>al</strong> rod rotat<strong>in</strong>g<br />
with an angular velocity ID. Another str<strong>in</strong>g of the same length<br />
as the first and carry<strong>in</strong>g on its end another weight with a mass<br />
m is secured to the first weight.<br />
Prove that dur<strong>in</strong>g rotation the angle b<strong>et</strong>ween the first str<strong>in</strong>g<br />
and the vertic<strong>al</strong> will be sm<strong>al</strong>ler than the angle b<strong>et</strong>ween the<br />
vertic<strong>al</strong> and the second str<strong>in</strong>g. Disregard the weight of the<br />
str<strong>in</strong>g.<br />
199. A weightless rod carries two weights of mass m and JW.<br />
The rod is h<strong>in</strong>ge-jo<strong>in</strong>ted to vertic<strong>al</strong> axis 00' (Fig. 81), which<br />
rotates with an angular velocity ID. D<strong>et</strong>erm<strong>in</strong>e the angle cp formed<br />
by the rod and the vertic<strong>al</strong>.<br />
200. A horizont<strong>al</strong> straight bar rotates with a constant angular<br />
velocity around a vertic<strong>al</strong> axis. A body can move without friction<br />
over the bar. Initi<strong>al</strong>ly, the body is r<strong>et</strong>a<strong>in</strong>ed <strong>in</strong> equilibrium by<br />
a spr<strong>in</strong>g (Fig. 82). What will happen to the body if an <strong>in</strong>iti<strong>al</strong><br />
velocity is imparted to it <strong>al</strong>ong the bar? The length of the spr<strong>in</strong>g<br />
<strong>in</strong> an unstr<strong>et</strong>ehed state can be neglected. .<br />
201. A m<strong>et</strong><strong>al</strong>lic cha<strong>in</strong> with a length of 1=62.8 em and whose<br />
ends are jo<strong>in</strong>ed tog<strong>et</strong>her is fitted onto a wooden disk (Fig. 83).<br />
The disk rotates with a speed of n = 60 rps. F<strong>in</strong>d the tension<br />
of the cha<strong>in</strong> T if its mass is m = 40 g.<br />
202. Water flows with a velocity v <strong>al</strong>ong a rubber tube hav<strong>in</strong>g<br />
the form of a r<strong>in</strong>g (Fig. 84). The radius of the r<strong>in</strong>g is Rand<br />
the diam<strong>et</strong>er of the tube d~R. What force is the rubber tube<br />
str<strong>et</strong>ched with?<br />
203. A homogeneous rod with a length I and a mass m rotates<br />
with an angular velocity (J) <strong>in</strong> a horizont<strong>al</strong> plane around an axis<br />
pass<strong>in</strong>g through its end. F<strong>in</strong>d the tension of the rod at a distance x<br />
from its axis of rotation.<br />
204. A b<strong>al</strong>l with the mass m secured on a weightless rod rotates<br />
with a constant velocity v <strong>in</strong> a horizont<strong>al</strong> plane (Fig. 85).<br />
Its k<strong>in</strong><strong>et</strong>ic energy <strong>in</strong> a coord<strong>in</strong>ate system that is stationary with<br />
respect to the axis of rotation is constant and equ<strong>al</strong> to mv'/2.<br />
The k<strong>in</strong><strong>et</strong>ic energy changes with time from zero to 4mv"/2 with<br />
respect to a read<strong>in</strong>g system that moves rectil<strong>in</strong>early <strong>in</strong> a horizont<strong>al</strong><br />
plane with a velocity v relative to the axis. What is the<br />
reasIn for this change <strong>in</strong> the energy?<br />
4-2042
50 PROBLEMS<br />
Fig. 84 Fig. 85<br />
205. A th<strong>in</strong> homogeneous hoop rolls over a horizont<strong>al</strong> surface<br />
with a constant velocity v. How and under the action of what<br />
forces does the tot<strong>al</strong> energy of a sm<strong>al</strong>l section AB that is at the<br />
highest po<strong>in</strong>t of the loop at the given moment change?<br />
206. A heavy spool with a thread wound on it lies on a rough<br />
horizont<strong>al</strong> surface over which it can roll without slipp<strong>in</strong>g. If the<br />
thread is pulled to the left <strong>in</strong> a horizont<strong>al</strong> direction, the spool<br />
will <strong>al</strong>so move to the left. If the direction of the thread is changed<br />
(Fig. 86), the spool will beg<strong>in</strong> to roll to the right at a certa<strong>in</strong><br />
angle a b<strong>et</strong>ween the direction of the thread and a vertic<strong>al</strong> l<strong>in</strong>e.<br />
D<strong>et</strong>erm<strong>in</strong>e this angle. What will happen to the spool at the<br />
given v<strong>al</strong>ue of this angle? The radius of the spool heads is R<br />
and that of its core is r.<br />
207. F<strong>in</strong>d the k<strong>in</strong><strong>et</strong>ic energy of a hoop with a mass M and a<br />
radius R if it moves uniformly with a velocity v and rotates<br />
~, ,<br />
with an angular velocity (J)<br />
around an axis pass<strong>in</strong>g through<br />
its centre.<br />
208. F<strong>in</strong>d the k<strong>in</strong><strong>et</strong>ic energy<br />
of the crawler of a tractor mo-<br />
Fig. 86<br />
.,<br />
Fig. 81
MECHANICS<br />
61<br />
,<br />
Fig. 88 Fig. 89<br />
v<strong>in</strong>g with a speed v. The distance b<strong>et</strong>ween the centres of the<br />
wheels which the crawler is placed onto is I. The radius of the<br />
wheels is r. A unit of crawler length weighs G.<br />
209. Two cyl<strong>in</strong>ders of equ<strong>al</strong> mass and size are made of unknown<br />
materi<strong>al</strong>s of different density. How is it possible to tell which<br />
of the two is hollow?<br />
210. A flexible cable is wound <strong>in</strong> one row around a drum<br />
with a radius R (Fig. 87). The weight of a unit of cable length<br />
is p. The entire cable weighs G. The drum moves by <strong>in</strong>ertia<br />
without slipp<strong>in</strong>g <strong>al</strong>ong a horizont<strong>al</strong> surface, and the cable is<br />
wound otT onto it.<br />
At the <strong>in</strong>iti<strong>al</strong> moment, when the cable was compl<strong>et</strong>ely wound<br />
on the drum, the velocity of the drum centre was v.<br />
Appraise the velocity of the drum centre at the moment of<br />
time when a part of the cable with a length x lies on the surface,<br />
neglect<strong>in</strong>g the radius of the cable cross section (<strong>in</strong> comparison<br />
with R) and the mass of the drum.<br />
What force changes the momentum of the cable?<br />
211. A friction force f is applied to a pulley with a radius r<br />
rotat<strong>in</strong>g around a stationary axis (Fig. 88). D<strong>et</strong>erm<strong>in</strong>e the change<br />
<strong>in</strong> the angular velocity of the pulley with time if this velocity<br />
is W o at the <strong>in</strong>iti<strong>al</strong> moment. The mass of the pulley is m and<br />
the mass of the spokes can be neglected.<br />
212. A hoop of radius, rotat<strong>in</strong>g with an angular velocity 6).<br />
is placed on a rough horizont<strong>al</strong> surface. D<strong>et</strong>erm<strong>in</strong>e the velocity v<br />
of the centre of the hoop when the hoop ceases to slip. At the<br />
<strong>in</strong>iti<strong>al</strong> moment the velocity of the centre of the hoop is zero.<br />
213. A translation<strong>al</strong> velocity V o is imparted <strong>in</strong> a 1iorizont<strong>al</strong><br />
direction to a hoop with a radius r placed on a rough horizont<strong>al</strong><br />
4·
52 PROBLEMS<br />
surface. F<strong>in</strong>d the angular velocity w of rotation of the hoop<br />
after it stops slipp<strong>in</strong>g.<br />
214. A hoop with a radius, rotat<strong>in</strong>g with an angular velocity<br />
(00<br />
is placed on a rough horizont<strong>al</strong> surface. A translation<strong>al</strong><br />
velocity V o is imparted to the hoop (Fig. 89). D<strong>et</strong>erm<strong>in</strong>e the<br />
nature of the motion of the hoop, assum<strong>in</strong>g that the force of<br />
slid<strong>in</strong>g friction is f.<br />
215. A cyl<strong>in</strong>dric<strong>al</strong> tube with a radius, is connected by means<br />
of spokes to two hoops with a radius R. The mass of both the<br />
hoops is M. The mass of the tube and the spokes <strong>in</strong> comparison<br />
with the mass M can be neglected. A str<strong>in</strong>g passed over a weightless<br />
pulley is wound around the tube. A weight with a mass m<br />
is attached to the end of the str<strong>in</strong>g (Flg. 90).<br />
F<strong>in</strong>d the acceleration of the weight, the tension of the str<strong>in</strong>g<br />
and the force of friction act<strong>in</strong>g b<strong>et</strong>ween the hoops and the surface.<br />
(Assume that the hoops do not slip.) Also d<strong>et</strong>erm<strong>in</strong>e the<br />
coefficient of friction at which the hoops will slip.<br />
216. A spool lies on an <strong>in</strong>cl<strong>in</strong>ed plane. A thread is wound on<br />
the spool and its free end carry<strong>in</strong>g a weight hav<strong>in</strong>g a mass m is<br />
thrown over a weightless pulley (Fig. 91). It is assumed that the<br />
mass of the spool M is uniformly distributed over a circle with<br />
a radius R and there is no friction. D<strong>et</strong>erm<strong>in</strong>e the angle of <strong>in</strong>cl<strong>in</strong>ation<br />
a, at which the centre of gravity of the spool will be<br />
at rest.<br />
217. A board with a mass M is placed on two identic<strong>al</strong> cyl<strong>in</strong>dric<strong>al</strong><br />
rolls with a radius R. The rolls rest on a horizont<strong>al</strong><br />
surface. At the <strong>in</strong>iti<strong>al</strong> moment the system is at rest. Then, a<br />
Fig. 90 Fig. 9/
MECHANICS 53<br />
~ force Q is applied to the board <strong>in</strong> a horizont<strong>al</strong><br />
direction. · F<strong>in</strong>d the acceleration of<br />
the board and the forces or friction act<strong>in</strong>g<br />
b<strong>et</strong>ween the rolls and the board and <strong>al</strong>so<br />
b<strong>et</strong>ween the rolls and the horizont<strong>al</strong> surface.<br />
The rolls have the form of th<strong>in</strong>-w<strong>al</strong>led<br />
cyl<strong>in</strong>ders with a mass m each and do not<br />
slip.<br />
218. A double-step pulley consists of two<br />
rigidly connected th<strong>in</strong> hoops with radii R<br />
and r and masses M 1 and M't respectively.<br />
Str<strong>in</strong>gs carry<strong>in</strong>g weights m 1<br />
and m, at<br />
their ends are passed around each step<br />
(Fig. 92).<br />
F<strong>in</strong>d the acceleration of the weights m 1<br />
~ and m" the tension of the str<strong>in</strong>gs and the<br />
force with which the system acts on the<br />
Fig. 92<br />
axis of the pulley.<br />
219. A homogeneous th<strong>in</strong>-w<strong>al</strong>led cyl<strong>in</strong>der<br />
with a radius R and a mass M rolls without slipp<strong>in</strong>g under the<br />
force of gravity down an <strong>in</strong>cl<strong>in</strong>ed plane form<strong>in</strong>g an angle a, with<br />
the horizon.<br />
Us<strong>in</strong>g the law of conservation of energy, d<strong>et</strong>erm<strong>in</strong>e the velocity<br />
of the centre of gravity and the angular velocity of cyl<strong>in</strong>der<br />
rotation when the time t elapses after motion beg<strong>in</strong>s. (It is<br />
assumed that the cyl<strong>in</strong>der is at rest at the <strong>in</strong>iti<strong>al</strong> moment.) Also<br />
f<strong>in</strong>d the acceleration of the centre of gravity of the cyl<strong>in</strong>der.<br />
1-9. The Law of Gra~itation<br />
220. \Vhy does the Earth impart to <strong>al</strong>l bodies the same deceleration<br />
irrespective of their mass?<br />
221. F<strong>in</strong>d the magnitude and dimension of the gravitation<strong>al</strong><br />
constant <strong>in</strong> the CGS system, bear<strong>in</strong>g <strong>in</strong> m<strong>in</strong>d that the mean radius<br />
of the Earth is R. -=6.4 X 10 8 em and its mass At = 6 X 10 27 g.<br />
222. In what conditions will bodies <strong>in</strong>side a spaceship be<br />
weightless, i. e., cease to exert any pressure on the w<strong>al</strong>ls of the<br />
cab<strong>in</strong>?<br />
223. A light pendulum consist<strong>in</strong>g of a rod and a disk (Fig. 93)<br />
is attached to a wooden block which can freely f<strong>al</strong>l <strong>al</strong>ong guid<strong>in</strong>g<br />
wires.
54<br />
PROBLEMS<br />
'-...<br />
' ..........-.......... "<br />
-------~<br />
....-.----41-J ~ ,81<br />
Fig. 93 Fig. 94<br />
The pendulum is brought out of equilibrium through an angle a<br />
and released. At the moment when the pendulum passed through<br />
its lowermost position, the block was released and began to f<strong>al</strong>l<br />
freely. How will the pendulum move with respect to the block?<br />
Disregard friction and the resi stance of the air.<br />
224. A plan<strong>et</strong> moves <strong>al</strong>ong an ellipse hav<strong>in</strong>g the Sun <strong>in</strong> its<br />
focus. Tak<strong>in</strong>g <strong>in</strong>to account the work of. the force of gravity,<br />
<strong>in</strong>dicate the po<strong>in</strong>ts on the trajectory at which the velocity of<br />
the plan<strong>et</strong> will be maximum and m<strong>in</strong>imum.<br />
225. An artifici<strong>al</strong> satellite of the Earth moves at an <strong>al</strong>titude<br />
of h = 670 km <strong>al</strong>ong a circular orbit. F<strong>in</strong>d the velocity of the<br />
satellite.<br />
226. How will the velocity of an artifici<strong>al</strong> satellite of the<br />
Earth change with time when it moves <strong>in</strong> the upper 1ayers of<br />
the atmosphere?<br />
227. Two satellites move <strong>al</strong>ong a circular orbit <strong>in</strong> the same<br />
direction at a sm<strong>al</strong>l distance from each other. A conta<strong>in</strong>er has<br />
to be thrown from the first sa tellite onto the second one. When<br />
will the conta<strong>in</strong>er reach the second satellite faster: if it is<br />
thrown <strong>in</strong> the direction of motion of the first satellite or <strong>in</strong><br />
the opposite direction? The velocity of the conta<strong>in</strong>er with respect<br />
to the satellite u is much less than that of the satellite v.<br />
228. D<strong>et</strong>erm<strong>in</strong>e the mass of the Sun M if the mean radius of<br />
the Earth's orbi t is R = 149 x 10 8 km.<br />
229. D<strong>et</strong>erm<strong>in</strong>e the m<strong>in</strong>imum distance h from the first Sovi<strong>et</strong><br />
artifici<strong>al</strong> satellite launched on October 4, 1957, to the Earth'S<br />
surface if the follow<strong>in</strong>g data are known: the maximum distance<br />
h<strong>et</strong>ween the satellite and the Earth H = 900 krn, the period<br />
of revolution around the Earth T = 96 m<strong>in</strong>utes, the major see
235. A block floats vertic<strong>al</strong>ly <strong>in</strong> a glass filled with water. How<br />
will the level of the water <strong>in</strong> the glass change if the block as<br />
sumes a horizont<strong>al</strong> posi tion?<br />
236. A vessel filled with water<br />
is placed on the edge of a board<br />
(Fig. 95). Will equilibrium be violated<br />
if a sm<strong>al</strong>l board carry<strong>in</strong>g a<br />
weight is placed on the surface of<br />
the water?<br />
237. A piece of ice fioa ts <strong>in</strong> a<br />
~ glass filled with water. How will<br />
Fig. 95 the level of the water <strong>in</strong> the<br />
o<br />
MECHANICS 55<br />
miaxis of the Moon's orbit R= 384,400 krn, the period of ro..<br />
tation of the Moon around the Earth T = 27.3 days and the<br />
Earth's radius R o = 6,370 km.<br />
230. An air bubble with a radius r and an iron b<strong>al</strong>l with the<br />
same radius are present <strong>in</strong> water. Will they be attracted or<br />
repulsed from each other? What is the force of <strong>in</strong>teraction b<strong>et</strong>ween<br />
them? The distance b<strong>et</strong>ween the centre of the b<strong>al</strong>l and the<br />
bubble is R.<br />
231. Two air bubbles with a radius, are present <strong>in</strong> water.<br />
Are the bubbles attracted or repulsed? What is the force of <strong>in</strong>teraction<br />
b<strong>et</strong>ween them? The distance b<strong>et</strong>ween the bubbles is R.<br />
232. A lead b<strong>al</strong>l with a radius of R= 50 em has <strong>in</strong>side a<br />
spheric<strong>al</strong> space with a radius r = 5 cm whose centre is at a<br />
distance of d = 40 cm from the centre of the b<strong>al</strong>l (Fig. 94).<br />
With what force will a materi<strong>al</strong> particle with a mass of<br />
m= 10 g at a distance of 1=80 em from the centre of the<br />
b<strong>al</strong>l be attracted to it if the l<strong>in</strong>e connect<strong>in</strong>~ the centres of the<br />
b<strong>al</strong>l and the space forms an angle a=60° with the l<strong>in</strong>e connect<strong>in</strong>g<br />
the centre of the b<strong>al</strong>l and the materi<strong>al</strong> particle?<br />
233. A body whose dimensions may be neglected is placed<br />
<strong>in</strong>side a th<strong>in</strong> homogeneous sphere. Prove that the force of attraction<br />
act<strong>in</strong>g from the sphere on the body is zero irrespective<br />
of its position <strong>in</strong> the sphere.<br />
234. What is the force with which a body with a mass m<br />
<strong>in</strong> a deep m<strong>in</strong>e will be attracted to the centre of the Earth<br />
if the distance b<strong>et</strong>ween the body and the Earth's centre is ,?<br />
The density of the Earth is the same everywhere and equ<strong>al</strong> to p,<br />
1·10. Hydro- and Aerostatics
56<br />
PROBLEMS<br />
~==-==~~~~----------<br />
------<br />
- - - -<br />
--<br />
-- - - - -=..-=---=----:-<br />
----<br />
7- ~_-----=<br />
Fig. 96<br />
glass change when the ice melts? Consider the follow<strong>in</strong>g cases:<br />
(1) the ice is absolutely homogeneous;<br />
(2) a stone is frozen <strong>in</strong> the ice;<br />
(3) the ice conta<strong>in</strong>s an air bubble.<br />
238. A solid homogeneous body submerged <strong>in</strong>to a liquid with<br />
a specific weight of '\'1 weighs 01' and <strong>in</strong> a liquid with a specific<br />
weight of 'Vt it weighs G•• D<strong>et</strong>erm<strong>in</strong>e the specific weight y<br />
of the body.<br />
239. A hole is cut <strong>in</strong> the ice <strong>in</strong> the middle of a large lake.<br />
The ice is 10 m<strong>et</strong>res thick. What length of rope is required to<br />
scoop up a buck<strong>et</strong>ful of water?<br />
240. A match-box with a sm<strong>al</strong>l stone on its bottom floats <strong>in</strong><br />
a cup filled with water. Wi)) the level of the water <strong>in</strong> the cup<br />
change if the stone is taken out of the box and dropped <strong>in</strong>to<br />
the water?<br />
241. A ship pass<strong>in</strong>g through a lock rises to a higher level <strong>in</strong><br />
a chamber of the lock <strong>in</strong>to which water is pumped from the<br />
side of the lower level (Fig. 96). When will the pumps perform<br />
more work: when a large ship is <strong>in</strong> the chamber, or a sm<strong>al</strong>l boat?<br />
242. A square with a side a and a rectangle with sides a and<br />
2u are cut out from two plates of equ<strong>al</strong> thickness with specific<br />
weights of 3.5 gf/cm' and 2 gf/cm 8 , the square be<strong>in</strong>g cut out of<br />
the heavier materi<strong>al</strong>. The square and the rectangle are fastened<br />
tog<strong>et</strong>her <strong>in</strong> the form of the l<strong>et</strong>ter L and placed upside down on<br />
the bottom of an empty vesseJ (Fig. 97). \Vhat will occur if the<br />
vessel is filled with water?
MECHANICS 57<br />
243. A tube floats vertic<strong>al</strong>ly <strong>in</strong> water (Fig. 98). The portion<br />
of the tube protrud<strong>in</strong>g from the water is h = 5 em. Oil with a<br />
specific weight of y =0.9 gf/cm 8 is poured <strong>in</strong>to the tube. What<br />
length can the tube have so that it can be compl<strong>et</strong>ely filled<br />
with oil?<br />
244. A vessel with water f<strong>al</strong>ls with an acceleration a < g.<br />
How does the pressure p <strong>in</strong> the vessel change with depth?<br />
245. A vessel with a body Boat<strong>in</strong>g <strong>in</strong> it f<strong>al</strong>ls with an acceleration<br />
a < g. Will the body rise to the surface?<br />
246. A cart supports a cubic tank filled with water up to its<br />
top (Fig. 99). The cart moves with a constant acceleration a.<br />
D<strong>et</strong>erm<strong>in</strong>e the pressure at po<strong>in</strong>t A which is at a depth h and a<br />
distance I from the front w<strong>al</strong>l, if the tank is tightly closed with<br />
a lid. In uniform motion the lid does not exert any pressureon<br />
the water.<br />
247. A rubber b<strong>al</strong>l with a mass m and a radius R is submer..<br />
ged <strong>in</strong>to water to a depth h and released. What height will the<br />
ban jump up to above the surface of the water? Disregard the<br />
resistance of the water and the air.<br />
248. Mercury is poured <strong>in</strong>to two communicat<strong>in</strong>g cyl<strong>in</strong>dric<strong>al</strong><br />
vessels, and water is poured <strong>in</strong> above it. The level of the water<br />
is the same <strong>in</strong> both vessels. Will the level of the water and the<br />
mercury be the same if a piece of wood is dropped <strong>in</strong>to one<br />
vessel and some water equ<strong>al</strong> <strong>in</strong> weight to this piece is added to<br />
the other? Consider cases when the cross sections of the vessels<br />
are the same or different.<br />
:t:::====;7<br />
=--~ =-.J.:--w---.:=<br />
,--I -:c- 1--<br />
=-F~~~<br />
-- L--~ ......._ ..;7 ~ _<br />
Fig. 97 Fig. 98
58<br />
PROBLEMS<br />
"/I I<br />
11<br />
L..-__----...,..~~<br />
Fig. 99<br />
Fig. 100<br />
249. Mercury is poured <strong>in</strong>to cyl<strong>in</strong>dric<strong>al</strong> communicat<strong>in</strong>g vessels<br />
with different cross sections. An iron block with a volume V o is<br />
dropped <strong>in</strong>to the broad vessel, and as a result the level of the<br />
mercury <strong>in</strong> it rises. Then water is poured <strong>in</strong>to the vessel until<br />
the mercury reaches the previous level. F<strong>in</strong>d the height of the<br />
water column h if the cross section of the narrow veSsel is AI.<br />
250•.One end of a board with a length 1 is placed on top of<br />
a stone protrud<strong>in</strong>g from water. Part a is above the po<strong>in</strong>t of<br />
support (Fig. 100). What part of the board is below the surface<br />
of the water if the specific weight of wood is y?<br />
251. A conic<strong>al</strong> vessel without a bottom tightly stands on a<br />
table. A liquid is poured <strong>in</strong>to the vessel and as soon as its level<br />
reaches the height h, the pressure of the liquid raises the vessel.<br />
The radius of the bottom greater base of the vessel is R, the<br />
angle b<strong>et</strong>ween the cone generatrix and a vertic<strong>al</strong> is a, and the<br />
weight of the vessel is G. What is the density of the liquid?<br />
252. Can water be pumped over a<br />
w<strong>al</strong>l 20 m<strong>et</strong>res high with the aid of<br />
a syphon?<br />
- - --<br />
---<br />
-<br />
-- --<br />
A<br />
FiB- 101 Fi,. 102
MECHANICS<br />
59<br />
c<br />
F<br />
----- --- -- - ~ ---<br />
Fig. 103 Fig~ 104<br />
253. The vessel shown <strong>in</strong> Fig. 101 is entirely filled with water.<br />
What will happen if plug A is removed? The radius of the<br />
hole is about 0.5 em,<br />
254. Four piston pumps are made of pipe sections with a large<br />
and a sm<strong>al</strong>l diam<strong>et</strong>ers. The pumps lifted water to the same<br />
height H+h (Fig. 102). Which of the pistons should be pulled<br />
with a greater force to keep it <strong>in</strong> equilibrium? Disregard the<br />
weight of the pistons.<br />
255. A piston weigh<strong>in</strong>g G= 3 kgf has the form of a circular<br />
disk with a radius R = 4 em. The disk has a hole <strong>in</strong>to which a<br />
th<strong>in</strong>-w<strong>al</strong>led pipe with a radius ·r = 1 em is <strong>in</strong>serted. The piston<br />
can enter a cyl<strong>in</strong>der tightly and without friction, and is <strong>in</strong>iti<strong>al</strong>ly<br />
at the bottom of the cyl<strong>in</strong>der. What height H will the piston<br />
rise to if m = 700 g of water is poured <strong>in</strong>to the pipe?<br />
256. A vessel with a hole <strong>in</strong> its bottom is fastened on a cart.<br />
The mass of the vessel and the cart is M and the area of the<br />
vessel base is A. What force F should the cart be pulled with<br />
so that a maximum amount of water rema<strong>in</strong>s <strong>in</strong> the vessel? The<br />
dimensions of. the vessel are shown <strong>in</strong> Fig. 103. There is no<br />
friction.<br />
257. The follow<strong>in</strong>g design of a perp<strong>et</strong>uum mobile was suggested<br />
(Fig. 104). A herm<strong>et</strong>ic vessel is divided <strong>in</strong>to two h<strong>al</strong>ves by an<br />
air-tight partition through which .a tube and a water turb<strong>in</strong>e of<br />
a speci<strong>al</strong> design are passed. The turb<strong>in</strong>e is provided with chambers<br />
hav<strong>in</strong>g covers which close and open automatic<strong>al</strong>ly. The<br />
pressure PI <strong>in</strong> the lower part of the vessel is greater than the<br />
pressure P2 <strong>in</strong> the upper part, and the water rises <strong>al</strong>ong the tube<br />
fill<strong>in</strong>g an open chamber of the turb<strong>in</strong>e. After this the chamber<br />
closes and the disk turns. In the lower part of the vessel the<br />
chamber opens automatic<strong>al</strong>ly and r<strong>et</strong>urns the water. After this
60 PROBLEMS<br />
the chamber closes herm<strong>et</strong>ic<strong>al</strong>ly, <strong>et</strong>c. Why will this mach<strong>in</strong>e not<br />
function perp<strong>et</strong>u<strong>al</strong>ly?<br />
258. The follow<strong>in</strong>g <strong>al</strong>ternative of the mach<strong>in</strong>e described <strong>in</strong><br />
Problem 257 was suggested. Air-tight chambers (Fig. 105) are<br />
filled with water <strong>in</strong> the right-hand side of the disk and lower.<br />
At the bottom the chambers open and, <strong>in</strong> contrast to the mach<strong>in</strong>e<br />
<strong>in</strong> Problem 257, the w<strong>al</strong>ls of the chambers are automatic<strong>al</strong>ly<br />
r<strong>et</strong>racted <strong>in</strong>to the disk. In the upper portion of the vessel<br />
the w<strong>al</strong>ls are automatic<strong>al</strong>ly pushed out and the chambers are<br />
filled with water. Otherwise this perp<strong>et</strong>uum mobile is designed<br />
on the same l<strong>in</strong>es as the previous one. Why will it <strong>al</strong>so fail<br />
to work?<br />
259. Three vessels with attached bottoms are submerged <strong>in</strong>to<br />
water to the same depth (Fig. 106). Each bottom will f<strong>al</strong>l<br />
off if the respective vessel is filled with 1 kgf of water. WiIJ<br />
the bottom f<strong>al</strong>loff if the vessels are filled with 1 kgf of oil?<br />
1 kgf of mercury? or if a weight of 1 kgf is placed <strong>in</strong>to each<br />
vessel?<br />
260. A body is weighed on an accurate an<strong>al</strong>ytic<strong>al</strong> b<strong>al</strong>ance<br />
placed under a glass hood. Will the read<strong>in</strong>g of the b<strong>al</strong>ance<br />
change if the air is pumped out from the hood?<br />
261. A man carries a tyre tube and decides to make it lighter<br />
by us<strong>in</strong>g the expulsive force of air (accord<strong>in</strong>g to Archimedes'<br />
pr<strong>in</strong>ciple). For this purpose he <strong>in</strong>flates the tube, thus<br />
<strong>in</strong>creas<strong>in</strong>g its volume. Will his aim be achieved?<br />
262. What error is made <strong>in</strong> weigh<strong>in</strong>g a body with a volume<br />
of V = 1 litre if when weighed <strong>in</strong> the air copper weights weigh<strong>in</strong>g<br />
0 1 = 800 gf are placed on the pan of the b<strong>al</strong>ance. The<br />
specific weightof copper '\'1 = 8.8 gf/cm ' and of air Yo = 1.29gf/litre.<br />
FI,. /O! PI,. 106
MECHANICS<br />
61<br />
Fig. lUI<br />
263. A cup and a V-shaped mercury barom<strong>et</strong>ers are brought<br />
<strong>in</strong>to equilibrium on a very sensitive b<strong>al</strong>ance (Fig. 107). The<br />
barom<strong>et</strong>ers are made of the same materi<strong>al</strong>, have tubes of the<br />
same diam<strong>et</strong>er and conta<strong>in</strong> the same amount of mercury. The<br />
distance b<strong>et</strong>ween the soldered ends of the tubes and the upper<br />
levels of the mercury <strong>in</strong> them is <strong>al</strong>so the same. How will the<br />
equilibrium of the b<strong>al</strong>ance change if the atmospheric pressure<br />
grows?<br />
264. Appraise the weight of the Earth's atmosphere.<br />
265. An air mattress is filled with air to a certa<strong>in</strong> pressure<br />
exceed<strong>in</strong>g the atmospheric pressure. When will the air pressure<br />
<strong>in</strong> the mattress be greater: when a man stands on it or when<br />
he lies on it?<br />
266. A wheel of a motor vehicle is designed as follows.<br />
A rubber tube enclosed <strong>in</strong> a tyre cas<strong>in</strong>g is placed onto a m<strong>et</strong><strong>al</strong>lic<br />
rim. The tube is then <strong>in</strong>flated with air. The air pressure<br />
<strong>in</strong> the top and the bottom of the tube is the same. Besides<br />
the air pressure, the rim is acted upon by the force of gravi ty<br />
(Fig. 108). Why does the rim not lower? What holds it <strong>in</strong> a<br />
state of equilibrium?<br />
267. A steam boiler consists of a cyl<strong>in</strong>dric<strong>al</strong> portion and two<br />
hemispheric<strong>al</strong> heads (Fig. 109), <strong>al</strong>l of the same radius. The cyl<strong>in</strong>der<br />
w<strong>al</strong>ls are 0.5 em thick. All the parts of the boiler are
62 PROBLEMS<br />
made of the same materi<strong>al</strong>. How thick should the w<strong>al</strong>ls of the<br />
heads be for the strength of <strong>al</strong>l the parts of the boiler to be<br />
identic<strong>al</strong>?<br />
268. What shape should a steam boiler be given to obta<strong>in</strong><br />
the maximum strength with the given w<strong>al</strong>l thickness?<br />
269. Why is b<strong>al</strong>last <strong>al</strong>ways taken on a stratosphere b<strong>al</strong>loon,<br />
<strong>al</strong>though extra weight <strong>in</strong>evitably decreases its ceil<strong>in</strong>g?<br />
1-11. Hydro- and Aerodynamics<br />
270. Two holes with an area of A = 0.2 em! each are drilled<br />
one above the other <strong>in</strong> the w<strong>al</strong>l of a vessel filled with water ..<br />
The distance b<strong>et</strong>ween the holes H = 50 em. Every second<br />
Q= 140 em" of water is poured <strong>in</strong>to the vessel.. F<strong>in</strong>d the po<strong>in</strong>t<br />
where the streams flow<strong>in</strong>g out of the holes <strong>in</strong>tersect.<br />
271. A broad vessel with water stands on a smooth surface.<br />
The level of the water <strong>in</strong> the vessel is h. The vessel tog<strong>et</strong>her<br />
with the water weighs G.. The side w<strong>al</strong>l of the vessel has at the<br />
bottom a plugged hole (with rounded edges) with an area A. At<br />
what coefficient of friction b<strong>et</strong>ween the bottom and the surface<br />
will the vessel beg<strong>in</strong> to move if the plug is removed?<br />
272. When a stream of liquid flows out of a vessel through<br />
a hole with an area A o , the force that acts on the w<strong>al</strong>l with<br />
the hole is 2pA o sm<strong>al</strong>ler than the force act<strong>in</strong>g' on the opposite<br />
w<strong>al</strong>l (see Problem 271). .<br />
If a tube is <strong>in</strong>serted <strong>in</strong>to the hole as shown <strong>in</strong> Fig. 110, the<br />
difference b<strong>et</strong>ween the forces act<strong>in</strong>g on the opposite w<strong>al</strong>ls will<br />
approximately equ<strong>al</strong> pAo, s<strong>in</strong>ce the tube will prevent the motion<br />
of the liquid near the w<strong>al</strong>ls.<br />
On the other hand, the change <strong>in</strong> the momentum of the liquid<br />
flow<strong>in</strong>g out of the vessel <strong>in</strong> a unit of time is <strong>al</strong>ways 2pA,<br />
Fig. JOB<br />
Fig. J09
MECHANICS<br />
63<br />
1--- --- -<br />
h-- -<br />
Fig. 110 Fig. 111<br />
where A is the cross-section<strong>al</strong> area of the stream. How can these<br />
two facts be brought <strong>in</strong>to agreement?<br />
273. A stream of water flow<strong>in</strong>g out of a pipe with a diam<strong>et</strong>er<br />
of d = 1 ern at a velocity of v = 1 m/s strikes a vertic<strong>al</strong> w<strong>al</strong>l.<br />
D<strong>et</strong>erm<strong>in</strong>e the force act<strong>in</strong>g on the w<strong>al</strong>l, assum<strong>in</strong>g that the pipe<br />
is perpendicular to it and neglect<strong>in</strong>g splash<strong>in</strong>g of the water.<br />
274. A gas flows with a velocity v <strong>al</strong>ong a pipe with a cross<br />
section A bent through 90 degrees. The density of the gas is p.<br />
What force does the gas act on the pipe with? Disregard the<br />
compression of the gas and friction.<br />
275. F<strong>in</strong>d the force act<strong>in</strong>g on the blade of an undershot wheel<br />
(Fig. 111) if the stream after imp<strong>in</strong>g<strong>in</strong>g on the blade cont<strong>in</strong>ues<br />
to move with the velocity of the blade.<br />
The height of the water head is h, the radius of the wheel<br />
is R, the angular velocity of the wheel is (t) and the cross-sec<br />
.tion<strong>al</strong> area of the stream is A.<br />
r ------<br />
- -----~-~<br />
--<br />
FiR. 112<br />
U:<br />
Fig. 113
---<br />
--r-- -<br />
Fig. 114<br />
~--l---I""<br />
Fig. 115<br />
276. A ship g<strong>et</strong>s a large hole <strong>in</strong> its underwater portion<br />
(Fig. 112). In what direction will it beg<strong>in</strong> to move as a result<br />
of this?<br />
277. A liquid ftows out of a broad vessel through a narrow<br />
pipe (Fig. 113). How are the pressure and velocity of the liquid<br />
<strong>in</strong> the vessel and <strong>in</strong> the pipe distributed <strong>al</strong>ong a vertic<strong>al</strong>?<br />
278. The vessel with water described <strong>in</strong> the previous problem<br />
is suspended from a spr<strong>in</strong>g b<strong>al</strong>ance. The lower end of the pipe<br />
is closed with a plug. How will the read<strong>in</strong>g of the b<strong>al</strong>ance<br />
change at the first moment after the plug is removed and the<br />
liquid beg<strong>in</strong>s to ftow out?<br />
279. A vessel with water is placed on one pan of a b<strong>al</strong>ance<br />
<strong>in</strong> equilibrium (Fig. 114). Will the equilibrium change if the<br />
cock is opened? The outftow<strong>in</strong>g water g<strong>et</strong>s onto the same pan<br />
on which the vessel is placed.<br />
280. Figure 115 shows a self-act<strong>in</strong>g water lift<strong>in</strong>g device c<strong>al</strong>led<br />
a hydraulic ram. The pr<strong>in</strong>ciple of its operation is based on<br />
the phenomenon of a hydraulic impact-a sharp <strong>in</strong>crease <strong>in</strong> the<br />
pressure of a liquid Bow<strong>in</strong>g <strong>al</strong>ong a tube when its flow is suddenly<br />
stopped, for example, by the shutt<strong>in</strong>g of a v<strong>al</strong>ve that<br />
discharges the water from the tube.<br />
A tube with a length of 1= 2 m<strong>et</strong>res and a diam<strong>et</strong>er of<br />
d = 20 cm is lowered <strong>in</strong>to a brook with a current velocity of<br />
0=400 em/s.
MECHANICS<br />
65<br />
Fig. 116<br />
(a)<br />
(0)<br />
Fig. 117<br />
First l<strong>et</strong> v<strong>al</strong>ve V 2<br />
be open and v<strong>al</strong>ve V 1 be shut. A sharp<br />
<strong>in</strong>crease <strong>in</strong> pressure will cause v<strong>al</strong>ve VI to open (v<strong>al</strong>ve V 2 will<br />
close at the same time) and the water will flow up <strong>in</strong>to vessel<br />
A. The pressure drops. v<strong>al</strong>ve VI shuts and V 2 opens. The<br />
water <strong>in</strong> the tube assumes its course and the phenomenon is<br />
repeated <strong>in</strong> the previous sequence.<br />
F<strong>in</strong>d the amount of water raised by the ram <strong>in</strong> one hour to<br />
a height of h = 30 m<strong>et</strong>res. if each v<strong>al</strong>ve opens thirty times a<br />
m<strong>in</strong>ute.<br />
281. Dur<strong>in</strong>g storms, when the velocity of the w<strong>in</strong>d is very<br />
high, it tears off the roofs of build<strong>in</strong>gs. Two cases can be dist<strong>in</strong>guished:<br />
(I) if the roof is fastened more firmly at po<strong>in</strong>ts A<br />
and B than at ridge C, the w<strong>in</strong>d will break the roof <strong>al</strong>ong<br />
ridge C and raise both h<strong>al</strong>ves up (Fig. 116a); (2) if the roof is<br />
secured more firmly at the ridge and less firmly at po<strong>in</strong>ts A<br />
and B, the w<strong>in</strong>d will first lift the roof up and then carry it<br />
aside (Fig. 116b). How do you expla<strong>in</strong> this phenomenon?<br />
282. Why will a light celluloid b<strong>al</strong>l placed <strong>in</strong> a stream of<br />
a gas or water issu<strong>in</strong>g at a high velocity from a tube with a<br />
narrow neck freely hover <strong>in</strong> this stream (Fig. 117)?<br />
283. The demonstration device shown <strong>in</strong> Fig. 118 consists of<br />
two disks A and B. The hole <strong>in</strong> the centre of disk A is connected<br />
by a pipe to a cyl<strong>in</strong>der conta<strong>in</strong><strong>in</strong>g compressed air. Disk<br />
B hangs on three p<strong>in</strong>s <strong>al</strong>ong which it can move freely up and<br />
down. If a stream of compressed air is passed through the pipe,<br />
the lower disk will beg<strong>in</strong> to knock aga<strong>in</strong>st the upper one.<br />
Expla<strong>in</strong> the cause of this phenomenon.<br />
5-2042
66<br />
PROBLEMS<br />
~<br />
~<br />
I<br />
A<br />
~<br />
• B<br />
Fig. 118 Fig. /19<br />
f<br />
284. The bottom of a broad vessel is provided with a narrow<br />
tube through which the water can flow out of the vessel<br />
(Fig. 119). A screen is placed b<strong>et</strong>ween the vessel and the tube.<br />
If a light b<strong>al</strong>l is submerged to the bottom of the vessel the<br />
water flows out of it, and the b<strong>al</strong>l will not rise to the surface.<br />
If the outflow of the water is stopped, the b<strong>al</strong>l will immediately<br />
rise to the surface. Why?<br />
(This experiment can easily be conducted <strong>in</strong> a s<strong>in</strong>k with a<br />
p<strong>in</strong>g pong b<strong>al</strong>l.)<br />
285. A pump is designed as a horizont<strong>al</strong> cyl<strong>in</strong>der with a piston<br />
hav<strong>in</strong>g an area of A and an outl<strong>et</strong> orifice hav<strong>in</strong>g an area<br />
of a arranged near the cyl<strong>in</strong>der axis. D<strong>et</strong>erm<strong>in</strong>e the velocity of<br />
outflow of a liquid from the pump if the piston moves with a<br />
constant velocity under the action of a force F. The density of<br />
the liquid is p.<br />
FiB. 120 Fig. /21
MECHANICS 67<br />
286. In Problem 285, the velocity v may become <strong>in</strong>f<strong>in</strong>itely<br />
high even with a sm<strong>al</strong>l force F when a ~ A. Expla<strong>in</strong> this paradoxic<strong>al</strong><br />
phenomenon...<br />
287. A water clock (clepsydra) used <strong>in</strong> ancient Greece is designed<br />
as a vessel with a sm<strong>al</strong>l orifice 0 (Fig. 120). The time<br />
is d<strong>et</strong>erm<strong>in</strong>ed accord<strong>in</strong>g to the level of the water <strong>in</strong> the vessel.<br />
What should the shape of the vessel be for the time sc<strong>al</strong>e to<br />
be uniform?<br />
288. A cyl<strong>in</strong>dric<strong>al</strong> vessel with a liquid rotates with an angular<br />
velocity ro around a vertic<strong>al</strong> axis (Fig. 121). D<strong>et</strong>erm<strong>in</strong>e the<br />
change of pressure <strong>in</strong> the horizont<strong>al</strong> cross section of the vessel<br />
depend<strong>in</strong>g on the distance to the axis of rotation.<br />
Note. Apply the m<strong>et</strong>hod described <strong>in</strong> solv<strong>in</strong>g Problem 203.<br />
289. F<strong>in</strong>d the shape of the surface of a liquid <strong>in</strong> a cyl<strong>in</strong>dric<strong>al</strong><br />
vessel rotat<strong>in</strong>g with an angular velocity ro around a vertic<strong>al</strong><br />
axis (Le., the height of the liquid level depend<strong>in</strong>g on the distance<br />
to the axis of rotation).<br />
290. Why do the tea leaves gather <strong>in</strong> the middle of a glass<br />
after stirr<strong>in</strong>g?
CHAPTER 2<br />
HEAT.<br />
MOLECULAR PHYSICS<br />
2-1. Therm<strong>al</strong> Expansion of Solids and Liquids<br />
291. An iron tyre is to be fitted onto a wooden wheel 100 ern<br />
<strong>in</strong> diam<strong>et</strong>er. The diam<strong>et</strong>er of the tyre is 5 mm sm<strong>al</strong>ler than<br />
that of the wheel. How much should the temperature of the<br />
tyre be <strong>in</strong>creased for this purpose? The coefficient of l<strong>in</strong>ear expansion<br />
of iron ~1 = 12x 10-· degr'.<br />
292. Why is only iron or steel used as re<strong>in</strong>forcement <strong>in</strong> concr<strong>et</strong>e<br />
structures, while other m<strong>et</strong><strong>al</strong>s, dur<strong>al</strong>um<strong>in</strong> for example, are<br />
never employed?<br />
293. Why is a greater time required to measure the temperature<br />
of a human body with a thermom<strong>et</strong>er than to shake it down?<br />
294. The height of a mercury column measured with a brass<br />
sc<strong>al</strong>e at a temperature of t 1 is HI. What height H; will the<br />
mercury column have at to = O°C? The coefficients of l<strong>in</strong>ear expansion<br />
of brass a and of volume expansion of mercury yare<br />
known.<br />
295. How can the temperature of a human body be measured<br />
with a thermom<strong>et</strong>er if the temperature of the ambient air is<br />
+ 42°C?<br />
296. D<strong>et</strong>erm<strong>in</strong>e the lengths of an iron and a copper ruler l~ and<br />
Z; at t = O°C if the difference <strong>in</strong> their lengths at t 1<br />
= 50°C and<br />
t 2<br />
= 450°C is the same <strong>in</strong> magnitude and equ<strong>al</strong> to 1=2 em. The<br />
coefficient of l<strong>in</strong>ear expansion of iron (1,1 = 12X 10- 8 deg-l and<br />
of copper (%2 = 17X 10-· deg-l.<br />
297. The period of oscillations of a pendulum depends on the<br />
length, which changes with the temperature. How should the<br />
pendulum be suspended so that its length does not change with<br />
the temperature?<br />
298. A glass cyl<strong>in</strong>der can conta<strong>in</strong> m o = 100 g of mercury at<br />
a temperature of to = O°C. When t 1 = 20°C, the cyl<strong>in</strong>der can conta<strong>in</strong><br />
m 1 = 99.7 g of mercury. In both cases the temperature of<br />
the mercury is assumed to be equ<strong>al</strong> to that of the cyl<strong>in</strong>der. Use<br />
these data to f<strong>in</strong>d the coefficient of l<strong>in</strong>ear expansion of glass a,<br />
bear<strong>in</strong>g <strong>in</strong> m<strong>in</strong>d that the coefficient of volume expansion of<br />
mercury 11== 18X 10- 1 deg-l.
HEAT. MOLECULAR PHYSICS 69<br />
299. A clock with a m<strong>et</strong><strong>al</strong>lic pendulum is 't 1<br />
= 5 seconds fast<br />
each day at a temperature of t, = + 15°C and 't 2 = 10 seconds<br />
slow at a temperature of t 2<br />
= + 30°C. F<strong>in</strong>d the coefficient -a of<br />
therm<strong>al</strong> expansion of the pendulum m<strong>et</strong><strong>al</strong>, remember<strong>in</strong>g that the<br />
period of oscillations of a pendulum T = 2nV ~. where I is<br />
the length of the pendulum and g is the gravity acceleration.<br />
2...2. The Law of Conservation of Energy.<br />
Therm<strong>al</strong> Conductivity<br />
300. A brick with a mass m is lowered onto a cart with a<br />
mass M that moves rectil<strong>in</strong>early with a constant velocity of 00.<br />
F<strong>in</strong>d the amount of heat liberated <strong>in</strong> this case.<br />
301. An iron washer slides down a weightless rubber cord<br />
with a length lo (Fig. 122). The force of friction b<strong>et</strong>ween the<br />
cord and the washer is constant and equ<strong>al</strong> to f. The coefficient<br />
of elasticity of the cord k is known.<br />
0.<br />
F<strong>in</strong>d the amount of heat Q evolved.<br />
~<br />
What part of the work produced by friction<br />
on the cord will be converted <strong>in</strong>to<br />
heat?<br />
302. A refrigerator consum<strong>in</strong>g P watts<br />
converts q litres of water <strong>in</strong>to ice <strong>in</strong> 't<br />
m<strong>in</strong>utes at a temperature t .<br />
..<br />
~ ~h /~h:I<br />
1 0<br />
Fig. 122 Fig. /23
70 PROBLEMS<br />
What quantity of heat will be liberated by the refrigerator<br />
<strong>in</strong> a room dur<strong>in</strong>g this time if the heat capacity of the refrigerator<br />
can be neglected?<br />
303. Will the temperature <strong>in</strong> a room drop if the door of a<br />
work<strong>in</strong>g refrigerator is opened?<br />
304. It is easiest to heat premises by means of electric appliances.<br />
Is this m<strong>et</strong>hod the most advantageous from the energ<strong>et</strong>ic<br />
po<strong>in</strong>t of view?<br />
305. Equ<strong>al</strong> quantities of s<strong>al</strong>t are dissolved <strong>in</strong> two identic<strong>al</strong><br />
vessels filled with water. In one case the s<strong>al</strong>t is one large cryst<strong>al</strong><br />
and <strong>in</strong> the other-powder.<br />
In which case will the temperature of the solution be higher<br />
after the s<strong>al</strong>t is compl<strong>et</strong>ely dissolved, if <strong>in</strong> both<br />
cases the s<strong>al</strong>t<br />
and the water orig<strong>in</strong><strong>al</strong>ly had the same temperatures?<br />
306. It is known that if water is heated or cooled with certa<strong>in</strong><br />
care it can r<strong>et</strong>a<strong>in</strong> its liquid state at temperatures below<br />
O°C and higher than + 100°C.<br />
A c<strong>al</strong>orim<strong>et</strong>er with a heat capacity of q = 400 c<strong>al</strong>/deg conta<strong>in</strong>s<br />
m, = 1 kg of water cooled to t 1<br />
= -10°C. Next m. = 100g<br />
of water overheated to /2 = + 120°C is added to it.<br />
What is the temperature <strong>in</strong> the c<strong>al</strong>orim<strong>et</strong>er?<br />
307. An <strong>in</strong>candescent lamp consum<strong>in</strong>g P = 54 watts is immersed<br />
<strong>in</strong>to a transparent c<strong>al</strong>orim<strong>et</strong>er conta<strong>in</strong><strong>in</strong>g'V = 650 em! of<br />
water. Dur<strong>in</strong>g ~ = 3 m<strong>in</strong>utes the water is heated by t = 3.4°C.<br />
What part Q of the energy consumed by the lamp passes out<br />
of the c<strong>al</strong>orim<strong>et</strong>er <strong>in</strong> the form of radiant energy? .<br />
308. The area of a brick w<strong>al</strong>l fac<strong>in</strong>g a stre<strong>et</strong> is A = 12 m S<br />
and its thickness d = 1 m<strong>et</strong>re. The temperature of the outside<br />
air T o=-15°C and that of the air <strong>in</strong> the room T= + 15°C.<br />
What amount of heat is lost from the room <strong>in</strong> 24 hours? The<br />
coefficient of therm<strong>al</strong> conductivity of brick is A= 0.003 c<strong>al</strong>/ern X<br />
X s-deg,<br />
309. A w<strong>al</strong>l consists of two adjo<strong>in</strong><strong>in</strong>g panels made of different<br />
materi<strong>al</strong>s. The coefficients of therm<strong>al</strong> conductivity and the<br />
thicknesses of the panels are At' d 1<br />
and A 2 , d 2<br />
, respectively<br />
(Fig. 123). The temperatures of the extern<strong>al</strong> surfaces of the<br />
w<strong>al</strong>l are T 1 and To (where To> T.) and are kept constant.<br />
F<strong>in</strong>d the temperature T" on the plane b<strong>et</strong>ween the panels.<br />
.<br />
310. Assum<strong>in</strong>g <strong>in</strong> Problem 309 that the panels have the same<br />
thickness d, d<strong>et</strong>erm<strong>in</strong>e the<br />
of the w<strong>al</strong>l.<br />
coefficient of therm<strong>al</strong> conductivity
HEAT. MOLECULAR PHYSICS 71<br />
- d<br />
AI<br />
A2<br />
A,<br />
d<br />
Z<br />
d<br />
Z<br />
~ d -<br />
A,<br />
A<br />
AS<br />
A Z<br />
A<br />
A,<br />
A,<br />
A<br />
A Z<br />
}.,8<br />
Fig. 124 Fig. 125 Fig. 126<br />
31t. A w<strong>al</strong>l consists of <strong>al</strong>ternat<strong>in</strong>g blocks with a length d<br />
and coefficients of therm<strong>al</strong> conductivity x, and A,2 (Fig. 124).<br />
The cross-section<strong>al</strong> areas of the blocks are the same. D<strong>et</strong>erm<strong>in</strong>e<br />
the coefficient of therm<strong>al</strong> conductivity of the w<strong>al</strong>l.<br />
312. Two w<strong>al</strong>ls I and II of the same thickness are made of<br />
h<strong>et</strong>erogeneous m<strong>et</strong><strong>al</strong>s, as shown <strong>in</strong> Figs. 125 and 126. In what<br />
case will the coefficient of therm<strong>al</strong> conductivity be greater?<br />
313. Dur<strong>in</strong>g one second m grammes of water boils away <strong>in</strong> a<br />
pan. Assum<strong>in</strong>g that heat is received by the water only through<br />
the bottom of the pan and neglect<strong>in</strong>g the transfer of heat from<br />
the pan w<strong>al</strong>ls and the water surface to the ambient air, d<strong>et</strong>erm<strong>in</strong>e<br />
the temperature T of the pan bottom <strong>in</strong> contact with the<br />
heater. The area of the pan bottom is A and its thickness d.<br />
The coefficient of therm<strong>al</strong> conductivity is A.<br />
2-3. Properties of Gases<br />
314. The cap of a founta<strong>in</strong>-pen is usu<strong>al</strong>ly provided with a<br />
sm<strong>al</strong>l orifice. If it is clogged, the <strong>in</strong>k beg<strong>in</strong>s to leak out of the<br />
pen. What is the cause of this phenomenon?
72 PROBLEMS<br />
315. A barom<strong>et</strong>er gives wrong read<strong>in</strong>gs because some air IS<br />
present above the mercury column. At a pressure of POI = 755 mm<br />
Hg the barom<strong>et</strong>er shows PI = 748 mrn, and at POt = 740 mm it<br />
shows P2 = 736 mm. F<strong>in</strong>d the length 1 of the barom<strong>et</strong>er tube<br />
(Fig. 127).<br />
316. A glass tube with a length of 1= 50 ern and a cross section<br />
of A = 0.5 ern- is soldered at one end and is submerged<br />
<strong>in</strong>to water as shown <strong>in</strong> Fig. 128.<br />
What force F should be applied to hold the tube<br />
under the<br />
water if the distance from the surface of the water to the soldered<br />
end ish = IOcm and the atmospheric pressure Po = 760 mm Hg?<br />
The weight of the tube G= 15 gf.<br />
317. A narrow tube open at both ends is passed through the<br />
cork of a vessel filled with water. The tube does not reach the<br />
bottom of the vessel (Mariotte's vessel shown <strong>in</strong> Fig. 129). Draw<br />
a diagram show<strong>in</strong>g how the pressure p of the air <strong>in</strong> the vessel<br />
depends on the quantity of water Q that has flowed out.<br />
318. Upon each double stroke a piston pump sucks <strong>in</strong> a volume<br />
V o of air. When this pump is used to evacuate the air<br />
from a vessel with a volume V it performs n double strokes.<br />
-,<br />
•<br />
-- ----- - -<br />
-_11 -_-_-<br />
I<br />
h -<br />
Fig. 127 Fig. 128<br />
Fig. 129
HEAT. MOLECULAR PHYSICS 73<br />
The <strong>in</strong>iti<strong>al</strong> pressure <strong>in</strong>side the vessel Po is equ<strong>al</strong> to atmospheric<br />
pressure.<br />
After that. another pump with the same active volume V o<br />
beg<strong>in</strong>s to suck <strong>in</strong> the atmospheric air, <strong>al</strong>so mak<strong>in</strong>g n double<br />
strokes. What will the pressure <strong>in</strong> the vessel be?<br />
319. A mercury column with a length 1 is <strong>in</strong> the middle of<br />
a horizont<strong>al</strong> tube with a length L closed at both ends. If the<br />
tube is placed vertic<strong>al</strong>ly, the mercury column will shift through<br />
the distance ~l from its <strong>in</strong>iti<strong>al</strong> position.<br />
At what distance will the centre of the column be from the<br />
middle of the tube if one end of the tube placed horizont<strong>al</strong>ly<br />
is opened? if the upper end of the tube placed vertic<strong>al</strong>ly is<br />
opened? if the lower end of the tube placed vertic<strong>al</strong>ly is<br />
opened?<br />
The atmospheric pressure is H ern Hg. The temperature rema<strong>in</strong>s<br />
the same.<br />
320. Bear<strong>in</strong>g <strong>in</strong> m<strong>in</strong>d that, accord<strong>in</strong>g to Avogadro's law, the<br />
volume of one gramme-molecule of any gas under standard conditions<br />
(temperature ooe and pressure 1 atm) is 22.4 litres,<br />
f<strong>in</strong>d the constant <strong>in</strong> the equation of state of an ide<strong>al</strong> gas (the<br />
Clapeyron-Mendeleyev equation) for a quantity of a gas equ<strong>al</strong><br />
to one mole and prove that this constant is the same for <strong>al</strong>l<br />
gases.<br />
321. Write the equation of state for an arbitrary mass of an<br />
ide<strong>al</strong> gas whose molecular weight J.1 is known.<br />
322. How would the pressure <strong>in</strong>side a fluid change if the<br />
forces of attraction b<strong>et</strong>ween the molecules suddenly disappeared?<br />
323. A vessel conta<strong>in</strong>s one litre of water at a temperature of<br />
27° C. What would the pressure <strong>in</strong> the vessel be if the forces<br />
of <strong>in</strong>teraction b<strong>et</strong>ween the water molecules suddenly disappeared?<br />
324. Is the pressure the same <strong>in</strong>side a gas and at the w<strong>al</strong>l<br />
of the vessel conta<strong>in</strong><strong>in</strong>g this gas?<br />
325. Is the concentration of gas molecules <strong>in</strong>side a vessel and<br />
at its w<strong>al</strong>l the same?<br />
326. F<strong>in</strong>d the temperature of a gas conta<strong>in</strong>ed <strong>in</strong> a closed vessel<br />
if its pressure <strong>in</strong>creases by 0.4 per cent of the <strong>in</strong>iti<strong>al</strong> pressure<br />
when it is heated by 1° C.<br />
327. A th<strong>in</strong>-w<strong>al</strong>led rubber sphere weigh<strong>in</strong>g G= 50 gf is filled<br />
with nitrogen and submerged <strong>in</strong>to a lake to a depth of h = 100<br />
m<strong>et</strong>res.<br />
F<strong>in</strong>d the mass of the nitrogen m if the sphere is <strong>in</strong> a position<br />
of equilibrium. Will equilibrium be stable? The atmospheric
74 PROBLEMS<br />
pressure Po = 760 mm Hg. The temperature <strong>in</strong> the lake t = +4° C.<br />
Disregard the tension of the rubber.<br />
328. Two hollow glass b<strong>al</strong>ls are connected by a tube with a<br />
drop of mercury <strong>in</strong> the middle. Can the temperature of the ambient<br />
air be appraised from the position of the drop?<br />
329. A cyl<strong>in</strong>der closed at both ends is separated <strong>in</strong>to two<br />
equ<strong>al</strong> (42 em each) parts by a piston impermeable to heat. Both<br />
the parts conta<strong>in</strong> the same masses of gas at a temperature of<br />
27°C and a pressure ofl atm.<br />
How much should the gas be heated <strong>in</strong> one part of the cyl<strong>in</strong>der<br />
to shift the piston by 2 cm? F<strong>in</strong>d the pressure p of the gas<br />
after shift<strong>in</strong>g of the piston.<br />
330. Dry atmospheric air consists of nitrogen (78.09 per cent<br />
by volume), oxygen (20.95 per cent), argon (0.93 per cent) and<br />
carbon dioxide (0.03 per cent). Disregard<strong>in</strong>g the negligible admixtures<br />
of other gases (helium, neon, krypton, xenon), d<strong>et</strong>erm<strong>in</strong>e<br />
the composition of the air (<strong>in</strong> per cent) by weight.<br />
331. F<strong>in</strong>d the mean (effective) molecular weight of dry atmospheric<br />
air if the composition of the air <strong>in</strong> per cent is known<br />
(see Problem 330).<br />
332. The density of the vapour of a compound of carbon and<br />
hydrogen is 3 g/lit at 43°C and 820 mm Hg. What is the molecular<br />
formula of this compound?<br />
333. When will the change <strong>in</strong> the pressure of a gas be greater-if<br />
it is compressed to a certa<strong>in</strong> extent <strong>in</strong> a heat-impermeable<br />
envelope or upon isotherm<strong>al</strong> compression?<br />
334. A gas that occupies a volume of VI = 1 lit at a pressure<br />
of PI = I atm expands isotherm<strong>al</strong>ly to a volume of VI = 2 lit.<br />
Then, the pressure of the gas iii h<strong>al</strong>ved at the same volume.<br />
Next the gas expands at a constant pressure to V4 = 4 litres.<br />
Draw a diagram of p versus V and use it to d<strong>et</strong>erm<strong>in</strong>e the<br />
process <strong>in</strong> which the gas performs the greatest work. How does<br />
the temperature change?<br />
335. A cyclic process 1-2-3-1 depicted on a V-t diagram<br />
(Fig. 130) is performed with a certa<strong>in</strong> amount of an ide<strong>al</strong> gas, .<br />
Show the same process on a p-V diagram and <strong>in</strong>dicate the<br />
stages when the gas received and when it rejected heat.<br />
336. A gas heated geyser consumes V o = 1.8 m" of m<strong>et</strong>hane<br />
(CH c ) an hour. F<strong>in</strong>d the temperature tt of the water heated<br />
by the geyser if the water flows out at a rate of v = 50 crn/s.<br />
The diam<strong>et</strong>er of the stream D = I ern, the <strong>in</strong>iti<strong>al</strong> temperature<br />
of the water and the gas t 1 = 11 0 C and the c<strong>al</strong>orific v<strong>al</strong>ue of
HEAT. MOLECULAR PHYSICS 75<br />
v<br />
....---...2<br />
Fig. 130 Fig. 131<br />
Fig. 132<br />
r<br />
-~--~---~-----..<br />
------<br />
--n---<br />
--M- --<br />
,<br />
I<br />
I<br />
I<br />
I<br />
I<br />
I<br />
I<br />
I<br />
If ----:------------ I<br />
m<strong>et</strong>hane qo = 13,000 c<strong>al</strong>rg. The gas <strong>in</strong> the tube is under a pressure<br />
of p = 1.2 atm. The efficiency of the heater 11 =:: 60 per cent.<br />
337. A closed vessel impermeable to heat conta<strong>in</strong>s ozone (0 3<br />
)<br />
at a temperature of t 1 = 527 0 C. After some time the ozone is<br />
compl<strong>et</strong>ely converted <strong>in</strong>to oxygen (0 2<br />
) ,<br />
F<strong>in</strong>d the <strong>in</strong>crease of the pressure <strong>in</strong> the vessel if q = 34,000<br />
c<strong>al</strong> have to be spent to form one gramme-molecule of ozone<br />
from oxygen.<br />
Assume that the heat capacity of one gramme-molecule of<br />
oxygen at a constant volume is equ<strong>al</strong> to Cv ::= 5 c<strong>al</strong>/deg- mole.<br />
338. Twenty grammes of helium <strong>in</strong> a cyl<strong>in</strong>der under a piston<br />
are transferred <strong>in</strong>f<strong>in</strong>itely slowly from a state with a volume of<br />
V1 == 32 lit and a pressure of PI ==<br />
I<br />
I<br />
4.1 atm to a state with V2 =-= 9 lit and<br />
P2 == 15.5 atm. What maximum temperature<br />
will the gas reach <strong>in</strong> this<br />
process if it is depicted on the p-l!<br />
diagram as a straight l<strong>in</strong>e (Fig. 131)?<br />
339. Will the energy of the air <strong>in</strong><br />
a room <strong>in</strong>crease if a stove is heated<br />
<strong>in</strong> it?<br />
Note. Assume the energy of a unit<br />
of mass of the -air u to be proportion<strong>al</strong><br />
to the absolute temperature, Le.,<br />
u=cT.<br />
340. The temperature <strong>in</strong> a room<br />
with a volume of 30 m" rose from<br />
15°C to 25°C. How much will the mass<br />
I<br />
I
76 PROBLEMS<br />
of the air <strong>in</strong> the room change if the atmosphere pressure p = 1<br />
atm? The molecular (mean) weight of the air is J..t = 28.9 g/mole.<br />
341. A sm<strong>al</strong>l tube filled with air and open at the bottom is<br />
placed <strong>in</strong>to a water-filled open vessel with a screen on the top<br />
(Fig. 132). The tube cannot be turned over. Draw a diagram<br />
show<strong>in</strong>g how the depth of submergence of the tube depends on<br />
the temperature of the water if the temperature first slowly<br />
<strong>in</strong>creases and then gradu<strong>al</strong>ly decreases.<br />
342. A cyl<strong>in</strong>der conta<strong>in</strong>s m = 20 g of carbon dioxide under<br />
a heavy piston. The gas is heated from the temperature t, = 20° C<br />
to t 2<br />
= 108°C. What work is performed by the gas?<br />
343. What quantity of heat should be imparted to carbon<br />
dioxide (see Problem 342) expand<strong>in</strong>g at a constant pressure as<br />
a result of heat<strong>in</strong>g? The molar heat of the carbon dioxide (heat<br />
capacity of one gramme-molecule) at a constant volume is<br />
C v=6.864 c<strong>al</strong>/mole-deg,<br />
2-4. Properties of Liquids<br />
344. Is it more difficult to compress a litre of air to three<br />
atmospheres, or a litre of water?<br />
345. How can a m<strong>in</strong>imum and a maximum thermom<strong>et</strong>ers be<br />
made us<strong>in</strong>g the phenomena of w<strong>et</strong>tability and unw<strong>et</strong>tability?<br />
346. The surface layer of a liquid is frequently likened to a<br />
str<strong>et</strong>ched rubber film. In what respect does this an<strong>al</strong>ogy disagree<br />
with re<strong>al</strong>ity?<br />
347. To remove a grease spot from a fabric it is good to<br />
apply some p<strong>et</strong>rol to the edges of the spot, while the spot itself<br />
should never be w<strong>et</strong>ted with p<strong>et</strong>rol immediately. Why not?<br />
348. Why is moisture r<strong>et</strong>a<strong>in</strong>ed longer <strong>in</strong> soil if it is harrowed?<br />
349. Ski<strong>in</strong>g boots are usu<strong>al</strong>ly warmed up to ensure that they<br />
absorb grease b<strong>et</strong>ter. Should the boots be warmed up outside<br />
or <strong>in</strong>side?<br />
350. Why can an iron be used to remove greasy spots from<br />
cloth<strong>in</strong>g?<br />
351. Why do drops of water appear at the end of a piece<br />
of firewood <strong>in</strong> the shade when it is be<strong>in</strong>g dried <strong>in</strong> the sun?<br />
352. A vessel whose bottom has round holes with a diam<strong>et</strong>er<br />
of d = 0.1 mm is filled with water. F<strong>in</strong>d the maximum height<br />
of the water level h at which the water does not flow out. The<br />
water does not w<strong>et</strong> the bottom of the vessel.
HEAT. MOLECULAR PHYSICS 77<br />
353. A soapy film is str<strong>et</strong>ched over a rectangular vertic<strong>al</strong> wire<br />
frame (Fig. 133). What forces hold section abed <strong>in</strong> equilibrium?<br />
354. A cube with a mass m = 20 g w<strong>et</strong>table by water floats<br />
on the surface of water. Each face of the cube a is 3 em long.<br />
What is the distance b<strong>et</strong>ween the lower face of the cube and<br />
the surface of the water?<br />
355. The end of a capillary tube with a radius r is immersed<br />
<strong>in</strong>to water. What amount of heat will be evolved when the<br />
water rises <strong>in</strong> the tube?<br />
356. A capillary tube is lowered <strong>in</strong>to a vessel with a liquid<br />
whose vapour pressure may be neglected. The density of the<br />
liquid is p. The vessel and the tube are <strong>in</strong> a. vacuum under the<br />
bell of an air pump (Fig. 134). F<strong>in</strong>d the pressure <strong>in</strong>side the<br />
liquid <strong>in</strong> the capillary tube at a height h above the level of<br />
the liquid <strong>in</strong> the vessel.<br />
357. The course of reason<strong>in</strong>g- given below is usu<strong>al</strong>ly followed<br />
to prove that the molecules of the surface layer of a liquid have<br />
surplus potenti<strong>al</strong> energy. A molecule <strong>in</strong>side the liquid is acted<br />
upon by the forces of attraction from the other molecules which<br />
compensate each other on the average. If a molecule is s<strong>in</strong>gled<br />
out on the surf ace, the result<strong>in</strong>g force of attraction from the<br />
other molecule is directed <strong>in</strong>to the liquid. For this reason the<br />
molecule tends to move <strong>in</strong>to the liquid. and def<strong>in</strong>ite work should<br />
a ~--------------- 6<br />
C 1-0----____________ tl<br />
.: h<br />
Fig. /33<br />
Fig. 134
78<br />
PROBLEMS<br />
--<br />
II<br />
-- - --- --<br />
- - - -.-<br />
-------<br />
Fig. 135 Fig. 136<br />
be done to br<strong>in</strong>g it to the surface. Therefore, each molecule of<br />
the surface layer has excess potenti<strong>al</strong> energy equ<strong>al</strong> to this work.<br />
The average force that acts on any molecule from the side of<br />
<strong>al</strong>l the others, however, is <strong>al</strong>ways equ<strong>al</strong> to zero if the liquid is<br />
<strong>in</strong> equilibrium. This is why the work done to move the liquid<br />
from a depth to the surface should <strong>al</strong>so be zero. What is the<br />
orig<strong>in</strong>, <strong>in</strong> this case, of the surface energy?<br />
358. One end of a glass capillary tube with a radius r = 0.05 cm<br />
is immersed <strong>in</strong>to water to a depth of h:=. 2 ern. What pressure<br />
is required to blow an air bubble out of the lower end of<br />
the tube?<br />
359. A glass capillary tube with an <strong>in</strong>tern<strong>al</strong> diam<strong>et</strong>er of 0.5 mm<br />
is immersed <strong>in</strong>to water. The top end of the tube projects by<br />
2 em above the surface of the water. What is the shape of<br />
the meniscus?<br />
360. Water rises to a height h <strong>in</strong> a capillary tube lowered<br />
vertic<strong>al</strong>ly <strong>in</strong>to water to a depth I (Fig. 135). The lower end<br />
of the tube is closed, the tube is then taken out of the water<br />
and opened aga<strong>in</strong>. D<strong>et</strong>erm<strong>in</strong>e the length of the water column<br />
rema<strong>in</strong><strong>in</strong>g <strong>in</strong> the tube.<br />
361. Two capillary tubes of the same cross section are lowered<br />
<strong>in</strong>to a vessel with water (Fig. 136). The water <strong>in</strong> the straight
HEAT. MOLECULAR PHYSICS 79<br />
Fig. 137 Fig. 138<br />
tube rises to. a height h.<br />
What will the level of the water be<br />
<strong>in</strong> the bent tube and what form will the meniscus take? The<br />
lower end of the bent tube is at a depth H· below the level of<br />
the water <strong>in</strong> the vessel.<br />
Consider the follow<strong>in</strong>g five cases:<br />
(1) H > h<br />
(2) H =h<br />
(3) 0 < H
80 PROBLEMS<br />
angle will be different, and therefore the horizont<strong>al</strong> component<br />
of the force of surface tension F will <strong>al</strong>so be different on each<br />
side of an arm (Fig. 139). Will this cause the cross to rotate?<br />
364. Light bodies w<strong>et</strong>ted by water (for example, two matches)<br />
and float<strong>in</strong>g on its surface are mutu<strong>al</strong>ly attracted. This <strong>al</strong>so<br />
occurs if bodies are not w<strong>et</strong>ted (matches coated with a th<strong>in</strong><br />
layer of paraff<strong>in</strong>e). If one body is w<strong>et</strong>ted and the other not,<br />
they will be repulsed. How can these phenomena be expla<strong>in</strong>ed?<br />
2-5. Mutu<strong>al</strong> Conversion of Liquids and Solids<br />
365. Water <strong>in</strong> a glass freezes at O°C. If this water is separated<br />
<strong>in</strong>to f<strong>in</strong>e drops, the water <strong>in</strong> them can be overcooled<br />
to -40 0 C. For example, water drops which clouds are composed<br />
of usu<strong>al</strong>ly beg<strong>in</strong> to freeze at a temperature below -17° C. How<br />
can these facts be expla<strong>in</strong>ed?<br />
366. A vessel with 100 g. of water at a temperature of 0°C<br />
is suspended <strong>in</strong> the middle of a room. In 15 m<strong>in</strong>utes the temperature<br />
of the water rises to 2°C. When ice equ<strong>al</strong> <strong>in</strong> weight<br />
to the water is placed <strong>in</strong>to the vessel, it melts dur<strong>in</strong>g ten hours.<br />
May these data be used to appraise the specific heat of fusion<br />
of ice H?<br />
367. Two identic<strong>al</strong> pieces of ice fly toward each other with<br />
equ<strong>al</strong> velocities and are converted <strong>in</strong>to vapour upon impact.<br />
F<strong>in</strong>d the m<strong>in</strong>imum possible velocities of the<br />
pieces before the<br />
impact if their temperature is -12°C.<br />
368. A c<strong>al</strong>orim<strong>et</strong>er conta<strong>in</strong>s ice. D<strong>et</strong>erm<strong>in</strong>e the heat capacity<br />
of the c<strong>al</strong>orim<strong>et</strong>er if Q 1<br />
= 500 c<strong>al</strong> are required to heat it tog<strong>et</strong>her<br />
with its contents from 270 to 27T K, and Q, = 16,600 c<strong>al</strong> from<br />
272 to 274° K.<br />
369. A c<strong>al</strong>orim<strong>et</strong>er conta<strong>in</strong>s 400 g of water at a temperature<br />
of +5°C. Then, 200 g of water at a temperature of +IO°C are<br />
added and 400 g of ice at a temperature of -60° C are put <strong>in</strong>.<br />
What is the temperature <strong>in</strong> the c<strong>al</strong>orim<strong>et</strong>er?<br />
370. Ice with a mass of m t = 600 g and at a temperature of<br />
t,=-IO°C is placed <strong>in</strong>to a copper vessel heated to t 1=350°C.<br />
As a result, the vessel now conta<strong>in</strong>s m a<br />
=::; 550 g of ice mixed<br />
with water. F<strong>in</strong>d the mass of the vessel. The specific heat of<br />
copper C 1<br />
= 0.1 c<strong>al</strong>/deg-g.<br />
371. When a sm<strong>al</strong>l ice cryst<strong>al</strong> is placed <strong>in</strong>to overcooled water<br />
it beg<strong>in</strong>s to freeze <strong>in</strong>stantaneously.
HEAT. MOLECULAR PHYSICS 81<br />
(1) What amount of ice is formed from M = 1 kg of water<br />
overcooled to t = _8°C?<br />
(2) What should the temperature of the overcooled water be<br />
for <strong>al</strong>l of it to be converted <strong>in</strong>to ice?<br />
Disregard the relation b<strong>et</strong>ween the heat capacity of the water<br />
and the temperature.<br />
372. One hundred grammes of ice at a temperature of O°C<br />
are placed <strong>in</strong>to a heat-impermeable envelope and compressed to<br />
p = 1,200 atm. F<strong>in</strong>d the mass of the melted part of the ice if<br />
the melt<strong>in</strong>g po<strong>in</strong>t decreases <strong>in</strong> direct proportion to the pressure,<br />
and if it lowers by 1°C when the pressure is <strong>in</strong>creased by 138 atm.<br />
2-6. Elasticity and Strength<br />
373. A copper r<strong>in</strong>g with a radius of r = 100 ern and a crosssection<strong>al</strong><br />
area of A = 4 mm' is fitted onto a steel rod with<br />
a radius R = 100.125 ern. With what force F will the r<strong>in</strong>g be<br />
expanded if the modulusof elasticity of copper E = 12,000 kgf/mm 2 ?<br />
Disregard the deformation of the rod. .<br />
374. What work can be performed by a steel rod with a<br />
length I and a"cross-section<strong>al</strong> area A when heated by t degrees?<br />
375. A wire with a length of 21 is str<strong>et</strong>ched b<strong>et</strong>ween two<br />
posts. A lantern with a mass M is suspended exactly from the<br />
middle of the wire. The cross-section<strong>al</strong> area of the wire is A<br />
and its modulus of elasticity E.D<strong>et</strong>erm<strong>in</strong>e the angle a of<br />
sagg<strong>in</strong>g of the wire, consider<strong>in</strong>g it to be sm<strong>al</strong>l (Fig. 140).<br />
376. A steel rod with a cross section A = 1 ern' is tigh tlY<br />
fitted b<strong>et</strong>ween two stationary absolutely rigid w<strong>al</strong>ls. What<br />
2/<br />
6-2042
82<br />
PROBLEMS<br />
~ Fig. 141<br />
force F will the rod act with on the w<strong>al</strong>ls if it is heated<br />
by /)"t = 5°C?<br />
The coefficient of l<strong>in</strong>ear therm<strong>al</strong> expansion of steel<br />
a=I.lxI0-'deg- 1 and its modulus of elasticity E=<br />
= 20,000 kgf/rnrn",<br />
377. Two rods made of different materi<strong>al</strong>s are placed b<strong>et</strong>ween<br />
massive w<strong>al</strong>ls (Fig. 141). The cross section of the rods is A<br />
and their respective lengths 11 and l.. The rods are heated by t<br />
degrees.<br />
F<strong>in</strong>d the force with which the rods act on each other if their<br />
coefficients of l<strong>in</strong>ear therm<strong>al</strong> expansion a l and a! and the moduli<br />
of elasticity of their materi<strong>al</strong>s E; and £. are known. Disregard<br />
the deformation of the w<strong>al</strong>ls.<br />
I<br />
I 1<br />
Fig. 142<br />
Fig. 143
HEAT. MOLECULAR PHYSICS 83<br />
Steel<br />
Copper<br />
Steel<br />
Fig. 144<br />
378. A homogeneous block with a mass m =7100 kg hangs on<br />
three vertic<strong>al</strong> wires of equ<strong>al</strong> length arranged symm<strong>et</strong>ric<strong>al</strong>ly<br />
(Fig. 142). F<strong>in</strong>d the tension of the wires if the middle wire<br />
is of steel and the other two are of copper. All the wires have<br />
the same cross section. Consider the modulus of elasticity of<br />
steel to be double that of copper.<br />
379. A re<strong>in</strong>forced-concr<strong>et</strong>e column is subjected to compression<br />
by a certa<strong>in</strong> load. Assum<strong>in</strong>g that the modulus of elasticity of<br />
concr<strong>et</strong>e Ec is one-tenth of that of iron Eb and the crosssection<strong>al</strong><br />
area of the iron is one-twenti<strong>et</strong>h of that of concr<strong>et</strong>e;<br />
f<strong>in</strong>d the portion of the load act<strong>in</strong>g on the concr<strong>et</strong>e.<br />
380. A steel bolt is <strong>in</strong>serted <strong>in</strong>to a copper tube as shown<br />
<strong>in</strong> Fig. 143. F<strong>in</strong>d the forces <strong>in</strong>duced <strong>in</strong> the bolt and <strong>in</strong> the<br />
A<br />
B<br />
Fig. 145<br />
6 *
84 PROBLEMS<br />
tube when the nut is turned through one revolution if the<br />
length of the tube is I, the pitch of the bolt thread is hand<br />
the cross-section<strong>al</strong> areas of the bolt and the tube are A b and<br />
At, respectively.<br />
381. A copper plate is soldered to two steel plates as shown<br />
<strong>in</strong> Fig. 144. What tensions will arise <strong>in</strong> the plates if the<br />
temperature is <strong>in</strong>creased by to C? All three plates have the same<br />
cross sections.<br />
382. F<strong>in</strong>d the maximum permissible l<strong>in</strong>ear velocity of a<br />
rotat<strong>in</strong>g th<strong>in</strong> lead r<strong>in</strong>g if the ultimate strength of lead<br />
au= 200 kgf/cm' and its density p = 11.3 g/cm".<br />
383. An iron block AB has both ends fixed. Hook H is<br />
fastened with two nuts <strong>in</strong> a hole <strong>in</strong> the middle of the block<br />
(Fig. 145). The block is clamped by the nuts with a force Fe-<br />
What forces will act on the upper and lower nuts from the<br />
side of the block if the hook carries a load whose weight can<br />
change from zero to a= 2F o ? Disregard sagg<strong>in</strong>g of the block<br />
and the weight of the hook.<br />
2-7. Properties of Vapour,<br />
384. Water vapour amount<strong>in</strong>g to 150g at a temperature of<br />
+100° C is admitted <strong>in</strong>to a c<strong>al</strong>orim<strong>et</strong>er conta<strong>in</strong><strong>in</strong>g 100g of ice<br />
at a temperature of -20 0 C. What will the temperature <strong>in</strong>side<br />
the c<strong>al</strong>orim<strong>et</strong>er be if its heat capacity is 75 c<strong>al</strong>/deg?<br />
385. Why does a strong j<strong>et</strong> of steam burst out from a<br />
boil<strong>in</strong>g k<strong>et</strong>tle when the gas burner is switched off, while no<br />
steam was visible before?<br />
386. Prove that the density of water vapour at a temperature<br />
near room temperature, expressed <strong>in</strong> g/m 3 , is approximately<br />
equ<strong>al</strong> to the pressure of water vapour expressed <strong>in</strong> millim<strong>et</strong>res<br />
of mercury.<br />
387. The pressure of saturated water vapours <strong>in</strong> a herm<strong>et</strong>ic<br />
vessel <strong>in</strong>creases with the temperature as shown <strong>in</strong> Fig. 146.<br />
The pressure of an ide<strong>al</strong> gas at constant volume is directly<br />
proportion<strong>al</strong> to the temperature.<br />
Us<strong>in</strong>g a table of _the properties of saturat<strong>in</strong>g water vapours<br />
(Table 2), f<strong>in</strong>d wh<strong>et</strong>her the equation of state of an ide<strong>al</strong> gas<br />
can be used to c<strong>al</strong>culate the density or the specific volume of<br />
saturated water vapours. Expla<strong>in</strong> the result obta<strong>in</strong>ed.<br />
388. N<strong>in</strong>e grammes of water vapour at a temperature of 30°C
HEAT. MOLECULA~ PHYSICS<br />
85<br />
Table 2. Properties of Saturat<strong>in</strong>g<br />
Water Vapours<br />
t (OC)<br />
Specific<br />
Specific<br />
Pressure volume<br />
Pressure<br />
t<br />
volume<br />
(kgf/crn 2 COC)<br />
) of vapour (kgf/ern l ) of vapour<br />
(m 3 j k g ) (rnl/kg)<br />
17.2 0.02 68.3 151.1 5 0.3818<br />
45.4 0.1 14.96 158.1 6 0.3214<br />
59.7 0.2 7.80 164.2 7 0.2778<br />
75.4 0.4 4.071 169.6 8 0.2448<br />
85.45 0.6 2.785 174.5 9 0.2189<br />
93.0 0.8 2.127 179.0 10 0.1980<br />
96.2 0.9 1.905 187.1 12 0.1663<br />
99.1 1 1.726 194.1 14 0.1434<br />
100 1.0333 1.674 200.4 16 0.1261<br />
116.3 1.8 0.996 206.2 18 0.1125<br />
119.6 2 0.902 211 .... 20 0.1015<br />
132.9 3 0.617 232.8 30 0.0679<br />
142.9 4 0.4708 249.2 40 0.0506<br />
are compressed isotherm<strong>al</strong>ly <strong>in</strong> a cyl<strong>in</strong>der under a piston. At<br />
what volume will the vapour beg<strong>in</strong> to condense?<br />
Note. Use Table 2.<br />
389. The relative humidity <strong>in</strong> a room is 10 per cent at a temperature<br />
of 15°C. How will the relative humidity change if the<br />
temperature <strong>in</strong> the room gradu<strong>al</strong>ly <strong>in</strong>creases by 10° C?<br />
390. A cold autumn drizzle s<strong>et</strong>tled down for the day. Wash<strong>in</strong>g<br />
is hung up to dry <strong>in</strong> a room. Will it dry faster if the w<strong>in</strong>dow<br />
is opened?<br />
391. Two vessels connected by tubes with cocks are filled<br />
with water to different levels (Fig. 147). The air has been pumped<br />
out of the vessels. What will occur if the vessels are connected<br />
(1) by open<strong>in</strong>g the cock <strong>in</strong> the lower tube, (2) by open<strong>in</strong>g the<br />
cock <strong>in</strong> the upper tube?<br />
392. What is the relative humidity of air at a temperature of<br />
t} = 10°C if the moisture from this air, when the latter has been<br />
heated to t 2<br />
= 30°C, beg<strong>in</strong>s to condense after isotherm<strong>al</strong> compression<br />
from 1 to 10 at.<br />
Note. Use Table 2.<br />
393. A porous body is placed for- dry<strong>in</strong>g under the bell of a<br />
vacuum pump. The pressure under the bell is ma<strong>in</strong>ta<strong>in</strong>ed at 6.5 rom
86<br />
PROBLEMS<br />
p<br />
- ---<br />
Fig. 146<br />
Fig. 147<br />
Hg for one hour, after which it sharply drops. The capacity of<br />
the pump is 60 lit/m<strong>in</strong>. The temperature under the pump bell<br />
is t = 5°C. What amount of water did the body conta<strong>in</strong>?<br />
394. Water with a mass of m == 30 g at a temperature of 0°C<br />
is <strong>in</strong> a heat-<strong>in</strong>sulated cyl<strong>in</strong>der under a weightless piston. The<br />
area of the piston A = 512 em- and the extern<strong>al</strong> pressure p = 1 atm.<br />
What height will the piston rise to if an electric heater conta<strong>in</strong>ed<br />
<strong>in</strong> the cyl<strong>in</strong>der supplies Q = 5,760 c<strong>al</strong>?
CHAPTER 3<br />
ELECTRICITY<br />
AND MAGNETISM<br />
3-1. Electrostatics<br />
395. What is' the force of <strong>in</strong>teraction b<strong>et</strong>ween po<strong>in</strong>t charges<br />
of one coulomb at a distance of 1 km from each other?<br />
Can a sm<strong>al</strong>l (sever<strong>al</strong> centim<strong>et</strong>res) body have an electrostatic<br />
charge of one coulomb?<br />
396. Three identic<strong>al</strong> sm<strong>al</strong>l b<strong>al</strong>ls, each weigh<strong>in</strong>g 0.1 gf, are<br />
suspended at one po<strong>in</strong>t on silk threads hav<strong>in</strong>g a length of l == 20 em.<br />
What charges should be imparted to the b<strong>al</strong>ls for each thread to<br />
form an angle of a = 30° with the vertic<strong>al</strong>?<br />
397. Two identic<strong>al</strong> b<strong>al</strong>ls are suspended on threads at a distance<br />
from each other. The b<strong>al</strong>ls are given equ<strong>al</strong> charges and immersed<br />
<strong>in</strong> kerosene. D<strong>et</strong>erm<strong>in</strong>e the density of the materi<strong>al</strong> of the b<strong>al</strong>ls<br />
if the threads do not deflect from the vertic<strong>al</strong> <strong>in</strong> a vacuum or<br />
<strong>in</strong> kerosene. The density of kerosene Po == 0.8 g/cm 3 and its permittivity<br />
(dielectric constant) 8 = 2.<br />
398. Two sm<strong>al</strong>l b<strong>al</strong>ls with equ<strong>al</strong> but opposite charges are secured<br />
<strong>in</strong> a horizont<strong>al</strong> plane at a distance a from each other. A third<br />
charged b<strong>al</strong>l is suspended on a str<strong>in</strong>g. The po<strong>in</strong>t of suspension<br />
is first so moved that the third b<strong>al</strong>l, when <strong>in</strong> a state of equilibrium,<br />
is precisely above the first b<strong>al</strong>l at a distance a from it,<br />
and then it is so moved that the third b<strong>al</strong>l is above the second<br />
one. F<strong>in</strong>d the angles through which the str<strong>in</strong>g is deflected from<br />
the vertic<strong>al</strong> if the angle of deflection above one of the b<strong>al</strong>ls is<br />
twice that above the other.<br />
399. In classic<strong>al</strong> experiments performed to measure the charge<br />
of an electron, a charged drop of oil is placed b<strong>et</strong>ween the horizont<strong>al</strong><br />
plates of a plane capacitor. Under the action of an electrostatic<br />
field, the drop moves uniformly upward, cover<strong>in</strong>g a certa<strong>in</strong><br />
distance dur<strong>in</strong>g the time t l<br />
, or downward, when the sign of the<br />
charge on the plates changes, cover<strong>in</strong>g the same distance dur<strong>in</strong>g<br />
the time tIe<br />
Assum<strong>in</strong>g the force of friction b<strong>et</strong>ween the drop and the air<br />
to be proportion<strong>al</strong> to the velocity of the drop, f<strong>in</strong>d the time t<br />
dur<strong>in</strong>g which the drop travels the same distance after the field<br />
is switched off.
88 PROBLEMS<br />
400. If only one charged body is available, can it be used to obta<strong>in</strong><br />
a charge exceed<strong>in</strong>g many times <strong>in</strong> absolute magnitude that<br />
which it itself has?<br />
401. A charged body has some energy. What is the source of<br />
the energy <strong>in</strong> a body ·which receives a charge as a result of the<br />
process described <strong>in</strong> the solution to Problem 400?<br />
402. Can two likely-charged b<strong>al</strong>ls be attracted to each<br />
other?<br />
403. A th<strong>in</strong> wire r<strong>in</strong>g with a radius R carries an electric<br />
charge q. The centre of the r<strong>in</strong>g conta<strong>in</strong>s a charge Q of the<br />
same sign as q, and Q ~ q. F<strong>in</strong>d the force which the r<strong>in</strong>g is<br />
str<strong>et</strong>ched with.<br />
404. Two po<strong>in</strong>t charges Q 1 and Q 2<br />
are at a distance d from<br />
each other. F<strong>in</strong>d the <strong>in</strong>tensity of the electric field at a po<strong>in</strong>t at<br />
a distance 'I from the charge Q1 and '2 from the charge Q2.<br />
Consider the cases of opposite and like charges.<br />
405. Three identic<strong>al</strong> positive charges Q are arranged at the<br />
vertices of an equilater<strong>al</strong> triangle. The side of the triangle is a.<br />
F<strong>in</strong>d the <strong>in</strong>tensity of the field at the vertex of a regular t<strong>et</strong>rahedron<br />
of which the triangle is the base.<br />
406. A positive charge Q is uniformly distributed over a th<strong>in</strong><br />
wire r<strong>in</strong>g with a radius R. F<strong>in</strong>d the <strong>in</strong>tensi ty of the electric<br />
field at po<strong>in</strong>ts on the axis of the r<strong>in</strong>g at a distance , from its<br />
centre.<br />
407. F<strong>in</strong>d the po<strong>in</strong>ts on the axis of the charged r<strong>in</strong>g (see<br />
Problem 406) at which the <strong>in</strong>tensity of the electric field is maximum.<br />
D<strong>et</strong>erm<strong>in</strong>e the <strong>in</strong>tensity of the field at these po<strong>in</strong>ts.<br />
408. Two par<strong>al</strong>lel m<strong>et</strong><strong>al</strong> plates, each with an area A, carry<br />
the charges Q 1<br />
and Qt. The distance b<strong>et</strong>ween the plates is much<br />
less than their l<strong>in</strong>ear dimensions. D<strong>et</strong>erm<strong>in</strong>e the <strong>in</strong>tensity of the<br />
electric field at po<strong>in</strong>ts A, B, and C (Fig. 148).<br />
409. Two large current-conduct<strong>in</strong>g plates are arranged par<strong>al</strong>lel<br />
to each other. The distance b<strong>et</strong>ween them is much less than their<br />
dimensions. One of the plates is given a charge + Q. What are<br />
the charges <strong>in</strong>duced on the surfaces of the other plate?<br />
410. A molecule is at a distance, from the axis of a charged<br />
<strong>in</strong>f<strong>in</strong>itely long m<strong>et</strong><strong>al</strong>lic cyl<strong>in</strong>der. F<strong>in</strong>d the force act<strong>in</strong>g on the<br />
molecule if the <strong>in</strong>tensity of the cyl<strong>in</strong>der field is expressed by<br />
the formula E = 2q (q is the charge .per unit of cyl<strong>in</strong>der length)<br />
r<br />
and the molecule has the form of a "dumb-bell"<br />
I and with charges + Q and - Q at its ends.<br />
with a length
ELECTRICITY AND MAGNETISM 89<br />
411. Two molecules of equ<strong>al</strong> mass are at a certa<strong>in</strong> distance<br />
from the axis of a charged cyl<strong>in</strong>der. One molecule has a constant<br />
electric moment p = Ql (see Problem 410). An "elastic" force<br />
acts b<strong>et</strong>ween the charges of the other molecule,<br />
Le., the distance<br />
1 is d<strong>et</strong>erm<strong>in</strong>ed from the expression QE = kl, where E is<br />
the mean <strong>in</strong>tensity of the field act<strong>in</strong>g on the molecule and k is<br />
a proportion<strong>al</strong>ity factor.<br />
First, the electric moments of the molecules are the same and<br />
their velocities are zero. .<br />
Which molecule will reach the surface of the cyl<strong>in</strong>der quicker<br />
under the action of the force of attraction?<br />
412. A charge + Q is imparted to a rectangular m<strong>et</strong><strong>al</strong> plate<br />
with sides a and b. The thickness c of the plate is much sm<strong>al</strong>ler<br />
than a and b. F<strong>in</strong>d the <strong>in</strong>tensity of the field created by this<br />
charged plate at po<strong>in</strong>ts <strong>in</strong> space close to the centre of the plate.<br />
, 413. A po<strong>in</strong>t electric charge + Q is at a distance d from a<br />
large current-conduct<strong>in</strong>g plate. F<strong>in</strong>d the force with which the<br />
plate acts on the po<strong>in</strong>t charge.<br />
414. A po<strong>in</strong>t charge + Qt is brought up to a distance d from<br />
the centre of the m<strong>et</strong><strong>al</strong> plate described <strong>in</strong> Problem 412. The<br />
distance d is much sm<strong>al</strong>ler than sides a and b of the plate.<br />
D<strong>et</strong>erm<strong>in</strong>e the force with which the plate acts on the charge<br />
+ Qt. When will a positively charged plate attract a positive<br />
po<strong>in</strong>t charge?<br />
415. What potenti<strong>al</strong> can an isolated m<strong>et</strong><strong>al</strong> sphere with a radius<br />
R <strong>in</strong> the air be charged to if the <strong>in</strong>tensity of the electric<br />
field that causes a puncture <strong>in</strong> the air is Eo = 30,000 Vfern?<br />
416. A body with a charge -q is <strong>in</strong>troduced through a sm<strong>al</strong>l<br />
orifice <strong>in</strong>to a hollow current-conduct<strong>in</strong>g sphere with a radius R<br />
carry<strong>in</strong>g a charge + ,Q. What is the potenti<strong>al</strong> of a po<strong>in</strong>t <strong>in</strong> space<br />
at a distance of R > , from the centre of the sphere?<br />
A 8 C<br />
• ••<br />
Fig. 148 Fig. 149<br />
Fig. 150<br />
.
90 PROBLEMS<br />
417. A m<strong>et</strong><strong>al</strong> leaf is attached to the <strong>in</strong>tern<strong>al</strong> w<strong>al</strong>l of an electrom<strong>et</strong>er<br />
<strong>in</strong>sulated from the earth (Fig. 149). The rod and the<br />
hous<strong>in</strong>g of the electrom<strong>et</strong>er are connected by a conductor, and<br />
then a certa<strong>in</strong> charge is imparted to the hous<strong>in</strong>g. Will the leaves<br />
of the electrom<strong>et</strong>er deflect? What will happen to the leaves if<br />
the conductor is removed and the rod is then earthed?<br />
418. The hous<strong>in</strong>g of the electrom<strong>et</strong>er described <strong>in</strong> Problem 417<br />
is given a charge (the conductor is absent). Will the leaves of<br />
the electrom<strong>et</strong>er deflect <strong>in</strong> this case? Will the angle. of deflection<br />
of the leaves change if the rod is earthed?<br />
419. By touch<strong>in</strong>g different po<strong>in</strong>ts on a m<strong>et</strong><strong>al</strong> buck<strong>et</strong> hav<strong>in</strong>g<br />
a narrow bottom with a test b<strong>al</strong>l connected by a wire to an<br />
earthed electrom<strong>et</strong>er (Fig. 150), we can observe an identic<strong>al</strong><br />
deflection of the leaves of the electrom<strong>et</strong>er at any position of<br />
the b<strong>al</strong>l. If the wire is removed, the deflection of the leaves of<br />
the electrom<strong>et</strong>er, whose rod the b<strong>al</strong>l is made to contact, will<br />
depend on what po<strong>in</strong>t of the buck<strong>et</strong> surface (extern<strong>al</strong> or <strong>in</strong>tern<strong>al</strong>)<br />
we touched previously. Why? .<br />
420. Why does an electrom<strong>et</strong>er connected by a wire to the<br />
m<strong>et</strong><strong>al</strong> body shown <strong>in</strong> Fig. 151 make it possible to measure the<br />
potenti<strong>al</strong> of the body? Why do the leaves deflect <strong>in</strong> proportion<br />
to the density of the charge on separate portions of the body<br />
when the charge is transferred from the body to the electrom<strong>et</strong>er<br />
with the aid of an <strong>in</strong>sulated current-conduct<strong>in</strong>g b<strong>al</strong>l?<br />
421. An uncharged current-conduct<strong>in</strong>g sphere with a radius R<br />
is at a distance d from a po<strong>in</strong>t charge Q. What is the potenti<strong>al</strong><br />
of the sphere?<br />
422. An isolated current-conduct<strong>in</strong>g sphere with a radius R<br />
carries a charge + Q. What is the energy of the sphere?<br />
423. A m<strong>et</strong><strong>al</strong> sphere two m<strong>et</strong>res <strong>in</strong> diam<strong>et</strong>er is <strong>in</strong> the centre<br />
of a large room and charged. to a potenti<strong>al</strong> of 100,000 V. What<br />
quantity of heat will be liberated if the sphere is connected to<br />
the earth with a conductor?<br />
424. Two m<strong>et</strong><strong>al</strong> b<strong>al</strong>ls with radii of '1 = 1 ern and '2 = 2 em<br />
at a distance of R= 100 em from eaeh other are connected to<br />
a battery with an electromotive force of
ELECTRICITY AND MAGNETISM 91<br />
92<br />
PROBLEMS<br />
Fig. 153<br />
Fig. 154<br />
431. Appraise the capacitance of a human body <strong>in</strong> the order<br />
of its magnitude.<br />
432. Will the read<strong>in</strong>gs of an electrom<strong>et</strong>er jo<strong>in</strong>ed to a g<strong>al</strong>vanic<br />
cell change if a capacitor is connected <strong>in</strong> par<strong>al</strong>lel with it?<br />
Will the capacitance of the capacitor have any significance?<br />
433. Four identic<strong>al</strong> plane capacitors with an air dielectric are<br />
connected <strong>in</strong> series. The <strong>in</strong>tensity of the field at which the air<br />
is punctured is Ea = 3 X lOt VIcm. The distance b<strong>et</strong>ween the<br />
plates d=O.7 em.<br />
(1) What maximum voltage can be fed to this battery of capacitors?<br />
(2) What will this maximum voltage be if one of the capacitors<br />
is replaced by a similar one <strong>in</strong> which glass is used as a<br />
dielectric?<br />
The permittivity of glass e = 7 and the punctur<strong>in</strong>g, field <strong>in</strong>tensity<br />
for glass E g=9x 10& V/cm.<br />
434. D<strong>et</strong>erm<strong>in</strong>e the voltages Uland U 2 on the capacitors<br />
(Fig. 153) if 8 1 = 12 kV and
ELECTRICITY AND MAGNETISM<br />
93<br />
Fig. 156<br />
435. F<strong>in</strong>d the capacitance Co of the battery of identic<strong>al</strong> capacitors<br />
shown <strong>in</strong> Fig. 154.<br />
436. Each edge of a cube made of wire conta<strong>in</strong>s a capacitor<br />
C (Fig. 155). F<strong>in</strong>d the capacitance of this battery if it is connected<br />
to the circuit by means of conductors jo<strong>in</strong>ed to the opposite<br />
apices A and B of the cube.<br />
437. The spark-capacitor transformer designed by Arkadyev can<br />
be used to obta<strong>in</strong> short-time high voltages. A diagram of the<br />
device is shown <strong>in</strong> Fig. 156.<br />
A group of capacitors connected <strong>in</strong> par<strong>al</strong>lel by conductors AB<br />
and CD hav<strong>in</strong>g a very high resistance is jo<strong>in</strong>ed to a high-voltage<br />
SOUTce.<br />
The upper plate of each capacitor is connected through a spark<br />
gap to the lower plate of the follow<strong>in</strong>g capacitor (gaps 1, 2, 3<br />
and 4). Each follow<strong>in</strong>g gap is greater than the preced<strong>in</strong>g one.<br />
A discharge occurs when the potenti<strong>al</strong> difference b<strong>et</strong>ween the<br />
plates atta<strong>in</strong>s the punctur<strong>in</strong>g voltage of the first gap. After this<br />
the second, third, <strong>et</strong>c.,<br />
gaps will be punctured, What will the<br />
potenti<strong>al</strong> difference be when the last gap is punctured if there<br />
are n capacitors and the voltage applied is Yo?<br />
438. The plates of a charged plane capacitor are <strong>al</strong>ternately<br />
earthed. Will the capacitor be discharged?<br />
439. A plane capacitor is charged to a potenti<strong>al</strong> difference U.<br />
Both plates are arranged symm<strong>et</strong>ric<strong>al</strong>ly with respect to the earth<br />
so that their potenti<strong>al</strong>s relative to it are +U/2 and - U/2,<br />
respectively.<br />
How will the potenti<strong>al</strong>s of the plates change relative to the<br />
earth if the first plate is earthed, then disconnected from the<br />
earth, after which the second plate is earthed?<br />
440. One of the plates of a capacitor connected to a battery<br />
with an e.m.I. of
94 PROBLEMS<br />
Prove that if these capacitors are connected <strong>in</strong> par<strong>al</strong>lel, their<br />
tot<strong>al</strong> electrostatic energy dim<strong>in</strong>ishes. Expla<strong>in</strong> the drop <strong>in</strong> the<br />
energy.<br />
442. A dielectric <strong>in</strong> the form of a sphere is <strong>in</strong>troduced <strong>in</strong>to a<br />
homogeneous electric field. How will the <strong>in</strong>tensity of the field<br />
change at po<strong>in</strong>ts A, Band C (Fig. 158).<br />
443. One of the plates of a plane capacitor with a mica dielectric<br />
carries a positive charge of Q = 1.4 X 10& CGS Q<br />
• The other<br />
plate isolated from earth is not charged. The area of each plate<br />
A :=: 2,500 em", The permittivity of mica 8 r = 7.<br />
F<strong>in</strong>d the <strong>in</strong>tensity of the field <strong>in</strong> the space b<strong>et</strong>ween the plates.<br />
444. What is the force of <strong>in</strong>teraction of the b<strong>al</strong>ls connected<br />
to the battery (Problem 424) if they are immersed<br />
<strong>in</strong> kerosene?<br />
The permittivity of kerosene e, = 2.<br />
445. The plates of a plane capacitor are connected to a storage<br />
battery whose e.rn.I. is
.---<br />
ELECTRICITY AND MAGNETISM<br />
~l/~<br />
I~ l--~ ~I<br />
(a)<br />
Fig. 159<br />
plate is A and the permittivity of the dielectric is ere F<strong>in</strong>d the<br />
capacitance of the capacitor <strong>in</strong> both cases.<br />
448. D<strong>et</strong>erm<strong>in</strong>e the energy of a plane capacitor hav<strong>in</strong>g the<br />
space b<strong>et</strong>ween its plates filled with a dielectric.<br />
449. A plane capacitor is filled with a dielectric whose permittivity<br />
is 8,.. The <strong>in</strong>tensity of the field <strong>in</strong> the dielectric is E.<br />
What is the <strong>in</strong>tensity of the field <strong>in</strong> a space made <strong>in</strong>side the<br />
dielectric and hav<strong>in</strong>g the form of a long th<strong>in</strong> cyl<strong>in</strong>der directed<br />
<strong>al</strong>ong the field or the form of a par<strong>al</strong>lelepiped one side of<br />
which is much sm<strong>al</strong>ler than the other two? The sm<strong>al</strong>ler side is<br />
directed <strong>al</strong>ong the field.<br />
450. Two rectangular plates with a length I and an area A<br />
are arranged par<strong>al</strong>lel to each other at a distance d. They are<br />
charged to a potenti<strong>al</strong> difference U (plane capacitor). A dielectric<br />
with a permittivity e, whose thickness is d and whose width<br />
is equ<strong>al</strong> to that of the plates is drawn <strong>in</strong>to the space b<strong>et</strong>ween<br />
the latter. The length of the dielectric is greater than 1<br />
(Fig. 160). F<strong>in</strong>d the result<strong>in</strong>g force F act<strong>in</strong>g on the dielectric<br />
from the side of the field depend<strong>in</strong>g on the distance x.<br />
451. Solve Problem 450 if the<br />
capacitor is connected to a battery<br />
whose e.m.f. is U. Disregard the resistance<br />
of the connect<strong>in</strong>g wires.<br />
452. The follow<strong>in</strong>g design of a perp<strong>et</strong>uum<br />
mobile has been suggested.<br />
Kerosene is poured <strong>in</strong>to commu- ;!J<br />
nicat<strong>in</strong>g vessels (Fig. 161). One part tJ<br />
~--l ·1<br />
95<br />
Fig. 160 Fig. 161
96 PROBLEMS<br />
of the vessel is placed <strong>in</strong>to a strong electric field b<strong>et</strong>ween<br />
the plates of a capacitor ow<strong>in</strong>g to which the level of the kerosene<br />
<strong>in</strong> this part is higher than <strong>in</strong> the other. A cha<strong>in</strong> of b<strong>al</strong>ls<br />
is passed over two pulleys.<br />
The specific weight of the materi<strong>al</strong> of the b<strong>al</strong>ls is less than<br />
that of the kerosene.<br />
The lift<strong>in</strong>g force act<strong>in</strong>g on the b<strong>al</strong>ls will be greater <strong>in</strong> the<br />
left-hand part than <strong>in</strong> the right-hand one because more b<strong>al</strong>ls<br />
are immersed <strong>in</strong> the kerosene <strong>in</strong> the left-hand part. For this<br />
reason the <strong>in</strong>ventor believes that the cha<strong>in</strong> should start rotat<strong>in</strong>g<br />
clockwise. Why will there actu<strong>al</strong>ly be no rotation?<br />
453. The space b<strong>et</strong>ween the plates of a plane capacitor is<br />
filled with a dielectric whose permittivity is ere One plate is<br />
given a charge +Q and the other -Q. D<strong>et</strong>erm<strong>in</strong>e the density<br />
of the bound electric charges that appear on the surface of the<br />
dielectric and the forces that are exerted by the field on the<br />
dielectric.<br />
454. The space b<strong>et</strong>ween the plates of a plane capacitor is<br />
filled with a dielectric. Each molecule of the dielectric is assumed<br />
to have the form of a "dumb-bell" with a length I whose<br />
ends carry charges +Q and -Q. The number of molecules <strong>in</strong><br />
a unit of volume (1 ern") is n.<br />
L<strong>et</strong> us assume further that <strong>al</strong>l the molecules have turned <strong>al</strong>ong<br />
the electric field under its action. F<strong>in</strong>d the <strong>in</strong>tensity E of the<br />
field <strong>in</strong>side the capacitor filled with the dielectric if before fill<strong>in</strong>g<br />
it the <strong>in</strong>tensity of the field was Eo.<br />
455. A dielectric consists of molecules each of which can be<br />
represented as two charges +Q and -Q b<strong>et</strong>ween which an<br />
"elastic force" acts. The latter term should be understood to mean<br />
that x (the distance b<strong>et</strong>ween the charges +Q and -Q) can be<br />
found from the equ<strong>al</strong>ity kx= QE, where E is the <strong>in</strong>tensity<br />
of the field act<strong>in</strong>g on the charges, and k is a proportion<strong>al</strong>ity<br />
factor.<br />
Assume that a unit of volume (1 ern") of the dielectric conta<strong>in</strong>s<br />
n molecules. Solve Problem 454, assum<strong>in</strong>g that the space<br />
b<strong>et</strong>ween the plates of the plane capacitor is filled with a dielectric<br />
of this type.<br />
D<strong>et</strong>erm<strong>in</strong>e the permittivity of the dielectric.<br />
456. A capacitor is filled with the dielectric whose properties<br />
are described <strong>in</strong> Problem 455.<br />
F<strong>in</strong>d the energy stored <strong>in</strong> the dielectric ow<strong>in</strong>g to its polarization.<br />
.
ELECTRICITY AND MAGNETISM 97<br />
Fig. 162 Fig. 163<br />
3-2. Direct Current<br />
457. Is there an electric field near the surface of a conductor<br />
carry<strong>in</strong>g direct current?<br />
458. Draw approximately the arrangement of the force l<strong>in</strong>es<br />
of an electric field around a homogeneous conductor bent to<br />
form an arc (Fig. 162). The conductor carries direct current.<br />
459. D<strong>et</strong>erm<strong>in</strong>e the resistance, if an amm<strong>et</strong>er shows a current<br />
of J = 5 A and a voltm<strong>et</strong>er 100 V (Fig. 163). The <strong>in</strong>tern<strong>al</strong><br />
resistance of the voltm<strong>et</strong>er R =2,500 O.<br />
460. What resistance , should be used to shunt a g<strong>al</strong>vanom<strong>et</strong>er<br />
with an <strong>in</strong>tern<strong>al</strong> resistance of R= 10,000 ohm to reduce its<br />
sensitivity n = 50 times?<br />
461. D<strong>et</strong>erm<strong>in</strong>e the voltage across a resistance R us<strong>in</strong>g a voltm<strong>et</strong>er<br />
connected to its ends. What relative error 'will be made<br />
if the read<strong>in</strong>gs of the voltm<strong>et</strong>er are taken as the voltage applied<br />
before it was switched on? The current <strong>in</strong>tensity .<strong>in</strong> the circuit<br />
is constant. ' .<br />
462. An amm<strong>et</strong>er is connected to measure the current <strong>in</strong>tensity<br />
<strong>in</strong> a circuit with a resistance R. What relative error will<br />
be made if connection of the amm<strong>et</strong>er does not change the current<br />
<strong>in</strong>tensity <strong>in</strong> the cir- .<br />
cuit? The voltage across<br />
the ends of the circuit<br />
kept constant.<br />
is<br />
c"<br />
C'<br />
Fig. /64 ne. 165<br />
7-2042
A<br />
Fig. /66 Fig. /67<br />
463. Two conductors with temperature coefficients of resistance<br />
a l and at have resistances R Ol and R OI<br />
at O°C. F<strong>in</strong>d the<br />
temperature coefficient of a circuit consist<strong>in</strong>g of these conductors<br />
if they are connected <strong>in</strong> series and <strong>in</strong> par<strong>al</strong>lel.<br />
464. F<strong>in</strong>d the resistance of the circuit shown <strong>in</strong> Fig. 164. Disregard<br />
the resistance of the connect<strong>in</strong>g wires AC'C and BC"D.<br />
465. F<strong>in</strong>d the resistance of the hexagon shown <strong>in</strong> Fig. 165 if<br />
it is connected to a circuit b<strong>et</strong>ween po<strong>in</strong>ts A and B. The resistance<br />
of each conductor <strong>in</strong> the. diagram is R.<br />
466. F<strong>in</strong>d the resistance of a wire cube when it is connected<br />
to a circuit b<strong>et</strong>ween po<strong>in</strong>ts A and B (Fig. 166). The resistance<br />
of each edge .of the cube is R.<br />
467. Resistances R 1 and R" each 600, are connected <strong>in</strong> series<br />
(Fig. 167). The potenti<strong>al</strong> difference b<strong>et</strong>ween po<strong>in</strong>ts A and<br />
B is U = 120V. F<strong>in</strong>d the read<strong>in</strong>g VI of a voltm<strong>et</strong>er connected<br />
to po<strong>in</strong>ts C and D if its <strong>in</strong>tern<strong>al</strong> resistance r = 120Q.<br />
468. Wires identic<strong>al</strong> <strong>in</strong> cross section A and resistivity pare<br />
soldered <strong>in</strong>to a- rectangle ADBC (Fig. 168) with the diagon<strong>al</strong><br />
AB of the same cross section and materi<strong>al</strong>. F<strong>in</strong>d the resistance<br />
b<strong>et</strong>ween po<strong>in</strong>ts A and<br />
II<br />
B and b<strong>et</strong>ween C and<br />
D if AD=BC=a<br />
and AC=BD=b.<br />
P 8<br />
11<br />
A<br />
IJ<br />
Fi,.168<br />
A 8<br />
Flg.169<br />
,.
ELECTRICITY AND MAGNETISM .99<br />
A<br />
R R H R<br />
B<br />
D Fig. 110<br />
469. Figure 169 shows the diagram of a Wheatstone bridge<br />
used to measure resistances. Here R; is the unknown resistance.<br />
R o a standard resistance, G a g<strong>al</strong>vanom<strong>et</strong>er connected by a slid<strong>in</strong>g<br />
contact D to a homogeneous conductor AB with a high resistance<br />
(a slide wire).<br />
Prove that the equation :: = ~: is true when no current flows<br />
through the g<strong>al</strong>vanom<strong>et</strong>er. Disregard the resistance of the connect<strong>in</strong>g<br />
wires.<br />
470. What resistance should be connected b<strong>et</strong>ween po<strong>in</strong>ts C<br />
and D (Fig. 170) so that the resistance of the entire circuit<br />
(b<strong>et</strong>ween po<strong>in</strong>ts A and B) does not depend on the number of<br />
elementary cells?<br />
471. The output voltage can be reduced <strong>in</strong> the output circuits<br />
of generators as desired by means of an attenuator designed as<br />
the voltage divider shown <strong>in</strong> Fig. 171.<br />
A speci<strong>al</strong> selector switch makes it possible to connect the<br />
output term<strong>in</strong><strong>al</strong> either to the po<strong>in</strong>t with a potenti<strong>al</strong> U0 produced<br />
by the generator, or to any ·of the po<strong>in</strong>ts Ul' UI' ••• J Un'<br />
each hav<strong>in</strong>g a potenti<strong>al</strong> k times sm<strong>al</strong>ler (k > 1) than the previous<br />
one. The second output term<strong>in</strong><strong>al</strong> and the lower ends of<br />
the resistances are earthed..<br />
F<strong>in</strong>d the ratio b<strong>et</strong>ween the resistances R 1 : R.: R, with any<br />
number of cells <strong>in</strong> the attenuator.<br />
472. What devices are needed to verify Ohm's law experiment<strong>al</strong>ly,<br />
i.e. t to show that the current <strong>in</strong>tensity is directly proportion<strong>al</strong><br />
to the potenti<strong>al</strong> difference?<br />
473. A charge Q is imparted to two identic<strong>al</strong> plane capacitors<br />
connected <strong>in</strong> par<strong>al</strong>lel.<br />
lIa ~ /!, Un<br />
1*<br />
Fig. 171
100 PROBLEMS<br />
It<br />
At the moment of time t = 0 the distance<br />
b<strong>et</strong>ween the plates of the first capacitor<br />
beg<strong>in</strong>s to <strong>in</strong>crease uniformly accord<strong>in</strong>g to<br />
the law d 1 = do+ ot, and the distance b<strong>et</strong>ween<br />
the plates of the second capacitor to<br />
+r· <br />
Fig. 172<br />
decrease uniformly accord<strong>in</strong>g to the law<br />
d;=do-vt. Neglect<strong>in</strong>g the resistance of<br />
the feed<strong>in</strong>g wires. f<strong>in</strong>d the <strong>in</strong>tensity of the<br />
current <strong>in</strong> the circuit when the plates of<br />
the capacitors move .<br />
. 474. F<strong>in</strong>d the work performed by an electrostatic field .(see<br />
Problem 473) when the distance b<strong>et</strong>ween. the plates of the first<br />
capacitor <strong>in</strong>creases and that b<strong>et</strong>ween the plates of the second<br />
capacitor simultaneously decreases by a.<br />
475. A curious phenomenon was observed by an experimenter<br />
work<strong>in</strong>g with a very sensitive g<strong>al</strong>vanom<strong>et</strong>er while sitt<strong>in</strong>g on<br />
a chair at a table. (The g<strong>al</strong>vanom<strong>et</strong>er was secured on a w<strong>al</strong>l<br />
and the ends of its w<strong>in</strong>d<strong>in</strong>g connected to an open key on the<br />
table.) Upon ris<strong>in</strong>g from the chair and touch<strong>in</strong>g the table with<br />
his hand, the experimenter observed an appreciable deflection<br />
of the g<strong>al</strong>vanom<strong>et</strong>er po<strong>in</strong>ter." If he touched the table while sitt<strong>in</strong>g<br />
on the chair there was no deflection. Also, the g<strong>al</strong>vanom<strong>et</strong>er<br />
showed no deflection when he touched the table without<br />
first sitt<strong>in</strong>g down. Expla<strong>in</strong> this phenomenon.<br />
476. The follow<strong>in</strong>g effect was observed <strong>in</strong> a very sensitive<br />
g<strong>al</strong>vanom<strong>et</strong>er when the circuit was opened. If a charged body<br />
is brought up to one end of the w<strong>in</strong>d<strong>in</strong>g of the g<strong>al</strong>vanom<strong>et</strong>er.<br />
its po<strong>in</strong>ter deflects. If the body is brought up to the other end<br />
of the w<strong>in</strong>d<strong>in</strong>g, the deflection is <strong>in</strong> the same direction. Expla<strong>in</strong><br />
this phenomenon.<br />
477. How is the potenti<strong>al</strong> distributed <strong>in</strong> a Daniell cell when<br />
the extern<strong>al</strong> circuit is opened?<br />
S, 8~ s, &z &, A s,<br />
cl~~ B 8 8<br />
~~~ ~:;~ ~~~<br />
r, l'z Ii Ii Ii IZ Ii J!<br />
Fill. 173<br />
&1>Sz G,a::8 8 &;>&z 8r=&z<br />
'i
ELECTRICITY AND MAGNETISM 101<br />
478. Show graphic<strong>al</strong>ly the distribution of a potenti<strong>al</strong> <strong>al</strong>ong<br />
the closed circuit illustrated <strong>in</strong> Fig. 172 and on this basis deduce<br />
Ohm's law for a closed circuit.<br />
479. Show graphic<strong>al</strong>ly the approximate distribution of a potenti<strong>al</strong><br />
<strong>al</strong>ong the closed circuits depicted <strong>in</strong> Fig. 173.<br />
D<strong>et</strong>erm<strong>in</strong>e the current <strong>in</strong>tensity for each circuit and the potenti<strong>al</strong><br />
difference b<strong>et</strong>ween po<strong>in</strong>ts A and B. Disregard the resistance<br />
of the connect<strong>in</strong>g wires.<br />
480. Prove that an electromotive force <strong>in</strong> a circuit conta<strong>in</strong><strong>in</strong>g<br />
a g<strong>al</strong>vanic cell<br />
is equ<strong>al</strong> to the work of forces of non-electrostatic<br />
orig<strong>in</strong> when a s<strong>in</strong>gle positive charge moves <strong>al</strong>ong a<br />
closed circuit.<br />
481. About 106,000 c<strong>al</strong>ories are evolved when one mole of<br />
z<strong>in</strong>c comb<strong>in</strong>es with sulphuric acid and about- 56,000 c<strong>al</strong>ories are<br />
consumed when a mole of copper is Iiberated from blue vitriol.<br />
Use these data to f<strong>in</strong>d the e. m. f. of a Daniell cell.<br />
482. Two Daniell cells with <strong>in</strong>tern<strong>al</strong> resistances of r 1 = 0.80<br />
and r2 = 1.3Q and the same e. rn. f. s are connected <strong>in</strong> par<strong>al</strong>lel<br />
and across an extern<strong>al</strong> resistance R. F<strong>in</strong>d the ratio b<strong>et</strong>ween the<br />
quantities of the z<strong>in</strong>c dissolved <strong>in</strong> these elements dur<strong>in</strong>g a def<strong>in</strong>ite<br />
<strong>in</strong>terv<strong>al</strong> of time.<br />
483. A Daniell cell is made of absolutely pure materi<strong>al</strong>s.<br />
F<strong>in</strong>d the consumption of z<strong>in</strong>c and cryst<strong>al</strong>s of blue vitriol<br />
CuSO t<br />
• 5H 2<br />
0 if the cell produces a current of 0.1 A <strong>in</strong> eight hours.<br />
484. In a Daniell cell, copper is replaced by wax coated<br />
with a layer of graphite. Describe the phenomena that will<br />
occur <strong>in</strong> such a cell if the z<strong>in</strong>c is connected to the graphite<br />
layer by a wire.<br />
485. How will the e. m. f of the battery shown <strong>in</strong> Fig. 174<br />
change if the partition b<strong>et</strong>ween the vessels is removed? A solution<br />
of sulphuric acid is used as the electrolyte.<br />
486. A homogeneous carbon rod lies on the bottom of a<br />
vessel filled with electrolyte. A voltm<strong>et</strong>er with a high resistance<br />
is connected to the ends of the rod. A z<strong>in</strong>c rod bears aga<strong>in</strong>st<br />
the middle of the carbon rod (Fig. 175).<br />
What will the voltm<strong>et</strong>er show if the z<strong>in</strong>c rod is placed vertic<strong>al</strong>ly?<br />
How will the read<strong>in</strong>gs of the voltm<strong>et</strong>er change if the<br />
z<strong>in</strong>c rod is <strong>in</strong>cl<strong>in</strong>ed to the right or the left?<br />
487. A hollow current-conduct<strong>in</strong>g sphere with a radius of<br />
R = 5 em is placed <strong>in</strong>to an electrolytic bath filled with a solution<br />
of blue vitriol. The surface of the sphere has an orifice<br />
with a radius of r = 0.5 <strong>in</strong>m.
102 .<br />
PROBLEMS<br />
. Fig. 174 Fig. /75<br />
How much will the weight of, the sphere <strong>in</strong>crease if the<br />
copper is deposited dur<strong>in</strong>g t = 30 m<strong>in</strong> with a current density<br />
<strong>in</strong> the electrolyte of j =0.01 A/em'.<br />
488. If a capacitor carry<strong>in</strong>g a charge Q is discharged through<br />
an electrolytic bath with acidified water, m grammes of d<strong>et</strong>onat<strong>in</strong>g<br />
gas will be liberated. Accord<strong>in</strong>g to Faraday's law, the<br />
quantity of substance evolved dur<strong>in</strong>g electrolysis depends only<br />
on the amount of electricity passed through the electrolyte.<br />
This means that if the capacitor is discharged through N series-connected<br />
baths, then mN grammes of d<strong>et</strong>onat<strong>in</strong>g gas will<br />
be liberated. N can be made as great as required to obta<strong>in</strong> any<br />
quantity of the gas. Combustion of this gas can produce any<br />
amount of energy, which is obviously <strong>in</strong>consistent with the law<br />
of conservation of energy, s<strong>in</strong>ce the <strong>in</strong>iti<strong>al</strong> energy of a charged<br />
capacitor is not <strong>in</strong>f<strong>in</strong>itely great. Expla<strong>in</strong> this fact.<br />
489. When d<strong>et</strong>onat<strong>in</strong>g gas explodes, 34,500 c<strong>al</strong> are liberated<br />
per gramme of reacted hydrogen. Use these data to f<strong>in</strong>d the<br />
m<strong>in</strong>imum e. m. f. of a battery at which the electrolysis of water<br />
is possible.<br />
490. In electrolysis, positive and negative ions are cont<strong>in</strong>uously<br />
neutr<strong>al</strong>ized on the respective electrodes. What ma<strong>in</strong>ta<strong>in</strong>s<br />
the concentration of the ions <strong>in</strong> the electrolytes at a constant<br />
level? In what sections of the electrolyte is the reduction <strong>in</strong> the<br />
number of the ions compensated?<br />
491. The tot<strong>al</strong> density of the current <strong>in</strong> electrolytes is d<strong>et</strong>erm<strong>in</strong>ed<br />
as the sum of the current of the positive ions and that<br />
of the negative ions:
ELECTRICITY AND MAGNETISM 103<br />
where e is the charge of an ion. and n and v are the concentrations<br />
and velocities of the positive and negative ions.<br />
Why is the amount of substance evolved. for example, on<br />
the cathode considered to be proportion<strong>al</strong> to the full current,<br />
and not to the current en+v+?<br />
492. What m<strong>in</strong>imum change <strong>in</strong> temperature can be d<strong>et</strong>erm<strong>in</strong>ed<br />
with the aid of an iron-constantan thermocouple, if the<br />
measur<strong>in</strong>g <strong>in</strong>strument (g<strong>al</strong>vanom<strong>et</strong>er) has a sensitivity of 10- 9 A<br />
and a resistance of R = 20Q? The e. m. f. of the thermocouple<br />
is 50 microvolts (50 X 10- 8 V) per degree and its resistance<br />
r=5Q.<br />
493. The temperature of the hot jo<strong>in</strong>ts of a thermoelectric<br />
battery /1= 127°C and of the cold ones t a = 27°C. The e. m. f.<br />
of the battery tB = 4 V. Two c<strong>al</strong>ories of heat are supplied to the<br />
heated jo<strong>in</strong>ts <strong>in</strong> a unit of time to ma<strong>in</strong>ta<strong>in</strong> a constant temperature.<br />
An electrolytic bath with a solution of blue vitriol is connected<br />
to the. battery. What maximum (theor<strong>et</strong>ic<strong>al</strong>) amount of<br />
copper can be deposited on the cathode <strong>in</strong> a unit of. time?<br />
494. A current with an <strong>in</strong>tensity of I flows through a storage<br />
battery with an <strong>in</strong>tern<strong>al</strong> resistance of r and an e. m. f.<br />
of
104 PROBLEMS<br />
length of L = 5.6 km to f<strong>in</strong>d an <strong>in</strong>sulation breakdown b<strong>et</strong>ween<br />
the wires. It was found that if the conductors at the other end<br />
of the l<strong>in</strong>e were opened, the current flow<strong>in</strong>g through the battery<br />
was J1 = 1.5 A, and if short-circuited the current was I I = 2 A.<br />
The short-circuit current of the battery 1 8 = 96 A and the resistance<br />
of each conductor of the l<strong>in</strong>er = 70.. F<strong>in</strong>d the resistance<br />
of the <strong>in</strong>sulation R at the po<strong>in</strong>t of breakdown.<br />
500. G<strong>al</strong>vanic cells with e. m. f. s of
ELECTRICITY AND MAGNETISM 105<br />
resistance of , = 0.30 by connect<strong>in</strong>g them <strong>in</strong>to separate identic<strong>al</strong><br />
groups?<br />
505. An electric stove designed for 220 V is to be adapted<br />
for 110 V without chang<strong>in</strong>g or shorten<strong>in</strong>g the coil so that its<br />
power rema<strong>in</strong>s the same. What should be done for this?<br />
506. A lamp the resistance of whose filament <strong>in</strong> a heated<br />
state is R = 2.9Q is placed <strong>in</strong>to a c<strong>al</strong>orim<strong>et</strong>er conta<strong>in</strong><strong>in</strong>g a mixture<br />
of water and ice. In what time will the amount of water<br />
<strong>in</strong> the c<strong>al</strong>orim<strong>et</strong>er <strong>in</strong>crease by m = 15 g if the lamp is connected<br />
to ma<strong>in</strong>s with a voltage of u = 220 V? The specific heat of<br />
fusion of ice H = 80 c<strong>al</strong>/g.<br />
507. An electric lamp with a tungsten filament consumes<br />
50 watts. When the lamp burns, the temperature of its filament<br />
is 2,500°C. What power will be consumed by the lamp<br />
at the first moment after it is switched on? The temperature<br />
coefficient of resistance of tungsten a = 4.5 X 10- 3 deg- 1 •<br />
508. Why does the <strong>in</strong>candescence of the lamps <strong>in</strong> a room<br />
noticeably drop as soon as a high-power device (for example,<br />
an electric iron) is switched on, and after a brief <strong>in</strong>terv<strong>al</strong><br />
of time <strong>in</strong>crease, reach<strong>in</strong>g about the same v<strong>al</strong>ue as before?<br />
509. The wires lead<strong>in</strong>g from the ma<strong>in</strong>s <strong>in</strong>to a build<strong>in</strong>g have<br />
a resistance of R o = 0.50. The voltage <strong>in</strong> the ma<strong>in</strong>s is constant<br />
and equ<strong>al</strong> to U 0 = 127 V. What is the maximum permissible<br />
power of the electric energy consumed <strong>in</strong> the build<strong>in</strong>g if the<br />
voltage across the devices connected to the ma<strong>in</strong>s should not<br />
drop below U = 120 V?<br />
510. An electric tea-k<strong>et</strong>tle has two w<strong>in</strong>d<strong>in</strong>gs. When one of<br />
them is switched on the k<strong>et</strong>tle beg<strong>in</strong>s to boil <strong>in</strong> t} m<strong>in</strong>utes,<br />
and when the other is switched on-<strong>in</strong> t, m<strong>in</strong>utes. In what<br />
time will the k<strong>et</strong>tle beg<strong>in</strong> to boil if both w<strong>in</strong>d<strong>in</strong>gs are switched<br />
on simultaneously <strong>in</strong> series and <strong>in</strong> par<strong>al</strong>lel?<br />
511. An electric heater has three w<strong>in</strong>d<strong>in</strong>gs. If two w<strong>in</strong>d<strong>in</strong>gs<br />
are connected <strong>in</strong> par<strong>al</strong>lel and the third is connected to them<br />
<strong>in</strong> series, then with various comb<strong>in</strong>ations of the w<strong>in</strong>d<strong>in</strong>gs the<br />
water <strong>in</strong> the tank will beg<strong>in</strong> to boil <strong>in</strong> 20, 40 and 16 m<strong>in</strong>utes,<br />
respectively.<br />
In what time will the water beg<strong>in</strong> to boil if <strong>al</strong>l the w<strong>in</strong>d<strong>in</strong>gs<br />
are connected (1) <strong>in</strong> series? (2) <strong>in</strong> par<strong>al</strong>lel?<br />
512. When a direct current flows through a conductor, the<br />
amount of energy liberated is QU, where Q is the charge pass<strong>in</strong>g<br />
through the conductor and U the potenti<strong>al</strong> difference, while
106 PROBLEMS<br />
an energy of QU/2 is liberated when a capacitor is discharged.<br />
Why?<br />
513. When electric energy is transmitted over great distances,<br />
a transformer is used to so <strong>in</strong>crease the voltage as to reduce<br />
the current <strong>in</strong>tensity at the same power. Accord<strong>in</strong>g to the Joule<br />
Lenz law, the amount of heat evolved <strong>in</strong> the wires Q =<br />
= O.24/ 2 Rt, and therefore the losses due to heat evolution will<br />
be sm<strong>al</strong>l with sm<strong>al</strong>l currents.<br />
U<br />
On the other hand, Q 2<br />
= 0.24 R t, i. e., the amount of heat<br />
evolved grows with an <strong>in</strong>crease <strong>in</strong> the voltage. Expla<strong>in</strong> why an<br />
<strong>in</strong>crease <strong>in</strong> the voltage saves electric energy when it is transmitted<br />
over great distances.<br />
514. When two identic<strong>al</strong> lamps are connected <strong>in</strong> series to the<br />
circuit of a battery, the voltage drop across the <strong>in</strong>tern<strong>al</strong> resistance<br />
is k% of the e. m. f. The rated voltage of the lamp is<br />
U volts and the rated power p watts. D<strong>et</strong>erm<strong>in</strong>e the <strong>in</strong>tern<strong>al</strong><br />
resistance r of the battery.<br />
515. A storage battery with an e. m. f. of 11= 10 V and an<br />
<strong>in</strong>tern<strong>al</strong> resistance of r = 1Q is connected across an extern<strong>al</strong><br />
resistance R and liberates <strong>in</strong> it a power P ~ 9 watts. F<strong>in</strong>d the<br />
potenti<strong>al</strong> difference V across the term<strong>in</strong><strong>al</strong>s of the battery. What<br />
is the cause of the ambiguity of the result?<br />
516. What maximum useful power (evolved on an extern<strong>al</strong><br />
resistance) can be produced by a storage battery with an e. m. f.<br />
of
ELECTRICITY AND MAGNETISM 107<br />
C<br />
Fig. 179<br />
521. A conductor carries a current of I = 10 A.<br />
The cross-section<strong>al</strong> area of the conductor A = 5 ems<br />
----4t--t1~a and the number of free electrons <strong>in</strong> 1 ern" of the<br />
conductor n = 10 23 • F<strong>in</strong>d the directed velocity of<br />
the electrons v assum<strong>in</strong>g it to be the same for <strong>al</strong>l<br />
the electrons.<br />
522. A m<strong>et</strong><strong>al</strong> rectangular par<strong>al</strong>lelepiped with<br />
sides d, b, C (d~c, b~c) moves with an acceleration<br />
a <strong>in</strong> the direction shown by the arrow<br />
<strong>in</strong> Fig. 179. F<strong>in</strong>d the <strong>in</strong>tensity of the electric field produced by<br />
the accelerated motion of the m<strong>et</strong><strong>al</strong> block and the density of<br />
the electric charges on its surfaces that are perpendicular to the<br />
direction of motion.<br />
523. A. solid m<strong>et</strong><strong>al</strong> cyl<strong>in</strong>der whose radius is R rotates with<br />
a constant angular velocity roo F<strong>in</strong>d how the field <strong>in</strong>tensity<br />
depends on the distance to the axis of the cyl<strong>in</strong>der and d<strong>et</strong>erm<strong>in</strong>e<br />
the potenti<strong>al</strong> difference b<strong>et</strong>ween the surface of the cyl<strong>in</strong>der<br />
and the axis.<br />
3-3. Electric Current <strong>in</strong> Gases and a Vacuum<br />
524. Will a glow discharge occur if an anode is placed <strong>in</strong>to<br />
a cathode dark space (the region of the cathode drop)?<br />
525. Figure 180 shows a diagram of an X-ray tube with a<br />
cold cathode: C is the cathode, A the anode and A c the anticathode.<br />
A high voltage is created b<strong>et</strong>ween the anode and the<br />
cathode. The electrons accelerated near the cathode (region of<br />
the cathode drop) bombard the anticathode with a high velocity<br />
and <strong>in</strong>itiate X-rays. Why is the tube provided with two electrodes,<br />
an anode and an anticathode, <strong>in</strong>stead of one?<br />
A<br />
A<br />
c".... -"<br />
Fig. 180<br />
Fig. 181
108<br />
....-------------<br />
PROBLEJ\\S<br />
Fig. 182<br />
Fig. 183<br />
~~------<br />
~<br />
Fig. 184<br />
526. Why are the anode and the anticathode of an X-ray<br />
tube connected by a wire (see Problem 525)? What will occur<br />
if the anticathode is isolated?<br />
527. Figure 181 shows a Geiger-Muller counter of elementary<br />
particles. A high voltage is. produced b<strong>et</strong>ween the hous<strong>in</strong>g of<br />
tube A and a th<strong>in</strong> wire ab that is only slightly sm<strong>al</strong>ler than<br />
the critic<strong>al</strong> voltage necessary to ignite the charge.<br />
When a fast charged particle enters the counter, the molecules<br />
of the gas are ionized and a discharge beg<strong>in</strong>s. The flow of<br />
current through the circuit is accompanied by a drop of voltage<br />
across the large resistance R. This voltage drop is recorded<br />
after amplification by correspond<strong>in</strong>g <strong>in</strong>struments.<br />
For the counter to answer its purpose, the discharge caused<br />
by the particle should be quickly ext<strong>in</strong>guished. What ext<strong>in</strong>guishes<br />
the discharge <strong>in</strong> the circuit <strong>in</strong> Fig. 181?<br />
528. A capacitor with a capacitance C = 8 ern and a distance<br />
b<strong>et</strong>ween the plates of<br />
d = 3 mm is connected to a high-voltage<br />
source through a resistance of R = l03Q (Fig. 182).<br />
The air <strong>in</strong> the space b<strong>et</strong>ween the plates of the capacitor is<br />
ionized by X-rays so that n = 10· ionic pairs form <strong>in</strong> one cubic<br />
centim<strong>et</strong>re per second. The charge of each ion is equ<strong>al</strong> to that<br />
of an electron.<br />
F<strong>in</strong>d the vol tage drop across the resistance R assum<strong>in</strong>g that<br />
<strong>al</strong>l the ions reach the plates before they recomb<strong>in</strong>e.<br />
529. What will occur to a burn<strong>in</strong>g electric arc if the negative<br />
carbon is <strong>in</strong>tensively cooled? What will. occur if the positive<br />
carbon is cooled?<br />
530. Why is an electric iron with a thermocontroller used<br />
only with <strong>al</strong>ternat<strong>in</strong>g current?<br />
531. What energy <strong>in</strong> ergs will an electron acquire after pass<strong>in</strong>g<br />
through a potenti<strong>al</strong> difference of 1 V <strong>in</strong> a vacuum? (In atomic<br />
physics this energy is taken as the unit "electron-volt")
ELECTRICITY AND MAGNETISM 109<br />
532. Does the trajectory of a charged particle <strong>in</strong> an electrostatic<br />
field co<strong>in</strong>cide with a force l<strong>in</strong>e?<br />
533. A broad m<strong>et</strong><strong>al</strong> plate is connected to earth through a<br />
g<strong>al</strong>vanom<strong>et</strong>er. A charged b<strong>al</strong>l flies <strong>al</strong>ong a straight l<strong>in</strong>e above<br />
the plate at a distance much less than the l<strong>in</strong>ear dimensions<br />
of the plate (Fig. 183).<br />
Draw an approximate diagram show<strong>in</strong>g how the current flow<strong>in</strong>g<br />
through the g<strong>al</strong>vanom<strong>et</strong>er depends on time.<br />
534. An electron moves <strong>al</strong>ong a m<strong>et</strong><strong>al</strong> tube with a variable<br />
cross section (Fig. 184). How will its velocity change when it<br />
approaches the neck of the tube?<br />
535. A potenti<strong>al</strong> difference U is created b<strong>et</strong>ween a filament<br />
emitt<strong>in</strong>g electrons and a current-conduct<strong>in</strong>g r<strong>in</strong>g (Fig. 185). The<br />
electrons move with an acceleration <strong>al</strong>ong the axis of the r<strong>in</strong>g.<br />
Their k<strong>in</strong><strong>et</strong>ic energy <strong>in</strong>creases while the battery produc<strong>in</strong>g the<br />
potenti<strong>al</strong> difference U performs no work, s<strong>in</strong>ce no current flows<br />
through the circuit. (It is assumed that the electrons do not<br />
imp<strong>in</strong>ge upon the r<strong>in</strong>g.) How can this be brought <strong>in</strong>to agreement<br />
with the law of conservation of energy?<br />
536. A charge +Q is uniformly distributed over a th<strong>in</strong> r<strong>in</strong>g<br />
with a radius R. F<strong>in</strong>d the velocity of a negative po<strong>in</strong>t charge<br />
- Q at the moment it passes through the centre 0 of the r<strong>in</strong>g<br />
if the charge - Q was <strong>in</strong>iti<strong>al</strong>ly at rest at po<strong>in</strong>t A sufficiently<br />
removed from the r<strong>in</strong>g (Fig. 186). The mass of the charge - Q<br />
is equ<strong>al</strong> to m. The r<strong>in</strong>g is stationary.<br />
537. The plates of a plane capacitor with a capacitance C<br />
at a distance of 1 from each other carry charges +Q and - Q.<br />
An electron flies <strong>in</strong>to the middle of the capacitor with a<br />
velocity Vo directed par<strong>al</strong>lel to the plates.<br />
What is the velocity of the electron at a sufficiently great<br />
distance from the capacitor?<br />
What is the nature of the change <strong>in</strong> the velocity of the<br />
electron (<strong>in</strong> absolute magnitude) when it moves <strong>in</strong>side and outside<br />
of the capacitor?<br />
----~~-------~<br />
Fig. /85<br />
Fig. 186
110<br />
PROBLEMS<br />
Fig. 187<br />
Fig. 188<br />
Consider the follow<strong>in</strong>g three cases:<br />
(1) at the <strong>in</strong>iti<strong>al</strong> moment the electron is at the same distance<br />
from both plates of the capacitor;<br />
(2) at the <strong>in</strong>iti<strong>al</strong> moment the electron is at a distance of 1/4<br />
from the positive plate;<br />
(3) the electron is at a distance of 1/4 from the negative plate.<br />
538. A battery (directly heated) triode is connected to the<br />
circuit shown <strong>in</strong> Fig. 187. The e. m. f. of the B battery IJ 1 = 80 V,<br />
of the A battery G'I. = 6 V and of the C battery t8 a = 2 V.<br />
With what energies will the electrons reach the anode? How<br />
will the energy of the electrons reach<strong>in</strong>g the anode change if<br />
the e. m. f.
ELECTR ICITY AND MAGNETISM<br />
III<br />
541. An electronic v<strong>al</strong>ve (one of the triodes 6H8C) is connected<br />
to the circuit of a battery with an e. m. f. of tB = 250 V<br />
<strong>in</strong> series with a resistance of R = 10· Q (Fig. 189).<br />
The v<strong>al</strong>ve grid is connected to the negative pole of a sm<strong>al</strong>l<br />
battery ( 0<br />
where k = 0.12 mA/V, are connected to a circuit as shown <strong>in</strong><br />
Fig. 190.<br />
Draw a diagram show<strong>in</strong>g how the current I <strong>in</strong> the circuit<br />
depends on the voltage V if eft1 = 2 V, itI = 5 V, G8 = 7 V, and V<br />
can change from -10 V to + 10V.<br />
543. C<strong>al</strong>culate the sensitivity of a cathode-ray tube to voltage,<br />
i. e., the deflection of the light spot on the screen caused by a<br />
potenti<strong>al</strong> difference of 1V on the control grids. The length of.<br />
the control grids is l, the distance b<strong>et</strong>ween them is d, the distance<br />
from the end of the grids to the screen is L, and the accelerat<strong>in</strong>g<br />
potenti<strong>al</strong> difference is Ue-<br />
Fig. 189 Fig. 190
112 PROBLEMS<br />
3-4. Magn<strong>et</strong>ic Field of a Current. Action of a Magn<strong>et</strong>ic<br />
Field on a Current and Mov<strong>in</strong>g Charges<br />
544. D<strong>et</strong>erm<strong>in</strong>e the dimension and magnitude of the coefficient k<br />
<strong>in</strong> the expression for the <strong>in</strong>tensity of the magn<strong>et</strong>ic fie-ld of a<br />
solenoid H = k 4nJ ~. if H is measured <strong>in</strong> oersteds and I <strong>in</strong><br />
egs electrostatic units.<br />
The dimension of the oersted co<strong>in</strong>cides with that of the electric<br />
field <strong>in</strong>tensity <strong>in</strong> cgs units.<br />
545. Two w<strong>in</strong>d<strong>in</strong>gs connected as shown <strong>in</strong> Fig. 191 are wound<br />
around a th<strong>in</strong> iron r<strong>in</strong>g with a radius R = 10 ern. The first w<strong>in</strong>d<strong>in</strong>g<br />
has 2,000 turns and the second 1,000 turns. F<strong>in</strong>d the <strong>in</strong>tensityof<br />
the magn<strong>et</strong>ic field <strong>in</strong>side the r<strong>in</strong>g if a current of I = IDA<br />
flows through the w<strong>in</strong>d<strong>in</strong>gs. .<br />
546. A current I flows through an <strong>in</strong>f<strong>in</strong>itely long conductor<br />
ABC bent to form a right angle (Fig. 192).<br />
How many times will the <strong>in</strong>tensity of the magn<strong>et</strong>ic field change<br />
at po<strong>in</strong>t M if an <strong>in</strong>f<strong>in</strong>itely long straight conductor BD is so<br />
connected to po<strong>in</strong>t B that the current I branches at po<strong>in</strong>t B<br />
<strong>in</strong>to two equ<strong>al</strong> parts and the current <strong>in</strong> the conductor AB rema<strong>in</strong>s<br />
the same?<br />
Note. Take <strong>in</strong>to account the fact thai the <strong>in</strong>tensity of a magn<strong>et</strong>ic<br />
field <strong>in</strong>duced at a certa<strong>in</strong> po<strong>in</strong>t by a sm<strong>al</strong>l element of<br />
current is perpendicular to the plane conta<strong>in</strong><strong>in</strong>g this element<br />
and a radius-vector drawn from this current element to the<br />
given po<strong>in</strong>t.<br />
547. A current flows through a conductor arranged <strong>in</strong> one plane<br />
as shown <strong>in</strong> Fig. 193. F<strong>in</strong>d the <strong>in</strong>tensity of the magn<strong>et</strong>ic field<br />
TN<br />
J<br />
J<br />
I<br />
"A __- ...-~I ~_---<br />
P 0<br />
C<br />
Fig. 191 Fig. 192 Fig. 193
I<br />
I<br />
I<br />
ELECTRICITY AND MAGNETISM 113<br />
J<br />
JI<br />
J j<br />
I<br />
J<br />
I<br />
J<br />
I<br />
....---+-t-..J...<br />
J ~"'::J ~<br />
I -<br />
Fig. 194 Fig. 195 Fig. 196<br />
at an arbitrary po<strong>in</strong>t on l<strong>in</strong>e AB, which is the axis of symm<strong>et</strong>ry<br />
0-1 the conductor.<br />
548. How will a magn<strong>et</strong>ic po<strong>in</strong>ter be positioned if it is placed<br />
<strong>in</strong> the centre of a s<strong>in</strong>gle-layer toroid<strong>al</strong> solenoid through which<br />
a direct current flows?<br />
549. A current I flows <strong>al</strong>ong an <strong>in</strong>f<strong>in</strong>ite straight th<strong>in</strong>-w<strong>al</strong>led<br />
pipe. Bear<strong>in</strong>g <strong>in</strong> m<strong>in</strong>d that the <strong>in</strong>tensity of the magn<strong>et</strong>ic field<br />
of an <strong>in</strong>f<strong>in</strong>ite straight conductor at a distance , from it is proportion<strong>al</strong><br />
to I [r, f<strong>in</strong>d the <strong>in</strong>tensity of the magn<strong>et</strong>ic field at an<br />
arbitrary po<strong>in</strong>t <strong>in</strong>side the pipe.<br />
550. Remember<strong>in</strong>g that the <strong>in</strong>tensity of a magn<strong>et</strong>ic field <strong>in</strong>side<br />
a long cyl<strong>in</strong>dric<strong>al</strong> conductor H = k 2njr, where j is the current<br />
density and r is the distance from the conductor axis, f<strong>in</strong>d the<br />
<strong>in</strong>tensity of the field at an arbitrary po<strong>in</strong>t on a long cyl<strong>in</strong>dric<strong>al</strong><br />
space <strong>in</strong>side the conductor (Fig. 194) through which a current<br />
with a density j flows. The axis of, the space is par<strong>al</strong>lel to the<br />
axis of the conductor and is ata distance d from it.<br />
551. Draw the distribution of the force l<strong>in</strong>es of a magn<strong>et</strong>ic<br />
field <strong>in</strong> the space of the cyl<strong>in</strong>dric<strong>al</strong> conductor descri bed In<br />
Problem 550.<br />
552. D<strong>et</strong>erm<strong>in</strong>e the dimension and the magnitude of the coefficient<br />
k <strong>in</strong> the expression for the force F = kH I1s<strong>in</strong> cp act<strong>in</strong>g<br />
from a magn<strong>et</strong>ic field on a current if H is <strong>in</strong> oersteds and I <strong>in</strong><br />
egs electrostatic units.<br />
8-2042
114 PROBLEMS<br />
553. Will the density of a direct current flow<strong>in</strong>g <strong>in</strong> a cyl<strong>in</strong>dric<strong>al</strong><br />
conductor be constant across the entire cross section of the<br />
conductor?<br />
554. A lightn<strong>in</strong>g arrester is connected to earth by a circular<br />
copper pipe. After lightn<strong>in</strong>g strikes, it is discovered that the<br />
pipe became a circular rod. Expla<strong>in</strong> the cause of this phenomenon.<br />
555. A very great current is made to flow for a short time<br />
through a thick w<strong>in</strong>d<strong>in</strong>g of 8 solenoid. Describe the deformation<br />
of the w<strong>in</strong>d<strong>in</strong>g from the viewpo<strong>in</strong>t of qu<strong>al</strong>ity.<br />
556. The magn<strong>et</strong>ic system of a g<strong>al</strong>vanom<strong>et</strong>er consists of a<br />
magn<strong>et</strong>, pole shoes A and B, and a cyl<strong>in</strong>der made of soft iron<br />
(Fig. 195). The magn<strong>et</strong>ic force l<strong>in</strong>es <strong>in</strong> the gap b<strong>et</strong>ween the<br />
shoes and the cyl<strong>in</strong>der are perpendicular to the surface of the<br />
cyl<strong>in</strong>der. The <strong>in</strong>tensity of the magn<strong>et</strong>ic field is H. A rectangular<br />
coil with n turns is placed <strong>in</strong> the gap on axis O. The sides of<br />
the coil are par<strong>al</strong>lel to the diam<strong>et</strong>er and the generatrix of the<br />
cyl<strong>in</strong>der. The area of each turn is A. One end of a spir<strong>al</strong> spr<strong>in</strong>g<br />
is so attached to the axis of the coil that when the latter rotates<br />
through an angle a, the deformation of the spr<strong>in</strong>g creates<br />
a rotation<strong>al</strong> moment ka that tends to turn the coil to a position<br />
of equilibrium. D<strong>et</strong>erm<strong>in</strong>e the angle through which the coil<br />
will turn if a current I passes through it.<br />
557. A current of I = 1 A flows through a wire r<strong>in</strong>g with a<br />
radius R = 5 em suspended on two flexible conductors. The r<strong>in</strong>g<br />
is placed <strong>in</strong> a homogeneous magn<strong>et</strong>ic field with an <strong>in</strong>tensity of<br />
H = 10Oe whose force l<strong>in</strong>es are horizont<strong>al</strong>. What force will the<br />
r<strong>in</strong>g be tensioned with?<br />
558.· A wire r<strong>in</strong>g with a radius R = 4 ern is placed <strong>in</strong>to a<br />
h<strong>et</strong>erogeneous magn<strong>et</strong>ic field whose force l<strong>in</strong>es at the po<strong>in</strong>ts of<br />
<strong>in</strong>tersection with the r<strong>in</strong>g form an angle of a = 10 0<br />
with a norm<strong>al</strong><br />
to the plane of the r<strong>in</strong>g (Fig. 196). The <strong>in</strong>tensity of the magn<strong>et</strong>ic<br />
field act<strong>in</strong>g on the r<strong>in</strong>g H = 100 Oe. A current of I = 5 A<br />
flows through the r<strong>in</strong>g. What force does the magn<strong>et</strong>ic field act<br />
on the r<strong>in</strong>g with?<br />
559. A rectangular circuit ABeD with sides a and b placed<br />
<strong>in</strong>to a homogeneous magn<strong>et</strong>ic field with an <strong>in</strong>tensity H can<br />
revolve around axis 00' (Fig. 197). A direct current I constantly<br />
flows through the circuit.<br />
D<strong>et</strong>erm<strong>in</strong>e the work performed by the magn<strong>et</strong>ic field when<br />
the circuit is turned through 180 0<br />
if <strong>in</strong>iti<strong>al</strong>ly the plane of the<br />
circuit is perpendicular to the magn<strong>et</strong>ic field and arranged<br />
as shown <strong>in</strong> Fiz. 197.
ELECTRICITY AND MAGNETISM<br />
115<br />
o<br />
a<br />
lJ<br />
H<br />
o<br />
0'<br />
Fig. 197<br />
H<br />
A<br />
Fig /98<br />
8 *<br />
B<br />
Fig. /99
116 PROBLEMS<br />
tic field <strong>in</strong> the direction of side a with a velocity e. The <strong>in</strong>tensity<br />
of the magn<strong>et</strong>ic field H is perpendicular to the base of<br />
the block with the sides a and c (Fig. 199). .<br />
D<strong>et</strong>erm<strong>in</strong>e the <strong>in</strong>tensity of the electric field <strong>in</strong> the block and<br />
the density of the electric charges on the surfaces of the par<strong>al</strong>lelepiped<br />
formed by sides a and b.<br />
567. An uncharged m<strong>et</strong><strong>al</strong> cyl<strong>in</strong>der with a radius r revolves about<br />
its axis <strong>in</strong> a magn<strong>et</strong>ic field with an angular velocity ro. The<br />
<strong>in</strong>tensity of the magn<strong>et</strong>ic field is directed <strong>al</strong>ong the axis of the<br />
cyl<strong>in</strong>der.<br />
What should the <strong>in</strong>tensity of the magn<strong>et</strong>ic field be for no<br />
electrostatic field to appear <strong>in</strong> the cyl<strong>in</strong>der?<br />
3-5. Electromagn<strong>et</strong>ic Induction. Alternat<strong>in</strong>g<br />
Current<br />
568. D<strong>et</strong>erm<strong>in</strong>e the direction of the <strong>in</strong>tensity of an electric<br />
field <strong>in</strong> a turn placed <strong>in</strong> a magn<strong>et</strong>ic field (Fig. 200) directed away<br />
from us <strong>in</strong> a direction perpendicular to the plane of the turn.<br />
The <strong>in</strong>tensity of the magn<strong>et</strong>ic field grows with time.<br />
569. A rectangular circuit ABeD moves translation<strong>al</strong>ly <strong>in</strong> the<br />
magn<strong>et</strong>ic field of a current flow<strong>in</strong>g <strong>al</strong>ong straight long conductor<br />
00' (Fig. 201). F<strong>in</strong>d the direction of the current <strong>in</strong>duced <strong>in</strong> the<br />
circuit if the turn moves away from the conductor.<br />
570. A non-magn<strong>et</strong>ized iron rod flies through a coil connected<br />
to a battery and an amm<strong>et</strong>er (Fig. 202). Draw an approximate<br />
diagram of the change of current <strong>in</strong> the coil with time as the<br />
rod flies through it.<br />
571. A current <strong>in</strong> a coil grows directly with time. What is the<br />
nature of the relation b<strong>et</strong>ween the current and time <strong>in</strong> another<br />
coil <strong>in</strong>ductively connected to the first one?<br />
0'<br />
B..---.O<br />
o<br />
+<br />
A 0<br />
Fig. 200<br />
a<br />
Fig. 201 Fig. 202
ELECTRICITY AND MAGNETISM<br />
117<br />
A<br />
B<br />
a<br />
2 E<br />
a<br />
Fig. 20,1<br />
D<br />
Fig. 204<br />
tZ<br />
a<br />
F<br />
572. Will the result of Problem 571 change if an iron core is<br />
<strong>in</strong>serted <strong>in</strong>to the second coil?<br />
573. A wire r<strong>in</strong>g with a radius r is placed <strong>in</strong>to a homogeneous<br />
magn<strong>et</strong>ic field whose <strong>in</strong>tensity is perpendicular to the plane of<br />
the r<strong>in</strong>g and changes with time accord<strong>in</strong>g to the law H = kt.<br />
F<strong>in</strong>d the <strong>in</strong>tensity of the electric field <strong>in</strong> the turn<br />
574. A r<strong>in</strong>g of a rectangular cross section (Fig. 203) is made<br />
of a materi<strong>al</strong> whose resistivity is p, The r<strong>in</strong>g is placed <strong>in</strong> a homogeneous<br />
magn<strong>et</strong>ic field. The <strong>in</strong>tensity of the magn<strong>et</strong>ic field is<br />
directed <strong>al</strong>ong the axis of the r<strong>in</strong>g and <strong>in</strong>creases directly with<br />
time, H" = kt. F<strong>in</strong>d the <strong>in</strong>tensity of the current <strong>in</strong>duced <strong>in</strong> the<br />
r<strong>in</strong>g.<br />
575. A coil hav<strong>in</strong>g n turns, each with an area of A, is connected<br />
to a b<strong>al</strong>listic g<strong>al</strong>vanom<strong>et</strong>er. (The latter measures the quantity<br />
of electricity pass<strong>in</strong>g through it.) The resistance of the entire<br />
circuit is R. First the coil is b<strong>et</strong>ween the poles of a magn<strong>et</strong> <strong>in</strong><br />
a region where the magn<strong>et</strong>ic field H is homogeneous and its<br />
<strong>in</strong>tensity is perpendicular to the area of the turns. Then the coil<br />
is placed <strong>in</strong>to a space with no .magn<strong>et</strong>ic field.<br />
Fig. 205<br />
of the circuit shown <strong>in</strong> Fig. 204 if the <strong>in</strong>tensi-<br />
of a homogeneous magn<strong>et</strong>ic field is perpendicular<br />
to the plane of the draw<strong>in</strong>g and changes<br />
<strong>in</strong> time accord<strong>in</strong>g to the law H = kt, The resistance<br />
of a unit- of length of the conductors is r.<br />
577. The w<strong>in</strong>d<strong>in</strong>g of a laboratory regulat<strong>in</strong>g<br />
the<br />
~<br />
h<br />
j<br />
ty<br />
What quantity of the electricity passes through<br />
g<strong>al</strong>vanom<strong>et</strong>er? (Express the answer <strong>in</strong> coulombs.)<br />
576. D<strong>et</strong>erm<strong>in</strong>e the current <strong>in</strong> the conductors<br />
autotransformer is wound around an iron core ha-
118<br />
PROBLEMS<br />
Fig. 206 Fig. 207<br />
v<strong>in</strong>g the form of a rectangular toroid (Fig. 205). For protection<br />
aga<strong>in</strong>st eddy (Foucault) currents the core is assembled of th<strong>in</strong><br />
iron lam<strong>in</strong>as <strong>in</strong>sulated from one another by a layer of varnish.<br />
This can be done <strong>in</strong> various ways:<br />
(1) by assembl<strong>in</strong>g the core of th<strong>in</strong> r<strong>in</strong>gs piled on one another;<br />
(2) by roll<strong>in</strong>g up a long band with a width h;<br />
(3) by assembl<strong>in</strong>g the core of rectangular lam<strong>in</strong>as 1X h <strong>in</strong> size,<br />
arrang<strong>in</strong>g them <strong>al</strong>ong the radii of the cyl<strong>in</strong>der.Which is the best way?<br />
578. A direct <strong>in</strong>duced current I is generated <strong>in</strong> a homogeneous<br />
circular wire r<strong>in</strong>g. The variable magn<strong>et</strong>ic field produc<strong>in</strong>g this current<br />
is perpendicular to the plane of the r<strong>in</strong>g, concentrated near<br />
its axis and has an axis of symm<strong>et</strong>ry pass<strong>in</strong>g through the centre<br />
of the r<strong>in</strong>g (Fig. 206).<br />
What is the potenti<strong>al</strong> difference b<strong>et</strong>ween po<strong>in</strong>ts A and B? What<br />
is the read<strong>in</strong>g of an electrom<strong>et</strong>er connected to these po<strong>in</strong>ts?<br />
579. A variable magn<strong>et</strong>ic field creates a constant e.rn.f. C <strong>in</strong><br />
a circular conductor ADBKA (see Problem 578). The resistances<br />
of the conductors ADB, AKB and ACB (Fig. 207) are equ<strong>al</strong> to<br />
R 1<br />
, R 2<br />
and R a , respectively. What current will be shown by<br />
amm<strong>et</strong>er C? The magn<strong>et</strong>ic field is concentrated near the axis of<br />
the circular conductor.<br />
580. The resistance of conductor ACB (see Problem 579) is<br />
R a = O. F<strong>in</strong>d the currents 11' I, and /a and the potenti<strong>al</strong> difference<br />
V A-VB"<br />
581. A medic<strong>al</strong> <strong>in</strong>strument used to extract <strong>al</strong>ien particles from<br />
an eye has the form of a strong permanent magn<strong>et</strong> or an electromagn<strong>et</strong>.<br />
When brought close to the eye (without touch<strong>in</strong>g it) it<br />
extracts iron and steel particles (fil<strong>in</strong>gs, chips, <strong>et</strong>c.),<br />
What current should flow through the electromagn<strong>et</strong> to extract,<br />
without touch<strong>in</strong>g the eye, m<strong>et</strong><strong>al</strong> objects made of non-ferromagn<strong>et</strong>ic<br />
materi<strong>al</strong>s (<strong>al</strong>um<strong>in</strong>ium, copper, <strong>et</strong>c.)?
ELECTRICITY AND MAGNETISM 119<br />
582. A wire r<strong>in</strong>g secured on the axis pass<strong>in</strong>g through its centre<br />
and perpendicular to the force l<strong>in</strong>es is placed <strong>in</strong> a homogeneous<br />
magn<strong>et</strong>ic field (Fig. 208). The <strong>in</strong>tensity of the field beg<strong>in</strong>s to<br />
grow. F<strong>in</strong>d the possible positions of equilibrium of the r<strong>in</strong>g and<br />
show the position of stable equilibrium. What will happen if the<br />
<strong>in</strong>tensity of the field decreases?<br />
583. A conductor with a length l and mass m can slide without<br />
friction <strong>al</strong>ong two vertic<strong>al</strong> racks AB and CD connected<br />
by a resistor R. The system is <strong>in</strong> a homogeneous magn<strong>et</strong>ic field<br />
whose <strong>in</strong>tensity H is perpendicular to the plane of the draw<strong>in</strong>g<br />
(Fig. 209).<br />
•<br />
How will the movable conductor travel <strong>in</strong> the field of gravity<br />
if the resistance of the conductor itself and the racks is neglected?<br />
584. A conductor with a mass m and length I can move without<br />
friction <strong>al</strong>ong two m<strong>et</strong><strong>al</strong>lic par<strong>al</strong>lel racks <strong>in</strong> a horizont<strong>al</strong> plane<br />
and connected across capacitor C. The entire system is <strong>in</strong> a<br />
homogeneous magn<strong>et</strong>ic field whose <strong>in</strong>tensity H is directed upward.<br />
A force F is applied to the middle of the conductor perpendicular<br />
to it and par<strong>al</strong>lel to the racks (Fig. 210).<br />
D<strong>et</strong>erm<strong>in</strong>e the acceleration of the conductor if the resistance<br />
of the racks, feed<strong>in</strong>g wires and conductor is zero. What k<strong>in</strong>ds<br />
of energy will the work of the force F be converted <strong>in</strong>to? Assume<br />
that the velocity of the conductor is zero at the <strong>in</strong>iti<strong>al</strong> moment.<br />
585. Consider<strong>in</strong>g the motion of a straight magn<strong>et</strong> <strong>in</strong> a plane<br />
perpendicular to a wire and us<strong>in</strong>g the law of conservation of<br />
energy, prove that the field of a long forward current dim<strong>in</strong>ishes<br />
with the distance from the wire as I/R.<br />
586. A cyl<strong>in</strong>der made of a non-magn<strong>et</strong>ic materi<strong>al</strong> has N turns<br />
of a wire (solenoid) wound around it. The radius of the cyl<strong>in</strong>der<br />
.m.<br />
H<br />
A ..<br />
R<br />
0N<br />
I<br />
m<br />
t-C<br />
B 0<br />
ct<br />
0N<br />
Fig. 209 Fig. 210<br />
E
120 PROBLEMS<br />
is , and its length l (r ~ I). The resistance of the wire is R.<br />
What should the voltage at the ends of the solenoid be for the<br />
current flow<strong>in</strong>g <strong>in</strong> it to <strong>in</strong>crease directly with time, i.e., I =kt?<br />
587. A solenoid (see Problem 586) is connected to a battery<br />
whose e.m.I. is cC. The key is closed at the moment t = O. What<br />
is the <strong>in</strong>tensity of the current flow<strong>in</strong>g through the circuit of<br />
the solenoid if the resistance R of the solenoid, battery and<br />
feed<strong>in</strong>g wires is neglected?<br />
588. C<strong>al</strong>culate the work of the battery (see Problem 587)<br />
dur<strong>in</strong>g the time 1". What k<strong>in</strong>d of energy is this work converted<br />
<strong>in</strong>to?<br />
589. A r<strong>in</strong>g made of a superconductor is placed· <strong>in</strong>to a homogeneous<br />
magn<strong>et</strong>ic field whose <strong>in</strong>tensity grows from zero to H o •<br />
The plane of the r<strong>in</strong>g is perpendicular to the force l<strong>in</strong>es of the<br />
field. F<strong>in</strong>d the <strong>in</strong>tensity of the <strong>in</strong>duction current appear<strong>in</strong>g <strong>in</strong><br />
the r<strong>in</strong>g. The radius of the r<strong>in</strong>g is , and its <strong>in</strong>ductance L.<br />
590. A superconductive r<strong>in</strong>g with a radius r is <strong>in</strong>.a homogeneous<br />
magn<strong>et</strong>ic field with an <strong>in</strong>tensity H~ The force l<strong>in</strong>es of the<br />
field are perpendicular to the plane of the r<strong>in</strong>g. There is no<br />
current <strong>in</strong> the r<strong>in</strong>g.<br />
F<strong>in</strong>d the magn<strong>et</strong>ic flux pierc<strong>in</strong>g the r<strong>in</strong>g after the magn<strong>et</strong>ic<br />
field is swi tched off.<br />
591. F<strong>in</strong>d the <strong>in</strong>ductance of a coil wound onto the iron core<br />
shown <strong>in</strong> Fig. 211. The number of turns of the coil N, the<br />
cross-section<strong>al</strong> area A, the perim<strong>et</strong>er of the core (medium l<strong>in</strong>e)<br />
I and the permeability of the core l-t, are known.<br />
Note. Take <strong>in</strong>to account the fact that .the <strong>in</strong>tensity of the<br />
magn<strong>et</strong>ic field <strong>in</strong>side the core is practic<strong>al</strong>ly constant and can<br />
be approximately expressed by the formula H = OAn ~ I.<br />
592. Estimate approximately the coefficient of mutu<strong>al</strong> <strong>in</strong>ductance<br />
of the w<strong>in</strong>d<strong>in</strong>gs of a transformer. Consider the w<strong>in</strong>d<strong>in</strong>gs<br />
Fig. 211 Fig. 212
ELECTRICITY AND MAGNETISM 121<br />
as coils of identic<strong>al</strong> cross section. Disregard the dispersion of<br />
the force l<strong>in</strong>es of the magn<strong>et</strong>ic field.<br />
Note. The coefficient of mutu<strong>al</strong> <strong>in</strong>ductance of two circuits is<br />
the ratio b<strong>et</strong>ween the magn<strong>et</strong>ic flux
122 PROBLEMS<br />
-~<br />
t<br />
Fig. 218<br />
597. An <strong>al</strong>ternat<strong>in</strong>g s<strong>in</strong>usoid<strong>al</strong> current flows through a coil<br />
without any ohmic resistance. Draw a diagram show<strong>in</strong>g the product<br />
of the current and the voltage (<strong>in</strong>stantaneous power) changes<br />
depend<strong>in</strong>g on time. Expla<strong>in</strong> the nature of the curve.<br />
What is the average power consumed by the coil per period?<br />
598. Like an electric arc, daylight lamps have a dropp<strong>in</strong>g<br />
current-voltage characteristic, and for this reason a coil with a<br />
high <strong>in</strong>ductance (choke) is connected <strong>in</strong> series with the lamp as<br />
a b<strong>al</strong>last resistance for stable burn<strong>in</strong>g. Why are ord<strong>in</strong>ary less<br />
expensive resistors not used?<br />
599. Why are capacitors connected <strong>in</strong> par<strong>al</strong>lel with electric<br />
devices hav<strong>in</strong>g a high <strong>in</strong>ductance (chokes, for example), if there<br />
are many of them <strong>in</strong> 8-C ma<strong>in</strong>s?<br />
600. (a) A tap C is made from the middle of a coil with an<br />
iron core (the w<strong>in</strong>d<strong>in</strong>g is a thick copper wire with many turns)<br />
(Fig. 214). A constant potenti<strong>al</strong> difference U 1 is created b<strong>et</strong>ween<br />
po<strong>in</strong>ts Band C. F<strong>in</strong>d the voltage UI b<strong>et</strong>ween po<strong>in</strong>ts A and B.<br />
(b) A variable potenti<strong>al</strong> difference with an amplitude V 1 (for<br />
example, from town ma<strong>in</strong>s) is applied b<strong>et</strong>ween po<strong>in</strong>ts Band C.<br />
F<strong>in</strong>d the ampli tude UI of the variable potenti<strong>al</strong> difference b<strong>et</strong>ween<br />
po<strong>in</strong>ts A and B.<br />
601. Why does the presence of<br />
a very high voltage <strong>in</strong> the secondary<br />
w<strong>in</strong>d<strong>in</strong>g of a step-up transformer<br />
(see Problem 513) not lead<br />
Ma<strong>in</strong>s<br />
127Va.c.<br />
A fII-.......-------<br />
Fig. 214 Fig. 215
ELECTRICITY AND MAGNETISM 123<br />
to great losses of energy<br />
due to the evolution of heat<br />
L<br />
<strong>in</strong> the w<strong>in</strong>d<strong>in</strong>g itself?<br />
602. Show that the proportion<br />
::= Z:. where I I<br />
and Jt are the currents <strong>in</strong><br />
the W<strong>in</strong>d<strong>in</strong>gs and N 1 and<br />
Nt the numbers of turns<br />
Fig. 216 <strong>in</strong> them, exists, if the<br />
no-load current of a transformer<br />
and the ohmic resistance of its w<strong>in</strong>d<strong>in</strong>gs are neglected.<br />
Consider the w<strong>in</strong>d<strong>in</strong>gs as coils with the same cross section.<br />
603. What punctur<strong>in</strong>g voltages should capacitor C and diode D<br />
be c<strong>al</strong>culated for if the rectifier (Fig. 215) can operate both with<br />
or without load?<br />
604. An <strong>al</strong>ternat<strong>in</strong>g voltage with an amplitude of U = 600 V<br />
is excited at the ends of the secondary w<strong>in</strong>d<strong>in</strong>g of a transformer<br />
supply<strong>in</strong>g a biphase rectifier (Fig. 216). The capacitance of<br />
the capacitor C is so high that the current J flow<strong>in</strong>g through<br />
the resistance R = 5 kQ can be considered as approximately constant<br />
(J = 40 rnA).<br />
Assum<strong>in</strong>g that each of the diodes passes no current <strong>in</strong> the<br />
opposite direction, f<strong>in</strong>d the share of the period T dur<strong>in</strong>g which<br />
no current flows through the v<strong>al</strong>ve.<br />
3-6. Electric<strong>al</strong> Mach<strong>in</strong>es<br />
605. The resistance <strong>in</strong> the load circuit of an a-c generator<br />
<strong>in</strong>creases. How should the power of the motor revolv<strong>in</strong>g the<br />
generator change for the frequency of the <strong>al</strong>ternat<strong>in</strong>g current to<br />
rema<strong>in</strong> the same?<br />
606. The force act<strong>in</strong>g on a mov<strong>in</strong>g charged particle from<br />
the side of a magn<strong>et</strong>ic field (the Lorentz force) is <strong>al</strong>ways perpendicular<br />
to the velocity. Therefore, this force performs no<br />
work (see Problem 561).<br />
Why does an electric motor operate <strong>in</strong> this case? We know<br />
that the force act<strong>in</strong>g on a conductor carry<strong>in</strong>g a current is caused<br />
by the action of the field on separate particles whose motion<br />
produces a current.<br />
607. Can a d-e series motor connected to ma<strong>in</strong>s with a<br />
voltage of U = 120V develop a power of P = 200 watts if the<br />
resistance of its w<strong>in</strong>d<strong>in</strong>gs is R = 20 Q?
124 PROBLEMS<br />
608. D<strong>et</strong>erm<strong>in</strong>e the efficiency of a series and a shunt-wound<br />
motors if they develop the maximum power. The voltages across<br />
the term<strong>in</strong><strong>al</strong>s U and the resistances of the w<strong>in</strong>d<strong>in</strong>gs of the<br />
rotor R 1<br />
and the stator R 2 are the same <strong>in</strong> both motors and<br />
are known.<br />
609. The rotor of a model of a d-e motor consists of one<br />
turn <strong>in</strong> the form of a rectangle. The <strong>in</strong>tensity of the magn<strong>et</strong>ic<br />
field H produced by a permanent magn<strong>et</strong> (north at the left and<br />
south at the right) is directed <strong>al</strong>ong the radius because the gap<br />
b<strong>et</strong>ween the pole shoes and iron cyl<strong>in</strong>der C is very sm<strong>al</strong>l<br />
(Fig. 217).<br />
A potenti<strong>al</strong> difference U is applied to the turn, whose area<br />
is A and resistance is R.<br />
F<strong>in</strong>d the power of the motor as a function of the angular<br />
velocity roe At what angular velocity ro will the power be rnaximum,<br />
and what will the current be at this power?<br />
610. D<strong>et</strong>erm<strong>in</strong>e how the rotation<strong>al</strong> moment (torque) M depends<br />
on the angular veloci ty, us<strong>in</strong>g the condition of the<br />
previous problem.<br />
611. F<strong>in</strong>d the nature of the relation b<strong>et</strong>ween the power of<br />
a d-c motor (see Problem 609) and the <strong>in</strong>tensity of the magn<strong>et</strong>ic<br />
field H at a given speed. At what v<strong>al</strong>ue of H does the<br />
power reach its maximum?<br />
612. D<strong>et</strong>erm<strong>in</strong>e the <strong>in</strong>tensity of the magn<strong>et</strong>ic field <strong>in</strong> a d-e<br />
motor (see Problem 609) at which the torque M is maximum.<br />
The speed of the armature is known.<br />
613. A d-e shunt-wound motor develops a mechanic<strong>al</strong> power<br />
of P = 160 watts with a voltage across the term<strong>in</strong><strong>al</strong>s of<br />
U = 120 V. The armature rotates at n = 10 rev/sf D<strong>et</strong>erm<strong>in</strong>e the<br />
maximum possible speed of<br />
the motor at this voltage.<br />
The resistance of the armature<br />
is R=20Q.<br />
614. A d-c shunt-wound<br />
N S motor has an angular velocity<br />
of the armature rotation of<br />
ro = 100 rad/s at a voltage of<br />
U = 120 V across the term<strong>in</strong><strong>al</strong>s.<br />
The resistance of ·the motor<br />
armature w<strong>in</strong>d<strong>in</strong>g is R = 20Q.<br />
Fig. 217<br />
What electromotive force will<br />
this motor develop when used
ELECTRICITY AND MAGNETISM<br />
125<br />
......------/<br />
~~-----o<br />
~----2 --..r.,<br />
~--------9 -..~<br />
Fig. 218 Fig. 219<br />
as a generator if it is rotated with the same angular velocity?<br />
The voltage <strong>in</strong> the stator w<strong>in</strong>d<strong>in</strong>gs is kept constant and equ<strong>al</strong><br />
to 120 V. At the velocity <strong>in</strong>dicated, the mechanic<strong>al</strong> moment on<br />
the motor shaft is M = 1.6x 10 7 dyne-ern.<br />
615. How will the speed· n of a shunt-wound motor change<br />
when the current <strong>in</strong> the stator w<strong>in</strong>d<strong>in</strong>gs grows if the voltage across<br />
the armature U and the mechanic<strong>al</strong> moment M applied to the<br />
armature axis rema<strong>in</strong> constant?<br />
616. What param<strong>et</strong>ers of ma<strong>in</strong>s would d<strong>et</strong>erm<strong>in</strong>e the power<br />
of a d-e series motor connected to them if the w<strong>in</strong>d<strong>in</strong>g of the<br />
motor were made of a superconductor?<br />
617. Prove that if the w<strong>in</strong>d<strong>in</strong>gs of a three...phase generator<br />
are star-connected (Fig, 218), the voltages b<strong>et</strong>ween l<strong>in</strong>ear conductors<br />
U12' U18 and U2S are Va times greater than the phase<br />
voltages VOl' V 02 and Uos-<br />
618. Prove that when the w<strong>in</strong>d<strong>in</strong>gs of a three-phase generator<br />
and the load resistances are star-connected (Fig. 219), the cur-<br />
ClJ<br />
I<br />
to<br />
Fig. 220 Fig. 221
126 PROBLEMS<br />
rent I. flow<strong>in</strong>g through the neutr<strong>al</strong> conductor is zero if<br />
R 1=R2=Ra=R.<br />
619. Prove that if the <strong>in</strong>tensities of the magn<strong>et</strong>ic field generated<br />
by three pairs of electromagn<strong>et</strong>s are equ<strong>al</strong> <strong>in</strong> amplitude<br />
and shifted <strong>in</strong> phase through ~ n (Fig. 220). the resultant magn<strong>et</strong>ic<br />
field can be described by a vector rotat<strong>in</strong>g with a constant<br />
angular velocity ro about po<strong>in</strong>t O.<br />
Each pair of electromagn<strong>et</strong>s creates magn<strong>et</strong>ic helds directed<br />
<strong>al</strong>ong the respective diam<strong>et</strong>ers of the r<strong>in</strong>g: HI' H 2 and H 8<br />
• The<br />
electromagn<strong>et</strong>s are fed with an <strong>al</strong>ternat<strong>in</strong>g current hav<strong>in</strong>g the<br />
frequency 0>.<br />
620. Two identic<strong>al</strong> coils perpendicular to each other are<br />
divided <strong>in</strong> h<strong>al</strong>f and connected to a circuit as shown <strong>in</strong> Fig. 221.<br />
The <strong>in</strong>ductance of the choke Ch and the ohmic resistance R<br />
are so selected that the <strong>in</strong>tensities of the currents <strong>in</strong> the coils<br />
are the same. The ohmic resistance and <strong>in</strong>ductive reactance of<br />
the coils are much less than the <strong>in</strong>ductive reactance of the<br />
choke.<br />
What will occur if an <strong>al</strong>um<strong>in</strong>ium cyl<strong>in</strong>der A secured on an<br />
axis is <strong>in</strong>troduced <strong>in</strong>to the space b<strong>et</strong>ween the poles of the coils?
CHAPTER 4<br />
OSCILLATIONS<br />
AND WAVES<br />
4-1. Mechanic<strong>al</strong> Oscillations<br />
621. A weight suspended from a long str<strong>in</strong>g oscillates <strong>in</strong> a<br />
vertic<strong>al</strong> plane and is deflected through an angle a, from the vertic<strong>al</strong><br />
(a mathematic<strong>al</strong>' pendulum). The same weight can rotate<br />
over a circumference so that the str<strong>in</strong>g describes a cone (a conic<strong>al</strong><br />
pendulum). When will the tension of the str<strong>in</strong>g deflected<br />
through an angle a, from the vertic<strong>al</strong> be greater?<br />
622. A clock with an oscillation period of the pendulum of 1<br />
second keeps accurate time on the surface of the Earth.<br />
When<br />
will the clock go slower <strong>in</strong> a day: if it is raised to an <strong>al</strong>titude<br />
of 200 m<strong>et</strong>res or lowered <strong>in</strong>to a m<strong>in</strong>e to a depth of 200 m<strong>et</strong>res?<br />
623. Two sm<strong>al</strong>l spheres each with a mass of In = I g are<br />
secured to the ends of a weightless rod with a length d = 1 m<strong>et</strong>re.<br />
The rod is so suspended from a h<strong>in</strong>ge that it can rotate without<br />
friction around a vertic<strong>al</strong> axis pass<strong>in</strong>g through its middle. Two<br />
large spheres with masses M = 20 kg are fastened on one l<strong>in</strong>e<br />
with the rod. The distance b<strong>et</strong>ween the centres of a large<br />
sphere and a sm<strong>al</strong>l one L = 16 em (Fig. 222). F<strong>in</strong>d the period<br />
of sm<strong>al</strong>l oscillations of this torsion pendulum.<br />
624. What is the period of oscillations of a mathematic<strong>al</strong><br />
pendulum <strong>in</strong> a railway carriage mov<strong>in</strong>g horizont<strong>al</strong>ly with an<br />
acceleration a?<br />
~~---- rl----~t-<br />
FIg. 222
PROBLEMS<br />
625. F<strong>in</strong>d the period of oscillations of a pendulum <strong>in</strong> a lift<br />
mov<strong>in</strong>g vertic<strong>al</strong>ly with an acceleration a.<br />
626. A block performs sm<strong>al</strong>l oscillations <strong>in</strong> a vertic<strong>al</strong> plane<br />
while mov<strong>in</strong>g without friction over the <strong>in</strong>tern<strong>al</strong> surface of a<br />
spheric<strong>al</strong> cup. D<strong>et</strong>erm<strong>in</strong>e the period of oscillations of the block<br />
if the <strong>in</strong>tern<strong>al</strong> radius of the cup is R and the face of the<br />
block is much sm<strong>al</strong>ler than R.<br />
627. How will the period of oscillations of the block <strong>in</strong> the<br />
cup change (see Problem 626) lf, besides the force of gravity,<br />
the cup is acted upon by a force F directed vertic<strong>al</strong>ly upward?<br />
The mass of the cup M is much greater than that of the<br />
block m.<br />
628. How will the period of oscillations of the block <strong>in</strong> the<br />
cup change (see Problem 626) if the cup is placed onto a smooth<br />
horizont<strong>al</strong> surface over which it can move without friction?<br />
629. A hoop with a mass m and a radius r can roll without<br />
slipp<strong>in</strong>g over the <strong>in</strong>tern<strong>al</strong> surface of a cyl<strong>in</strong>der with a radius R<br />
(Fig. 223). D<strong>et</strong>erm<strong>in</strong>e the period of motion of the hoop centre<br />
if the angle
OSCILLATIONS AND WAVES 129<br />
city k and a pulley with a mass M. A str<strong>in</strong>g passed over the<br />
pulley' connects the weight to the spr<strong>in</strong>g. F<strong>in</strong>d the period of<br />
oscillations of the weight if the pulley is a th<strong>in</strong>-w<strong>al</strong>led cyl<strong>in</strong>der.<br />
633. With what frequency will a bottle with a mass of<br />
m = 200 g and a cross-section<strong>al</strong> area of A = 50 em- oscillate if<br />
it floats on the surface of water <strong>in</strong> a vertic<strong>al</strong> position?<br />
Note. Remember that the period of oscillations of a weight<br />
on a spr<strong>in</strong>g is T = 2n V; . where k is the coefficient of elasticity<br />
of the spr<strong>in</strong>g.<br />
634. Mercury is poured <strong>in</strong>to communicat<strong>in</strong>g cyl<strong>in</strong>dric<strong>al</strong> vessels.<br />
F<strong>in</strong>d the oscillation period of the mercury if the crosssection<strong>al</strong><br />
area of each vessel is A = 0.3 ern! and the mass of the<br />
mercury m = 484 g. The specific weight of mercury is<br />
'V = 13.6 gfjcm s .<br />
635. A m<strong>in</strong>e pierces the Earth <strong>al</strong>ong one of its diam<strong>et</strong>ers. In<br />
what time will a body thrown <strong>in</strong>to the m<strong>in</strong>e reach the centre<br />
of the Earth? There is no resistance to motion.<br />
636. A str<strong>in</strong>g secured at its ends is .str<strong>et</strong>ched with a force f.<br />
A po<strong>in</strong>t weight with a mass m is attached to the middle of the<br />
str<strong>in</strong>g (Fig. 227). D<strong>et</strong>erm<strong>in</strong>e<br />
the period of sm<strong>al</strong>l<br />
oscillations of the weight.<br />
The mass of the str<strong>in</strong>g<br />
and the force of gravity<br />
can be neglected.<br />
637. How will the period<br />
of vertic<strong>al</strong> oscilla-<br />
A<br />
Fig. 225 Fig. 226<br />
9-2042
130 PROBLEMS<br />
Fig. 227<br />
tions of a weight hang<strong>in</strong>g on two identic<strong>al</strong> spr<strong>in</strong>gs change if<br />
the series connection of the spr<strong>in</strong>gs is changed to a par<strong>al</strong>lel one?<br />
638. Two mathematic<strong>al</strong> pendulums each with a length 1 are<br />
connected by a weightless spr<strong>in</strong>g as shown <strong>in</strong> Fig. 228. The<br />
coefficient of elasticity of the spr<strong>in</strong>g is k. In equilibrium the<br />
pendulums occupy a vertic<strong>al</strong> position and the spr<strong>in</strong>g is not deformed.<br />
D<strong>et</strong>erm<strong>in</strong>e the frequencies of sm<strong>al</strong>l oscillations of the two<br />
l<strong>in</strong>ked pendulums when they are deflected <strong>in</strong> one plane through<br />
equ<strong>al</strong> angles <strong>in</strong> one direction (oscillations <strong>in</strong> phase) and <strong>in</strong><br />
opposite directions (oscillations <strong>in</strong> antiphase).<br />
639. A force of 5 kgf should be applied to the handle of an<br />
open sw<strong>in</strong>g<strong>in</strong>g door to keep it <strong>in</strong> equilibrium (the door is r<strong>et</strong>urned<br />
to its usu<strong>al</strong> closed position of equilibrium by spr<strong>in</strong>gs). Can<br />
the door be opened by a force of 10 gf applied to the same<br />
handle? Disregard friction <strong>in</strong> the door h<strong>in</strong>ges.<br />
640. A weightless rod with a length I is rigidly attached to<br />
a weightless pulley with a radius r. The end of the rod carries<br />
a mass m (Fig..229).. A str<strong>in</strong>g to whose free end a mass M is<br />
I<br />
z<br />
111 v<br />
Fig. 228
OSCILLATIONS AND WAVES<br />
131<br />
N<br />
Fig. 229<br />
fastened is passed over the pulley. In what condition will the<br />
motion of the system be oscillatory if the angle ex b<strong>et</strong>ween the<br />
rod and the vertic<strong>al</strong> is zero at the <strong>in</strong>iti<strong>al</strong> moment?<br />
4-2. Electric<strong>al</strong> Oscillations<br />
641. Why is a permanent magn<strong>et</strong> needed <strong>in</strong> a telephone receiver?<br />
Why must the <strong>in</strong>tensity of the magn<strong>et</strong>ic field of this<br />
magn<strong>et</strong> be greater than the maximum <strong>in</strong>tensity of the magn<strong>et</strong>ic<br />
field generated by the current flow<strong>in</strong>g through the w<strong>in</strong>d<strong>in</strong>g<br />
of the telephone coil?<br />
642. F<strong>in</strong>d the frequency of natur<strong>al</strong> oscillations <strong>in</strong> a circuit<br />
consist<strong>in</strong>g of a solenoid with a length of 1= 3 ern, a cross-section<strong>al</strong><br />
area of 8 1<br />
= 1 ern! and a plane capacitor with a plate<br />
area of 8 2<br />
= 30 ern", the distance b<strong>et</strong>ween them be<strong>in</strong>g d = 0.1 em.<br />
The number of solenoid turns is N = 1,000.<br />
643. An electric circuit consists of a capacitor with a constant<br />
capacitance and a coil <strong>in</strong>to which a core can move. One<br />
core is made of ferrite and is used as an <strong>in</strong>sulator, and the<br />
other is made of copper. How will the frequency of natur<strong>al</strong><br />
oscillations of the circuit change if (a) the copper core, (b) the<br />
ferri te core, is moved <strong>in</strong>to the coil?<br />
9*
132<br />
PROBLEMS<br />
1;<br />
~ -----<br />
v<br />
Fig. 230<br />
Fig. 231<br />
644. The capacitors described <strong>in</strong> Problem. 594 are connected<br />
by a superconductor. No heat is liberated. How can the decrease<br />
<strong>in</strong> the energy of the electric field be expla<strong>in</strong>ed <strong>in</strong> this<br />
case from a qu<strong>al</strong>itative po<strong>in</strong>t of view? .<br />
645. A vol tage of VI = V 10 cos rot is applied to the vertic<strong>al</strong>ly<br />
deflect<strong>in</strong>g plates of an oscillograph and a voltage of V 2<br />
= V 20<br />
cos (rot-cp) to the horizont<strong>al</strong>ly deflect<strong>in</strong>g plates. F<strong>in</strong>d the<br />
trajectory of the cathode beam on the oscillograph screen if<br />
the phase difference b<strong>et</strong>ween the voltage on the plates is<br />
CP. =~ and cP, = n.<br />
646. Figure 230 shows a circuit consist<strong>in</strong>g of a battery E, a<br />
neon tube N, a capaci tor C and a resistor R.<br />
The characteristic of the neon tube (current <strong>in</strong> the tube<br />
versus voltage) is shown <strong>in</strong> Fig. 231. Current does not flow<br />
through the tube when the voltage is low. When the potenti<strong>al</strong><br />
<strong>in</strong> the tube reaches V, (strik<strong>in</strong>g potenti<strong>al</strong>), the tube ignites,<br />
the current jumps to I, and then grows <strong>in</strong> proportion to V. If<br />
the voltage drops, the current drops at a slower rate than that<br />
at which it <strong>in</strong>creased. The tube goes out at an ext<strong>in</strong>guish<strong>in</strong>g<br />
potenti<strong>al</strong> V e •<br />
Draw the approximate relation b<strong>et</strong>ween the change of voltage<br />
.ecross the capacitor and the time when the key K is<br />
closed.<br />
647. How will the period of relaxation oscillations change <strong>in</strong><br />
a circuit with a neon tube (see Problem 646) if the capacitance<br />
of the capaci tor C and the resistance R change?
OSCILLATIONS AND WAVES 133<br />
648. A plane capacitor <strong>in</strong>corporated <strong>in</strong>to an oscIllatory circuit<br />
is so designed that its plates can move relative to each other.<br />
How can the circuit be made to oscillate param<strong>et</strong>ric<strong>al</strong>ly by<br />
mov<strong>in</strong>g the plates?<br />
4-3. Waves<br />
649. A th<strong>in</strong> str<strong>in</strong>g is replaced by another made of the same<br />
materi<strong>al</strong>, but with a double diam<strong>et</strong>er. How many times should<br />
the tension of the str<strong>in</strong>g be changed to r<strong>et</strong>a<strong>in</strong> the previous<br />
frequency of asci11atlons?<br />
650. F<strong>in</strong>d the frequencies of natur<strong>al</strong> oscillations of a steel<br />
str<strong>in</strong>g with a length of l = 50 em and 8<br />
diam<strong>et</strong>er of d = 1 mm<br />
if the tension of the str<strong>in</strong>g is T = 2,450 dynes. The density of<br />
steel is p = 7.8 g/cm".<br />
651. F<strong>in</strong>d the frequencies of the natur<strong>al</strong> oscillations of an air<br />
column <strong>in</strong> a pipe with a length of l = 3.4 m<strong>et</strong>res closed at both<br />
ends.<br />
652. A tun<strong>in</strong>g fork with a frequency of natur<strong>al</strong> oscillations<br />
of v = 340 5- 1 sounds above a cyl<strong>in</strong>dric<strong>al</strong> vessel one m<strong>et</strong>re high.<br />
Water is slowly poured <strong>in</strong>to the vessel. At what level of the<br />
water <strong>in</strong> the vessel will the sound of the tun<strong>in</strong>g fork be appreciably<br />
<strong>in</strong>tensified?<br />
653. What is the shape of the front of the shock wave produced<br />
<strong>in</strong> air when a bull<strong>et</strong> flies at a supersonic velocity?<br />
654. A j<strong>et</strong> airplane flies at a velocity of 500 mls at a distance<br />
of 6 km from a man. At what distance from the man was<br />
the plane when the man heard its sound?<br />
655. If a source of sound and a man are about at the same<br />
<strong>al</strong>titude the sound is heard b<strong>et</strong>ter <strong>in</strong> the direction of the w<strong>in</strong>d<br />
than <strong>in</strong> the opposite direction. How can you expla<strong>in</strong> this phenomenon?<br />
656. Why can TV programmes be seen only with<strong>in</strong> the range<br />
of direct visibility?<br />
657. A radio direction f<strong>in</strong>der operates <strong>in</strong> pulse duty. The pulse<br />
frequency is f = 1,700 cps and its duration is 1: ~ 0.8 J.LS. D<strong>et</strong>erm<strong>in</strong>e<br />
the maximum and m<strong>in</strong>imum range of this f<strong>in</strong>der.<br />
658. Besides the wave transmitted directly from the station<br />
(po<strong>in</strong>t A), a TV aeri<strong>al</strong> (po<strong>in</strong>t C <strong>in</strong> Fig. 232) receives the wave<br />
reflected from the iron roof of a build<strong>in</strong>g (po<strong>in</strong>t B). As a result<br />
a double image appears. How many centim<strong>et</strong>res are the images<br />
shifted with respect to each other if the aeri<strong>al</strong> and the roof are
134 PROBLEMS<br />
00<br />
DC-<br />
Fig. 232 Fig. 233<br />
located as shown <strong>in</strong> Fig. 232? The width of the TV screen is<br />
I =50 em.<br />
Note. Remember that the image is 'resolved <strong>in</strong>to 625 l<strong>in</strong>es,<br />
and 25 frames are transmitted a second.<br />
659. A vibrator with a length of 1= 0.5 m<strong>et</strong>re is immersed<br />
<strong>in</strong>to a vessel with kerosene (8,. = 2). What is the length of the<br />
electromagn<strong>et</strong>ic wave radiated by .the vibrator <strong>in</strong> a vacuum as<br />
it emerges from the vessel?<br />
660. Figure 233 depicts a TV aeri<strong>al</strong>. How is the plane of<br />
oscillations of the magn<strong>et</strong>ic vector of the wave com<strong>in</strong>g from the<br />
TV centre orientated?
CHAPTER 5<br />
GEOMETRICAL<br />
OIJTICS<br />
5-1. Photom<strong>et</strong>ry<br />
661. A round h<strong>al</strong>l with a diam<strong>et</strong>er of D = 30 m<strong>et</strong>res is illum<strong>in</strong>ated<br />
by a lamp secured <strong>in</strong> the centre of the ceil<strong>in</strong>g. F<strong>in</strong>d<br />
the height h of the h<strong>al</strong>l if the m<strong>in</strong>imum illum<strong>in</strong>ation of the<br />
w<strong>al</strong>l is double that of the floor.<br />
662. A lamp rated at /1 = 100 cd hangs above the middle of<br />
8 round table with a diam<strong>et</strong>er of D = 3 m<strong>et</strong>res at a height of<br />
H = 2 m<strong>et</strong>res. It is replaced by a lamp with I, = 25 cd and<br />
the distance to the table is changed so that the illum<strong>in</strong>ation<br />
of the middle of the table rema<strong>in</strong>s as before. How will the illum<strong>in</strong>ation<br />
of the edge of the table change?<br />
663. Sources of light 8 1 and S, of equ<strong>al</strong> <strong>in</strong>tensity are arranged<br />
at the vertices of an isosceles right triangle (Fig. 234). How<br />
should a sm<strong>al</strong>l plate A be positioned for its illum<strong>in</strong>ation to be<br />
maximum? The sides of the triangle AS 1 = AS, = a.<br />
664. An attempt to use a photom<strong>et</strong>er to measure the lum<strong>in</strong>ous<br />
<strong>in</strong>tensity of a certa<strong>in</strong> source of light failed s<strong>in</strong>ce the lum<strong>in</strong>ous<br />
. <strong>in</strong>tensity was very high and the illum<strong>in</strong>ation of the photom<strong>et</strong>er<br />
fields with the aid of a standard source could not be equ<strong>al</strong>ized<br />
even when the source be<strong>in</strong>g <strong>in</strong>vestigated was placed on the very<br />
edge of the photom<strong>et</strong>er bench. Then a third source was employed<br />
with a lum<strong>in</strong>ous <strong>in</strong>tensity lower than that of the one be<strong>in</strong>g <strong>in</strong>vestigated.<br />
At a distance of r1 = 10 cm from the photom<strong>et</strong>er the<br />
standard source produced the same illum<strong>in</strong>ation of the fields as<br />
~ ......_---------_...... sz<br />
Fig. 284
136 PROBLEMS<br />
the third one that was placed at a distance of r 2 = 50 ern. After<br />
that, the standard source was replaced by the one be<strong>in</strong>g <strong>in</strong>vestigated<br />
and equ<strong>al</strong> illum<strong>in</strong>ations were obta<strong>in</strong>ed at distances from<br />
the photom<strong>et</strong>er r 8 = 40 em (source be<strong>in</strong>g <strong>in</strong>vestigated) and<br />
r. = 10 ern (auxiliary source). F<strong>in</strong>d how many times the lum<strong>in</strong>ous<br />
<strong>in</strong>tensity of the source be<strong>in</strong>g <strong>in</strong>vestigated is greater than<br />
that of the standard source.<br />
665. The ray of 8 searchlight f<strong>al</strong>ls on the w<strong>al</strong>l of a house and<br />
produces a' bright spot with a radius of r = 40 em. How many<br />
times will the illum<strong>in</strong>ation of the w<strong>al</strong>l of a remote house be<br />
sm<strong>al</strong>ler if the radius of the spot on it is 2 m<strong>et</strong>res?<br />
666. A lamp with a lum<strong>in</strong>ous <strong>in</strong>tensity of I = 100 cd is 'fastened<br />
to the ceil<strong>in</strong>g of a room. D<strong>et</strong>erm<strong>in</strong>e the tot<strong>al</strong> lum<strong>in</strong>ous flux<br />
f<strong>al</strong>l<strong>in</strong>g onto <strong>al</strong>l the w<strong>al</strong>ls and the floor of the room.<br />
667. What part of the energy radiated by the Sun reaches<br />
the Earth? The .radlus of the Earth is 6,400 km and the average<br />
distance from the Earth to the Sun is 149, 000, 000 km.<br />
668. A hot glow<strong>in</strong>g wire is placed on the axis of a hollow<br />
cyl<strong>in</strong>der with a radius R. The length of the wire is much greater<br />
than the height of the cyl<strong>in</strong>der. How many times will the illum<strong>in</strong>ation<br />
of the <strong>in</strong>tern<strong>al</strong> surface of the cyl<strong>in</strong>der change if its<br />
radius is R" (assume that R 2 < R 1 )?<br />
669. At what height should a lamp be hung above the centre<br />
of a round table to obta<strong>in</strong> the maximum illum<strong>in</strong>ation on its<br />
edges?<br />
670. Why can 8 text be read through tissue paper only if<br />
the paper is placed directly on the page?<br />
6 -2. Fundament<strong>al</strong> Laws of Optics<br />
671. Why is the shadow of a man's legs on the ground sharp<br />
and that of his head diffused? In what condltions will the<br />
shadow be equ<strong>al</strong>ly dist<strong>in</strong>ct everywhere?<br />
672. How should a pencil be held above a table to obta<strong>in</strong><br />
a sharp shadow if the source of light is a daylight lamp <strong>in</strong> the<br />
form of a long tube secured on the ceil<strong>in</strong>g?<br />
673. In autumn, when the trees have shed <strong>al</strong>l their leaves,<br />
one can frequently see shadows from two par<strong>al</strong>lel branches.<br />
The lower branch produces a sharp dark shadow and the<br />
upper branch a broader and lighter one. If two such shadows<br />
are accidently superimposed, one can see a bright light stripe'<br />
<strong>in</strong> the middle of the darker shadow, so that the shadow seems
GEOMETRICAL OPTICS<br />
137<br />
.0<br />
Fig. 235<br />
Fig. 236<br />
to be a double one (Fig. 235). How do you expla<strong>in</strong> this phenomenon?<br />
674. When sunrays pass through a sm<strong>al</strong>l open<strong>in</strong>g In the foliage<br />
at the top of a high tree, they produce an elliptic<strong>al</strong> spot on the<br />
ground. The major and m<strong>in</strong>or axes of the ellipses are a =<br />
12 cm<br />
and b = 10 em, respectively. What is the height of the tree H?<br />
The angular dimensions of the Sun's disk are ~ == 1/108<br />
rad.<br />
675. A periscope is designed with two reflect<strong>in</strong>g prisms. D<strong>et</strong>erm<strong>in</strong>e<br />
the ratio of the wid ths of these prisms if the distance b<strong>et</strong>ween<br />
them AB = L and the distance from the lower prism to the eye<br />
of the observer Be = 1 (Fig. 236). The objects viewed through<br />
the periscope are at a great distance from it.<br />
676. What m<strong>in</strong>imum height should a flat mirror secured vertic<strong>al</strong>ly<br />
on a w<strong>al</strong>l have for a man to see his reflection full size<br />
without chang<strong>in</strong>g the position of his head? D<strong>et</strong>erm<strong>in</strong>e <strong>al</strong>so the<br />
required distance b<strong>et</strong>ween the floor and the lower edge of the<br />
mirror.<br />
677. Sunrays are reflected from a horizont<strong>al</strong> mirror and f<strong>al</strong>l<br />
on a vertic<strong>al</strong> screen. An oblong object is placed on the mirror<br />
(Fig. 237). Describe the shadow on the screen.<br />
678. In what conditions does the shape of the reflection of<br />
a sunbeam from a sm<strong>al</strong>l mirror not depend on the shape of the<br />
mirror?<br />
679. How can a re<strong>al</strong> landscape on a photograph be dist<strong>in</strong>guished<br />
from its reflection <strong>in</strong> qui<strong>et</strong> water?
138 PROBLEMS<br />
680. F<strong>in</strong>d graphic<strong>al</strong>ly the positions of an observer's eye that<br />
will <strong>al</strong>low him to see <strong>in</strong> a mirror of f<strong>in</strong>ite dimensions the image<br />
of a straight l<strong>in</strong>e arranged as shown <strong>in</strong> Fig. 238.<br />
681. A flat mirror is arranged par<strong>al</strong>lel to 8 w<strong>al</strong>l at a distance<br />
l from it. The light produced by a po<strong>in</strong>t source fastened on<br />
the w<strong>al</strong>l f<strong>al</strong>ls on the mirror, is reflected and produces a light<br />
spot on the w<strong>al</strong>l. With what velocity will the light spot move<br />
<strong>al</strong>ong the w<strong>al</strong>l if the mirror is brought up to the w<strong>al</strong>l with a<br />
velocity v? How will the dimensions of the light spot change?<br />
682. Us<strong>in</strong>g the condition of Problem 681, f<strong>in</strong>d wh<strong>et</strong>her the<br />
illum<strong>in</strong>ation of the w<strong>al</strong>l at the po<strong>in</strong>t where the light spot is<br />
will change when the mirror is moved. The dimensions of the<br />
mirror are much sm<strong>al</strong>ler than the distance from the mirror to<br />
the source of light.<br />
683. A flat mirror revolves at a constant angular velocity,<br />
mak<strong>in</strong>g n = 0.5 revolution per second. Wi th what veloci ty will<br />
a light spot move <strong>al</strong>ong a spheric<strong>al</strong> screen with a radius of<br />
10 m<strong>et</strong>res if the mirror is at the centre of curvature of the screen?<br />
684. When the Russian scientist A. A. Belopolsky <strong>in</strong>vestigated<br />
the Doppler optic<strong>al</strong> phenomenon he observed light repeatedly<br />
reflected from mov<strong>in</strong>g mirrors (Fig. 239). The mirrors were placed<br />
on disks revolv<strong>in</strong>g <strong>in</strong> different directions.<br />
(a) The angular velocity 0) of rotation of the disks be<strong>in</strong>g known,<br />
f<strong>in</strong>d the angular velocity Q of rotation of a beam that is conse..<br />
cutively reflected n times from the mirrors.<br />
(b) D<strong>et</strong>erm<strong>in</strong>e the l<strong>in</strong>ear velocity of the n-th image at the moment<br />
when the mirrors are par<strong>al</strong>lel to each other and their<br />
reflect<strong>in</strong>g portions move with a velocity v <strong>in</strong> different directions.<br />
685. Solve Problem 684 if the disks rotate <strong>in</strong> the<br />
tion.<br />
same direc<br />
Mlrrol' Fig. 237
GEOMETRICAL OPTICS 139<br />
W//#/////7//'?'<br />
Fig. 238 Fig. 239<br />
686. A narrow beam of light S is <strong>in</strong>cident on a dihedr<strong>al</strong><br />
angle a, = 60 0 formed by identic<strong>al</strong>. flat mirrors OM and ON<br />
secured on axis 0 (Fig. 240). After "be<strong>in</strong>g reflected from the<br />
mirrors, the light is focussed by lens L and g<strong>et</strong>s <strong>in</strong>to stationary<br />
receiver R. The mirrors rotate with a constant angular velocity.<br />
What part of the light energy of the beam will reach the<br />
receiver dur<strong>in</strong>g a time that greatly exceeds the period of rotation<br />
if the beam passes at a distance a from an axis equ<strong>al</strong> to h<strong>al</strong>f<br />
the length of mirror OM?<br />
687. Can a flat mirror be used <strong>in</strong>stead of an ord<strong>in</strong>ary c<strong>in</strong>ema<br />
screen?<br />
688. A project<strong>in</strong>g camera stand<strong>in</strong>g near a w<strong>al</strong>l <strong>in</strong> a room<br />
produces an image with an area of A = 1 rot on the opposite<br />
w<strong>al</strong>l. What will the area of the image be if a flat mirror is so<br />
hung on the w<strong>al</strong>l opposite the camera that the image is obta<strong>in</strong>ed<br />
on the w<strong>al</strong>l near which the camera stands?<br />
Fig. 240
140 PROBLEMS<br />
W 689. Two flat mirrors AO<br />
~ and OB form a dihedr<strong>al</strong> angle<br />
A • q> =-; ~ were h n IS · an ·t In e-<br />
• B ger. A po<strong>in</strong>t source of light<br />
S is placed b<strong>et</strong>ween the mirrors<br />
at the same distance from<br />
both of them. F<strong>in</strong>d the num-<br />
~~~~~~~~~~~~ her of images of the source<br />
<strong>in</strong> the mirrors.<br />
Fig. 241<br />
690. Two flat mirrors AOand<br />
OB form an arbitrary dihedr<strong>al</strong><br />
angle cp = 2n, where a is any number greater than 2. A po<strong>in</strong>t<br />
a<br />
source of light S is b<strong>et</strong>ween the mirrors at equ<strong>al</strong> distances from<br />
them. F<strong>in</strong>d the number of images of the source <strong>in</strong> the mirrors.<br />
691. In what direction should a beam of light be sent from<br />
po<strong>in</strong>t A (Fig. 241) conta<strong>in</strong>ed <strong>in</strong> a mirror box for it to f<strong>al</strong>l onto<br />
po<strong>in</strong>t B after be<strong>in</strong>g reflected once from <strong>al</strong>l four w<strong>al</strong>ls?<br />
Po<strong>in</strong>ts A and B are <strong>in</strong> one plane perpendicular to the w<strong>al</strong>ls<br />
of the box (i.e., <strong>in</strong> the plane of the draw<strong>in</strong>g).<br />
692. Why does the water seem much darker directly below an<br />
airplane fly<strong>in</strong>g over a sea than at the horizon?<br />
693. Over what distance will a beam pass<strong>in</strong>g through a planepar<strong>al</strong>lel<br />
plate be displaced if the thickness of the plate is d,<br />
the refraction Index is n and the angle of <strong>in</strong>cidence is i?<br />
Can the beam be displaced by more than the thickness of the<br />
plate?<br />
n<br />
A<br />
A _--6--___.<br />
c........_ .....---......a<br />
Fig. 242<br />
Fig. 243<br />
C ----
GEOMETRICAL OPTICS 141<br />
~,<br />
8<br />
Fig. 244<br />
a<br />
694. At what v<strong>al</strong>ues of the refraction<br />
<strong>in</strong>dex of a rectangular prism can<br />
a ray travel as shown <strong>in</strong> Fig. 242?<br />
The section of the prism is an isosceles<br />
triangle and the ray is norm<strong>al</strong>ly<br />
<strong>in</strong>cident onto AB.<br />
695. A rectangular glass wedge is<br />
lowered <strong>in</strong>to water. The refraction <strong>in</strong>dex<br />
of glass is n 1 = 1.5. At what angle<br />
a (Fig. 243) will the beam of light<br />
norm<strong>al</strong>ly <strong>in</strong>cident on AB reach AC entirely?<br />
696. On bright sunny days drivers frequently see puddles on<br />
some parts of asph<strong>al</strong>t country highways at a distance of 80-]00<br />
m<strong>et</strong>res ahead of the car. As the driver approaches such places,<br />
the puddles disappear and reappear aga<strong>in</strong> <strong>in</strong> other places approximately<br />
at the same distance away. Expla<strong>in</strong> this phenomenon.<br />
697. A thick plate is made of a transparent materi<strong>al</strong> whose<br />
refraction <strong>in</strong>dex changes from n 1 on its upper edge to n" on its<br />
lower edge. A beam enters the plate at the angle a. At what<br />
angle will the beam leave the plate?<br />
698. A cubic<strong>al</strong> vessel with non-transparent w<strong>al</strong>ls is so located<br />
that<br />
the eye of an observer does not see its bottom, but sees<br />
<strong>al</strong>l of the w<strong>al</strong>l CD (Fig. 244).<br />
What amount of water should be poured <strong>in</strong>to the vessel for<br />
-the observer to see an object F arranged at a distance of b= 10 em<br />
from corner D? The face of the vessel is a = 40 em.<br />
699. A man <strong>in</strong> a boat is look<strong>in</strong>g at the bottom of a lake.<br />
How does the seem<strong>in</strong>g depth of the lake h depend on the angle l<br />
formed by the l<strong>in</strong>e of vision with the vertic<strong>al</strong>? The actu<strong>al</strong> depth<br />
of the lake is everywhere the same and equ<strong>al</strong> to H.<br />
700. The cross section of a glass prism has the form of an<br />
equilater<strong>al</strong> triangle. A ray is <strong>in</strong>cident onto one of the faces<br />
perpendicular to it. F<strong>in</strong>d the angle tp b<strong>et</strong>ween the <strong>in</strong>cident ray<br />
and the ray that leaves the prism. The refraction <strong>in</strong>dex of glass<br />
is n = 1.5.<br />
701. The cross section of a glass prism has the form of an<br />
isosceles triangle. One of the equ<strong>al</strong> faces is coated with silver.<br />
A ray is norm<strong>al</strong>ly <strong>in</strong>cident on another unsilvered face and, be<strong>in</strong>g<br />
reflected twice, emerges through the base of the prism perpendicular<br />
to it. F<strong>in</strong>d the angles of the prism.
142 PROBLEMS<br />
702. A ray <strong>in</strong>cident on the face of a prism is refracted and<br />
escapes through an adjacent face. What is the maximum permissible<br />
angle of refraction of the prism ex if it is made of glass with<br />
a refraction <strong>in</strong>dex of n = 1.5?<br />
703. A beam of light enters a glass prism at an angle a and<br />
emerges <strong>in</strong>to the air at an angle p. Hav<strong>in</strong>g passed through the<br />
prism, the beam<br />
is reflected from the orig<strong>in</strong><strong>al</strong> direction by an<br />
angle y. F<strong>in</strong>d the angle of refraction of the prism cp and the<br />
refraction <strong>in</strong>dex of the materi<strong>al</strong> which it is made of.<br />
704. The faces of prism ABeD made of glass with a refraction<br />
<strong>in</strong>dex n form dihedr<strong>al</strong> angles: L A = 90°, L B = 75°, L C = 135 0<br />
and L D = 60° (the Abbe prism). A beam of light f<strong>al</strong>ls on face<br />
AB and after compl<strong>et</strong>e <strong>in</strong>tern<strong>al</strong> reflection from face Be escapes<br />
through face AD. F<strong>in</strong>d the angle of <strong>in</strong>cidence a of the beam<br />
onto face AB if a beam that has passed through the prism is<br />
perpendicular to the <strong>in</strong>cident beam.<br />
705. If a she<strong>et</strong> of paper is covered with glue or water the<br />
text typed on the other side of the she<strong>et</strong> can be read. Expla<strong>in</strong><br />
why?<br />
5·3. Lenses and Spheric<strong>al</strong> Mirrors<br />
706. F<strong>in</strong>d the refraction <strong>in</strong>dex of the glass which a symm<strong>et</strong>ric<strong>al</strong><br />
convergent lens is made of if its foc<strong>al</strong> length is equ<strong>al</strong> to the<br />
radius of curvature of its surface.<br />
707. A plano-convex convergent lens is made of glass with a<br />
refraction <strong>in</strong>dex of n = 1.5. D<strong>et</strong>erm<strong>in</strong>e the relation b<strong>et</strong>ween the<br />
foc<strong>al</strong> length of this lens f and the radius of curvature of its convex<br />
surface R.<br />
708. F<strong>in</strong>d the radii of curvature of a convexo-concave convergent<br />
lens made of glass with a refraction <strong>in</strong>dex of n = 1.5<br />
hav<strong>in</strong>g a foc<strong>al</strong> length of f = 24 em. One of the radii of curvature<br />
is double the other.<br />
709. A convexo-convex lens made of glass with a refraction<br />
<strong>in</strong>dex of n = 1.6 has a foc<strong>al</strong> length of f = 10 em. What will the<br />
foc<strong>al</strong> length of this lens be if it is placed <strong>in</strong>to a transparent<br />
medium with a refraction <strong>in</strong>dex of n 1<br />
= 1.5? Also f<strong>in</strong>d the foc<strong>al</strong><br />
length of this lens <strong>in</strong> a medium with a refraction <strong>in</strong>dex of<br />
n t = 1.7.<br />
710. A th<strong>in</strong> glass lens has an optic<strong>al</strong> power of D = 5 diopters.<br />
When this lens is immersed <strong>in</strong>to a liquid with a refraction<br />
<strong>in</strong>dex nIt the lens acts as a divergent one with a foc<strong>al</strong> length
GEOMETRICAL OPTICS 143<br />
B of f = lOa em. F<strong>in</strong>d the refraction<br />
<strong>in</strong>dex n s of the liquid if<br />
that of the lens glass ls n 1 =<br />
1.5.<br />
F 0 F 711. The distance b<strong>et</strong>ween an<br />
object and a divergent lens is<br />
m times greater than the foc<strong>al</strong><br />
length of the lens. How many<br />
times will the image be sm<strong>al</strong>ler.<br />
than the object?<br />
712. The hot filament of a lamp and its image obta<strong>in</strong>ed with<br />
Fig. 245<br />
the aid of a lens hav<strong>in</strong>g an optic<strong>al</strong> power of four diopters are<br />
equ<strong>al</strong> <strong>in</strong> size. Over what distance should the lamp be moved<br />
away from the lens to decrease its image five times?<br />
713. The distance b<strong>et</strong>ween two po<strong>in</strong>t sources of light is 1= 24 cm.<br />
Where should a convergent lens with a foc<strong>al</strong> length of f = 9 ern<br />
be placed b<strong>et</strong>ween them to obta<strong>in</strong> the images of both sources at<br />
the same po<strong>in</strong>t?<br />
714. The height of a candle flame is 5 ern. A lens produces<br />
an image of this flame 15 cm high on a screen. Without touch<strong>in</strong>g<br />
the lens, the candle is moved over a distance of 1= 1.5 ern<br />
away from the lens, and a sharp image of the flame 10 cm<br />
high is obta<strong>in</strong>ed aga<strong>in</strong> after shift<strong>in</strong>g the screen. D<strong>et</strong>erm<strong>in</strong>e the<br />
ma<strong>in</strong> foc<strong>al</strong> length of the lens.<br />
715. A converg<strong>in</strong>g beam of rays is <strong>in</strong>cident onto a divergent<br />
lens so that the cont<strong>in</strong>uations of <strong>al</strong>l the rays <strong>in</strong>tersect at a<br />
po<strong>in</strong>t ly<strong>in</strong>g on the optic<strong>al</strong> axis of the lens at a distance of<br />
b= 15 em from it.<br />
F<strong>in</strong>d the foc<strong>al</strong> length of the lens <strong>in</strong> two cases:<br />
(1) after be<strong>in</strong>g refracted <strong>in</strong> the lens, the rays are assembled<br />
at a po<strong>in</strong>t at a distance of a 1 = 60 em from the lens;<br />
(2) the cont<strong>in</strong>uations of the refracted rays <strong>in</strong>tersect at a po<strong>in</strong>t<br />
at a distance of a 2<br />
= 60 em <strong>in</strong> front of the lens.<br />
716. The distance b<strong>et</strong>ween an electric lamp and a screen is<br />
d = 1 m<strong>et</strong>re. In what positions of a convergent lens with a foc<strong>al</strong><br />
.B<br />
Fig. 246
144<br />
PROBLEMS<br />
.9'<br />
Fig. 247<br />
length of f = 21 em will the image of the lamp filament be<br />
sharp?<br />
Can an image be obta<strong>in</strong>ed if the foc<strong>al</strong> length is I' = 26 em?<br />
717. A th<strong>in</strong> convergent lens produces the image of a certa<strong>in</strong><br />
object on a screen. The height of the image is hi. Without<br />
chang<strong>in</strong>g the distance b<strong>et</strong>ween the object and the screen, the<br />
lens is shifted, and it is found that the height of the second<br />
sharp image is h 2<br />
• D<strong>et</strong>erm<strong>in</strong>e the height of the object H.<br />
718. What is the radius R of a concave spheric<strong>al</strong> mirror at<br />
a distance of a = 2 m<strong>et</strong>res from the face of a man if he sees<br />
<strong>in</strong> it his image that is one and a h<strong>al</strong>f times greater than on<br />
a fiat mirror placed at the same distance from the face?<br />
719. Figure 245 shows ray AB that has passed through a<br />
divergent lens. Construct the path of the ray up to the lens<br />
if the position of its foci F is known.<br />
720. Figure 246 shows 8 lum<strong>in</strong>escent po<strong>in</strong>t and its image<br />
produced by a lens with an optic<strong>al</strong> axis NIN 2. F<strong>in</strong>d the position<br />
of the lens and its foci.<br />
721. F<strong>in</strong>d by construction the optic<strong>al</strong> centre of a lens and<br />
its ma<strong>in</strong> foci on the given optic<strong>al</strong> axis N IN 2 if the positions<br />
of the source S and the image S' are known (Fig. 247).<br />
722. The position of the optic<strong>al</strong> axis N IN2' the path of ray<br />
AB <strong>in</strong>cident upon a lens and the refracted ray Be are known<br />
(Fig. 248). F<strong>in</strong>d by construction the position of the ma<strong>in</strong> foci<br />
of the lens.<br />
723. A convergent lens produces the image of a source at<br />
po<strong>in</strong>t S' on the ma<strong>in</strong> optic<strong>al</strong> axis. The positions of the centre<br />
of the lens 0 and its foci F are known, and OF < OS'. F<strong>in</strong>d<br />
by construction the position of source S.<br />
A~t'<br />
B
GEOMETRICAL OPTICS<br />
145<br />
-8' ~ Fig. 249<br />
724. Po<strong>in</strong>t S' is the image of a po<strong>in</strong>t source of light S <strong>in</strong><br />
a spheric<strong>al</strong> mirror whose optic<strong>al</strong> axis is N1N,. (Fig. 249). F<strong>in</strong>d<br />
by construction the position of the centre of the mirror and<br />
its focus.<br />
725. The positions of optic<strong>al</strong> axis N 1<br />
N 2 of a spheric<strong>al</strong> mirror<br />
t the source and the image are known (Fig. 250). F<strong>in</strong>d by<br />
construction the positions of the centre of the mirror, its focus<br />
and the pole for the cases: (a) A-source, B- image; (b) B<br />
source, A-image.<br />
726. A po<strong>in</strong>t source of light placed at a distance from a<br />
screen creates an illum<strong>in</strong>ation of 2.25 Ix at the centre of the<br />
screen. How will this illum<strong>in</strong>ation change if on the other side<br />
of the source and at the same distance from it we place:<br />
(a) an <strong>in</strong>f<strong>in</strong>ite flat mirror par<strong>al</strong>lel to the screen?<br />
(b) a concave mirror whose centre co<strong>in</strong>cides wi th the centre<br />
of the screen?<br />
(c) a convex mirror with the same radius of curvature as<br />
the concave mirror?<br />
I<br />
727. A man wish<strong>in</strong>g to g<strong>et</strong> a picture of a zebra photographed<br />
a white donkey after fitt<strong>in</strong>g a glass with black streaks onto<br />
the objective of his camera. What will be on the photograph?<br />
728. The layered lens shown <strong>in</strong> Fig. 251 is made of two<br />
k<strong>in</strong>ds of glass. What image will be produced by this lens with<br />
./1<br />
Fig. 250<br />
Fig. 251<br />
10-2042
146 PROBLEMS<br />
a po<strong>in</strong>t source arranged on the optic<strong>al</strong> axis? Disregard the reflec-·<br />
tion of light on the boundary b<strong>et</strong>ween layers.<br />
729. The visible dimensions of the disks of the Sun and the<br />
Moon at the horizon seem magnified as compared with their<br />
visible dimensions ·at the zenith. How can it be proved experiment<strong>al</strong>ly<br />
with the aid of a lens that this magnification is<br />
apparent?<br />
5-4. Optic<strong>al</strong> Systems and Devices<br />
730. A source of light is located at double foc<strong>al</strong> length from<br />
a convergent lens. The foc<strong>al</strong> length of the lens is f = 30 em.<br />
At what distance from the lens should a flat mirror be placed<br />
so that the rays reflected from the mirror are par<strong>al</strong>lel after<br />
pass<strong>in</strong>g through the lens for the second time?<br />
731. A par<strong>al</strong>lel beam of rays is <strong>in</strong>cident on a convergent<br />
lens with a foc<strong>al</strong> length of 40 em. Where should a divergent<br />
lens with a foc<strong>al</strong> length of 15 em be placed for the beam of<br />
rays to rema<strong>in</strong> par<strong>al</strong>lel after pass<strong>in</strong>g through the two lenses?<br />
732. An object is at a distance of 40 em from a convex<br />
spheric<strong>al</strong> mirror with a radius of curvature of 20 em. At what<br />
distance from the object should a steel plate be placed for the<br />
image of the object <strong>in</strong> the spheric<strong>al</strong> mirror and the plate to<br />
be <strong>in</strong> one plane?<br />
733. At what distance from a convexo-convex lens with a<br />
foc<strong>al</strong> length of f = 1 m<strong>et</strong>re should a concave spheric<strong>al</strong> mirror<br />
with a radius of curvature of R= 1 m<strong>et</strong>re be placed for a beam<br />
<strong>in</strong>cident on the lens par<strong>al</strong>lel to the major optic<strong>al</strong> axis of the<br />
system to leave the lens, rema<strong>in</strong><strong>in</strong>g par<strong>al</strong>lel to the optic<strong>al</strong> axis,<br />
after be<strong>in</strong>g reflected from the mirror? F<strong>in</strong>d the image of the<br />
object produced by the given optic<strong>al</strong> system.<br />
734. An optic<strong>al</strong> system consists of two convergent lenses with<br />
foc<strong>al</strong> lengths '1 = 20 em and t, == 10 ern. The distance b<strong>et</strong>ween<br />
the lenses is d = 30 em. An object is placed at a distance of<br />
at = 30 cm from the first lens. At what distance from the second<br />
lens will the image be obta<strong>in</strong>ed?<br />
735. D<strong>et</strong>erm<strong>in</strong>e the foc<strong>al</strong> length of an optic<strong>al</strong> system consist<strong>in</strong>g<br />
of two th<strong>in</strong> lenses: a divergent one with a foc<strong>al</strong> length t.<br />
and a convergent one with a foc<strong>al</strong> length {20 The lenses are<br />
fitted tightly aga<strong>in</strong>st each other, so that the distance b<strong>et</strong>ween<br />
them can be neglected. The optic<strong>al</strong> axes of the lenses co<strong>in</strong>cide.
GEOMETRICAL OPTICS 147<br />
A~a<br />
B<br />
f<br />
Fig. 252<br />
736. A par<strong>al</strong>lel beam of light is <strong>in</strong>cident on a system consist<strong>in</strong>g<br />
of three th<strong>in</strong> lenses with a common optic<strong>al</strong> axis. The foc<strong>al</strong><br />
lengths of the lenses are equ<strong>al</strong> to 11 = + 10 em, 12 = -20 em<br />
and 13= +9 ern, respectively. The distance b<strong>et</strong>ween the first<br />
and the second .lenses is 15 ern and b<strong>et</strong>ween the second and<br />
the third 5 ern. F<strong>in</strong>d the position of the po<strong>in</strong>t at which the<br />
beam converges when it leaves the system of lenses.<br />
737. A lens with a foc<strong>al</strong> length of f = 30 em produces on a<br />
screen a sharp image of an object that is at a distance of<br />
a = 40 cm from the lens. A plane-par<strong>al</strong>lel plate with a thickness<br />
of d = 9 ern is placed b<strong>et</strong>ween the lens and the object perpendicular<br />
to the optic<strong>al</strong> axis of the lens. Through what distance<br />
should the screen be shifted for the image of the object to<br />
rema<strong>in</strong> dist<strong>in</strong>ct? The refraction <strong>in</strong>dex of the glass of the plate<br />
is n = 1.8. .<br />
738. An object AB is at a distance of a = 36 em from a lens<br />
with a foc<strong>al</strong> length of f = 30 em. A fla t mirror turned through 45°<br />
with respect to the optic<strong>al</strong> axis of the lens is placed beh<strong>in</strong>d<br />
it at a distance of 1= 1 m<strong>et</strong>re (Fig. 252).<br />
At what distance H from the optic<strong>al</strong> axis should the bottom<br />
of a tray with water be placed to obta<strong>in</strong> a sharp image of the<br />
object on the bottom? The thickness of the water layer ·<strong>in</strong> the<br />
tray is d = 20 cm. .<br />
739. A glass wedge with a sm<strong>al</strong>l angle of refraction ex is placed<br />
at a certa<strong>in</strong> distance from a convergent lens with a foc<strong>al</strong><br />
length I, one surface of the wedge be<strong>in</strong>g perpendicular to the<br />
optic<strong>al</strong> axis of the lens. A po<strong>in</strong>t source of light is on the other<br />
side of the lens <strong>in</strong> its focus. The rays reflected from the wedge<br />
produce, after refraction <strong>in</strong> the lens, two images of the source<br />
10 *
148 PROBLEMS<br />
displaced with respect to each other by d. F<strong>in</strong>d the refraction<br />
<strong>in</strong>dex of the wedge glass.<br />
740. A concave mirror has the form of a hemisphere with<br />
a radius of R = 55 em. A th<strong>in</strong> layer of an unknown transparent<br />
liquid is poured <strong>in</strong>to this mirror, and it was found that the<br />
given optic<strong>al</strong> system produces, with the source <strong>in</strong> a certa<strong>in</strong><br />
position, two re<strong>al</strong> images, one of which co<strong>in</strong>cides with the<br />
source and the other is at a distance of 1= 30 em from it.<br />
F<strong>in</strong>d the refraction <strong>in</strong>dex n of the liquid.<br />
741. A convexo-convex lens has a foc<strong>al</strong> length of II = 10 em.<br />
One of the lens surfaces hav<strong>in</strong>g a radius of curvature of<br />
R == 10 ern is coated with silver. Construct the image of the<br />
object produced by the given optic<strong>al</strong> system and d<strong>et</strong>erm<strong>in</strong>e the<br />
position of the image if the object is at a distance of a= 15 em<br />
frorn the lens.<br />
742. A recess <strong>in</strong> the form of a spheric<strong>al</strong> segment is made <strong>in</strong><br />
the flat surface of a massive block of glass (refraction <strong>in</strong>dex n).<br />
The piece of glass taken out of the recess is a th<strong>in</strong> convergent<br />
lens with a foc<strong>al</strong> length f. F<strong>in</strong>d the foc<strong>al</strong> lengths fl and '2 of<br />
the spheric<strong>al</strong> surface obta<strong>in</strong>ed.<br />
743. A narrow par<strong>al</strong>lel beam of light rays is <strong>in</strong>cident on a<br />
transparent sphere with a radius R and a refraction <strong>in</strong>dex<br />
n <strong>in</strong> the direction of one of the diam<strong>et</strong>ers. At what<br />
distance f from the centre of the sphere will the rays be focussed?<br />
744. F<strong>in</strong>d the position of the ma<strong>in</strong> planes of a transparent<br />
sphere used as a lens.<br />
745. An object is at a distance of d = 2.5 ern from the surface<br />
of a glass sphere with a radius of R = 10 ern. F<strong>in</strong>d the position<br />
of the image produced by the sphere. The refraction <strong>in</strong>dex of<br />
the glass is n = 1.5.<br />
746. A spheric<strong>al</strong> flask is made of glass with a refraction <strong>in</strong>dex<br />
n. The thickness of the flask w<strong>al</strong>ls ~R is much less than<br />
its radius R.<br />
Tak<strong>in</strong>g this flask as an optic<strong>al</strong> system and consider<strong>in</strong>g only<br />
the rays close to the straight l<strong>in</strong>e pass<strong>in</strong>g through the centre of<br />
the sphere, d<strong>et</strong>erm<strong>in</strong>e the position of the foci and the ma<strong>in</strong><br />
planes of the system.<br />
747. A beam of light is <strong>in</strong>cident on a spheric<strong>al</strong> drop of water<br />
at an angle i. F<strong>in</strong>d the angle e through which the beam is<br />
deflected from the <strong>in</strong>iti<strong>al</strong> direction after a s<strong>in</strong>gle reflection from<br />
the <strong>in</strong>tern<strong>al</strong> surface of the drop.
GEOMETRICAL OPTICS 149<br />
748. A par<strong>al</strong>lel beam of rays is <strong>in</strong>cident on a spheric<strong>al</strong> drop<br />
of water.<br />
(1) C<strong>al</strong>culate the angles e through which the rays are deflected<br />
from the <strong>in</strong>i ti<strong>al</strong> posi tion for various angles of <strong>in</strong>cidence: 0, 20,<br />
40, 50, 55, 60, 65, and 70°.<br />
(2) Plot a diagram show<strong>in</strong>g 6 versus t and use it to f<strong>in</strong>d the<br />
approximate v<strong>al</strong>ue of the angle of m<strong>in</strong>imum deflection 6 m<strong>in</strong> •<br />
(3) D<strong>et</strong>erm<strong>in</strong>e the v<strong>al</strong>ues of the angle 9 near which the rays<br />
issu<strong>in</strong>g from the drop are approximately par<strong>al</strong>lel.<br />
The refraction <strong>in</strong>dex of water is n = 1.333. (This v<strong>al</strong>ue of n<br />
is true for red ra ys.)<br />
749. What magnification can be obta<strong>in</strong>ed with the aid of a<br />
project<strong>in</strong>g camera hav<strong>in</strong>g a lens with a ma<strong>in</strong> foc<strong>al</strong> length of<br />
40 em if the distance from the lens to the screen is 10 m<strong>et</strong>res?<br />
750. C<strong>al</strong>culate the condenser of a project<strong>in</strong>g camera, i.e., f<strong>in</strong>d<br />
its diam<strong>et</strong>er D and foc<strong>al</strong> length I, if the source of light has<br />
dimensions of about d = 6 mID, and the diam<strong>et</strong>er of the lens is<br />
Do = 2 em. The distance b<strong>et</strong>ween the. source of light and the<br />
lens is 1= 40 em. The size of a slide is 6 X 9 em.<br />
751. In some photographic cameras ground glass is used for<br />
focuss<strong>in</strong>g. Why is transparent glass not used for this purpose?<br />
752. Two lanterns of the same lum<strong>in</strong>ance are at different<br />
distances from an observer.<br />
(1) Will they appear to the observer as equ<strong>al</strong>ly bright?<br />
(2) Will their images on photographs be equ<strong>al</strong>ly bright if the<br />
lanterns are photographed on different frames so that their<br />
images are focussed?<br />
753. An object is photographed from a sm<strong>al</strong>l distance by two<br />
cameras with identic<strong>al</strong> lens speeds, but different foc<strong>al</strong> lengths.<br />
Should the exposures be the same?<br />
754. It can be noticed that a white w<strong>al</strong>l illum<strong>in</strong>ated by the<br />
s<strong>et</strong>t<strong>in</strong>g Sun seems brighter than the surface of the Moon at the<br />
same <strong>al</strong>titude above the horizon as the Sun. Does this mean that<br />
the surface of the Moon consists of dark rock?<br />
755. Why does a swimmer see only hazy contours of objects<br />
when he opens his eyes under water, while they are dist<strong>in</strong>ctly<br />
visible if he is us<strong>in</strong>g a mask?<br />
75~. A short-sighted man, the accommodation of whose eye<br />
is b<strong>et</strong>ween at = 12 ern and a 2 = 60 em, wears spectacles through<br />
which he dist<strong>in</strong>ctly sees remote 'objects. D<strong>et</strong>erm<strong>in</strong>e the m<strong>in</strong>imum<br />
distance as at which the man can read a book through his<br />
spectacles.
150 PROBLEMS<br />
757. Two men, a far-sighted and a short-sighted ones, see<br />
objects through their spectacles as a man with norm<strong>al</strong> eyesight.<br />
When the far-sighted man accidently put on the spectacles of<br />
his short-sighted friend, he found that he could see dist<strong>in</strong>ctly<br />
only <strong>in</strong>f<strong>in</strong>itely far objects.<br />
At what m<strong>in</strong>imum distance a can the short-sighted man read<br />
sm<strong>al</strong>l type if he wears the spectacles of the far-sighted man?<br />
758. An object is exam<strong>in</strong>ed by a naked eye from a distance D.<br />
What is the angular magnification if the same object is viewed<br />
through a magnify<strong>in</strong>g glass held at a distance r from the eye<br />
and so arranged that the image is at a distance L from the eye?<br />
The foc<strong>al</strong> length of the lens is f.<br />
Consider the cases:<br />
(I)L=oo<br />
(2) L=D.<br />
759. The objective is taken out of a telescope adjusted to<br />
<strong>in</strong>f<strong>in</strong>ity and replaced by a diaphragm with a diam<strong>et</strong>er D. A screen<br />
shows a re<strong>al</strong> image of the diaphragm hav<strong>in</strong>g a diam<strong>et</strong>er d at<br />
a certa<strong>in</strong> distance from the eyepiece. What was the magnification<br />
of the telescope?<br />
760. The double-lens objective of a photographic camera is<br />
made of a divergent lens with a foc<strong>al</strong> length of fl = 5 em <strong>in</strong>st<strong>al</strong>led<br />
at 8 distance of 1= 45 em from the film. Where should<br />
a convergent lens with a foc<strong>al</strong> length of t, = 8 cm be placed to<br />
obta<strong>in</strong> a sharp image of remote objects on the film?<br />
761. C<strong>al</strong>culate the diam<strong>et</strong>er D of the Moon's image on a negative<br />
for the three different positions of the lenses described <strong>in</strong><br />
Problem 760.<br />
The diam<strong>et</strong>er of the Moon is seen from the Earth at an angle<br />
of «p=31'5" ~O.9x 10- 2 rad.<br />
762. The ma<strong>in</strong> foc<strong>al</strong> length of the objective of a microscope<br />
is fob = 3 mm and of the eyepiece feye = 5 em. An object is at<br />
a distance of a = 3.1 mm from the objective. F<strong>in</strong>d the magnification<br />
of the microscope for a norm<strong>al</strong> eye.
CHAPTER 6<br />
PHYSICAL<br />
OPTICS<br />
•<br />
6-1. Interference of Light<br />
763. Two light waves are superposed <strong>in</strong> a certa<strong>in</strong> section of<br />
space and ext<strong>in</strong>guish each other. Does this mean that a quantity<br />
of light is converted <strong>in</strong>to other k<strong>in</strong>ds of energy?<br />
764. Two coherent sources of light 8 1<br />
and 8, are at a distance 1<br />
from each other. A screen is placed at a distance D ~ l from<br />
the sources (Fig. 253). F<strong>in</strong>d the distance b<strong>et</strong>ween adjacent <strong>in</strong>terference<br />
bands. near the middle of the screen (po<strong>in</strong>t A) if the<br />
sources send light with a wavelength 'A.<br />
765. Two flat mirrors form an angle close to 180 0 (Fig. 254).<br />
A source of light S is placed at equ<strong>al</strong> distances b from the<br />
mirrors. F<strong>in</strong>d the <strong>in</strong>terv<strong>al</strong> b<strong>et</strong>ween adjacent <strong>in</strong>terference bands<br />
on screen MN at a distance OA = a from the po<strong>in</strong>t of <strong>in</strong>tersection<br />
of the mirrors. The length of the light wave is known and<br />
equ<strong>al</strong> to 'A. Shield C does not <strong>al</strong>low the light to pass directly<br />
from the source to the screen.<br />
766. Lloyd's <strong>in</strong>terference experiment consisted <strong>in</strong> obta<strong>in</strong><strong>in</strong>g<br />
on a screen a pattern from source S and from its virtu<strong>al</strong> image S'<br />
<strong>in</strong> mirror AO (Fig. 255). How will the <strong>in</strong>terference pattern<br />
obta<strong>in</strong>ed from sources S and Sf differ from the pattern considered<br />
<strong>in</strong> Problem 764?<br />
t ·oS;<br />
I .....CE-------.D -----~...A<br />
l ..sz<br />
Fig. 253
152 PROBLEMS<br />
H 767. Two po<strong>in</strong>t sources<br />
with the same phases of<br />
oscillation are on a straight<br />
l<strong>in</strong>e perpendicular to a<br />
screen. The nearest source<br />
A is at a distance of D ~ A<br />
from the screen. What<br />
shape will the <strong>in</strong>terference<br />
bands have on the screen?<br />
What is the distance on<br />
the screen from the per-<br />
N pendicular to the nearest<br />
bright band if the distance<br />
b<strong>et</strong>ween the sources is<br />
Fig. 254<br />
1= n'A ~ 'A (n is an <strong>in</strong>teger)?<br />
768. F<strong>in</strong>d the radius 'k of the k..th bright r<strong>in</strong>g (see Problem<br />
767) if D=l=nA, n~l, and k=n, n-I, n-2, <strong>et</strong>c.<br />
769. How can the experiment described <strong>in</strong> Problem 767 be<br />
carried out <strong>in</strong> practice?<br />
770. Light from source S is <strong>in</strong>cident on the Fresnel biprism<br />
shown <strong>in</strong> Fig. 256. The light beams refracted by the different<br />
faces of the prism partly overlap and produce an <strong>in</strong>terference<br />
pattern on a screen on its section AB.<br />
F<strong>in</strong>d the distance b<strong>et</strong>ween adjacent <strong>in</strong>terference bands if the<br />
distance from the source to the prism is a = 1 m<strong>et</strong>re and from<br />
the prism to the screen b= 4 m<strong>et</strong>res. The angle of refraction<br />
of the prism is ex = 2 X 10- 8 rad.<br />
The glass which the prism is made of has a refraction <strong>in</strong>dex<br />
of n = 1..5. The length of the light wave A= 6,000 A.<br />
771. How many <strong>in</strong>terference bands can be observed on a screen<br />
<strong>in</strong> an <strong>in</strong>st<strong>al</strong>lation with the biprism described <strong>in</strong> the previous<br />
problem?<br />
A<br />
Fig. 255
PHYSICAL OPTICS<br />
153<br />
H<br />
&7
154 PROBLEMS<br />
773. A convergent lens with a foc<strong>al</strong> length of f = 10 em is<br />
cut <strong>in</strong>to two h<strong>al</strong>ves that are then moved apart to a distance<br />
of d = 0.5 mm (8 double lens). Appraise the number of <strong>in</strong>terference<br />
bands on a screen at a distance of D = 60 em beh<strong>in</strong>d the<br />
lens if a po<strong>in</strong>t source of monochromatic light (A = 5,000 A) is<br />
placed <strong>in</strong> front of the lens at a distance of a= 15 em from it.<br />
774. A centr<strong>al</strong> portion with a width of d = 0.5 rom is cut out of<br />
a convergent lens hav<strong>in</strong>g a foc<strong>al</strong> length of f = 10 ern, as shown<br />
<strong>in</strong> Fig. 258. Both h<strong>al</strong>ves are tightly fitted aga<strong>in</strong>st each other.<br />
The lens receives monochromatic light (A. = 5,000 A) from a po<strong>in</strong>t<br />
source at a distance of a = 5 ern from it. At what distance<br />
should a screen be fixed on the opposite side of the lens to<br />
observe three <strong>in</strong>terference bands on it?<br />
What is the maximum possible number of <strong>in</strong>terference bands<br />
that can be observed <strong>in</strong> this <strong>in</strong>st<strong>al</strong>lation?<br />
775. F<strong>in</strong>d the distance b<strong>et</strong>ween the neighbour<strong>in</strong>g bands of an<br />
<strong>in</strong>terference pattern produced by a lens with a radius of R = 1 ern,<br />
described <strong>in</strong> Problem 774, if this distance does not depend on<br />
the position °of the screen.<br />
At what position of the screen will the number of <strong>in</strong>terference<br />
bands be maximum?<br />
The source sends monochromatic light with a wavelength of<br />
A,= 5,OOOA.<br />
776. What will occur with the <strong>in</strong>terference pattern <strong>in</strong> the<br />
<strong>in</strong>st<strong>al</strong>lation described<br />
<strong>in</strong> Problem 775 if a plane-par<strong>al</strong>lel glass<br />
plate with a thickness of d 1<br />
= 0.11 em is <strong>in</strong>troduced <strong>in</strong>to a light<br />
beam that has passed through the upper h<strong>al</strong>f of the lens and<br />
a plate d 2<br />
= 0.1 em thick <strong>in</strong>to a light beam that has passed<br />
through the lower h<strong>al</strong>f of the lens? The refraction <strong>in</strong>dex of glass<br />
is n = 1.5. The plates are arranged norm<strong>al</strong>ly to the light beams<br />
pass<strong>in</strong>g through them.<br />
777. Why ari Newton's r<strong>in</strong>gs formed only by the <strong>in</strong>terference<br />
of rays 2 and 3 reflected from the boundaries of the air layer<br />
b<strong>et</strong>ween the lens and the glass (Fig. 259), while ray 4 reflected<br />
from the flat face of the lens does not affect the nature of the<br />
<strong>in</strong>terference pattern?<br />
778. Will the nature of the <strong>in</strong>terference pattern change <strong>in</strong> the<br />
<strong>in</strong>st<strong>al</strong>lation described <strong>in</strong> Problem 765 if shield C is removed?<br />
Consider the distance a to be great (one m<strong>et</strong>re). 1 he waves<br />
radiated by the source are not monochromatic.<br />
779. Will Newton's r<strong>in</strong>gs be seen more dist<strong>in</strong>ctly <strong>in</strong> reflected<br />
light or <strong>in</strong> transmitted light?
PHYSICAL OPTICS<br />
155<br />
1429<br />
Fig. 259<br />
780. Dust prevents contact b<strong>et</strong>ween a plane-convex lens and<br />
the glass plate on which it is placed. The radius of Newton's<br />
fifth dark r<strong>in</strong>g is '1 = 0.08 em.<br />
If the dust is removed the radius of this r<strong>in</strong>g will <strong>in</strong>crease<br />
to '2 = 0.1 ern, F<strong>in</strong>d the thickness of the dust layer d if the<br />
radius of curvature of the convex surface of the lens is R = 10 em.<br />
781. A plane-convex lens with a radius of curvature R 2<br />
< R 1<br />
is placed with its convex side onto the surface of a plane-concave<br />
lens hav<strong>in</strong>g a radius of curvature of R r F<strong>in</strong>d the radii of Newton's<br />
r<strong>in</strong>gs that appear around the po<strong>in</strong>t of contact of the lenses<br />
if monochromatic light with a wavelength A, is norm<strong>al</strong>ly <strong>in</strong>cident<br />
onto the system.<br />
782. A th<strong>in</strong> coat of a transparent substance with a refraction<br />
<strong>in</strong>dex n lower than that of glass is applied to the surface of optic<strong>al</strong><br />
glass (coated glass) to reduce its reflection factor (reflectance).<br />
Estimate the thickness of this coat if light rays are <strong>in</strong>cident<br />
<strong>al</strong>most norm<strong>al</strong>ly onto the optic<strong>al</strong> glass.<br />
783. A norm<strong>al</strong> eye can discern various colour t<strong>in</strong>ts with a<br />
difference of wavelengths of 100 A. Bear<strong>in</strong>g this <strong>in</strong> m<strong>in</strong>d, estimate<br />
the maximum thickness of a th<strong>in</strong> air layer at which an<br />
iriterference pattern caused by superposition of the rays reflected<br />
from the boundaries of this layer can be observed <strong>in</strong> white light.<br />
784. A monochromatic flux from a remote source with a<br />
wavelength A, is <strong>al</strong>most norm<strong>al</strong>ly <strong>in</strong>cident on a th<strong>in</strong> glass wedge.<br />
A screen is placed at a distance d from the wedge. A lens with<br />
a foc<strong>al</strong> length f projects the <strong>in</strong>terference pat tern produced <strong>in</strong><br />
the wedge onto the screen. The distance b<strong>et</strong>ween the <strong>in</strong>terference<br />
bands on the screen ~l is known. F<strong>in</strong>d the angle a of the<br />
wedge if the refraction <strong>in</strong>dex of glass is n.
156 PROBLEMS<br />
6.. 2. Diffraction of Light<br />
785. C<strong>al</strong>culate the radii of the Fresnel zones of a spheric<strong>al</strong><br />
wave with a radius a for po<strong>in</strong>t B removed by a distance of<br />
a+b from a source of monochromatic waves with a length of A,<br />
bear<strong>in</strong>g <strong>in</strong> m<strong>in</strong>d that a ~ A and b~ A.<br />
786. C<strong>al</strong>culate the radii of the Fresnel zones of a plane wave<br />
for po<strong>in</strong>t B removed from the wave front by a distance of<br />
b~ At, where A is the wavelength of the source.<br />
787. A po<strong>in</strong>j source of monochromatic light wi th a wavelength<br />
of A= 5,000 A is at a distance of a = 6.75 m<strong>et</strong>res from a shield<br />
hav<strong>in</strong>g an aperture with a diam<strong>et</strong>er D = 4.5 rnm. A screen is<br />
placed at a distance of b = a from the shield (Fig. 260). How<br />
will the illum<strong>in</strong>ation change at po<strong>in</strong>t B of the screen ly<strong>in</strong>g<br />
on the axis of the beam if the diam<strong>et</strong>er of the aperture is<br />
<strong>in</strong>creased to D 1 = 5.2 mm?<br />
788. How can the fact that an <strong>in</strong>crease of the aperture (see<br />
Problem 787) may reduce the illum<strong>in</strong>ation on the axis of a beam<br />
be agreed with the law of conservation of energy? Indeed, the<br />
tot<strong>al</strong> lum<strong>in</strong>ous flux pen<strong>et</strong>rat<strong>in</strong>g beh<strong>in</strong>d the shield <strong>in</strong>creases<br />
when the aperture grows.<br />
789. A plane light wave (At = 6,000 A) is <strong>in</strong>cident on a shield<br />
with a circular diaphragm. A screen is placed at a distance of<br />
b-:.2 m<strong>et</strong>res beh<strong>in</strong>d the diaphragm. At wha t diam<strong>et</strong>er D of the<br />
diaphragm will the illum<strong>in</strong>ation of the screen at po<strong>in</strong>t B ly<strong>in</strong>g<br />
on the axis of the light beam be maximum?<br />
790. Assum<strong>in</strong>g the distances from a source to a shield and<br />
from the shield to a screen to be about the same and equ<strong>al</strong><br />
to a, f<strong>in</strong>d the condi tions <strong>in</strong> which the diffraction of light<br />
waves with a length A on the aperture <strong>in</strong> the shield 'will be<br />
sufficiently dist<strong>in</strong>ct (the <strong>in</strong>tensity on the axis of the beam will<br />
depend on the diam<strong>et</strong>er of the aperture).<br />
A 4<br />
~--------------------- B<br />
--4 -Itb<br />
Fig. 260
PHYSI CAL OPTICS<br />
157<br />
c<br />
1--------------1---------- B<br />
Fig. 261<br />
791. Prove that a bright spot will be observed at po<strong>in</strong>t B<br />
beh<strong>in</strong>d circular screen C (Fig. 261) if its dimensions are sufficiently<br />
sm<strong>al</strong>l.<br />
792. At wha t distance from each other should two men stand<br />
for an eye to dist<strong>in</strong>guish them from a distance of about 11 km?<br />
The resolv<strong>in</strong>g power of a norm<strong>al</strong> eye is approximately 1'.<br />
793. A plane light wave (with a length A) is norm<strong>al</strong>ly <strong>in</strong>cident<br />
on a narrow slit with a width b. D<strong>et</strong>erm<strong>in</strong>e the directions<br />
to the illum<strong>in</strong>ation m<strong>in</strong>ima.<br />
794. F<strong>in</strong>d the optimum dimensions of the aperture <strong>in</strong> a p<strong>in</strong><br />
hole camera depend<strong>in</strong>g on the wavelength, Le., the radius,<br />
of the aperture at which a po<strong>in</strong>t source will appear on the<br />
camera w<strong>al</strong>l as a circle of m<strong>in</strong>imum diam<strong>et</strong>er, if the distance<br />
from the source of light to the camera is great as compared<br />
with its depth d.<br />
Note. The directions to the illum<strong>in</strong>ation m<strong>in</strong>ima are d<strong>et</strong>erm<strong>in</strong>ed<br />
<strong>in</strong> the order of their magnitude by the same formula as<br />
<strong>in</strong> the case of a slit (see Problem 793), the diam<strong>et</strong>er of the<br />
aperture 2r be<strong>in</strong>g used <strong>in</strong>stead of the wid th of the slit b.<br />
795. A monochromatic wave is norm<strong>al</strong>ly <strong>in</strong>cident on a diffraction<br />
grat<strong>in</strong>g with a period of d = 4 X 10- 4 cm. F<strong>in</strong>d the wavelength<br />
A if the angle b<strong>et</strong>ween the spectra of the second and<br />
third orders is a = 2°30'. The angles of deflection are sm<strong>al</strong>l.<br />
796. A plane monochromatic wave (At = 5 X 10-& ern) is <strong>in</strong>cident<br />
on a diffraction grat<strong>in</strong>g with 500 l<strong>in</strong>es. D<strong>et</strong>erm<strong>in</strong>e the maximum<br />
order of the spectrum k that can be observed when the rays<br />
are norm<strong>al</strong>ly <strong>in</strong>cident on the grat<strong>in</strong>g.<br />
797. F<strong>in</strong>d the constant d of a grat<strong>in</strong>g that can an<strong>al</strong>yse <strong>in</strong>frared<br />
radiation with wavelengths up to At = 2 X 10- 1 em. The<br />
radiation is norm<strong>al</strong>ly <strong>in</strong>cident on the grat<strong>in</strong>g.<br />
798. A monochromatic wave is norm<strong>al</strong>ly <strong>in</strong>cident on a diffraction<br />
grat<strong>in</strong>g with a period of d = 4 X 10- 4 em. A lens with a
158 P~OBLEMS<br />
foc<strong>al</strong> length of f = 40 ern that produces an image of the diffraction<br />
pattern on a screen is arranged beh<strong>in</strong>d the grat<strong>in</strong>g.<br />
F<strong>in</strong>d the wavelength A if the first maximum is obta<strong>in</strong>ed at<br />
a distance of 1= 5'cm from the centr<strong>al</strong> one.<br />
799. A source of white light, a diffraction grat<strong>in</strong>g and a<br />
screen are immersed <strong>in</strong>to water. What will the changes <strong>in</strong> the<br />
diffraction pattern be if the angles by which the light rays are<br />
deflected by the grat<strong>in</strong>g are sm<strong>al</strong>l?<br />
800. Light passed through a light filter is norm<strong>al</strong>ly <strong>in</strong>cident<br />
on a diffraction grat<strong>in</strong>g with a period qf d=2x 10-' cpl. The<br />
filter passes wavelengths from At = 5,000 A to AI = 6,000 A. Will<br />
the spectra of different orders be superposed?<br />
801. Solve Problem 796, assum<strong>in</strong>g that a plane wave<br />
(At = 5 X 10- 6 ern) is <strong>in</strong>cident on the grat<strong>in</strong>g at an angle of 30°.<br />
802. Solve Problem 797, assum<strong>in</strong>g that the rays may f<strong>al</strong>lon<br />
the grat<strong>in</strong>g diagon<strong>al</strong>ly,<br />
803. F<strong>in</strong>d the condition that d<strong>et</strong>erm<strong>in</strong>es the directions to the<br />
pr<strong>in</strong>cip<strong>al</strong> maxima if light waves f<strong>al</strong>l diagon<strong>al</strong>ly on a grat<strong>in</strong>g<br />
with 8 period d~kA (k is the order of the spectrum).<br />
6-3. Dispersion of Light and Colours of Bodies<br />
804. A beam of white light f<strong>al</strong>ls at an angle of cx=30° on<br />
a prism with a refraction angle of q> = 45°. D<strong>et</strong>erm<strong>in</strong>e the angle<br />
e b<strong>et</strong>ween the extreme rays of the spectrum when they emerge<br />
from the lens if the refraction <strong>in</strong>dices of the prism for the<br />
extreme rays of the visible spectrum are n,= 1.62 and n v= 1.67.<br />
805. White light is <strong>in</strong>cident from a po<strong>in</strong>t source on the<br />
optic<strong>al</strong> axis of a convexo-convex lens at a distance of a = 50 cm<br />
from it. The radii of curvature of the lens are R 1<br />
= R 2<br />
= 40 ern.<br />
A diaphragm with a diam<strong>et</strong>er of D = 1 cm that restricts the<br />
cross section of the light beam is placed tightly <strong>in</strong> front of<br />
the lens. The refraction <strong>in</strong>dices for the extreme rays of the<br />
visible spectrum are n, = 1.74 and n v = 1.8, respectively. What<br />
pattern can be observed on a screen arranged at a distance of<br />
b= 50 em from the lens and perpendicular to its optic<strong>al</strong> axis?<br />
806. Us<strong>in</strong>g the results of Problem 748, construct the elementary<br />
theory of the ra<strong>in</strong>bow, i. e., show that the centre of a<br />
ra<strong>in</strong>bow lies on a straight l<strong>in</strong>e drawn from the Sun through the<br />
eye of an observer and that the arc of the ra<strong>in</strong>bow is a part<br />
of a circle <strong>al</strong>l of whose po<strong>in</strong>ts can be seen at an angle of 42°
PHYSICAL OPTICS 159<br />
(for red light) with reference to the straight l<strong>in</strong>e connect<strong>in</strong>g<br />
the eye of the observer and the centre of the ra<strong>in</strong>bow.<br />
807. Expla<strong>in</strong> from a qu<strong>al</strong>itative po<strong>in</strong>t of view the appearance<br />
of a double ra<strong>in</strong>bow. How do the colours <strong>al</strong>ternate <strong>in</strong> the primary<br />
and the reflection ra<strong>in</strong>bows?<br />
808. Can a ra<strong>in</strong>bow be observed at midday <strong>in</strong> Moscow dur<strong>in</strong>g<br />
the summer solstice (on June 22)?<br />
Note. At this time the Sun is the highest above the horizon<br />
<strong>in</strong> the northern hemisphere.<br />
809. The length of a wave <strong>in</strong> water dim<strong>in</strong>ishes n times,<br />
n be<strong>in</strong>g the refraction <strong>in</strong>dex. Does this mean that a diver cannot<br />
see surround<strong>in</strong>g objects <strong>in</strong> their natur<strong>al</strong> colours?<br />
810. The word "excellent" is written on a she<strong>et</strong> of white<br />
paper with a red pencil and the word "good" with a green<br />
pencil. A green and a red pieces of glass are available. Through<br />
which glass can the word "excellent n<br />
be seen?<br />
811. Why do coated lenses (see Problem 782) have a purpleviol<strong>et</strong><br />
(lilac) t<strong>in</strong>t?<br />
812. Why do the colours of th<strong>in</strong> films (for example, oil films on<br />
water) and the colours of a ra<strong>in</strong>bow have different t<strong>in</strong>ts?<br />
813. A th<strong>in</strong> soapy film is str<strong>et</strong>ched over a vertic<strong>al</strong> frame.<br />
When the film is illum<strong>in</strong>ated by white light it shows three<br />
bands coloured purple (crimson), yellow and light-blue (bluegreen).<br />
F<strong>in</strong>d the arrangement and the order of the bands<br />
814. Why does the Moon, purely white <strong>in</strong> the daytime, have<br />
a yellowish hue after suns<strong>et</strong>?<br />
815. Why does a column of smoke ris<strong>in</strong>g above the roofs of<br />
houses seem blue aga<strong>in</strong>st the dark background of the surround<strong>in</strong>g<br />
objects, and yellow or even reddish aga<strong>in</strong>st the background of<br />
a bright sky?<br />
816. Why do the colours of moist objects seem deeper and<br />
richer than those of dry ones?
ANSWERS AND SOLUTIONS<br />
CHAPTER 1<br />
MECHANICS<br />
1-1. K<strong>in</strong>ematics of Uniform Rectil<strong>in</strong>ear Motion<br />
I. Dur<strong>in</strong>g the first hour after the me<strong>et</strong><strong>in</strong>g, the boat travelled away from<br />
the rafts. Dur<strong>in</strong>g the next 30 m<strong>in</strong>utes, when the eng<strong>in</strong>e was be<strong>in</strong>g repaired,<br />
the distance b<strong>et</strong>ween the boat and the rafts did not <strong>in</strong>crease. The boat overtook<br />
the rafts <strong>in</strong> one hour because its speed with respect to the water and hence<br />
to the rafts was constant. Thus the velocity of the current<br />
fI= ~ l+~·.~+l km/h=3 km/h<br />
2. As can be seen from Fig. 262, S=H H h s. S<strong>in</strong>ce the man moves uniformly,<br />
H<br />
s=vt. Hence S=H-h ot. The shadow moves with a constant velocity<br />
HHfIh • greater than the speed of the man. For this reason the velocity<br />
diagram has the form of a straight l<strong>in</strong>e par<strong>al</strong>lel to the axis of abscissas.<br />
3. The time of the me<strong>et</strong><strong>in</strong>g was 8 a. m. The man's speed was 4 km/h.<br />
The other questions can easily be answered with the aid of the chart <strong>in</strong> Fig.<br />
263-the man m<strong>et</strong> the twelfth bus at a distance of 10.7 km from the mill;<br />
the cyclist was overtaken by four buses.<br />
..,<br />
I,<br />
Fig. 262
MECHANICS<br />
161<br />
s,km<br />
90<br />
7:80 8 B:JO 9 9:30 10<br />
8:30 7 7:30 8 8:30 9 9:30 10~h<strong>al</strong>r8 Fig. 263<br />
4. The distance b<strong>et</strong>ween the tra<strong>in</strong>s is s=vt; on the other hand, s=vr+u't.<br />
Hence u=v(t-'t')=45 km/h.<br />
l'<br />
5. In Fig. 264, AA1N shows the usu<strong>al</strong> trip of the car. SC-the eng<strong>in</strong>eer's<br />
w<strong>al</strong>k until he m<strong>et</strong> the car at po<strong>in</strong>t C, CB- the motion of the car after it<br />
m<strong>et</strong> the eng<strong>in</strong>eer.<br />
Accord<strong>in</strong>g to the conditions, BN =KM = 10 m<strong>in</strong>utes. The time dur<strong>in</strong>g<br />
which the eng<strong>in</strong>eer had w<strong>al</strong>ked before he m<strong>et</strong> the car is<br />
SD=SM-DM=SM -~=55m<strong>in</strong>utes<br />
6. The chart <strong>in</strong> Fig. 265 shows the movement of the launches b<strong>et</strong>ween<br />
the land<strong>in</strong>g-stages M and K. It follows from the chart that the land<strong>in</strong>g-stages<br />
are served by eleven launches. A launch travell<strong>in</strong>g from M to K me<strong>et</strong>s eight<br />
launches, as does a launch travell<strong>in</strong>g. from K to M.<br />
11-2042<br />
Fig. 264
162 ANSWERS AND SOLUTIONS<br />
I 2 9 5<br />
G<br />
10<br />
5<br />
B 7 ist; hlJlJrS Fig. 265<br />
7. Each tourist will w<strong>al</strong>k h<strong>al</strong>f the distance and ride the bicycle the other<br />
h<strong>al</strong>f. The entire distance will be covered -<strong>in</strong> t=-2 5 +2~=5 hours 20 mtn-<br />
utes.<br />
Hence, the mean speed is v=T=7.S krn/h, The bicycle rema<strong>in</strong>s unused<br />
- dur<strong>in</strong>g h<strong>al</strong>f the time of motion, i.e., dur<strong>in</strong>g 2 hours 40 m<strong>in</strong>utes.<br />
8. Assume that the first candle burns down by the amount ~hl and the<br />
other by ~h2 dur<strong>in</strong>g the time At (Fig. 266). The shadow on the left w<strong>al</strong>l<br />
(from the first candle) will then lower over the distance<br />
tlx= ~hl+(l1h l -l1h 2 ) = 2l1h l -l1hs<br />
and that on the right w<strong>al</strong>l over the distance·<br />
l1y= aha - (~hl - Ah 2 ) = 2Ah a -l1h!<br />
h<br />
Remember<strong>in</strong>g that I1h l = t; flt and ~hs=-r; 4t, we g<strong>et</strong><br />
ax 2h h h<br />
Vl= I1t=7;-7;= tlt a<br />
h<br />
VI<br />
(2t,-t1)<br />
Ay 2h h h<br />
Va= At=t;-t;= t<br />
1ta (2t1-t,.><br />
If t 2 > tt' then VI > 0 and Va may be negative, l.e., the shadow on the<br />
right w<strong>al</strong> may move upward.<br />
0. The bus is at po<strong>in</strong>t A and the man at po<strong>in</strong>t B (Fig. 267). Po<strong>in</strong>t C is<br />
the spot where the man me<strong>et</strong>s the bus, a is the angle b<strong>et</strong>ween the direction<br />
towards the bus and the direction <strong>in</strong> which the man should run, AC=v1t lt<br />
BC=V2t2, where tt and t 2 are the times required for the bus and the man,<br />
respectively, to reach po<strong>in</strong>t C.<br />
Vt
MECHANICS<br />
163<br />
Fig. 266<br />
b s<strong>in</strong> a<br />
A glance at triangle ABC shows that AC= s<strong>in</strong> p , where s<strong>in</strong> ~= Be.<br />
Consequently. s<strong>in</strong> (1,= a b<br />
~t • Accord<strong>in</strong>g to the condition, t 1 ~ t., and<br />
Va 2<br />
therefore s<strong>in</strong> a. ~ :~ = 0.6. Hence 36°45' C;;;; a. OS;;;; 143°15'.<br />
The directions <strong>in</strong> which the man can move are with<strong>in</strong> the limits of the<br />
angle DBE. Upon runn<strong>in</strong>g <strong>in</strong> the directions BD or BE, the man will reach<br />
the highway at the same time as the bus. He will reach any po<strong>in</strong>t on the<br />
highway b<strong>et</strong>ween D and E before the bus.<br />
10. The m<strong>in</strong>imum speed can be d<strong>et</strong>erm<strong>in</strong>ed from the conditions<br />
. aVl 1<br />
t 1 = t 2, 81n a = oo<br />
l<br />
a<br />
Hence VS=TVl=2.4 mte.<br />
Here a=900. Therefore, the man should run <strong>in</strong> a direction perpendicular<br />
to the <strong>in</strong>iti<strong>al</strong> l<strong>in</strong>e (AB, Fig. 267) b<strong>et</strong>ween him and the bus.<br />
II. S<strong>in</strong>ce the man's speed <strong>in</strong> water is lower than that <strong>al</strong>ong the shore,<br />
the route AB will not necessarily take the shortest time.<br />
Assume that the man follows the route ADB (Fig. 268). L<strong>et</strong> us d<strong>et</strong>erm<strong>in</strong>e<br />
the distance x at which the time will be m<strong>in</strong>imum.<br />
The time of motion t is<br />
y d 2 +Xi +_s-_% = V-=2_V_t1_2.....;.+_X_2__~Ul;;:..%_+'"'-Vl __S<br />
=<br />
VI . VI VIVS<br />
This time will be m<strong>in</strong>imum if Y=VI Y d2 + X2 - VIX has the sm<strong>al</strong>lest v<strong>al</strong>ue.<br />
obvi ously, the v<strong>al</strong>ue of x that corresponds to the m<strong>in</strong>imum time t does not<br />
a<br />
II *<br />
~~<br />
8 Fig. 267
164 ANSWERS AND SOLUTIONS<br />
depend on the distance s. To f<strong>in</strong>d the v<strong>al</strong>ue of x correspond<strong>in</strong>g to the m<strong>in</strong>imum<br />
v<strong>al</strong>ue of y, l<strong>et</strong> us express x through Y and obta<strong>in</strong> the quadratic equation<br />
2YVt V~d2 - y 2<br />
x t - I 2 x+ 2 2 ~O<br />
V2- V J V2- Vl<br />
Its solution leads to the follow<strong>in</strong>g expression<br />
x<br />
VtY ± V2 Vy2+ d2V~ - v:d 2<br />
v:-v~<br />
S<strong>in</strong>ce % cannot be 8<br />
complex number, y2+d2v~ ~ v:d 2 •<br />
Th~ m<strong>in</strong>imum v<strong>al</strong>ue of y. is equ<strong>al</strong> to Ym<strong>in</strong>=dV v:-v:, and X=<br />
V. dVl corresponds to it.<br />
vi-vl<br />
If s~ dVt ,the man should Immediately swim to po<strong>in</strong>t B <strong>al</strong>ong AB.<br />
V v:-v~<br />
Otherwise, the man should run <strong>al</strong>ong the shore over the distance AD=sdVI<br />
, and then swim to B.<br />
V (J~-VI2<br />
L<strong>et</strong> us note<br />
that s<strong>in</strong> a=~ for the route correspond<strong>in</strong>g to the shortest<br />
°2<br />
time.<br />
12. (I) Graphic<strong>al</strong>ly, it is easier to solve the problem <strong>in</strong> a coord<strong>in</strong>ate system<br />
related to the water. The speed of a raft equ<strong>al</strong> to the velocity of the river<br />
current is zero <strong>in</strong> this system, and the speed of the ship upstream and downstream<br />
will be the same <strong>in</strong> magnitude. For this reason tan ex.!: tan ('X2=Vl<br />
on the chart show<strong>in</strong>g the motion of the motor-ship (Fig. 269). When the ship<br />
stops at the land<strong>in</strong>g-stage, its speed with respect to the water will be equ<strong>al</strong><br />
to the river current velocity Va. Hence tan a=v2.<br />
B<br />
s<br />
B<br />
"<br />
a<br />
Fig. 269<br />
a<br />
I
MECHANICS<br />
s<br />
165<br />
J 2<br />
B<br />
j t; hours Fig. 270<br />
It can<br />
be seen from Fig. 269 that<br />
t<br />
BF<br />
tan ~·t3-tan(Xl·t l<br />
V2= an a.=n= t. 2.5 km/h<br />
(2) From the moment the ship me<strong>et</strong>s the rafts to the moment It overtakes<br />
them, the rafts will cover a distance equ<strong>al</strong> to<br />
s=v. (t 1+t.+t 8 )<br />
On the other hand, this distance is equ<strong>al</strong> to the difference b<strong>et</strong>ween the distances<br />
travelled by the ship upstream and downstream:<br />
Hence,<br />
s=t 3 (Vi+V2)-<br />
t 1 (VI-VI)<br />
and<br />
13. The motion of the launches leav<strong>in</strong>g their land<strong>in</strong>g-stages at the same<br />
time is shown by l<strong>in</strong>es MEB and KEA, where E is their me<strong>et</strong><strong>in</strong>g po<strong>in</strong>t<br />
(Fig. 210). S<strong>in</strong>ce the speeds of the launches relative to the water are the same,<br />
MA and KB are straight l<strong>in</strong>es.<br />
Both launches will travel the same time if they me<strong>et</strong> at the middle b<strong>et</strong>ween<br />
the land<strong>in</strong>g-stages. Po<strong>in</strong>t 0 where they me<strong>et</strong> Iles on the <strong>in</strong>tersection of<br />
l<strong>in</strong>e KB with a perpendicular erected from the middle of distance KM. The<br />
motion of the launches is shown by l<strong>in</strong>es KOD and COB. It can be seen from<br />
Fig. 270 that AM AF is similar to I1COF, and, therefore, the sought time<br />
MC=45 m<strong>in</strong>utes..<br />
14. The speed of the launches with respect to the water VI and the velocity<br />
of the river current v 2 can be found from the equations s= t1 (VI +VI) and<br />
s=t 2 (V l - V 2<br />
) , where it and t 2 are the times of motion of the launches downstream<br />
and upstream. It follows from the condition that 11= 1.5 hours and<br />
t,=3 hours.
166<br />
ANSWERS AND SOLUTIONS<br />
Hence<br />
and therefore<br />
5 (t 1+12 ) 15 k h<br />
VI = 2/<br />
1<br />
t 2<br />
m/<br />
5 (t -t )<br />
V 2 = 2 1 = 5 km/h<br />
2t 1 t 2<br />
The po<strong>in</strong>t of the me<strong>et</strong><strong>in</strong>g is at a distance of 20 km from land<strong>in</strong>g-stage M.<br />
15. L<strong>et</strong> us assume that the river flows from C to T with a velocity of V o<br />
S<strong>in</strong>ce the duration of motion of the boat and the launch is the same, we can<br />
•<br />
write the equation<br />
_5=2(_5+_5)<br />
VI + Vo V2+ Vo v 2 - Vo<br />
where s is the distance b<strong>et</strong>ween the land<strong>in</strong>g-stages. Hence,<br />
v~+4v2Vo+4V 2 V1 - v: = 0<br />
Vo= -20 2 ± V 5v:-4vIV2= -20 ± 19.5<br />
The solution Vo = - 39.5 krn/h should he discarded, s<strong>in</strong>ce with such a current<br />
velocity neither the boat nor the launch could go upstream.<br />
For this reason. Vo=- 0.5 krn/h, i. e.• the river flows from T to C.<br />
16. The speed of the boat v with respect to the bank is directed <strong>al</strong>ong AB<br />
(Fig. 271). Obviously, v=vo+u. We know the direction of vector v and the<br />
magnitude and direction of vector Yo. A glance at the draw<strong>in</strong>g shows that<br />
vector u will be m<strong>in</strong>imum when u1-v.<br />
Consequently.<br />
b<br />
Um<strong>in</strong> =Vo cos a. where cos a<br />
Ya 2 +b 2<br />
17. L<strong>et</strong> the speed u be directed at an angle a. to the bank (Fig. 272). Hence<br />
t (u cos a-v)=BC=a. and tu s<strong>in</strong> a=AC=b<br />
where t is the time the boat is <strong>in</strong> motion.<br />
By exclud<strong>in</strong>g a from these equations. we g<strong>et</strong><br />
t 2 (u 2_v2)-2vat-(a2+b2)=O<br />
whence t= 15/21 hour. It is therefore impossible to cover the distance AB <strong>in</strong><br />
30 m<strong>in</strong>utes.<br />
c a B<br />
c<br />
6<br />
A<br />
Fig. 271<br />
Fig. 272
MECHANICS 167<br />
Fig. 273<br />
A<br />
S<strong>in</strong>(~+~-~)<br />
and therefore v=u . ( A) •<br />
B<br />
sIn It-..,<br />
It is impossible to d<strong>et</strong>erm<strong>in</strong>e the<br />
18. L<strong>et</strong> Uo be the velocity of the w<strong>in</strong>d<br />
relative to the launch. Hence the flag on<br />
the mast will be directed <strong>al</strong>ong Uo' If<br />
v is the speed of the launch with respect<br />
to the bank, then u=uo+v (Fig. 273).<br />
In triangle FCDthe angleDCF=~+Ct-~<br />
and the angle FDC= n - p. Accord<strong>in</strong>g<br />
to the s<strong>in</strong>e theorem,<br />
v<br />
s<strong>in</strong> (~+~- ~ )<br />
u<br />
s<strong>in</strong>(n- P)<br />
velocity of the current from the known<br />
speed of the launch with respect to the bank, s<strong>in</strong>ce we do not know the direction<br />
of the mov<strong>in</strong>g launch with respect to the water.<br />
19. (1) If the speed of the plane relative to the air is constant and equ<strong>al</strong><br />
to u, then its speed with respect to the Earth with a tail w<strong>in</strong>d (<strong>al</strong>ong side Be)<br />
is VBC=V+U, with a head w<strong>in</strong>d vDA=v-u and with a side w<strong>in</strong>d VAB=vCD=<br />
= YV 2 _ U2 (Fig. 274 a and b).<br />
Hence, the time required to fly around the square is<br />
a a 2a v+ YV 2 _ U2<br />
2a---......--<br />
1 V+U v-u Vv2 u2 V 2_U2<br />
t =--+--+<br />
(2) If the w<strong>in</strong>d blows <strong>al</strong>ong the diagon<strong>al</strong> of the square from A to C, then<br />
(see Fig. 274c)<br />
V2=v~B+U2_2uVAB cos 45 0<br />
U<br />
~<br />
B C V V<br />
Fig. 274<br />
A 11<br />
u<br />
l/<br />
(d) (b) (c) (d)
168 ANSWERS AND SOLUTIONS<br />
\<br />
\\\\\\\<br />
The speed <strong>al</strong>ong sides AB and<br />
\ \<br />
\ \ U.<br />
'-______ ~ Fig. 275<br />
Be Is<br />
V2 r: u2<br />
VAR=VRC=-2- u+ V V 2_ 2<br />
The speed <strong>al</strong>ong sides CD and AD (Fig. 274d)<br />
2<br />
vCD=vDA=-V/ u+ YV - u;<br />
L<strong>et</strong> us leave only the plus sign before the root <strong>in</strong> both solutions to preserve<br />
a clockwise direction of the flight. The time required to fly around the square is<br />
t2<br />
4a -(V 2 - ~2<br />
V2_U2<br />
Fig. 276
MECHANICS 169<br />
20. L<strong>et</strong> us use the follow<strong>in</strong>g notation; u 12=speed of the second vehicle<br />
with respect to the first one; U 21 =speed of the first vehicle with respect to<br />
the second one.<br />
Obviously, U 12=U21 and u~2=v~+v~+2vIV2 cos a (Fig. 275). The time<br />
sought is t = 3- .<br />
U 12<br />
21. Dur<strong>in</strong>g the time ~t the straight l<strong>in</strong>e AB will travel a distance ol~t<br />
and the straight l<strong>in</strong>e CD a distance u2~t. The po<strong>in</strong>t of <strong>in</strong>tersection of the<br />
l<strong>in</strong>es will travel to position 0' (Fig. 276). The distance 00' can be found<br />
from triangle OFO' or OEO', where OF= v~ At =£0' and OE= v~ At =FO', i. e.,<br />
s<strong>in</strong> a s<strong>in</strong> ~<br />
whence<br />
00' = YOF2+OE2+20F· OE cos ~=V &t<br />
1-2. K<strong>in</strong>ematics of Non-Uniform and Uniformly<br />
Variable Rectil<strong>in</strong>ear Motion<br />
22. The mean speed over the entire dis lance Om = t1+ :2 + t3' where tl'<br />
1 2<br />
and is are the times dur<strong>in</strong>g which the vehicle runs at the speeds U I • o:<br />
and Os respectively. Obviously,<br />
Consequently,<br />
18 krn/h<br />
23. The path s travelled by the po<strong>in</strong>t <strong>in</strong> five seconds is equ<strong>al</strong> numeric<strong>al</strong>ly<br />
to the area enclosed b<strong>et</strong>ween Oabcd and the time axis (see Fig. 6): 51 = 10.5 em.<br />
The mean velocity of the po<strong>in</strong>t <strong>in</strong> five seconds is VI = SI/t1 = 2.1 cm/~<br />
and the mean acceleration of the po<strong>in</strong>t dur<strong>in</strong>g the same time is<br />
Av<br />
at =7; =0.8 cm/s"<br />
The path travelled <strong>in</strong> 10 seconds is 52= 25 em.<br />
Therefore, the mean velocity and the mean acceleration are<br />
u 2 = ~2 =2.5 cm/s, a 2 = O.2 em/5 2<br />
2<br />
24. Dur<strong>in</strong>g a sm<strong>al</strong>l time <strong>in</strong>terv<strong>al</strong> ~t the bow of the boat will move from<br />
po<strong>in</strong>t A to po<strong>in</strong>t B (Fig. 277). The distance AB=v l ~t, where VI is the speed<br />
of the boat. A rope length of OA-OB=CA=v At will be taken up dur<strong>in</strong>g<br />
this time. The triangle ABC may be considered as a right one, s<strong>in</strong>ce AC ~ OA.<br />
v<br />
Therefore VI=--.<br />
COSeL
o<br />
s<br />
Fig. 278<br />
.D<br />
A<br />
B<br />
B'<br />
Fig. 277<br />
25. Assume that at the <strong>in</strong>iti<strong>al</strong> moment t = 0 the object was at po<strong>in</strong>t S<br />
(Fig. 278), and at the moment t occupied the position CD. The similarity<br />
of the triangles SeD and SBA <strong>al</strong>lows us to write the equation<br />
AB =~=!!!.-<br />
3D Vii<br />
The velocity of po<strong>in</strong>t B at a given moment of time is v 2 = ~~' if the<br />
time ~t dur<strong>in</strong>g which the edge of the shadow is shifted by the distance BB'<br />
tends to zero.<br />
• I , hl ( 1 '1) hi ~t hl<br />
SlnceBB=AB-AB=v;- T-I+~t t(t+At),thenv2 Vlt(t+~t)'<br />
hl<br />
or, remember<strong>in</strong>g that ~t~t, we have U2= - t 2'<br />
VI<br />
26. For uniformly accelerated motion x=xo+Vol + a~2 • Therefore Vo=<br />
= 35 crn/s, a=82 cm/s 2 , and xo= 11 em is the <strong>in</strong>iti<strong>al</strong> coord<strong>in</strong>ate of the po<strong>in</strong>t.<br />
27. It follows from the velocity chart (see Fig. 8) that the <strong>in</strong>iti<strong>al</strong> velocity<br />
vo=4 crn/s (OA=4 cm/s). The acceleration a=g~= I crn/s''. First the velocity<br />
of the body decreases. At the moment t 1 =4 s it is zero and then<br />
grows <strong>in</strong> magnitude.<br />
The second chart (see Fig. 9) <strong>al</strong>so shows uniformly variable motion. Before<br />
the body stops, it travels a distance of h= 10 em. Accord<strong>in</strong>g to the first<br />
chart, the distance to the stop<br />
equ<strong>al</strong> to the area of triangle<br />
:r:<br />
DAB is 8 em. Therefore, the<br />
charts show different motions.<br />
A different <strong>in</strong>iti<strong>al</strong> velocity<br />
v' = 2h =5 cm/s and a differ<br />
11<br />
ent acceleration a'= 2h2= 1.25<br />
t 1<br />
em/s 2 correspond to the second<br />
chart.<br />
28. The mean speeds of both<br />
the motor vehicles are the same
MECHANICS<br />
171<br />
and equ<strong>al</strong> to<br />
m= CJ 2VIV2<br />
01+ v 2<br />
=36 krn/h<br />
Therefore, the distance b<strong>et</strong>ween po<strong>in</strong>ts A and B is 72 km. The ftrst vehicle<br />
travelled h<strong>al</strong>f of this distance <strong>in</strong> t' = 6/5 h, and the other h<strong>al</strong>f <strong>in</strong> t"= 4/5 h.<br />
The second vehicle travelled <strong>al</strong>l the time with the acceleration<br />
25<br />
a=-=36 krn/h''<br />
t~<br />
and reached a speed of v/;:=ato= 72 krn/h at the end of its trip. It acquired<br />
a speed of 30 krn/h <strong>in</strong><br />
t1=V~o=~ h<br />
and a speed of 45 km/h <strong>in</strong> t 2 = V 45 = 5/4 h after the moment of departure.<br />
a<br />
At these moments the first vehicle moved at the same speed as the second.<br />
At the moment when one vehicle overtook the other, both of them travelled<br />
the same distance, and therefore the follow<strong>in</strong>g equ<strong>al</strong>ities should be true<br />
at 2 6<br />
vl t = 7 for t~5 hand<br />
at 2 6<br />
V1t'+V2 (t - t' )= 2 for "5 h
172<br />
ANSWERS AND SOLUTIONS<br />
'0;<br />
U2<br />
OIr----~---1-~-+--...lIII~t-~-+---'IIIr-----'~<br />
Fig. 280<br />
It follows from the velocity chart (Fig. 280) that the mmimum time<br />
't'=0.3 s. Besides. the second b<strong>al</strong>l may beg<strong>in</strong> to f<strong>al</strong>l <strong>in</strong> 0.6, 0.9. 1.2 s, <strong>et</strong>c.,<br />
after the first b<strong>al</strong>l beg<strong>in</strong>s to drop.<br />
The time t dur<strong>in</strong>g which the velocities of the' two b<strong>al</strong>ls are -the same is<br />
0.3 s. The process periodic<strong>al</strong>ly repeats every 0.6 second.<br />
31. The <strong>in</strong>iti<strong>al</strong> equations are<br />
gt l g (t-'f)2 n-l<br />
T=n, 2<br />
where 't' is the duration of motion of the body over the<br />
its path.<br />
Hence,<br />
t= Y2;. t-t:= y2(n;1)<br />
n-th centim<strong>et</strong>re of<br />
~= y; CVn-Yn-I}<br />
32. Upon denot<strong>in</strong>g the coord<strong>in</strong>ate and the velocity of the first body with<br />
respect to the tower by Xl and VI and those of the second by X2 and V2_ we<br />
can write the follow<strong>in</strong>g equations<br />
Xa= -Vo (/-1')<br />
VI=Vo-gt;<br />
(The upward direction is considered to be positive here.)<br />
The velocity of the first body relative to the second is U=VI-V2=2vO-g-r<br />
and it does not change with time.<br />
The distance b<strong>et</strong>ween the bodies is<br />
g-r l<br />
S=XI-X2=(2vO- g't') l-vo't'+ 2<br />
The bodies move uniformly with respect to each other and therefore the<br />
distance b<strong>et</strong>ween them changes <strong>in</strong> proportion to the time.
MECHANICS<br />
173<br />
Fig. 281<br />
33. Accord<strong>in</strong>g to the condition, AA'=vt and cc= a~2 (Fig. 281). From<br />
the similarity of the triangles AA'O, BB'O and GG'D we have<br />
AA' BB' cc<br />
AD = BO = CO<br />
A glance at Fig. 281 shows that AO=AB+BD and CO=BC-BO.<br />
These ratios <strong>al</strong>low us to d<strong>et</strong>erm<strong>in</strong>e<br />
AA'-Ge' vt at 2<br />
BB' 2 2-4<br />
Hence, po<strong>in</strong>t B moves with the <strong>in</strong>iti<strong>al</strong> velocity ~ directed upward and a<br />
constant acceleration ~ directed downward. After reach <strong>in</strong>g the height<br />
v 2<br />
h =4a ' the po<strong>in</strong>t will move downward.<br />
34. L<strong>et</strong> us denote the speed of the l<strong>et</strong>t-hand truck at a certa<strong>in</strong> moment of<br />
time by VI' of the right-hand one by V2' and of the towed one by Va. Then,<br />
'dur<strong>in</strong>g the time t the left-hand truck will cover the distance<br />
a l t 2<br />
51=V11+ - 2-<br />
the right-hand truck the distance<br />
a2t'~<br />
s2=vat+-2-<br />
and the towed truck the distance<br />
Q<br />
st<br />
5s=03t+- 2-<br />
2<br />
At the same time it is easy to see that S3= 51t S2<br />
must be true at any v<strong>al</strong>ue of t , then<br />
S<strong>in</strong>ce this equ<strong>al</strong>ity<br />
01 + U2 and <strong>al</strong> + a2<br />
va= --2"- aa= -2-
174 ANSWERS AND SOLUTIONS<br />
35. The acceleration of the book with respect to the floor of the lift<br />
depends on the direction of acceleration of the lift and not on the direction<br />
of its motion (the direction of its velocity).<br />
If the acceleration of the lift is directed upward, the<br />
acceleration of the<br />
book will be g+a. If it is directed downward. the acceleration of the book<br />
will be g-a.<br />
36. The acceleration of the stone with respect to the Earth is g and with<br />
respect to the railway carriage Y a 2 +g 2 •<br />
37. If the speed of the 11ft did not change, the b<strong>al</strong>l would jump up to a<br />
height H.<br />
In a system of read<strong>in</strong>g hav<strong>in</strong>g a constant speed equ<strong>al</strong> to that of the lift<br />
at the moment the b<strong>al</strong>l beg<strong>in</strong>s to f<strong>al</strong>l, the lift will rise dur<strong>in</strong>g the time t to<br />
the height h l = a~~ , and dur<strong>in</strong>g the next time <strong>in</strong>terv<strong>al</strong> t to the height h 2 =<br />
=at 2 -<br />
t 2 .<br />
a 2<br />
. Its tot<strong>al</strong> height of ris<strong>in</strong>g is h=h 1 +h 2 = at 2 •<br />
The sought height which the b<strong>al</strong>l will jump up to above the floor of the<br />
lift is x=H-h=H-at 2 •<br />
38. The time dur<strong>in</strong>g which the load is lifted to the height h is t=V:~.<br />
The speed of the load relative to the crane <strong>in</strong> a vertic<strong>al</strong> direction at this<br />
moment is VI =att and <strong>in</strong> a horizont<strong>al</strong> one v2=a.t.<br />
The tot<strong>al</strong> speed of the load with respect to the ground is<br />
v=y v~+v:+v~<br />
39. In free f<strong>al</strong>l<strong>in</strong>g, body A will travel a vertic<strong>al</strong> distance 51 =g~2 dur<strong>in</strong>g<br />
the time t. Dur<strong>in</strong>g the same ti me the wedge should move over a distance<br />
52 = a 2<br />
t 2 • If the body is constantly <strong>in</strong> contact with the wedge, then ~ =<br />
. 51<br />
=cot a. as can be seen from Fig. 282.<br />
Therefore, the acceleration sought a=gcot ct.<br />
If the acceleration of the wedge <strong>in</strong> a horizont<strong>al</strong> direction is greater than<br />
g cot a. the body will move away from the wedge.<br />
1-3. Dynamics of Rectil<strong>in</strong>ear Motion<br />
40. The force F applied to the sphere d<strong>et</strong>erm<strong>in</strong>es, accord<strong>in</strong>g to Newton's<br />
second law, the magnitude and direction of acceleration of the sphere, but<br />
does not d<strong>et</strong>erm<strong>in</strong>e its velocity. For this reason the sphere can move <strong>in</strong> any<br />
direction under the force F, and may <strong>al</strong>so have a velocity equ<strong>al</strong> to zero.<br />
41. The resultant of <strong>al</strong>l the forces is 6 kgf and co<strong>in</strong>cides <strong>in</strong> direction with<br />
the force 5 kgf. Therefore, the acceleration of the sphere is a=~= 14.7 m/s<br />
m<br />
2<br />
and is directed towards the force 5 kgf.<br />
Noth<strong>in</strong>g can be said about the direction of motion (see the solution<br />
to Problem 40).<br />
42. In the MKS (or 51) system the weight of the body G=mg=9.8 N.<br />
The unit of force <strong>in</strong> the technic<strong>al</strong> system is 1 kgf, I, e. t the force wi th which
, Fig. 283<br />
the Earth attracts a body with a mass of 1 kg. In the cas system the weight<br />
of the body is 980,000 dynes. When Newton's second law is used to f<strong>in</strong>d the<br />
force <strong>in</strong> the technic<strong>al</strong> system of units, it should be borne <strong>in</strong> m<strong>in</strong>d that the<br />
mass shoul d <strong>al</strong>so be expressed <strong>in</strong> technic<strong>al</strong> units.<br />
43.. The sought angle Cl can be d<strong>et</strong>erm<strong>in</strong>ed from the ratio (F ig. 283):<br />
tan Cl=~g<br />
44. The body slid<strong>in</strong>g <strong>al</strong>ong the vertic<strong>al</strong> diam<strong>et</strong>er AB will cover the entire<br />
distance <strong>in</strong> the time tAB= {2~B. For an arbitrary groove arranged at<br />
2AC<br />
an angle a. to the diam<strong>et</strong>er AB, the time of motion is tAc-<br />
- { --<br />
gcos cx·<br />
S<strong>in</strong>ce AC= AB cos a, then tAC= tAB. All the bodies will reach the edge of<br />
the disk simultaneously.<br />
45. The force of air resistance F will reach its m<strong>in</strong>imum after the<br />
parachutist's speed becomes constant, and we have F=mg=75 kgf.<br />
46. Accord<strong>in</strong>g tv Newton's second law, N-mg= ± rna. Therefore,<br />
N =mg+ma if the acceleration of the lift is directed upward, and N=mg-ma<br />
if downward, irrespective of the direction of the speed. .<br />
When a=g, then N=0. (Here and below N denotes the force of norm<strong>al</strong><br />
pressure. or the force of norm<strong>al</strong> reaction.)<br />
47. Accord<strong>in</strong>g to Newton's second law, ma=kmg. Hence, the coefficient of<br />
friction k=!!:.... S<strong>in</strong>ce <strong>in</strong> an elastic impact only the direction of the velocity<br />
g<br />
changes, irrespective of the angle, then a= ~ , where s= 12.5 m<strong>et</strong>res is the<br />
tot<strong>al</strong> path travelled by the puck before it stops.<br />
v<br />
Therefore, k=-2=0.102.<br />
2<br />
gs<br />
48. Assum<strong>in</strong>g the acceleration of a motor vehicle to be constant, we can<br />
write a= ~ . S<strong>in</strong>ce the maximum force of friction <strong>in</strong> brak<strong>in</strong>g is<br />
accord<strong>in</strong>g to Newton's second<br />
motor vehicle.<br />
v 2<br />
kmg, then,<br />
law, m 2i=kmg, where m is the mass of the<br />
Hence, k= 2 V2 • Upon <strong>in</strong>sert<strong>in</strong>g the v<strong>al</strong>ues of u and s from the table <strong>in</strong><br />
gs<br />
this formula, we can f<strong>in</strong>d the coefficients of friction for various pavements.
176 ANSWERS AND SOLUTIONS<br />
ice<br />
dry snow<br />
w<strong>et</strong> wood block<br />
dry wood block<br />
w<strong>et</strong> asph<strong>al</strong>t<br />
dry asph<strong>al</strong>t<br />
dry concr<strong>et</strong>e<br />
k=O.l<br />
k=0.2<br />
k=O.3<br />
k=O.5<br />
k=O.4<br />
k=O.6<br />
k=O.7<br />
The coefficient of friction does not depend on the speed if the accuracy to<br />
the first digit after the decim<strong>al</strong> po<strong>in</strong>t is wanted.<br />
49. When the motor vehicle accelerates, the rear w<strong>al</strong>l of the fuel tank<br />
imparts an acceleration vlt to the p<strong>et</strong>rol. Accord<strong>in</strong>g to Newton's second law,<br />
the force F required for this acceleration is Alp ~ J where A is the area of<br />
the rear w<strong>al</strong>l of the fuel tank. In conformity with Newton's third law, the<br />
p<strong>et</strong>rol will act with the same force on the w<strong>al</strong>l. The hydrostatic pressure of<br />
the p<strong>et</strong>rol on both w<strong>al</strong>ls is the same. Hence, the difference of pressures exerted<br />
on the w<strong>al</strong>ls is Ap= ~ = Ip -7-.<br />
M<br />
50. The mass of the left..hand part of the rod mt=T I and of the<br />
right-hand part m2= ~ (L-I), where M is the mass ()f the entire rod.<br />
Under the action of the appl ied forces each part of the rod moves with the<br />
same acceleration a. Therefore,<br />
Ft-F=mta<br />
F-F 1 = m2a<br />
Hence the force F is<br />
B<br />
Fig. 284<br />
F<br />
51. The motion of the b<strong>al</strong>l will be uniform. The images of the b<strong>al</strong>l on the<br />
film appear at <strong>in</strong>terv<strong>al</strong>s of t = 1/24 s.<br />
The distance b<strong>et</strong>ween the positions A and B of the b<strong>al</strong>l <strong>in</strong> space that<br />
correspond to the positions C and D of the images on the film is AB=CD ~; •<br />
as shown <strong>in</strong> Fig. 284. The foc<strong>al</strong> length of the lens OF= 10 em, OE= 15 m<strong>et</strong>res,<br />
and CD=3 rnrn, The velocity of the b<strong>al</strong>l<br />
A<br />
AB<br />
vt = - t- = 10.8 m/s.<br />
When the b<strong>al</strong>l is <strong>in</strong> uniform motion. mg=<br />
=Iro~. In the second case 4 mg=kv:. Hence,<br />
E v:=4v~ and v 2=21.6 tnls.<br />
G<br />
k=- ~ 3.9XIO- o kgf·s 2/m2<br />
v~<br />
52. Figure 285 shows the forces that act<br />
on the weights. The equations of motion for
MECHANICS 177<br />
the weights can be written as follows:<br />
mta=T-mtR and m2a=mte-T<br />
where T is the tension of the str<strong>in</strong>g and a is the acceleration. (The accelerations<br />
of the weights are the same s<strong>in</strong>ce the str<strong>in</strong>g is considered to be<br />
unstr<strong>et</strong>chable. The weightlessness of the str<strong>in</strong>g and the pulley d<strong>et</strong>erm<strong>in</strong>e the<br />
constancy of T.)<br />
Therefore,<br />
m -m 1 a 2 g=327 cm/s 2<br />
m J+m2<br />
T=m t (a+g)=130,700 dynes=133 gf<br />
The time of motion t= y2: :::: 1 s.<br />
53. If the mass of the pulleys and the rope is negligibly sm<strong>al</strong>l, then<br />
(Fig. 286) 2F-T=O, and T-G=ma.<br />
Hence. F= ~ (I+;). When a=O, we have F= ~ .<br />
54. If. the mass of the second weight is much greater than 200 g, both<br />
weights will move with an acceleration somewhat below g, the acceleration<br />
of the lighter weight be<strong>in</strong>g directed upwards. To make a weight of mass m<br />
move upward with an acceleration g, a force of 2mg should be applied to it.<br />
For this reason the str<strong>in</strong>g should withstand a tension of about 400 gf.<br />
55. The dynamom<strong>et</strong>er first shows F= 3 kgf. If the read <strong>in</strong>g of the dynamom<strong>et</strong>er<br />
does not change, the weight 2 kgf is acted upon by the upward force<br />
of the str<strong>in</strong>g tension equ<strong>al</strong> to 3 kgf. Therefore, this weight rises with an<br />
acceleration of a= ~. The other weight lowers with the same acceleration.<br />
F<br />
Fig. 285<br />
Fig. 286<br />
12-2042
178 ANSWERS AND SOLUTIONS<br />
The addition<strong>al</strong> weight on the second pan 'can be found from the<br />
equation<br />
Fig. 287<br />
Hence 0 1 =3 kgf.<br />
56. The sphere is acted upon by three forces: the force of<br />
gravlty, the force of tension of the upper rope and the force<br />
applied to the lower one when it is pulled (Fig. 287).<br />
The acceleration imparted to the sphere by the pull can be<br />
found from the equation ma=F 1 +mg-F 2 •<br />
For the lower rope to break, the force applied to it should<br />
be greater than the tension of the upper one, L<strong>et</strong>t F 1 > F 2 •<br />
For this condition, the acceleration imparted to the sphere<br />
is greater than that of gravity, i.e, t a > g.<br />
57. Accord<strong>in</strong>g to Newton's second law,<br />
(ml +m 2 ) a=l1hg s<strong>in</strong> a-m28 s<strong>in</strong> ~-kmlgcos a-km~cos ~<br />
The weights will be at the same height after travell<strong>in</strong>g the distance s,<br />
which conforms to the follow<strong>in</strong>g equations: s s<strong>in</strong> a.=h-s s<strong>in</strong> ~ and s=a;s.<br />
Upon elim<strong>in</strong>at<strong>in</strong>g s and a from the three equations we obta<strong>in</strong><br />
ml g-c 2 (s<strong>in</strong> a+s<strong>in</strong> ~) (k cos ~+s<strong>in</strong> ~)+2h<br />
m2 g-r:2 (s<strong>in</strong> a+s<strong>in</strong> ~) (s<strong>in</strong> a-k cos a)-2h<br />
58. The equations of motion give the follow<strong>in</strong>g formulas for the acceleration<br />
of the stone:<br />
<strong>al</strong>=g(s<strong>in</strong>a+kcosa) <strong>in</strong> upward motion<br />
a 2<br />
= g (s<strong>in</strong> a-k cos a) <strong>in</strong> downward motion<br />
The k<strong>in</strong>ematic equations can be written as follows:<br />
l=v o t 1 -<br />
<strong>al</strong>t~<br />
a2t:<br />
2 ; l=2; vo-a tt 1=O<br />
We f<strong>in</strong>d from these five equations that<br />
2l-gt 2 s<strong>in</strong> a 1<br />
k ~ 0.37<br />
gt~ cos a<br />
t 2 = t 1V gt~si~a.-l 4.2s<br />
59. For this case the equations of dynamics can be written as<br />
mg-T=ma, and T=Ma<br />
where T is the tension of the str<strong>in</strong>g.<br />
Hence.<br />
m 2<br />
a=--g=-g<br />
M+m 7
MECHANICS 179<br />
The equations of k<strong>in</strong>ematics give x=vot- a~2. and vt=vo-at. Upon<br />
solv<strong>in</strong>g this system of equations, we f<strong>in</strong>d that <strong>in</strong> 5 seconds the cart will<br />
stay at the same place (x=O) and will have a speed vt=7 mls directed to<br />
the right. The car t will cover the distance<br />
5=2 {v o ~ a(tr} = 17.5 m<strong>et</strong>res<br />
60. The ice-boat can move only <strong>in</strong> the direction of its runners. When the<br />
speed of the boat exceeds that of the w<strong>in</strong>d, the velocity of the latter with<br />
respect to the boat has a component directed backward. If the velocity of<br />
the w<strong>in</strong>d with respect to the boat <strong>al</strong>so has a component perpendicular to<br />
the direction of motion, the sail can be so s<strong>et</strong> that the force F act<strong>in</strong>g on<br />
it will push the boat forward (Fig. 288).<br />
Therefore, the speed cf the boat can exceed that of the w<strong>in</strong>d. In practice<br />
It can be two or three times greater.<br />
61. (1) At the <strong>in</strong>iti<strong>al</strong> moment the acceleration is a o = ~ ~ 13.1 mjs2. it<br />
changes with time accord<strong>in</strong>g to the law a= M F f1t'<br />
where f1=200 kgjs is<br />
the mass of the fuel consumed by the rock<strong>et</strong> <strong>in</strong> a unit of time. A diagram<br />
of the acceleration is shown <strong>in</strong> Fig. 289. In 20 seconds the velocity is numeric<strong>al</strong>ly<br />
equ<strong>al</strong> to the hatched area, v === 300 m/s.<br />
(2) Newton's second law can be written as<br />
(M-Ilt) a=F-(M -Ilt)g-f<br />
Accord<strong>in</strong>g to the <strong>in</strong>iti<strong>al</strong> conditions, t=20 sand a=O.8 g. Hence, the force<br />
0/111/3 3<br />
Veloclt!l 0/ w<strong>in</strong>d relative<br />
toice-boat<br />
Fig. 288<br />
Fig. 289<br />
12 *
180<br />
ANSWERS AND SOLUTIONS<br />
--+-<br />
0<br />
9<br />
2g<br />
3g<br />
4g<br />
5g<br />
Fig. 290 Fig. 29/<br />
of air resistance is<br />
f=F-(M -Jit) g-(M -~li) 0.8 g= 12,800 kgf<br />
(3) Newton's equation for the weight gives mla=kx-m1g, where m 1 is<br />
the mass of the weight at the end of the spr<strong>in</strong>g, a is the acceleration of<br />
the rock<strong>et</strong>, k is the coefficient of elasticity of the spr<strong>in</strong>g, x is the elongation<br />
of the spr<strong>in</strong>g. Accord<strong>in</strong>g to the condition, m1g=kl o • Therefore,<br />
x=!JL (a+g). The sc<strong>al</strong>e of the device should be graduated uniformly<br />
g<br />
(Fig. 290). An acceleration of g corresponds to a division of one centim<strong>et</strong>re.<br />
62. The only force act<strong>in</strong>g on the bead is the reaction force of the rod N<br />
directed at right angles to the rod. The absolute acceleration W a of the bead<br />
(relative to a stationary observer) will be directed <strong>al</strong>ong the l<strong>in</strong>e of action<br />
of the reaction force N. The relative acceleration w,. is directed <strong>al</strong>ong the<br />
rod (Fig. 291)<br />
wa=a+w r<br />
It follows from the triangle of accelerations that W r =a cos Cl and<br />
wa=as<strong>in</strong> Cl.<br />
From Newton's second law, the reaction force is N = ma s<strong>in</strong> Cl.<br />
The time 't' dur<strong>in</strong>g which the bead moves <strong>al</strong>ong the rod can be found from<br />
. a cos Cl·",2 (-2-[-<br />
the equation [= 2 . Hence, '1'= --•<br />
a cos a.<br />
63. When the bead moves it is acted upon by the friction force kN and<br />
the reaction force N.<br />
The absolute acceleration will be directed <strong>al</strong>ong the resultant force F. It<br />
follows from Fig. 292 that"<br />
N =rna s<strong>in</strong> . a, and w,.=a cos ex--m=a kN ( cos k .<br />
(X- SIn ex)
MECHANICS<br />
181<br />
Hence (see Problem 62).<br />
.r 2l<br />
't= V a (cos a.-k s<strong>in</strong> a)<br />
If k ~ cot a, the bead will not move with respect to the rod, and<br />
of friction is rna cos a..<br />
the force<br />
64. The equations of motion of the block and the body have the form:<br />
ma=f (1)<br />
Mb::xF-f (2)<br />
where f Is the force of friction; a and b are accelerations.<br />
L<strong>et</strong> us assume that there is no slid<strong>in</strong>g, then a=b. The acceleration and<br />
the force of friction can be found from the equations of motion. The force<br />
of friction is f=m M:m' For the body not to slide, the force of friction<br />
should satisfy the <strong>in</strong>equ<strong>al</strong>ity f
182 ANSWERS AND SOLUTIONS<br />
65. Initi<strong>al</strong>ly the wagon is <strong>in</strong> uniformly r<strong>et</strong>arded motion at a speed of<br />
v= vo- kt, where f is the force of friction equaI to kmg. The body is <strong>in</strong><br />
uniformly accelerated motion at a speed of u=L t.<br />
m<br />
If the wagon is long, the speeds of the body and the wagon may become<br />
equ<strong>al</strong>. This will occur at the moment of time 1'= f V o f . After this, both<br />
m+JW<br />
the body and the wagon will beg<strong>in</strong> to move at a constant speed equ<strong>al</strong> to<br />
+m<br />
M MVo<br />
• By this time the wagon will have covered a distance<br />
and the body a distance<br />
s=vo1'- 2~ 1'1<br />
s=L't 2<br />
2m<br />
The distance covered by the body relative to the cart is 8-5. It should<br />
be shorter than 1. Thus, the body will not slip off the wagon if S-5 ~ I,<br />
Le.,<br />
Mv~<br />
2gk (M +m) -ct<br />
66. L<strong>et</strong> us consider the str<strong>in</strong>g element <strong>in</strong> the slit. Assume that the str<strong>in</strong>g<br />
moves downward. Then, the str<strong>in</strong>g element will be acted upon by the force<br />
of str<strong>in</strong>g tension on both sides and the force of friction (Fig. 293).<br />
S<strong>in</strong>ce the mass of this str<strong>in</strong>g element is neglected, T1-F - T 2 =0.<br />
The equations of dynamics can be written as follows:<br />
m l g- T 1=mIa<br />
m~-T2=-m2a<br />
Hence<br />
(ml-m2) g'-F<br />
a<br />
mt+ m2<br />
67. S<strong>in</strong>ce the weights move uniformly, the tension of the str<strong>in</strong>g is equ<strong>al</strong> to<br />
the weight nu. Therefore, the force with which the pulley acts on the bar<br />
is 2m.g, i.e., <strong>in</strong> the first case it is 2 kgf and <strong>in</strong> the second 6 kgf. In both<br />
cases the b<strong>al</strong>ance will show the sum of the first and second weights, Le.,<br />
4 kgf. The force of friction equ<strong>al</strong> to 2 kgf is applied to the bar at the side<br />
of the second weight. In the first case it is added to the force of pressure<br />
exerted by the pulley on the bar and <strong>in</strong> the second it is subtracted from it.<br />
68. S<strong>in</strong>ce the masses of the pulleys and the str<strong>in</strong>g may be neglected, the<br />
tension of the str<strong>in</strong>g will be the same everywhere (Fig. 294). Therefore,<br />
mlg-T=m 1 <strong>al</strong><br />
mJI - 2T = m2a 'J<br />
m~-T=m3a3<br />
a1+aa<br />
a2=--2-
MECHANICS<br />
183<br />
Fig. 294<br />
Fig. 295<br />
(see Problem 34). Hence,<br />
at<br />
a 2<br />
4ml m3- 3m2m3+ ml m2<br />
g<br />
4ml m3+m 2m3+mtm t<br />
mlm2-4mlm3+m2m3<br />
g<br />
4m1m 3+ m2mS+ m 1m2<br />
4mlmS-3mlm2+m2m3<br />
as= 4mlmS+m2mS+mlm2 g<br />
69. The second monkey will be at the same height as the first.<br />
If the mass of the pulley and the weight of the rope are disregarded,<br />
the force T tension<strong>in</strong>g both ends of the rope will be the same and, therefore,<br />
the forces act<strong>in</strong>g on each monkey will equ<strong>al</strong> F=T-mg. Both monkeys have<br />
the same accelerations <strong>in</strong> magnitude and direction and will reach the pul..<br />
ley at the same time.<br />
70. S<strong>in</strong>ce the mass of the pulleys and the str<strong>in</strong>g is negligibly sm<strong>al</strong>l, the<br />
tension of the thread is the same everywhere.<br />
Therefore,<br />
mlg-T=m1a1<br />
mgg- 2T= m 2a2<br />
2T-T=O<br />
Hence, T=O and a1=a2= g.<br />
Both weights f<strong>al</strong>l freely with an acceleration g. Pulleys Band C rotate<br />
counterclockwise and pulley A clockwise.<br />
71. (1) The forces act<strong>in</strong>g on the table and the weight are shown <strong>in</strong> Fig. 295.<br />
The equations of horizont<strong>al</strong> motion have the follow<strong>in</strong>g form:<br />
for the table with the pulley<br />
and for' {h'c weight<br />
0 1<br />
F-F+F/r=-a t g
184 ANSWERS AND SOLUTIONS<br />
L<strong>et</strong> us assume that the force F is so sm<strong>al</strong>l that the weight does not<br />
slide over the table. Hence a 1=a2 and F1r=F Gl~G2 ·<br />
By gradu<strong>al</strong>ly <strong>in</strong>creas<strong>in</strong>g the force F we sh<strong>al</strong>l thereby <strong>in</strong>crease the force of<br />
friction F1" If the table and the weight are immovable relative to each other,<br />
however, the force of friction b<strong>et</strong>ween them cannot exceed the v<strong>al</strong>ue Ffr.max=kG 2 •<br />
For this reason the weight will beg<strong>in</strong> to slide over the table when<br />
01+01 G2<br />
F F 0<br />
> 'f.max-a--=ka<br />
( I+G2)=10 kgf<br />
. 1 1<br />
In our case F=8 kgf, consequently, the weight will<br />
<strong>al</strong>=~=Gl~G2 g=:5g~ 314 em/5 2<br />
not slide, and<br />
(2) In this case the equations of motion fo the table with the pulleys and<br />
the weight will have the form:<br />
-F+F,,=G 1 a<br />
g l<br />
F-F,,=G2 a 2<br />
g<br />
The accelerations of the table and the weight are directed oppositely, and the<br />
weight will be sure to slide.<br />
Hence, F1,=k0 2 • .<br />
The acceleration of the table is<br />
-F+kG 2 2<br />
01= g=--g=- 131 em/s<br />
0 2<br />
1 15<br />
and it will move to the left.<br />
72. Accord<strong>in</strong>g to Newton's second law, the change <strong>in</strong> the momentum of<br />
the system "cannon-b<strong>al</strong>l It dur<strong>in</strong>g the duration of the shot -r should be equ<strong>al</strong><br />
to the impulse of the forces act<strong>in</strong>g on the system.<br />
Along a horizont<strong>al</strong> l<strong>in</strong>e<br />
mvocos a-MvI=FIf'"<br />
where FIf" is the impulse of the forces of friction.<br />
Along a vertic<strong>al</strong> l<strong>in</strong>e<br />
mvo s<strong>in</strong> a=N-r-(Mg+mg) '"<br />
where NT. is the impulse of forces of norm<strong>al</strong> pressure (reactions of a horizont<strong>al</strong><br />
area), (Mg+mg) -r is the impulse of the forces of gravity, Remember<strong>in</strong>g that<br />
Ftr =kN. we obta<strong>in</strong><br />
m m. M+m<br />
[)t=M Vo cos a-k ¥vosln a-k~g-r<br />
or s<strong>in</strong>ce g'f ~ Vo<br />
VI. ZVo (cos a,-k s<strong>in</strong> a,)<br />
This solution is suitable for k cot a the cannon will rema<strong>in</strong>
MECHANICS<br />
185<br />
1-4. The Law of Conservation of Momentum<br />
73. The momentum of the m<strong>et</strong>eorite is transmitted to the air molecules<br />
and, <strong>in</strong> the f<strong>in</strong><strong>al</strong> run, to the Earth.<br />
74. L<strong>et</strong> us divide the mass of the disk <strong>in</strong>to pairs of identic<strong>al</strong> elements<br />
ly<strong>in</strong>g on one straight l<strong>in</strong>e at equ<strong>al</strong> distances from the centre. The momentum<br />
of each pair is zero, s<strong>in</strong>ce the momenta of both masses are equ<strong>al</strong> but oppositely<br />
directed. Therefore, the momentum of the entire disk is zero.<br />
75. The propeller of 8 convention<strong>al</strong> helicopter is rotated by an eng<strong>in</strong>e<br />
mounted <strong>in</strong>side the fuselage. Accord<strong>in</strong>g to Newton's third law, oppositely<br />
directed forces are applied from the propeller to the eng<strong>in</strong>e. These forces<br />
create a torque that tends to rotate the fuselage <strong>in</strong> a direction opposite to<br />
rotation of the propeller. The tail rotor is used to compensate for this torque.<br />
In a j<strong>et</strong> helicopter the forces from the propeller are applied to the outflow<strong>in</strong>g<br />
gases and for this reason do not create any torque. .<br />
76. The velocity of the boat u can be found with the aid of the law of<br />
conservation of momentum. In a horizont<strong>al</strong> direction<br />
Mu=mv cos a.<br />
Hence, u=8 cm/s,<br />
77. At the highest po<strong>in</strong>t the rock<strong>et</strong> reaches, its velocity is zero. The change<br />
<strong>in</strong> the tot<strong>al</strong> momentum of the rock<strong>et</strong> fragments under the action of extern<strong>al</strong><br />
forces (the force of gravity) is negligible s<strong>in</strong>ce the impulse of these forces is<br />
very sm<strong>al</strong>l <strong>in</strong> view of the <strong>in</strong>stantaneous nature of the explosion. For this<br />
reason the tot<strong>al</strong> momentum of the rock<strong>et</strong> fragments before and immediately<br />
after the explosion rema<strong>in</strong>s constant and equ<strong>al</strong> to zero. At the same time, the<br />
sum of the three vectors (m1v 1l m2v2, mava) may be zero only when they are<br />
<strong>in</strong> one plane. Hence it follows that the vectors Vi' v2 and Va <strong>al</strong>so lie <strong>in</strong> one<br />
plane.<br />
78. L<strong>et</strong> the mass of the man be m and that of the boat M. If the man<br />
moves with a speed v relative to the boat the latter will move at a speed - u<br />
with respect to the bank. and therefore the man moves with a speed VI=v-u<br />
with respect to the bank.<br />
Accord<strong>in</strong>g to the law of conservation of momentum,<br />
m (v-u)-Mu=O<br />
m<br />
Hence, u=m+ M v and the speed of the man relative to the bank is<br />
M<br />
vt=m+M u.<br />
S<strong>in</strong>ce the signs of VI and v co<strong>in</strong>cide, the distance b<strong>et</strong>ween the man and<br />
the bank will <strong>in</strong>crease whatever the ratio b<strong>et</strong>ween the masses m and M.<br />
79. The speed of the boat u with respect to the shore is related to the<br />
speed of the man v with respect to the boat by the equation u=m~M v (see<br />
Problem 78). The relation b<strong>et</strong>ween the speeds rema<strong>in</strong>s constant dur<strong>in</strong>g motion.<br />
For this reason the relation b<strong>et</strong>ween the distances travelled will be equ<strong>al</strong> to<br />
the relation b<strong>et</strong>ween the speeds<br />
s m<br />
-Y=m+M
186 ANSWERS AND SOLUTIONS<br />
where s is the distance travelled by the boat and 1is its length (the distance<br />
covered by the man <strong>in</strong> the boat).<br />
Therefore, for the boat to reach the shore its length should be at least<br />
1=--s=2.5<br />
m+M<br />
m<strong>et</strong>res.<br />
m<br />
80. When the spr<strong>in</strong>g extends, it will act on both weights. The weight at<br />
the w<strong>al</strong>l will first rema<strong>in</strong> stationary while the second weight will beg<strong>in</strong> to<br />
move. When the spr<strong>in</strong>g extends compl<strong>et</strong>ely (I.e., is no longer deformed), the<br />
. second weight will have a certa<strong>in</strong> velocity. Therefore, the system will acquire<br />
momentum <strong>in</strong> a horizont<strong>al</strong> direction that will be r<strong>et</strong>a<strong>in</strong>ed <strong>in</strong> the follow<strong>in</strong>g<br />
period, s<strong>in</strong>ce the extern<strong>al</strong> forces will not act <strong>in</strong> this direction. Thus, the<br />
system as a whole will move away from the w<strong>al</strong>l, the weights <strong>al</strong>ternately<br />
converg <strong>in</strong>g and diverg<strong>in</strong>g.<br />
81. The speed of the cart will not depend on the po<strong>in</strong>t of impact. The<br />
momentum of the revolv<strong>in</strong>g cyl<strong>in</strong>der is zero, irrespective of its direction and<br />
velocity of rotation (see Problem 74). For this reason the bull<strong>et</strong> will impart<br />
to the "cyl<strong>in</strong>der-cart" system the same momentum as it would to a cyl<strong>in</strong>der<br />
rigidly secured on the cart.<br />
82. L<strong>et</strong> us denote the velocity of the rock<strong>et</strong> at the end of the k-th second<br />
by Vk. Gas with a mass m is ejected from the rock<strong>et</strong> at the end of the (k+ l)-th<br />
second, and it carries away a momentum equ<strong>al</strong> to<br />
m(-u+v,,)<br />
It follows from the law of conservation of momentum that<br />
(M -km) v,,=[M-(k+ 1) m] t'k+l +m (-u+v,,)<br />
The change <strong>in</strong> the velocity of the rock<strong>et</strong> per second is<br />
mu<br />
M-(k+l)m<br />
If we know how the velocity changes <strong>in</strong> one second, we can write the<br />
expression for the velocity at the end of the n-th second<br />
vn=vo+U(M m +M m 2<br />
+...+-Mm )<br />
-m - m -nm<br />
83. The velocity of the rock<strong>et</strong> will grow. This becomes obvious if we pass<br />
over to a read<strong>in</strong>g system with respect to which the rock<strong>et</strong> is at rest at the given<br />
moment. The pressure of the ejected gases will push the rock<strong>et</strong> forward.<br />
84. L<strong>et</strong> the mass of the boat be M, that of a sack m and the velocity of<br />
the boats Vo. When a sack is thrown out of a boat the latter is acted upon<br />
by a certa<strong>in</strong> force <strong>in</strong> a direction perpendicular to vo. It should be noted,<br />
however, that the boat does not change its velocity, s<strong>in</strong>ce the resistance of<br />
the water prevents later<strong>al</strong> motion of the boats. The velocity of a boat will<br />
change only when a sack is dropped <strong>in</strong>to it. .<br />
Apply<strong>in</strong>g the law of conservation of momentum to the "sack-boat" system,<br />
we can write <strong>in</strong> the first case:<br />
for one boat (M+m)vo-mvo= (M + 2m) VI<br />
for the other one -Mvo+mvt=(M+rn)v2<br />
Here ~Vl and V 2 are the f<strong>in</strong><strong>al</strong> velocities of the boats. From these simultaneous<br />
equatiIons M<br />
VI=-t'2= M+2m vo-
MECHANICS 187<br />
Fig. 296<br />
..,-----........<br />
»< "<br />
/' ,<br />
/ A '.<br />
I \<br />
I \<br />
\<br />
IJJ<br />
'......<br />
o<br />
......-----------"<br />
,,/J<br />
I<br />
(71'<br />
When the sacks are exchanged simultaneously,<br />
the f<strong>in</strong><strong>al</strong> velocities of the<br />
boats v~ and 0; can be found from the<br />
equations:<br />
Mvo-mvo=(M +m) v~ ;<br />
-Mvo+mvo=(M·+m) v~<br />
, , M-m<br />
Hence. VI =-V2 = M +m Vo' Thus, the<br />
f<strong>in</strong><strong>al</strong> velocity of the boats will be higher<br />
<strong>in</strong> the first case.<br />
85. Extern<strong>al</strong> forces do not act <strong>in</strong> a horizont<strong>al</strong><br />
direction on the "hoop-be<strong>et</strong>le" systern.<br />
For this reason the centre of gravity<br />
of the system (po<strong>in</strong>t C <strong>in</strong> Fig. 296) will not<br />
move <strong>in</strong> a horizont<strong>al</strong> plane. The distance<br />
from the centre of gravi ty of the syst em<br />
.. to the centre of the hoop is CO=m;M R.<br />
S<strong>in</strong>ce this distance is constant, the centre of the hoop 0 will describe a circle<br />
with the radius CO about the stationary po<strong>in</strong>t C. It is easy to see t hat the tra ..<br />
jectory of the be<strong>et</strong>le is a circle with the radius AC=m~M R.<br />
The mutu<strong>al</strong> positions and the direction of motion of the be<strong>et</strong>le and the<br />
hoop are shown <strong>in</strong> Fig. 296.<br />
86. S<strong>in</strong>ce no extern<strong>al</strong> forces act on the system <strong>in</strong> a horizont<strong>al</strong> direction.<br />
the projection of the tot<strong>al</strong> momentum of the "wedge-weights" system onto the<br />
horizont<strong>al</strong> direction must. rema<strong>in</strong> constant (equ<strong>al</strong> to zero). It thus follows that<br />
the wedge will beg<strong>in</strong> to move only if the weights move.<br />
For the weight m 2 to move to the right, the condition<br />
should be observed.<br />
Therefore. ml E;; s<strong>in</strong> a-k cos a. Here the wedge will move to the left. For<br />
the weight m~to move to the leit.ithe follow<strong>in</strong>g condition should be observed:<br />
or<br />
mlg~ m~ s<strong>in</strong> ex+km~cos ex<br />
Here the wedge will move to the right.<br />
Hence, for the wedge to be <strong>in</strong> equillbrtum,<br />
of the weights should satisfy the <strong>in</strong>equ<strong>al</strong>ity<br />
the ratio b<strong>et</strong>ween the masses<br />
s<strong>in</strong> a.-k cos a~ ml ~ s<strong>in</strong> a+k cos a.<br />
mj
188<br />
ANSWERS AND SOLUTIONS<br />
1-5. Statics<br />
k<br />
87. ll=lk+l.<br />
88. In the position of equilibrium (Fig. 291)mg-2mg cos a=O. Therefore,<br />
a=600. The sought distance h=l cot a= ~)_<br />
. Equilibrium will s<strong>et</strong> <strong>in</strong> after<br />
J' 3 ,<br />
the oscillations caused by the weight be<strong>in</strong>g lowered attenuate.<br />
89. The equ<strong>al</strong>ity of the projections of the forces onto the direction of the<br />
vertic<strong>al</strong> (Fig. 298) gives the equation<br />
2N s<strong>in</strong> i -2F/ T cos ; =0<br />
where N is the force of norm<strong>al</strong> pressure and FIT~ kN is the force of friction.<br />
The weight of the wedge may usu<strong>al</strong>ly be neglected.<br />
a<br />
Hence, tan 2"
MECHANICS<br />
189<br />
Fig. 299<br />
92. To turn the beam, the moment of the forces applied to its ends should<br />
be greater than the moment of the forces of friction when they reach their<br />
maximum.<br />
The forces of friction are distributed uniformly <strong>al</strong>ong the beam (Fig. 299).<br />
The mean arm of the forces of friction act<strong>in</strong>g on the left- or right-hand part<br />
of the beam is l/4, if the length of the entire beam is 1.<br />
The moment of <strong>al</strong>l the forces of friction with respect to the beam centre<br />
is 2 k~ ~ • Consequently, to turn the beam around, the applied forces F<br />
should satisfy the <strong>in</strong>equ<strong>al</strong>ity<br />
kQ<br />
whence F >4'<br />
To move the beam translation<strong>al</strong>ly, 2F should be greater<br />
fore, it is easier to turn the beam.<br />
than kG. There<br />
93. The equation of motion of the load is Goa=F - Go (Fig. 300). The<br />
g<br />
sum of the forces act<strong>in</strong>g on the crane vertic<strong>al</strong>ly is zero. Therefore, G 1 +G 2 =<br />
= G+F. S<strong>in</strong>ce the sum of the moments of the forces relative to po<strong>in</strong>t A<br />
L<br />
is zero, we have Fl+G 2=LG t .<br />
Solution of these simultaneous equations gives<br />
G 1 =:: 2.23 toni, and O 2 === 1.77 tonf<br />
94. For the lever to be <strong>in</strong> equilibrium. the force applied at po<strong>in</strong>t D should<br />
produce a moment equ<strong>al</strong> to G· AB. The force will be m<strong>in</strong>imum when the<br />
maximum arm is equ<strong>al</strong> to BD.<br />
0, ....-----l<br />
Ar=======t===r========IB<br />
B<br />
Fig. 300 Fig. 301
190 ANSWERS AND SOLUTIONS<br />
Hence,<br />
F=G ~g=;2' and it is directed at right angles to BD.<br />
95. If there is no friction b<strong>et</strong>ween the floor and the boxes, the latter will<br />
move simultaneously. If the coefficient of friction is not zero, the right..hand<br />
box will move first (see Fig. 35), s<strong>in</strong>ce the" force applied to it by the rod<br />
will be greater than the force applied to the left-hand box.<br />
Indeed, the rod is acted upon from the side of the right-hand box by<br />
the force F1 directed opposite to F, and from the side of the left-hand box<br />
by the force F2 directed <strong>al</strong>ong F I The sum of the forces <strong>in</strong> equilibrium is<br />
zero. Therefore, F1 = F+F2' and the force F1 will reach the maximum force<br />
of friction of rest before F!I<br />
96. The equ<strong>al</strong>ity to zero of the sum of the moments of the forces act<strong>in</strong>g<br />
on the sphere with respect to po<strong>in</strong>t A (Fig. 301) gives us the equation<br />
F'rR-NR=O<br />
S<strong>in</strong>ce FIr ~ kN, then k ~ 1.<br />
97. For a body to be at rest, the tot<strong>al</strong> moment of the forces that tends<br />
to turn the body clockwise should be equ<strong>al</strong> to the moment of the forces<br />
that tends to turn the body counterclockwise around a po<strong>in</strong>t (around the<br />
centre of gravity t for example). In our case the moment of the forces of<br />
friction that turns the brick clockwise should be equ<strong>al</strong> to the moment of the<br />
forces of the pressure exerted by the plane on the brick. It follows that<br />
the force of pressure exerted by the plane on the right h<strong>al</strong>f of the brick<br />
should be greater than on the left one. Accord<strong>in</strong>g to Newton's third law,<br />
the force with which the right h<strong>al</strong>f of the brick presses aga<strong>in</strong>st the plane<br />
should be greater than that of the left h<strong>al</strong>f.<br />
98. To lift the roller onto the step, the moment of the forces turn<strong>in</strong>g the<br />
roller around po<strong>in</strong>t A (Fig. 302) counterclockwise should at least be equ<strong>al</strong><br />
to the moment of the forces turn<strong>in</strong>g it clockwise, i.e.,<br />
Hence, h<br />
2 ±rr<br />
R. S<strong>in</strong>ce h < R, then<br />
h=(t_ V 22<br />
)R~O.29R<br />
G (R-h)=G VR2_(R-h)2<br />
99. S<strong>in</strong>ce the force of friction is zero on one of the planes, it is <strong>al</strong>so zero<br />
on the other one. Otherwise, the sphere would rotate around its centre, for<br />
Fig. 902 Fig. 303
MECHANICS 191<br />
the moment of <strong>al</strong>l the other forces relative to this centre is zero (because<br />
the arm of each of these forces with reference to the centre of the sphere<br />
Is zero).<br />
The sums of the projections of the forces on the vertic<strong>al</strong> and horizont<strong>al</strong><br />
directions are equ<strong>al</strong> to zero (Fig. 303). For this reason<br />
where Nt and N 2<br />
Nt cos CX 2 - N2 cos at :z:: 0<br />
G-N t s<strong>in</strong> cx 2 - N 2 s<strong>in</strong> (Xl =0<br />
are the sought pressure forces. Hence,<br />
N!. G S!i 2.6 kgf<br />
SIn a 2 +cos f t +f2 = 2kN.<br />
For the last <strong>in</strong>equ<strong>al</strong>ity to be satisfied, k should be sm<strong>al</strong>ler than ~ .<br />
101. A board on a rough log form<strong>in</strong>g an angle cx with a horizont<strong>al</strong> plane<br />
Is similar to a body r<strong>et</strong>a<strong>in</strong>ed by the forces of friction on an <strong>in</strong>cl<strong>in</strong>ed plane<br />
with an angle cx at its base. Therefore, <strong>in</strong> equilibrium, FIr==mg s<strong>in</strong> cx. Bea..<br />
r<strong>in</strong>g <strong>in</strong> m<strong>in</strong>d that FIr ~ kmg cos a, we have tane~ k.<br />
~<br />
...-.....----...~ ... B<br />
a<br />
lJ<br />
•<br />
mg<br />
Fig. 804 Fig. 305<br />
B
o • • •<br />
~.<br />
Fig. 306 Fig. 907<br />
.,<br />
102. The forces applied to the ladder are shown <strong>in</strong> Fig. 305. In equilibrium<br />
the sums of the projections of the forces <strong>al</strong>ong a vertic<strong>al</strong> and a horlzont<strong>al</strong><br />
l<strong>in</strong>es are equ<strong>al</strong> to zero.<br />
Therefore, N1= F/' and N2= mg.<br />
S<strong>in</strong>ce the sum of the moments of the forces relative to po<strong>in</strong>t B is zero,<br />
we can write another equation<br />
s<strong>in</strong> a,<br />
Nt cos a=mg- 2<br />
-<br />
Hence, Ffr=mg ta~ ~.<br />
F/r ~ kN 2' the ladder will be <strong>in</strong> equilibrium if<br />
tan C%
MECHANICS 193<br />
culations:<br />
cot a ~ cot 2P or a.e::;; 2~<br />
104. If at the moment when end B of the rod beg<strong>in</strong>s to rise. the force<br />
of friction F/r ..;; kN proves sufficient for end A not to slip. the rod will<br />
beg<strong>in</strong> to rotate around po<strong>in</strong>t A. Otherwise, end A will beg<strong>in</strong> to slip until<br />
the force of friction FIT =kN can keep the rod <strong>in</strong> equilibrium (Fig. 307).<br />
After this the rod will beg<strong>in</strong> to rotate around end A.<br />
L<strong>et</strong> us f<strong>in</strong>d the coefficient of friction k at which slipp<strong>in</strong>g stops with<br />
8 def<strong>in</strong>i te angle a b<strong>et</strong>ween the rod and the str<strong>in</strong>g.<br />
The equ<strong>al</strong>ity of the forces at the moment when the rod is <strong>al</strong>most horizont<strong>al</strong><br />
gives us the equations:<br />
F,,=Tcosa<br />
G=N+Tsln ex<br />
The equ<strong>al</strong>ity of the moments of the forces with respect to po<strong>in</strong>t A can<br />
be written as<br />
I<br />
G2=~lslna<br />
By us<strong>in</strong>g this system of equations, we f<strong>in</strong>d that<br />
FJr<br />
k=-;:r=cota<br />
For the rod not to slip at <strong>al</strong>l it is necessary that k;;:;:: cot 60"= ,; .<br />
-, J' 3<br />
105. The sum of the moments of the forces act<strong>in</strong>g on the man relative to<br />
his centre of gravity is zero. For this reason the force F act<strong>in</strong>g from the<br />
Earth is <strong>al</strong>ways directed towards the man's centre of gravity C (Fig. 308).<br />
The horizont<strong>al</strong> component of this force cannot be greater than the maximum<br />
force of friction of rest: F s<strong>in</strong> ex ~ kF cos a. Hence, k ~ tan a.<br />
,<br />
0 1 / N<br />
--1---<br />
II<br />
,I/1<br />
J,<br />
IJ<br />
M<br />
---/l~<br />
, I<br />
I I<br />
/ I I•<br />
III<br />
Fig. 308 Fig. 309 Fig. 310<br />
13-2042
194<br />
ANSWERS AND SOLUTIONS<br />
A<br />
C~~7;7l'l;~~W<br />
'lr 0<br />
Fig. 31J<br />
F,r<br />
B<br />
N<br />
Fig. 312<br />
106. The ladder is acted upon by three forces: its weight G t the force<br />
from the Earth F and the reaction of the support N (Fig. 309). S<strong>in</strong>ce the<br />
w<strong>al</strong>l is smooth, the force N is perpendicular to it.<br />
It will be the easiest to d<strong>et</strong>erm<strong>in</strong>e the force F if we f<strong>in</strong>d the po<strong>in</strong>t with<br />
respect to which the moments of the forces G and N are equ<strong>al</strong> to zero. This<br />
will be the po<strong>in</strong>t at which the straight l<strong>in</strong>es ON and 00 <strong>in</strong>tersect. The<br />
moment of the force F relative to this po<strong>in</strong>t should <strong>al</strong>so be zero. Therefore,<br />
the force should be directed so that it cont<strong>in</strong>ues past po<strong>in</strong>t O.<br />
Figure 309 shows that the direction of the force F forms with the ladder<br />
1<br />
the angle p=3
MECHANICS 195<br />
109. In equilibrium the sum of the forces act<strong>in</strong>g on the picture is zero<br />
(Fig. 312). Therefore, G=F/r+T cos a, and N=T s<strong>in</strong> (1,. The force of friction<br />
should satisfy the <strong>in</strong>equ<strong>al</strong>ity<br />
FIr<br />
Flr~kN or k~N<br />
The equ<strong>al</strong>ity of the moments relative to po<strong>in</strong>t B gives us the equation<br />
~ I s<strong>in</strong> a=T (l cos a+ Vdl-I I s<strong>in</strong>l a) s<strong>in</strong> a<br />
and<br />
Hence,<br />
FIr<br />
N<br />
l cos a+2 V d 2 -1 2 s<strong>in</strong> 2 a<br />
l s<strong>in</strong> a<br />
110. L<strong>et</strong> us first f<strong>in</strong>d the direction of the force f with which rod BC acts<br />
on rod CD. Assume that this force has a vertic<strong>al</strong> component directed upward.<br />
Then. accord<strong>in</strong>g to Newton's third law, rod CD acts on rod BC with a force<br />
whose vertic<strong>al</strong> component is directed downward. This contradicts the symm<strong>et</strong>ry<br />
of the problem, however. Therefore, the vertic<strong>al</strong> component of the force f<br />
should be equ<strong>al</strong> to zero. The force act<strong>in</strong>g on rod CD from rod DE will have<br />
both a horizont<strong>al</strong> and a vertic<strong>al</strong> components, as shown <strong>in</strong> Fig. 313a.<br />
S<strong>in</strong>ce <strong>al</strong>l the forces act<strong>in</strong>g on CD are equ<strong>al</strong> to zero, F=mg and f=f'.<br />
The equa lity to zero of the moment of the forces with respect to D gi yes us:<br />
or<br />
f s<strong>in</strong> ~<br />
CD=mg co; ~ CD<br />
mg<br />
tan~=2T<br />
Figure 313b shows the forces act<strong>in</strong>g on rod DE. S<strong>in</strong>ce the moment of the<br />
Of,<br />
(a)<br />
/<br />
.0<br />
(b)<br />
Fig. 313<br />
13 :+:
196 ANSWERS AND SOLUTIONS<br />
N<br />
F;r<br />
9 m,yt<br />
F<br />
Ffr<br />
"-<br />
Fig. 314 Fig. 315<br />
forces with respect to E is equ<strong>al</strong> to zero, it follows that<br />
. rosa<br />
f s<strong>in</strong> aDE=F cos a DE+mg--r DE<br />
or<br />
3mg<br />
tana=2f<br />
Consequently, tan a=3 tan p.<br />
111. The forces act<strong>in</strong>g on the box are shown <strong>in</strong> Fig. 314. The conditions<br />
of equilibrium have the form: F cos a=F 1, and G=N+Fs<strong>in</strong>a.<br />
At the moment equilibrium is disturbed, the force of friction reaches its<br />
maximum: F ,,=kN. Hence, F ~k'. The v<strong>al</strong>ue of F will be rnl-<br />
JI cosa s<strong>in</strong> a<br />
nimum at an angle a correspond<strong>in</strong>g to the maximum denom<strong>in</strong>ator. To f<strong>in</strong>d<br />
the maximum l<strong>et</strong> us transform the denom<strong>in</strong>ator, <strong>in</strong>troduc<strong>in</strong>g a new quantity<br />
q> <strong>in</strong>stead of k so that tan q>= k.<br />
or<br />
Then,<br />
cos a+ks<strong>in</strong> a<br />
cos (a-q»<br />
cos cp<br />
cos a+k s<strong>in</strong> a= YI +k" cos (~-q»<br />
S<strong>in</strong>ce the maximum v<strong>al</strong>ue of cos(a- q»<br />
F . _ kG<br />
eun>: YI +ki<br />
F<br />
Hence, k= ..r<br />
0.75.<br />
,. <strong>al</strong>_FI<br />
112. The forces act<strong>in</strong>g on the cyl<strong>in</strong>der are shown <strong>in</strong> Fig. 315. S<strong>in</strong>ce the<br />
cyl<strong>in</strong>der does not move translation<strong>al</strong>ly,<br />
F1,-Fcosa=O<br />
F s<strong>in</strong> a-mg+N=O<br />
is unity, then
MECHANICS 197<br />
The force of friction FIr = kN- Hence,<br />
F<br />
kmg<br />
cos a+k s<strong>in</strong> a<br />
The denom<strong>in</strong>ator of this expression can be written as A s<strong>in</strong> (a+ cp).<br />
(see Problem Ill).<br />
where<br />
A=Yl+k 2<br />
Therefore, the m<strong>in</strong>imum force with which the str<strong>in</strong>g should be pulled Is<br />
kmg<br />
Fm<strong>in</strong> "Ir •<br />
f 1+k 2<br />
The angle a l can be found from the equation cos a1+k s<strong>in</strong> <strong>al</strong>= Y 1+ k Z ,<br />
and tan a1=k.<br />
113. The forces act<strong>in</strong>g on the piston and the rear lid of the cyl<strong>in</strong>der are<br />
F 1=Fz=pA (Fig. 316a). Po<strong>in</strong>t C of the wheel is <strong>al</strong>so acted upon <strong>in</strong> a horizont<strong>al</strong><br />
direction by the force F2 transmitted from the piston through the<br />
crank gear.<br />
The sum of the moments of the forces act<strong>in</strong>g on the wheel with respect<br />
to its axis is zero. (The mass of the wheel is neglected.) Therefore. FIrR = F2"<br />
where Ftr is the force of friction. S<strong>in</strong>ce the sum of the forces which act on<br />
the wheel is <strong>al</strong>so zero, the force F3 applied to the axis by the bear<strong>in</strong>gs of<br />
the locomotive is F3 ~ FIr+F2- Accord<strong>in</strong>g to Newton's third law. the force<br />
F.=F s acts on the locomotive from the side of the axis. Hence, the tractive<br />
r<br />
effort F=F.-FI=Fjr=pA R'<br />
In the second position of the piston and the crank gear, the forces we are<br />
<strong>in</strong>terested <strong>in</strong> are illustrated <strong>in</strong> Fig. 316b. For the same reason as <strong>in</strong> the prer<br />
vious case, Flr=F. R'<br />
The tractive effort F=F.-F.=F/,=pA ~ ,<br />
As could be expected. the tractive effort is equ<strong>al</strong> to the force of friction,<br />
for the latter is the only extern<strong>al</strong> force that acts on the locomotive.<br />
114. The maximum length of the extend<strong>in</strong>g part of the top brick is 1/2.<br />
The centre of gravity of the two upper bricks C, is at a distance of 1/4 from<br />
the edge of the second brick (Fig. 317). Therefore. the second brick may<br />
extend by this length relative to the third one.<br />
The centre of gravity of the three upper bricks C 3 Is d<strong>et</strong>erm<strong>in</strong>ed by the<br />
equ<strong>al</strong>ity of the moments of the forces of gravity with respect to C 3 ; namely.<br />
G ( ~ -x) =2Gx. Hence x= ~, i.e. the third brick may extend over the<br />
&-<br />
fa)<br />
Fig. 316
198<br />
ANSWERS AND SOLUTIONS<br />
.0<br />
Fig. 317<br />
99<br />
Fig. 318<br />
,%,<br />
----------<br />
J(<br />
fourth one by not more than 1/6. Similarly it can be found that the fourth<br />
brick extends over the fifth one by L/B, <strong>et</strong>c. The nature of the change <strong>in</strong> the<br />
length of the extend<strong>in</strong>g part with an <strong>in</strong>crease <strong>in</strong> the number of bricks is<br />
obvious. The maximum distance over which the right-hand edge of the upper<br />
brick can extend over the right-hand edge of the lowermost brick can be<br />
written as the series<br />
L=f(l+ ;+~+++ ...)<br />
When the number of the bricks is <strong>in</strong>creased <strong>in</strong>f<strong>in</strong>itely this sum tends to<br />
<strong>in</strong>f<strong>in</strong>ity.<br />
Indeedt<br />
the sum of the series<br />
1 1 1 1 1 1 1<br />
1+2+3'+4 +5+6+7+8+ ...<br />
is greater than that of the series<br />
~ Y2<br />
,..-"'-. ....---"'--~<br />
1 1 1 1 1 I 1<br />
1+"2+4+4+8+8"+8+8+ ...<br />
and the latter sum will be <strong>in</strong>f<strong>in</strong>itely great if the number of terms is <strong>in</strong>f<strong>in</strong>ite.<br />
The centre of gravity of <strong>al</strong>l the bricks passes through the right-hand edge<br />
of the lowermost brick. Equilibrium will be unstable. The given example<br />
would be possible if the Earth were flat.<br />
115. L<strong>et</strong> us <strong>in</strong>scribe a right polygon <strong>in</strong>to a circle with a radius, (Fig. 318).<br />
L<strong>et</strong> us then f<strong>in</strong>d the moment of the forces of gravity (with respect to the
MECHANICS 199<br />
axis AK) applied to the middles of<br />
the sides of the polygon AB, BC,<br />
CD, DE, <strong>et</strong>c., assum<strong>in</strong>g that the<br />
force of gravity acts at right angles<br />
to the draw<strong>in</strong>g. This moment is<br />
equ<strong>al</strong> to pg (ABx 1 +8Cx<br />
+ DEx.+ EFx"+<br />
2+CDxs+<br />
FKxe), where p<br />
is the mass of 8 unit of the wire<br />
length.<br />
o From the similarity of the cor-<br />
. 319 respond<strong>in</strong>g triangles it can be shown<br />
F ~g.<br />
that the products ABx1' BCx,. CDxa.<br />
<strong>et</strong>c., are equ<strong>al</strong> respectively to AB'h,<br />
B'C'h, C' D'h, <strong>et</strong>c., where h is the apothem of the polygon.<br />
Therefore, the moment is equ<strong>al</strong> to<br />
pgh(AB' +B'C'+C'D'+D'E'+E'F'+F'K)=pgh2r<br />
If the number of sides <strong>in</strong>creases <strong>in</strong>f<strong>in</strong>itely, the v<strong>al</strong>ue of h will tend to r<br />
and the moment to 2r 2pg. On the other hand, the moment is equ<strong>al</strong> to the<br />
product of the weight of the wire nrpg and the distance x from the centre<br />
of gravity to axis AK. Therefore, 2r 2pg=itrpgx, whence x=.! r,<br />
n<br />
116. L<strong>et</strong> us divide the semicircle <strong>in</strong>to triangles and segments, as shown <strong>in</strong><br />
Fig. 319. The centre of gravity of a triangle, as is known, is at the po<strong>in</strong>t<br />
of <strong>in</strong>tersection of its medians. In our case the centre of gravity of each<br />
triangle is at a distance of ~ h from po<strong>in</strong>t 0 (h Is the apothem). When the<br />
sides are <strong>in</strong>creased <strong>in</strong>f<strong>in</strong>itely <strong>in</strong> number the centres of gravity of the triangles<br />
will lie on a circle with a radius of f r, while the areas of the segments<br />
wi11 tend to zero.<br />
Thus, the problem consists <strong>in</strong> d<strong>et</strong>erm<strong>in</strong><strong>in</strong>g the centre of gravity of a se..<br />
micircle with a radius of ~ r,<br />
It follows from the solution of Problem 115 that x, which is the distance<br />
b<strong>et</strong>ween the centre of gravity of the semicircle and po<strong>in</strong>t 0, is equ<strong>al</strong> to<br />
2 2 4<br />
x="1t3 r = 3n ' ·<br />
117. By apply<strong>in</strong>g the m<strong>et</strong>hod used to solve <strong>Problems</strong> 115 and 116, it can<br />
be shown that the centre of gravity is at po<strong>in</strong>t C at a distance<br />
2 81n"2 · ex<br />
CO= r from the centre of curvature of the arc (see Fig. 45).<br />
a<br />
lt8. From the solutions of <strong>Problems</strong> 115, 116 and 117, It can be shown<br />
• ex<br />
4 810'2<br />
that the centre of gravity is at po<strong>in</strong>t C at a distance of CO==3-a- r from<br />
po<strong>in</strong>t O.<br />
119. When the centre of gravity is d<strong>et</strong>erm<strong>in</strong>ed, the plate with a cut-out<br />
portion can form<strong>al</strong>ly be considered as a solid one if we consider that a se-
200 ANSWERS AND SOLUTIONS<br />
rnicircle with a negative mass equ<strong>al</strong> <strong>in</strong> magnitude to the mass of the cut..out<br />
portion Is superposed on it.<br />
The moment of the gravity forces of the positive and negative masses with<br />
respect to axis AB is equ<strong>al</strong> to<br />
r nr' 4) 1<br />
pg ( 2r B ---- r =-r 3 pg<br />
2 2 3n 3<br />
if the force of gravity acts at right angles to the draw<strong>in</strong>g (see Fig. 47),<br />
where p is the mass of a unit area of the plate (see the solution to Problem<br />
116). On the other hand, this moment is equ<strong>al</strong> to the product of the<br />
weight of the plate and the distance x=OC from its centre of gravity to<br />
axis AB.<br />
nrl) 1<br />
Hence, xPi ( 2r 2 -<br />
2<br />
=3" r 3 pg.<br />
2<br />
Therefore, %=3 (4-n) r.<br />
1-6. Work and Energy<br />
120. The work of a force does not depend on the mass of the body acted<br />
upon by the given force. A force of 3 kgf will perform the work<br />
W = Fh= 15 kgl-m. This work is used to <strong>in</strong>crease the potenti<strong>al</strong> energy<br />
(5 kgf-m) and the k<strong>in</strong><strong>et</strong>ic energy (10 kgf-m) of the load.<br />
121. k=O.098 J/kgf·cm.<br />
122. First l<strong>et</strong> us f<strong>in</strong>d the force with which the air presses on one of the<br />
hemispheres. Assume that its base is covered with a flat lid <strong>in</strong> the form of<br />
a disk with a radius R. If the air is pumped out from this vessel the force<br />
of pressure on the flat cover will be PI =pA=prtR2. Obviously, the same<br />
pressure will be exerted on the hemisphere by the air t otherwise the forces<br />
will not be mutu<strong>al</strong>ly b<strong>al</strong>anced and the vessel will perp<strong>et</strong>u<strong>al</strong>ly move <strong>in</strong> the<br />
direction of the greater force. The number of horses should be FI/F, s<strong>in</strong>ce<br />
the other hemisphere may simply be tied to a post. The tensioned rope will<br />
produce the same force as the team of horses pull<strong>in</strong>g at the other side.<br />
123. The change <strong>in</strong> the momentum of a body Is equ<strong>al</strong> to the impulse of<br />
the force of gravity. S<strong>in</strong>ce the forces act<strong>in</strong>g on the stone and the Earth are<br />
the same and act dur<strong>in</strong>g the same time, the changes <strong>in</strong> the momenta of<br />
these bodies are <strong>al</strong>so the same.<br />
The change <strong>in</strong> the k<strong>in</strong><strong>et</strong>ic energy of a body is equ<strong>al</strong> to the work of the<br />
. forces of gravitation<strong>al</strong> attraction. The forces are equ<strong>al</strong>, bu t the paths traversed<br />
by the stone and the Earth are <strong>in</strong>versely proportion<strong>al</strong> to their masses.<br />
This is why the law of conservation of energy may be written <strong>in</strong> a form<br />
which disregards the change <strong>in</strong> the k<strong>in</strong><strong>et</strong>ic energy of the Earth: mt+Ep=const.<br />
where m is the mass of the stone and Ep the potenti<strong>al</strong> energy of <strong>in</strong>teraction.<br />
124. Accord<strong>in</strong>g to the law of conservation of energy,<br />
mlv~<br />
m l gh= - 2-<br />
where ml is the mass of the pile driver, h the height from which it drops,<br />
and VI its velocity before the impact.
MECHANICS 201<br />
S<strong>in</strong>ce the Impact is Instantaneous, the force of resistance cannot appreciably<br />
change the tot<strong>al</strong> momentum of the system.<br />
See<strong>in</strong>g that the Impact is <strong>in</strong>elastic<br />
mlvl =(ml+m.) v.<br />
where m2 is the mass of the pile, v~ the velocity of the driver and the pile<br />
at the first moment after the impact.<br />
The mechanic<strong>al</strong> energy of the driver and the pile is spent on work done<br />
to overcome the resistance of the soil F:<br />
(m +m )v l<br />
1<br />
2<br />
2<br />
I<br />
+(ml+m2)gs=Fs<br />
where s is the depth which the pile is driven to <strong>in</strong>to the soil.<br />
Hence,<br />
ml h<br />
F= + ·- mtg+mlR+m2g=32,500 kg!<br />
ml m2 s<br />
125. As a result of the <strong>in</strong>elastic impact, the l<strong>in</strong>ear velocity of the box<br />
with the bull<strong>et</strong> at the first moment will be equ<strong>al</strong> to u=M~m' where u is<br />
the velocity of the bull<strong>et</strong>. On the basis of the law of conservation of energy,<br />
the angle of deflection a is related to the velocity v by the expression.<br />
whence<br />
(M+m) u l m 2v2 (M+m) L (I-cos a)g<br />
2 2(M+m)<br />
. aM+m ..rv=2stn2"-m<br />
y Lg<br />
126. S<strong>in</strong>ce the explosion is <strong>in</strong>stantaneous, the extern<strong>al</strong> horizont<strong>al</strong> forces<br />
(forces of friction) cannot appreciably change the tot<strong>al</strong> momentum of the<br />
system dur<strong>in</strong>g the explosion. This momentum is zero both before and directly<br />
after the explosion.<br />
Therefore, mtVt+m2v2 =0.<br />
Hence, VI = _ m2 •<br />
V2 ml<br />
S<strong>in</strong>ce the carts f<strong>in</strong><strong>al</strong>ly stop, their <strong>in</strong>iti<strong>al</strong> k<strong>in</strong><strong>et</strong>ic energies are spent on work<br />
aga<strong>in</strong>st the forces of friction:<br />
mlv~<br />
m2v:<br />
-2-=km t gsl' and -2-=km~s2<br />
vr 51<br />
Hence,.=-,and, therefore, s2=2 m<strong>et</strong>res.<br />
VI 81<br />
127. L<strong>et</strong> us denote the speed of the body and the wagon after they stop<br />
mov<strong>in</strong>g with respect to each other by u: Accord<strong>in</strong>g to the law of conserv<strong>et</strong>ion<br />
of momentum,<br />
(M +m) U= Mvo (1)
202 ANSWERS AND SOLUTIONS<br />
The wagon loses its k<strong>in</strong><strong>et</strong>ic energy <strong>in</strong> view of the fact that the force of<br />
friction f actlng on it performs negative work<br />
MV~_Mu'_fS<br />
2 2-<br />
where S is the distance travelled by the wagon.<br />
The body acquires k<strong>in</strong><strong>et</strong>ic energy because the force of friction act<strong>in</strong>g on it<br />
performs positive work<br />
mu l<br />
2=ls<br />
Here s is the distance travelled by the body.<br />
It is easy to see that the change <strong>in</strong> the k<strong>in</strong><strong>et</strong>ic energy of the system<br />
Mv~ _<br />
[MU 2 2]<br />
+ mU =/ (S-s) (2)<br />
2 2 2<br />
is equ<strong>al</strong> to the force of friction multiplied by the motion of the body relattve<br />
to the cart.<br />
It follows from equations (1) and (2) that<br />
mMv~<br />
S<strong>in</strong>ce S-s ~ I. then 1~ 2/ (M +m)<br />
mMv~<br />
S-s=2f(M +m)<br />
Bear<strong>in</strong>g <strong>in</strong> m<strong>in</strong>d that f=kmg. we have l;a. 2kg7~~+m) ·<br />
128. The combustion of the second portion <strong>in</strong>creases the velocity of the<br />
rock<strong>et</strong> v by Av. S<strong>in</strong>ce combustion is <strong>in</strong>stantaneous, then accord<strong>in</strong>g to the law<br />
of conservation of momentum,<br />
(M +m) v=M (v+t\v)+m (v-u)<br />
where m is the mass of a fuel portion, M the mass of the rock<strong>et</strong> without<br />
fuel, u the outflow velocity of the gases relative to the rock<strong>et</strong>.<br />
The velocity <strong>in</strong>crement of the rock<strong>et</strong> Av=Z u does not depend on the<br />
veloci ty v before the second portion of the fuel burns. On the contrary, the<br />
<strong>in</strong>crement <strong>in</strong> the k<strong>in</strong><strong>et</strong>ic energy of the rock<strong>et</strong> (without fuel)<br />
se, M(vtliV)1 _M;I=mu(2~u+v)<br />
wi11 be the greater, the higher is v.<br />
The maximum <strong>al</strong>titude of the rock<strong>et</strong> is d<strong>et</strong>erm<strong>in</strong>ed by the energy it receives.<br />
For this reason the second portion of the fuel can be burnt to the greatest<br />
advantage when the rock<strong>et</strong> atta<strong>in</strong>s its maximum velocity. i. e.• directly after<br />
the first portion is ejected. Here the greatest part of the mechanic<strong>al</strong> energy<br />
produced by the combustion of the fuel will be imparted to the rock<strong>et</strong>, while<br />
the mechanic<strong>al</strong> energy of the combustion products will be m<strong>in</strong>imum.
MECHANICS 203<br />
129. It will be sufficient to<br />
consider the consecutive combustion<br />
of two portions of fuel. L<strong>et</strong><br />
the mass of the rock<strong>et</strong> with the<br />
fuel first be equ<strong>al</strong> to M +2m.<br />
dh After combustion of the first<br />
portion, the velocity of the rockmUl<br />
.<br />
my <strong>et</strong> v= M +m .where Ul IS the<br />
velocity of the gases with respect<br />
to the rock<strong>et</strong>. The <strong>in</strong>iti<strong>al</strong> velo-<br />
,,11 city of the rock<strong>et</strong> is assumed to<br />
be zero.<br />
Fig. 320<br />
The <strong>in</strong>crement <strong>in</strong> the velocity<br />
of the rock<strong>et</strong> after the second<br />
portion burns is ~v = m;;2 t where U2 is the new velocity of the gases with<br />
respect to the rock<strong>et</strong>.<br />
Combustion of the first portion produces the mechanic<strong>al</strong> energy dEl =<br />
(M +m) tJ2 mu 2<br />
and of the second portion the energy<br />
= 2 +T t<br />
Accord<strong>in</strong>g to the <strong>in</strong>iti<strong>al</strong> condition, ~El = ~E2'<br />
(M +m) v 2<br />
2<br />
Hence,<br />
2 ! m 2 m \ _ 2 (m 2 m )<br />
Ul \ 2 (M + m) +2 ) - U2 2M + "2<br />
Therefore, Ut > u 2 ; the velocity of the gases with respect to the rock<strong>et</strong> dim<strong>in</strong>ishes<br />
because the mass of the rock<strong>et</strong> decreases as the fuel burns.<br />
130. Both. slopes may be broken <strong>in</strong>to any arbitrary number of sm<strong>al</strong>l <strong>in</strong>cl<strong>in</strong>ed<br />
planes with various angles of <strong>in</strong>cl<strong>in</strong>ation. L<strong>et</strong> us consider one of<br />
them (Fig. 320).<br />
The work done to lift the body up such an <strong>in</strong>cl<strong>in</strong>ed plane is equ<strong>al</strong> to<br />
the work aga<strong>in</strong>st the forces of gravity mg ~h plus the work aga<strong>in</strong>st the forces<br />
of friction FIr~s.<br />
/,l<br />
But FIr= kmg cos ex and ~s =_0__ . Therefore, F Jrt1s = kmg/1l. The tot<strong>al</strong><br />
cos ex.<br />
work L\W = mg (L\h+k81). If we consider <strong>al</strong>l the <strong>in</strong>cl<strong>in</strong>ed surfaces and sum<br />
up the elementary works, the tot<strong>al</strong> work will be<br />
W = ~ ~w =mg (~L\h+k~ 8l)=mgh+kmgl<br />
The work is d<strong>et</strong>erm<strong>in</strong>ed only by the height of the mounta<strong>in</strong> h and the length l<br />
of its foot.<br />
131. The force appl ied to the handle will be m<strong>in</strong>imum if it forms a right<br />
angle with the handle. Denot<strong>in</strong>g the force sought by F, we sh<strong>al</strong>l have from<br />
the golden rule of mechanics: 2nRF=Gh. Hence, F=2~~'
204 ANSWERS AND SOLUTIONS<br />
132. Accord<strong>in</strong>g to the def<strong>in</strong>ition, the efficiency f) WI'.;W2 ,where W I =<br />
= GH is the work done to lift the load G to a height H, and W 2 is the ~ork<br />
done aga<strong>in</strong>st the forces of friction. S<strong>in</strong>ce the force of friction is capable of<br />
hold<strong>in</strong>g the load In equilibrium, the work of this force cannot be less than<br />
the work WI. The m<strong>in</strong>imum work of the forces of friction is WI= WI- Therefore,<br />
rr ~50%.'<br />
133. As the man climbs the ladder the b<strong>al</strong>loon will descend by a height h.<br />
Therefore, the work done by the man will be spent to <strong>in</strong>crease his potenti<strong>al</strong><br />
energy by the amount mg (I-h) and that of the b<strong>al</strong>loon by mgh (the b<strong>al</strong>loon<br />
without the man will be acted upon by the lift<strong>in</strong>g force mg directed upwards).<br />
Hence,<br />
W =mg(l-h)+mgh=mgl<br />
This result can be obta<strong>in</strong>ed at once <strong>in</strong> c<strong>al</strong>culat<strong>in</strong>g the work done by the man<br />
<strong>in</strong> the system related to the ladder.<br />
If the man climbs with a velocity v with respect to the ladder, he moves<br />
at V-VI with respect to the Earth, where VI is the velocity of the ascend<strong>in</strong>g<br />
b<strong>al</strong>loon.<br />
Accord<strong>in</strong>g to the law of conservation of momentum, (v-v.) m = Mv.,<br />
Hence,<br />
m<br />
°1= M+m U<br />
134. To deliver twice as much water <strong>in</strong> a unit of time, a velocity two<br />
times greater should be imparted to the double mass of the water. The work<br />
of the motor is spent to impart a k<strong>in</strong><strong>et</strong>ic energy m;2 to the water. Therefore,<br />
the power of the motor should be <strong>in</strong>creased eight times.<br />
135. (1) The work done to raise the water out of the pit is<br />
H 3 3<br />
Wi= pg2 AX4'lf=g pgAH2<br />
where p is the density of water.<br />
The work<br />
1 H<br />
W.=2"P 2 Av 2<br />
Is required to impart k<strong>in</strong><strong>et</strong>ic energy to the water.<br />
The velocity (I with which the water flows out of the pipe onto the<br />
ground can be found from the ratio ~ A = nR 2 ut.<br />
The tot <strong>al</strong> work is<br />
w-! 2 1 H8A3<br />
- 8 pg AH + 16P n2R'T2<br />
(2) In the second case the work required to raise the water is less than<br />
WI by &W;=pgAlh(H-~). The work required to impart k<strong>in</strong><strong>et</strong>ic energy
MECHANICS<br />
205<br />
to the water is<br />
The tot<strong>al</strong> work W'=Wl-~W~+W~.<br />
136. It is the simplest to solve the problem <strong>in</strong> a coord<strong>in</strong>ate system rela..<br />
ted to the esc<strong>al</strong>ator. The man will w<strong>al</strong>k a distance l=_.h_+ut relative<br />
sm a<br />
to the esc<strong>al</strong>ator, where ur is the distance covered by the esc<strong>al</strong>ator. He should<br />
perform the work W=(-.h_+v-r) mgs<strong>in</strong>cx, s<strong>in</strong>ce dur<strong>in</strong>g the ascent the<br />
s<strong>in</strong> ex<br />
force mg is applied over the distance 1 and forms an angle of 90 0 - a with it.<br />
Part of the work mgh is spent to <strong>in</strong>crease the potenti<strong>al</strong> energy of the<br />
man, and the rema<strong>in</strong>der mgucs<strong>in</strong> a. tog<strong>et</strong>her with the work of the motor<br />
that drives the esc<strong>al</strong>ator is spent to overcome the forces of friction.<br />
137. The relation b<strong>et</strong>ween the elastic force and the deformation is shown<br />
<strong>in</strong> Fig. 321. The work done to str<strong>et</strong>ch (or compress) the spr<strong>in</strong>g by a sm<strong>al</strong>l<br />
amount 8X is shown by the area of the hatched rectangle ~W =F~x. The tot<strong>al</strong><br />
work <strong>in</strong> str<strong>et</strong>ch<strong>in</strong>g or compress<strong>in</strong>g the spr<strong>in</strong>g by the amount I, equ<strong>al</strong> to its<br />
potenti<strong>al</strong> energy Ep• is shown by the area of triangle OBC: W=Ep=k~2.<br />
L<strong>et</strong> us rec<strong>al</strong>l that an expression for the distance covered <strong>in</strong> uniformly accelerated<br />
motion can usu<strong>al</strong>ly be obta<strong>in</strong>ed by similar reason<strong>in</strong>g.<br />
138. The man act<strong>in</strong>g with force F on the spr<strong>in</strong>g does the work WI =- FL.<br />
At the same time, the floor of the railway carriage is acted upon from the<br />
side of the man by the force of friction F. The work of this force W2=FL.<br />
Therefore, the tot<strong>al</strong> work performed by the man <strong>in</strong> a coord<strong>in</strong>ate system<br />
related to the Earth is zero, <strong>in</strong> the same way as <strong>in</strong> a system related to the tra<strong>in</strong>.<br />
139. In the system of the tra<strong>in</strong> the work done is equ<strong>al</strong> to the potenti<strong>al</strong><br />
kl<br />
energy of the str<strong>et</strong>ched spr<strong>in</strong>g (see Problem 137) W= T' i<br />
s<strong>in</strong>ce the force of<br />
Fig. 321<br />
friction b<strong>et</strong>ween the man and the floor of the railway carriage does not do<br />
work <strong>in</strong> this system.<br />
In the system related to the Earth,<br />
the work the man does to str<strong>et</strong>ch<br />
F<br />
the spr<strong>in</strong>g is equ<strong>al</strong> to the product<br />
of the mean force k~ and the distance<br />
L-l, i. e., W l=k~<br />
(L-l). The<br />
man acts on the floor of the carriage<br />
with the same mean force ~ . The<br />
kl<br />
work of this force W 2 = 2' L. The<br />
o ~-----~~~--f---!....<br />
IC tot<strong>al</strong> work <strong>in</strong> the given coord<strong>in</strong>ate<br />
klt<br />
system W=W 1+W2 = T is the<br />
same as <strong>in</strong> the system or the carriage.
206 ANSWERS AND SOLUTiONS<br />
140. On the basis of the laws of conservation of momentum and energy<br />
we can write the follow<strong>in</strong>g equations:<br />
mtvt +m2v2=mlv~ + m2V~<br />
2 2 '2 '2<br />
mtVl + m 2 V 2 = m1Vl + m2 V2<br />
2 2 2 2<br />
where v~ and v; are the velocities of the spheres after the collision.<br />
Upon solv<strong>in</strong>g these simultaneous equations, we obta<strong>in</strong><br />
, (mt-m2)vl+2m2v2. d ,_(m2-mt)v2+2mtVl<br />
Vt ,an V2 - -------..<br />
ml+m2<br />
mt+m2<br />
(1) If the second sphere was at rest before the collision (v 2 =0), then<br />
(m 1 - m 2 ) VI • ' 2m 1 V1<br />
v. V2<br />
m 1 +m 2 ' mt +m 2<br />
If ml > m2' the first sphere cont<strong>in</strong>ues to move <strong>in</strong> the same direction as before<br />
the collision, but at a lower velocity.<br />
If m 1 < ms. the first sphere will jump back after the collision. The se..<br />
cond sphere will move <strong>in</strong> the same direction as the first sphere before the<br />
collision.<br />
, 2mv 2 ' 2mv t<br />
(2) If m 1=m2 , then Vl= 2m =V 2 and V2= 2m =Vl' Upon collision<br />
the spheres exchange their velocities.<br />
141. The elastic impact imparts a velocity v to the left-hand block. At<br />
this moment the right-hand block is still at rest s<strong>in</strong>ce the spr<strong>in</strong>g is not deformed.<br />
L<strong>et</strong> us denote the velocities of the l<strong>et</strong>t-and right-hand blocks at an arbit..<br />
rary moment of time by U 1 and U 2 , and the absolute elongation of the spr<strong>in</strong>g<br />
at this moment by x.<br />
Accord<strong>in</strong>g to the laws of conservation of momentum and energy t we have<br />
or<br />
Upon substitut<strong>in</strong>g u t + U 2<br />
kx 2<br />
Therefore, U1U 2 = 2m<br />
m (u 1 +u 2)=mv<br />
mu~ + mu~ + kx 2 _ mv 2<br />
222-2<br />
kx 2=m [t12-(ui+u~)]<br />
for v <strong>in</strong> the last equation, we obta<strong>in</strong><br />
kx 2=2mu 1u2<br />
and U 1+U2=V.<br />
It can be seen from the last two equations that u 1 and U2 will have the<br />
same sign and both blocks will move <strong>in</strong> the same direction.<br />
The quantity x 2 will be maximum when the product of the velocities u 1<br />
and U 2 is <strong>al</strong>so maximum. Hence, to f<strong>in</strong>d the answer to the second question,<br />
it is necessary to d<strong>et</strong>erm<strong>in</strong>e the maximum product U 1U2 assum<strong>in</strong>g that the<br />
sum "1 +"2 is constant and equ<strong>al</strong> to v.
MECHANICS<br />
.----------j<br />
; L.. ...1t<br />
--------..,tZ2 1<br />
a<br />
207<br />
b<br />
Fig. 322<br />
L<strong>et</strong> us consider the obvious <strong>in</strong>equ<strong>al</strong>ity (UI-U2)2~O, or U~-2UIU2+U~;;a:O.<br />
L<strong>et</strong> us add 4U 1 U 2 to the right and left sides of the <strong>in</strong>equ<strong>al</strong>ity. Hence,<br />
U~+2UIU2+U~~4uIU2 or (Ul+U2)2~4uIU2' S<strong>in</strong>ce Ul+U2=0, then<br />
4U I U 2 ~ v 2 •<br />
Therefore, the maximum v<strong>al</strong>ue of "lUI is equ<strong>al</strong> to 0 2 /4 and it is atta<strong>in</strong>ed<br />
v<br />
when<br />
U 1 = u2= 2 '<br />
At this moment the distance b<strong>et</strong>ween the blocks is<br />
I ± xmax=l ± CIV ~<br />
142. The lower plate will rise if the force of elasticity act<strong>in</strong>g on it is<br />
greater than its weight: kX2 > m 2 g. Here XI is the deformation of the spr<strong>in</strong>g<br />
str<strong>et</strong>ched to the maximum (position c <strong>in</strong> Fig. 322). Position a shows an undeformed<br />
spr<strong>in</strong>g. .<br />
For the spr<strong>in</strong>g to expand over a distance of X2 it should be compressed<br />
by Xl (position b <strong>in</strong> Fig. 322), which can be found on the basis of the law<br />
of conservation of energy:<br />
Hence,<br />
kx~ kx:<br />
T=T+mJR(Xl +X2)<br />
2mtg+ mzg<br />
xI>-k- T<br />
To compress the spr<strong>in</strong>g by Xl' the weight of the plate should be supplemented<br />
with a force which satisfies the equation F+mtg=kxl'<br />
Therefore, the sought force F > mlg+m2g.<br />
143. In a read<strong>in</strong>g system related to the w<strong>al</strong>l, the velocity of the b<strong>al</strong>l is<br />
v+u. After the impact the velocity of the b<strong>al</strong>l will be-(v+u) <strong>in</strong> the same<br />
read<strong>in</strong>g system. The velocity of the b<strong>al</strong>l after the impact with respect to a<br />
stationary read<strong>in</strong>g system will be<br />
-(o+u)-u=- (v+2u)
208 ANSWERS AND SOLUTION<br />
The k<strong>in</strong><strong>et</strong>ic energy after the impact is ~ (v+2U)2 and before the im pact<br />
~V2<br />
2 ·<br />
The change <strong>in</strong> the k<strong>in</strong><strong>et</strong>ic energy is equ<strong>al</strong> to 2mu (u+v).<br />
L<strong>et</strong> us now c<strong>al</strong>culate the work of the elastic forces act<strong>in</strong>g on the b<strong>al</strong>l<br />
dur<strong>in</strong>g the impact. L<strong>et</strong> the collision cont<strong>in</strong>ue for l' seconds. For the sake of<br />
simplicity we assume that dur<strong>in</strong>g the impact the elastic force is constant<br />
(gener<strong>al</strong>ly speak<strong>in</strong>g the result does not depend on this assumption). S<strong>in</strong>ce<br />
the impact changes the momentum by 2m (v+ u), the elastic force is<br />
The work of th is force is<br />
F<br />
2m (v+u)<br />
1:<br />
W=F5=Ful:=2m (o+u) U1:<br />
l'<br />
2m (v+u) u<br />
It is easy to see that this work is equ<strong>al</strong> to the change <strong>in</strong> the k<strong>in</strong><strong>et</strong>ic<br />
energy.<br />
144. (I) Up to the moment when the rope is tensioned the stones will<br />
f<strong>al</strong>l freely<br />
g (t-'t')2<br />
2<br />
The moment of tension<strong>in</strong>g of the rope can be d<strong>et</strong>erm<strong>in</strong>ed from the condition<br />
I =5 1-52. Hence, t =3 seconds, 51 =44.1 m<strong>et</strong>res, s2=4.9 m<strong>et</strong>res. The<br />
time is counted from the moment the first stone beg<strong>in</strong>s to f<strong>al</strong>l. When the<br />
rope is tensioned, there occurs an elastic impact and the stones exchange<br />
their velocities (see Problem 140). At the moment of impact VI=gt = 29.4 tnt«,<br />
and v2=g(t-'t)=9.8 m/s.<br />
The duration of f<strong>al</strong>l<strong>in</strong>g of the first stone t 1 (after the rope is tensioned)<br />
can be found from the condition<br />
gt:<br />
h-S 1 = V1t 1+-2-<br />
and of the second stone t 2 from the condition<br />
gt:<br />
h-5 2 = V 1t l +- 2<br />
-<br />
Therefore t 1 e:i 1.6 seconds, and t, ~ 1.8 seconds.<br />
The first stone f<strong>al</strong>ls dur<strong>in</strong>g 4.6 seconds and the second dur<strong>in</strong>g 2.8 seconds.<br />
(2) If the rope is <strong>in</strong>elastic, the veloci ties of the stones after it is tensioned<br />
are equ<strong>al</strong>ized (<strong>in</strong>elastic impact): v= VIt V2 = 19.6 mts. The duration of I<strong>al</strong>l<strong>in</strong>g<br />
of the stones with the rope tensioned is d<strong>et</strong>erm<strong>in</strong>ed from the equations;<br />
, gt~2 , ut~2<br />
h-51= vt1+- 2-<br />
and h-s2= ut2+- 2-<br />
51 and S2 are the same as <strong>in</strong> the first case.
MECHANICS 209<br />
Hence, t;:=!: 1.2 seconds, and t~ ~ 3.3 seconds.<br />
The first stone f<strong>al</strong>ls dur<strong>in</strong>g 4.2 seconds and the second dur<strong>in</strong>g 4.3 seconds.<br />
145. If only the right-hand b<strong>al</strong>l is moved aside, the extreme left-hand<br />
b<strong>al</strong>l will bounce off after the impact through an angle equ<strong>al</strong> to that by<br />
which the right-hand b<strong>al</strong>l was deflected.<br />
If two right-hand b<strong>al</strong>ls are simultaneously moved aside and released, then<br />
two extreme left-hand b<strong>al</strong>ls will bounce off, and so on.<br />
When the first b<strong>al</strong>l strikes the second one, the first b<strong>al</strong>l stops and transmits<br />
its momentum to the second one (see the solution to Problem 140), the<br />
second transmits the same momentum to the third, <strong>et</strong>c. S<strong>in</strong>ce the extreme<br />
left-hand b<strong>al</strong>l has no "neighbour" at its left, it will bounce off (provided<br />
there is no friction or losses of energy) through the same angle by which<br />
the extreme right-hand b<strong>al</strong>l was deflected.<br />
When the left-hand b<strong>al</strong>l ~ after deviation through the maximum angle,<br />
strikes the b<strong>al</strong>l at its right, the process of transmission of momentum will<br />
take the reverse course.<br />
When two right-hand b<strong>al</strong>ls are deflected at the same time, they wiJl<br />
transmit their momenta to the row <strong>in</strong> turn after a very sm<strong>al</strong>l period of time.<br />
In this way· the other b<strong>al</strong>ls will receive two impulses that will be propagated<br />
<strong>al</strong>ong the row at a certa<strong>in</strong> time <strong>in</strong>terv<strong>al</strong>. The extreme left-hand b<strong>al</strong>l will<br />
bounce off after it receives the "first portion" of the momentum. Next, its<br />
"neighbour" will bounce off after receiv<strong>in</strong>g the next portion of the momentum<br />
from the extreme right-hand b<strong>al</strong>l.<br />
If three right-hand b<strong>al</strong>ls are deflected, the row of b<strong>al</strong>ls will receive<br />
portions of momentum from the third, second and first b<strong>al</strong>l follow<strong>in</strong>g each<br />
other <strong>in</strong> very sm<strong>al</strong>l <strong>in</strong>terv<strong>al</strong>s of time. If four b<strong>al</strong>ls are deflected and released<br />
at the same time, four b<strong>al</strong>ls will bounce off at the left while the other two<br />
will rema<strong>in</strong> immobile.<br />
146. The strik<strong>in</strong>g b<strong>al</strong>l will jump back and the other b<strong>al</strong>ls up to the steel<br />
one will rema<strong>in</strong> <strong>in</strong> place. The steel b<strong>al</strong>l and <strong>al</strong>l the others after it will<br />
beg<strong>in</strong> to move to the left with different velocities.<br />
The fastest<br />
velocity will be imparted to the extreme left-hand b<strong>al</strong>l. The<br />
next one will move slower, <strong>et</strong>c. The b<strong>al</strong>ls will move apart (see the solutions<br />
to <strong>Problems</strong> 140 and 145).<br />
147. Assume that the weight 2m lowers through a distance of H. The<br />
weights m will accord<strong>in</strong>gly rise to the height h (Fig. 323).<br />
. 2mv 2 2mv ll<br />
On the basis of the law of conservation of energy, 2mgh+-r+--r=<br />
=2mgH, or v~+v:=2g(H-h), where VI is the velocity of the weights m<br />
and V2 that of the weight 2m.<br />
When the weight 2m lowers, its velocity v 2 approaches the velocity Vi'<br />
s<strong>in</strong>ce the angles b<strong>et</strong>ween the parts of the str<strong>in</strong>g thrown over the pulleys tend<br />
to zero. In the limit, v 2 d VI- At the same time, H-h ~ I.<br />
Therefore, the maximum velocity of the weights is<br />
v=ygl<br />
148. The velocities of the weights are the same If they cover identic<strong>al</strong><br />
distances ~s <strong>in</strong> equ<strong>al</strong>ly sm<strong>al</strong>l <strong>in</strong>terv<strong>al</strong>s of time. These distances will be the<br />
same at such an angle AN B at which lower<strong>in</strong>g of the weight ml through<br />
~s = NK (Fig. 324) is attended by an <strong>in</strong>crease <strong>in</strong> part AN B of the str<strong>in</strong>g<br />
<strong>al</strong>so by the amount Ss. Therefore, when the velocities are equ<strong>al</strong>, HK ::::s<br />
14-204~
210 ANSWERS AND SOLUTIONS<br />
I<br />
:h<br />
I<br />
o<br />
m<br />
Zm<br />
m<br />
Fig. 323<br />
As<br />
~s.<br />
~ BK-BN=2 and FK=AK-AN="2- The triangles NHK and NFK<br />
wi 11 be the closer to right ones, the sm<strong>al</strong>ler is the distance 6.5. When<br />
6.s-+ 0, the angles NHK and NFK tend to be right ones, while the angles<br />
K NHand /( NF tend to 30°. Therefore, the velocities will be equ<strong>al</strong> when<br />
L. ANB= 120°.<br />
L<strong>et</strong> us use the law of conservation of energy to f<strong>in</strong>d these velocities<br />
Hence,<br />
mlgh=2 (2-V3) mzgh+ mlt m2 v2<br />
The weights will oscillate near the<br />
VI = 2gh ml- 2 (2-ya) ms ~ 0<br />
ml+m2<br />
position of equilibrium, which corresponds<br />
to the angle ANB=2 arc cos 2 ml ~ 149°. The maximum deviation from<br />
m2<br />
the position of equilibrium corresponds to the angle ANB= 1200.<br />
149. S<strong>in</strong>ce there is no slipp<strong>in</strong>g of the board on the rollers and of the<br />
rollers on the horizont<strong>al</strong> surface, the distance b<strong>et</strong>ween the axes of the rollers<br />
A<br />
C<br />
rna Fig. 824
MECHANICS<br />
211<br />
Fig. 825<br />
<strong>in</strong> motion rema<strong>in</strong>s constant. For this reason the board will move translation<strong>al</strong>ly<br />
<strong>in</strong> a horizont<strong>al</strong> direction and at the same time down <strong>al</strong>ong the rollers.<br />
When the rollers move through a certa<strong>in</strong> distance it each po<strong>in</strong>t on the<br />
board (<strong>in</strong> particular, its centre of gravity A) will move <strong>in</strong> a horizont<strong>al</strong><br />
direction through the same distance I and will <strong>al</strong>so move the same distance<br />
<strong>al</strong>ong the rol1ers: AB=BC=l (Fig. 325). (This is particularly obvious if<br />
we consider the motion of the rollers <strong>in</strong> a coord<strong>in</strong>ate system travell<strong>in</strong>g<br />
tog<strong>et</strong>her with the rollers.) As a result, the centre of gravity of the board<br />
will move <strong>al</strong>ong straight l<strong>in</strong>e AC <strong>in</strong>cl <strong>in</strong>ed to the horizon at an angle a/2,<br />
s<strong>in</strong>ce ABC is an isosceles triangle. The motion will be uniformly accelerated.<br />
The board acquires k<strong>in</strong><strong>et</strong>ic energy ow<strong>in</strong>g to the reduction of its potenti<strong>al</strong><br />
energy m;! =mgl s<strong>in</strong> a, or v 2=2g1 s<strong>in</strong> a. On the other hand, <strong>in</strong> uniformly<br />
accelerated motion v 2=2as, where s=AC=2l cos ~<br />
. Hence, the acceleration<br />
v 2 • a<br />
a=2i=gslnT· .<br />
150. L<strong>et</strong> us c<strong>al</strong>culate the difference b<strong>et</strong>ween the potenti<strong>al</strong> energies for the<br />
two positions of the cha<strong>in</strong>-when it lies entirely on the table and when a<br />
part of it x hangs from the table. This difference is equ<strong>al</strong> to the weight<br />
~ xg of the hang<strong>in</strong>g part multiplied by x/2, s<strong>in</strong>ce the cha<strong>in</strong> is homogeneous<br />
and the centre of gravity of the hang<strong>in</strong>g end is at a distance of x/2 from.<br />
the edge of the table.<br />
Ivfv 2 _ MR 2<br />
On the basis of the law of conservation of energy, -2-- 4f x or<br />
u= -Vg:z2 . At this moment of time the acceleration can be found from<br />
N ton' diM M Th r gx<br />
ew on s secon aw: a=2/ gx. ere.ore, a=2l .<br />
To c<strong>al</strong>culate the reaction of the table edge, l<strong>et</strong> us first f<strong>in</strong>d the tension<br />
of the cha<strong>in</strong> at the po<strong>in</strong>t of its contact with the table. It is equ<strong>al</strong> to the<br />
change <strong>in</strong> the momentum of the part of the cha<strong>in</strong> ly<strong>in</strong>g on the table<br />
M M Mg<br />
F =21 (2l-x) a-2f u 2 = 2l 3<br />
(l-x) ~<br />
14 *
212<br />
ANSWERS AND SOLUTIONS<br />
Fig. 326<br />
Fig. 327<br />
L<strong>et</strong> us now consider a very sm<strong>al</strong>l element of the cha<strong>in</strong> <strong>in</strong> contact with<br />
the table edge. This element is acted upon by the three forces (Fig. 326).<br />
S<strong>in</strong>ce this element is <strong>in</strong>f<strong>in</strong>itely sm<strong>al</strong>l, the sum of the three forces which act<br />
on it should be zero.<br />
Therefore, the force of reaction is<br />
N= F Y2= Y2 M (l;;Xl x x<br />
When x > I, the cha<strong>in</strong> no longer touches the edge of the table.<br />
151. L<strong>et</strong> us denote the .velocity of the wagon by v. The horizont<strong>al</strong><br />
component of the velocity of the pendulum with respect to the wagon is<br />
u cos ~ (Fig. 327). S<strong>in</strong>ce the wagon moves, the velocity of the pendulum<br />
with respect to the rails is v+ u cos p. The extern<strong>al</strong> forces do not act on<br />
the system <strong>in</strong> a horizont<strong>al</strong> direction. Therefore, on the basis of the law of<br />
conservation of momentum, we have<br />
m(v+ucos~)+Mv=O (1)<br />
s<strong>in</strong>ce the system was <strong>in</strong>iti<strong>al</strong>ly at rest.<br />
The vertic<strong>al</strong> component of the pendulum velocity with respect to the<br />
wagon and the rails is u s<strong>in</strong> ~.<br />
Accord<strong>in</strong>g to the Pythagorean theorem, the square of the pendulum<br />
veloci ty relative to the rails is (v+u cos ~)2+u 2 s<strong>in</strong> 2 p.<br />
With the aid of the law of conservation of energy, we can obta<strong>in</strong> a second<br />
equation <strong>in</strong>terrelat<strong>in</strong>g the velocities v and u:<br />
; [(ucosP+v)2+u2s<strong>in</strong> 2 pl+ ~ v 2=mgl(cosp-cosa) (2)<br />
It can be found from equations (1) and (2) that<br />
v 2<br />
2m 2g1 (cos ~ - cos a) cos 2 ~<br />
(M+m)· (M+ms<strong>in</strong>2~)<br />
In a particular case, when p=O (assum<strong>in</strong>g ~ ~ I)<br />
or<br />
m 2<br />
v 2 = 2 M2 gl (1- cos ex)
J\\ECHAN ICS<br />
213<br />
Fig. 828<br />
152. L<strong>et</strong> us denote the velocity of the wedge by u, and the horizont<strong>al</strong><br />
and vertic<strong>al</strong> components of the velocity u of the block with reference to a<br />
stationary read<strong>in</strong>g system by Ux and u y (Fig. 328).<br />
On the basis of the laws of conservation of momentum and energy we<br />
can write<br />
Mv 2 m (2 2)<br />
-Mv+mux=O, and -2-+2 ux+Uy =mgh<br />
It should be noted that the angle a with the horizont<strong>al</strong> surface is formed<br />
by the relative velocity urel' i.e., the velocity of the block with respect<br />
to the mov<strong>in</strong>g wedge, and not the absolute velocity of the block u, by<br />
which is meant the velocity relative to a stationary horizont<strong>al</strong> surface.<br />
u<br />
It follows from the velocity diagram (Fig. 329) that +Y =tan cx. Upon<br />
v Ux<br />
solv<strong>in</strong>g these equations with respect to v, we obta<strong>in</strong><br />
, ;r 2mgh<br />
v= V M+m[(~r+(~+lrtan2a] ==<br />
, ;r 2gh<br />
= V .~ +( ~ r +(: +Ir tan2 a<br />
At the same moment of time the absolute .velocity of the block is<br />
u=Vu~+u;=V2gh" ;r1- M m ( Mr<br />
V 1+ m+M<br />
I +m tan2a<br />
When the mass of the wedge is much greater than that of the block<br />
u tends, as should be expected, to V2gh. t<br />
153. The velocity of the rod with respect to the mov<strong>in</strong>g wedge is directed<br />
at a~ angle ex. to the horizon. If the velocity of the wedge is added to this<br />
relative velocity, the result will be the absolute velocity u of the rod
214<br />
ANSWERS AND SOLUTIONS<br />
Fig. 329<br />
Fig. 330<br />
(Fig. 330). The relation b<strong>et</strong>ween the velocities is obviously equ<strong>al</strong> to<br />
~=tan a<br />
v<br />
Mv 2 milt.<br />
It follows from the law of conservation of energy that -2-+T = mgh.<br />
Upon cancell<strong>in</strong>g u {rom these t\VO equations, we g<strong>et</strong> an expression for v:<br />
.. / 2mgh<br />
v= V M+mtan2 a<br />
We can "then wri te for the relative velocity of the rod:<br />
The<br />
velocity of the rod is<br />
u=tana .. /<br />
1 .. / 2mgh<br />
Urel = cos fl. V M +m tan 2 ex<br />
__2mgh__= .. /2 mgta~h<br />
V m+Mtan 2 a V m+Mtan 2 a<br />
It can be seen from this formula that the velocity of the rod changes with<br />
the path h travelled accord<strong>in</strong>g to the law of uniformly accelerated motion<br />
u=V ~ah. Therefore. the acceleration of the rod is<br />
m tan 2 ex<br />
a m+M tan 2 ex<br />
1-7. K<strong>in</strong>ematics of Curvil<strong>in</strong>ear Motion<br />
154. The driv<strong>in</strong>g pulley rotates with an angular velocity of WI = 2nn 1 and<br />
the driven one with a velocity of w2=2nn 2 • The velocity of the drive belt<br />
is v=w.'.=(J)'l.'2.<br />
Hence, ~=W2=n2.<br />
'2 00, n l<br />
The sought diam<strong>et</strong>er is<br />
D 1=D2 2=100 mm.<br />
n 1<br />
155. (l) L<strong>et</strong> us denote lhe length of the crawler by L= na. Hence,<br />
L- 2nR is the distance b<strong>et</strong>ween the axes of the wheels.<br />
2
MECHANICS 215<br />
The number of l<strong>in</strong>ks tak<strong>in</strong>g part <strong>in</strong> translation<strong>al</strong> motion is n l = ~ =<br />
L-2nR Th b f I· k · ttl ti t th E h<br />
= 2a • e same num er 0 In s IS a res re a ve 0 e art.<br />
In rotary motion there are n.= 2nR l<strong>in</strong>ks.<br />
a<br />
(2) The time dur<strong>in</strong>g which the tractor moves is t o =!.. . Dur<strong>in</strong>g a full rea<br />
. v<br />
volution of the crawler the l<strong>in</strong>k will move translation<strong>al</strong>ly a distance 21 at a<br />
speed of 2t1. The duration of motion of one l<strong>in</strong>k dur<strong>in</strong>g a revolution is :~.<br />
Altog<strong>et</strong>her, the crawler will make N= ~<br />
revolutions. Therefore, the duration<br />
of translation<strong>al</strong> motion of the l<strong>in</strong>k is 1 1 = Nl . This is the time dur<strong>in</strong>g which<br />
v<br />
the l<strong>in</strong>k is at rest.<br />
The l<strong>in</strong>k will participate <strong>in</strong> rotary motion dur<strong>in</strong>g the time<br />
2Nl s+2nRN-NL<br />
I'}. =/ 0 v v<br />
If s ~ L, the number of revolutions may be assumed as a whole number,<br />
neglect<strong>in</strong>g the duration of an <strong>in</strong>compl<strong>et</strong>e revolution.<br />
156. The duration of flight of a molecule b<strong>et</strong>ween the cyl<strong>in</strong>ders is t= R-r .<br />
v<br />
Dur<strong>in</strong>g this time the cyl<strong>in</strong>ders will turn through the angle wt and, therefore,<br />
I=R(J)t=wR R-r<br />
v<br />
6>R (R-r)<br />
Hence, v= 1 •<br />
157. L<strong>et</strong> us denote the sought radius by R and the angular speed of the<br />
tractor over the arc by (a). Hence (Fig. 331),<br />
til= CJ) ( R- ~ ) , and tlo = CJ) ( R+ : )<br />
Thus<br />
R-.!.<br />
VI =__2 and R=!!. vo+v1 = 6 m<br />
VO R+.!. 2 VO-Vl<br />
2<br />
158. First the observer is at the pole (po<strong>in</strong>t 0 <strong>in</strong> Fig. 332). The axis of<br />
the Earth passes through po<strong>in</strong>t 0 perpendicular to the draw<strong>in</strong>g. L<strong>in</strong>e 0 A<br />
(par<strong>al</strong>lel to BC) is directed toward the star. The mounta<strong>in</strong> is at the right<br />
of po<strong>in</strong>t A. The angle (%= ooat is the angle through which the Earth rotates<br />
dur<strong>in</strong>g the time at with the angular velocity w. To see the start the observer<br />
OC<br />
should run a distance OC.OAcoat. The observer's speed v=dt =OAw=O.7 m/s.
216<br />
ANSWERS AND SOLUTIONS<br />
Fig.3d1<br />
Fig. 882<br />
159. L<strong>et</strong> us take po<strong>in</strong>t A from which the boat departs as the orig<strong>in</strong> of the<br />
coord<strong>in</strong>ate system. The direction of the axes is shown <strong>in</strong> Fig. 333. The boat<br />
moves <strong>in</strong> a direction perpendicular to the current at a constant velocity u.<br />
For this reason the boat will be at a distance g from the bank <strong>in</strong> the time<br />
t=.1L after departure. L<strong>et</strong> us consider the motion of the boat up to the<br />
u<br />
middle of the river (y
MECHANICS<br />
217<br />
1'-'~~---------""2<br />
2<br />
9<br />
46------------..lIIII~-9 Fig. 334<br />
river <strong>in</strong> the time T=;u and will be carried downstream dur<strong>in</strong>g the same<br />
time over a distance of s= a~2 = ~~.<br />
river (po<strong>in</strong>t D) to the opposite bank,<br />
When mov<strong>in</strong>g from the middle of the<br />
the boat will aga<strong>in</strong> be carried away<br />
over the same dist~nce s. Thus, the sought distance is ;:. When the boat<br />
moves to the middle of the river X= a~2=(It t 2<br />
and y=ut.<br />
L<strong>et</strong> us use these ratios to d<strong>et</strong>erm<strong>in</strong>e the trajectory of the boat from A to D.<br />
We g<strong>et</strong> y2=:!f.. x (a parabola). The other h<strong>al</strong>f of the trajectory (DB) is of the<br />
Vo .<br />
same nature as the first one.<br />
160. It is obvious from considerations of symm<strong>et</strong>ry that at any moment<br />
of time the tortoises will be at the corners of a square whose side gradu<strong>al</strong>ly<br />
dim<strong>in</strong>ishes (Fig. 334). The speed of each tortoise can be resolved <strong>in</strong>to a radi<strong>al</strong><br />
(directed towards the centre) and a perpendicular components. The radi<strong>al</strong><br />
speed wiII be equ<strong>al</strong> to (I, = ";2. Each tortoise has to w<strong>al</strong>k a distance of<br />
a<br />
1=V2 to the centre.<br />
Therefore, the tortoises will me<strong>et</strong> at the centre of the square after the time<br />
l a<br />
t=-=- elapses.<br />
v, v<br />
161. Ship B moves toward ship A with the speed 0, At the same time<br />
ship A sails away from ship B with the speed v cos a (Fig. 335). Therefore,
218 ANSWERS AND SOLUTIONS<br />
a<br />
Fig. 335<br />
the distance AB reduces with a speed of v (I-cos a). Po<strong>in</strong>t C (the projection<br />
of po<strong>in</strong>t B onto the trajectory of ship A) moves with a speed of v (I-cos a).<br />
For this reason distance AC <strong>in</strong>creases with a speed of v cosa. Therefore, the<br />
sum of the distances s= AB + AC rema<strong>in</strong>s constant as the ships move. At the<br />
<strong>in</strong>iti<strong>al</strong> moment po<strong>in</strong>t C co<strong>in</strong>cides with A, and therefore s=AB=a. After<br />
a sufficiently great <strong>in</strong>terv<strong>al</strong> of time po<strong>in</strong>t C will co<strong>in</strong>cide with B, and<br />
AB= AC= ; =; ,and the ships will move at a distance of 1.5 km from<br />
each other.<br />
162. With respect to the read<strong>in</strong>g system shown <strong>in</strong> Fig. 336, the coord<strong>in</strong>ates<br />
and the velocities of the body can be d<strong>et</strong>erm<strong>in</strong>ed at any moment of time<br />
from the follow<strong>in</strong>g formulas:<br />
x=voxt (1)<br />
y=voyt- gt2<br />
(2)<br />
T<br />
~=~ ~<br />
vy=voy-gt (4)<br />
Here Vox=Vo cos'a and voy=vo s<strong>in</strong> a are the projections of the <strong>in</strong>iti<strong>al</strong> velocities<br />
on the axes x and g. Equations (1), (2), (3) and (4) provide an answer<br />
to <strong>al</strong>l the questions stipulated <strong>in</strong> the problem.<br />
The duration of flight T can be found from equation (2). When y = 0, we<br />
. gTl 2v s<strong>in</strong> t%<br />
have Vo s<strong>in</strong> aT-2=0. Hence, T 0g<br />
6 :c Fig. 396
MECHANICS 219<br />
V~ s<strong>in</strong> 2a<br />
The distance of the flight is L=vo cos ex T= . This distance<br />
g<br />
will be maximum when a=45°:<br />
VI<br />
Lmax=-o g<br />
The height at which the body will be after the time -r elapses is equ<strong>al</strong><br />
g'f<br />
to h=vo s<strong>in</strong> (X-r-T.<br />
2<br />
The velocity of the body at the moment 'f is equ<strong>al</strong> to v=V~,'where<br />
vx=vocosa and vy=vos<strong>in</strong>a-g'f. Hence, v=Vv~+g2't'2_2voR'1's<strong>in</strong>a and<br />
it forms an angle of ~ with the vertic<strong>al</strong> that can be found from the equation<br />
tan ~= v.o cos ex .<br />
Vo s<strong>in</strong> a-g-r<br />
163. The coord<strong>in</strong>ates of the body x and y change with time accord<strong>in</strong>g to<br />
the law<br />
. gt 2<br />
y=Vo s<strong>in</strong> at-To<br />
x=vo cos at<br />
Upon exclud<strong>in</strong>g the time from these expressions, we obta<strong>in</strong> an equation<br />
for the trajectory y= - ~ g x 2 + tan a·x. This is an equation of a pa-<br />
2vo cost a.<br />
rabola. By denot<strong>in</strong>g the coord<strong>in</strong>ates of the vertex of the parabola (po<strong>in</strong>t A<br />
<strong>in</strong> Fig. 336) by Xo and Yo, the equation of the trajectory can be written as<br />
y-Yo =k (X-XO)2, where<br />
and<br />
k= g . Y<br />
2v~ cos> ex ' 0<br />
v~ s<strong>in</strong> 2a<br />
Xo= 2g<br />
v~ s<strong>in</strong> 2 a.<br />
2g<br />
164. The trajectory of the b<strong>al</strong>l takes the form of a parabola pass<strong>in</strong>g through<br />
a po<strong>in</strong>t with the coord<strong>in</strong>ates hand s. Therefore (see the solution to Pro-<br />
blem 163),<br />
Hence,<br />
. h= g s2+tao as<br />
2v~ cos 2 a<br />
gs2<br />
V2-_~-~--~<br />
n-2 cos> a (tan as -h) (s s<strong>in</strong> 2a.- h cos 2a)-h y 52 + h's<strong>in</strong> (2a-q»-h<br />
where tan cp=h/s. The m<strong>in</strong>imum velocity<br />
V<br />
gs2<br />
s2+h'l.-h<br />
V<br />
tlo= V = g (h+ Vhl+S I )
220<br />
ANSWERS AND SOLUTIONS<br />
" ...--<br />
, ~tI' "<br />
, I~<br />
Fig. 837 Fig. 338<br />
is atta<strong>in</strong>ed when<br />
m 1t h+ Ys2+h 2 s<br />
a=..I+-4=arctan s arctan 2 ¥h2 + s 2-h<br />
(Fig. 337.)<br />
165. The coord<strong>in</strong>ates and the velocities of the body at any moment of<br />
time with respect to the read<strong>in</strong>g system shown <strong>in</strong> Fig. 338 are d<strong>et</strong>erm<strong>in</strong>ed<br />
by the same equations as <strong>in</strong> Problem 162.<br />
At the moment when the body f<strong>al</strong>ls <strong>in</strong>to the water its coord<strong>in</strong>ate y=- H.<br />
For this reason the duration of flight T can be found from the equation<br />
gT2<br />
-H=v o s<strong>in</strong> aT-T<br />
Hence,<br />
Vo s<strong>in</strong> a ± V v~ s<strong>in</strong> 2 a.+2gH<br />
T= ~---------<br />
g<br />
S<strong>in</strong>ce T > Ot we sh<strong>al</strong>l r<strong>et</strong>a<strong>in</strong> the plus sign. The distance from the bank is<br />
sa • 2<br />
VoSID a.+vocosaV sa· I +2 H<br />
L=vo cos a.T 2g --g- Vo s<strong>in</strong> a g<br />
The body will be at a height h above the water after the time<br />
't= v s<strong>in</strong>a ± V v 2s<strong>in</strong>2a+2g(H-h)<br />
-.,;;o~<br />
_<br />
g<br />
If Ih I < I H I, only the plus sign has a physic<strong>al</strong> mean<strong>in</strong>g. When h ~ n,<br />
both solutions have a mean<strong>in</strong>g. Dur<strong>in</strong>g its motion the body will be twice at<br />
the same height above the water.<br />
It is the simplest to f<strong>in</strong>d the f<strong>in</strong><strong>al</strong> velocity v with the aid of the law of<br />
conservation of energy<br />
2 2<br />
mvo + H mv<br />
2 mg =T<br />
Therefore,<br />
v=V v:+2gH
MECHANICS<br />
221<br />
O~----s------..t Fig. 339<br />
166. In the read<strong>in</strong>g system depicted" <strong>in</strong> Fig. 339, the coord<strong>in</strong>ates of the<br />
stone are d<strong>et</strong>erm<strong>in</strong>ed at any moment of time by the follow<strong>in</strong>g equations:<br />
X=Vocos ai<br />
gt<br />
y=ho+vos<strong>in</strong> aJ- 2<br />
T<br />
At the moment when the stone f<strong>al</strong>ls, y=O and X=S, where s is the distance<br />
covered by the stone.<br />
Upon solv<strong>in</strong>g these equations with respect to the angle a, we obta<strong>in</strong><br />
tana= v~<br />
(I ± V1+2;hO_g;:3)<br />
gs 0 0<br />
This expression has a mean<strong>in</strong>g when<br />
1+ 2gho _g2s2 ~ 0<br />
v~ v~<br />
Vo V v 2 + 2gho<br />
PoV v~+2gho<br />
Hence, 5 ~ 0 • Its maximum v<strong>al</strong>ue is smax<br />
g<br />
g<br />
When s is sm<strong>al</strong>ler, two v<strong>al</strong>ues of the angle a correspond to each of its v<strong>al</strong>ues,<br />
the difference b<strong>et</strong>ween which is the less, the nearer s is to its maximum<br />
v<strong>al</strong>ue.<br />
Fig. 340
222 ANSWERS AND SOLUTIONS<br />
Therefore, for the maximum distance of Oight,<br />
v: 00 1<br />
tan 1%= V ,r-; and 1%=30"<br />
gSm" v~+2gho f 3<br />
187. The components of the velocities of the bodies <strong>al</strong>ong x and y at any<br />
moment of time are d<strong>et</strong>erm<strong>in</strong>ed as<br />
011/=V O s<strong>in</strong> <strong>al</strong>-gt ; (121/=00 s<strong>in</strong> ~ -gt;<br />
vtx=uo cos at; and Vlx=- Vo cos a 2<br />
L<strong>et</strong> u be the velocity of the second body with respect to the firstone. Hence<br />
u,=oo s<strong>in</strong> <strong>al</strong>-gt-v o s<strong>in</strong> CXs+gt=vo (s<strong>in</strong> at-s<strong>in</strong> ( 2)<br />
uJ'=vo (cos at +COSCXs)<br />
Therefore, the velocity u is equ<strong>al</strong> to<br />
u=V u~+u;=2cos (1%Itl%l) (/0<br />
The bodies move with respect to each other at a constant velocity. After the<br />
time l' the distance b<strong>et</strong>ween them will be<br />
s=2vocos (1%1t lXs ) "C<br />
168. The horizont<strong>al</strong> path of the bomb s=Yl2_h 2 = v cos at, where t is<br />
the duration of f<strong>al</strong>l<strong>in</strong>g of the bomb. The vertic<strong>al</strong> path is h=v s<strong>in</strong> I%t+ g~1<br />
(Fig. 340).<br />
Upon exclud<strong>in</strong>g the time from these equations, we f<strong>in</strong>d<br />
tan a=- Vi ± .. I(VI)2 + 2hv t -1<br />
gs JI gs gsl<br />
The solution with the plus sign has a mean<strong>in</strong>g. The m<strong>in</strong>us sign corresponds<br />
to ex < 0, i.e., the bomb is dropped when the dive-bomber is fly<strong>in</strong>g upward.<br />
169. It will be convenient to solve this problem <strong>in</strong> a read<strong>in</strong>g system rela·<br />
ted to the uniformly mov<strong>in</strong>g vehicles.<br />
In this system, the highway moves back with the speed of 0=50 km/h,<br />
the vehicles are at rest with respect to each other and their wheels rotate.<br />
The l<strong>in</strong>ear speed of po<strong>in</strong>ts on the circumference of the wheel and tha t of the<br />
stuck stone IS <strong>al</strong>so v. The stone will fly the maximum distance if it flies out<br />
when its speed forms an angle of 45 0 with the horizon. L<strong>et</strong> us f<strong>in</strong>d this<br />
distance. Neglect<strong>in</strong>g the fact that the stone is somewhat above the level of<br />
2 • 2 I<br />
the highway when it is thrown out, we obta<strong>in</strong> 1 v s<strong>in</strong> a v = 19.6 m<strong>et</strong>.<br />
g g<br />
res. The m<strong>in</strong>imum distance b<strong>et</strong>ween the vehicles should be 19.6 m<strong>et</strong>res.<br />
170. It will be much easier to solve the problem if the axes of coord<strong>in</strong>ates<br />
are directed <strong>al</strong>ong the <strong>in</strong>cl<strong>in</strong>ed plane and perpendicular to it (Fig. 341).<br />
In this case the components of the acceleration of the b<strong>al</strong>l on the axes x<br />
and y will be ~pectively equ<strong>al</strong> to as=g.=g s<strong>in</strong> ex and a 1l<br />
= g J/= - gcosa.
,~\ECHANICS<br />
223<br />
Fig. 341<br />
Fig. 342<br />
Upon the first impact with the <strong>in</strong>cl<strong>in</strong>ed plane, the velocity of the b<strong>al</strong>l will<br />
be Vo= Y2gh. The <strong>in</strong>iti<strong>al</strong> velocity of the b<strong>al</strong>l after the first impact is Vo<br />
and forms an angle a with the y-axis (Fig. 341).<br />
The distance b<strong>et</strong>ween the po<strong>in</strong>ts of the first and second impacts is<br />
. t2<br />
11 =v o s<strong>in</strong> at)+ g Sl~ a ,where it is the duration of flight and is d<strong>et</strong>erm<strong>in</strong>ed<br />
by the equation<br />
gcos ai~<br />
2<br />
Hence, i1= 2v o and i1=8h s<strong>in</strong> a. The velocity of the b<strong>al</strong>l at the second<br />
g<br />
impact can be found from the equations<br />
v)x=vOX+aXil =vo s<strong>in</strong> a+g s<strong>in</strong> at l =3vo s<strong>in</strong> a<br />
Vly=Voy + ayt l =0 0 cos a,-g cos at l =- Vo cos a.<br />
After the impact these velocities are equ<strong>al</strong> to<br />
o<br />
v 2X=vt x , and V2y=- v 1Y<br />
The distance b<strong>et</strong>ween the po<strong>in</strong>ts of the second and third impacts is equ<strong>al</strong> to<br />
. g s<strong>in</strong> at~<br />
1 2 = 30 0 s<strong>in</strong> at 2 + 2<br />
where 1 2 .is the time dur<strong>in</strong>g which the b<strong>al</strong>l is <strong>in</strong> flight. S<strong>in</strong>ce the <strong>in</strong>iti<strong>al</strong><br />
velocity <strong>al</strong>ong the y' axis is the same as dur<strong>in</strong>g the first impact, 1 2 = / }.<br />
Therefore, 1 2 = 16hs<strong>in</strong> a.<br />
Similarly, it can be shown that the distance b<strong>et</strong>ween the next po<strong>in</strong>ts<br />
la=24h s<strong>in</strong> a.<br />
Consequently. 1 1:l2 : [3 ••• =1:2:3, <strong>et</strong>c.
224 ANSWERS AND SOLUTIONS<br />
171. The motion of the body can be considered as superrosftion of movement<br />
<strong>al</strong>ong a circumference With a radius R <strong>in</strong> a horizonta plane and vertic<strong>al</strong><br />
f<strong>al</strong>l<strong>in</strong>g. Accord<strong>in</strong>gly, the velocity of the body v at the given moment<br />
can be represented as the geom<strong>et</strong>ric<strong>al</strong> sum of two components: VI =v cos a<br />
directed horizont<strong>al</strong>ly and v 2=v s<strong>in</strong> a. directed vertic<strong>al</strong>ly (Fig. 342). Here a<br />
is the angle formed by the helic<strong>al</strong> l<strong>in</strong>e of the groove with the horizon.<br />
In curvil<strong>in</strong>ear motion the acceleration of a body is equ<strong>al</strong> to the geom<strong>et</strong>ric<strong>al</strong><br />
sum of the tangenti<strong>al</strong> and norm<strong>al</strong> accelerations. The norm<strong>al</strong> acceleration<br />
that corresponds to movement <strong>al</strong>ong the circumference is<br />
v~ v 2 cost ex<br />
<strong>al</strong> n = 7[= R<br />
The vertic<strong>al</strong> motion is rectil<strong>in</strong>ear, and therefore a2n = O.<br />
The sought acceleration a = V Q~"t+a:'t+ a~n, where a 1t and a2't are the<br />
tangenti<strong>al</strong> accelerations that correspond to motion <strong>al</strong>ong the circumference<br />
and <strong>al</strong>ong the vertic<strong>al</strong>. The tot<strong>al</strong> tangenti<strong>al</strong> acceleration Q't Is obviously<br />
equ<strong>al</strong> to a't=V a~1'+a:1'..<br />
The v<strong>al</strong>ue of a't can be found by ment<strong>al</strong>ly developIng the surface of the<br />
cyl<strong>in</strong>der with the helic<strong>al</strong> groove <strong>in</strong>to a plane. In this case the groove will<br />
become an <strong>in</strong>cl<strong>in</strong>ed plane with B height nh and a length of Its base 2nRn.<br />
h<br />
Apparently, a~=g s<strong>in</strong> a.=g y .<br />
h 2+4n2 R2<br />
To d<strong>et</strong>erm<strong>in</strong>e DIn' l<strong>et</strong> us f<strong>in</strong>d v from the law of conservation of energy:<br />
mv 2 2 Bn"nhgR .<br />
T=mghn. Consequently, () =2ghn and a 1n h 2+4n2<br />
R" Upon Insert<strong>in</strong>g<br />
Fig. 343
MECHANICS 225<br />
sought accele<br />
the found accelerations a'f and <strong>al</strong> n <strong>in</strong>to the expression for the<br />
ration, we g<strong>et</strong><br />
ghYh?+4n I RI+ 64n 4l n I Rs<br />
a= h2 + 4n2RI<br />
172. As usu<strong>al</strong>, the motion of the b<strong>al</strong>l can be considered as the' result<br />
of summation of vertic<strong>al</strong> (uniformly accelerated) and horizont<strong>al</strong> (uniform) motions.<br />
The simplest m<strong>et</strong>hod of solution is to plot a diagram show<strong>in</strong>g how the<br />
coord<strong>in</strong>ates of the b<strong>al</strong>l <strong>al</strong>ong the horizont<strong>al</strong> depend on the time for the limit<strong>in</strong>g<br />
velocities 267 cmls and 200 cmls (Fig. 343). The lower broken l<strong>in</strong>e corresponds<br />
to the maximum velocity and the upper one to the m<strong>in</strong>imum<br />
velocity. In the course of time, as can be seen from the diagram, the <strong>in</strong>de..<br />
f<strong>in</strong>iteness of the b<strong>al</strong>l coord<strong>in</strong>ate x shown by the section of the horizont<strong>al</strong><br />
straight l<strong>in</strong>e b<strong>et</strong>ween the l<strong>in</strong>es of the diagram <strong>in</strong>creases. The vertic<strong>al</strong> hatch<strong>in</strong>g<br />
<strong>in</strong> .Fig. 343 shows the movement of the b<strong>al</strong>l from M to N, and the horizont<strong>al</strong><br />
hatch<strong>in</strong>g-from N to M. The cross..hatched areas correspond to <strong>in</strong>def<strong>in</strong>iteness<br />
<strong>in</strong> the direction of the horizont<strong>al</strong> velocity.<br />
(I) The diagram shows that after the b<strong>al</strong>l bounces once from slab N the<br />
direction of its horizont<strong>al</strong> velocity will be <strong>in</strong>def<strong>in</strong>ite when the duration of<br />
f<strong>al</strong>l<strong>in</strong>g OK< t ~ OL or t > AB (where OK=0.15 s, OL=O.2 sand AB=0.225 s).<br />
gt 2<br />
Hence, 10 em
226 ANS\VERS AND SOLUTIONS<br />
A<br />
2u<br />
(6)<br />
c<br />
c Fig. 344<br />
observer. The po<strong>in</strong>ts which are at 8 distance equ<strong>al</strong> to the disk radius from<br />
the <strong>in</strong>stantaneous axis (po<strong>in</strong>t C) will have the same velocity (<strong>in</strong> absolute<br />
magnitude) as that of the axis, i.e., v (Fig. 344b).<br />
174. The angle b<strong>et</strong>ween adjacent spokes of the front wheel is cp= ~: .<br />
The wheel will seem stationary on the screen if it turns through the angle<br />
a=kq> dur<strong>in</strong>g the time b<strong>et</strong>ween the film<strong>in</strong>g of two successive frames T= 1/24 s.<br />
Here k is a positive <strong>in</strong>teger. On the other hand, the angle of rotation of the<br />
wheel dur<strong>in</strong>g the time T is equ<strong>al</strong> to a=co"t', where CI) is the angular speed of<br />
the wheel. Therefore. the front wheel will seem stationary if CJ)~2 and N1tk, IT<br />
the speed of the cart V= (J)T = 2MT • The m<strong>in</strong>imum speed of the cart 0m<strong>in</strong> =<br />
N l't'<br />
2n.r<br />
= N=8.8 rn/s.<br />
1<br />
The rear wheel will <strong>al</strong>so seem stationary if<br />
NR<br />
Hence, N 2 = _ I _ = 9 when k l=k2=1.<br />
r<br />
175. (1) The spokes will seem to rotate counterclockwise if dur<strong>in</strong>g the<br />
time 'T (see Problem 174) the wheel turns through the angle ~1 which satisfies<br />
the condition ker > ~l > kff- ~ , where k= 1, 2, 3, ...• The consecutive positions<br />
of the wheel spokes are -I.own for this case· <strong>in</strong> Fig. 345a.<br />
It seems to<br />
the audience that each spoke t 'ns through the angle a < ~ counterclockwise.<br />
The possible angular velocit ies lie with<strong>in</strong> the <strong>in</strong>terv<strong>al</strong><br />
kq> (2k- 1) q><br />
-;- > 001 > --2-:r-
MECHANICS 227<br />
S<strong>in</strong>ce the front and rear wheels have the same number of spokes, the wheels<br />
will seem to revolve counterclockwise if the speed of the cart is<br />
kepr kfPr cpr<br />
-:r>v>-:r-2't (1)<br />
kcpR<br />
> v > kcpR _ fPR<br />
't 't' 2-r<br />
(2)<br />
S<strong>in</strong>ce R= 1.5r, the second <strong>in</strong>equ<strong>al</strong>ity can be rewritten as foHows:<br />
I 5 kcpr 1.5kfPr I.SepT<br />
· T>v>-'t--~<br />
Both Inequ<strong>al</strong>ities, which are congruent only when k= 1, give the permissible<br />
speeds of the cart <strong>in</strong> the form<br />
cpr > v > 0.75 q>r<br />
't<br />
't<br />
2n .<br />
Or, s<strong>in</strong>ce CP=6' we have 8.8 m/s > v > 6.6 m/s.<br />
(2) The spokes of the rear wheel will seem to revolve clockwise if dur<strong>in</strong>g<br />
the time r the wheel turns through the angle ~2 which satisfies the condition<br />
(2k-l) : > Pi > (k - 1) cp (Fig. 345b). Hence, the follow<strong>in</strong>g <strong>in</strong>equ<strong>al</strong>ity is true<br />
for the speed of the cart:<br />
1 5 (2k-l) cpt 1.5(k-l) cpr<br />
• 2't >v> 't<br />
At the same time <strong>in</strong>equ<strong>al</strong>ity (1) should be complied with. When k= 1, both<br />
<strong>in</strong>equ<strong>al</strong>ities are congruent if 0.75 cpr > v > 0.5 fPr . When k=2 they are can-<br />
't<br />
't'<br />
gruent if 2cpr > v > 1.5 cpr • If k > 2 the <strong>in</strong>equ<strong>al</strong>ities are <strong>in</strong>congruent.<br />
T "<br />
F'II<br />
D'O<br />
(h)<br />
Fig..845<br />
15 *
228 ANSWERS AND SOLUTIONS<br />
Fig. 346<br />
_.......~~--~<br />
~ Therefore,<br />
or<br />
6.6 m/s > v > 4.4 mls<br />
17.6 mls > t1 > 14.2 mls<br />
176. The <strong>in</strong>stantaneous axis of rotation<br />
(see Prob tern 173) passes through po<strong>in</strong>t C<br />
(Fig. 346). For this reason the velocity<br />
of po<strong>in</strong>t A relative to the block is<br />
R+,<br />
VA=V-,- .<br />
Po<strong>in</strong>t B has<br />
R-r<br />
the velocity vB=v --.<br />
r<br />
Po<strong>in</strong>ts on a circle with the radius r whose<br />
centre is po<strong>in</strong>t C have an <strong>in</strong>stantaneous<br />
velocity equ<strong>al</strong> to that of the spool core.<br />
177. The trajectories of po<strong>in</strong>ts A, Band<br />
c are shown <strong>in</strong> Fig. 347. Po<strong>in</strong>t B describes<br />
a curve usu<strong>al</strong>ly c<strong>al</strong>led an ord<strong>in</strong>ary cycloid. Po<strong>in</strong>ts A and C describe an<br />
elongated and a shortened cycloids.<br />
178. The l<strong>in</strong>ear velocity of po<strong>in</strong>ts on the circumference of the shaft<br />
Ul=OO ~ and that of po<strong>in</strong>ts on the race u 2 = Q ~. S<strong>in</strong>ce the b<strong>al</strong>Is do not<br />
slip, the same <strong>in</strong>stantaneous velocities will be imparted to the po<strong>in</strong>ts on the<br />
b<strong>al</strong>l bear<strong>in</strong>g that at this moment are <strong>in</strong> contact with the shaft and the race.<br />
The <strong>in</strong>stantaneous velocity of any po<strong>in</strong>t on the b<strong>al</strong>l can be regarded as the<br />
sum of two velocities: the velocity of motion of its centre Vo and the l<strong>in</strong>ear<br />
velocity of rotation around the centre. The b<strong>al</strong>l will rotate with a certa<strong>in</strong><br />
angular velocity (00 (Fig. 348).<br />
Therefore,<br />
VI =vo-wo'<br />
V2=VO+OOO'<br />
Hence,<br />
1 1<br />
vO=T (v t +v2)=4 (
MECHANICS 229<br />
lIQ<br />
1I2<br />
I<br />
\{JiLl<br />
I III<br />
B<br />
{}I<br />
U;<br />
I<br />
I<br />
A<br />
Fig. 348 Fig. 349<br />
179. S<strong>in</strong>ce the cone rolts without slipp<strong>in</strong>g, the po<strong>in</strong>ts on the generatrix<br />
OA (Fig. 349) should be stationary. This fact is used to d<strong>et</strong>erm<strong>in</strong>e the velocity<br />
Q with which the cone rotates around its axis.<br />
For po<strong>in</strong>t A we have ~=Qhtan c%,. Hence, Q= ~. The veloc.ity<br />
cos a<br />
sm a<br />
of an arbitrary po<strong>in</strong>t D 1 on diam<strong>et</strong>er AB of the cone base is the sum of two<br />
veloci ties:<br />
VI =(1) (h cos a -,s<strong>in</strong> a)+ ~.'CJ)<br />
~lna<br />
where , is the distance from the centre of the base C to the gIven po<strong>in</strong>t.<br />
For po<strong>in</strong>t D 2 below centre C we have<br />
v 2 =<br />
(I) (h cos a+, s<strong>in</strong> a)-....:.!!!<br />
s<strong>in</strong> a,<br />
The velocity of the lowermost po<strong>in</strong>t is zero and of the uppermost po<strong>in</strong>t<br />
v=2(1)h cos (l,.<br />
180. The l<strong>in</strong>ear velocities should be the same where bevel gears E and C,<br />
and E and D, mesh. S<strong>in</strong>ce gears E rotate with the velocity ro around axle A,<br />
and the axle rotates <strong>in</strong> another plane with the velocity a, the follow<strong>in</strong>g<br />
equation is true for gears E and C to mesh:<br />
'1001='00+'1°<br />
For mesh<strong>in</strong>g of gears E and D, a similar equation has the form<br />
'100,=- 'CO+'I Q<br />
Hence,<br />
2Q=(01+002<br />
'1 200=, (001-002)<br />
When wheel B rotates at a speed of 0, the angular velocities of the driv<strong>in</strong>g<br />
wheels may differ from each other from zero to 2Q.
230 ANSWERS AND SOLUTIONS<br />
T=(M +m) g ( 4 510 2 ~ + I)<br />
182. T 2l; 1=10mro T 2l; 2=9mCJ) T s=7mCJ) 2l. and T4,=4mCJ) 2 l .<br />
183. The distances from the centre of gravity to the masses ml and m 2<br />
are equ<strong>al</strong>, respectively, to<br />
m m x= 1<br />
2 land y= 1<br />
mt+m 2 ml+m2<br />
L<strong>et</strong> us denote the velocity of the centre of gravity by u and the angular<br />
velocity of rotation by co. Thus U+WX=Vl and U-WY=fJ 2 •<br />
Hence,<br />
Vt- V2 d mtv• +m2v2<br />
CJ)=--, an U=-~-~<br />
1 mt +m 2<br />
184. The velocity of rotation will be r<strong>et</strong>arded. The platform imparts to<br />
the shell an addi tion<strong>al</strong> momentum <strong>al</strong>ong a tangent to the trajectory of the<br />
end of the cannon barrel. Accord<strong>in</strong>g to Newton's third law, the shell ejected<br />
from the barrel will press aga<strong>in</strong>st its <strong>in</strong>ner part <strong>in</strong> a direction opposite to rotation.<br />
185. When the body touches the horizont<strong>al</strong> plane, the vertic<strong>al</strong> and horizont<strong>al</strong><br />
components of its velocity will be V'Oer= V 2gH s<strong>in</strong> a and vhor =<br />
= y 2gH cos cx. If the impact is absolutely elastic, the vertic<strong>al</strong> component<br />
will change its sign. while the horizont<strong>al</strong> component will rema<strong>in</strong> the same.<br />
The trajectory of the body will take the form of sections of parabolas<br />
(Fig. 350), with h=H s<strong>in</strong> 2 cx and l=2H s<strong>in</strong> 2a.<br />
If the impact is absolutely <strong>in</strong>elastic. the vertic<strong>al</strong> component of the velocity<br />
will be zero, and the body will move uniformly over the horizont<strong>al</strong><br />
plane with the velocity fJ= V 2gH cos cx.<br />
186. The Earth acts on the motor..cycle with two forces (Fig. 351), namely,<br />
N - the reaction of the support, and f - the force of friction. The sum of<br />
H<br />
Fig. 350
MECHANICS 231<br />
N<br />
A<br />
Fig. 951<br />
T<br />
With the given v<strong>al</strong>ue of a, it will be the sm<strong>al</strong>ler, the greater is R.<br />
Hence, the rod will f<strong>al</strong>l faster if it is placed on end B.<br />
188. Accord<strong>in</strong>g to Newton's second law, moolR =mg cos a,-N, where N Is<br />
the force which the deformed rod acts on the sphere with.<br />
The deformation of the rod will disappear when the rod no longer presses<br />
on the floor, and N=O. As shown <strong>in</strong> Problem 187, 0=2 V~ s<strong>in</strong> ; ·<br />
Upon <strong>in</strong>sert<strong>in</strong>g this v<strong>al</strong>ue of co <strong>in</strong> the equation of motion, we f<strong>in</strong>d that<br />
cos tI= ~ . Hence, tI=48°10'.<br />
For the rod not to slip, the condition N s<strong>in</strong> a ~ 5 , the rod will not slip until N is zero, l.e., until tI';;;;<br />
2<br />
arc cos ~ , the equation m0 1 R=mg cos tI- N will give us<br />
N
232 ANSWERS AND SOLUTIONS<br />
9-"-,,~, a<br />
I "-<br />
I<br />
II<br />
I<br />
IIlII11%<br />
Ns<strong>in</strong>~<br />
A<br />
.0<br />
B<br />
Fig. 352<br />
will beg<strong>in</strong> to f<strong>al</strong>l freely from the moment when the angle ex reaches the<br />
2<br />
v<strong>al</strong>ue ao= arc cos 3".<br />
At this moment u=CJlR=vi gR forms an angle CXo with the horizon<br />
and the height of the sphere above the ground is CD= ~<br />
us f<strong>in</strong>d the distance sought by us<strong>in</strong>g the laws of free f<strong>al</strong>l<strong>in</strong>g:<br />
AB=AD+DB=R 5 )15+4 V23 ~ 1.12R<br />
27 -<br />
R (Fig. 353). L<strong>et</strong><br />
190. The bead moves <strong>al</strong>ong section ADB under the action of the force of<br />
gravity (Fig. 354). For the bead to reach po<strong>in</strong>t B after it leaves po<strong>in</strong>t A,<br />
the horizont<strong>al</strong> path which it travels should be 2R s<strong>in</strong>
MECHANICS 233<br />
From equations (I) and (2) we g<strong>et</strong> h=} R. The velocity at po<strong>in</strong>t Acan be<br />
found from the law of conservation of energy<br />
5 mv 2<br />
mg 2<br />
R = 2A+mgR(1+cosa) (3)<br />
The body thrown at an angle a to the horizon will fly a horizont<strong>al</strong> distance<br />
of<br />
v~ s<strong>in</strong> 2a<br />
AB=---<br />
(4)<br />
g<br />
On the other hand,<br />
AB=2R·s<strong>in</strong>a<br />
(5)<br />
It follows from equations (4) and (5) that<br />
Rg<br />
A-cos<strong>et</strong><br />
v2 _<br />
Upon <strong>in</strong>sert<strong>in</strong>g this v<strong>al</strong>ue <strong>in</strong>to equation (3), we obta<strong>in</strong>:<br />
mg- 2<br />
5 R=2<br />
mgR +mgR+mgRcosa<br />
cos ex,<br />
3 ± 1<br />
Therefore, cos a=-4- and, consequently, CXt=O, and cx 2=60°.<br />
It is easy to see that if ex > 60°, the body will f<strong>al</strong>l <strong>in</strong>side the loop; if<br />
a. < 60° it will fly out.<br />
192. L<strong>et</strong> us consider the forces that act on the str<strong>in</strong>g thrown over the<br />
left nail (Fig. 356). The vertic<strong>al</strong> components of the forces of tension T act<strong>in</strong>g<br />
on the weights are mg if the str<strong>in</strong>g is secured on the nail. Accord<strong>in</strong>g to<br />
Fig. 354 Fig. 355
234 . ANSWERS AND SOLUTIONS<br />
Newton's third law t the knot (po<strong>in</strong>t 0) is acted upon by the same forces T.<br />
Their sum is directed vertic<strong>al</strong>ly downward and is equ<strong>al</strong> to 2mg.<br />
If only one weight. is rotat<strong>in</strong>g, the vertic<strong>al</strong> component of the str<strong>in</strong>g tension<br />
T' is equ<strong>al</strong> to 2mg (if the weight does not move downward). The tension<br />
of the str<strong>in</strong>g itself, however, is T' > 2mg (Fig. 356). Therefore, the<br />
system will not be <strong>in</strong> equilibrium. The right-hand weight will have a greater<br />
pull.<br />
193. The direction of the acceleration co<strong>in</strong>cides with that of the resultant<br />
force. The acceleration is directed downward when the b<strong>al</strong>l is <strong>in</strong> its two<br />
extreme upper positions Band C (Fig. 357). The acceleration will be directed<br />
upward if the b<strong>al</strong>l is <strong>in</strong> its extreme bottom position A and horizont<strong>al</strong>ly<br />
<strong>in</strong> positions D and L d<strong>et</strong>erm<strong>in</strong>ed by the angle cx.<br />
L<strong>et</strong> us f<strong>in</strong>d ex. Accord<strong>in</strong>g to Newton's second law, the product of the<br />
mass and the centrip<strong>et</strong><strong>al</strong> acceleration is equ<strong>al</strong> to the sum of the projections<br />
of the forces on the direction of the radius of rotation:<br />
mv 2<br />
-1- = T - mg cos ex<br />
On the other hand, as can be seen from Fig. 357, we have T= mg . On<br />
cos ex<br />
the basis of the law of conservation of energy:<br />
mol<br />
T=mgl cos a,<br />
1<br />
We can f<strong>in</strong>d from these equations that cosa= ya' and therefore<br />
ex ~ 54 045'.<br />
194. L<strong>et</strong> us denote the angular velocity of the rod by co at the moment<br />
when it passes through the vertic<strong>al</strong> position. In conformity with the law of<br />
conservation of energy:<br />
co 2 t 2<br />
2 (miT!+m,T.)=g (I-cos a) (mITt +m2's)<br />
or
MECHANICS 235<br />
Fig. 357<br />
whence<br />
Fig. 358<br />
Vl=ror 1= 2rt sln~'" /gmlr~+m2r:<br />
2 V m1' 2 + m 2' S<br />
. a. ml'l+m2'2<br />
vs=r2=2r2sln- g 2 2<br />
2 mt'2+ m2'2<br />
195. The resultant of the forces applied to the b<strong>al</strong>l F=mg tan a. should<br />
build up a centrip<strong>et</strong><strong>al</strong> acceleration a=oo 2 r, where r=1 s<strong>in</strong> a. (Fig. 358).<br />
Hence,<br />
mg tan a-=moo 2 l s<strong>in</strong> (I<br />
This equation has two solutions:<br />
CXt=O<br />
cx! = arc cos ro~ I<br />
Both solutions are v<strong>al</strong>id <strong>in</strong> the second case: a1=0 (here the b<strong>al</strong>l is <strong>in</strong> a<br />
state of unstable equilibrium) and ~ = 60 0 •<br />
In the first case the only solution is <strong>al</strong>=0.<br />
198. L<strong>et</strong> us resolve the force F act<strong>in</strong>g from the side of the rod on the<br />
weight m <strong>in</strong>to mutu<strong>al</strong>ly perpendicular components T and N (Fig. 359).<br />
L<strong>et</strong> us project the forces ontoa vertic<strong>al</strong> and a horizont<strong>al</strong> l<strong>in</strong>es and write<br />
Newton's equations for these directions<br />
mooll s<strong>in</strong> cp=T s<strong>in</strong> cp- N cos q><br />
mg=T cos cp+N s<strong>in</strong>,<br />
L<strong>et</strong> us d<strong>et</strong>erm<strong>in</strong>e T and N from these equations<br />
T=m (CJ)11 s<strong>in</strong> t cp+g cos cp)<br />
N=m 19-CJ) 2 1cos cp) s<strong>in</strong> cp
236<br />
o<br />
ANSWERS AND SOLUTIONS<br />
a<br />
A<br />
H<br />
Fig. 859<br />
0'<br />
Fig. 360<br />
0'<br />
Therefore.<br />
F =YT2+N2=m y g2+ro4 l 2 s<strong>in</strong> 2 cp<br />
197. The forces act<strong>in</strong>g on the bead are shown <strong>in</strong> Fig. 360: f is the<br />
of friction, mg the weight and N the force of the norm<strong>al</strong> reaction.<br />
force<br />
Newton's equations for the projection of the forces on a horizont<strong>al</strong><br />
vertic<strong>al</strong> directions will have the form<br />
and a<br />
f s<strong>in</strong> q> 1= N cos q>=mro2 l s<strong>in</strong> q><br />
f cos q> ± N s<strong>in</strong> q>-mg=O<br />
The upper sign refers to the case shown <strong>in</strong> Fig. 360 and the lower one to<br />
the case when the force N acts <strong>in</strong> the opposite direction. We f<strong>in</strong>d from these<br />
equations that<br />
f=mro 2 l s<strong>in</strong> 2 cp+mg cos q><br />
N= ± (mg s<strong>in</strong> q> - mro 2 l s<strong>in</strong> cp cos q»<br />
In equilibrium f ~ kN or<br />
k s<strong>in</strong> cp-cos cp g<br />
1<br />
(k<br />
cos cP+.) s<strong>in</strong> q> • 2 when k ~ cot cp<br />
(J)<br />
and<br />
k s<strong>in</strong> g<br />
I ~. (k .) • 2 when k ~<br />
s<strong>in</strong> q> cos cp-sln<br />
tan q><br />
q> co<br />
198. Figure 361 shows the forces act<strong>in</strong>g on the weights. Here T 1 and T 2<br />
are the tensions of the str<strong>in</strong>g. L<strong>et</strong> us write Newton's equations for the proj<br />
ectlons onto a horizont<strong>al</strong> and a vertic<strong>al</strong> directions.
MECHANICS 237<br />
For the first weight<br />
T I s<strong>in</strong> q>-T 2 s<strong>in</strong> 'i'=mooll s<strong>in</strong>,<br />
T1 cos q> - T t COS'\1'-mg=O (1)<br />
For the second weight<br />
mro 21 (s<strong>in</strong> q>+ s<strong>in</strong> 'i')=T2 s<strong>in</strong>", (2)<br />
T 2 cos"i'=mg<br />
Upon exclud<strong>in</strong>g T 1 and T 2 from the system of equations (I) and (2), we<br />
obta<strong>in</strong> the equations<br />
a s<strong>in</strong> q>=2 tan cp- tan",<br />
a (s<strong>in</strong> q>+s<strong>in</strong> "p) =tan",·<br />
00 2 l<br />
where a=-'<br />
g<br />
From these equations we g<strong>et</strong> 2 tan q>- tan", < tan'P and, therefore, q> < W.<br />
199. The forces act<strong>in</strong>g on the weights are shown <strong>in</strong> Fig. 362. Here T I' N I<br />
and T 2' N 2 are the components of the forces act<strong>in</strong>g from the side of the rod<br />
on the weights m and M.<br />
The forces NI and N2 act <strong>in</strong> opposite directions, s<strong>in</strong>ce the sum of the<br />
moments of the forces act<strong>in</strong>g on the rod with respect to po<strong>in</strong>t 0 is zero because<br />
the rod is weightless: N 1b-N2 (b+a)=O. The equations of motion of<br />
the masses m and M for projections on a horizont<strong>al</strong> and a vertic<strong>al</strong> directions<br />
have the form<br />
mro 2 b s<strong>in</strong> q>=T 1 s<strong>in</strong> q>-N I cos +N 1 s<strong>in</strong> q>=mg<br />
Mw 2 (b+a) s<strong>in</strong> q>=T 2 s<strong>in</strong> cp+N 2 cos q>; T 2 cos cp-N 2 s<strong>in</strong> cp=Mg<br />
Upon exclud<strong>in</strong>g the unknown quantities T 1. T 2' Nt and N 2 from the system,<br />
We f<strong>in</strong>d that<br />
g mb+M (a+b)<br />
(1) qJ=Ot and (2) cos q>= 002 mb 2+M (a+b)2<br />
o<br />
Fig. 361 Fig. 362<br />
0'
238 ANSWERS AND SOLUTIONS<br />
Fig. 963<br />
The first solution is true for<br />
any angular velocities of rotation,<br />
and the second when 0) ~<br />
.. / mb+M (a+b) (see the<br />
V g mb 2+M (a+b)2<br />
solution to Problem 195).<br />
200. In the state of equilibrium<br />
mro 2 x=kx, where x is the<br />
distance from the body to the<br />
axis.<br />
It is thus obvious that with<br />
any v<strong>al</strong>ue of x the spr<strong>in</strong>g imparts<br />
the centrip<strong>et</strong><strong>al</strong> acceleration necessary<br />
for rotation to the body.<br />
For this reason the latter will<br />
move after the imp<strong>et</strong>us with a constant velocity up to stop A or as<br />
long as the law of proportion<strong>al</strong>ity b<strong>et</strong>ween the force act<strong>in</strong>g on the spr<strong>in</strong>g and<br />
its deformation is v<strong>al</strong>id.<br />
201. L<strong>et</strong> us write Newton's second law for a sm<strong>al</strong>l portion of the cha<strong>in</strong><br />
hav<strong>in</strong>g the mass ~ R Aa and shown <strong>in</strong> Fig. 363:<br />
7R Aa(2nn)'R=2Ts<strong>in</strong> A;<br />
S<strong>in</strong>ce the angle Aa is sm<strong>al</strong>l, s<strong>in</strong> A; ~ A; , whence T=mln'=9.2 kgf.<br />
202. L<strong>et</strong> us take a sm<strong>al</strong>l element of the tube with the length R Aa<br />
(Fig. 364). The str<strong>et</strong>ched w<strong>al</strong>ls of the tube impart an acceleration a=~ to the<br />
water flow<strong>in</strong>g <strong>al</strong>ong this element. Accord<strong>in</strong>g to Newton's third law, tne water<br />
will act on the element of the tube with the force<br />
Y
MECHANICS<br />
239<br />
t-4----- 7j------......<br />
i+T<br />
?i+l------~<br />
Fig. 865<br />
sider an arbitrary section with the number i (Fig. 365). The acceleration of the<br />
various po<strong>in</strong>ts <strong>in</strong> this section will be different, s<strong>in</strong>ce the distances from the<br />
po<strong>in</strong>ts to the axis of rotation are not the same. If the difference ';+1-'1 is<br />
sm<strong>al</strong>l, however, the acceleration of the i-th section may be assumed as equ<strong>al</strong><br />
to ro2 'i+<br />
2+'1<br />
, and this will be the more accurate, the sm<strong>al</strong>ler is the<br />
length of the section.<br />
The i-th section is acted upon by the elastic force T;+l from the side of<br />
the deformed section i+ 1 and the force Ti from the side of the section<br />
i-I. S<strong>in</strong>ce the mass of the i-th section is ~ ('1+1 -rd, on the<br />
Newton's second law we can write that<br />
m<br />
'·+1+,.<br />
T;-Ti+l=T (ri+l- f ; ) 00 2 I 2 t<br />
or<br />
mID? ( 2 2)<br />
Ti+1-T;=-y r'+l-r/<br />
basis of<br />
L<strong>et</strong> us write the equations of motion for the sections from k to n, <strong>in</strong>clusive,<br />
assum<strong>in</strong>g that 'n+i=l and ',,=x:<br />
moo? a<br />
-Tn=--(12_ r n)<br />
. 21<br />
moo? I I<br />
Tn-T n-l = -Y ~rn-fn-l)<br />
mID! t 2<br />
Tk+I-Tk+ 1=-21 (rk+I-'k+J<br />
mID'}, 2<br />
Tk+l-T,,=-21(rk+l-X!)<br />
;;<br />
The first equation <strong>in</strong> this system takes <strong>in</strong>to account the fact that the<br />
elastic force does not act on the end of the rod, te., Tn+l =0. Upon<br />
summ<strong>in</strong>g up the equations of the system, we f<strong>in</strong>d that the sought tension<br />
• moo'/,<br />
IS equ<strong>al</strong> to T"=2(11_X 2 ) .
240 ANSWERS AND SOLUTIONS<br />
The closer the sections of the rod are to the axis of rotation, the more<br />
wi11 they be str<strong>et</strong>ched.<br />
204. In the read<strong>in</strong>g system that is stationary with respect to the axis, the<br />
force of tension of the rod does not perform any work, s<strong>in</strong>ce this force constantly<br />
rema<strong>in</strong>s 'perpendicular to the velocity of the b<strong>al</strong>l. In the mov<strong>in</strong>g system<br />
this force performs work other than zero, and this changes the k<strong>in</strong><strong>et</strong>ic energy<br />
of the b<strong>al</strong>l.<br />
205. Section AB of the hoop with a mass m at the highest po<strong>in</strong>t has an<br />
energy of mg 2R+m ~V)2. Dur<strong>in</strong>g motion the k<strong>in</strong><strong>et</strong>ic and potenti<strong>al</strong> energies<br />
of section AB beg<strong>in</strong> to decrease ow<strong>in</strong>g to the work of the forces of the elastic<br />
deformation of the hoop whose resultant produces a centrip<strong>et</strong><strong>al</strong> force <strong>al</strong>ways<br />
directed towards the centre. The velocity of section AB forms an obtuse angle a<br />
with the force F (Fig. 366). For this reason the work of the force W 1=F!1S cos a<br />
is negative, and, consequently, the energy of the section with the mass m<br />
dim<strong>in</strong>ishes.<br />
After section AB passes through the lowermost position, it is easy to see<br />
that the work of the force F becomes positive and the energy of section AB<br />
will grow.<br />
206. L<strong>et</strong> us draw a tangent to the <strong>in</strong>ner circumference of the spool (Fig. 367)<br />
from po<strong>in</strong>t A, which is the <strong>in</strong>stantaneous axis of rotation (see Problem 173).<br />
If the directions of the thread and the<br />
tangent AC co<strong>in</strong>cide, the moment<br />
of the forces that rotate the spool around the <strong>in</strong>stantaneous axis will be zero.<br />
Therefore, if the spool is at rest, it will not rotate around the <strong>in</strong>stantaneous<br />
axis and, consequently, it will not move translation<strong>al</strong>ly.<br />
The angle a at which the motion of the spool is reversed can be found from<br />
triangle AOB; namely, s<strong>in</strong> a = ; . If the thread is <strong>in</strong>cl<strong>in</strong>ed more than a,<br />
the spool will roll to the right, If a Is sm<strong>al</strong>ler it will move to the left on<br />
condition that it does not slip. If the tension of the thread T satisfies the<br />
condition Tr ez.iR, where f is the force of friction, the spool will rema<strong>in</strong> <strong>in</strong><br />
place. Otherwise, when s<strong>in</strong> a= ~ it will beg<strong>in</strong> to rotate counterclockwise<br />
around po<strong>in</strong>t o.<br />
207. Break the hoop <strong>in</strong>to equ<strong>al</strong> sm<strong>al</strong>l sections each with a mass of L\m.<br />
Consider two symm<strong>et</strong>ric<strong>al</strong> sections (relative to the centre). All the particles of<br />
the hoop simultaneously participate <strong>in</strong> translation<strong>al</strong> motion with the velocity<br />
Fig. 366 Fig. 367
MECHANICS<br />
Fig. 368<br />
two sections, we can write<br />
241<br />
v and <strong>in</strong> rotation<strong>al</strong> motion with the<br />
velocity VI =roR. The resultant velocity<br />
V2 of the upper section of the hoop<br />
can be found as the geom<strong>et</strong>ric<strong>al</strong> sum of<br />
the velocities v and VI (Fig. 368):<br />
v:=vr+v 2 +2wI cos ex<br />
For a symm<strong>et</strong>ric<strong>al</strong> section<br />
v:=vl+v~-2vvl cos ex<br />
The tot<strong>al</strong> k<strong>in</strong><strong>et</strong>ic energy of both sections<br />
is<br />
~E,,=<br />
A I I<br />
= mvs +~= Amv 2 + f1m O)' R2<br />
2 2<br />
S<strong>in</strong>ce this expression is v<strong>al</strong>id for any<br />
for the entire hoop<br />
Mvs MR20)2<br />
EII=-2-+-2-<br />
If the hoop rolls without slipp<strong>in</strong>g, v=roR and, therefore, Ek=Mv l •<br />
2Gv 2<br />
208. Ek=-g- (",,+1).<br />
209. The cyl<strong>in</strong>der made of a denser materi<strong>al</strong> will obviously be hollow. At<br />
the same translation<strong>al</strong> velocities without slipp<strong>in</strong>g the k<strong>in</strong><strong>et</strong>ic energy of rotation<br />
will be greater <strong>in</strong> the hollow cyl<strong>in</strong>der, s<strong>in</strong>ce the particles of its mass are<br />
further from the centre and, therefore, have higher velocities.<br />
For this reason the hollow cyl<strong>in</strong>der will roll down an <strong>in</strong>cl<strong>in</strong>ed plane without<br />
slipp<strong>in</strong>g slower than the solid one. At the end of the plane, the tot<strong>al</strong><br />
k<strong>in</strong><strong>et</strong>ic energies of both cyl<strong>in</strong>ders are the same. This is possible only when<br />
the velocities are different, s<strong>in</strong>ce when the velocities are the same, the energies<br />
of translation<strong>al</strong> motion are identic<strong>al</strong>, while the energy of rotation<strong>al</strong> motion<br />
of a solid cyl<strong>in</strong>der will <strong>al</strong>ways be sm<strong>al</strong>ler than that of a hollow one.<br />
210. When the drum moves, the force of friction performs no work, s<strong>in</strong>ce<br />
the cable and the drum do not slip. Hence, the energy of the system does<br />
not change:<br />
!!-v'+GR=G-px u2+
242<br />
ANSWERS AND SOLUTIONS<br />
time t:<br />
W=f ooort oor t<br />
The change <strong>in</strong> the k<strong>in</strong><strong>et</strong>ic energy of the pulley is equ<strong>al</strong> to this work:<br />
mr 2 I [rt<br />
-2- (000- 2 (0 ) =2 (000+ 00)<br />
It<br />
Hence, co=coo--;nr.<br />
212. S<strong>in</strong>ce the force of friction I is constant, the change <strong>in</strong> the momentum<br />
of the hoop dur<strong>in</strong>g the time t is mv=ft. If the hoop rolls without slipp<strong>in</strong>g,<br />
the velocity of the po<strong>in</strong>t on the hoop which the friction force is applied to<br />
is zero. -<br />
Upon equat<strong>in</strong>g the work of the friction forces to the difference b<strong>et</strong>ween<br />
the k<strong>in</strong><strong>et</strong>ic energies, we have<br />
I<br />
2-1 coo,+O t<br />
mooo, _<br />
2 mv-- 2<br />
(see Problem 207).<br />
Upon solv<strong>in</strong>g the equations with respect to 0, we f<strong>in</strong>d:<br />
wor<br />
v=2<br />
213. The equations which show the change <strong>in</strong> the momentum and the k<strong>in</strong><strong>et</strong>ic<br />
energy of the hoop have the form:<br />
m (vo-v)=lt<br />
mv~ -mv2 = f vo+O t<br />
2 2<br />
where V=CJ)f is the velocity of the centre of the hoop when it rolls without<br />
slipp<strong>in</strong>g.<br />
Upon solv<strong>in</strong>g these equations with respect to V, we have:<br />
v=~ 2<br />
Therefore, the sought v<strong>al</strong>ue is w= ;~ .<br />
214. The equations that show the change <strong>in</strong> the momentum and the k<strong>in</strong><strong>et</strong>ic<br />
energy of the hoop have the form:<br />
mv: + mCJ)~rl<br />
m(vo-v)=ft<br />
_~_ m6)2r2 = I (vo+6)or)+(v+cur) t<br />
22222<br />
where v is the velocity of the hoop centre at any subsequent moment of time.<br />
Upon solv<strong>in</strong>g this system of equations, we f<strong>in</strong>d that<br />
f<br />
It<br />
t.t=vo-tn t, and oo=c.>o--:-rn;
MECHANICS 243<br />
If Vo < CiloT, the hoop will stop at the moment of time 't'= m;o when rotat<strong>in</strong>g<br />
with the angular velocity<br />
Vo<br />
00=(00--. Then the hoop beg<strong>in</strong>s to move<br />
r<br />
with slipp<strong>in</strong>g <strong>in</strong> the reverse direction. In a certa<strong>in</strong> time the hoop will stop<br />
slipp<strong>in</strong>g and will roll without slipp<strong>in</strong>g to the left with a translation<strong>al</strong> velocity<br />
wor-vo<br />
v 2 (see Problem 213).<br />
If Vo > CiloT, then after the time 't'= m~Cilo<br />
elapses the hoop will stop rotat<strong>in</strong>g<br />
and will move to the right with a translation<strong>al</strong> velocity v=vo-rroo. Next<br />
the hoop will rotate <strong>in</strong> the reverse direction, and <strong>in</strong> some time it will roll<br />
without slipp<strong>in</strong>g to the right. Its angular velocity will be<br />
vo-rwo<br />
00= 2r<br />
Practice shows that the loop will <strong>al</strong>so be braked when it does not sli p,<br />
We did not g<strong>et</strong> such a result s<strong>in</strong>ce the specific roll<strong>in</strong>g friction was neglected.<br />
215. S<strong>in</strong>ce the hoops do not slip, Vo (velocity of the centre of gravity of<br />
the hoops) and v (velocity of the weight) are related by the expression<br />
R<br />
vo=v R-r<br />
Assume that the weight lowers through the distance h. If the system was at<br />
rest at the <strong>in</strong>iti<strong>al</strong> moment. from the law of conservation of energy we have<br />
mv 2 2<br />
mgh=-2-+ Mvo·<br />
(see Problem 207).<br />
This expression can be used to f<strong>in</strong>d the velocit y of the weight:<br />
V<br />
v= 2mgh<br />
R 2<br />
m+2M ( R-r )<br />
Hence, the acceleration of the weight is<br />
a mg<br />
Tr<br />
m+2M ( RR<br />
The weight lowers with the acceleration a under the action of the force of<br />
gravity mg and the tension T of the str<strong>in</strong>g.<br />
The sought tension T is<br />
16 *<br />
T=m(g-a)<br />
2mMg(h)'<br />
m+2M (RRrr
244 ANSWERS AND SOLUTIONS<br />
S<strong>in</strong>ce the centre of gravity of the hoop moves with an acceleration equ<strong>al</strong> to<br />
a -RR under the action of the force T and the force of friction F, Newton's<br />
-r<br />
second law gives us the follow<strong>in</strong>g equation for the force F<br />
R<br />
F=T-Ma--<br />
R-r<br />
or<br />
Mm g (h Y(2- ¥ ) Mmg('+i)<br />
F<br />
m+2M ( RR rr m ( 1-;r+2M<br />
The friction force of rest cannot exceed the v<strong>al</strong>ue kMg. For this reason<br />
slipp<strong>in</strong>g occurs when<br />
Mmg(hr(1+N--)<br />
or<br />
m+2M (<br />
R )2 > kMg<br />
-R-<br />
-r<br />
l+!..<br />
R<br />
2~+(1-;r<br />
k
MECHANICS 245<br />
Therefore, the acceleration of the weight is<br />
a= mg RI<br />
m+M-;r<br />
If we know the acceleration
246 ANSWERS AND SOLUTIONS<br />
F F q<br />
equation of motion of the centre of gravity of a roll will have the form:<br />
m2..=F-/<br />
2<br />
It follows from this equation that 1=0.<br />
218. L<strong>et</strong> us assume, to <strong>in</strong>troduce d<strong>et</strong>erm<strong>in</strong>ancy, that mtR > m2'. In this<br />
case the first weight will lower and the second one wil rise. If the first<br />
weight lowers through a distance of h, the other one will rise<br />
The decrease <strong>in</strong> the potenti<strong>al</strong> energy will be<br />
m1gh-magh; .gh ( ml-ma; )<br />
through hi.<br />
If the absolute velocity of the first weight is v, that of the second weight<br />
will be v ~ .<br />
All the po<strong>in</strong>ts of the first step of the pulley move with the velocity v,<br />
and those of the second step with the velocity v ~. The k<strong>in</strong><strong>et</strong>ic energy of<br />
the system will be<br />
or<br />
ml+Mt 2+ m2+ M9 ,2 2<br />
2 v 2 Ri v<br />
It follows from the law of conservation of energy that<br />
ml+M1 2+ m2+Ms ,9 2- ( _ !.) h<br />
2 v 2 RIV - ml ma R g<br />
V= JI<br />
..<br />
(ml+M1)+(ma+Ma) ;2<br />
r 2<br />
( m1-mai )gh<br />
Therefore, the acceleration of the first weight is<br />
,<br />
ml-m2l[
MECHANICS<br />
247<br />
Fig. 370<br />
From the ratio Ul =!i, where a2 is the acceleration of the second weight,<br />
a2 '<br />
we can f<strong>in</strong>d that<br />
On the basis of Newton's second law, the tensions of the str<strong>in</strong>gs T 1 and Ta<br />
are equ<strong>al</strong> to:<br />
, ,2<br />
M 1 +m2 R+ R2(m2+ M2)<br />
,a<br />
ml+M1 +(ma+ M2) R2<br />
ml+M1 +-k (m1+Mz-k)1<br />
. ,2 .:<br />
ml+Ml+(ms+Ms) RZ t<br />
The force F which the system acts on the axis of the pulley with is<br />
F=T 1+T.+(M1+Ms) g<br />
219. L<strong>et</strong> the path travelled by the centre of gravity of the cyl<strong>in</strong>der dur<strong>in</strong>g<br />
he time t be equ<strong>al</strong> to s, and the velocity of the centre of gravity be v at<br />
his moment (see Fig. 370).<br />
Hence, from the law of conservation of energy we have<br />
Mvl=Mgs s<strong>in</strong> ex,<br />
Thus. the velocity is v=Vgs s<strong>in</strong> a and the acceleration a=g s~n a .<br />
The velocity of the centre of gravity of the cyl<strong>in</strong>der and the angular velocity<br />
of its rotation will be<br />
o=gs~nat and fJ)=g~~at
248 ANSWERS AND SOLUTIONS<br />
1-9. The Law of Gravitation<br />
220. Accord<strong>in</strong>g to Newton's second law, mig==F, where m, is the <strong>in</strong>erti<strong>al</strong><br />
mass-a quantity that characterizes the ability of bodies to acquire an acceleration<br />
under the <strong>in</strong>fluence of a def<strong>in</strong>ite force.<br />
mM<br />
On the other hand, accord<strong>in</strong>g to the law of gravitation F='V ~2 g, where<br />
y is the gravity constant and m g and Mg are the gravitation<strong>al</strong> masses of<br />
the <strong>in</strong>teract<strong>in</strong>g bodies. The gravitation<strong>al</strong> mass d<strong>et</strong>erm<strong>in</strong>es the force of gravity<br />
and, <strong>in</strong> this sense, can be referred to as a gravitation<strong>al</strong> charge.<br />
It is not obvious <strong>in</strong> advance that mi=m g • If this equation (proportion<strong>al</strong>ity<br />
is sufficient) is satisfied, however, the gravity acceleration is the same for <strong>al</strong>l<br />
bodies s<strong>in</strong>ce, when the gravity force is <strong>in</strong>troduced <strong>in</strong>to Newton's second law,<br />
the masses mj and m g can be cancelled, and g will be equ<strong>al</strong> to y ; .<br />
Identic<strong>al</strong> accelerations are imparted to <strong>al</strong>l bodies irrespective of their<br />
masses, only by the force of gravity.<br />
221. The gravitation<strong>al</strong> acceleration g=y ~ (see Problem 220). Assum<strong>in</strong>g<br />
that g=982 em/s, we f<strong>in</strong>d that y=6.68X 10- 8 ems,-1 S-2.<br />
222. The bodies <strong>in</strong>side the spaceship will cease to exert any pressure on<br />
the w<strong>al</strong>ls of the cab<strong>in</strong> if they have the same acceleration as the spaceship.<br />
Only the force of gravity can impart an identic<strong>al</strong> acceleration <strong>in</strong> this space<br />
to <strong>al</strong>l the bodies irrespective of their mass. Consequently, it is essenti<strong>al</strong> that<br />
the eng<strong>in</strong>e of the spaceship be shut off and there be no resistance of the<br />
extern<strong>al</strong> medium. The spaceship may move <strong>in</strong> any direction with respect to<br />
that of the force of gravity.<br />
223. The force of gravity imparts the same acceleration to the pendulum<br />
and the block. Gravity does not cause any deformations <strong>in</strong> the system dur<strong>in</strong>g<br />
free f<strong>al</strong>l<strong>in</strong>g. For this reason the pendulum will so move with respect to the<br />
block as if there is no gravitation (see the solution to Problem 222). The<br />
pendulum will move with a constant angular velocity as long as the block<br />
f<strong>al</strong>ls.<br />
224. On section Be A (Fig. 371) the force of gravity performs positive work<br />
(the angle 9 1 is acute) and the velocity of the plan<strong>et</strong> will <strong>in</strong>crease, reach<strong>in</strong>g<br />
its maximum at po<strong>in</strong>t A.<br />
~<br />
C<br />
B<br />
"<br />
"\\\<br />
,<br />
I<br />
I<br />
I I<br />
, /'<br />
Fig. 311 Fig. 372
MECHANICS 249<br />
On section ADB the force of gravity performs negative work (the angle a 2<br />
is obtuse) and the velocity of the plan<strong>et</strong> will decrease and reach its m<strong>in</strong>imum<br />
at po<strong>in</strong>t B.<br />
225. For the satellite to move <strong>al</strong>ong a closed orbit (a circle with a radius<br />
R+h) it should be acted upon by a force directed toward the centre. In<br />
our case this is the force of the Earth's attraction. Accord<strong>in</strong>g to Newton's<br />
second law,<br />
mv 2 mM<br />
R+h =1' (R+h)2<br />
where M<br />
is the mass of the Earth, R=6,370 km is the radius of the Earth,<br />
and y is the gravitation<strong>al</strong> constant.<br />
At the Earth's surface<br />
mM<br />
"W=mg<br />
Therefore,<br />
.. j--gti.2<br />
v= V R+h ~7.5 krn/s<br />
226. The resistance of the atmosphere will cause the satell ite to gradu<strong>al</strong>ly<br />
approach the Earth and the .radius of its orbit dim<strong>in</strong>ishes.<br />
S<strong>in</strong>ce this resistance is sm<strong>al</strong>l <strong>in</strong> the upper layers, the decrease of the ra..<br />
dlus is <strong>in</strong>significant dur<strong>in</strong>g one revolution. Consider<strong>in</strong>g the orbit to be approximately<br />
circular, we can write<br />
mv 2 mM<br />
y=yw<br />
where R is the radius of the orbit. Therefore, v= y Y: • i.e., the velocity<br />
of the satellite <strong>in</strong>creases with a reduction of R.<br />
This can be illustrated as follows. In view of the atmospheric resistance.<br />
a satellite placed, for example, mto a circular orbit (dotted l<strong>in</strong>e <strong>in</strong> Fig. 372)<br />
will actu<strong>al</strong>ly move <strong>al</strong>ong a certa<strong>in</strong> helix (solid l<strong>in</strong>e <strong>in</strong> Fig. 372). For this<br />
reason the projection of the force of gravity F onto the direction of the sa..<br />
tellite velocity v differs from zero. It is the work of the force F (greater<br />
than the atmospheric resistance f) that <strong>in</strong>creases the velocity.<br />
When the satellite moves <strong>in</strong> the atmosphere, its tot<strong>al</strong> mechanic<strong>al</strong> energy<br />
dim<strong>in</strong>ishes but, as the Earth is approached, the potenti<strong>al</strong> energy drops faster<br />
than the tot<strong>al</strong> energy, caus<strong>in</strong>g the k<strong>in</strong><strong>et</strong>ic energy to grow.<br />
It should be stressed that the high force of resistance <strong>in</strong> the dense layers<br />
of the atmosphere does not <strong>al</strong>low us to consider, even approximately, the<br />
motion of the satellite as rotation <strong>al</strong>ong a circle, and our conclusion is not<br />
correct. .<br />
227. If the conta<strong>in</strong>er is thrown <strong>in</strong> the direction opposite to the motion of<br />
satellite A. it will beg<strong>in</strong> to move <strong>al</strong>ong a certa<strong>in</strong> ellipse 2 <strong>in</strong>side the orbit<br />
of the satellite (Fig. 373). The period of revolution of the conta<strong>in</strong>er will be<br />
slightly less than that of satellite B. Therefore they can me<strong>et</strong> at the po<strong>in</strong>t<br />
of contact b<strong>et</strong>ween the orbits only after a great number of revolutions.<br />
The conta<strong>in</strong>er should be thrown <strong>in</strong> the direction of motion of satellite A.<br />
It will beg<strong>in</strong> to move <strong>al</strong>ong ellipse 3. The velocity u should be such that
250<br />
ANSWERS AND SOLUTIONS<br />
",'<br />
,~<br />
I<br />
I<br />
f<br />
I<br />
I<br />
I ț<br />
, •<br />
\,<br />
',2 ,<br />
' ...._-,<br />
Fig. 373<br />
Fig. 874<br />
dur<strong>in</strong>g one revolution of the conta<strong>in</strong>er. satellite B <strong>al</strong>so makes one revolution<br />
and <strong>in</strong> addition covers the distance AB This is quite possible. s<strong>in</strong>ce the<br />
period of revolution <strong>al</strong>ong ellipse 3 is somewhat greater than <strong>al</strong>ong circular<br />
orbit 1. The conta<strong>in</strong>er will me<strong>et</strong> the satellite at the po<strong>in</strong>t where orbits 3<br />
and 1 co<strong>in</strong>cide.<br />
228. Assum<strong>in</strong>g the Earth's orbit to be approximately circular, the force of<br />
gravity can be d<strong>et</strong>erm<strong>in</strong>ed by the equation F=mCJ)2R. where m is the mass<br />
of the Earth, and 6)=2.; is its angular velocity (T=365 days). On the other<br />
hand, accord<strong>in</strong>g to the law of gravitation, F=V m;: , where M is the mass<br />
of the Sun. Hence,<br />
or<br />
mM<br />
"/f2=m0l 2 R<br />
Cl)2R3<br />
M=--~2Xl()3s g<br />
y<br />
229. S<strong>in</strong>ce both the Moon and the satellite move <strong>in</strong> the field of gravity<br />
of the Earth, l<strong>et</strong> us use Kepler's third law<br />
T~ (h+H +2R o )3<br />
r: = BRa<br />
(Fig. 374). Therefore,<br />
h=2R (~:r/3-H-2Ro=220 km<br />
230. S<strong>in</strong>ce the mass of the b<strong>al</strong>l is greater than the mass of the water <strong>in</strong><br />
the same volume, the field of gravity will be greater near the b<strong>al</strong>l than away
MECHANICS<br />
251<br />
--! :-_t~l ---,'!...- -~-- ==- __--:t =~-= ).:=<br />
_ .. _------~-<br />
~- - ~ -.-, - - - -<br />
- -- -- - ----<br />
Fig. 375<br />
R R R<br />
attraction b<strong>et</strong>ween the air and the<br />
circle a (Fig. 375).<br />
from it. Therefore, the water near<br />
the b<strong>al</strong>l will be compressed addition<strong>al</strong>ly.<br />
The pressure of the water<br />
act<strong>in</strong>g on the bubble from the left<br />
will be sm<strong>al</strong>ler than the pressure<br />
that acts from the right. On the<br />
other hand, the force of gravity b<strong>et</strong>ween<br />
the air <strong>in</strong> the bubble and the<br />
b<strong>al</strong>l is greater than the· force of<br />
volume of the water shown by dotted<br />
S<strong>in</strong>ce the mass of the air <strong>in</strong> the bubble is very sm<strong>al</strong>l, the first factor<br />
becomes decisive. The bubble will be repulsed from the b<strong>al</strong>l.<br />
Conversely, the motion of the iron baII will be d<strong>et</strong>erm<strong>in</strong>ed by the fact<br />
t hat the force of attraction b<strong>et</strong>ween the air <strong>in</strong> the bubble and the b<strong>al</strong>l is<br />
much less than that b<strong>et</strong>ween the ba 11 and the volume of the water shown by<br />
dotted circle b.<br />
To c<strong>al</strong>culate the force, l<strong>et</strong> us reason as follows. A homogeneous medium<br />
(water) conta<strong>in</strong>s a sphere <strong>al</strong>most devoid of mass (the bubble) and a sphere<br />
with an excess mass (the b<strong>al</strong>l). From a form<strong>al</strong> standpo<strong>in</strong>t, this can be regarded<br />
as the presence of negative and positive masses.<br />
The force of <strong>in</strong>teraction b<strong>et</strong>ween the spheres <strong>in</strong> the water is equ<strong>al</strong><br />
to the <strong>in</strong>teraction <strong>in</strong> vacuum of a negative mass equ<strong>al</strong> to the mass of the<br />
water <strong>in</strong> the bubble and a positive mass equ<strong>al</strong> to the mass of the<br />
that exceeds the mass of the water <strong>in</strong> the same volume. .<br />
Therefore,<br />
ml (m<br />
F 2-ml)<br />
=-y R2<br />
iron b<strong>al</strong>l<br />
Here ml is the mass of the water <strong>in</strong> a sphere with a radius r, and m2 is the<br />
mass of the iron baII.<br />
231. The field of gravity is sm<strong>al</strong>ler near the bubble than <strong>in</strong> a homogeneous<br />
liquid, and the liquid is compressed less. For this reason one bubble will<br />
move <strong>in</strong>to the volume of liquid near the other one, and the bubbles will be<br />
mutu<strong>al</strong>ly attracted.<br />
Two bubbles <strong>in</strong> a homogeneous liquid with negligibly sm<strong>al</strong>l masses can<br />
be considered from a form<strong>al</strong> standpo<strong>in</strong>t as negative masses superimposed upon<br />
the positive mass m of the medium <strong>in</strong> the volume of a bubble:<br />
(-m)(-m) m 2<br />
F=V R2 y R2<br />
232. If the b<strong>al</strong>l were solid, the· force of gravity F 1 =1' ~:" where<br />
M= ~ :tR3p is the mass of the b<strong>al</strong>l without a spheric<strong>al</strong> space. The presence<br />
of this space is equiv<strong>al</strong>ent to the appearance of a force of repulsion<br />
F2=y~~m , where m'=: :trsp, and s is the distance b<strong>et</strong>ween the centre of<br />
the .space and the materi<strong>al</strong> particle.<br />
The sought force F is the geom<strong>et</strong>ric<strong>al</strong> sum of the forces F 1 and Fa<br />
(Fig. 376).
(1<br />
A6000-"""'------------..li.....-<br />
1<br />
Accord<strong>in</strong>g to the cos<strong>in</strong>e rule,<br />
F=V F~+F:-2FIF2 cos p=<br />
Fig. 976<br />
4 .. /~R~8---r6---2""!!!'R..... 8r-S-c-o-s-=p<br />
=3"nymp V zr+(12-d 2) 2 11(lI-d 2) ~5.8XIO-t gf<br />
233. The sought force of attraction is the geom<strong>et</strong>ric<strong>al</strong> sum of the forces<br />
created by separate elements of the sphere. The sm<strong>al</strong>l elements 01 and 0'2<br />
(Fig. 377) are cut out of the sphere as cones with vertices at po<strong>in</strong>t A obta<strong>in</strong>ed<br />
when the generatrix Be revolves around axis 8 182 • The areas of the elements<br />
( AS )2ro (AS )1ro (AS )2 rop<br />
are 1 and 2 respectively and their masses are _ .....1_~<br />
cos (Xl cos (X2 " cos (Xl<br />
and (AS 2 )2 rop , where (J) is the solid angle at which both elements can be<br />
cos (Xl<br />
seen from po<strong>in</strong>t A; p is the surface density of the sphere (the mass<br />
per unit<br />
of surface); L. (Xl= L. (X2' s<strong>in</strong>ce StOSI is an isosceles triangle. The forces of<br />
attraction created by the elements are equ<strong>al</strong>, respectively, to:<br />
and<br />
m (AS 1 )2
MECHANICS<br />
253<br />
Fig. 377<br />
Therefore, the sought force<br />
4n<br />
"'3 omr<br />
This force decreases <strong>in</strong> proportion to , as the centre of the Earth is approached.<br />
1-10. Hydro- and Aerostatics<br />
235. The level of the water will not change because the quantity of water<br />
displaced will rema<strong>in</strong> the same.<br />
236. Equilibrium will not be violated, s<strong>in</strong>ce accord<strong>in</strong>g to Pasc<strong>al</strong>'s law the<br />
pressure on the bottom of the vessel will be the same everywhere.<br />
237. (1) S<strong>in</strong>ce the piece of ice floats, the weight of the water displaced by<br />
it is equ<strong>al</strong> to the weight of the ice itself or the weight of the water it produces<br />
upon melt<strong>in</strong>g. For this reason the water formed by the piece of ice<br />
Will occupy a volume equ<strong>al</strong> to that of the submerged portion, and the level<br />
of the water will not change.<br />
(2) The volume of the submerged portion of the piece with the stone is<br />
greater than the sum of the volumes of the stone and the water produced by<br />
the melt<strong>in</strong>g ice. Therefore, the level of the water <strong>in</strong> the glass will drop.<br />
(3) The weight of the displaced water is equ<strong>al</strong> to that of the ice (the<br />
weight of the air <strong>in</strong> the bubble may be neglected). For this reason, as <strong>in</strong><br />
case (1), the level of the water will not change.<br />
238. In the first case the weight of the body submerged <strong>in</strong>to the liquid is<br />
0 1 =(1'-1'1) V and <strong>in</strong> the second Q9~('Y-1'2) V, where V is the volume of<br />
the body.<br />
Therefore,<br />
0 21'1 - 0 11'2<br />
",=~-~----<br />
G 2 - Q 1<br />
239. The ice can be supported by the edge of the shore only <strong>in</strong> sm<strong>al</strong>l<br />
ponds. It will <strong>al</strong>ways float <strong>in</strong> the middle of a large lake. The ratio b<strong>et</strong>ween<br />
the densities of the ice and the water is 0.9. Therefore, n<strong>in</strong>e-tenths of the
254 ANSWE~S AND SOLUTIONS<br />
entire thickness of the ice will be <strong>in</strong> the water. The distance from the surface<br />
of the ice to the water is one m<strong>et</strong>re.<br />
240. After the stone is taken out, the match-box becomes lighter by the<br />
weight of the stone, and, consequently, the volume of water· the box displaces<br />
decreases by VI =G/'\'I' where G is the weight of the stone and '\'1 the specific<br />
weight of the water. When dropped <strong>in</strong>to the water,<br />
the stone will displace<br />
a volume of water equ<strong>al</strong> to its own volume V 2=GI'\'2' where '\'2 is the<br />
specific weight of the materi<strong>al</strong> of the stone. S<strong>in</strong>ce V2 > VI' then VI > V2.<br />
Therefore, the level of the water <strong>in</strong> the cup will lower.<br />
241. In both cases the pumps perform identic<strong>al</strong> work, s<strong>in</strong>ce the same<br />
amount of pumped <strong>in</strong> water rises to the same level.<br />
242. The <strong>in</strong>verted L will be stable on the bottom of the empty vessel,<br />
s<strong>in</strong>ce a perpendicular dropped from the centre of gravity of the figure is<br />
with<strong>in</strong> the limits of the support<strong>in</strong>g area. As water is poured <strong>in</strong>to the vesseI,<br />
the expulsion force act<strong>in</strong>g on the rectangle wiIJ grow <strong>in</strong> magnitude (it is<br />
assumed that the water can flow underneath the figure). When the depth<br />
of the water <strong>in</strong> the vessel is O.5a, the sum of the moments of the forces<br />
which tend to turn the body clockwise will be equ<strong>al</strong> to the sum of those<br />
act<strong>in</strong>g <strong>in</strong> the opposite direction. As the vessel is filled further, the figure<br />
will f<strong>al</strong>l.<br />
243. The length of the tube x can be found from the condition<br />
iX=Yo (x-h), which shows the equ<strong>al</strong>ity of the pressures at the depth of<br />
the lower end of the tube. Here Vo is the specific weight of the water.<br />
Hence,<br />
x=-.1i!-= 50 ern<br />
Vo-y<br />
244. L<strong>et</strong> us separate a column with a height of h <strong>in</strong>side the water<br />
(Fig. 378). The equation of motion of this column has the form: ma=mg-pA,<br />
where m= pAh is the mass of the water and p is the pressure at the depth h.<br />
Therefore, p=ph (g-a).<br />
245. In accordance with the solution of Problem 244, the force of expulsion<br />
can be written as follows: F=pV (g-a), where V is the volume of the<br />
submerged portion of the body. The equation of motion of a float<strong>in</strong>g body<br />
with a mass M has the form: Ma=Mg-pV (g-a).<br />
A,<br />
A-<br />
1/---<br />
---8<br />
~~~;';~..i~~·----D<br />
Flg.378<br />
Fig. 979
MECHANICS 255<br />
Hence, V= ~, as <strong>in</strong> a stationary vessel, and the body does not rise to<br />
p<br />
the surface.<br />
246. If the tank were at rest or moved uniformly, the pressure at the<br />
depth h would be equ<strong>al</strong> to PI =pgh.<br />
On the other hand, if the tank moved with an acceleration and the<br />
force of gravity were absent, the pressure at po<strong>in</strong>t A would be equ<strong>al</strong> to<br />
P2 =p<strong>al</strong>. It is this pressure that would impart, <strong>in</strong> conformity with Newton's<br />
second law, the required acceleration a to the column of water with a length i.<br />
In accelerated motion of the tank, both the pressure PI and the pressure<br />
P2 appear <strong>in</strong> the field of gravity. Accord<strong>in</strong>g to Pasc<strong>al</strong>'s law, the pressure<br />
<strong>in</strong> the water is the same <strong>in</strong> <strong>al</strong>l directions. For this reason the pressures PI<br />
and P2 are summated, and the result <strong>in</strong>g pressure at po<strong>in</strong>t A is P== P(gh+at).<br />
247. Us<strong>in</strong>g the law of conservation of energy and Archimedes' pr<strong>in</strong>ciple.<br />
we obta<strong>in</strong> the follow<strong>in</strong>g equation<br />
mgx= ( : nR3p-m) gh<br />
Where p is the density of the water and x is the sought height.<br />
Hence,<br />
( ~ nR3p-m) h<br />
x= ~---------<br />
m<br />
248. If the level of the water <strong>in</strong> the vessel is the same, the level of the<br />
mercury will <strong>al</strong>so be the same before the piece of wood is dropped.<br />
Dropp<strong>in</strong>g of the piece of wood gives the same result as add<strong>in</strong>g of the<br />
amount of water that will be displaced by this piece, i. e., the amount of<br />
water equ<strong>al</strong> to it by weight. Therefore, if the cross sections of the vessels<br />
are the same, the levels of the water and the mercury <strong>in</strong> both vessels wi II<br />
co<strong>in</strong>cide.<br />
If the cross sections are different, the water will be higher and the<br />
mercury lower <strong>in</strong> the vessel with a sm<strong>al</strong>ler cross section. This will occur<br />
because the pressure on the surface of the mercury will <strong>in</strong>crease differently<br />
when amounts of water equ<strong>al</strong> <strong>in</strong> weight and volume are added to the<br />
vessels with various cross sections.<br />
249. After the block is dropped <strong>in</strong>to the broad vessel, the level of the<br />
mercury <strong>in</strong> both vessels will rise by the amount x and occupy position AB<br />
(Fig. 379).<br />
The required height of the water column <strong>in</strong> the broad vessel is d<strong>et</strong>erm<strong>in</strong>ed<br />
by the equ<strong>al</strong>ity of the pressures, for example, at level CD<br />
(y+x) PIg=hp2g<br />
where PI is the density of the mercury and P2 that of the water. The vaIue<br />
of y can be found by us<strong>in</strong>g the condition that the volume of the mercury<br />
is constant: .<br />
(x+ y) Al =V 2<br />
Where V2 is the volume of the mercury displaced by the block after the<br />
water is poured <strong>in</strong>.
256 ANSWERS AND SOLUTIONS<br />
If the water covers the block entirely,<br />
then accord<strong>in</strong>g to Archimedes'<br />
pr<strong>in</strong>ciple,Vopog=V2Plg+ (Vo- V2) P2i,<br />
where Po is the density of iron.<br />
Solv<strong>in</strong>g the equations, we obta<strong>in</strong><br />
h- PI (PO-P2) Vo<br />
- P2 (Pt-P2) At<br />
If the water does not cover the<br />
block, Archimedes' pr<strong>in</strong>ciple can be<br />
Fig. 380<br />
written as VoPog=V2Plg+hAp~,<br />
where A=V:/ 3 is the area of a block<br />
face. In this case the sought height h PoV o<br />
P2 (At +V:/ 8 )<br />
The first solution is true when A
MECHANICS<br />
257<br />
Fig. 381 Fig. 382<br />
252. A liquid moves <strong>in</strong> a syphon by the action of the forces of cohesion<br />
b<strong>et</strong>ween the elements of the liquid. The liquid <strong>in</strong> the long elbow outweighs<br />
the liquid <strong>in</strong> the short one, and pumps it over. It could be assumed on this<br />
basis that water can be pumped over a w<strong>al</strong>l of any height with the ski of<br />
a syphon. This is not so, however. At a lift<strong>in</strong>g height of 10 m<strong>et</strong>res the<br />
pressure <strong>in</strong>side the liquid becomes zero. The air bubbles <strong>al</strong>ways present <strong>in</strong><br />
water will beg<strong>in</strong> to expand, and the water column will be broken, thus<br />
stopp<strong>in</strong>g the action of the syphon.<br />
253. The device will first act as a syphon and the water will flow through<br />
the narrow pipe <strong>in</strong>to the reservoir. Then an air bubble will slip through A<br />
and divide the liquid <strong>in</strong> the upper part <strong>in</strong>to two portions. After this the<br />
liquid will no longer flow out. .<br />
254. The pressure of the water directly under the piston of each pump<br />
is less than atmospheric pressure by pg (H +h), where p is the density of<br />
water. Therefore, to keep the piston <strong>in</strong> equilibrium, it should be pulled<br />
upward with the force F = pg (H+h) A, where A is the area of the piston.<br />
Hence, a greater force should be applied to the pistons with a greater area.<br />
255. The pressure on the bottom is P=pgbH+h) (Fig. 382). On the<br />
other hand, s<strong>in</strong>ce the vessel is a cyl<strong>in</strong>der, p= :R~.<br />
The height h can be d<strong>et</strong>erm<strong>in</strong>ed if we equate the forces act<strong>in</strong>g on the piston<br />
pghn (RI-r 2 )= G<br />
Hence,<br />
1 ( G r 2 )<br />
nR2p g R2~r2<br />
H=-- m----- ~ 10 cm<br />
256. To prevent flow<strong>in</strong>g out of the water, the vessel should be given such<br />
an acceleration at which the surface of the water takes the position shown<br />
<strong>in</strong> Fig. 383. The maximum volume of the water is ~: and the mass of the<br />
entire system is M+be'; p. The required acceleration can be found from the<br />
condition that the sum of the forces act<strong>in</strong>g on a sm<strong>al</strong>l element of the water<br />
with a mass 11m near the surface is directed horizont<strong>al</strong>ly (Fig. 383).<br />
17-2042
258 ANSWERS AND SOLUTIONS<br />
N Accord<strong>in</strong>g to Newton's second law,<br />
~ma==t1mg tan a.<br />
Therefore, the sought force is<br />
F = ( M + b~~ P) g ~<br />
257. The lower part of a chamber<br />
is filled with denser air. The air leaves<br />
the chambers <strong>in</strong> the upper part. The<br />
pressure is gradu<strong>al</strong>ly equa lized. The<br />
Fig. 383<br />
mach<strong>in</strong>e will cont<strong>in</strong>ue to function only<br />
as long as the difference <strong>in</strong> the pressures<br />
b<strong>et</strong>ween the two ha1ves of the<br />
vessel is sufficient to raise the water <strong>al</strong>ong the" tube to the upper portion.<br />
258. Here the disk is not asymm<strong>et</strong>ric<strong>al</strong> and the air pressure on the righthand<br />
side of the disk will be greater than on the left-hand one. The surplus<br />
force of the pressure act<strong>in</strong>g on the right-hand side of the disk is F=(Pt-P2) A,<br />
where A is the cross-section<strong>al</strong> area of the chamber. The weight of the chambers<br />
filled with water cannot exceed 0 = pgAh. S<strong>in</strong>ce h ~ 0 1 - O 2 , then F ~ G.<br />
pg<br />
The disk will beg<strong>in</strong> to rotate counterclockwise, and the chambers will rise<br />
from the bottom of the vessel to the top filled with air. The disk will rotate<br />
counterclockwise until the reduced difference of the pressure can no longer<br />
lift the water to the height h.<br />
259. The bottom of the cyl<strong>in</strong>dric<strong>al</strong> vessel will f<strong>al</strong>l off <strong>in</strong> <strong>al</strong>l three cases,<br />
s<strong>in</strong>ce the pressure exerted on the bottom from the top will <strong>al</strong>ways be equ<strong>al</strong><br />
to 1 kgf. In the vessel narrow<strong>in</strong>g upward the bottom will f<strong>al</strong>l off only when<br />
oil is poured <strong>in</strong>, s<strong>in</strong>ce its level here will be higher than <strong>in</strong> the cyl<strong>in</strong>dric<strong>al</strong><br />
vessel. In the vessel widen<strong>in</strong>g toward the top the bottom will f<strong>al</strong>l off when<br />
the mercury is poured <strong>in</strong>, because its level will be higher than <strong>in</strong> the cyl<strong>in</strong>dric<strong>al</strong><br />
vessel. This will <strong>al</strong>so occur when the weight is put <strong>in</strong>. In this case<br />
the weight will be distributed over a sm<strong>al</strong>ler area than <strong>in</strong> the other two cases.<br />
260. The read<strong>in</strong>g of the b<strong>al</strong>ance will <strong>in</strong>crease if the mean density of the<br />
body be<strong>in</strong>g weighed is less than the density of the weights. The read<strong>in</strong>g will<br />
decrease if the mean density of the body is greater than that of the weights.<br />
The equilibrium of the b<strong>al</strong>ance will not be disturbed if the weights and the<br />
body have the same mean density. .<br />
261. The man's aim will not be achieved, s<strong>in</strong>ce, while <strong>in</strong>creas<strong>in</strong>g the expulsive<br />
force. he will <strong>al</strong>so considerably <strong>in</strong>crease the weight of the tube (the<br />
density of the compressed air <strong>in</strong> the tube is greater than that of the atrnospheric<br />
air).<br />
262. The true weight of the body is<br />
The error is equ<strong>al</strong> to<br />
G=Gt+yo (v- ~:)<br />
d! 801.16 gf<br />
G-O t -0-100% ~O.14%<br />
263. When the atmospheric pressure changes, the Archimedean force act<strong>in</strong>g<br />
on the barom<strong>et</strong>ers from the side of the air will change ow<strong>in</strong>g to the change
MECHANICS 259<br />
<strong>in</strong> the density of the air and <strong>in</strong> the volume of the barom<strong>et</strong>ers when the level<br />
of the mercury changes <strong>in</strong> their open parts.<br />
If <strong>al</strong>l the conditions <strong>in</strong> the problem are taken <strong>in</strong>to account, the barom<strong>et</strong>ers<br />
have not only the same weight, but <strong>al</strong>so the same volume. For this reason<br />
the change <strong>in</strong> the expulsion force due to the first reason will be identic<strong>al</strong> for<br />
each of them. The change <strong>in</strong> the volumes will be different. To change the<br />
difference <strong>in</strong> the levels by a certa<strong>in</strong> amount <strong>in</strong> a U-shaped barom<strong>et</strong>er, the<br />
level of the mercury <strong>in</strong> each elbow should change by only h<strong>al</strong>f of this amount.<br />
In a cup barom<strong>et</strong>er the level of the mercury <strong>in</strong> the cup changes negligibly,<br />
and <strong>in</strong> the tube practic<strong>al</strong>ly by the entire change <strong>in</strong> the difference of the<br />
levels. Here the change of the volume of the mercury <strong>in</strong> the tube should be<br />
the same as <strong>in</strong> the cup.<br />
Therefore, the change of the volume <strong>in</strong> the cup barom<strong>et</strong>er will <strong>al</strong>so become<br />
less, and it will therefore outweigh the U-shaped barom<strong>et</strong>er.<br />
264. The norm<strong>al</strong> atmospheric pressure is approximately equ<strong>al</strong> to 1 kgf/cm 2 •<br />
This means that an atmospheric column of air with an area of I em 2 weighs<br />
1 kgf. If we know the surface of the Earth, we can f<strong>in</strong>d the weight of its<br />
atmosphere.<br />
The Earth's surface A=4nR', where R=6,370 km is the mean radius of<br />
the Earth.<br />
ThE' weight of the atmosphere a 2:: 4nRzx t kgf/cm 2 "E!!5 5X lOll ton f.<br />
265. If the man stands on the mattress, his weight will be distributed<br />
over a sm<strong>al</strong>ler area (that of his fe<strong>et</strong>) than if he lies down. Therefore, the<br />
state of equilibrium will s<strong>et</strong> <strong>in</strong> <strong>in</strong> the first case at a higher air pressure <strong>in</strong><br />
the mattress than <strong>in</strong> the second.<br />
286. L<strong>et</strong> us first consider the tube <strong>in</strong>flated with air (Fig. 384a shows a<br />
cross section of the tube). For sections AB and CD of the tube to be <strong>in</strong><br />
equilibrium it is obviously necessary that the tension of the expanded w<strong>al</strong>ls<br />
of the tube T b<strong>al</strong>ance the excess pressure <strong>in</strong>side the tube p,<br />
L<strong>et</strong> us now consider the forces that act on sections AB and CD when the<br />
tube is fitted onto a loaded wheel (Fig. 384b). The distribution of the forces<br />
'0<br />
I<br />
Fig. 384<br />
Ia)<br />
(b)<br />
'~<br />
Fig. 385<br />
17 *
260 ANSWERS AND SOLUTIONS<br />
act<strong>in</strong>g on AB does not. appreciably change <strong>in</strong> the top of the tube. There<br />
will be a difference <strong>in</strong> the bottom. Section CD will be acted upon by an<br />
elastic force from the side of the rim equ<strong>al</strong> to the load applied to the wheel<br />
(the weight of the wheel and one-fourth of the weight of the motor vehicle).<br />
This addition<strong>al</strong> force flattens the tube and the angle b<strong>et</strong>ween the forces T<br />
tension<strong>in</strong>g the rubber <strong>in</strong>creases. The tot<strong>al</strong> force of tension act<strong>in</strong>g on CD<br />
dim<strong>in</strong>ishes and the excess pressure of the air <strong>in</strong> the tube equ<strong>al</strong>izes- the force<br />
of tension, and <strong>al</strong>so the weight of the wheel and of part of the motor vehic Ie.<br />
Thus, the rim does not lower<br />
because it is supported by the excess pressure<br />
of the air <strong>in</strong> the tube. In the top of the tube the excess pressure is<br />
equ<strong>al</strong>ized by the tension of the tube w<strong>al</strong>ls, and <strong>in</strong> the bottom it equ<strong>al</strong>izes<br />
both the reduced tension of the w<strong>al</strong>ls and the force applied to the wheel.<br />
267. The force per unit of length with which the cyl<strong>in</strong>dric<strong>al</strong> portion of<br />
the boiler is str<strong>et</strong>ched <strong>in</strong> a direction perpendicular to axis 00 1 is<br />
2Rl<br />
fl=21 P= pR<br />
where 2RI Is the cross-section<strong>al</strong> area ABeD of the boiler and p is the pressure<br />
<strong>in</strong>side it (Fig. 385); 2Rlp Is the force act<strong>in</strong>g on one h<strong>al</strong>f of the cyl<strong>in</strong>der<br />
(see Problem 122).<br />
The maximum force per unit length of the hemispheric<strong>al</strong> heads can be found<br />
from the formula<br />
nRt pR II<br />
"=2nR P =T =2'<br />
Consequently, the heads can withstand twice the pressure that the cyl<strong>in</strong>dric<strong>al</strong><br />
portion of the boiler does if their w<strong>al</strong>ls are equ<strong>al</strong>ly thick. For the<br />
strength of <strong>al</strong>l the parts of the boiler to be identic<strong>al</strong>, the thickness of the<br />
head w<strong>al</strong>ls may be h<strong>al</strong>f that of the cyl<strong>in</strong>dric<strong>al</strong> w<strong>al</strong>ls, l. e., 0.25 em.<br />
268. The shape of the boiler should be such that the force applied per<br />
unit length of its cross section is m<strong>in</strong>imum. This force is f=P~ , where A<br />
is the cross-section<strong>al</strong> area of the boiler, 1 the peri m<strong>et</strong>er of the section and p<br />
the pressure of the steam.<br />
The force f will be m<strong>in</strong>imum with the sm<strong>al</strong>lest ratio b<strong>et</strong>ween the crosssection<strong>al</strong><br />
area and the perim<strong>et</strong>er.<br />
As is known, this ratio will be m<strong>in</strong>imum for a circle. It is <strong>al</strong>so known<br />
that a circle can be obta<strong>in</strong>ed by cutt<strong>in</strong>g a sphere with any plane. Therefore,<br />
a sphere is the most advantageous shape for the boiler.<br />
269. The ceil<strong>in</strong>g of a stratosphere b<strong>al</strong>loon is d<strong>et</strong>erm<strong>in</strong>ed not by the maximum<br />
<strong>al</strong>titude which it can ascend to, but by the <strong>al</strong>titude ensur<strong>in</strong>g a safe<br />
velocity of land<strong>in</strong>g. The envelope of a stratosphere b<strong>al</strong>loon is filled with a<br />
light gas (hydrogen or helium) only partly, s<strong>in</strong>ce when the b<strong>al</strong>loon ascends,<br />
the gas <strong>in</strong> the envelope expands and forces out the air, mak<strong>in</strong>g it possible<br />
to ma<strong>in</strong>ta<strong>in</strong> the lift<strong>in</strong>g force approximately constant. At a certa<strong>in</strong> <strong>al</strong>titude,<br />
the gas will fill the entire envelope. Even after this the lift<strong>in</strong>g force of the<br />
b<strong>al</strong>loon cont<strong>in</strong>ues to <strong>in</strong>crease at the expense of the gas flow<strong>in</strong>g out from the<br />
bottom hole <strong>in</strong> the envelope. The weight of the b<strong>al</strong>loon decreases and it will<br />
reach its ceil<strong>in</strong>g only after a certa<strong>in</strong> amount of gas has leaked out.<br />
For the b<strong>al</strong>loon to descend, some gas should addition<strong>al</strong>ly be expelled through<br />
the upper v<strong>al</strong>ve <strong>in</strong> the envelope, so that the lift<strong>in</strong>g force is only slightly<br />
sm<strong>al</strong>ler than the weight of the b<strong>al</strong>loon. At a sm<strong>al</strong>l <strong>al</strong>titude the velocity of
MECHANICS 261<br />
descent will be too high, s<strong>in</strong>ce the volume of the gas decreases and less of<br />
it rema<strong>in</strong>s <strong>in</strong> the b<strong>al</strong>loon than dur<strong>in</strong>g the ascent. The b<strong>al</strong>last is dropped to<br />
reduce the velocity of descent.<br />
1·11. Hydro- and Aerodynamics<br />
270. L<strong>et</strong> us denote the distance from the level of the water to the upper<br />
hole by h, the sought distance from the vessel to the po<strong>in</strong>t where the streams<br />
<strong>in</strong>tersect by x, and the distance from the level of the water <strong>in</strong> the vessel to<br />
this po<strong>in</strong>t by y (Fig. 386).<br />
The po<strong>in</strong>t of <strong>in</strong>tersection will rema<strong>in</strong> at the same place if the level of<br />
the water <strong>in</strong> the vessel does not change. This will occur if Q= AVI+ Av 2 ,<br />
where VI = Y 2gh and V2 = Y 2g (H +h) are the outflow velocities of the<br />
streams from the holes.<br />
On the basis of the laws of k<strong>in</strong>ematics,<br />
where t 1 and t 2 are the times dur<strong>in</strong>g which the water f<strong>al</strong>ls from the holes<br />
to the po<strong>in</strong>t of <strong>in</strong>tersection.<br />
Hence,<br />
1 ( Q2 2gA2) .<br />
%=2 2gA2 - H2~ = 120 em<br />
I ( Q2 2gA2)<br />
Y='2 2gAI+ H2 Q1 =130 em<br />
271. The velocity of water outflow from a hole is v= Y2gh. The impulse<br />
of the force act<strong>in</strong>g from the side of the vessel on the outftow<strong>in</strong>g water<br />
FAt= Amv, where Am=pAvAt is the mass of the water ftow<strong>in</strong>g out dur<strong>in</strong>g<br />
the time At. Hence, F=pv 2 A= 2pghA. The pressure at the bottom p=pgh<br />
and therefore F=2pA. The same force acts on the vessel from the side of<br />
the stream.<br />
-<br />
-----1-<br />
• --.....=--- ..s..~<br />
J.;<br />
-~_ .... - II<br />
------.._-<br />
- ~~ ~ ......<br />
-------...:.....-- H<br />
--..- ............. -<br />
!I<br />
_-.-.-....... ... ~<br />
Fig. 886
262 ANSWERS AND SOLUTIONS<br />
Thus, the water acts on the w<strong>al</strong>l with the hole with a force 2pA sm<strong>al</strong>ler<br />
than that act<strong>in</strong>g on the opposite w<strong>al</strong>l, and not with a force sm<strong>al</strong>ler by pA<br />
as might be expected. This is due to a reduction <strong>in</strong> the pressure act<strong>in</strong>g on<br />
the w<strong>al</strong>l with the hole, s<strong>in</strong>ce the water flows faster at this w<strong>al</strong>l.<br />
The vessel will beg<strong>in</strong> to move if kG < 2pA or<br />
k < 2pghA<br />
G<br />
272. Accord<strong>in</strong>g to Newton's second law, the equ<strong>al</strong>ity pA o = 2pA should<br />
exist. Therefore, if the liquid flows out through the tube, the cross-section<strong>al</strong><br />
area of the stream should be h<strong>al</strong>ved<br />
A _ A o<br />
-2<br />
This compression of the stream can be expla<strong>in</strong>ed as follows.<br />
The extreme streaml<strong>et</strong>s of the Iiquid approach<strong>in</strong>g the tube from above<br />
cannot, <strong>in</strong> view of <strong>in</strong>ertia, flow around the edge of the tube directly adher<strong>in</strong>g<br />
to its w<strong>al</strong>ls, and move towards the centre of the stream. Under the pressure<br />
of, the particles nearer to the centre of the stream, the l<strong>in</strong>es of flow straighten<br />
out and a contracted stream of the liquid flows <strong>al</strong>ong the tube.<br />
273. By neglect<strong>in</strong>g splash<strong>in</strong>g of the water, we thus assume the impact of<br />
the stream aga<strong>in</strong>st the w<strong>al</strong>l to be absolutely <strong>in</strong>elastic. Accord<strong>in</strong>g to Newton's<br />
second law. the change <strong>in</strong> the momentum of the water dur<strong>in</strong>g the time ~t is<br />
limv=FM, where lim=p n:' vM is the mass of the water flow<strong>in</strong>g dur<strong>in</strong>g<br />
the time ~t through the cross section of the pipe.<br />
Hence,<br />
plld 2<br />
F=--v 2 === 8 gi<br />
4<br />
274. When the gas flows <strong>al</strong>ong the pipe (Fig. 387), its momentum changes<br />
<strong>in</strong> direction, but not <strong>in</strong> magnitude.<br />
A mass of pAv passes <strong>in</strong> a unit of time through cross section I of the<br />
vertic<strong>al</strong> part. This mass br<strong>in</strong>gs <strong>in</strong> the momentum PI=pAUVl where VI is the<br />
J<br />
p<br />
Fig. 387<br />
H-h<br />
A=Po-pgh<br />
Fig. 388<br />
H
MECHANICS<br />
263<br />
vector of the velocity with which the gas flows <strong>in</strong> the vertic<strong>al</strong> part, numeric<strong>al</strong>ly<br />
equ<strong>al</strong> to the given velocity v.<br />
Dur<strong>in</strong>g the same time the momentum P2=pAvv 2 is carried away through<br />
cross section II. Here v 2 is the vector of .velocity <strong>in</strong> the horizont<strong>al</strong> part,<br />
<strong>al</strong>so numeric<strong>al</strong>ly equ<strong>al</strong> to v.<br />
The change <strong>in</strong> the momentum is equ<strong>al</strong> to the impulse of the force F that<br />
acts from the side of the pipe on the gas: F=pAv (Vi-VI)' In magnitude<br />
F=pAv 2 Y 2.<br />
Accord<strong>in</strong>g to Newton's third law, the gas acts on the pipe with the same<br />
force. This force is directed oppositely to the pipe bend.<br />
275. The <strong>in</strong>iti<strong>al</strong> velocity of the water with respect to the blade is<br />
v= Y2gh-roR. Therefore, a mass of water m=pA (Y2gh-roR) imp<strong>in</strong>geson<br />
the blade <strong>in</strong> a unit of time. After the impact, the velocity of the water<br />
with reference to the blade is zero, and for this reason the change <strong>in</strong> the<br />
momentum of the water <strong>in</strong> a unit of time is mv. Accord<strong>in</strong>g to Newton's<br />
second law, the sought force is<br />
F=pA (y 2gh-roR)2<br />
276. At the first moment the ship will beg<strong>in</strong> to move to the right, s<strong>in</strong>ce<br />
the pressure on the starboard side dim<strong>in</strong>ishes by 2pA, where p is the pressure<br />
at the depth h of the hole, and A is its area (see Problem 271). As soon as<br />
the stream of water reaches the opposite w<strong>al</strong>l. this w<strong>al</strong>l will be acted upon<br />
by the force F=pAv', where v is the velocity of the stream with respect to<br />
the ship (see Problem 273). The force F is slightly greater than 2pA, s<strong>in</strong>ce<br />
tJ > Y2gh because the ship moves towards the stream. As a result, the motion<br />
will beg<strong>in</strong> to r<strong>et</strong>ard.<br />
277. The velocity of the liquid <strong>in</strong> the pipe is constant <strong>al</strong>ong the entire<br />
cross section because the liquid has a low compressibility and the stream is<br />
cont<strong>in</strong>uous. This velocity is v=y 2gH.<br />
The velocity of the liquld <strong>in</strong> the vessel is practic<strong>al</strong>ly zero, s<strong>in</strong>ce its area<br />
is much greater than the cross-section<strong>al</strong> area of the pipe.<br />
Therefore, a pressure jump which we sh<strong>al</strong>l denote by PI - p" should exist<br />
on the vessel-pipe boundary. The work of the pressure forces causes the veloelt'i<br />
to change from zero to Y 2gB.<br />
On the basis of the law of conservation of energy.<br />
Amv<br />
-2-= t<br />
(Pl - PI) AAh<br />
where A is the cross-section<strong>al</strong> area of the pipe, Ah is the height of a sm<strong>al</strong>l<br />
element of the liquid, and Am=pAAh is the mass of this element.<br />
Hence, .<br />
pv 2<br />
T=PI-pf,=pgH<br />
S<strong>in</strong>ce the flow velocity is constant, the pressure <strong>in</strong> the pipe changes accord<strong>in</strong>g<br />
to the law<br />
P=Po-pg(h-x)<br />
as <strong>in</strong> a liquid at rest. Here Po is the atmospheric pressure and x is the<br />
distance from the upper end of the pipe.
264 ANSWERS AND SOLUTiONS<br />
The change of pressure <strong>in</strong> height is shown In Fig. 388. The pressure is<br />
laid off <strong>al</strong>ong the axis of ord<strong>in</strong>ates, and the distance from the surface of the<br />
liquid <strong>in</strong> the vessel <strong>al</strong>ong the axis of abscissas.<br />
278. The water flow<strong>in</strong>g out of the pipe dur<strong>in</strong>g a sm<strong>al</strong>l <strong>in</strong>terv<strong>al</strong> of time At<br />
will carrywith it the momentum Ap=pAvIAt, where v= Y2gH is the velocity<br />
of the outflowlng stream (see Problem 277). Accord<strong>in</strong>g to Newton's<br />
law, FAt = 2pgHAAt. The same force will act on the vessel with the water<br />
from the side of the outflow<strong>in</strong>g stream. Therefore, the read<strong>in</strong>g of the b<strong>al</strong>ance<br />
will decrease by 2pgHA at the <strong>in</strong>iti<strong>al</strong> moment.<br />
279. At the first mcment when the stream has not y<strong>et</strong> reached the pan,<br />
equilibrium will be violated. The pan will sw<strong>in</strong>g upwards s<strong>in</strong>ce the water<br />
flow<strong>in</strong>g out of the vessel no longer exerts pressure on its bottom.<br />
When the stream reaches the pan, equilibrium will be restored. L<strong>et</strong> us<br />
consider an element of the stream with the mass Am. This element, when<br />
f<strong>al</strong>l<strong>in</strong>g onto the pan, imparts to it an impulse AmY2gh In a vertic<strong>al</strong> direction,<br />
where h is the height of the cock above the pan. On the other hand,<br />
after leav<strong>in</strong>g the vessel, this element will cease to exert pressure on its<br />
bottom and on the pan duri~g the time of f<strong>al</strong>l<strong>in</strong>g t= Y2ii. This is equl-<br />
. g<br />
v<strong>al</strong>ent to the appearance of an impulse of force act<strong>in</strong>g on the vessel vertic<strong>al</strong>ly<br />
upward when the element of the liquid is f<strong>al</strong>l<strong>in</strong>g. The mean v<strong>al</strong>ue of<br />
this impulse dur<strong>in</strong>g the duration of the f<strong>al</strong>l is<br />
.. /2h ,r<br />
~mg V g=Am f 2gh<br />
Thus, each element of the tiquid Am is accompanied dur<strong>in</strong>g Its f<strong>al</strong>l<strong>in</strong>g by<br />
the appearance of two equ<strong>al</strong> and oppositely directed impulses of force. S<strong>in</strong>ce<br />
the stream is cont<strong>in</strong>uous, the b<strong>al</strong>ance will be <strong>in</strong> equilibrium.<br />
At the moment when the stream stops flow<strong>in</strong>g, the pan will sw<strong>in</strong>g down,<br />
s<strong>in</strong>ce the last elements of the liquid f<strong>al</strong>l<strong>in</strong>g on the pan act on it with<br />
a force that exceeds the weight of the elements, and there will no longer be<br />
a reduction <strong>in</strong>- the pressure on the bottom of the vessel.<br />
280. On the basis of the law of conservation of energy we can write<br />
Mv 2<br />
-2-=mgh<br />
where M is the mass of the water In the tube stopped by v<strong>al</strong>ve V. and m<br />
is the mass of the water raised to the height h.<br />
Therefore,<br />
pbtd 2 X ~=pVoRh<br />
4 2<br />
where V o is the volume of mass m.<br />
The average volume raised <strong>in</strong> two seconds is<br />
lnd 2v2<br />
Vo= 8gh =1.7 X to- 3 m 3<br />
One hour of ram operation wi11 raise<br />
V=1.7 X 10- 3 X 30 X 60=::= 3m 3
MECHANICS<br />
F<br />
265<br />
::::::::a<br />
vt~ Fig. 989<br />
281. The pressure of the air stream<strong>in</strong>g over the roof is' Jess than of air at<br />
rest. It is this surplus pressure of the stationary air under the roof that<br />
causes the described phenomena.<br />
282. S<strong>in</strong>ce the gas <strong>in</strong> the stream has a high velocity, the pressure <strong>in</strong>side<br />
the stream is below atmospheric. The b<strong>al</strong>l will be supported from the bottom<br />
by the thrust of the stream, and on the sides by the static atmospheric<br />
pressure.<br />
283. When air flows b<strong>et</strong>ween the disks, its velocity dim<strong>in</strong>ishes as it<br />
approaches the edges of the disks, and is m<strong>in</strong>imum at the edges. The pressure<br />
<strong>in</strong> a j<strong>et</strong> of air Is the lower, the higher its velocity. For this reason the<br />
pressure b<strong>et</strong>ween the disks is lower than atmospheric.<br />
The atmospheric pressure presses the lower disk aga<strong>in</strong>st the upper one,<br />
and the flow of the air is stopped. After this the static pressure of the air<br />
aga<strong>in</strong> moves the disk awayI and the process is repeated.<br />
284. The pressure dim<strong>in</strong>ishes <strong>in</strong> a stream of a flow<strong>in</strong>g liquid with an<br />
<strong>in</strong>crease <strong>in</strong> its velocity. The velocity with which the water flows <strong>in</strong> the<br />
vessel is much sm<strong>al</strong>ler than <strong>in</strong> the tube and, therefore, the pressure of the<br />
water <strong>in</strong> the vessel is greater than <strong>in</strong> the tube. The velocity Increases at the<br />
boundary b<strong>et</strong>ween the vessel and tube, and the pressure drops. For this<br />
reason the b<strong>al</strong>l is pressed aga<strong>in</strong>st the screen and does not rise.<br />
285. The piston will cover the distance ut dur<strong>in</strong>g the time t (Fig. 389).<br />
The force F will perform the work W;:::Fut. The mass of the liquid flow<strong>in</strong>g<br />
out dur<strong>in</strong>g the time t is pAut. The outflow velocity of the liquid v can be<br />
found from the equation Au=av. The change <strong>in</strong> the k<strong>in</strong><strong>et</strong>ic energy of the<br />
liqui d dur<strong>in</strong>g the time t is<br />
VI U' )<br />
pAut ( 2-2<br />
This change should be equ<strong>al</strong> to the work performed by the force F:<br />
(<br />
V' U 2)<br />
Fut=pAut 2-2<br />
Upon elim<strong>in</strong>at<strong>in</strong>g u, we f<strong>in</strong>d that<br />
2F I<br />
vZ=-x--<br />
Ap . as<br />
1- AI<br />
.. /2F<br />
If a~ A, then.v= V Ap·<br />
286. It was assumed <strong>in</strong> solv<strong>in</strong>g Problem 285 that the velocity of any<br />
element of a liquid <strong>in</strong> the pump is constant. The velocity changes from u<br />
to v when the liquid leaves the pump. This does not occur immediately after<br />
the force beg<strong>in</strong>s to act on the piston, however. The process requires some
266 ANSWERS AND SOLUTIONS<br />
time<br />
to become stable. i. e., after the particles of the liquid <strong>in</strong> the cyl<strong>in</strong>der<br />
acquire a constant velocity. When a -+ A this time tends to <strong>in</strong>f<strong>in</strong>ity and.<br />
for this reason, the velocity acquired by the liquid under the action of the<br />
constant force becomes <strong>in</strong>f<strong>in</strong>itely high.<br />
287. L<strong>et</strong> us <strong>in</strong>troduce the coord<strong>in</strong>ate system depicted <strong>in</strong> Fig. 390. Accord<strong>in</strong>g<br />
to Torricelli's formula, the outflow velocity of a liquid is V= Y 2gy,<br />
where y is the thickness of the water layer <strong>in</strong> the upper vessel. S<strong>in</strong>ce water<br />
is <strong>in</strong>compressible, aV= Av, where v is the velocity with which the upper<br />
layer of the water lowers, A is its area, and a is the area of the orifice.<br />
If we assume that the vessel is axi<strong>al</strong>ly symm<strong>et</strong>ric<strong>al</strong>, then A=nx 2 , where x<br />
is the horizont<strong>al</strong> coord<strong>in</strong>ate of the vessel w<strong>al</strong>l.<br />
Therefore,<br />
nx 2 a<br />
--=-=-:- =const<br />
V2gy v<br />
s<strong>in</strong>ce <strong>in</strong> confor mity with the <strong>in</strong>iti<strong>al</strong> condition, the water level should lower<br />
with a constant velocity. Hence, the shape of the vessel can be d<strong>et</strong>erm<strong>in</strong>ed<br />
from the equation<br />
where<br />
:1;2 V2<br />
k=<br />
2ga 2<br />
288. The pressure changes <strong>in</strong> a horizont<strong>al</strong> cross section depend<strong>in</strong>g on the<br />
dis tance to the axis , accord<strong>in</strong>g to the law<br />
p0)2<br />
P=PO+T ,2<br />
where Po is the pressure on the axis of the vessel and p the density of the<br />
liquid.<br />
The compressive deformation of the liquid will be maximum near the w<strong>al</strong>ls<br />
of the vessel, while the tensile deformation of the revolv<strong>in</strong>g rod (Problem 203)<br />
is maximum at the axis.<br />
--1--<br />
-r- --<br />
-- 1--<br />
_I<br />
I<br />
Fig. 390 Fig. 391
MECHANICS 267<br />
289. The excess pressure at a distance r from the<br />
axis of rotation is p=p;2,S (see the solution to Problem<br />
288). On the other hand, this pressure is d<strong>et</strong>erm<strong>in</strong>ed<br />
by the elevation of the liquid level <strong>in</strong> this section<br />
above the level on the axis: p=pgh (Fig. 391).<br />
Upon equat<strong>in</strong>g these expressions, we have<br />
(02<br />
h=-r<br />
2g<br />
2<br />
This is the equation of a parabola, and the surface<br />
of the liquid <strong>in</strong> the rotat<strong>in</strong>g vessel takes the form of<br />
.<br />
Ftg. 392<br />
a paraboloid of revolution.<br />
290. The stirr<strong>in</strong>g imparts a certa<strong>in</strong> angular velocity<br />
(0 to the particles of the water <strong>in</strong> the glass. The<br />
pressures <strong>in</strong> the liquid will be distributed <strong>in</strong> about the same way as <strong>in</strong> the so<br />
Iution to Problem 288. The excess pressure <strong>in</strong>side the liquid b<strong>al</strong>ances the<br />
pressure due to a higher level of the liquid at the edges of the glass (see<br />
Problem 289). ,<br />
When stirr<strong>in</strong>g is stopped, the velocity of rotation of the liquid near the<br />
bottom will beg<strong>in</strong> to decrease ow<strong>in</strong>g to friction, the greater the farther the<br />
elements of the liquid are from the centre.<br />
Now the excess pressure caused by rotation wiII no .longer b<strong>al</strong>ance the weight<br />
of the liquid column near the edge of the vessel. This will cause the liquid<br />
to circulate as shown <strong>in</strong> Fig. 392. This is why the tea leaves will gather <strong>in</strong><br />
the middle of the glass.
CHAPTER 2<br />
HEAT.<br />
MOLECULAR PHYSICS<br />
2-1. Therm<strong>al</strong> Expansion of Solids and Liquids<br />
291. l\t=420 c.<br />
0<br />
292. Re<strong>in</strong>forced-concr<strong>et</strong>e structures are very strong because the expansion<br />
coefficient of concr<strong>et</strong>e is very close to that of iron and steel.<br />
293. The quantity of heat transferred from one body to another <strong>in</strong> a unit<br />
of time is proportion<strong>al</strong> to the difference b<strong>et</strong>ween the temperatures of these<br />
bodies. When the temperatures of the thermom<strong>et</strong>er and the surround<strong>in</strong>g objects<br />
differ appreciably, the volume of the mercury will change at a fast ra teo<br />
If the temperature of the thermom<strong>et</strong>er is nearly the same as that of the surround<strong>in</strong>g<br />
bodies, the volume of the mercury changes slowly.<br />
For this reason it takes longer for the thermom<strong>et</strong>er to take the temperature<br />
of a human body. I f the warm thermom<strong>et</strong>er is brought <strong>in</strong> contact with relatively<br />
cool air <strong>in</strong> the room, the mercury column "drops" so fast due to the<br />
great temperature difference that it can be shaken down <strong>in</strong> a moment.<br />
294. When the sc<strong>al</strong>e cools down from t l to to=O° C, the v<strong>al</strong>ue of each<br />
graduation dim<strong>in</strong>ishes. Therefore, the height of the mercury column read off<br />
the sc<strong>al</strong>e with a temperature of to=0 0<br />
C will be different and equ<strong>al</strong> H =HIX<br />
X (1+ at 1)' The heights of the mercury columns at different temperatures and<br />
identic<strong>al</strong> pressures are <strong>in</strong>versely proportion<strong>al</strong> to the densities:<br />
n, P I<br />
HI =~=l+ytl<br />
Hence,<br />
H =H1(1+at 1 ) ::::: H (l+at _lUt)<br />
o 1+ yt 1 1 r 1<br />
l<br />
295. The thermom<strong>et</strong>er can be precooled <strong>in</strong> a refrigerator and shaken. If no<br />
refrigerator is available, put the thermom<strong>et</strong>er <strong>in</strong>to your mouth or <strong>in</strong> your<br />
arm-pit for a time sufficient for the entire thermom<strong>et</strong>er to reach the body<br />
temperature, then take it out and shake it immediately. The thermom<strong>et</strong>er will<br />
show the temperature of the body.<br />
296. The difference <strong>in</strong> the lengths of the rulers at a temperature t 1 is<br />
At a temperature t 2<br />
l~(l +a 1 tt)-l; (1+ r.J. 2 t 1)=l<br />
this difference is equ<strong>al</strong> to<br />
L~ (1 +<strong>al</strong>t2)~l; (1+Cltt2)= ± 1<br />
The plus sign corresponds to the case when the difference In lengths is<br />
constant (see Fig. 393a). The relation b<strong>et</strong>ween the lengths and the temperature<br />
shown <strong>in</strong> Fig. 39311 corresponds to the m<strong>in</strong>us sign.<br />
In the first case the system of equations gives<br />
1;(u=~l=6.8 em; 1;(u=~l=4.8cm<br />
"s-~<br />
a.-<strong>al</strong>
HEAT. MOLECULAR PHYSICS 269<br />
(0) .<br />
t<br />
/l ~<br />
(b)<br />
t<br />
Fig. 393<br />
In the second case ~<br />
cm;<br />
When I=()O C, the iron ruler should be longer than the copper one.<br />
297. A possible way of suspension is depicted <strong>in</strong> Fig. 394, where / and 2<br />
are rods with a sm<strong>al</strong>l coefficient of l<strong>in</strong>ear expansion (11 (e. g., steel rods), and 3<br />
are rods with a high coefficient of expansion a, (e. g., z<strong>in</strong>c or brass rods).<br />
The lengths of the rods can be so selected that the length of the pendulum<br />
does not change with -the temperature.' With this aim <strong>in</strong> view it is essenti<strong>al</strong><br />
that ClJ (11+11) = (1, ' , .<br />
298. When the cyl<strong>in</strong>der is heated, its volume <strong>in</strong>creases accord<strong>in</strong>g to the<br />
same law as that of the glass: (11=00 (1+yl/) , where y is the coefficient of<br />
volume expansion of glass. If the densities o. mercury at the temperatures to<br />
and '1 are denoted by Po and PI' we can write that mo=vopo and ml=vIP1,<br />
where
270<br />
ANSWERS AND SOLUTIONS<br />
!<br />
This system of equations will give the follow<strong>in</strong>g<br />
expression for 1':<br />
Fig,394<br />
Z 9<br />
e<br />
l' m 1 (1+1'l t 1)- mO~ 3X 10-~ deg- 1<br />
mot l<br />
The coefficient of l<strong>in</strong>ear expansion a= ; ===10-&<br />
deg- 1 •<br />
299. L<strong>et</strong> the pendulum of an accurate clock<br />
perform N oscillations a day. At the temperature<br />
t l the pendulum of our clock will perform N<br />
oscillations <strong>in</strong> n - 5 seconds (where n = 86.400<br />
is the number of seconds <strong>in</strong> a day) and at the<br />
temperature t 2 <strong>in</strong> n+10 seconds. The periods of<br />
oscillations will respectively be equ<strong>al</strong> to<br />
n-5 n+10<br />
Tl=~ and T2=-N-'<br />
T 1 n-5 15<br />
Hence, T =n+IO~ I-n° On the other<br />
2<br />
hand, bear<strong>in</strong>g <strong>in</strong> m<strong>in</strong>d that the period of renrrr:<br />
we obta<strong>in</strong><br />
dulum oscillations T=2n V ~.<br />
T fl+aL l ,r~~--<br />
T: = l 1+aL<br />
2<br />
~ y 1+a(tl-t2)~1+<br />
Upon equat<strong>in</strong>g the expressions<br />
the periods. we f<strong>in</strong>d that<br />
a<br />
+"2(t1-t:!)<br />
for the ratio of<br />
30<br />
a ~ ~ 2.3X 10- 6 deg- I<br />
(t 2- t1) n<br />
2-2. The Law of Conservation of Energy.<br />
Therm<strong>al</strong> Conductivity<br />
I<br />
300. Accord<strong>in</strong>g to the Jaw of conservation of energy. the liberated heat is<br />
equ<strong>al</strong> to the loss of k<strong>in</strong><strong>et</strong>ic energy<br />
, Q_Mv~ (M+m)v 2<br />
- 2 2<br />
where v is the velocity of the cart after the brick has been lowered onto it.<br />
This velocity can be found from the law of conservation of momentum:<br />
Mv o<br />
v=M+m·<br />
I hani 1 it Q . Mmv~ d I th 1 u Q J Mmv~<br />
n mec anica urn s =2 (M +m) an In erma urn s = 2 (M +m)'<br />
where j is the therm<strong>al</strong> equiv<strong>al</strong>ent of work.
HEAT. MOLECULAR PHYSICS 271<br />
301. On the basis of the law of conservation of energy,<br />
mgl =mv 2 + k (1-/0 ) 2 + Q<br />
2 2<br />
where l is the length of the cord at the moment when the washer leaves it.<br />
• On the other hand, we can write that<br />
mv 2<br />
mgl=T+W l + W 2<br />
where WI = flo is the work of the force of friction act<strong>in</strong>g, on the washer (the<br />
washer travels a path of lo relative to the cord), and W s=f (1-1 0<br />
) is the work<br />
of the force of friction act<strong>in</strong>g on the cord. Therefore,<br />
Us<strong>in</strong>g Hooke's Law<br />
we f<strong>in</strong>d that<br />
k (1-1 0 )2<br />
Q=W 1+W2 2<br />
f=k (l-Io><br />
f2<br />
Q=flo+ 2k<br />
The work W1 is used entirel y to liberate heat. Only h<strong>al</strong>f of the work W2 = ~ ,<br />
however, is converted <strong>in</strong>to heat, the other h<strong>al</strong>f produc<strong>in</strong>g the potenti<strong>al</strong> energy<br />
k (/-1 0 )2<br />
2 .<br />
302. The electric current performs the work W =PT. At the expense of<br />
this work the refrigerator will lose the beat QI=qH+ qct ; where c is the<br />
heat capacity of water and H is the heat of fusion of ice. Accord<strong>in</strong>g to the<br />
law of conservation of energy, the amount of heat liberated <strong>in</strong> the room will be<br />
Q. = W+Q2=Pt:+qct+qH<br />
s<strong>in</strong>ce <strong>in</strong> the f<strong>in</strong><strong>al</strong> run the energy of the electric current is converted <strong>in</strong>to heat.<br />
303. The temperature <strong>in</strong> the room will rise. The quantity of heat liberated<br />
<strong>in</strong> a unit of time will be equ<strong>al</strong> to' the power consumed by the refrigerator,<br />
s<strong>in</strong>ce <strong>in</strong> the f<strong>in</strong><strong>al</strong> run the energy of the electric current is converted <strong>in</strong>to heat,<br />
and the heat removed from the refrigerator is r<strong>et</strong>urned aga<strong>in</strong> <strong>in</strong>to the room.<br />
304. It is more advantageous to use a refrigerator that removes heat from<br />
the outside air .and liberates it <strong>in</strong> the room. The heat liberated <strong>in</strong> the room<br />
<strong>in</strong> a unit of time is P+Q2' where P is the power consumed by the refrigerator<br />
and Q, is the heat removed from the outside air <strong>in</strong> a unit of time (see<br />
Problem 302).<br />
It is only the high cost and complicated equipment that prevent the use<br />
of such therm<strong>al</strong> pumps for heat<strong>in</strong>g at present.<br />
305. When s<strong>al</strong>t is dissolved, its cryst<strong>al</strong> lattice is destroyed. The process<br />
requires a certa<strong>in</strong> amount of energy that can be obta<strong>in</strong>ed from the solvent.<br />
In the second case, part of the <strong>in</strong>termolecular bonds of the cryst<strong>al</strong> lattice<br />
have <strong>al</strong>ready been destroyed <strong>in</strong> crush<strong>in</strong>g the cryst<strong>al</strong>. For this reason, less<br />
energy is required to dissolve the powder and the water will be higher <strong>in</strong><br />
temperature <strong>in</strong> the second vessel. The effect, however, will be extremely<br />
negligible.
272 ANSWERS AND SOLUTIONS<br />
306. The quantity of heat removed from the water be<strong>in</strong>g cooledis m 2c (tt-8),<br />
where 6 is the f<strong>in</strong><strong>al</strong> temperature. The cold water receives the heat mlc (6-tl).<br />
The heat imparted to the c<strong>al</strong>orim<strong>et</strong>er Is q(a-t l ) . On the basis of the law<br />
of conservation of energy,<br />
mlc (6-t1)+ q(6-t t )= mgc (t 2- 6)<br />
whence<br />
8=(mItt +m"t,,) c+ qt l ~ 40 C<br />
(m l +m 2 ) c+q -<br />
307. The power spent to heat the water <strong>in</strong> the c<strong>al</strong>orim<strong>et</strong>er is<br />
Pl=pVctJ<br />
l'<br />
where p is the density of the water, C is its specific heat, and J =4.18 Jzc<strong>al</strong><br />
is the mechanic<strong>al</strong> equiv<strong>al</strong>ent of heat. The sought v<strong>al</strong>ue is<br />
P-P t pVctJ<br />
Q=-p-=1--W~5 per cent<br />
A<br />
308. Q=l[ (T1-To)At ~ 9,331 kc<strong>al</strong><br />
309. The quantity of heat Q pass<strong>in</strong>g through the first pane) a second is<br />
Q= Al T I ~T1 A, where A is the area of a panel. S<strong>in</strong>ce the process is statio<br />
To-T<br />
nary, the sameamount of heat passes through the second panel: Q ="" T A.<br />
T,,-T I To-T"<br />
We f<strong>in</strong>d from the condition A t - - A=A<br />
d- 2 ~ A that<br />
t<br />
T _AtdlTo+Atd,Tt<br />
"- A 2dl +Atd 2<br />
310. Upon <strong>in</strong>sert<strong>in</strong>g the temperature T" <strong>in</strong>to the expression for Q (see Problem<br />
309) when d l=d2=d, we f<strong>in</strong>d that<br />
Q= 2"'I A 2 To-T1 A<br />
x, +A.:a 2d<br />
Therefore, the coefficient of therm<strong>al</strong> conductivity of the w<strong>al</strong>l is<br />
A= 2AI A t<br />
At+ A"<br />
311. The quantity of heat pass<strong>in</strong>g a secon d through the cross-section<strong>al</strong> areas<br />
of the blocks with coefficients of therm<strong>al</strong> conductivity Al and A2 is equ<strong>al</strong>,<br />
respectively, to<br />
Q Al A2 T A<br />
t=cr(TI-To) A and Q2={[ (T1- 0)<br />
The quantity of heat pass<strong>in</strong>g through two blocks whose entire cross-section<strong>al</strong><br />
area is 2A is<br />
Q=Ql+ Q 2<br />
= Al+A2 Tt-T 0 2A<br />
2 d
HEAT. MOLECULAR PHYSICS<br />
273<br />
Hence, the coefficient of therm<strong>al</strong> conductivity of the w<strong>al</strong>l is equ<strong>al</strong> to<br />
~ _AI+A 2<br />
"'- 2<br />
312. The coefficients of therm<strong>al</strong> conductivity of w<strong>al</strong>ls I and II are equ<strong>al</strong> to<br />
'1 Al+A 2 d ~ 2AIAI<br />
""1=-2- an "'II = At+A 2<br />
(see the solutions to <strong>Problems</strong> 310 and 311). It follows from the obvious <strong>in</strong>equ<strong>al</strong>ity<br />
(A.I - At)! > 0 that<br />
(AI+A 2 )2 > 4"'1AI<br />
Hence,<br />
At+A 2 2AIA 2 • Il 'l<br />
-2- > Al +1<br />
2<br />
,I.e., "'I > "'II<br />
313. The quantity of heat supplied by the heater <strong>in</strong>to the water through<br />
the pan bottom is<br />
A<br />
Q=7f(T-T t ) A=mr<br />
where T 1 is the boil<strong>in</strong>g po<strong>in</strong>t of the water and, is the specific heat of vaporization.<br />
Therefore,<br />
2·3. Prope'rties of Gases<br />
314. The removable cap acts as a pump and under it a rarefied space is<br />
formed that sucks out the <strong>in</strong>k. The orifice serves to ma<strong>in</strong>ta<strong>in</strong> a constant pressure<br />
under the cap.<br />
315. Assum<strong>in</strong>g that the temperature rema<strong>in</strong>s constant, l<strong>et</strong> us app ly Boyle's<br />
law to the volume of air above the mercury:<br />
(POI-PI) (1-748 mm)=(P02-P2) (1-736 mm)<br />
whence 1=764 mm.<br />
316. In a position of eqnilibrium f-6--F=O, where f is the force of<br />
expulsion equ<strong>al</strong> to yh1A (here y is the specific weight of the water t and h l<br />
the height of the air column <strong>in</strong> the tube after submergence). In our case the<br />
force of expulsion Is built up by the difference of pressure on the soldered<br />
end of the tube from below and lrom above: f=ptA-(lJo+'Vh) A, where Pi<br />
is the air pressure <strong>in</strong> the tube after submergence. Accord<strong>in</strong>g to Boyle's law,<br />
PolA= PlhiA.<br />
It follows from this system of equations that<br />
E= : [y (po+yh)2+4Poyl.:.....(po+yh»)-G=8.65 gf<br />
18-2042
274<br />
ANSWERS AND SOLUTIONS<br />
P<br />
A<br />
~ ---------------------<br />
B<br />
1/ Fig. 395<br />
317. First. the pressure p of the air will decrease approximately isotherm<strong>al</strong>ly<br />
ow<strong>in</strong>g to the drop of the level of the water <strong>in</strong> the vessel. This will<br />
cont<strong>in</strong>ue until the tot<strong>al</strong> pressure at the level of the lower end of the tube<br />
becomes equ<strong>al</strong> to the atmospheric pressure Po; i, e., p+pgh = Po. where h is<br />
the height of the water column <strong>in</strong> the vessel above the level of the lower<br />
end of the tube. From this moment on air bubbles will beg<strong>in</strong> to pass <strong>in</strong>to the<br />
vessel. The pressure at the level of the lower end of the tube will rema<strong>in</strong><br />
equ<strong>al</strong> to the atmospheric pressure, while the air pressure P=Po-pgh will<br />
grow l<strong>in</strong>early with a drop <strong>in</strong> the water level. The water will flow out from<br />
the vessel at a constant velocity.<br />
The relation b<strong>et</strong>ween p and Q is shown <strong>in</strong> Fig. 395. The negligible fluctuations<br />
of pressure when separate bubbles pass <strong>in</strong> are not shown <strong>in</strong> Fig. 395.<br />
318. When the air is be<strong>in</strong>g pumped out of the vessel, the pressure <strong>in</strong> the<br />
vessel after one double stroke will become equ<strong>al</strong> to Pt =vP+o~ • After the<br />
o V)2<br />
second double stroke PtV=p z (V +00) and, consequently, Pz=Po (v +00 •<br />
<strong>et</strong>c<br />
Ȧfter n double strokes the pressure <strong>in</strong> the vessel will be<br />
p'=Po (V:vJ"<br />
When air is be<strong>in</strong>g delivered <strong>in</strong>to the vessel. after n double strokes the<br />
pressure will be<br />
_ ,+Ponvo_ {( V )n+nuo}<br />
p-p ---Po -- -<br />
V V+vo V<br />
Here p > Po at any n, s<strong>in</strong>ce dur<strong>in</strong>g delivery the pump dur<strong>in</strong>g each double<br />
stroke sucks <strong>in</strong> air with a pressure (Jo, and dur<strong>in</strong>g evacuation (10 of the air is<br />
pumped out at pressures below Po.<br />
319. Apply<strong>in</strong>g Boyle's law to the two volumes of gas <strong>in</strong> the closed tube,<br />
we obta<strong>in</strong><br />
P-2-<br />
L-I<br />
A=Pl<br />
(L-l<br />
-2--A1 )<br />
A<br />
L-l<br />
(L-l)<br />
p -2- A=tJ2 -2-+~1 A<br />
Pl=P2+Vl
HEAT. MOLECULAR PHYSICS 275<br />
Here p is the pressure <strong>in</strong> the tube placed<br />
horizont<strong>al</strong>ly, Pl and P2 are the<br />
pressures <strong>in</strong> the lower and upper ends of the tube placed vertic<strong>al</strong>ly when its<br />
ends are closed, y is the specific weight of the mercury, A is the cross-section<strong>al</strong><br />
area of the tube.<br />
Hence, the <strong>in</strong>iti<strong>al</strong> pressure <strong>in</strong> the tube is<br />
p=y.!- (l!._&1)<br />
2 III 1 0<br />
Here. for the sake of brevity we have designated L 2 1 by 1 0<br />
,<br />
If one end of the horizont<strong>al</strong> tube is opened', the pressure of the gas Jn the<br />
tube will become equ<strong>al</strong> to the atmospheric pressure.<br />
Accord<strong>in</strong>g to Boyle's law. ploA = yH 11 A (here H is the atmospheric pressure),<br />
and, therefore,<br />
l-~(~-~)<br />
1- 2H &l 1 0<br />
The mercury column will shift through the distance<br />
L\11= lo- lt = l!JL [2H -(~-~)J<br />
2H I L\l 1 0<br />
For the mercury not to flow out of the tube, the follow<strong>in</strong>g condition is required<br />
\Vhen the upper end of the vertic<strong>al</strong><br />
Hence,<br />
&~ ~ V (~r+l+ ~<br />
tube is opened<br />
ploA ="1 (H + 1)l2A<br />
tlI2 = 1 0-12 2(:+1) [27- ( ~~ - ~:) +2]<br />
The mercury will not flow out of the tube. if<br />
When the lower end is opened<br />
whence<br />
~~ ~ j1{2(~+l)r+I+2(Hl+l)<br />
ploA =Y (H-I) 13A<br />
M3= 10- 13= 2 (~~l) [27- (~~ - ~:) -2]<br />
The follow<strong>in</strong>g condition should be satisfied to prevent the mercury column<br />
from be<strong>in</strong>g forced out of the tube<br />
18 *.<br />
~ 1//-4 (H -1)2 1+2 (H -I)<br />
t1l ~ r 12 + 1
276 ANSWERS AND SOLUTIONS<br />
320. S<strong>in</strong>ce for one gramme-molecule of any. gas at P= I atm and T =237 0 K<br />
we have V,.= 22.4 lltres, then for one mole C=P~"=O.082lit.atm/mole.deg.<br />
This constant is usu<strong>al</strong>ly denoted by R and c<strong>al</strong>led the univers<strong>al</strong> gas constant.<br />
The v<strong>al</strong>ues of R <strong>in</strong> the various systems of units are:<br />
R=0.848 kgl-rn/mole- deg=8.3X 10 7 erg/mole- deg= 1.986 c<strong>al</strong>/rnole-deg<br />
321. At a fixed pressure and temperature the volume occupied by the gas<br />
is proportion<strong>al</strong> to its mass. A volume of VfA- corresponds to one gramme-molecule<br />
and a volume of V to an arbitrary mass m. Obviously. VIi = V ~. where I..t<br />
is the molecular weight expressed <strong>in</strong> grammes.<br />
Upon <strong>in</strong>sert<strong>in</strong>g this expression <strong>in</strong>to the equation of state for one grammemolecule.<br />
we have<br />
pV=!!!...RT<br />
J.I.<br />
322. If the attraction b<strong>et</strong>ween the molecules suddenly disappeared. the<br />
pressure shou ld <strong>in</strong>crease. To prove this, l<strong>et</strong> us ment<strong>al</strong>ly s<strong>in</strong>gle out two layers<br />
J and IJ <strong>in</strong>side a fluid (Fig. 396). The molecules pen<strong>et</strong>rat<strong>in</strong>g from layer I<br />
<strong>in</strong>to layer I J ow<strong>in</strong>g to therm<strong>al</strong> motion collide with the molecules <strong>in</strong> layer<br />
II, and as a result this layer is acted upon by the pressure forces Pt that<br />
depend on the temperature. The forces of attraction act on layer I I from the<br />
side of the molecules <strong>in</strong> layer I <strong>in</strong> the opposite direction. The resultant pressure<br />
of layer I on layer I J P=Pt-Pi, where Pi is the pressure 'caused by<br />
the <strong>in</strong>tern<strong>al</strong> forces of attraction. When PI disappears, the pressure grows.<br />
323. If the forces of attraction b<strong>et</strong>ween<br />
the molecules disappeared the water would<br />
"'<br />
/1.<br />
be converted <strong>in</strong>to an ide<strong>al</strong> gas. The pressure<br />
can be found from the equation of state of<br />
an idea 1 gas:<br />
m RT<br />
p=~ V~ 1,370 atm<br />
~<br />
~<br />
--<br />
.:<br />
.-<br />
--<br />
.-<br />
.,<br />
~<br />
•<br />
..<br />
..<br />
~<br />
~<br />
I<br />
Fig. 396<br />
Jr<br />
~<br />
Fig. 397
HEAT. MOLECULAR PHYSICS 271<br />
324. L<strong>et</strong> us separate a cyl<strong>in</strong>dric<strong>al</strong> volume of the gas <strong>in</strong> direct contact<br />
with the w<strong>al</strong>l (Fig. 397). The forces act<strong>in</strong>g on the side surface of the cyl<strong>in</strong>..<br />
der are mutu<strong>al</strong>ly b<strong>al</strong>anced. S<strong>in</strong>ce the volume is <strong>in</strong> equilibrium, the pressure<br />
on the gas from the side of the w<strong>al</strong>l should <strong>al</strong>ways be equ<strong>al</strong> to the pressure<br />
on the .other base of the cyl<strong>in</strong>der from the side of the gas. We can conclude,<br />
on the basis of Newton's third law, that the pressure of the gas on the w<strong>al</strong>l<br />
is equ<strong>al</strong> to the pressure <strong>in</strong>side the vessel.<br />
325. The pressure <strong>in</strong> the gas depends on the forces of <strong>in</strong>teraction b<strong>et</strong>ween<br />
the molecules (see Problem 322). The forces of mutu<strong>al</strong> <strong>in</strong>teraction of the<br />
molecules and of the molecules with the wa11 are different, however. Hence<br />
the pressures <strong>in</strong>side the gas and at the w<strong>al</strong>ls of the vessel (see Problem 324)<br />
can be identic<strong>al</strong> only if the concentrations are different.<br />
326. S<strong>in</strong>ce the volume is constant<br />
Hence,<br />
~=.~, or P2-Pl=T2- T 1 = O.004<br />
PI T 1 PI T 1<br />
=T 2-T1=2500K<br />
T<br />
1 0.004<br />
327. From Archimedes' law, mg+G=yV, where y is the specific weight<br />
of water and V is the volume of the sphere. The equation of state gives<br />
(Po+ yh) V=!!!:.... RT<br />
~<br />
Upon del<strong>et</strong><strong>in</strong>g V from these equations, we f<strong>in</strong>d that<br />
m GJ-t (Po +yh) ~ 0.666 g<br />
VRT - fJ-g (Po + yh)<br />
and equilibrium will be unstable.<br />
328. When the tube is horizont<strong>al</strong>, the device cannot be used as a thermom<strong>et</strong>er,<br />
s<strong>in</strong>ce the pressures exerted on the drop from the right and from ·the<br />
left will be b<strong>al</strong>anced at any temperature.<br />
If the tube is placed vertic<strong>al</strong>ly, the pressure of the gas <strong>in</strong> the lower b<strong>al</strong>l<br />
will be higher than <strong>in</strong> the ufper one by a constant magnitude. If the volume<br />
is the same, the pressure wil grow with a rise <strong>in</strong> the temperature the faster,<br />
the higher is the <strong>in</strong>iti<strong>al</strong> pressure. To ma<strong>in</strong>ta<strong>in</strong> a constant difference of the<br />
pressures <strong>in</strong> the b<strong>al</strong>ls, the drop will beg<strong>in</strong> to move upward, and <strong>in</strong> this case<br />
the device can be employed as a thermom<strong>et</strong>er.<br />
329. S<strong>in</strong>ce the masses of the gas are the same <strong>in</strong> both ends and the piston<br />
is <strong>in</strong> equilibrium,<br />
Hence,<br />
T 2 =!.!. T 1 =330 0<br />
VI<br />
Apply<strong>in</strong>g Boyle's law to the volume of the gas whose temperature does not<br />
change, we obta<strong>in</strong><br />
p= Po Vo= l. 05 atm<br />
VI<br />
K
278 ANSWERS AND SOLUTIONS<br />
330. When the extern<strong>al</strong> conditions are the same, equ<strong>al</strong> volumes of various<br />
gases conta<strong>in</strong> an equ<strong>al</strong> number of molecules (Avogadro's law). Therefore,<br />
Vl:V,:Vs:V.=Nt:N2:Ns:Nc, where Vi is the volume of a gas and N,. the<br />
number of molecules of this gas.<br />
The mass of a certa<strong>in</strong> amount of a gas is proportion<strong>al</strong> to the nurnber of<br />
its molecules and the molecular weight of the gas:<br />
ml: m2: ma:m. = N1""'1: N2""2: N8""3: N.J.t.<br />
On the other hand, denot<strong>in</strong>g the relative volume of this gas <strong>in</strong><br />
V·<br />
by nj=V 100%, we have<br />
. .. VI. V2 • Vs • Vc N1.N2.N3 • N.<br />
nt·n 2 .na·n,=V·V·V·V=7j·N'N·!i<br />
per cent<br />
If the composition of air <strong>in</strong> per cent is described by ni= mi 100% (comm<br />
position by weight), we can obta<strong>in</strong> from the previous ratios that<br />
n ' 'n' ·n''n' _m1.m2.m3, m4 _ N1Jll . Na ""2 N af.ls . N.Jl4_ II.<br />
n ·n II. • n II. <strong>in</strong> u<br />
1 • 2· 3· • - m · m · m · m - N · N . N · N - lr1' 2r2' 3r3· 4r4<br />
Hence,<br />
, n~+n;+n;+n;<br />
n, n .,.,..<br />
n 1Jll+nl J.L 2 + naJla+n.J1. I I<br />
Remember<strong>in</strong>g that n; +n;+n;+n; = 100 per cent, we obta<strong>in</strong><br />
Therefore,<br />
, nilJi1OO%<br />
n1<br />
nl""'l + n2J.t2 -l- n<strong>al</strong>-'a+ n.J.t4<br />
n;==75.52%; n;=23.15%; n;= 1.28%;<br />
n~=O.05%<br />
331. For each gas, the equation of state can be written as follows:<br />
Hence,<br />
PIV=.!!!.L RT<br />
J.ll<br />
P2V=~RT<br />
fJ2<br />
P3V=~RT<br />
""3<br />
P4V=~ RT<br />
f.14<br />
(PI +P2+P3+P4)V= (~+~+~+.!!!!.)RT<br />
f.tl ""2 f.La fJ4<br />
On the other hpnd, for a mixture of gases pV=~ RT t<br />
f.1<br />
+ m 2+m3+tn4 and Il. is the sought molecular weight.<br />
where m=ml+
HEAT. MOLECULAR PHYSICS 279<br />
I/otm<br />
1 ---- !<br />
0.5 Z<br />
/lZ6 ------ 4<br />
.3<br />
I<br />
II<br />
J Z J 4 .J{lit Fig. 398<br />
Accord<strong>in</strong>g to D<strong>al</strong>ton's law, p=p,+p!+Ps+P,. Therefore,<br />
11= ml+ mS+ ma+ m• n~+n;+n~+n~ =28.966<br />
.!!!!..+~+~+~ !!i+ n; + n~ + n;<br />
""'1 fi! J,Ls J,L. ftl fl.' fJ.af.t.<br />
where ni=~ 100% is the composition of the air <strong>in</strong> per cent by weight.<br />
m<br />
The resul t obta<strong>in</strong>ed <strong>in</strong> the previous problem <strong>al</strong>lows us to f<strong>in</strong>d .... from the<br />
known composition of the air. by volume<br />
f..L ....l n t + ""2 n2 +""sna+ J.t.n. = 28.966<br />
nl +n 2+nS+n.<br />
332. On the basis of Clapeyron's equation,<br />
J1 = mRT=pRT=72 g/mole<br />
pV p<br />
The sought formula is C&H 1 2 (one of the pentane isomers).<br />
333. When the gas is compressed <strong>in</strong> a heat ..impermeable envelope, the<br />
work performed by the extern<strong>al</strong> force is spent to <strong>in</strong>crease the <strong>in</strong>tern<strong>al</strong> energy<br />
of the gas. and its temperature <strong>in</strong>creases. The pressure <strong>in</strong> the gas will <strong>in</strong>crease<br />
both ow<strong>in</strong>g to a reduction <strong>in</strong> its volume and an <strong>in</strong>crease <strong>in</strong> its tempe..<br />
rature. In isotherm<strong>al</strong> compression the pressure rises only ow<strong>in</strong>g to a reduction<br />
<strong>in</strong> the volume.<br />
Therefore. the pressure will <strong>in</strong>crease more <strong>in</strong>' the first case than <strong>in</strong> the<br />
second.<br />
334. A diagram of p versus V is shown <strong>in</strong> Fig. 398. The greatest work<br />
equ<strong>al</strong> to the hatched area <strong>in</strong> Fig. 398 is performed dur<strong>in</strong>g the isotherm<strong>al</strong><br />
process (1..2). .<br />
The temperature does not change on section 1-2, and is h<strong>al</strong>ved on section<br />
2-3. After this the temperature rises. and, T,=T 1 when V 4=4 lit.<br />
335. L<strong>in</strong>e 1-2 is an isobaric l<strong>in</strong>e (Fig. 399). The gas is heated at a constant<br />
pressure, absorb<strong>in</strong>g heat.
280<br />
ANS,WERS AND SOLUTIONS<br />
p<br />
1.....---+---....<br />
v<br />
L<strong>in</strong>e 2-3 is an isochoric l<strong>in</strong>e. The<br />
gas is cooled at a constant volume,<br />
the pressure drops and heat is Iibera...<br />
ted.<br />
L<strong>in</strong>e 3..l is an isotherm<strong>al</strong> l<strong>in</strong>e. The<br />
volume of the gas dim<strong>in</strong>ishes at a<br />
constant temperature. The pressure ri ...<br />
sese The gas is not heated, <strong>al</strong>though it<br />
is subjected to the work of extern<strong>al</strong><br />
forces. Hence, the gas rejects heat on<br />
this section.<br />
338. The amount of heat liberated<br />
per hour upon combustion of the m<strong>et</strong>hane<br />
is<br />
Fig. 399<br />
Q _90PVof.'<br />
1- RT<br />
where 11=16 g/mole is the mass of one mole of the gas and T=t+273° =<br />
= 284 0 K is its temperature. The amount of heat received by the water <strong>in</strong><br />
one hour is<br />
nDI<br />
Q'=Tvpc(tt-t l ) 3,600<br />
where p=1 g/cms, is the density of the water and c= 1 c<strong>al</strong>/deg- g is the<br />
specific heat.<br />
Accord<strong>in</strong>g to the condition,<br />
Q: ='1=0.6<br />
Upon solv<strong>in</strong>g these simultaneous equations, we f<strong>in</strong>d that<br />
t t +<br />
qopVoJ.LTJ - 930r<br />
2= 1 900nD2vpcRT = -\J<br />
337. In the <strong>in</strong>iti<strong>al</strong> state plV =~. RT l' where J.tl is the molecular weight<br />
~1<br />
of the ozone. In the f<strong>in</strong><strong>al</strong> state, P2V = m RT 2' where .... 2 is the molecular<br />
1-12<br />
weight of the oxygen. The heat b<strong>al</strong>ance equation gives<br />
~ q =Cvm (T 2 - T I)<br />
J.Ll J.L2<br />
Upon solv<strong>in</strong>g these simultaneous equations, we f<strong>in</strong>d that<br />
P'I.=-q...-+J.LI=lO<br />
PI CvTt ""2<br />
338. In view of the l<strong>in</strong>ear dependence of pressure on volume we can write:<br />
p==aV+b.
HEAT. MOLECULAR PHYSICS<br />
r<br />
281<br />
I \<br />
I \<br />
I \<br />
I \ \<br />
I \<br />
I \<br />
I \<br />
, \<br />
I \<br />
I \<br />
\<br />
I \<br />
va ~O:G If II Fig. 400<br />
The constants a and b can be found from the condition of the problem:<br />
a=P V<br />
l - P v 2 :!!:-0.5atm/lit<br />
1- 2<br />
b plV1-plV 2 S!!E 20 atm<br />
VI-V'<br />
Upon <strong>in</strong>sert<strong>in</strong>g the expression for p <strong>in</strong>to the equation of state of an ide<strong>al</strong><br />
m<br />
gas pV=-RT=const T, we f<strong>in</strong>d that<br />
J.L<br />
aVI+bV=const T (1)<br />
The relation b<strong>et</strong>ween T and V (see Fig. 400) is a parabola. The curve<br />
reaches its maximum at Vmax= - :a 2E 20 lit when the roots of quadratic<br />
equation (1) co<strong>in</strong>cide. Here<br />
b<br />
Pmax=aVmax+b=2 e: 10 atm<br />
Therefore,<br />
T<br />
_Pmax V maxJ.L ~ 490 0 K<br />
mQX- mR -<br />
339. The energy of 8 unit volumeof gas "1=CTp, where p Is the density ofthe<br />
air. Accord<strong>in</strong>g to the equation of state of an Ide<strong>al</strong>gas, Pi=mB (B is a constant).<br />
S<strong>in</strong>ce p=~ , then pT= ; . Therefore, Ul=~ P is d<strong>et</strong>erm<strong>in</strong>ed only by the<br />
pressure. The energy of <strong>al</strong>l the air <strong>in</strong> the room is <strong>al</strong>so d<strong>et</strong>ermi ned only by<br />
the pressure. The pressure <strong>in</strong> the room is equ<strong>al</strong> to the atmospheric pressure
282 ANSWERS AND SOLUTIONS<br />
Fig. 401<br />
and does not change when the air is heated. For this reason the energy of<br />
the air <strong>in</strong> the room <strong>al</strong>so does not change. As the air is heated, some of it<br />
flows outside through cracks, and this ensures a constancy of energy despite<br />
the heat<strong>in</strong>g. The energy would <strong>in</strong>crease with heat<strong>in</strong>g only <strong>in</strong> a herm<strong>et</strong>ic<strong>al</strong>ly<br />
closed room.<br />
340. On the basis of the equation of state. the sought mass of the air<br />
will be<br />
A _ f.LpV T 2 - T 1<br />
um------<br />
~ 13k<br />
g<br />
R T 1T2 •<br />
341. L<strong>et</strong> the tube first be near the bottom <strong>in</strong> a state of stable equilibrium.<br />
Upon heat<strong>in</strong>g, the air pressure <strong>in</strong> the tube and, correspond<strong>in</strong>gly. the force of<br />
expulsion <strong>in</strong>crease. At a certa<strong>in</strong> temperature TIthe tube beg<strong>in</strong>s to rise to the<br />
surface. S<strong>in</strong>ce the pressure of the water gradu<strong>al</strong>ly decreases upwards from the<br />
bottom, the volume of the air <strong>in</strong> the tube and, therefore. the force of expulsion<br />
cont<strong>in</strong>ue to <strong>in</strong>crease. The tube will quickly reach the surface of the<br />
water. Upon a further <strong>in</strong>crease <strong>in</strong> the temperature, the tube will be at the<br />
surface. If the temperature lowers, the tube will not s<strong>in</strong>k at T 1. because it<br />
has a great reserve of buoyancy caused by an appreciable <strong>in</strong>crease <strong>in</strong> the<br />
force of expulsion as the tube rises. It is only when T 2 < T 1 that the tube<br />
beg<strong>in</strong>s to s<strong>in</strong>k. Here the force of expulsion will drop because, as submergence<br />
cont<strong>in</strong>ues, the air <strong>in</strong> the tube will occupy a sm<strong>al</strong>ler volume. The tube<br />
will reach the bottom very quickly.<br />
The relation b<strong>et</strong>ween the depth of submergence h of the tube and the<br />
temperature T is shown <strong>in</strong> Fig. 401.<br />
The tube will <strong>al</strong>ways be at the bottom when T < T 2 and at<br />
the surface<br />
when T > T 1. If T 2 < T < Tit the tube will be either at the bottom or at<br />
the surface, depend<strong>in</strong>g on the previous temperatures.<br />
342. The gas expands at a certa<strong>in</strong> constant pressure p built up by the<br />
piston. The work W=p (V2-VI). where Viand V2 are the Initi<strong>al</strong> and f<strong>in</strong><strong>al</strong><br />
volumes of the gas. By us<strong>in</strong>g the equation of state. l<strong>et</strong> us express the product<br />
pV through the temperature T. Then.<br />
W=~ R (T 2 - Td ~ 33.9 kgf-m<br />
LL
HEAT. MOLECULAR PHYSICS 283<br />
343. The heat imparted to the gas is used to heat it and perform mecha..<br />
nic<strong>al</strong> work. Accord<strong>in</strong>g to the law of conservation of energy,<br />
2-4. Properties of Liquids<br />
344. It is more difficult to compress a litre of air, s<strong>in</strong>ce more work has<br />
to be done <strong>in</strong> this case.<br />
Water has a sm<strong>al</strong>l compressibility, and a sm<strong>al</strong>l reduction <strong>in</strong> volume is<br />
required to <strong>in</strong>crease the pressure <strong>in</strong>side it to three atmospheres.<br />
345. A maximum thermom<strong>et</strong>er can be made as follows. A sm<strong>al</strong>l unweUable<br />
freely mov<strong>in</strong>g body is placed <strong>in</strong>side the tube of a horizont<strong>al</strong> thermom<strong>et</strong>er<br />
(Fig. 402). The position of the body will show the maximum temperature,<br />
s<strong>in</strong>ce the body will move <strong>al</strong>ong the tube when the liquid expands and will<br />
rema<strong>in</strong> <strong>in</strong> place when the liquid <strong>in</strong> the tube is compressed.<br />
To make a m<strong>in</strong>imum thermom<strong>et</strong>er, a w<strong>et</strong>table body should be placed<br />
<strong>in</strong>side the liquid <strong>in</strong> the tube.<br />
346. When an elastic rubber film is str<strong>et</strong>ched, the force of tension depends<br />
on the amount of deformation of the film. The force of surface tension is<br />
d<strong>et</strong>erm<strong>in</strong>ed only by the properties of the liquid and does not change with<br />
an <strong>in</strong>crease of its surface.<br />
347. The surface tension of pure p<strong>et</strong>rol is less than that of p<strong>et</strong>rol <strong>in</strong><br />
which grease is dissolved. For this reason the p<strong>et</strong>rol applied to the edges<br />
will contract the spot towards the centre. If the spot itself is w<strong>et</strong>ted, it will<br />
spread over the fabric.<br />
348. Capillaries of the type shown <strong>in</strong> Fig. 403 form <strong>in</strong> a compact surface<br />
layer of soil. They converge towards the top, and the water <strong>in</strong> them rises to<br />
the surface, from which it is <strong>in</strong>tensively evaporated. Harrow<strong>in</strong>g destroys this<br />
structure of the capillaries and the moisture is r<strong>et</strong>a<strong>in</strong>ed <strong>in</strong> the soil longer.<br />
349. Leather conta<strong>in</strong>s a great number of capillaries. A drop of a w<strong>et</strong>t<strong>in</strong>g<br />
liquid <strong>in</strong>side a capillary hav<strong>in</strong>g a constant cross section will be <strong>in</strong> equllibrium.<br />
When the liquid is heated, the surface tension dim<strong>in</strong>ishes and the<br />
liquid is drawn towards the cold part of the capillary. The grease will be<br />
drawn <strong>in</strong>to the leather if it is heated outside.<br />
350. The grease melts, and capillary forces carry it to the surface of the<br />
cold fabric placed under the cloth<strong>in</strong>g (see Problem 349).<br />
351. The end of the piece of wood <strong>in</strong> the shade is colder, and the caplltary<br />
forces move the water <strong>in</strong> th is direction.<br />
352. The hydrostatic pressure should be b<strong>al</strong>anced by the capillary pres<br />
4~<br />
sure: pgh={f' Hence, h=30 em.<br />
353. The follow<strong>in</strong>g forces act vertic<strong>al</strong>ly on section abed of the film: weight,<br />
surface tension F ab applied to l<strong>in</strong>e ab and surface tension Fcd applied to cd.<br />
Fig. 402
284 ANSWERS AND SOLUTIONS<br />
Fig. 403<br />
mg+4aa. ~ 2 3<br />
x a 2pg -. em<br />
Equilibrium is possible only if F<strong>al</strong>l<br />
is greater than Fcd by an amount<br />
equ<strong>al</strong> to the weight of the section<br />
of the film be<strong>in</strong>g considered.<br />
The difference b<strong>et</strong>ween the forces<br />
of surface tension can be expla<strong>in</strong>ed<br />
by the difference <strong>in</strong> the concentration<br />
of the soap <strong>in</strong> the surface<br />
layers of the film.<br />
354. The force of expulsion b<strong>al</strong>ances<br />
the weight of the cube mg<br />
and the force of surface tension 4aa.<br />
i.e., a 2 xpg = mg+ 4aa , where x is<br />
the sought distance. Therefore,<br />
The forces of surface tension <strong>in</strong>troduce a correction of about 0.1 em.<br />
355. The water rises to a height h = 2a . The potenti<strong>al</strong> energy of the wapgr<br />
ter colurnn is<br />
E = mgh = 2na;2<br />
p 2 pg<br />
4n(£2<br />
The forces of surface tension perform the work W=2nrah=--. One h<strong>al</strong>f<br />
Pi<br />
of this work goes to <strong>in</strong>crease the potenti<strong>al</strong> energy, and the other h<strong>al</strong>f to<br />
evolve heat. Hence,<br />
2na<br />
Q=pg<br />
2<br />
356. The pressure <strong>in</strong>side the liquid at a po<strong>in</strong>t that is at a height h above<br />
a certa<strong>in</strong> level is less than the pressure at this level by pgh. The pressure<br />
is zero at the level of the liquid <strong>in</strong> the vessel. Therefore, the pressure at<br />
the height h is negative (the liquid is str<strong>et</strong>ched) and is equ<strong>al</strong> to p= -pgh.<br />
357. The forces of attraction act<strong>in</strong>g on a molecule <strong>in</strong> the surface layer<br />
from <strong>al</strong>l the other molecules produce a resultant directed downward. The<br />
closest neighbours, however, exert a force of repulsion on the molecule which<br />
is therefore <strong>in</strong> equi librium.<br />
Ow<strong>in</strong>g to the forces of attraction and repulsion, the density of the liquid<br />
is sm<strong>al</strong>ler <strong>in</strong> the surface layer than <strong>in</strong>side. Indeed, molecule / (Fig. 404) is<br />
acted upon by the force of repulsion from molecule 2 and the forces of<br />
attract ion from <strong>al</strong>l the other molecules (3, 4, ...). Molecule 2 is acted upon<br />
by the forces of repulsion from 3 and / and the forces of attraction from the<br />
molecules <strong>in</strong> the deep layers. As a result, distance /-2 should be greater<br />
than 2-3, <strong>et</strong>c.<br />
This course of reason<strong>in</strong>g is quite approximate (therm<strong>al</strong> motion, <strong>et</strong>c., is<br />
disregarded), but nevertheless it gives a qu<strong>al</strong>itatively correct result.<br />
An <strong>in</strong>crease <strong>in</strong> the surface of the liquid causes new sections of the rarefied<br />
surface layer to appear. Here work should be performed aga<strong>in</strong>st the forces<br />
of attraction b<strong>et</strong>ween the molecules. It is this work that constitutes the<br />
surface energy.
HEAT. MOLECULAR PHYSICS 285<br />
358. The required pressure should exceed the atmospheric pressure by an<br />
amount that can b<strong>al</strong>ance the hydrostatic pressure of the water column and<br />
the capillary pressure <strong>in</strong> the air bubble with a radius r,<br />
The excess pressure is p+pgh+ 2a =4,840 dyne/ems,<br />
r<br />
359. S<strong>in</strong>ce <strong>in</strong> this case p gh < 2a, , the water rises to the top end of the<br />
r<br />
tube. The meniscus will be a part of a spheric<strong>al</strong> segment (Fig. 405): The<br />
radius of curvature of the segment is d<strong>et</strong>erm<strong>in</strong>ed from the condition that the<br />
forces of surface tension ba lance the weight of the water column:<br />
2Jtra cos q:> = rtr 2hpg.<br />
rhpa<br />
Hence. cos p = 20.'. It is obvious from Fig. 405 that the radius of curr<br />
2a<br />
vature of the segment R=--=h-=O.74 mm.<br />
cos q> gp<br />
360. When the tube is opened. a convex meniscus of the same shape as<br />
on the top is formed at its lower end. For this reason the length of the<br />
water column rema<strong>in</strong><strong>in</strong>g <strong>in</strong> the tube will be 2h if I ~ h. and 1+h if t «; h.<br />
361. (1) The forces of surface tension can r<strong>et</strong>a<strong>in</strong> a water column with a<br />
height not over h <strong>in</strong> this capillary tube. Therefore. the water will flow out<br />
of the tube.<br />
(2) The water does not flow out.<br />
The meniscus is convex. and will<br />
be a hemisphere for an absolutely<br />
w<strong>et</strong>t<strong>in</strong>g liquid.<br />
(3) The water does not flow out.<br />
The meniscus is convex and is less<br />
curved than <strong>in</strong> the second case.<br />
(4) The water does not flow out.<br />
The meniscus is Oat.<br />
----------------<br />
@<br />
-------@-------<br />
-------@-------<br />
------@-------<br />
Fig. 404 Fig. 405<br />
---- --
286<br />
ANSWERS AND SOLUT IONS<br />
JI!!!I!IIII!!!!~~~""-------<br />
--_.....---_...<br />
-------~~~~I!!!!!!!<br />
-..------ - - -<br />
(aj<br />
-------.-#-- -~<br />
~ .r.__~ ...<br />
- --.- --<br />
(b) Fig. 406<br />
(5) The water does not flow out. The meniscus is concave.<br />
362. The pressure p <strong>in</strong>side the soap-bubble with a radius R exceeds the<br />
atmospheric pressure by the amount of the double capillary pressure, s<strong>in</strong>ce<br />
the bubble film is double: P=Po+ ~ •<br />
The pressure <strong>in</strong>side the bubble with a radius R tog<strong>et</strong>her with the pressure<br />
of the section of the film b<strong>et</strong>ween the bubbles should b<strong>al</strong>ance the pressure<br />
<strong>in</strong>side the sm<strong>al</strong>ler bubble. Therefore, ~a. +::= ~, where u, is the radius<br />
of curvature of section AB. Hence, R x = RRr •<br />
-r<br />
At any po<strong>in</strong>t of contact the forces of surface tension ba lance each other<br />
and are mutu<strong>al</strong>ly equ<strong>al</strong>. This is possible only when the angles b<strong>et</strong>ween them<br />
are equ<strong>al</strong> to 120°.<br />
363. Accord<strong>in</strong>g to the law of conservation of energy, the cross will not<br />
rotate. The components of the forces of surface tension are b<strong>al</strong>anced by the<br />
forces of hydrostatic 'pressure, s<strong>in</strong>ce the hydrostatic pressure of the water.<br />
higher than the level <strong>in</strong> the vessel is negative (see Problem 356).<br />
364. If the bodies are w<strong>et</strong>ted by water, its surface will take the form shown<br />
<strong>in</strong> Fig. 4000. B<strong>et</strong>ween the matches, above level MN, the water is tensioned<br />
by the capillary forces, and the pressure <strong>in</strong>side the water is less than the<br />
atmospheric pressure. The matches will be attracted toward each other, s<strong>in</strong>ce<br />
they are subjected to the atmospheric pressure on their sides.<br />
For unw<strong>et</strong>ted matches, the form of the surface is shown <strong>in</strong> Fig. 40Sb. The<br />
pressure b<strong>et</strong>weE'n the matches is equ<strong>al</strong> to the atmospheric pressure and is<br />
greater than the latter on the sides below level M N.
.;~<br />
HEAT. MOLECULAR PHYSICS 287<br />
=M~ ~<br />
----<br />
- -~-_-~---== -::-.!L- -_-= ~L -=<br />
(a)<br />
w- -- ---<br />
------~~~~------~~~~~~<br />
N~<br />
---<br />
(h)<br />
Fig. 407<br />
In the last case two various forms of the surface correspond to the w<strong>et</strong>t<strong>in</strong>g<br />
angles when the matches approach each other (Fig. 407). One of them, however,<br />
cannot be obta<strong>in</strong>ed <strong>in</strong> practice (Fig. 407a). The pressure at level KL<br />
should be the same everywhere. In particular, the pressure of columns AB<br />
and CD of different height should be the same. But this is impossible, s<strong>in</strong>ce<br />
the position of the column can be so selected that their surfaces are identicaJ<br />
<strong>in</strong> form. In this case the addition<strong>al</strong> pressure of the surface forces will be the<br />
same, and' the hydrostatic pressure different. As a result, when the matches<br />
approach each other, the surface of the water b<strong>et</strong>ween them will tend to<br />
assurne a horizont<strong>al</strong> form (Fig. 407b). In this case, as can be seen from the<br />
figure, the pressure b<strong>et</strong>ween the matches at level MN is equ<strong>al</strong> to the atmospheric<br />
pressure. The pressure exerted from the left on the first match is <strong>al</strong>so<br />
equ<strong>al</strong> to the atmospheric pressure below level MN. The pressure act<strong>in</strong>g<br />
on the second match from the right is less than the atmospheric pressure<br />
above level JWN. As a result, the matches will be repulsed.<br />
2-5. Mutu<strong>al</strong> Conversion of Liquids and Solids<br />
385. Water will freeze at zero only <strong>in</strong> the presence of centres of cryst<strong>al</strong>lization.<br />
of the<br />
Any<br />
water<br />
<strong>in</strong>soluble<br />
is great,<br />
particles can serve as such centres. When the mass<br />
it will <strong>al</strong>ways conta<strong>in</strong> at least one such centre. This<br />
will be enough for <strong>al</strong>l the water to freeze. If the mass of the water is divided<br />
<strong>in</strong>to very f<strong>in</strong>e drops, centres of cryst<strong>al</strong>lization will be present only <strong>in</strong> a<br />
comparatively sm<strong>al</strong>l number of the drops, and only they will freeze.
288 ANSWERS AND SOLUTIONS<br />
386. The water and the ice receive about the same amount of heat <strong>in</strong> a<br />
unit of time, s<strong>in</strong>ce the difference b<strong>et</strong>ween the temperatures of the water and<br />
the air <strong>in</strong> the room is approximately the same as that for the ice and the<br />
air. In 15 m<strong>in</strong>utes the water receives 200 c<strong>al</strong>ories. Therefore, the ice receives<br />
8,000 c<strong>al</strong>ories <strong>in</strong> ten hours. Hence, H =80 c<strong>al</strong>/g.<br />
367. 0=2,464 m/s.<br />
368. The heat b<strong>al</strong>ance equations have the form<br />
c<br />
Q 1 =mtcIL\#+CAt<br />
~t At<br />
Q2=mlclT+mtH+mlc2T+C&t<br />
where ml and Cl are the mass and heat capacity of the ice, C Is the heat<br />
capacity of the c<strong>al</strong>orim<strong>et</strong>er, c 2 is the heat capacity of the water, and<br />
~t=2°C.<br />
Hence,<br />
Q ( C2<br />
HI) Q<br />
1 -+--+- - 2<br />
2c1 cl~t 2 150 lid<br />
CI ~t At H :z: ca eg<br />
"C;T-T+C;-<br />
369. The amount of heat that can be liberated by the water when it is<br />
cooled to O°C is 4,000 c<strong>al</strong>. Heat<strong>in</strong>g of the ice to O°C requires 12,000 c<strong>al</strong>.<br />
Therefore, the ice can be heated only by the heat liberated when the water<br />
freezes. .One hundred grammes of water should be frozen to produce the<br />
lack<strong>in</strong>g 8,000 c<strong>al</strong>ories.<br />
As a result, the c<strong>al</strong>orim<strong>et</strong>er will conta<strong>in</strong> a mixture of 500 g of water and<br />
500 g of ice at a temperature of O°C.<br />
370. The f<strong>in</strong><strong>al</strong> temperature of the contents <strong>in</strong> the vessel is O=O°C. The<br />
heat b<strong>al</strong>ance equation has the form<br />
m1c1(tt-8)=mtcl(8-t 2 )+ (m 2 - ma) H<br />
where m 1 Is the sought mass of the vessel and c, is the heat capacity of<br />
the ice. Therefore,<br />
m,Ct (9-t t )+ (m t - ms) H 200<br />
ml = CI (tl - 6) g<br />
371. (1) The sought mass of the ice m can be found from the equation<br />
mH =Mc(-t). Hence, m= 100 g.<br />
(2) The heat b<strong>al</strong>ance equation can bewritten <strong>in</strong> this case as MH= Me(-t).<br />
Hence, t=-80°C.<br />
372. The melt<strong>in</strong>g po<strong>in</strong>t of the ice compressed to 1,200 atm will drop by<br />
At =8.8° C. The ice will melt until it cools to -8.8° C. The amount of heat<br />
Q=mtH is absorbed, where m\ is the mass of the melted ice and H is the<br />
specific heat of fusion. From the heat b<strong>al</strong>ance equation m1H =mc~t, where<br />
c is the heat capacity of the ice.<br />
Hence,<br />
cmtu oc:<br />
m1=n - 5.6 g
HEAT. MOLECULAR PHYSICS 289<br />
2 -6. Elasticity and Strength<br />
373. F AE (R-r) 60 kgf.<br />
r<br />
•<br />
374. When the rod with fastened ends is heated by t degrees, it develops<br />
an elastic force F equ<strong>al</strong>, accord<strong>in</strong>g to Hooke's law, to<br />
F = AEAI = AEaJ<br />
I<br />
where E is the modulus of elasticity of steel and (X is its coefficient of therrna<br />
I expansion.<br />
If one of the rod ends is gradu<strong>al</strong>ly released, the length of the rod. will<br />
<strong>in</strong>crease by Al= lat. The force will decrease l<strong>in</strong>early from F to zero and its<br />
average magnitude will be F/2.<br />
F I<br />
The sought work W=2 Al = 2 AEla 2 t 2 •<br />
375. The tension of the wire T=2 ~g • It follows from Hooke's law<br />
s<strong>in</strong>a<br />
Ai<br />
that T =2f EA.<br />
S<strong>in</strong>ce Al=2 (_/__1) , then T I-cosa AE=2~g . At sm<strong>al</strong>l ancos<br />
ex cos ex. S<strong>in</strong> a<br />
gles, s<strong>in</strong> a ~ a, and cos a=1-2 s<strong>in</strong>S ~ e; 1-~s • Bear<strong>in</strong>g this <strong>in</strong> m<strong>in</strong>d, we<br />
obta<strong>in</strong><br />
cx= V Mg<br />
AE<br />
376. The rod heated by At would extend by Al=loaAt <strong>in</strong> a free state,<br />
where L o is the <strong>in</strong>iti<strong>al</strong> length of the rod. To fit the heated rod b<strong>et</strong>ween the<br />
w<strong>al</strong>ls, it should be compressed by 111. In conformity with Hooke's law,<br />
IF<br />
Al=EA<br />
Therefore, F = E AaAt= 110 kgf.<br />
377. When the rods are heated <strong>in</strong> a free state, their tot<strong>al</strong> length will<br />
<strong>in</strong>crease by Al=All+Al,=(<strong>al</strong>'l+a21z)t. .<br />
Compression by the same amount ~l will reduce the lengths of the rods<br />
by ~l; and AI;, where AI; + ~l;=L11. This requires the force<br />
F - ElA AI' EzAA1;<br />
- '1 t 1 2<br />
Upon solv<strong>in</strong>g this system of equations, we f<strong>in</strong>d that<br />
F <strong>al</strong>lt+asll At<br />
.!L+~<br />
E) £2<br />
The rods will act upon each other with this force.<br />
19-2042
290 ANSWERS AND SOLUTIONS<br />
378. It is obvious from cons iderations of symm<strong>et</strong>ry that the wires will<br />
elongate equ<strong>al</strong>ly. L<strong>et</strong> us denote this elongation by ~I. On the basis of<br />
Hooke's law, the tension of a steel wire Fs=~' AEs and of a copper one<br />
Al<br />
Fe=-l- AEc·<br />
It follows that the ratio b<strong>et</strong>ween the tensions is equ<strong>al</strong> to the ratio b<strong>et</strong>ween<br />
he respective Young's moduli .<br />
Fe Ee 1<br />
p;= E$ ="2<br />
In equilibrium 2F c + Fs=mg.<br />
Therefore,<br />
Fc= "'4 g = 25 kgf and Fs = 2Fc = 50 kgf.<br />
379. On the basis of Hooke's law,<br />
~l ~l<br />
Fe=-l- AcEc and Fi=-l- AiE;<br />
F<br />
It follows that F~ = 2.<br />
Thus, two-third~ of the load are resisted by the concr<strong>et</strong>e and one-third by<br />
the iron.<br />
FI<br />
380. The compressive force F shortens the tube by A E and the tensile<br />
FI c C<br />
force F extends the bolt by A E ·<br />
s s<br />
The sum A~~s + A~. is equ<strong>al</strong> to the motion of the nut <strong>al</strong>ong the bolt:<br />
Hence,<br />
..f!.-+~=h<br />
AsEs AcEc<br />
F -!!:.. AsEsAcEc<br />
- I AsEs+AcEe<br />
381. S<strong>in</strong>ce the coefficient of l<strong>in</strong>ear therm<strong>al</strong> expansion of copper a c Is greater<br />
than that of steel as,. the <strong>in</strong>crease <strong>in</strong> temperature will lead to compression<br />
of the copper plate and tension of the steel ones. In view of symm<strong>et</strong>ry, the<br />
relative elongations of <strong>al</strong>l the three plates are the same. Denot<strong>in</strong>g the compressive<br />
force act<strong>in</strong>g on the copper plate from the sides of the steel plates<br />
by F, we sh<strong>al</strong>l have for the relative elongation of the copper plate: ~'=<br />
F<br />
= act- AE .<br />
c<br />
Either steel plate is subjected to the tensile force F/2 from the side of<br />
the copper one. Upon equat<strong>in</strong>g the relative elongation of the plates, we<br />
obta<strong>in</strong>:
HEAT. MOLECULAR PHYSICS 291<br />
F<br />
On upper nut<br />
Fpllf----.-,<br />
o<br />
O=Fg<br />
On lower nut<br />
8 Fig. 408<br />
Hence<br />
F _2AE cEs «(Xc-as) t<br />
- 2E s+Ec<br />
mvl<br />
382. When the r<strong>in</strong>g rotates, the tension T = 2- appears In It (see Probnr<br />
tern 201). For a th<strong>in</strong> r<strong>in</strong>g m=2nrAp, where A is the cross section of the<br />
r<strong>in</strong>g. Therefore. ~ =pv·.<br />
Hence. the maximum<br />
velocity v= V~Q!;~l m/s.<br />
:$83. Initi<strong>al</strong>ly, an elastic force Po acts on each nut from the side of the<br />
extended bolt. .<br />
The load G~ F0 cannot <strong>in</strong>crease the length of the part of the bolt b<strong>et</strong>ween<br />
the nuts and change its tension. For this reason the force act<strong>in</strong>g on the<br />
upper nut from the side of the block will not change as long as G~ Fe-<br />
The lower nut is acted upon by the force F0 from the side of the top part<br />
of the bolt and by the force G from the bottom part. S<strong>in</strong>ce the nut is <strong>in</strong><br />
equilibrium, the force exerted on it from the block is F=Fo-G. Thus the<br />
action of the load G~ F0 consists only <strong>in</strong> reduc<strong>in</strong>g the pressure of the lower<br />
nut on the block.<br />
When G > F OI the length of the bolt will <strong>in</strong>crease and the force act<strong>in</strong>g<br />
on the lower nut from the side of the block will disappear. The upper nut<br />
will be acted on by the force G.<br />
The relation b<strong>et</strong>ween the forces act<strong>in</strong>g on the nuts and the weight of the<br />
load G is shown <strong>in</strong> Fig. 408. .<br />
2·7. Properties of Vapours<br />
384. The c<strong>al</strong>orim<strong>et</strong>er will conta<strong>in</strong> 142 g of water and 108 g of vapour at<br />
a temperature 100 0 C.<br />
385. By itself, water vapour cr steam is <strong>in</strong>visible. We can observe only<br />
a sm<strong>al</strong>l cloud of the f<strong>in</strong>est drops appear<strong>in</strong>g after condensation. When the gas<br />
burner is switched off, the streams of heated air that previously enveloped<br />
19 ':
292 ANSWERS AND SOLUTIONS<br />
the k<strong>et</strong>tle disappear, and the steam com<strong>in</strong>g out of the k<strong>et</strong>tle is cooled and<br />
condenses.<br />
386. On the basis of the equation of state of an ide<strong>al</strong> gas p= ~ = ;';. .<br />
If the pressure is expressed <strong>in</strong> mm Hg and the volume <strong>in</strong> m 3 , then R:::;::<br />
760X 0.0224 mm Hg·m s<br />
= 273 deg.mole·<br />
273<br />
Therefore, p= 1.06PT' At temperatures near room temperature, p~p g/m s .<br />
387. It seems at first sight that the equation of state of an ide<strong>al</strong> gas cannot<br />
give v<strong>al</strong>ues of the density or specific volume of saturated vapours close<br />
to the actu<strong>al</strong> ones. But this is not so. If we c<strong>al</strong>culate the density of a vapour<br />
by the formula p= ~ = k~ and compare the v<strong>al</strong>ues obta<strong>in</strong>ed with<br />
those <strong>in</strong> Table 2 (p. 85), we sh<strong>al</strong>l observe good agreement.<br />
This is expla<strong>in</strong>ed as. follows. The pressure of an ide<strong>al</strong> gas grows <strong>in</strong> direct<br />
proportion to the temperature at a constant volume of the gas and, therefore,<br />
at a constant density. The relation b<strong>et</strong>ween the pressure of saturated<br />
vapours and the temperature depicted <strong>in</strong> Fig. 146 corresponds to a constant<br />
volume of a saturated vapour and the liquid which it is <strong>in</strong> equilibrium<br />
with. As the temperature <strong>in</strong>creases, the density of the vapeur grows, s<strong>in</strong>ce<br />
the liquid parti<strong>al</strong>ly transforms <strong>in</strong>to a vapour. An appreciable <strong>in</strong>crease <strong>in</strong> the<br />
mass of the vapour corresponds to a sm<strong>al</strong>l change <strong>in</strong> the volume it occupies.<br />
The pressure-density ratio becomes approximately proportion<strong>al</strong> to the temperature,<br />
as with an ide<strong>al</strong> gas.<br />
The Clapeyron-Mendeleyev equation ma<strong>in</strong>ly gives a correct relationship<br />
b<strong>et</strong>ween p, V and T for water vapour up to the v<strong>al</strong>ues of these param<strong>et</strong>ers<br />
that correspond to the beg<strong>in</strong>n<strong>in</strong>g of condensation. This equation, however,<br />
cannot describe the process of transition of 8 vapour <strong>in</strong>to a liquid and <strong>in</strong>dicate<br />
the v<strong>al</strong>ues of p, 1> and T at which this transition beg<strong>in</strong>s.<br />
388. At 300C the pressure of saturated vapours p= 31.82 mm Hg. Accord<strong>in</strong>g<br />
to the equation of state of an ide<strong>al</strong> gas,<br />
V=.!!!. RT e: 296 lit<br />
1-1 p<br />
389. When the temperature gradu<strong>al</strong>ly <strong>in</strong>creases, the pressure of the water<br />
vapours <strong>in</strong> the room may be considered constant.<br />
The vapour pressure p= ~;;o corresponds to a humidity of Wo= 10 per<br />
cent, where Po= 12.79 mm Hg is the pressure of the saturated vapours<br />
-at 15° C. At a temperature of 25°C the pressure of the saturated vapours is<br />
PI =23.76 mm Hg. For this reason the sought relative humidity is<br />
w=L l00o/o=~=5.40/o<br />
PI PJ'<br />
390. Accord<strong>in</strong>g to the conditions of the problem, the relative humidity<br />
outside and <strong>in</strong> the room is close to 100 per cent. The pressure of saturated<br />
water vapours outside, however, is much sm<strong>al</strong>ler than <strong>in</strong> the room, because<br />
the temperature of the air <strong>in</strong> the room is higher and much time is required<br />
to equ<strong>al</strong>ize the pressures ow<strong>in</strong>g to pen<strong>et</strong>ration of the vapours outside through
HEAT. MOLECULAR PHYSICS 293<br />
slits. Therefore, if the w<strong>in</strong>dow is opened, the vapours will quickly flow out<br />
from the room and the wash<strong>in</strong>g will be dried faster.<br />
391. (1) The water levels will become the same as <strong>in</strong> communicat<strong>in</strong>g<br />
vessels. The water vapours <strong>in</strong> the left-hand vessel will partly condense. and<br />
some water will evaporate <strong>in</strong> the right-hand vessel.<br />
(2) The levels will become the same because the vapours will flow from one<br />
vessel <strong>in</strong>to the other.<br />
At a given temperature the pressure of saturated vapours is identic<strong>al</strong> <strong>in</strong><br />
both vessels at the surface of the water and will decrease at the same rate<br />
with height. For this reason the pressure of the vapours <strong>in</strong> the vessels at the<br />
same level is different. This causes the vapour to flow over and condense <strong>in</strong><br />
the vessel with the lower water level.<br />
392. When t 2 = 30° C, the pressure of the vapours is equ<strong>al</strong> to the pressure<br />
P20 of saturated vapours (P20=31.8 mm Hg) only if the air pressure is 10 at.<br />
Upon isotherm<strong>al</strong> reduction of the air pressure to one-tenth, the volume of<br />
the air will <strong>in</strong>crease ten times. Hence, at atmospheric pressure and a temperature<br />
of 30° C, the pressure of the water vapour is p=3.18 mm Hg. It follows<br />
from the Clapeyron equation that at a temperature of 1 1 = 10°C, the<br />
T<br />
vapour pressure PI = P T:' where T I = 283° K and T1=303° K.<br />
The sought relative humidity is<br />
w=Pt 1000/0 = P.. TTl 1000/ 0<br />
~ 32.60/0<br />
Po Po 2<br />
where Po=9.2 mm Hg is the pressure of saturated vapours at t 1 = 10°C·<br />
393. The pressure p=6.5 mm Hg is the pressure of saturated water vapours<br />
at t=5° C. A sharp drop of the pressure shows that <strong>al</strong>l the water has<br />
been converted <strong>in</strong>to vapour. The volume of the vapour pumped out until the<br />
water is evaporated compl<strong>et</strong>ely is V=3,600 litres.<br />
On the basis of the Clapeyron-Mendeleyev equation of state, the sought<br />
mass of the water is<br />
v .<br />
m=P / !!!!S. 23.4 g<br />
R<br />
394. An amount of heat Q 1 =mc At =3,000 c<strong>al</strong> is required to heat the<br />
water to 100°C. Therefore, Q2= Q- QI =2,760 c<strong>al</strong> will be spent for vapour<br />
formation. The amount of the water converted <strong>in</strong>to vapour is ml =<br />
Q<br />
=J=5.1 g.<br />
r<br />
In conformity with the equation of state of an ide<strong>al</strong> gas, this vapour will<br />
occupy a volume of V= m 1 RT. Upon neglect<strong>in</strong>g the reduction of the YO-<br />
J.L P .<br />
lume occupied by the water, we can f<strong>in</strong>d the height which the piston is rai<br />
V<br />
sed to: h=;r= 17 ern.
CHAPTER 3<br />
ELECTRICITY<br />
AND MAGNETISM<br />
3-1. Electrostatics<br />
Q2<br />
395. F=7=918 kgf.<br />
The force is very great. It is impossible to impart a charge of one coulomb<br />
to a sm<strong>al</strong>l body s<strong>in</strong>ce the electrostatic forces of repulsion are so high that<br />
the charge cannot be r<strong>et</strong>a<strong>in</strong>ed on the body.<br />
396. The b<strong>al</strong>ls will be arranged at the corners of an equilater<strong>al</strong> triangle<br />
with a side ~31. The force act<strong>in</strong>g from any two b<strong>al</strong>ls on the third is<br />
4Q2<br />
£=12 Y3'<br />
The b<strong>al</strong>l will be <strong>in</strong> equilibrium if tan a,=f- (where<br />
mg<br />
a=300). Hence,<br />
I<br />
Q=2 Y mg~ 100 CGS Q _<br />
397. S<strong>in</strong>ce the threads do not deflect from the vertic<strong>al</strong>, the coulombian<br />
force of repulsion is b<strong>al</strong>anced by the force of attraction b<strong>et</strong>ween the b<strong>al</strong>ls<br />
<strong>in</strong> conformity with the law of gravitation.<br />
Therefore, <strong>in</strong> a vacuum<br />
Q2 p2V.<br />
-;:2=Y7<br />
and <strong>in</strong> kerosene (tak<strong>in</strong>g <strong>in</strong>to account the results of Problem 230)<br />
Q2 (p-PO)2 Vi<br />
E r2<br />
r = 'V r 2<br />
where V is the volume of the b<strong>al</strong>ls.<br />
Hence,<br />
P= Po ye r ~ 2.74 g/cm8<br />
Ye,.-l<br />
398. The conditions of equilibrium of the suspended b<strong>al</strong>l give the follow<strong>in</strong>g<br />
equations for the two cases be<strong>in</strong>g considered:<br />
- QQs Y2 0<br />
T 1 s<strong>in</strong> (Xt - 2a 2 X -2-=<br />
T +QQs Y2 QQ s 0<br />
1 cos <strong>al</strong> 2a 2 X -2--7-mg=<br />
· QQs V2 0<br />
T 2 srn (X,2- 2a 2 X -2-=<br />
T2cosa2+QQs-QQsx V2 -mg=O<br />
a 2 2a 2 2
ELECTRICITY AND MAGNETISM<br />
295<br />
Fig. 409 Fig. 410<br />
where T 1 and T 2 are the tensions of the thread, <strong>al</strong> and ~ the angles of<br />
deflection of the thread, + Q and ~ Q the charges of the fixed b<strong>al</strong>ls, + Qs<br />
the charge of the suspended b<strong>al</strong>l, and mg is the weight of the suspended<br />
b<strong>al</strong>l (Fig. 409).<br />
Upon exclud<strong>in</strong>g the unknowns from the above simultaneous equations, we<br />
g<strong>et</strong><br />
cot a 1 -cot a 2=cot ai-cot 2<strong>al</strong> =2(V2-1)<br />
whence,<br />
cot <strong>al</strong>=2 (2 Y2 - 1) ± V35-16 Y 2<br />
Thus, C&J. =7°56' and a,=15°52' when mg > ~~s ( 1--':2 ), and<br />
<strong>al</strong>=82°04' and a 2=164°OB' when mg < ~~s (1--':2 ).<br />
399. In uniform motion the drop is acted upon by the force of gravity G.<br />
the expulsive force of the air (Archimedean force) F, the force of the electrostatic<br />
field eE and the force of friction aga<strong>in</strong>st the air kv=k ~ . All the forces<br />
are b<strong>al</strong>anced. Therefore,<br />
G-F-eEO+k..!..=O<br />
t 1<br />
G-F+eE-k~=O<br />
t;<br />
G-F-k!..=O<br />
t<br />
where e is the charge of the drop, E the <strong>in</strong>tensity of the electric field, and s<br />
the distance covered by the drop.
296 ANSWERS AND SOLUTIONS<br />
Upon solv<strong>in</strong>g the equations, we g<strong>et</strong><br />
t= 2tlt2<br />
t l-t9<br />
400. It can, if we use the phenomenon<br />
of electrostatic <strong>in</strong>duction. Br<strong>in</strong>g a<br />
conductor on an <strong>in</strong>sulated support up<br />
to the charged body and connect the<br />
conductor to the earth for a short time.<br />
The conductor will r<strong>et</strong>a<strong>in</strong> a charge opposite<br />
<strong>in</strong> sign to the given one, while<br />
Fig. 411<br />
the like charge will pass <strong>in</strong>to the<br />
earth.<br />
The charge can be removed from the conductor by. <strong>in</strong>troduc<strong>in</strong>g the latter<br />
<strong>in</strong>to a m<strong>et</strong><strong>al</strong>lic space. The operation may be repeated many times with a<br />
charge of any magnitude.<br />
Electrostatic mach<strong>in</strong>es operate on a similar pr<strong>in</strong>ciple.<br />
401. The energy is produced by the mechanic<strong>al</strong> work that has to be performed<br />
<strong>in</strong> mov<strong>in</strong>g the conductor from the oppositely charged body to the<br />
body .that accumulates the charge.<br />
402. They can, if the charge of one b<strong>al</strong>l is much greater than that of the<br />
other. The forces of attraction caused by the <strong>in</strong>duced charges may exceed the<br />
forces of repulsion.<br />
403. S<strong>in</strong>ce Q ~ q, the <strong>in</strong>teraction b<strong>et</strong>ween the separate elements of the r<strong>in</strong>g<br />
can be neglected. L<strong>et</strong> us take a sm<strong>al</strong>l element of the r<strong>in</strong>g with a length RA.a,<br />
(Fig. 410). From the side of the charge Q it is acted upon by the force<br />
IJ.F= ~~q • where t1q= q:na.. The tension forces of the r<strong>in</strong>g T b<strong>al</strong>ance IJ.F.<br />
From the condition of equilibrium, and remember<strong>in</strong>g that AeJ is sm<strong>al</strong>l. we<br />
have<br />
t1F=2T s<strong>in</strong> (IJ.;) • rs«<br />
The sought force is the tension T= ::.~. ·<br />
404. L<strong>et</strong> us consider the case of opposite charges Q 1 > 0 and Q, < O. The<br />
<strong>in</strong>tensities created by the charges Q 1 and Qz are equ<strong>al</strong>, respectively. to<br />
£1 = ~l<br />
,]<br />
and £.= Q.- . A glance at Fig. 411 shows that<br />
'1<br />
EI=Er+ £:-2E 1E, cos cp<br />
From triangle ABC<br />
Hence,<br />
cosq><br />
,I+,:-d l<br />
2'1"
ELECTRICITY AND MAGNETISM 297<br />
A 8<br />
a<br />
Fig. 412 Fig. 419<br />
If the charges are like<br />
E-VQ~ + Q: +QIQI( I 2<br />
- 4 • 8 a rt+'t-dl)<br />
'1 '1 '1'.<br />
405. Each charge creates at po<strong>in</strong>t D a field <strong>in</strong>tensity of E 1 = ~ . The<br />
a<br />
tot<strong>al</strong> <strong>in</strong>tensity will be the sum of three vectors (Fig. 412). The SUIO of. the<br />
horizont<strong>al</strong> components of these vectors will be zero, s<strong>in</strong>ce they are equ<strong>al</strong> <strong>in</strong><br />
magnitude and form angles ~of 120° with each other. The vectors form angles<br />
of 90°-a with the vertic<strong>al</strong>, where a is the angle b<strong>et</strong>ween the edge of the<br />
t<strong>et</strong>rahedron and the <strong>al</strong>titude h of triangle ABC.<br />
The vertic<strong>al</strong> components are identic<strong>al</strong>, each be<strong>in</strong>g equ<strong>al</strong> to ~ s<strong>in</strong>
298 ANSWERS AND SOLUTIONS<br />
407. The <strong>in</strong>tensity of the field at an arbitrary po<strong>in</strong>t A on the axis of the<br />
r<strong>in</strong>g is .<br />
E - Q - Q . z<br />
- R2+ r2<br />
cos Ct- R2s<strong>in</strong> ex cos a, (1)<br />
(see Problem 406). . .<br />
Obviously, E reaches its maximum at the same v<strong>al</strong>ues of a, as the<br />
. 2E2Rt<br />
expression ~. But<br />
2E2R4<br />
--v= 2 s<strong>in</strong>- a, cos s a,=2 cos 2 a, (l-cos 2 ex) (I-cos 2 ex) (2)<br />
is the product of three positive factors<br />
a=2 cos! ex, (3)<br />
b=1-cos- a, (4)<br />
c= I-cos s ex (5)<br />
whose sum is constant (a+b+c=2). and b=c.<br />
The product abc=ab' will be maximum if the factors are equ<strong>al</strong>, i.e.,<br />
and. therefore,<br />
L<strong>et</strong> us prove this. If<br />
2<br />
a=b=c=3 (6)<br />
8<br />
abc=ab 2 = 27 (7)<br />
2<br />
a=3+ 2d<br />
where d is a certa<strong>in</strong> number that. as follows from Eq. (3). can be with<strong>in</strong><br />
_1. < d < ~ (8)<br />
3 3<br />
then, on the basis of Eq. (4)<br />
2<br />
b=--d<br />
3<br />
The product<br />
ab l = (~ +2d) (~ -dr=2~+2tP(d-l)<br />
is maximum, as follows from Eq. (8), when d=O. Hence,<br />
2 V3<br />
a=3 ana cosa=-a-<br />
The maximum <strong>in</strong>tensity of the field<br />
will be observed c.t po<strong>in</strong>ts at a distance<br />
of r= ~2 R from the centre of the r<strong>in</strong>g. This <strong>in</strong>tensity is equ<strong>al</strong> to<br />
2V3 Q<br />
Emax = - 9- R2
ELECTRICITY AND MAGNETISM<br />
299<br />
408.<br />
2n<br />
2n<br />
EB= A (QI - Q2), EC=A(Ql+Q~)<br />
2n<br />
EA=-ff(Ql+Q2)<br />
The <strong>in</strong>tensity is positive if it is directed from the left fo the right.<br />
409. The charges <strong>in</strong>duced on the surfaces of the other plate are + ~<br />
and - ~ . Only <strong>in</strong> this case will the electric field <strong>in</strong>side the plate be zero,<br />
as it should be when the charges are <strong>in</strong> equilibrium.<br />
4tO. The molecule will be attracted to the charged cyl<strong>in</strong>der. The force of<br />
attraction is<br />
F-2 Q(.!.__I_)_ 2qQl<br />
- q r r + 1 - , (r+l)<br />
In this expression we can neglect the quantity I (I ~ 10- 8 ern) as compared<br />
with , (r cannot be sm<strong>al</strong>ler than the cyl<strong>in</strong>der radius).<br />
We f<strong>in</strong><strong>al</strong>ly obta<strong>in</strong><br />
F- 2qQI<br />
- r~<br />
411. At the <strong>in</strong>iti<strong>al</strong> moment the forces act<strong>in</strong>g on both molecules are identic<strong>al</strong>.<br />
When the molecules approach the cyl<strong>in</strong>der, the force P l act<strong>in</strong>g on the<br />
molecule with a constant electric moment grows <strong>in</strong> proportion to l/ r 2 :<br />
F = 2qQl<br />
l ,2<br />
(see Problem 410). The force F?, that acts on the "elastic" molecule grows<br />
faster, <strong>in</strong> proportion to 1/,3, ow<strong>in</strong>g to the cont<strong>in</strong>uous <strong>in</strong>crease <strong>in</strong> the electric<br />
moment of this molecule (F z = 4%;~Z) .<br />
The masses of the molecules are the same, and for this reason the acceleration<br />
of the second molecule when it approaches the cyl<strong>in</strong>der grows faster<br />
than that of the first one and it will reach the surface of the cyl<strong>in</strong>der<br />
quicker.<br />
412. S<strong>in</strong>ce the thickness of the plate is sm<strong>al</strong>l it may be assumed that<br />
the charge is uniformly distributed on both surfaces, each hav<strong>in</strong>g an area abe<br />
Thus, the surface density<br />
is zero, and the <strong>in</strong>tensity outside the m<strong>et</strong><strong>al</strong> is<br />
Q 2nQ<br />
E=4n 2ab = ab<br />
of the charge a=2'!w. The field <strong>in</strong>side the m<strong>et</strong><strong>al</strong><br />
4t3. The negative charges <strong>in</strong>duced on the surface of a conductor are so<br />
distributed that the field <strong>in</strong>side the conductor result<strong>in</strong>g from a positive po<strong>in</strong>t<br />
charge and <strong>in</strong>duced negative charges is zero. (The <strong>in</strong>duced positive charges
300<br />
ANSWERS AND SOLUTIONS<br />
.0 E<br />
+<br />
+<br />
+<br />
+ + +<br />
tI+ - d<br />
0.-- --.0<br />
+<br />
-(J + +Q +<br />
+<br />
+<br />
+ :t<br />
+ + +<br />
+ + +<br />
+ 1/ F<br />
Fig. 414 Fig. 415<br />
will move to the remote edges of the plate and tneir field may be. neglected.)<br />
This distribution of the <strong>in</strong>duced charges does not depend on the thickness<br />
of the plate.<br />
L<strong>et</strong> us place a charge -Q on the left of the plate and at the same<br />
distance d. Obviously, the <strong>in</strong>duced positive charges will be distributed on<br />
the left side of the plate <strong>in</strong> the same manner as the negative charges on<br />
the right side. The charge -Q placed on the left of the plate will not<br />
cause a change <strong>in</strong> the electric field at the right of the plate. Thus, at the<br />
right of the plate, the electric field <strong>in</strong>duced by the charge +Q and the<br />
<strong>in</strong>duced negative charges co<strong>in</strong>cides with the field produced by the charges<br />
+Q and - Q and the charges <strong>in</strong>duced on the surfaces of the plate (Fig. 414).<br />
If the thickness of the plate is sm<strong>al</strong>l as compared with d, the plate<br />
may be regarded as <strong>in</strong>f<strong>in</strong>itely th<strong>in</strong> and hence the field created by the <strong>in</strong>duced<br />
charges outside the plate is zero.<br />
It has been shown that the field on the right of the plate produced by<br />
the charge +Q and the <strong>in</strong>duced negative charges is equ<strong>al</strong> to the field<br />
caused by the po<strong>in</strong>t charges +Q and -Q.<br />
S<strong>in</strong>ce the <strong>in</strong>tensity of the field caused by the <strong>in</strong>duced negative charges<br />
at the po<strong>in</strong>t where the charge +Q is equ<strong>al</strong>s the <strong>in</strong>tensity of the field caused<br />
by the po<strong>in</strong>t charge -Q at a distance of 2d from + Q, the sought force<br />
of attraction will be F= ~: .<br />
414. S<strong>in</strong>ce a and .b are much greater than c and d. it can be assumed<br />
that the plate is <strong>in</strong>f<strong>in</strong>itely large. Remember<strong>in</strong>g that the <strong>in</strong>tensity of a field<br />
caused by sever<strong>al</strong> charges is equ<strong>al</strong> to the sum of the <strong>in</strong>tensities produced by<br />
each of these charges, and us<strong>in</strong>g the results of the solutions of <strong>Problems</strong><br />
412 and 413, we can obta<strong>in</strong> the sought force:<br />
Qt<br />
2nQQ 1<br />
F=----<br />
ab 4d l<br />
The plus sign corresponds to the force of repulsion and the m<strong>in</strong>us sign to<br />
the force of attraction.
ELECT R ICITY AND MAGNETISM 301<br />
The positively charged plate will attract the positive po<strong>in</strong>t charge if<br />
Q~ 2nQQ 1<br />
4d 2 > ---aror<br />
415. The maximum charge that can be imparted to the sphere Is d<strong>et</strong>erm<strong>in</strong>ed<br />
by the equation<br />
The potenti<strong>al</strong> wilt<br />
be V= ~<br />
Q<br />
EO= R2<br />
=EoR=30,OOORV if the radius of the s~here<br />
is <strong>in</strong> cen tim<strong>et</strong>res.<br />
416. If the charged body is placed <strong>in</strong>to the centre of the sphere, an<br />
addition<strong>al</strong> charge -q uniformly distributed over the surface will obviously<br />
appear on the extern<strong>al</strong> surface of the sphere, and a charge + q on its<br />
<strong>in</strong>tern<strong>al</strong> surface. The potenti<strong>al</strong> VR at a distance of R from the centre of the<br />
sphere will be<br />
When the body moves <strong>in</strong>side the sphere, the outside field will not change.<br />
Therefore, the potenti<strong>al</strong> will be VR at any position of the charged body <strong>in</strong><br />
the sphere.<br />
417. The hous<strong>in</strong>g and the rod connected by the conductor will have equ<strong>al</strong><br />
potenti<strong>al</strong>s. For this reason the leaves will not deflect.<br />
After the conductor is removed and the rod is earthed, both leaves will<br />
deflect. This is the result of a potenti<strong>al</strong> difference appear<strong>in</strong>g b<strong>et</strong>ween the<br />
rod and the hous<strong>in</strong>g, s<strong>in</strong>ce the work of the electrostatic field is zero when<br />
the charge moves <strong>al</strong>ong a closed path ABCDEFA shown by the dotted l<strong>in</strong>e<br />
<strong>in</strong> Fig. 415. The work on section AC is zero, and the work on section AB<br />
is equ<strong>al</strong> to that on BC taken with the reverse sign. The potenti<strong>al</strong> difference<br />
b<strong>et</strong>ween the earth and the hous<strong>in</strong>g is equ<strong>al</strong> to that b<strong>et</strong>ween the hous<strong>in</strong>g<br />
and the rod.<br />
418. When the hous<strong>in</strong>g of the electrom<strong>et</strong>er is given, for <strong>in</strong>stance, a positive<br />
charge, electrostatic <strong>in</strong>duction will charge the b<strong>al</strong>l of the electrom<strong>et</strong>er rod<br />
positively and the end of the rod negatively. A potenti<strong>al</strong> difference will<br />
appear b<strong>et</strong>ween the hous<strong>in</strong>g and the rod, and for this reason both leaves<br />
will deflect. The potenti<strong>al</strong>s of the hous<strong>in</strong>g and the rod are positive with<br />
respect to the earth,. the potenti<strong>al</strong> of the hous<strong>in</strong>g be<strong>in</strong>g higher (that of the<br />
earth may be considered as zero).<br />
When the rod is connected to the earth, the potenti<strong>al</strong> difference b<strong>et</strong>ween<br />
the rod and the hous<strong>in</strong>g will <strong>in</strong>crease, as can be proved by the m<strong>et</strong>hod<br />
used <strong>in</strong> Problem 417. Therefore, the angle of deflection of the leaves will<br />
be greater.<br />
419. The electrom<strong>et</strong>er measures the potenti<strong>al</strong> difference b<strong>et</strong>ween the given<br />
body and the earth. S<strong>in</strong>ce the surface of the wire is equipotenti<strong>al</strong>, the leaves<br />
will deflect <strong>in</strong> the first case through the same angle wi th the b<strong>al</strong>l <strong>in</strong> any<br />
position (if the capacitance of the wire is negligibly sm<strong>al</strong>l).
302 ANSWERS AND SOLUTIONS<br />
In the second case the deflection of the leaves is d<strong>et</strong>erm<strong>in</strong>ed by the<br />
potenti<strong>al</strong> of the b<strong>al</strong>l with respect to the earth at the moment when the b<strong>al</strong>l<br />
is brought <strong>in</strong> contact with the electrom<strong>et</strong>er. This potenti<strong>al</strong> depends on the<br />
charge of the b<strong>al</strong>l, its dimensions and the arrangement of the surroundi ng<br />
objects. With a constant arrangement of the objects, the potenti<strong>al</strong> changes<br />
only when the charge of the b<strong>al</strong>l changes. When the b<strong>al</strong>l touches the buck<strong>et</strong>,<br />
the former acquires the potenti<strong>al</strong> of the latter, but the charge of the b<strong>al</strong>t<br />
will depend on the section of the surface be<strong>in</strong>g touched. If the <strong>in</strong>tern<strong>al</strong><br />
surface of the buck<strong>et</strong> is touched, the charge of the b<strong>al</strong>l will be zero, and<br />
if the extern<strong>al</strong> surface is touched, the charge will obviously be other<br />
than zero.<br />
As the charge is transferred by the b<strong>al</strong>l, its potenti<strong>al</strong> cont<strong>in</strong>uouslychanges,<br />
s<strong>in</strong>ce Its position with respect to the surround<strong>in</strong>g objects <strong>al</strong>so changes. As<br />
a result, this m<strong>et</strong>hod can b~ used to measure the distribution of the charge<br />
on the surface of the m<strong>et</strong><strong>al</strong>, but not its potenti<strong>al</strong>.<br />
420. The answer follows from the solution of Problem 419.<br />
In the second case the potenti<strong>al</strong> of the b<strong>al</strong>l and the read<strong>in</strong>gs of the<br />
electrom<strong>et</strong>er are d<strong>et</strong>erm<strong>in</strong>ed by the magnitude of the charge carried by the<br />
b<strong>al</strong>l from the surface of the wire be<strong>in</strong>g <strong>in</strong>vestigated. Upon contact with the<br />
wire. the potenti<strong>al</strong> of the b<strong>al</strong>l is the same irrespective of the po<strong>in</strong>t of contact.<br />
The capacitance of the b<strong>al</strong>l, however, depends on the shape of the section<br />
of the surface (particularly, on its curvature) which the b<strong>al</strong>l is brought <strong>in</strong><br />
contact with. Correspond<strong>in</strong>gly, the charge transferred to the b<strong>al</strong>l is <strong>al</strong>so<br />
d<strong>et</strong>erm<strong>in</strong>ed by the curvature of the section of contact.<br />
421. The potenti<strong>al</strong> of <strong>al</strong>l the po<strong>in</strong>ts of the sphere is the same. To solve<br />
the problem it is sufficient to f<strong>in</strong>d the potenti<strong>al</strong> of only one po<strong>in</strong>t. It is<br />
easiest to f<strong>in</strong>d the potenti<strong>al</strong> of the centre of the sphere. It Is equ<strong>al</strong> to the<br />
potenti<strong>al</strong> created at the centre of the sphere by a po<strong>in</strong>t charge U= ~ plus<br />
the potenti<strong>al</strong> created by the charges<br />
Q!<br />
We=-Qn =0, s<strong>in</strong>ce the<br />
potentia of an uncharged sphere is zero. The average potenti<strong>al</strong> of the sphere<br />
is cP=~.<br />
of a charged sphere:<br />
Upon multiply<strong>in</strong>g it by the <strong>in</strong>iti<strong>al</strong> charge, we obta<strong>in</strong> the energy
ELECTRICITY AND MAGNETISM 303<br />
The same result can be obta<strong>in</strong>ed if we use a diagram show<strong>in</strong>g the change<br />
<strong>in</strong> the potenti<strong>al</strong> of the sphere as the charge decreases. The diagram will<br />
have the form of a straight l<strong>in</strong>e pass<strong>in</strong>g at an angle to the axis of the<br />
charges, and the work will be numeric<strong>al</strong>ly equ<strong>al</strong> to the area limited by the<br />
diagram and the axes.<br />
Q2 RU"<br />
423. The energy of the charged sphere We=2R=-2-' where R is the<br />
radius of the sphere and U its potenti<strong>al</strong> (see Problem 422).<br />
Upon a discharge this energy will be liberated as heat. Express<strong>in</strong>g the<br />
energy <strong>in</strong> c<strong>al</strong>ories, we obta<strong>in</strong> We= 0.13 c<strong>al</strong>.<br />
424. The potenti<strong>al</strong> difference b<strong>et</strong>ween the b<strong>al</strong>ls should be WI' Besides, the heat Q is liberated<br />
<strong>in</strong> the conductor.<br />
Accord<strong>in</strong>g to the law of conservation of energy, however, the tot<strong>al</strong> amount<br />
of energy <strong>in</strong> the b<strong>al</strong>ls should be the same <strong>in</strong> both cases. S<strong>in</strong>ce the work WI<br />
and, correspond<strong>in</strong>gly, W 2' is the potenti<strong>al</strong> energy of the second b<strong>al</strong>l <strong>in</strong> the<br />
field of the first one <strong>in</strong> the first and second cases, then<br />
W 1+W<strong>et</strong> = W2+Q+W e2<br />
2 2<br />
where Wcl = ~: + ~; is the <strong>in</strong>tr<strong>in</strong>sic energy of the b<strong>al</strong>ls before connection<br />
2 q2<br />
and<br />
W e 2 = ;, + 2, is the <strong>in</strong>tr<strong>in</strong>sic energy of the b<strong>al</strong>ls after the charges are<br />
redistributed (see Problem 422).<br />
The energy liberated as heat is<br />
(qt -q2)2 (J.. _-!-)<br />
Q= Wel - We2+W 1 - W2 4 r I<br />
426. Assume that the radius of the envelope <strong>in</strong>creases by 6, which may<br />
be an <strong>in</strong>f<strong>in</strong>itely sm<strong>al</strong>l quantity. The expand<strong>in</strong>g force will perform the work<br />
W=4nR2f6, where f is the force per unit of area. This work is done at the<br />
expense of a reduction <strong>in</strong> the electrostatic energy. First the electrostatic
304 ANSWERS AND SOLUTIONS<br />
Fig. 416<br />
energy is 2 Q R 2 , and after expansion Q2<br />
2 (R+6)<br />
The change <strong>in</strong> the energy<br />
Q2 QI QI 6<br />
2R 2(R+6) 2 R(R+~)<br />
is equ<strong>al</strong> to the work W, I. e.,<br />
4nR'f~<br />
Q2~<br />
2R(R+6)<br />
Tak<strong>in</strong>g <strong>in</strong>to account the fact that 6 can be<br />
<strong>in</strong>f<strong>in</strong>itely sm<strong>al</strong>l, we obta<strong>in</strong> the follow<strong>in</strong>g expression<br />
for the force:<br />
f- Q2 -2'"'1'1'2<br />
-8nRt- ".v<br />
Here, (1= 4~~ is the density of the electricity, Le., the charge per unit<br />
of area.<br />
The sought force can <strong>al</strong>so be found directly. L<strong>et</strong> us consider a sm<strong>al</strong>l area<br />
a on the sphere (Fig. 416).<br />
L<strong>et</strong> us f<strong>in</strong>d the <strong>in</strong>tensity £1 of the electric field created on the area be<strong>in</strong>g<br />
considered by <strong>al</strong>l the charges except the ones on the area itself. To <strong>in</strong>troduce<br />
def<strong>in</strong>iteness, l<strong>et</strong> us consider the case when the sphere carries a positive<br />
charge.<br />
L<strong>et</strong> us denote by £2 the <strong>in</strong>tensity of the electric field created by the<br />
charges on the area itself. S<strong>in</strong>ce the result<strong>in</strong>g <strong>in</strong>tensity is zero <strong>in</strong>side the<br />
sphere, then £1 =£2·<br />
Q<br />
The result<strong>in</strong>g <strong>in</strong>tensity on the sphere E 1 +E J =W' and, therefore,<br />
Q<br />
2£1=W =4no. Hence, £1=2na.<br />
To f<strong>in</strong>d the force that acts from <strong>al</strong>l the charges outside t he area oft the<br />
charges on the area, the <strong>in</strong>tensity E 1 should be multiplied by the magnitude<br />
of the electric charge of the area aa:<br />
F=E 1oa=2no<br />
sa<br />
The force per unit of envelope area will be f=21tCJ 1 •<br />
427. For the charge q to be <strong>in</strong> equilibrium, the charges -Q should be<br />
at equ<strong>al</strong> distances a from it (Fig. 417). The. sum of the forces act<strong>in</strong>g on the<br />
charge - Q is <strong>al</strong>so zero:<br />
QS Qq<br />
4a2 -Qi'=O<br />
Hence, q= ~ . The distance a may have any v<strong>al</strong>ue. Equilibrium is unstable<br />
s<strong>in</strong>ce when the charge -Q is shifted <strong>al</strong>ong 00 1 to the left over a<br />
distance x, the force of attraction<br />
Q2<br />
F g = 4 (a+x)1
ELECTR ICITY AND MAGNETISM 305<br />
act<strong>in</strong>g from the side of the charge q is less than the force of repulsion<br />
Q2<br />
F Q=(2a+x)2<br />
and the charge -Q moves still farther from the position of equilibrium.<br />
When the charge -Q is shifted <strong>al</strong>ong 00 1 over a distance x towards the<br />
charge q, then F q > F Q for x
306 ANSWERS AND SOLUTIONS<br />
- -<br />
+<br />
+ +<br />
0 c + + +<br />
=<br />
A 9<br />
Fig. 418<br />
0'<br />
Fig. 419<br />
0' c'<br />
429. L<strong>et</strong> us first prove that the <strong>in</strong>tensity of the electric field at <strong>al</strong>l po<strong>in</strong>ts<br />
<strong>in</strong> the plane of section 00' is. directed perpendicular to this plane.<br />
L<strong>et</strong> us take an arbitrary po<strong>in</strong>t <strong>in</strong> the plane of the section and two sm<strong>al</strong>l<br />
areas arranged -arbitrarily but symm<strong>et</strong>ric<strong>al</strong>ly on the cyl<strong>in</strong>der with respect to<br />
section 00'. It is easy to see that the result<strong>in</strong>g <strong>in</strong>tensity of the field <strong>in</strong>duced<br />
by the charges on these areas will be directed <strong>al</strong>ong the axis of the cyl<strong>in</strong>der<br />
(Fig. 418). S<strong>in</strong>ce another element arranged symm<strong>et</strong>ric<strong>al</strong>ly with respect<br />
to the plane of the section can be found for each element. it follows that<br />
the <strong>in</strong>tensity produced by <strong>al</strong>l the elements will be par<strong>al</strong>lel to the axis of<br />
the cyl<strong>in</strong>der.<br />
L<strong>et</strong> us now prove that the <strong>in</strong>tensity will be the same at <strong>al</strong>l po<strong>in</strong>ts equidistant<br />
from the axis of the cyl<strong>in</strong>der.<br />
L<strong>et</strong> A and B be two such po<strong>in</strong>ts (Fig. 419). The <strong>in</strong>tensity of the field<br />
<strong>in</strong>side the cyl<strong>in</strong>der will not change if, apart from the available charge, each<br />
square centim<strong>et</strong>re of the cyl<strong>in</strong>der surface receives the same addition<strong>al</strong> negative<br />
charge so that the density of the charges at po<strong>in</strong>t C is zero. This is<br />
obvious from the fact that the field <strong>in</strong>side an <strong>in</strong>f<strong>in</strong>ite uniformly charged<br />
cyl<strong>in</strong>der is zero.<br />
In this case the densities of the charges will be distributed on the cyl<strong>in</strong>der<br />
surface as shown <strong>in</strong> Fig. 152. Therefore, the <strong>in</strong>tensities at po<strong>in</strong>ts A and<br />
B are the same.<br />
It now rema<strong>in</strong>s to prove that the <strong>in</strong>tensities of the fields at po<strong>in</strong>ts at<br />
different distances from the axis of the cyl<strong>in</strong>der are identic<strong>al</strong>.<br />
For this purpose l<strong>et</strong> us consider circuit BKLD (Fig. 420). With an electrostatic<br />
field the work <strong>in</strong> a closed circuit is known to be zero. The work is<br />
zero on sections KL and DB because the <strong>in</strong>tensity of the field is perpendicular<br />
to the path. The work on section BK is -EBI and on section LD it<br />
is EDl (as proved above, EB=EI(. and ED=E L ) .<br />
Hence, -EBl+EDl=O, i. e., E B =E D -<br />
It has thus been proved that the <strong>in</strong>tensity of the electric field at <strong>al</strong>l<br />
po<strong>in</strong>ts <strong>in</strong>side the cyl<strong>in</strong>der will be the same everywhere and directed <strong>al</strong>ong<br />
the axis of the cyl<strong>in</strong>der. It should be noted that such an arrangement of the<br />
charge on the surface of a conductor appears when direct current passes<br />
through it.<br />
430. The concept of capacitance can be used because the ratio b<strong>et</strong>ween<br />
the charge imparted to a conductor and the <strong>in</strong>crement of the potentia1 <strong>in</strong>du-
ELECTRICITY AND MAGNETISM<br />
307<br />
o<br />
=<br />
Fig. 420<br />
Hence, bear<strong>in</strong>g <strong>in</strong> m<strong>in</strong>d that the capacitance of the sphere is equ<strong>al</strong> to its<br />
radius, we f<strong>in</strong>d that<br />
c=V 3 X 6 X 10 4<br />
4n<br />
e: 25 cm<br />
Measurements give a close v<strong>al</strong>ue: C a: 30 em.<br />
432. The electrom<strong>et</strong>er will show the e. m. f. of the g<strong>al</strong>vanic cell irrespective<br />
of the capacitance.<br />
433. (I) U=4E ad=8.4 X lOC V.<br />
(2) A voltage of U 1 = E a d = 2.l X 10· V can be supplied to each "air" capacitor.<br />
The charge on the capacitor will be Q= C1U1= ~~. Upon series connection,<br />
the<br />
charge of <strong>al</strong>l the capacitors is the same. Therefore, the voltage<br />
on the capacitor with glass will be U 2 =c Q =~. The entire battery may<br />
2 e,.<br />
thus receive not more than U=3U 1+U2=6.6Xl04V.<br />
If the voltage exceeds 6.6X 10. V, <strong>al</strong>l four capaci tors will be punctured.<br />
The capacitor with the glass dielectric will be punctured last.<br />
434. When a charge moves <strong>in</strong> a closed circuit, the work of the forces of<br />
an electrostatic field is zero. Therefore<br />
The charges on the capacitors are the same, s<strong>in</strong>ce the sum of the charges<br />
present on the upper and the lower conductors is zero. Hence, Q=C1U 1 =<br />
= C2U2•<br />
20 *
308 ANSWERS AND SOLUTIONS<br />
+ -<br />
+0-~a.------....------.<br />
Fig. 421<br />
Therefore,<br />
u 1=e C2 c (cCl+cC2)=17.5 kV<br />
1+ 2<br />
u2= C C t e
ELECTRICITY AND MAGNETISM 309<br />
ted (Fig. 155). The result will be the equiv<strong>al</strong>ent diagram shown <strong>in</strong><br />
Fig. 422.<br />
The capacitance of the separate sections is" act 6C and 3C. The tot<strong>al</strong><br />
capacitance can be found from the formula<br />
121<br />
Co =3C + 6C<br />
Hence, Co= 1.2C.<br />
437. When the spark gaps are punctured, the par<strong>al</strong>lel connection of the<br />
capacitors automatic<strong>al</strong>ly changes to series connection; the voltage. b<strong>et</strong>ween<br />
the correspond<strong>in</strong>g plates of the capacitors grows, s<strong>in</strong>ce the capacitance of the<br />
system drops.<br />
Indeed, the high resistance of conductors AB and CD makes it possible<br />
to neglect the currents flow<strong>in</strong>g through them dur<strong>in</strong>g the time of discharge<br />
and consider them as <strong>in</strong>sulators through which the capacitors are not discharged.<br />
An equiv<strong>al</strong>ent diagram after the first spark gap is punctured is shown <strong>in</strong><br />
Fig. 423.<br />
Upon punctur<strong>in</strong>g of the first gap, the potenti<strong>al</strong> difference across the second<br />
gap will be equ<strong>al</strong> to the sum of the voltages across the first and the second<br />
capacitors. i. e., it will double. This will cause punctur<strong>in</strong>g of the second gap.<br />
When the n-th gap is punctured, the voltage <strong>in</strong> it will reach V=nVo.<br />
The resistances of conductors AB and CD should be high enough so as<br />
not to <strong>al</strong>low the capacitors to be discharged through them when the gaps<br />
are punctured with the plates connected <strong>in</strong> series.<br />
438. Yes, it will. Each plate has a def<strong>in</strong>ite, usu<strong>al</strong>ly sm<strong>al</strong>l,capacitance<br />
with respect to the earth (the force l<strong>in</strong>es are distorted near the edges of the<br />
plates and reach the earth).<br />
An equiv<strong>al</strong>ent diagram is shown <strong>in</strong> Fig. 424. The capacitance of the plates<br />
with respect to the earth is shown <strong>in</strong> the form of sm<strong>al</strong>l capacitances C 1 and<br />
C 2 •<br />
When the left-hand plate is short-circuited, part of the charge present <strong>in</strong><br />
It is neutr<strong>al</strong>ized. This will <strong>al</strong>so occur if the right-hand plate is short-circuited.<br />
The capacitor will cont<strong>in</strong>ue to be discharged the slower, the higher is the<br />
capacitance of the capacitor as compared with that of the plates relative to<br />
earth.<br />
439. The <strong>in</strong>iti<strong>al</strong> state of the system is illustrated by an equiv<strong>al</strong>ent diagram<br />
(Fig. 425a). The full charge of the capacitor is Q+ q. The force l<strong>in</strong>es<br />
of the charge + Q term<strong>in</strong>ate on the<br />
other plate of the capacitor. The +<br />
force l<strong>in</strong>es of the charges + q and +<br />
-q term<strong>in</strong>ate or start on the earth.<br />
S<strong>in</strong>ce C ~ C 1 =C~h then Q~ q.<br />
+<br />
+<br />
c<br />
Fig. 423 Fig. 424
310 ANSWERS AND SOLUTIONS<br />
+/l<br />
U<br />
-1/<br />
~u<br />
+1/-1/ -1/+1/<br />
8 r:<br />
C<br />
(b) (c) Fig. 425<br />
When the capacitor C 2 is short-circuited, the charge -q is neutr<strong>al</strong>ized.<br />
The potenti<strong>al</strong> difference b<strong>et</strong>ween the plates of the capacitor C should rema<strong>in</strong><br />
approximately equ<strong>al</strong> to U, s<strong>in</strong>ce q ~ Q. The work required to move the charge<br />
<strong>al</strong>ong the circuit ABCDA is zero. Therefore, the voltage across the capacitor<br />
C t should become equ<strong>al</strong> to ~ U and the charges on it to +2q and<br />
-2q. The charges on the plates of the capacitor C will be +Q -q and<br />
- Q + q, respectively (Fig. 425b).<br />
When the capacitor C is disconnected from the earth, the distribution of<br />
the charges and, hence, of the potenti<strong>al</strong>s will not change.<br />
If the capacitor C 1 is shorted, the charges will be redistributed as shown<br />
<strong>in</strong> Fig. 425c. It is only <strong>in</strong> this case that the required potenti<strong>al</strong>s relative to<br />
the earth will be obta<strong>in</strong>ed: zero on the left plate and ~ U on the right<br />
one. When the plates are earthed <strong>al</strong>ternately, the potenti<strong>al</strong> difference b<strong>et</strong>ween<br />
the plates will gradua lly drop because the charge decreases.<br />
440. No, they will not. When the plates are <strong>al</strong>ternately earthed the same<br />
processes will take place as <strong>in</strong> the absence of the battery (see Problem 439).<br />
The only difference is that the potenti<strong>al</strong> difference b<strong>et</strong>ween the plates is<br />
<strong>al</strong>ways kept constant.<br />
441. The tot<strong>al</strong> energy of the two capacitors before connection is<br />
1 (~ 2)<br />
Weo= 2 CIU1+C2U~<br />
and after connection<br />
W _..!..~_..!..(CIUl+C2U2)2<br />
er: 2 C 1+C2<br />
- 2 C1+C 9<br />
It is easy to see that Weo> We. The difference <strong>in</strong> the energies is<br />
Weo-We=c~f~s (U~+U:-2U1U2) > 0<br />
When U 1=U2 , we have W eo - We=O, and when C 1=C2 and U 2=O, then<br />
Weo=2We·<br />
The electrostatic energy dim<strong>in</strong>ishes because when the capacitors are connected<br />
by conductors, the charges flow from one capacitor to the other. Heat<br />
is liberated <strong>in</strong> the connect<strong>in</strong>g conductors. The quantity of heat evolved will<br />
not depend on the resistance of the conductors. When the resistance is low,<br />
the conductors will <strong>al</strong>low greater currents to flow through them, and vice<br />
versa.
ELECTRICITY AND MAGNETISM 311<br />
442. S<strong>in</strong>ce the dielectric is polarized, the <strong>in</strong>tensity will <strong>in</strong>crease at po<strong>in</strong>ts A<br />
and C and decrease at po<strong>in</strong>t B.<br />
2nQ<br />
443. E=-A-=50.2 CGSE.<br />
t,<br />
444. The capacitances and, therefore, the charges of the b<strong>al</strong>ls immersed <strong>in</strong><br />
kerosene <strong>in</strong>crease e, times:<br />
q~=e,ql' and q;=s,q.<br />
The force of <strong>in</strong>teraction of the charges <strong>in</strong> the dielectric, on the contrary,<br />
dim<strong>in</strong>ishes e, times.<br />
Hence,<br />
q q e q q e,~2rllr!<br />
F =_1_1_=.L...!....!.. ~. 0 0088 d<br />
8,R2 R2 R2(rl +r2)2 . yne<br />
The force of <strong>in</strong>teraction <strong>in</strong>creases e, times, while if the b<strong>al</strong>ls were disconnected<br />
from the battery it would decrease e, times.<br />
445. As a result of motion of the plates, the charge on the capacitor will<br />
be <strong>in</strong>creased by<br />
AQ=QI- Q,=!A (..!..._..!...)<br />
431 d. d 1<br />
tlA ( 1 1 )<br />
The battery will perform the work W = '~Q = 4n d. - d<br />
1<br />
The electrostatic energy of the capacitor will be <strong>in</strong>creased by<br />
AW,=Wes-W<strong>et</strong> =6~1_'~1=';~ (~I- ~J<br />
The mechanic<strong>al</strong> work WI was performed when the plates were moved closer<br />
.to each other. On the basis of the law of conservation of energy, W = W1+AWe.<br />
Therefore,<br />
W 1=W-AW<br />
=,IA (..!..._..!...)<br />
e 8n d, d 1<br />
At the expense of the work of the battery, the electrostatic energy of the<br />
capacitor <strong>in</strong>creased and the mechanic<strong>al</strong> work W1 was done.<br />
448. L<strong>et</strong> us consider for the sake of simplicity a dielectric <strong>in</strong> the form of<br />
a homogeneous very elongated par<strong>al</strong>lelepiped (Fig. 426).<br />
L<strong>et</strong> us resolve the field Eo <strong>in</strong> which the piece of dielectric (for example,<br />
8 rod) is placed <strong>in</strong>to components directed <strong>al</strong>ong the rod and perpendicular<br />
to it. These components will cause bound charges to appear on surfaces AB,<br />
CD, Be and AD. The field of the bound charges b<strong>et</strong>ween surfaces AD, BC,<br />
and AB. DC weakens the components' of the field Eo <strong>in</strong>side the dielectric,<br />
the component perpendicular to the rod be<strong>in</strong>g weakened more s<strong>in</strong>ce the bound<br />
. charges on surfaces AD and Be are close to each other and their field is<br />
similar to the homogeneous field of a plane capacitor, while the charges on<br />
the surfaces of the sm<strong>al</strong>l area are moved far apart. For this reason the full<br />
field <strong>in</strong>side the dielectric will not co<strong>in</strong>cide <strong>in</strong> direction with the field Eo.<br />
The dipoles appear<strong>in</strong>g will therefore be oriented not <strong>al</strong>ong Eo, but <strong>al</strong>ong a<br />
certa<strong>in</strong> direction OP form<strong>in</strong>g the angle p with Eo. (This refers to both ordi<br />
~ary and dipole molecules.) From an electric<strong>al</strong> standpo<strong>in</strong>t, a polarized dielectnc<br />
can be regarded as a large dipole form<strong>in</strong>g the angle ~ with the field Ea.
312 ANSWERS AND SOLUTIONS<br />
A<br />
JJ<br />
a<br />
Fig. 426<br />
The dielectric will rotate <strong>in</strong> this field until it occupies a position<br />
<strong>al</strong>ong the<br />
field. The field of the bound charges is an <strong>in</strong>tern<strong>al</strong> force and cannot cause<br />
rotation of the dielectric.<br />
447. (a) The capacitance of the capacitor will be equ<strong>al</strong> to that of capacitors<br />
connected <strong>in</strong> par<strong>al</strong>lel, one of which is filled with a dielectric, and the<br />
other not, i. e.,<br />
C=E rAl + 1 A (1-11) A {1+< -1)2}<br />
4n.dl 4ndl 4nd 8, I<br />
(b) The electric field b<strong>et</strong>ween the plates of the capacitor will not change,<br />
and, consequently, the capacitance will not change if the upper surface of<br />
the dielectric is coated with an <strong>in</strong>f<strong>in</strong>itely th<strong>in</strong> layer of a conductor. Therefore,<br />
the sought capacitance will be equ<strong>al</strong> to the capacitance of two capacitors<br />
connected <strong>in</strong> series:<br />
Hence,<br />
c<br />
COCl where Co A and C _ erA<br />
C O+C1<br />
' 4n (d-d 1<br />
) 1- 4M 1<br />
C<br />
erA<br />
4", {d 1 +e,. (d-d 1 ) }<br />
448. To simplify reason<strong>in</strong>g, assume that two par<strong>al</strong>lel m<strong>et</strong><strong>al</strong>lic plates carry<strong>in</strong>g<br />
the charges +Q and -f2 are placed <strong>in</strong>to a liquid dielectric. The <strong>in</strong>tensity<br />
of the electric field b<strong>et</strong>ween the plates is E=41tQ • The <strong>in</strong>tensity of<br />
e, A<br />
the field <strong>in</strong>duced by each plate will be.<br />
E<br />
Q<br />
1=E2=21t 8 r A<br />
L<strong>et</strong> us f<strong>in</strong>d the force act<strong>in</strong>g, for example, from the side of the first plate<br />
on the second. For this purpose the <strong>in</strong>tensity of the field <strong>in</strong>duced by the<br />
first plate should be multiplied by the charge on the second one. Thus,<br />
F= 2nQ2<br />
erA<br />
L<strong>et</strong> us assume that the first plate is fixed and the second can slowly move<br />
(we disregard the change <strong>in</strong> the mechanic<strong>al</strong> energy of the dielectric). The<br />
work that the electric field can perform <strong>in</strong> mov<strong>in</strong>g the plates up to direct<br />
contact Is equ<strong>al</strong> to the product of the force F (constant) by the distance d,
ELECTRICITY AND MAGNETISM 313<br />
+<br />
+<br />
+<br />
+<br />
i. e.,<br />
W=Pd=21t Q 2<br />
d<br />
8 r A<br />
This work is done at the expense<br />
of a reduction <strong>in</strong> the electric energy<br />
of the capacitor. Thus, the electrostatic<br />
energy will be equ<strong>al</strong> to<br />
2:nQ 2d Q2 QU<br />
we=e;r=2C o~ We=T<br />
where U is the potenti<strong>al</strong> difference.<br />
The above formula is true for any<br />
dielectric.<br />
449. The <strong>in</strong>tensity of the field<br />
. <strong>in</strong>side the space consists of E and<br />
Flg. 427<br />
the <strong>in</strong>tensity <strong>in</strong>duced by the charges<br />
that appear on the surfaces of the<br />
space ow<strong>in</strong>g to polarization of the dielectric (Fig. 421). In the first case the<br />
<strong>in</strong>fluence of the polariziof charges is negligibly sm<strong>al</strong>l, and E 1 =E. In the<br />
second case the action 0 the charges exerted on the surfaces of the space is<br />
fully compensated oy the action of the charges on the surfaces of the dielectric<br />
adjo<strong>in</strong><strong>in</strong>g the m<strong>et</strong><strong>al</strong> plates of the capacitor. For this reason the sought<br />
<strong>in</strong>tensity is equ<strong>al</strong> to that which would be <strong>in</strong>duced if the dielectric were removed<br />
<strong>al</strong>tog<strong>et</strong>her, i. e.,<br />
E 2 = 8, E<br />
450. If the dielectric is drawn <strong>in</strong>to the capacitor over the distance x, the<br />
energy of the capacitor will be<br />
QI AU2 1<br />
Wel =2C = 8nd<br />
x<br />
1+(e,-I)T<br />
s<strong>in</strong>ce C= 4~ {1+(8,-1) : } and Q= 4~ U (see Problem 447).<br />
If x is <strong>in</strong>creased by 6, the energy will be reduced and become equ<strong>al</strong> to<br />
AU' 1<br />
W'<strong>et</strong>=81td<br />
The difference b<strong>et</strong>ween the energies<br />
1+(8r-l)'~+6<br />
. I<br />
AUI<br />
6<br />
(8,-1)7<br />
11' e1 - 11' 61 = 8nd {1+(8,-I)xt6}{I+(8,-I)~}<br />
will be equ<strong>al</strong> to the work of the sought force F over the path 6. Gener<strong>al</strong>ly<br />
speak<strong>in</strong>g, the magnitude of the force will change over this path, but if a
314 ANSWERS AND SOLUTIONS<br />
sufficiently sm<strong>al</strong>l v<strong>al</strong>ue of ~<br />
It follows therefore that<br />
is taken we can write<br />
W e1 - W el=F6<br />
F_AU2 8,.-1<br />
-SMi {l+
ELECTRJCITY AND MAGNETISM 315<br />
the lift<strong>in</strong>g force act<strong>in</strong>g on the b<strong>al</strong>ls <strong>in</strong> the left-hand part will be the same<br />
as <strong>in</strong> the right-hand one despite the fact that the left-hand part conta<strong>in</strong>s<br />
more b<strong>al</strong>ls. .<br />
453. The <strong>in</strong>tensity of the field <strong>in</strong> the dielectric will be<br />
U Q 4nQ 4n<br />
E=cr=Cd=e,A =£;0 0<br />
Q<br />
where oo=A.<br />
This <strong>in</strong>tensity is created by the free charges on the plates of the capacitor,<br />
and by the bound charges <strong>in</strong>duced ow<strong>in</strong>g to polarization of the dielectric.<br />
The bound charges are on the surface of the dielectric. L<strong>et</strong> us denote the<br />
density of these charges by 01 (Fig. 428). The <strong>in</strong>tensity of the field created<br />
by the free charges is Eo=4noo, and that of the field created by the bound<br />
charges is E1=4nol.<br />
Thus the result<strong>in</strong>g <strong>in</strong>tensity E=E o-E1 =4n(°0-01)=41(;0 0 .<br />
2r<br />
This equation can be used to f<strong>in</strong>d 0 1 = 8,.-1 00.<br />
e,<br />
To d<strong>et</strong>erm<strong>in</strong>e the force act<strong>in</strong>g, for example, on the upper surface of the<br />
dielectric, it is necessary to c<strong>al</strong>culate the <strong>in</strong>tensity of the field on this surface<br />
created by <strong>al</strong>l the charges except for the charges on the surface itself. This<br />
<strong>in</strong>tensity will be equ<strong>al</strong> to<br />
E1 2 2 er +1<br />
Eo- - =4Mo- no!= noo--<br />
2 e,<br />
The force act<strong>in</strong>g on the upper surface of the dielectric will be directed upwards<br />
and be equ<strong>al</strong> to<br />
F 2<br />
8 r+18,.-1 A<br />
= 1«10----°0<br />
t,. e,<br />
An identic<strong>al</strong> force will act on the lower surface of the dielectric.<br />
Thus, the dielectric will be str<strong>et</strong>ched and a unit of area of the dielectric<br />
will be acted upon by a force equ<strong>al</strong> to<br />
t<br />
f =2n(Jo --I-<br />
e'-I<br />
e,<br />
454. S<strong>in</strong>ce the action of the field causes compl<strong>et</strong>e orientation of the molecules,<br />
the ends of the "dumb-bells" with negative electric charges will be <strong>in</strong><br />
___~ +<br />
A<br />
+ + ± + +<br />
+<br />
Fig. 42~<br />
~E"<br />
tEr<br />
61<br />
+ + .+ + §,<br />
(So<br />
+ + + + + f + +8<br />
Fig. 429
316 ANSWERS AND SOLUTIONS<br />
a layer with a thickness I near the positively charged plate, and the ends<br />
of the "dumb-bells" with a positive electric charge near the negatively charged<br />
plate (Fig. 429).<br />
•<br />
The quantity of negative and positive charges is -the same at a distance<br />
greater than 1 from the plates <strong>in</strong>side the dielectric.<br />
The <strong>in</strong>tensity of the field <strong>in</strong>side the dielectric produced by the charges <strong>in</strong><br />
layers A and B is E t=41tOt=41tQln and the tot<strong>al</strong> <strong>in</strong>tensity is £=£0<br />
- £1=Eo- 4n Qln.<br />
It should be noted, however, that the orient<strong>in</strong>g action of the field will<br />
<strong>al</strong>ways be hampered by the disorient<strong>in</strong>g therm<strong>al</strong> motion, which is not taken<br />
<strong>in</strong>to account <strong>in</strong> the c<strong>al</strong>culations.<br />
455. L<strong>et</strong> us denote the sought full <strong>in</strong>tensity of the field <strong>in</strong> the dielectric<br />
by E. The distance lover which the charges moved apart <strong>in</strong> each molecule<br />
can be found from the expression kl = qE. As <strong>in</strong> solv<strong>in</strong>g Problem 454" we<br />
obta<strong>in</strong><br />
and<br />
Q 2 n<br />
E=Eo-E I =Eo- 4n -k- E<br />
The permittivity 8,. can be d<strong>et</strong>erm<strong>in</strong>ed from the ratio E = Eo .<br />
e,<br />
Q2<br />
Hence, e,=1+4nTn.<br />
456. When the charges + Q and -Q move apart <strong>in</strong> the molecule over<br />
the distance I, the work k;2 is performed (see Problem 137).<br />
The energy storedIn the dielectric is<br />
kl 2<br />
We 1= T N<br />
where N = ALn= Vn is the number of molecules <strong>in</strong> the volume V of the dielectric<br />
located b<strong>et</strong>ween the plates of the capacitor. Thus.<br />
kl 2<br />
We1=nT V<br />
. QE Q2E2<br />
S<strong>in</strong>ce 1=-;;-, then W e1 = n ~V.<br />
nQ2 e -1 nQ2<br />
By express<strong>in</strong>g -k- through B,.. i.e., "4n =r we obta<strong>in</strong> for Wei<br />
W<br />
The tot<strong>al</strong> energy of the capacitor is<br />
el-sn<br />
_e,.-1 E 2V<br />
W _Q2 _ e,. E 2V<br />
e-2C-Sn
ELECTRICITY AND MAGNETISM 317<br />
This energy We can be represented as the sum of purely electrostatic<br />
energy<br />
£2<br />
Weo= BnV<br />
and the energy stored <strong>in</strong> the dielectric<br />
W =er-l EIV<br />
el 831<br />
3-2. Direct Current<br />
457. When a direct current flows <strong>al</strong>ong a conductor, the electric field <strong>in</strong>side<br />
it is constant and directed <strong>al</strong>ong it. If the charge moves <strong>in</strong> a closed circuit abed<br />
(Fig. 430) the work of the electric field is zero. L<strong>et</strong> us assume the sections<br />
ad and be to be <strong>in</strong>f<strong>in</strong>itely sm<strong>al</strong>l so that the work done on them may be<br />
neglected. Therefore, the work <strong>al</strong>ong ab is equ<strong>al</strong> to that a long de. For this<br />
reason the tangenti<strong>al</strong> component of the field near the surface of the conductor<br />
should be equ<strong>al</strong> to the field <strong>in</strong>side it.<br />
458. The arrangement of the force l<strong>in</strong>es is shown <strong>in</strong> Fig. 431. The <strong>in</strong>crease<br />
<strong>in</strong> the slope of the l<strong>in</strong>e near the bend can be expla<strong>in</strong>ed by the fact that the<br />
tangenti<strong>al</strong> component of the field at the surface of the conductor with an<br />
<strong>in</strong>variable cross section is constant (see Problem 457), while the norm<strong>al</strong> component<br />
decreases towards the curvature s<strong>in</strong>ce the potenti<strong>al</strong> difference b<strong>et</strong>ween<br />
the correspond<strong>in</strong>g sections ly<strong>in</strong>g on the opposite sides of the arc dim<strong>in</strong>ishes.<br />
459. By apply<strong>in</strong>g Ohm's law to section AB of the circuit, we have<br />
Hence,<br />
..!-=2-+J..<br />
V r R<br />
r-!.-__I- e::!.- (1-~)<br />
- I 1-~ I lR<br />
IR<br />
V '<br />
s<strong>in</strong>ce 'R =0.008 ~ 1: f<strong>in</strong><strong>al</strong>ly ,=20.16 Q.<br />
480. A reduction <strong>in</strong> the sensitivity n times means that the g<strong>al</strong>vanom<strong>et</strong>er<br />
carries a current /1 which is n times sm<strong>al</strong>ler than the current <strong>in</strong> the rest of<br />
FiS. 431
318 ANSWERS AND SOLUTIONS<br />
the circuit before the branch<strong>in</strong>g off. Therefore, the current /2 through the<br />
shunt is n -1 of the current J <strong>in</strong> the rest of the circuit. Hence,<br />
n<br />
r J1 1<br />
7[=J;,==n-l<br />
Therefore, r=-!l-l as 204C.<br />
n-<br />
Vo-V<br />
461. The error sought Is e=v;-, where Vo is the vol.tage across the<br />
resistance R before the voltm<strong>et</strong>er is switched on and V the voltage after it<br />
is switched OD. ..<br />
Accord<strong>in</strong>g to Ohm's law, Vo=IR and V-I R~R.' where Ro is the resistance<br />
of the voltm<strong>et</strong>er. Hence,<br />
RIRo<br />
£= R<br />
1+R o<br />
is d<strong>et</strong>erm<strong>in</strong>ed only by the ratio b<strong>et</strong>ween the resistances of the section of the<br />
circuit and the voltm<strong>et</strong>er. When Ro~ R, the error may be neglected.<br />
482. Before the amm<strong>et</strong>er is connected, 1 0 - ~ , and after It Is connected<br />
J= R~Ro' where Ro Is the resistance of the amm<strong>et</strong>er. The error is<br />
/0-1 1<br />
8=--=---<br />
/0 1+~ Ro<br />
When R o ~ R the error may be neglected.<br />
463. In series connection, the resistance of the circuit is<br />
R= Rot+ R02+«ZtRolt +aZRo2t<br />
On the other hand we can write R=Ro(l+a't), where Ro=Rol+Rol and<br />
'1/ is the sought temperature coefficient. Therefore,<br />
a/ R01(J,1+R0 2(J,2<br />
R01+Ro2<br />
where<br />
R 01Ro2 Ro<br />
R0 1+R02<br />
Omitt<strong>in</strong>g the terms proportion<strong>al</strong> to the products of the temperature coefficients<br />
as be<strong>in</strong>g sm<strong>al</strong>l. we obta<strong>in</strong><br />
rL"<br />
R0 2rL l + R01CXt<br />
R01+Ro2
ELECTRICITY AND MAGNETISM<br />
319<br />
9<br />
A<br />
a<br />
R<br />
4<br />
Fig. 432 Fig. 433<br />
464. Po<strong>in</strong>ts A and C have the same potenti<strong>al</strong>s because they are connected<br />
by a wire whose resistance can be neglected. The potenti<strong>al</strong>s of po<strong>in</strong>ts Band<br />
D are <strong>al</strong>so identic<strong>al</strong>.· For this reason the ends of resistors A, C and, correspond<strong>in</strong>gly,<br />
B, D can be assumed as connected tog<strong>et</strong>her. Thus, resistors AB,<br />
CB and CD are connected <strong>in</strong> par<strong>al</strong>lel. The equiv<strong>al</strong>ent diagram is shown <strong>in</strong><br />
Fig. 432.<br />
The tot<strong>al</strong> resistance is R13.<br />
465. Ow<strong>in</strong>g to symm<strong>et</strong>ry, it is obvious that the current <strong>in</strong> conductor J·7<br />
is equ<strong>al</strong> to that <strong>in</strong> 7-4, the current 2-7 is equ<strong>al</strong> to 7-3, and the current<br />
6..7 to 7-5 (see Fig. 165). For this reason the distribution of the currents<br />
and, hence, the resistance of the hexagon will not change if conductors 2..7,<br />
7..3, 6..7 and 5..7 are disconnected from the centre (Fig. 433). It is easy<br />
to c<strong>al</strong>cula te the resistance of this circuit, which is equivaJent to the <strong>in</strong>iti<strong>al</strong><br />
one. The resistance of the upper and lower parts is 8R/3.<br />
The tot<strong>al</strong> resistance R x can be found from the relation<br />
1 1 6<br />
R x = 2R + 8R<br />
4<br />
Hence, Rx=SR.<br />
468. Ow<strong>in</strong>g to symm<strong>et</strong>ry, it is obvious that the potenti<strong>al</strong>s of the cube<br />
vertices 2, 9 and 6 are the same, as are those of vertices 4, 5 and 7 (see<br />
Fig. 166).<br />
Therefore, vertices 2, 9, 6 and 4, 5, 7 can be connected by conductors<br />
without resistance, i.e., by bus-bars. This will not change the resistance of<br />
the cube. The bus-bars will thus be connected by six conductors 2-7, 2-4,<br />
3-5, 3-4, 6-7 and 6..5. The resistance of the circuit (Fig. 434) is equ<strong>al</strong><br />
to the sought resistance of the cube:<br />
Rx=!i+.B-+!i=~R<br />
3 636<br />
487. The resistance of section CD is equ<strong>al</strong> to RCD= R~~r=40Q, and<br />
that of the entire circuit is RAB=R 1 + RCD= 100Q. The current J=R U = 1.2 A.<br />
AD<br />
The voltage drop across section CD is U 1<br />
= JRcD= 48 V.
320 ANSWERS AND SOLUTIONS<br />
468. The resistance b<strong>et</strong>ween po<strong>in</strong>ts A and B is<br />
where<br />
R<br />
AB<br />
('a+'b) 'c<br />
_2_ +1.. 'a+'b+ 2,c<br />
'a+'b 'c<br />
The resistance b<strong>et</strong>ween po<strong>in</strong>ts C and D can be found if we consider the<br />
currents flow<strong>in</strong>g <strong>in</strong> the branches of the circuit (see Fig. 435). It is obvious<br />
from considerations of symm<strong>et</strong>ry that the currents <strong>in</strong> conductors DB and ACt<br />
and <strong>al</strong>so <strong>in</strong> AD and BCt are mutu<strong>al</strong>ly equ<strong>al</strong>, the current <strong>in</strong> AD be<strong>in</strong>g equ<strong>al</strong><br />
to it·-t- i 2 , s<strong>in</strong>ce the sum of the currents at junction A is equ<strong>al</strong> to zero.<br />
On section DAC<br />
and on section DARC<br />
(i 1 + i2) , a + i l'b= UDC<br />
Hence,<br />
The sought resistance is<br />
U CD<br />
'a+'c<br />
u<br />
2'a'b+'a'c + rv, DC<br />
'b-'a U<br />
2, a'b+'a'c+ 'b'c DC<br />
UCD<br />
RCD=-'- =2· +.<br />
£1 £2<br />
2'a'b+'c (ra+'b)<br />
Ta+'b+2 , c<br />
469. If no current flows through the g<strong>al</strong>vanom<strong>et</strong>er, the potenti<strong>al</strong>s of po<strong>in</strong>ts<br />
C and D are the same and the current /1 pass<strong>in</strong>g through the resistance Rx<br />
is equ<strong>al</strong> to the current flow<strong>in</strong>g through the resistance ROt while the current<br />
/2 <strong>al</strong>ong slide wire AB is the same <strong>in</strong> <strong>al</strong>l the cross sections.<br />
Fig. 434 Fig. 435
ELECTRICITY AND MAGNETISM 321<br />
Fig. 436 Fig. 437<br />
Accord<strong>in</strong>g to Ohm's law,<br />
and f<strong>in</strong><strong>al</strong>ly<br />
/lR x = / 2l 1 ~ and /lR o=/'},12 ~<br />
where p is the resistivity and A is the cross section of the slide wire.<br />
R 1<br />
Therefore, R: = 1:<br />
470. Such a resistance r should be connected b<strong>et</strong>ween po<strong>in</strong>ts C and D that<br />
the resistance of the last cell (Fig. 436) is <strong>al</strong>so r, In this case the last cell<br />
can be rep laced by the resistance r, then the same can be done with the next<br />
to last cell, <strong>et</strong>c. Now the tot<strong>al</strong> resistance of the circuit will not depend on<br />
the number of cells and will be equ<strong>al</strong> to r,<br />
The follow<strong>in</strong>g equation can be written for r:<br />
(2R+r) R r<br />
3R+r<br />
Hence, r=R(Y3-1)~O.73R.<br />
471. The last cell is a voltage divider that reduces the potenti<strong>al</strong> of the ~-th<br />
po<strong>in</strong>t k times as compared with the (n-l)th po<strong>in</strong>t. Hence, Un=<br />
U n -<br />
= R +R 1 R U n - 1 1 k 1 ( P' 437)<br />
3 =-k-' or<br />
1 s<br />
R = - see 19. ·<br />
3<br />
The relation Uj= U~-1 should be true for any cell. For this reason the<br />
'resistance of the last cell, of the last two. of the last three, <strong>et</strong>c .• cells should<br />
,<strong>al</strong>so be R 3 (see Problem 470). Therefore,<br />
..!.-=..!-+ 1<br />
e, R 2 R 1 +R 3<br />
R Rs (R1 +Rs) R<br />
2 a<br />
.s:<br />
R 1 k-l<br />
R 1 : R 2 : R s = (k- l )2:k :(k- l )<br />
472. Devices whose action is based, for example, on the deflection of a<br />
current-carry<strong>in</strong>g conductor <strong>in</strong> a magn<strong>et</strong>ic field cannot be used. The angle<br />
21-2042
322 ANSWERS AND SOLUTIONS<br />
through which the po<strong>in</strong>ter is deflected <strong>in</strong> such a device is proportion<strong>al</strong> to the<br />
current flow<strong>in</strong>g through it. D<strong>et</strong>erm<strong>in</strong>ation of the potenti<strong>al</strong> difference with the<br />
aid of this k<strong>in</strong>d of devices is based on Ohm's law: the current flow<strong>in</strong>g through<br />
a voltm<strong>et</strong>er is proportion<strong>al</strong> to the potenti<strong>al</strong> difference applied to it. To verify<br />
Ohm's law, an electrostatic voltm<strong>et</strong>er and an ord<strong>in</strong>ary amm<strong>et</strong>er are required.<br />
473. L<strong>et</strong> us denote the charges on the first and second capacitors at the<br />
moment of time t by ql and q2" The latter are related by the expressions<br />
S<strong>in</strong>ce<br />
then<br />
ql q2<br />
ql+ qz= Q and -C=-C<br />
A and c,<br />
4n (do+vt)<br />
1 2<br />
A<br />
4n (do-vt)<br />
and it therefore follows that<br />
do-vt<br />
ql=Q ---u;;- and<br />
do+vt<br />
qt=Q--u;-<br />
The reduction <strong>in</strong> the charge on the first capacitor is equ<strong>al</strong> to the <strong>in</strong>crease<br />
of the charge on the second capacitor. The current <strong>in</strong>tensity is<br />
1=_ ~ql = ~q2 =~<br />
~t 6.1 2d o<br />
The current will flow from the positively charged plate of the first capacitor<br />
to the positively charged plate of the second capacitor.<br />
474. The forces of attraction act<strong>in</strong>g b<strong>et</strong>ween the plates of the capacitors<br />
are equ<strong>al</strong>, respectively, to<br />
r.=2 n !!.L ql =~ QZ (dO- vt)2<br />
A 2 A d~<br />
for the first capacitor and to<br />
F _ 11; Q2 (d O + vt)2<br />
2-2 A d~<br />
for the second capacitor (see Problem 473).<br />
S<strong>in</strong>ce the plates of the first capacitor move apart, the forces of the electrostatic<br />
field perform the negative work WI. These forces perform the positive<br />
work W 2 <strong>in</strong> the second capacitor. The work AW done by the field when each<br />
plate moves over a sm<strong>al</strong>l distance Ax is<br />
2nQ2 x<br />
AW =f1W1 +AW2=(F 2 - F 1 ) 6.x=-A- do L\x<br />
where x=vt.<br />
Thus the work performed on a sm<strong>al</strong>l section is proportion<strong>al</strong> to th e displacement<br />
x, as when a spr<strong>in</strong>g is str<strong>et</strong>ched. Therefore, the tot<strong>al</strong> work can be found<br />
with the aid of the m<strong>et</strong>hod employed <strong>in</strong> solv<strong>in</strong>g Problem 137;<br />
nQ<br />
w=--<br />
2<strong>al</strong><br />
Ad:
ELECTRICITY AND MAGNETISM 323<br />
R<br />
The work W can <strong>al</strong>so be c<strong>al</strong>culated<br />
by another m<strong>et</strong>hod. S<strong>in</strong>ce the resistance<br />
of the connect<strong>in</strong>g wires is zero, the<br />
quantity of heat liberated is <strong>al</strong>so zero.<br />
For this reason the change <strong>in</strong> the electrostatic<br />
energies of the two capacitors<br />
will be equ<strong>al</strong> to the work of the electrostatic<br />
field.<br />
At the moment of time t, the energies<br />
of the first and the second capacitors<br />
will be<br />
We1= 2C<br />
and<br />
q~ n QI<br />
=T--.(do-vt)I(do+vt)<br />
1 Ado<br />
Fig. 498<br />
The tot<strong>al</strong> energy<br />
Q~ I<br />
We= Wei +We,= 1t Ado (do-a l )<br />
Therefore, the energy will drop by AWe = ~~: a l dur<strong>in</strong>g the time t. This<br />
change wi II be equ<strong>al</strong> to the work W of the electrostatic field.<br />
475. Rubb<strong>in</strong>g of the clothes aga<strong>in</strong>st the chair produces electrification, and<br />
the body of the experimenter and the chair fonn a sort of capacitor. When<br />
the experimenter stands up, the capacitance of this capacitor sharply decreases,<br />
and therefore the potenti<strong>al</strong> difference sharply <strong>in</strong>creases b<strong>et</strong>ween the chair<br />
(i. e., "earth") and the experimenter's body. The body should obviously be<br />
well <strong>in</strong>sulated from the earth (rubber soles).<br />
When the experimenter touches the table, the potenti<strong>al</strong> difference b<strong>et</strong>ween<br />
his hand and the earth levels out. An electric current is generated, a negligible<br />
part of which is branched off <strong>in</strong>to the g<strong>al</strong>vanom<strong>et</strong>er. To deflect the<br />
po<strong>in</strong>ter, the resistance b<strong>et</strong>ween one end of the g<strong>al</strong>vanom<strong>et</strong>er coil and the<br />
earth should be lower than that b<strong>et</strong>ween the other end and the earth.<br />
The path of the current is shown schematic<strong>al</strong>ly <strong>in</strong> Fig. 438. Here W is the<br />
W<strong>in</strong>d<strong>in</strong>g of the g<strong>al</strong>vanom<strong>et</strong>er, K is the key and R shows a very high but<br />
f<strong>in</strong>ite resistance b<strong>et</strong>ween one of the w<strong>in</strong>d<strong>in</strong>g ends and the earth.<br />
The po<strong>in</strong>ter of the g<strong>al</strong>vanom<strong>et</strong>er deflects despite the tremendous resistance<br />
of the circuit ow<strong>in</strong>g to the great potenti<strong>al</strong> difference appear<strong>in</strong>g when the<br />
capacitance is reduced.<br />
478. There is obviously a def<strong>in</strong>ite asymm<strong>et</strong>ry b<strong>et</strong>ween the conductors which<br />
the ends of the g<strong>al</strong> vanom<strong>et</strong>er w<strong>in</strong>d<strong>in</strong>g are connected to. This occurs when the<br />
resistance of the <strong>in</strong>sulation b<strong>et</strong>ween one end of the coil and the earth is less<br />
than b<strong>et</strong>ween the earth and the other end. It should <strong>al</strong>so be taken <strong>in</strong>to<br />
account that the resistance b<strong>et</strong>ween the conductors com<strong>in</strong>g from the coil is<br />
not <strong>in</strong>f<strong>in</strong>ite, despite good <strong>in</strong>sulation.<br />
The path of the current is illustrated <strong>in</strong> Fig. 439. W is the w<strong>in</strong>d<strong>in</strong>g of the<br />
g<strong>al</strong>vanom<strong>et</strong>er, C 1 and C" are the conductors lead<strong>in</strong>g from the ends of the<br />
21 *'
324<br />
ANSWERS AND SOLUTIONS<br />
w<br />
Fig. 439<br />
w<strong>in</strong>d<strong>in</strong>g, E is the earth and R lt R 2 and R s show sche~atic<strong>al</strong>ly very high<br />
but f<strong>in</strong>ite resistances appear<strong>in</strong>g because the <strong>in</strong>sulation IS not ide<strong>al</strong>; Rs ~<br />
~ R 1 + R?,. The dash l<strong>in</strong>e shows the path of the current if a negatively<br />
charged body is brought up to Ct. If the body is brought up to C l , the current<br />
path is shown by dots. In both cases the current will flow through the w<strong>in</strong>d<strong>in</strong>g<br />
of the g<strong>al</strong>vanom<strong>et</strong>er <strong>in</strong> the same direction. .<br />
This problem illustrates the presence of conductivity <strong>in</strong> <strong>al</strong>l bodies, which<br />
is especi<strong>al</strong>ly important <strong>in</strong> work<strong>in</strong>g with sensitive devices.<br />
477. Po<strong>in</strong>t A <strong>in</strong> Fig. 440 shows the potenti<strong>al</strong> of the positive (copper) electrode<br />
and po<strong>in</strong>t D that of the negative (z<strong>in</strong>c) electrode. In the ZnSO. solution,<br />
the z<strong>in</strong>c electrode is charged negatively as a result of evolution of<br />
positive ions of Zn, while the copper electrode <strong>in</strong> the CuS0 4 solution is·<br />
charged positively s<strong>in</strong>ce it receives positive ions of Cu. The potenti<strong>al</strong> of the<br />
electrolyte is depicted by the l<strong>in</strong>e BG. L<strong>in</strong>es AB=
ELECTRICITY AND MAGNETISM 325<br />
B<br />
(a)<br />
(h)<br />
B<br />
B<br />
(c)<br />
(d)<br />
Fig. 442<br />
not change. The correspond<strong>in</strong>g distribution of the potenti<strong>al</strong> is shown <strong>in</strong><br />
Fig. 441. The voltage drop on the <strong>in</strong>tern<strong>al</strong> resistance of the element occurs<br />
<strong>al</strong>ong CB and on the resistance R <strong>al</strong>ong AFD. The section CH' shows the<br />
voltage drop on the <strong>in</strong>tern<strong>al</strong> resistance ,<br />
equ<strong>al</strong> to Ir and the section AL the<br />
voltage drop on the resistance R.<br />
S<strong>in</strong>ce the l<strong>in</strong>e depict<strong>in</strong>g the potenti<strong>al</strong> ABCDF A is closed, the sum of the<br />
vol tage drops should be equ<strong>al</strong> to the sum of the potenti<strong>al</strong> jumps:<br />
eftl +c82=IR+Jr<br />
Hence, 1=R~"<br />
479. The potenti<strong>al</strong> is distributed as shown <strong>in</strong> Fig. 442a, b, c and d.<br />
(8) I ';\!cC2 ; VBA=VB-VA=lJl-I'1=-(cCa-I'2)=O<br />
'1 '2 '1 '2<br />
(b) /=
326 ANSWERS AND SOLUTiONS<br />
to the e. m. f., because <strong>in</strong> a state of equilibrium the chemic<strong>al</strong> forces that act<br />
<strong>in</strong> the layer of the electrolyte adjacent to the electrode are equ<strong>al</strong> to the<br />
electrostatic forces.<br />
S<strong>in</strong>ce the forces of non..electrostatic orig<strong>in</strong> do not act <strong>in</strong> the other sections<br />
of the circuit, the work performed by these forces is <strong>al</strong>so equ<strong>al</strong> to the e.m.I.<br />
of the battery when a s<strong>in</strong>gle positive charge moves <strong>al</strong>ong the closed circuit.<br />
(The work of electrostatic forces <strong>in</strong> a closed circuit is zero.)<br />
481. An energy of W e= 106,000-56,000=50,000 c<strong>al</strong>ories e: 2X 10 12 ergs<br />
is liberated per mole of the substances react<strong>in</strong>g <strong>in</strong> the cell. Ow<strong>in</strong>g to this<br />
energy, the electric current performs the work W = QtR, where B is the e.m.I,<br />
of the cell and Q is the quantity of transferred electricity. S<strong>in</strong>ce the copper<br />
and the z<strong>in</strong>c are biv<strong>al</strong>ent, the charges of their ions are equ<strong>al</strong> <strong>in</strong> magnitude<br />
to the doubled charge of an electron. One mole of the substance conta<strong>in</strong>s<br />
6.02Xl0 23 atoms. Therefore, Q=2X4.8XIO-loX6.02XI023CGSQ.<br />
W<br />
Hence, tG=~e: 3.5X 10- 3 CGS Q = 1.05V.<br />
482. The ratio b<strong>et</strong>ween the <strong>in</strong>tensities of the currents flow<strong>in</strong>g through the<br />
cells is 11 1 =!::L, s<strong>in</strong>ce the e.m.f.s of the cells are the same. Accord<strong>in</strong>g to<br />
I '1<br />
Faraday's law, the masses of the dissolved z<strong>in</strong>c are proportion<strong>al</strong> to the<br />
currents:<br />
483. As it passes <strong>in</strong>to solution <strong>in</strong> the form of an ion Zn + +, each atom<br />
of the z<strong>in</strong>c gives off to the extern<strong>al</strong> circuit two electrons carry<strong>in</strong>g a charge<br />
of q=2e=-3.2X 10- 19 C. At the same time the copper ions Cur + are depo..<br />
sited on the copper plate as neutr<strong>al</strong> atoms, ow<strong>in</strong>g to which the concentration<br />
of the CuS0 4 solution decreases. To ma<strong>in</strong>ta<strong>in</strong> the concentration constant, it is<br />
necessary to cont<strong>in</strong>u<strong>al</strong>ly dissolve cryst<strong>al</strong>s of CuSO,.5H 20 <strong>in</strong> an amount that<br />
will compensate for the pass<strong>in</strong>g of the ions Cur + and 50;- out of the solution.<br />
Accord<strong>in</strong>g to the <strong>in</strong>iti<strong>al</strong> conditions, a charge of Q =2,880 C passed through<br />
the cell. This corresponds to a transfer <strong>in</strong>to the solution of n =..9-=9X lOtI<br />
q<br />
atoms of z<strong>in</strong>c, i, e., about 0.98 g of the z<strong>in</strong>c. Correspond<strong>in</strong>gly, the same<br />
amount of copper atoms (about 0.95 g) will pass out of the solution and 3.73 g<br />
of cryst<strong>al</strong>s of blue vitriol will have to be dissolved to restore the concentration<br />
of the CuSO. solution.<br />
484. When the z<strong>in</strong>c is dissolved, the positive ions Zn+ + pass <strong>in</strong>to solution<br />
and the liberated electrons flow <strong>al</strong>ong the wire onto the graphite layer and<br />
neutr<strong>al</strong>ize the positive ions of copper <strong>in</strong> the CuS0 4 solution. Therefore, the<br />
graphite will be covered by a layer of copper.<br />
This phenomenon can be used <strong>in</strong> g<strong>al</strong>vanoplasty.<br />
485. The change <strong>in</strong> the e.m.I. of the battery depends on the ratio b<strong>et</strong>ween<br />
the dimensions of the electrodes and those of the vessel. If the two middle<br />
plates are <strong>al</strong>most equ<strong>al</strong> <strong>in</strong> size to the section of the vessel, the e.m.I, of the<br />
battery will change <strong>in</strong>significantly. If they are sm<strong>al</strong>l, the e.m.I, will be<br />
<strong>al</strong>most h<strong>al</strong>ved.<br />
486. The z<strong>in</strong>c rod forms a short-circuited g<strong>al</strong>vanic cell with each h<strong>al</strong>f of<br />
the carbon rod. The resistance of h<strong>al</strong>f of the carbon rod, the resistance of the
ELECTRICITY AND MAGNETISM<br />
327<br />
,. r z<strong>in</strong>c rod and the z<strong>in</strong>c-carbon contact are<br />
the extern<strong>al</strong> resistance of the cell (see the<br />
equiv<strong>al</strong>ent diagram <strong>in</strong> Fig. 443).<br />
When the z<strong>in</strong>c rod is vertic<strong>al</strong>, the currents<br />
i1 and i 2 <strong>in</strong> both h<strong>al</strong>ves of the carbon<br />
rod are identic<strong>al</strong> and the voltm<strong>et</strong>er will<br />
show zero. If the rod is <strong>in</strong>cl<strong>in</strong>ed, the <strong>in</strong>tern<strong>al</strong><br />
resistance of one of the cells will decrease<br />
and of the other <strong>in</strong>crease. The currents<br />
it and i 2 will be different and a<br />
Hearron potenti<strong>al</strong> difference will arise b<strong>et</strong>ween the<br />
ends of the carbon rod that will be registe-<br />
V<br />
red by the voltm<strong>et</strong>er.<br />
487. S<strong>in</strong>ce , ~ R, there wiII be practic<strong>al</strong>ly<br />
no field <strong>in</strong>side the sphere and no<br />
Fig. 443 current on its <strong>in</strong>tern<strong>al</strong> surface. Therefore,<br />
the mass of the deposited copper is<br />
m= : 4rr.~2 it ~ 1.86 g<br />
where Aln is the electrochemic<strong>al</strong> equiv<strong>al</strong>ent of copper (A is the atomic weight<br />
and n the v<strong>al</strong>ency) and F is Faraday's number.<br />
488. The matter is that the electrodes are polarized dur<strong>in</strong>g electrolysis and<br />
each bath acquires an e.rn.I. directed aga<strong>in</strong>st the current flow<strong>in</strong>g from the<br />
capacitor. For this reason the capacitor cannot be discharged compl<strong>et</strong>ely. The<br />
more baths we have, the greater will the tot<strong>al</strong> e.m.f. of polarization be, and<br />
the greater the charge rema<strong>in</strong><strong>in</strong>g on the capacitor. The energy of the d<strong>et</strong>onat<strong>in</strong>g<br />
gas will <strong>al</strong>ways be sm<strong>al</strong>ler than that of the charged capacitor.<br />
489. In the electrolysis of water, the electrodes are polarized and ail e.m. f.<br />
of polarization cCp appears that is directed aga<strong>in</strong>st the e.rn.I. of the battery.<br />
For this reason electrolysis will occur only if the e.m.I, of the battery is<br />
greater than ifp.<br />
i<br />
When a charge Q flows through the electrolyte, therbattery performs work<br />
aga<strong>in</strong>st the e.m.I. of polarization: W = cfip'Q. This work decomposes the water<br />
and d<strong>et</strong>onat<strong>in</strong>g gas is formed. On the basis of the law of conservation of<br />
energy, the chemic<strong>al</strong> energy of the d<strong>et</strong>onat<strong>in</strong>g gas W e liberated dur<strong>in</strong>g the<br />
flow of the charge Q is equ<strong>al</strong> to cfjpQ.<br />
In accordance with Faraday's law, the evolution of one gramme of hydrogen<br />
on the cathode is accompanied by the flow of electricity Q= m ~ F= 96,500 C.<br />
Therefore, 11p = ~e ::::: i.sV.<br />
The e.m.I. of the battery should be' greater than 1.5 V.<br />
490. A def<strong>in</strong>ite concentration of ions is the result of dynamic equilibrium:<br />
the number of ions produced by electrolytic dissociation is equ<strong>al</strong> to the reduction<br />
<strong>in</strong> the number of ions due to the reverse process-recomb<strong>in</strong>ation<br />
(when they collide, ions of opposite signs may form a neutr<strong>al</strong> molecule).<br />
Near the electrodes the concentration of the ions drops, and this equilibrium<br />
is violated. The number of ions that appear ow<strong>in</strong>g to dissociation is greater<br />
than the number of recomb<strong>in</strong>ed ions. It is this process that supplies the electrol~te<br />
with ions. The process takes place near the electrodes. Dynamic equiltbrlum<br />
<strong>in</strong>side the electrolyte is not violated.
328<br />
ANSWERS AND SOLUTIONS<br />
Fig. 444<br />
493. "The maximum theor<strong>et</strong>ic<strong>al</strong>ly pcssible efficiency of the thermoelectric<br />
battery is<br />
ELECTRICITY AND MAGNETISM 329<br />
497. At the first moment after the key K is closed, a potenti<strong>al</strong> difference<br />
will appear b<strong>et</strong>ween the plates of the capacitors C 1 and C 2 • The current will<br />
flow <strong>in</strong> the circuit until the capacitor C1 is charged. After this the potenti<strong>al</strong><br />
difference across the capacitor C 1 becomes equ<strong>al</strong> to the e.m.I. of the battery,<br />
and the potenti<strong>al</strong> difference b<strong>et</strong>ween the plates of the second capacitor will<br />
be zero.<br />
498. For two cells<br />
/= eel +/12<br />
'l+'2+ R<br />
where cC and r are the e.m.I, and the <strong>in</strong>tern<strong>al</strong> resistances of the cells. and R<br />
is the extern<strong>al</strong> resistance.<br />
For one cell (the first one. for example)<br />
/l=~<br />
'I+R<br />
Accord<strong>in</strong>g to the <strong>in</strong>iti<strong>al</strong> condition. / < /l' Le.•<br />
{Jt +
330<br />
t<br />
Fig. 445<br />
~<br />
+ - fj<br />
R<br />
-r<br />
~<br />
where<br />
ANSWERS AND SOLUTIONS<br />
condition, the potenti<strong>al</strong> of po<strong>in</strong>t A is lower than that<br />
of po<strong>in</strong>t B. Therefore, Ul=VB-VA > O.<br />
For the other circuit<br />
"2= V~ - V~ = (
ELECTRICITY AND MAGNETISM<br />
331<br />
Hence,<br />
Our condition will be fulfilled if /2 > O. Therefore the follow<strong>in</strong>g <strong>in</strong>equ<strong>al</strong>ities<br />
should exist<br />
or<br />
504. There are two m<strong>et</strong>hods of connect<strong>in</strong>g the storage battery cells. Either<br />
the batteries are connected <strong>in</strong> series <strong>in</strong> the separate groups, and the groups<br />
themselves <strong>in</strong> par<strong>al</strong>lel, or vice versa.<br />
Denot<strong>in</strong>g the tot<strong>al</strong> number of cells by N and the number of cells <strong>in</strong> a separate<br />
group by n, we sh<strong>al</strong>l have <strong>in</strong> the first case:<br />
J1 nCo Co<br />
R+,n 2<br />
.B..+ rn<br />
N n N<br />
s<strong>in</strong>ce the e. m. f. of one group is nCo, the resistance of the group is rn and<br />
the number of groups connected <strong>in</strong> par<strong>al</strong>lel is NIn. The current /. reaches<br />
its maximum if ~ + ~ n is m<strong>in</strong>imum. The m<strong>in</strong>imum of an expression of the<br />
type ax+!- can be found as follows<br />
x<br />
b<br />
y=ax+X-<br />
The relationship<br />
(I)<br />
Is shown graphic<strong>al</strong>ly by the curve <strong>in</strong> Fig. 446 which has its m<strong>in</strong>imum at<br />
po<strong>in</strong>t x o , at which the roots of quadratic equation (1) co<strong>in</strong>cide. For this<br />
reason,<br />
Therefore,<br />
and<br />
n=V~=4<br />
11 max=~O V:,=20 A
332<br />
ANSWERS AND SOLUTIONS<br />
9<br />
!/m<strong>in</strong><br />
Fig. 446<br />
Fig. 447<br />
IIOv<br />
In the second case<br />
N<br />
-
ELECTRICITY AND MAGNETISM 333<br />
509. The permissible drop of voltage on the feed<strong>in</strong>g wires is &U= IR o =7 V.<br />
where I is the maximum current. Hence. the maximum power is<br />
P=IU=AU X U<br />
Ro<br />
1,680 watts<br />
510. S<strong>in</strong>ce <strong>in</strong> <strong>al</strong>l cases the k<strong>et</strong>tle is connected to the same electric rnatns,<br />
it is more convenient to f<strong>in</strong>d the quantity of heat evolved from the formula<br />
U 2 U 2<br />
Q=0.24 Ii t, Hence, R=0.24Qt. S<strong>in</strong>ce U and Q are the same <strong>in</strong> <strong>al</strong>l cases,<br />
the latter equation can be rewritten as R=a.t, where (%=0.24 ~2.<br />
Denot<strong>in</strong>g the resistances of the w<strong>in</strong>d<strong>in</strong>gs by R 1 and R 2 , we have R 1 =at.<br />
and R 2 = at 2 " In par<strong>al</strong>lel connection of the w<strong>in</strong>d<strong>in</strong>gs<br />
R - R1R2 a, 2 t 1t 2<br />
a - R 1+R2 ex (t 1+t2) ata<br />
and <strong>in</strong> series connection<br />
Rb=R 1 + R 2 = a (t 1 +(2)=at b<br />
Therefore,<br />
511. (1) t a ~ 57 m<strong>in</strong>utes, (2) t b = 3 m<strong>in</strong>utes 30 seconds (see the solution to<br />
Problem 510). .<br />
512. When a direct current flows through a conductor, the potenti<strong>al</strong> difference<br />
does not change. When a capacitor is discharged, the potenti<strong>al</strong> difference<br />
changes from U to zero.<br />
513. When the losses of heat <strong>in</strong> high-voltage wires are c<strong>al</strong>culated by the<br />
formula Q=0.24 ~2 t, the v<strong>al</strong>ue of U is the potenti<strong>al</strong> difference at the ends<br />
of the l<strong>in</strong>e (voltage drop <strong>in</strong> the wires), but not the voltage <strong>in</strong> the secondary<br />
w<strong>in</strong>d<strong>in</strong>g of the .step-up transformer. This potenti<strong>al</strong> difference is sm<strong>al</strong>l (as<br />
dist<strong>in</strong>ct from the voltage <strong>in</strong> the w<strong>in</strong>d<strong>in</strong>g of the transformer) and decreases<br />
with a reduction of the current flow<strong>in</strong>g <strong>in</strong> the l<strong>in</strong>e.<br />
514. In conformity with the <strong>in</strong>iti<strong>al</strong> conditions, k= ~ 100, where til is the<br />
e. m. f. of the battery, and 1= ~ is the current <strong>in</strong> the circuit. Remember<strong>in</strong>g<br />
that cC=2U +1" we obta<strong>in</strong><br />
,<br />
2"U2<br />
P (lUO-k)·Q<br />
515. The power liberated <strong>in</strong> the extern<strong>al</strong> resistance R is P=IU. In our<br />
case U=IJ-I, and, therefore, l=cf}-U.<br />
Thus,<br />
,<br />
p =
334 ANSWERS AND SOLUTIONS<br />
Fig. 448<br />
U<br />
whence<br />
u=i+v~'-p,<br />
U 1=9Vor U 2=1 V.<br />
The ambiguity of the result is<br />
due to the fact that the same power<br />
can be liberated on various<br />
extern<strong>al</strong> resistances R, each R hav<strong>in</strong>g<br />
its own current:<br />
when U 1=9 V, we have /1=1 A<br />
P<br />
and R 1 =[i=9 Q<br />
1<br />
when U 2= 1 V, we have /2= 9 A<br />
P<br />
and R,,= /2 = 1/9 Q<br />
2<br />
516. The useful power (see Problem 515) is equ<strong>al</strong> to P
ELECTRICITY AND MAGNETISM 335<br />
'I<br />
1<br />
*2<br />
1 -----------------------<br />
p<br />
&----.a..----------R<br />
R=r<br />
fa)<br />
~-----a._---------R<br />
R=r<br />
(b) Fig. 449<br />
but <strong>in</strong> this case the useful power P= (f:t~)1 evot<br />
'I --+- I when R --+-CIO,<br />
ved tends (the same as the tot<strong>al</strong> power) to zero (Fig. 449).<br />
6R<br />
1)2=6R+,=O.9<br />
In per cent 1),==90%.<br />
519. Accord<strong>in</strong>g to Ohm's law, U'=8+1r. Hence, J=U-8.<br />
r<br />
Th'e useful power spent to charge the battery is<br />
r.=liJ U
336 ANSWERS AND SOLUTIONS<br />
"'---1/=1<br />
520. The usefuI power is<br />
PI 11 (U -cf})<br />
(see Problem 519). The heat evolved<br />
per unit of time is<br />
,<br />
,<br />
P2= (U _IJ)2<br />
Fig. 450<br />
Ord<strong>in</strong>arily, U -
ELECTRICITY AND MAGNETISM 337<br />
the region of the potenti<strong>al</strong> drop of the cathode. Otherwise, there will be no<br />
discharge.<br />
525. For a discharge to occur <strong>in</strong> the tube, the anode should be placed<br />
beyond the region of the cathode drop (see Problem 524). But <strong>in</strong> this case<br />
the electrons approach<strong>in</strong>g the anode upon collision with the molecules of the<br />
gas lose energy and will not generate X-rays when they imp<strong>in</strong>ge on the<br />
anode. Two electrodes are therefore necessary. The anode is located <strong>in</strong> the<br />
region of the glow and serves to ma<strong>in</strong>ta<strong>in</strong> the discharge. The anticathode <strong>in</strong><br />
the region of the cathode drop is bombarded by the electrons that did not<br />
lose their energ y.<br />
Tubes with a heater cathode have one electrode that acts as an anode and<br />
an anticathode.<br />
526. The electrons strik<strong>in</strong>g the anticathode flow <strong>al</strong>ong the wire onto the<br />
anode. If the wire is removed, the anticathode will be gradu<strong>al</strong>ly charged<br />
negatively and r<strong>et</strong>ard the electrons. In a certa<strong>in</strong> time after it "is switched<br />
on the X-ray tube will stop function<strong>in</strong>g.<br />
527. Before the discharge, the voltage on the counter is equ<strong>al</strong> to the<br />
e.m.I. of the source cfj. At the moment of discharge a current flows through<br />
the circuit and the voltage b<strong>et</strong>ween the hous<strong>in</strong>g and the wire becomes equ<strong>al</strong><br />
to U=c8-IR. The resistance R is very high, and the voltage drop IR is<br />
so great that the discharge stops.<br />
528. Accord<strong>in</strong>g to Ohm's law, the sought voltage drop U=IR, where J<br />
is the current <strong>in</strong> the circuit.<br />
The current is the same <strong>in</strong> <strong>al</strong>l the cross sections <strong>in</strong>side the capacitor.<br />
The positive plate owes its current only to the negative ions, and the negative<br />
plate to the positive ions. Some positive and negative ions pass<br />
through an arbitrary cross section <strong>in</strong>side the capacitor.<br />
J= enAd, where e is the charge of an electron and A is the area of the<br />
plates.<br />
For a plane capacitor Ad = 4nCd 2 •<br />
Therefore, U ==enx4nCd 2R ~ 1.4x 10- 11 V.<br />
529. If the negative carbon is cooled, the arc will be ext<strong>in</strong>guished, s<strong>in</strong>ce<br />
the" arc burns ow<strong>in</strong>g to strong thermionic emission from the cathode" that"<br />
ceases upon cool<strong>in</strong>g. Cool<strong>in</strong>g of the positive carbon will not affect function<strong>in</strong>g<br />
of the arc.<br />
530. When the contacts of the controller open, an electric arc may appear,<br />
s<strong>in</strong>ce the current <strong>in</strong> the iron becomes very high and the distance b<strong>et</strong>ween<br />
the contacts is sm<strong>al</strong>l. With an <strong>al</strong>ternat<strong>in</strong>g current the arc is unstable and<br />
. ext<strong>in</strong>guishes at once. A direct current produces a stable arc that at least<br />
will burn the contacts and put the iron out of commission.<br />
531. 1 eV (one electron-volt) = 1.6X 10- 12 erg.<br />
532. No, it does not. The tangents to the trajectory show the direction of<br />
the velocity of the particle and the tangents to the force l<strong>in</strong>e show the direction<br />
of the force act<strong>in</strong>g on the particle and, therefore, the direction of<br />
acceleration.<br />
The trajectory of the particle will co<strong>in</strong>cide with a force l<strong>in</strong>e ani y <strong>in</strong> a<br />
field with straight force l<strong>in</strong>es if the <strong>in</strong>iti<strong>al</strong> velocity of the particle is directed<br />
<strong>al</strong>ong a force l<strong>in</strong>e. .<br />
533. As the charge approaches the plate, electrostatic <strong>in</strong>duction causes the<br />
charges on the plate of the same sign as the fly<strong>in</strong>g one to pass <strong>in</strong>to the<br />
earth, while the charges of the opposite sign accumulate on the surface of<br />
the plate. A current pulse passes through the g<strong>al</strong>vanom<strong>et</strong>er (Fig. 451). No<br />
current flows through the g<strong>al</strong> vanom<strong>et</strong>er when the charge moves above the<br />
22-2042
338<br />
ANSWERS AND SOLUTIONS<br />
I<br />
t<br />
Fig. 451<br />
plate. An opposite current is generated when the charge moves away from<br />
the plate.<br />
534. Positive <strong>in</strong>duced charges will appear on the neck of the tube and<br />
accelerate the electron. The k<strong>in</strong><strong>et</strong>ic energy of the electron will <strong>in</strong>crease ow<strong>in</strong>g<br />
to the drop <strong>in</strong> the potenti<strong>al</strong> energy of the electron-tube system.<br />
535. The tot<strong>al</strong> energy of an electron is equ<strong>al</strong> to the sum of its k<strong>in</strong><strong>et</strong>ic and<br />
potenti<strong>al</strong> energies.<br />
As the electron approaches the r<strong>in</strong>g, its potenti<strong>al</strong> energy dim<strong>in</strong>ishes <strong>in</strong> the<br />
field"of the r<strong>in</strong>g, and as a result the k<strong>in</strong><strong>et</strong>ic energy <strong>in</strong>creases. After pass<strong>in</strong>g<br />
through the r<strong>in</strong>g, the electron moves away from it. The potenti<strong>al</strong><br />
energy of<br />
the electron <strong>in</strong>creases and the velocity gradu<strong>al</strong>ly drops to zero.<br />
536. The work done to move the charge - Q is' proportion<strong>al</strong> to the potenti<strong>al</strong><br />
difference b<strong>et</strong>ween po<strong>in</strong>t 0 and remote po<strong>in</strong>t A on the axis (see<br />
Fig. 186). At <strong>in</strong>f<strong>in</strong>ity the potenti<strong>al</strong> is taken equ<strong>al</strong> to zero. If the distance<br />
OA ~ R, the potenti<strong>al</strong> of po<strong>in</strong>t A can <strong>al</strong>so be assumed to be zero. The potenti<strong>al</strong><br />
at po<strong>in</strong>t 0 can be found by summ<strong>in</strong>g up the potenti<strong>al</strong>s produced by<br />
. the separate sm<strong>al</strong>l elements of the r<strong>in</strong>g: U0 =E ARQ = ~ ·<br />
. mv2 Q2<br />
On the basis of the law of conservation of energy 2 =7[ we f<strong>in</strong>d that<br />
Vz:::<br />
(<br />
2QI<br />
-<br />
mR<br />
537. As usu<strong>al</strong>, the potenti<strong>al</strong> <strong>in</strong> <strong>in</strong>f<strong>in</strong>ity is considered to be zero. The<br />
potenti<strong>al</strong>s of the plates are thus respectively equ<strong>al</strong> to + ~ and ~,<br />
where U = ~ . The potenti<strong>al</strong>s at the po<strong>in</strong>ts of the <strong>in</strong>iti<strong>al</strong> location of the<br />
electron are respectively: 0, +~ , and - ~ . The <strong>in</strong>iti<strong>al</strong> full energies of<br />
the electron are<br />
mv~<br />
(I) T'<br />
mv~ ell<br />
(2) T-T and<br />
,."<br />
(3) muo +eU<br />
2 4
ELECTRICITY AND MAGNETISM<br />
339<br />
The f<strong>in</strong><strong>al</strong> velocities VI' v 2 and V 3 are d<strong>et</strong>erm<strong>in</strong>ed<br />
from the law of con~erva tion of energy:<br />
(1)<br />
mv 2<br />
_0=:=_1.<br />
mv"<br />
2 ~ 2 t<br />
hence VI =Vo<br />
(2)<br />
(3)<br />
"u 2<br />
mvo+ ~== mV3.<br />
2 4 2 f<br />
Fig. 452<br />
In the first case the f<strong>in</strong><strong>al</strong> velocity is equ<strong>al</strong> to the<br />
<strong>in</strong>iti<strong>al</strong> one, <strong>in</strong> the second case it is lower than the<br />
<strong>in</strong>iti<strong>al</strong> velocity and <strong>in</strong> the third case higher.<br />
. In <strong>al</strong>l three cases the velocity first grows (dur<strong>in</strong>g motion <strong>in</strong> the capacitor)<br />
and then decreases.<br />
538. Electrons with energies rang<strong>in</strong>g from 80 eV to 74 eV reach the anode,<br />
s<strong>in</strong>ce a voltage drop of 6 V exists <strong>al</strong>ong the filament.<br />
The energy of the electrons at the anode is d<strong>et</strong>erm<strong>in</strong>ed only by the potenti<strong>al</strong><br />
difference passed by them and does not depend on the potenti<strong>al</strong> of the<br />
grid. The latter changes the distribution of the velocities of the electrons at<br />
<strong>in</strong>termediate po<strong>in</strong>ts of the path and affects the number of electrons reach<strong>in</strong>g<br />
the anode.<br />
539. On the basis of Ohm's law,<br />
The <strong>in</strong>tensity of the current<br />
Hence,<br />
cC=laRa+Ua (Fig. 452)<br />
la=AUa+BU~<br />
I =L+(ARa+l)-V(ARa+l)2+4CBRa =5 rnA<br />
4 Ra 28R~<br />
The second· root of the quadratic equation has no physic<strong>al</strong> mean<strong>in</strong>g, s<strong>in</strong>ce it<br />
corresponds to Ua < O.<br />
MO. The simultaneous equations d<strong>et</strong>erm<strong>in</strong><strong>in</strong>g the currents i 1 and i 2 have<br />
.the form:<br />
Hence,<br />
~U.- -<br />
i=i l +.i 2<br />
i' l =A 1U a+BIU~<br />
i2=A2Utl+B2U~<br />
Ua=l}-iR<br />
(AI+ Ai) R-l+ V(A 1 R+A 2 R + 1)2+4
340 ANSWERS AND SOLUTIONS<br />
/16'4<br />
D 2 7 10<br />
tfyolts<br />
Fig. 453<br />
The negative v<strong>al</strong>ue of Ua is discarded because it does not correspond to<br />
the sense of the problem. The sought currents are:<br />
i1<br />
1<br />
(Bt+B2)R(Bui?+(A.B2-A2BdRUa-B1Ua]=22.2 rnA<br />
i2=(B. +I B2)<br />
R [BtB+(AtB.-A.Bt) RUa-B2Ua!=37.8 rnA<br />
541 With the potenti<strong>al</strong> of the grid cC2=- 6 V, the current flow<strong>in</strong>g<br />
through the v<strong>al</strong>ve is 1 2 = ~ and with 111=-3 V it is 1.= ~l •<br />
Therefore, an <strong>in</strong>crease <strong>in</strong> the potenti<strong>al</strong> of the grid by cfJl -cfJ2=3 V raises<br />
the anode current of the v<strong>al</strong>ve by<br />
1<br />
11-J"=R (VI -U 2)= 3.5 rnA<br />
S<strong>in</strong>ce the grid characteristic of the v<strong>al</strong>ve <strong>in</strong> the region be<strong>in</strong>g considered<br />
is assumed to be l<strong>in</strong>ear, the addition<strong>al</strong> <strong>in</strong>crease <strong>in</strong> the potenti<strong>al</strong> of the grid<br />
relative to the cathode by 3 V (from-3 V to zero· with the short-circuited<br />
gri d and cathode) will<br />
<strong>in</strong>crease the anode current by another 3.5 rnA.<br />
The voltage drop across the resistance R will now <strong>in</strong>crease addition<strong>al</strong>ly<br />
by U t-Ut.=35 V, and become equ<strong>al</strong> to Uo=U. +(U 1-U2 )= 130 V. while<br />
the potenti<strong>al</strong> difference b<strong>et</strong>ween the anode and the cathode of the v<strong>al</strong>ve will<br />
be equ<strong>al</strong> to tB-Uo=120 V. .<br />
542. The first diode beg<strong>in</strong>s to conduct current only when U a > 0, I, e.,<br />
when V > cfJlt the second at V > li". and the third at V >
ELECTRICITY AND MAGNETISM 341<br />
v<br />
543. In Fig. 454, A and<br />
B are control grids, M N is<br />
the screen and OC the trajectory<br />
of an electron. The orig<strong>in</strong><br />
of the coord<strong>in</strong>ate system<br />
is at po<strong>in</strong>t O.<br />
An electron travell<strong>in</strong>g b<strong>et</strong>ween<br />
the grids moves <strong>in</strong> the<br />
o ....-...-:~----------i--+-..Iooo3I:c~ direction of the y-axis with<br />
a uniform acceleration of<br />
- eV<br />
a = md ' where U is the potenti<strong>al</strong><br />
difference b<strong>et</strong>ween A<br />
N<br />
and B. The electron covers<br />
the distance I <strong>al</strong>ong the x-<br />
Fig. 454<br />
rJ$ is the horizont<strong>al</strong> component of the<br />
from the condition<br />
mv~<br />
T=eUo<br />
axis <strong>in</strong> the time t 1 =!...., when<br />
Vx<br />
velocity of the electron d<strong>et</strong>erm<strong>in</strong>ed<br />
Dur<strong>in</strong>g the time t 1 the electron is deflected <strong>in</strong> the direction of the y-axis<br />
by Yl = a2t~= eUlI s<br />
' The electron moves outside the grids with a constant<br />
2dmv.(<br />
velocity dur<strong>in</strong>g the time t 2 = .!::-. The velocity <strong>al</strong>ong the y-axis is v,=aI 1 •<br />
Vx<br />
The deflection outside the grids is<br />
eVlL<br />
1J?=vut 2 = - -<br />
T dmv~<br />
The tot<strong>al</strong> deflection is<br />
eUl (I ) eVlL UIL<br />
Y=Yl+Y2=d 2 2+ L ~d----;=w:a<br />
mvx mo; 0<br />
and the sensitivity is<br />
y lL<br />
Q=U=2U o<br />
d<br />
3-4. Magn<strong>et</strong>ic Field of a Current. Action of a Magn<strong>et</strong>ic<br />
Field on a Current and Mov<strong>in</strong>g Charges<br />
. 644. If the current is expressed <strong>in</strong> amperes, the coefficient k· is numerl<br />
'~". c<strong>al</strong>ly equ<strong>al</strong> to 0.1. S<strong>in</strong>ce 1 A=3X 10' cgs electrostatic units (CGS/) the<br />
!L~ght v<strong>al</strong>ue of the coefficient will. be k=3X ~OlO if the current is rneasu<br />
~>. ted <strong>in</strong> these units. The dimension of the coefficient can be found directly
342 ANS'WERS AND SOLUTIONS<br />
.Irorn the formula for the <strong>in</strong>tensity H:<br />
_ [H] [l]<br />
[k]<br />
- [I]<br />
Bear<strong>in</strong>g <strong>in</strong> m<strong>in</strong>d that (H]=(E) and (E)= ~~~ =g". ocm-'I'os-l. (ll=cm.<br />
and [/]=gl/·.cm 3/ 1.s- 2 • we obta<strong>in</strong> [k]=s/cm.<br />
If a newconstant c is <strong>in</strong>troduced <strong>in</strong>stead of k. so that k =-.!.. . then c will be<br />
c<br />
equaI to the velocity of Hght <strong>in</strong> a vacuum.<br />
545. The <strong>in</strong>tensity of the field <strong>in</strong>duced by the first w<strong>in</strong>d<strong>in</strong>g is<br />
HI o.~:::: 1 400Oe<br />
The second w<strong>in</strong>d<strong>in</strong>g <strong>in</strong>duces the <strong>in</strong>tensity<br />
H O.4n1N2 -200 0<br />
2 2nR - e<br />
S<strong>in</strong>ce the fields HI and H'}, are directed oppositely. the sought field will be<br />
H=H 1-H2=200 Oe.<br />
546. The conductor Be does not <strong>in</strong>duce a field at po<strong>in</strong>t M ly<strong>in</strong>g on the<br />
cont<strong>in</strong>uation of Be. Accord<strong>in</strong>g to the rule giveil <strong>in</strong> the note, the magn<strong>et</strong>ic<br />
field produced by any elements of conductor Be should be perpendicular to<br />
l<strong>in</strong>e 8M. For this reason the presence of a field other than zero at po<strong>in</strong>t M<br />
would disagree with the symm<strong>et</strong>ry of the problem. because <strong>al</strong>l the directions<br />
perpendicular to 8M have equ<strong>al</strong> rights.<br />
S<strong>in</strong>ce the <strong>in</strong>tensity of the field is proportion<strong>al</strong> to the current. then H 1=kl<br />
without conductor BD. The fields from the conductors AB and BD are summated.<br />
Therefore after conductor BD is connected<br />
H 2 3<br />
whence HI ="2.<br />
kJ<br />
H 2=kl+7j<br />
547. At an arbitrary po<strong>in</strong>t on l<strong>in</strong>e AB any sm<strong>al</strong>l element 0 f current of<br />
conductor ACB <strong>in</strong>duces a magn<strong>et</strong>ic field perpendicular to the plane of the<br />
draw<strong>in</strong>g (see Problem 546). The element of conductor ADB symm<strong>et</strong>ric<strong>al</strong> to<br />
it <strong>in</strong>duces the same field, but directed oppositely. For this reason the field<br />
from any two elements arranged symm<strong>et</strong>ric<strong>al</strong>ly will be zero. Hence, the field<br />
at the arbitrary po<strong>in</strong>t on AB <strong>in</strong>duced by the entire conductor is zero. s<strong>in</strong>ce<br />
the straight sections of the conductor <strong>al</strong>so do not <strong>in</strong>duce a field on AB.<br />
548. In the ma<strong>in</strong>, the field of the solenoid will be concentrated <strong>in</strong>side<br />
the toroid<strong>al</strong> w<strong>in</strong>d<strong>in</strong>g and will not act on the magn<strong>et</strong>ic po<strong>in</strong>ter. A s<strong>in</strong>glelayer<br />
w<strong>in</strong>d<strong>in</strong>g, however. can simultaneously be regarded as one turn of a<br />
large radius that <strong>in</strong>duces a magn<strong>et</strong>ic field perpendicular to the plane of the<br />
torus.<br />
The magn<strong>et</strong>ic po<strong>in</strong>ter will be positioned <strong>al</strong>ong the axis of the torus. The<br />
direction of its poles can be d<strong>et</strong>erm<strong>in</strong>ed by the right-hand screw rule.<br />
549. The current flow<strong>in</strong>g <strong>al</strong>ong the pipe can be regarded as the sum of a<br />
great number of identic<strong>al</strong> straight currents uniformly distributed over the
ELECTRICITY AND MAGNETISM<br />
343<br />
Fig. 455<br />
Fig. 456<br />
surface of the pipe. The <strong>in</strong>tensity of the magn<strong>et</strong>ic field at any po<strong>in</strong>t of<br />
space can be represented as the sum of the <strong>in</strong>tensities of the fields <strong>in</strong>duced<br />
by these currents.<br />
Figure 455 shows a cross section of the pipe <strong>al</strong>ong which the current<br />
flows. L<strong>et</strong> us compare the" <strong>in</strong>tensities of the magn<strong>et</strong>ic fields HI and H2 created<br />
at po<strong>in</strong>t A by the straight currents /1 and /1 pass<strong>in</strong>g through the sm<strong>al</strong>l<br />
aRt<br />
a,R<br />
arcs 8 1 and 8 2 . The lengths of the arcs are 8 1 =-- and 8 2 = - _ 2_ ,<br />
cos'
344 .<br />
ANSWERS AND SOLUTIONS<br />
Fig. 457<br />
Fig. 458<br />
The field created by the current with a density I at an arbitrary po<strong>in</strong>t<br />
of space A is equ<strong>al</strong> to HI=k·2njR (Fig. 456). Here R is the distance from<br />
the axis of the conductor to po<strong>in</strong>t A. (It is assumed that the current flows<br />
toward us.)<br />
At the same po<strong>in</strong>t, the current flow<strong>in</strong>g through the volume. correspond<strong>in</strong>g<br />
to the space, but <strong>in</strong> the reverse direction, <strong>in</strong>duces a field of H 2=k·2njr.<br />
Figure 456 shows that the tot<strong>al</strong> <strong>in</strong>tensity of the field is<br />
Obviously,<br />
H=VH:+H:-2H 1Ht cos ex<br />
Therefore, the <strong>in</strong>tensity H =k·2njd is the same for <strong>al</strong>l the po<strong>in</strong>ts of the<br />
space. .<br />
551. Triangle AOC is similar to triangle BAD (Fig. 456) s<strong>in</strong>ce they have<br />
one equ<strong>al</strong> angle, and the sides conf<strong>in</strong><strong>in</strong>g these angles are proportion<strong>al</strong>.'<br />
Therefore, L AOC=L BAD. But R l..H 1 , and therefore H l..d.<br />
The <strong>in</strong>tensity of the magn<strong>et</strong>ic field at any ro<strong>in</strong>t of the space is perpendicular<br />
to the l<strong>in</strong>e that connects the centres 0 the conductor and the space.<br />
The distribution of the l<strong>in</strong>es is shown <strong>in</strong> Fig. 457.<br />
552. k=..!.., where c is the velocity of light <strong>in</strong> a vacuum.<br />
c<br />
553. No, it will not. Forces of attraction exist b<strong>et</strong>ween the separate elements<br />
of the current. As a result, the density of the current <strong>in</strong>creases somewhat<br />
toward the axis of the conductor. The effect is negligible.<br />
554. When lightn<strong>in</strong>g strikes. a very high current flows for an <strong>in</strong>stant<br />
through the pipe and the separate elements of the current are mutu<strong>al</strong>ly<br />
attracted with a high force. It is this force that converts the pipe <strong>in</strong>to<br />
a rod.
ELECTRICITY AND MAGNETISM 345<br />
555. The currents <strong>in</strong> the adjacent turns are par<strong>al</strong>lel and flow <strong>in</strong> the same<br />
direction. For this reason the turns will be mutu<strong>al</strong>ly attracted. At the same<br />
time the" currents <strong>in</strong> opposite sections of the turns flow <strong>in</strong> different directions.<br />
Therefore, opposite sections. are repulsed. .<br />
The turns of the w<strong>in</strong>d<strong>in</strong>g wi11 tend to <strong>in</strong>crease <strong>in</strong> diam<strong>et</strong>-er t and t he distance<br />
b<strong>et</strong>ween them <strong>al</strong>ong the axis of the solenoid will be reduced.<br />
HnAJ<br />
sse. a=-k-'<br />
567. The action of the magn<strong>et</strong>ic field will cause the r<strong>in</strong>g to so turn that<br />
the force l<strong>in</strong>es of the field are perpendicular to the plane of the r<strong>in</strong>g and<br />
form a right-hand screw with the direction of the current. The tension of<br />
the r<strong>in</strong>g will be maximum. Upon employ<strong>in</strong>g the m<strong>et</strong>hod used to solve Problem<br />
403, we obta<strong>in</strong><br />
F=kJRH=5 dynes<br />
558. The element of the r<strong>in</strong>g 111 is acted upon by the force I1F=kJHI1l<br />
(Fig. 458). L<strong>et</strong> us resolve it <strong>in</strong>to the components AF1 and AI. The component<br />
AF1 lies <strong>in</strong> the plane of the r<strong>in</strong>g and At = !iF s<strong>in</strong> a is norm<strong>al</strong> to the<br />
plane of the r<strong>in</strong>g. The resultant of the forces !iF1 that act on the separate<br />
elements of the r<strong>in</strong>g is zero. These forces only str<strong>et</strong>ch the r<strong>in</strong>g. The full<br />
force fact<strong>in</strong>g on the r<strong>in</strong>g is equ<strong>al</strong> to the sum of the forces ~f:<br />
f=~klHs<strong>in</strong> a.·lilj=klH.2nR s<strong>in</strong> a.~ 273 dynes<br />
&59. The forces act<strong>in</strong>g on BC and AD are perpendicular to the motion of<br />
these sides and, therefore, perform no work.<br />
The forces act<strong>in</strong>g on AB and CD are constant, form a right angle with<br />
the direction of the field and are numeric<strong>al</strong>ly equ<strong>al</strong> to f=kHla (Fig. 459).<br />
The sought work will be equ<strong>al</strong> to the double product of the force and the<br />
.motion of side AB or CD <strong>in</strong> the direction of the force. Wlren the circui t is<br />
turned through 18 0 , this motion is b.<br />
Therefore, W=2kHJab:<br />
&80. Assum<strong>in</strong>g that <strong>al</strong>l the electrons move with a velocit.y v, the <strong>in</strong>tensity<br />
of the current can be expressed as follows (see Problem 521):<br />
J=nAev<br />
. Upon <strong>in</strong>sert<strong>in</strong>g the v<strong>al</strong>ue of J Into the formula for F. we g<strong>et</strong><br />
F:::kHnAlev s<strong>in</strong> ex<br />
S<strong>in</strong>ce a piece of the conductor conta<strong>in</strong>s N = Aln electrons, the force act<strong>in</strong>g<br />
on one electron is f=kHev s<strong>in</strong> a.<br />
The force f is known as the Lorentz force.<br />
The direction of the Lorentz force is d<strong>et</strong>erm<strong>in</strong>ed by the left-hand rule<br />
(the magn<strong>et</strong>ic field <strong>in</strong>tersects the p<strong>al</strong>m, four f<strong>in</strong>gers are directed opposite to<br />
the motion of the electrons, or <strong>al</strong>ong the motion of a positively charged<br />
particle, and the thumb shows the direction of the Lorentz force).<br />
&81. The Lorentz force is <strong>al</strong>ways perpendicular to the velocity of a particle<br />
and therefore performs no work. The k<strong>in</strong><strong>et</strong>ic energy and, hence, the<br />
absolute velocity of the particle rema<strong>in</strong> constant.<br />
&82. The electron is acted upon by the force f=kevH. If H is measured<br />
In oersteds and the charge <strong>in</strong> CIlS electrostatic units, then k=: (see Prob-
346<br />
ANSWERS AND SOLUTIONS<br />
II<br />
/"---- .....,<br />
/ "<br />
~ I "<br />
I \.<br />
I ~<br />
\ I<br />
\ '.0<br />
\ /<br />
, /<br />
" fill'<br />
'---_.--"<br />
H<br />
H<br />
Fig. 459 Fig. 460<br />
lem 560). This force is constant <strong>in</strong> magnitude and perpendicular to the velocity<br />
v. For this reason the acceleration of the electron is <strong>al</strong>so constant <strong>in</strong><br />
magnitude and constantly rema<strong>in</strong>s perpendicular to the velocity. The velocity<br />
changes only <strong>in</strong> direction.<br />
With a constant acceleration perpendicular to the velocity, the motion at<br />
a velocity constant <strong>in</strong> magnitude is uniform motion <strong>al</strong>ong a circle.<br />
mv 2 e<br />
On the basis of Newton's second law, R=c·vH. Therefore, the electron<br />
will move <strong>al</strong>ong a circle with a radius R = n;:: ·<br />
563. L<strong>et</strong> us resolve the velocity of the electron <strong>in</strong>to the components v II<br />
par<strong>al</strong>lel to Hand V...L perpendicular to H (Fig. 460). The component vu does<br />
not change <strong>in</strong> magnitude or direction s<strong>in</strong>ce the Lorentz force does not act on<br />
a particle whose velocity is directed <strong>al</strong>ong the field. The component V.1. changes<br />
<strong>in</strong> direction <strong>in</strong> the same way as <strong>in</strong> Problem 562.<br />
Thus, rotation <strong>al</strong>ong a circle <strong>in</strong> a plane perpendicular to H is superposed<br />
on the uniform translation<strong>al</strong> motion <strong>al</strong>ong H. This produces motion <strong>al</strong>ong a<br />
helic<strong>al</strong> l<strong>in</strong>e with a constant pitch h=v u 1:, where 1: is the duration of one<br />
revolution of the electron <strong>al</strong>ong a circle with a radius R= mc:~n IX •<br />
. 2nR 2nmc 2nmc<br />
S<strong>in</strong>ce l'=--= -H ,then h= --;:;- v cos ex<br />
v~ e e'4<br />
564. The action of the Lorentz force (see Problem 560) will cause the<br />
electrons to move towards the edge of the band. For this reason one edge of<br />
the band will receive a negative charge and the other a positive one. An<br />
addition<strong>al</strong> electric field will be generated <strong>in</strong>side the band with an <strong>in</strong>tensity E<br />
directed perpendicular to the current. The electrons will cont<strong>in</strong>ue to move<br />
until the Lorentz force is equ<strong>al</strong>ized by the force act<strong>in</strong>g on the electron from<br />
the side of the electric field E: eE=kevH. Hence, E=kvH.<br />
The potenti<strong>al</strong> difference CPA-CPB=Ea=kvHa or, s<strong>in</strong>ce I = nevA, then<br />
I<br />
A-CPB= kHa<br />
ne Ā·<br />
565. q> A - CPB =:: 23JLV.<br />
566. The Lorentz force (see Problem 560) acts on both the free electrons<br />
and the positive ions at<br />
the po<strong>in</strong>ts of a cryst<strong>al</strong> lattice, s<strong>in</strong>ce both move <strong>in</strong> a<br />
-,
ELECTRICITY AND- MAGNETISM<br />
347<br />
a<br />
cl<br />
It 0H<br />
I~<br />
+ + + + + + + +<br />
Fig. 46/<br />
magn<strong>et</strong>ic field. In accordance with the left-hand rule, the force f that acts<br />
on the free electrons will be directed as shown <strong>in</strong> Fig. 461. The electrons are<br />
displaced with respect to the lattice, and one side of the -par<strong>al</strong>lelepiped is<br />
charged negatively and the other positively. An electric field is produced <strong>in</strong><br />
the block, and when the <strong>in</strong>tensity of this field satisfies the ratio eE=kevH,<br />
the electrons will no longer move with respect to the lattice.<br />
The sought <strong>in</strong>tensity E=kvH.<br />
The density of the charges (J can be found from the equation 41(
348 ANSWERS AND SOLUTIONS<br />
cord<strong>in</strong>g to Lens's law, the tot<strong>al</strong> current <strong>in</strong> the coil decreases when the rod<br />
enters it, and <strong>in</strong>creases when the rod leaves it.<br />
A diagram of the change <strong>in</strong> the current is shown <strong>in</strong> Fig. 463.<br />
571. The magn<strong>et</strong>ic flux changes at a constant rate, and therefore the e. m. f.<br />
of <strong>in</strong>duction <strong>in</strong> the second coil will <strong>al</strong>so be constant. If the coil is connected<br />
to a closed circuit, it will carry a direct current, which will s<strong>et</strong> In not at<br />
once, but depend<strong>in</strong>g on the coefficient of self..<strong>in</strong>duction of the second coil and<br />
its resistance.<br />
572. Yes, it will. The e.m.f. of <strong>in</strong>duction is proportion<strong>al</strong> to the rate of<br />
change of the magn<strong>et</strong>ic flux, while the magnitude of the magn<strong>et</strong>ic flux <strong>in</strong><br />
the iron core does not change directly with the current. The relationship will<br />
be more complicated.<br />
573. Accord<strong>in</strong>g to Faraday's law,<br />
A«D<br />
cCi= 10- S AT= 10- 8 kA<br />
The e.m.f, of <strong>in</strong>duction is numeric<strong>al</strong>ly equ<strong>al</strong> to the work performed by<br />
the electric field when a s<strong>in</strong>gle positive charge moves <strong>in</strong> the turn, i.e.,<br />
cCi=2nrE. Hence E= 2 8 ; .<br />
nr<br />
Thus, we f<strong>in</strong><strong>al</strong>ly obta<strong>in</strong>:<br />
E= 10_skn,r 2 =lO-skr<br />
2nf 2<br />
It should be noted that this electric field is <strong>in</strong>duced not by the electric<br />
charges, but by 8 magn<strong>et</strong>ic field vary<strong>in</strong>g with time. L<strong>et</strong> us rec<strong>al</strong>l that when<br />
an electric charge moves <strong>in</strong> a closed circuit <strong>in</strong> an electrostatic field the work<br />
is <strong>al</strong>ways equ<strong>al</strong> to zero. By an electrostatic field is meant an electric field<br />
<strong>in</strong>duced by electric charges.<br />
574. L<strong>et</strong> us divide the r<strong>in</strong>g <strong>in</strong>to n=b 6 a sm<strong>al</strong>l r<strong>in</strong>gs each with a width 6.<br />
L<strong>et</strong> us consider a r<strong>in</strong>g with a height h whose <strong>in</strong>tern<strong>al</strong> radius is x and extern<strong>al</strong><br />
radius is x+6. If 6 is sm<strong>al</strong>l as compared with x, the resistance of such a<br />
r<strong>in</strong>g can be expressed by the formula<br />
2nx<br />
R=p?;ji<br />
The e. m. f. of <strong>in</strong>duction act<strong>in</strong>g <strong>in</strong> this r<strong>in</strong>g (if 6 ~ x) is equ<strong>al</strong> to<br />
L\(!)<br />
8=10- 8 AT= 10-snx'k<br />
The <strong>in</strong>tensity of the current flow<strong>in</strong>g <strong>in</strong> such a r<strong>in</strong>g is<br />
Al =
B.LECTRICITY AND MAGNETISM<br />
349<br />
The expression <strong>in</strong> the braces is an arithm<strong>et</strong>ic<strong>al</strong> progression. Therefore,<br />
1= 10-8 kh (b- a) 2a+b-a-6<br />
2p 2<br />
This result will be the more accurate, the sm<strong>al</strong>ler is 6. Assum<strong>in</strong>g ~ as tend<strong>in</strong>g<br />
to zero, we obta<strong>in</strong><br />
kh<br />
1= 10- 8 - (b2-a~)<br />
4p<br />
575. On the basis of the law of electromagn<strong>et</strong>ic <strong>in</strong>duction and Ohm's law J<br />
we have for the quantity of electricity that passed through the g<strong>al</strong>vanom<strong>et</strong>er:<br />
or<br />
AcD<br />
AQ=/At=lO-slf<br />
10- 8<br />
Q=-r (cD-cDo)<br />
S<strong>in</strong>ce the <strong>in</strong>iti<strong>al</strong> magn<strong>et</strong>ic flux «1>o=H An and the f<strong>in</strong><strong>al</strong> ftux «%>=0, the quantity<br />
of electricity <strong>in</strong> coulombs will be Q= 10;8 HAn if R is measured <strong>in</strong><br />
ohms and H <strong>in</strong> oersteds.<br />
576. The e. m. f. of <strong>in</strong>duction tOt = 10- 8 kat acts <strong>in</strong> circuit ABeD, and<br />
8,= 1O- 8k a; <strong>in</strong> circuit BEFC.<br />
The simplest equiv<strong>al</strong>ent circuit with g<strong>al</strong>vanic cells used as the e. m. 1. of<br />
<strong>in</strong>duction wi II for our circuit have the form shown <strong>in</strong> Fig. 464.<br />
On the basis of Ohm's law,<br />
J3ar = tCI - / l 3ar= / 2 2ar- B 2<br />
S<strong>in</strong>ce the charge is r<strong>et</strong>a<strong>in</strong>ed, '1=1,+1a- All three currents can easily be<br />
found from the given system of equations:<br />
11 6Itl+ 2 i12 / _2
350<br />
ANSWERS AND SOLUTIONS<br />
; t<br />
B<br />
If there is no potenti<strong>al</strong> difference,<br />
the electrostatic field <strong>in</strong>side<br />
the r<strong>in</strong>g is zero. The current<br />
is produced by the e.m.I, of <strong>in</strong>duction<br />
uniformly distributed<br />
<strong>al</strong>ong the r<strong>in</strong>g:<br />
I=!l...=
ELECTRICITY AND MAGNETISM 351<br />
But the e. m. f. of <strong>in</strong>duction <strong>in</strong> circuit ACBDA is zero, s<strong>in</strong>ce the circuit is<br />
not pierced by a magn<strong>et</strong>ic field. Therefore,<br />
c8l=c83 and laRa-/1Rl=O<br />
The system of equations gives the follow<strong>in</strong>g v<strong>al</strong>ue for the sought current:<br />
I -
352 ANSWERS AND SOLUTIONS<br />
the r<strong>in</strong>g is arranged <strong>al</strong>ong the force l<strong>in</strong>es of the field (the <strong>in</strong>duced current is<br />
zero) or when the plane of the r<strong>in</strong>g is strictly perpendicular to the force l<strong>in</strong>es<br />
(the moment of the forces is zero).<br />
Accord<strong>in</strong>g to Lens's law, the first position of the r<strong>in</strong>g will be stable <strong>in</strong> an<br />
<strong>in</strong>creas<strong>in</strong>g magn<strong>et</strong>ic field, and the second unstable.<br />
In a decreas<strong>in</strong>g magn<strong>et</strong>ic field, on the contrary, equilibrium will be stable<br />
when the angle b<strong>et</strong>ween the plane of the r<strong>in</strong>g and. the force l<strong>in</strong>es is a right<br />
one, and unsta ble when the plane of the r<strong>in</strong>g is para lIel to the force l<strong>in</strong>es.<br />
583. L<strong>et</strong> the velocity of the conductor be v at a certa<strong>in</strong> moment of time.<br />
The e. m. f. (<strong>in</strong> volts) at the same moment of time will thus be cC= 10- 8 Hlo,<br />
and the current I=ir x 10- 8 Hlo,<br />
The action of the magn<strong>et</strong>ic field on the<br />
conductor carry<strong>in</strong>g a current will <strong>in</strong>duce a force f that prevents free dropp<strong>in</strong>g<br />
of the conductor:<br />
f= 10-' H~2fJ<br />
be<strong>in</strong>g considered, the acceleration can be de<br />
Hence, at the moment of time<br />
term<strong>in</strong>ed from the relationship<br />
H 2 l 2U<br />
ma=mg-f=mg-IO-8~<br />
It is easy to see that as the velocity <strong>in</strong>creases, the acceleration a will dlm<strong>in</strong>ish<br />
and become zero at the moment of equ<strong>al</strong>ity of the forces f=mg. From<br />
this moment on, the conductor will move with a constant velocity Vk equ<strong>al</strong> to<br />
mgRXI0 9<br />
Vk H 2l 2<br />
584. "The e. m. f. of <strong>in</strong>duction appear<strong>in</strong>g <strong>in</strong> the conductor (measured <strong>in</strong> volts)<br />
is It= 10- 8 Hlu.<br />
The charge on the plates of the capacitor can be found from the relationship<br />
Q=l1C= 10- 8 HlvC<br />
The current flow<strong>in</strong>g <strong>in</strong> the circuit is<br />
1=~~=10-8 HlC ~~=10-8HICa<br />
where a Is the sought acceleration.<br />
The <strong>in</strong>teraction of this current with the magn<strong>et</strong>ic field will produce a force F l<br />
act<strong>in</strong>g on the mov<strong>in</strong>g conductor. On the basis of Lenz's law, this force will<br />
be directed oppositely tc the force F.<br />
The force F I =kJHI = lo-eHt/laC, if C is measured <strong>in</strong> farads. The sought<br />
acceleration can be found from the equation ma=F-Fl.<br />
Hence,<br />
F<br />
is a constant quantity.
ELECTRICITY AND MAGNETISM<br />
353<br />
Fig. 466<br />
The work of the force F over the path S is spent to <strong>in</strong>crease the k<strong>in</strong><strong>et</strong>ic<br />
energy of the conductor and the electrostatic energy of the capacitor.<br />
585. L<strong>et</strong> the magn<strong>et</strong> be <strong>in</strong>iti<strong>al</strong>ly positioned as shown <strong>in</strong> Fig. 466. Its<br />
north end is at a distance R1 from the current and its south end at a distance<br />
R2' the length of the magn<strong>et</strong> be<strong>in</strong>g 1= R2 - R1. L<strong>et</strong> us now move the<br />
magn<strong>et</strong> <strong>in</strong> the plane ~ around the wire, keep<strong>in</strong>g the distances R 1 and R t<br />
unchanged until after one revolution the magn<strong>et</strong> occupies its <strong>in</strong>iti<strong>al</strong> position.<br />
S<strong>in</strong>ce dur<strong>in</strong>g this motion the tot<strong>al</strong> change <strong>in</strong> the magn<strong>et</strong>ic flux through the<br />
area restricted by the straight wire and the conductors that short-circuit the<br />
current at a great distance from the magn<strong>et</strong> is zero, the quantity of <strong>in</strong>duced<br />
electricity that has flown through the circuit is <strong>al</strong>so zero. On the basis of<br />
the law of conservation of energy, the work of the forces of the magn<strong>et</strong>ic<br />
field should <strong>al</strong>so be equ<strong>al</strong> to zero:<br />
2n R 1 H t m- 2n R 2 H 2 m= O<br />
where m is the magn<strong>et</strong>ic charge of the pole, and HI and H2 are the <strong>in</strong>tensities<br />
of the magn<strong>et</strong>ic field at the distances R1 and R t from the wire.<br />
Hence, Z:=~: which is possible only when H is proportion<strong>al</strong> to ~ .<br />
586. S<strong>in</strong>ce accord<strong>in</strong>g to the <strong>in</strong>iti<strong>al</strong> condition, the <strong>in</strong>tensity. of the magn<strong>et</strong>ic<br />
field is directly proportion<strong>al</strong> to time, i. e., H =0.4 n 7kt, then the e. m. 1.<br />
of self-<strong>in</strong>duction is equ<strong>al</strong> to<br />
N2<br />
354 ANSWERS AND SOLUTIONS<br />
I<br />
proportion to time, i. e., I =kt, where<br />
VI<br />
k== 4nx 10- 9 AN2' Therefore, / =<br />
= 10 9 N~ 1. If the resistance is f<strong>in</strong>ite<br />
4n A,<br />
and not zero, the current will <strong>in</strong>crease<br />
accord<strong>in</strong>g to the same law until the<br />
voltage drop I R across the resistance R<br />
becomes negligibly sm<strong>al</strong>l as compared<br />
with {Jio<br />
588. The work of the battery dur<strong>in</strong>g<br />
the time -r is W =
ELECTRICITY AND MAGNETISM 355<br />
590. The magn<strong>et</strong>ic flux through the r<strong>in</strong>g cannot change (see Problem 589).<br />
Therefore,
356<br />
ANSWERS AND SOLUTIONS<br />
u<br />
A<br />
u<br />
o<br />
Fig. 468<br />
IJ<br />
UB<br />
2<br />
given moment. S<strong>in</strong>ce xp changes from<br />
zero to U/2, the diagram show<strong>in</strong>g u<br />
versus
ELECTRICITY AND MAONETISM<br />
357<br />
Fig. 469<br />
The average v<strong>al</strong>ue of 2/ 1 / 0 s<strong>in</strong> rot a period Is equ<strong>al</strong> to zero and the ave..<br />
rage v<strong>al</strong>ue of<br />
. is equ<strong>al</strong> to the square of the effective v<strong>al</strong>ue of the <strong>al</strong>ternat<strong>in</strong>g current. Therefore,<br />
Q=kR (1:+1:)<br />
On the other hand, the quantity ol heat Q liberated In the amm<strong>et</strong>er per<br />
second is related to the effective <strong>in</strong>tensity of the current I flow<strong>in</strong>g through<br />
the amm<strong>et</strong>er by the formula Q=kIIR.<br />
Hence, the a-c therm<strong>al</strong> amm<strong>et</strong>er will show<br />
I=V 1~+I:==lO A<br />
597. S<strong>in</strong>ce R=O, the current I lags <strong>in</strong> phase beh<strong>in</strong>d the voltage U by<br />
rt/2. Diagrams show<strong>in</strong>g U = Uos<strong>in</strong> Cillo' I =1 0 s<strong>in</strong> ( Cilt - ~ )and the <strong>in</strong>stantaneous<br />
power P=IU are shown <strong>in</strong> Fig. 469. Tbe sign of P changes every<br />
quarter of a period. The supply of energy from the source to the coil corresponds<br />
to a positive v<strong>al</strong>ue of P. When P is negative, the energy r<strong>et</strong>urns from<br />
the coil to the source. On an average, the coil consumes no power dur<strong>in</strong>g a<br />
period. The mean power is equ<strong>al</strong> to zero.<br />
598. The <strong>in</strong>ductive reactance of a choke is much greater than its resistance:<br />
Loo ~ R. The advantage of a choke over an ord<strong>in</strong>ary resistor is that no heat<br />
is liberated on the <strong>in</strong>ductive reactance of a choke. For this reason a lamp<br />
with a choke is much more economic<strong>al</strong> than a lamp with a resistor connected<br />
<strong>in</strong> series.<br />
599. If LID ~ R, the phase shift b<strong>et</strong>ween the current and the voltage is<br />
great and the power consumed by the ma<strong>in</strong>s cannot be high. When capacitors<br />
are switched on, the phase shift is reduced, s<strong>in</strong>ce the current flow<strong>in</strong>g through<br />
a capacitor leads the voltage, compensat<strong>in</strong>g thereby for the lag of the current<br />
<strong>in</strong> phase <strong>in</strong> electric devices with a high <strong>in</strong>ductance. As a result, the power<br />
consumed by the ma<strong>in</strong>s <strong>in</strong>creases.<br />
800. (a) S<strong>in</strong>ce ends A and B are open, no current flows <strong>in</strong> section AC.<br />
Therefore, the voltage drop <strong>in</strong> AC is zero. For this reason U 2 = U 1·<br />
(b) When a variable potenti<strong>al</strong> difference is applied b<strong>et</strong>ween po<strong>in</strong>ts Band C,<br />
the current flow<strong>in</strong>g <strong>al</strong>ong BC creates a variable magn<strong>et</strong>ic flux that generates
358 ANSWERS AND SOLUTIONS<br />
an e.rn.I. of <strong>in</strong>duction <strong>in</strong> section AC. S<strong>in</strong>ce Ls» ~ R, the amplitude of this<br />
e.m.I. will <strong>al</strong>so equ<strong>al</strong> Ur- For this reason the amplitude of the voltage U2<br />
b<strong>et</strong>ween po<strong>in</strong>ts A and B will be equ<strong>al</strong> to 2U1 (step-up autotransformer).<br />
601. When an <strong>al</strong>ternat<strong>in</strong>g current flows <strong>in</strong> a conductor the amount of heat<br />
evolved is Q= l;ffRt. The expression for the evolved heat Q= ;ff t is true<br />
only when Ohm's law can be applied <strong>in</strong> the usu<strong>al</strong> form: 1= ~ .<br />
The w<strong>in</strong>d<strong>in</strong>g of a transformer has a very high <strong>in</strong>ductive reactance. For<br />
this reason Ohm's law <strong>in</strong> its usu<strong>al</strong> form and, therefore, the expression<br />
U 2<br />
Q = ~ff t cannot be applied. The amount of heat evolved is sm<strong>al</strong>l, s<strong>in</strong>ce<br />
the <strong>in</strong>tensity of the current and the ohmic resistance of the w<strong>in</strong>d<strong>in</strong>g are<br />
sm<strong>al</strong>l.<br />
602. If we neglect the ohmic resistance, the voltage across the term<strong>in</strong><strong>al</strong>s<br />
of the primary w<strong>in</strong>d<strong>in</strong>g U1 can be represented as the <strong>al</strong>gebraic sum of the<br />
e.m.I, of self-<strong>in</strong>duction of this w<strong>in</strong>d<strong>in</strong>g and the e.m.I. of <strong>in</strong>duction generated<br />
<strong>in</strong> it by the current flow<strong>in</strong>g through the secondary w<strong>in</strong>d<strong>in</strong>g<br />
U 1<br />
= L 1<br />
~/l -M ~/2<br />
~t ~t<br />
The m<strong>in</strong>us sign is due to the fact that the currents 11 and /2 have opposite<br />
phases.<br />
If the currents change accord<strong>in</strong>g to the laws /1 = 1 0 1 s<strong>in</strong> wt and 1 2<br />
=<br />
= 1 02 s<strong>in</strong> tat, then<br />
8/1 1 !1/ 2<br />
-s:r=oo 01 cos wt and M= 00/ 0 2 cos oot<br />
S<strong>in</strong>ce the voltage U1 is shifted <strong>in</strong> phase relative to the current 11 by n/2,<br />
we can write U1 = U10 cos wt.<br />
Upon divid<strong>in</strong>g the expression for U 1 by LICJ) cos wt, we g<strong>et</strong><br />
-LU-Q.l.<br />
U01 =/ 0 1<br />
-!!!'/ o<br />
..<br />
LICJ) L t -<br />
is the no-load current if the ohmic resistance of the w<strong>in</strong>d<strong>in</strong>g is neglected.<br />
1 00 Disregard<strong>in</strong>g the no-load current, we f<strong>in</strong>d that<br />
1 0 1 M<br />
T;;=L;<br />
By us<strong>in</strong>g the expressions for the coefficient of self-<strong>in</strong>duction and mutu<strong>al</strong><br />
<strong>in</strong>ductance from <strong>Problems</strong> 591 and 592. we obta<strong>in</strong><br />
1 1 J 0 1 N 2<br />
T; = 1 02<br />
= Nt<br />
603. The positive h<strong>al</strong>f-waves of the current will charge the capacitor to<br />
the amplitude voltage of the ma<strong>in</strong>s, equ<strong>al</strong> to 127 ·Y2" V= 180 V. When the<br />
diode carries no current, it receives the voltage of the ma<strong>in</strong>s (with an arnpli-<br />
U 2<br />
j
ELECTRICITY AND MAGNETISM 359<br />
A ......---...<br />
Fig. 470<br />
B<br />
tude of 180 V) plus the same vol..<br />
tage of the charged eapaeitor. The<br />
change <strong>in</strong> the potent ia I <strong>al</strong>ong the<br />
circuit at this moment of time is<br />
shown <strong>in</strong> Fig. 470.<br />
If the rectifier operates without<br />
load, the capacitor should be c<strong>al</strong>eufated<br />
for a punctur<strong>in</strong>g volta ge of at<br />
least 180 V, and the diode for a<br />
voltage of at least 360 V.<br />
604. The anode voltage of each<br />
diode is<br />
U a = ~<br />
s<strong>in</strong> rot-IR<br />
A current flows through the diode when Ua > 0 and does not flow through<br />
it when V a ~ O. In a quarter of a period the current will not flow dur<strong>in</strong>g<br />
the time <strong>in</strong>terv<strong>al</strong> 0 ~ t ~ t 1 (Fig. 471). where t 1 is d<strong>et</strong>erm<strong>in</strong>ed by the equation<br />
I/o<br />
-JH<br />
Fig. 471<br />
~ s<strong>in</strong> rot.-IR=0. Hence, t. = ~ arc s<strong>in</strong> 2~R . This is <strong>al</strong>so the time dur<strong>in</strong>g<br />
which the current does not flow <strong>in</strong> the follow<strong>in</strong>g quarters of the period.<br />
Altog<strong>et</strong>her <strong>in</strong> a period the current does not flow dur<strong>in</strong>g<br />
2T . 2/R<br />
"""it arc s<strong>in</strong> ---0-=0.465 T<br />
3-6. Electric<strong>al</strong> Mach<strong>in</strong>es<br />
605. If the frequency of the <strong>al</strong>ternat<strong>in</strong>g current rema<strong>in</strong>s the same, this<br />
means that the revolutions of the motor and the generator <strong>al</strong>so rema<strong>in</strong> as<br />
before. and the e.rn.I. of the generator wHI not change.<br />
When the extern<strong>al</strong> resistance <strong>in</strong> the circuit is high, the circuit will carry<br />
a sm<strong>al</strong>ler current and a lower power will be supplied. For this reason the<br />
power of the motor that revolves the generator should be reduced.<br />
606. The work performed by a field <strong>in</strong> mov<strong>in</strong>g conductors carry<strong>in</strong>g a current<br />
(armature w<strong>in</strong>d<strong>in</strong>gs) is not equ<strong>al</strong> to the tot<strong>al</strong> work of the field. Apart from<br />
the work spent to move the conductors, the magn<strong>et</strong>ic field performs work to
360 ANSWERS AND SOLUTIONS<br />
r<strong>et</strong>ard the electrons <strong>in</strong> the conductor, which produces an e.m.I, of <strong>in</strong>duction<br />
<strong>in</strong> the armature w<strong>in</strong>d<strong>in</strong>g. The first part of the work is positive and the second<br />
negative. The tot<strong>al</strong> work of the magn<strong>et</strong>ic field is zero.<br />
The electromotive force of the source that generates a current <strong>in</strong> the motor<br />
armature performs positive work, and the latter compensates for the negative<br />
work of the magn<strong>et</strong>ic field <strong>in</strong> r<strong>et</strong>ard<strong>in</strong>g the electrons.<br />
In essence, the motor does its work at the expense of the energy of the<br />
source feed<strong>in</strong>g it.<br />
607. The power consumed by the motor is P= JV; here V = tCi +f R ,<br />
where
ELECTRICITY AND MAGNETISM<br />
361<br />
Fig. 472<br />
1-4------ d -----11--.<br />
power is<br />
P=2kIHlv=klHAw<br />
The current , can be d<strong>et</strong>erm<strong>in</strong>ed<br />
from the formula<br />
1 _ U-c8i<br />
- R<br />
where tR;= IO-SHAw<br />
F<strong>in</strong><strong>al</strong>ly P can be written as follows:<br />
U<br />
H2A'"<br />
P=kHA R CIl-IO- 8 k-R- 00 2<br />
P reaches its max imum<br />
kU2 U 2<br />
Pmax = 4l[ X I08=4R J<br />
when 6>=2%A X lOS. Here tIli= ~ and 1=2~' In a unit of time the<br />
U2 ·<br />
battery performs the work 2R • From this amount, one h<strong>al</strong>f<br />
is converted <strong>in</strong>to<br />
mechanic<strong>al</strong> power and the other h<strong>al</strong>f is liberated as heat (see Problem 607).<br />
The relation b<strong>et</strong>ween P and 00 is shown <strong>in</strong> Fig. 473.<br />
_ kH AU _ k·lO~8H2A2<br />
610· M - R R CIl,<br />
U.10<br />
The moment will be equ<strong>al</strong> to zero when ro=---n;r 8<br />
(see Fig. 474). Here<br />
1=0, because tRi=U,<br />
611. The naTure of the relation b<strong>et</strong>ween P and H is shown <strong>in</strong> Fig. 475<br />
(see the solution to Problem 609). The power reaches its maximum when<br />
U U U2<br />
H=2ACI) X 10 8 • Here l1i=T and Pmax=4R J.<br />
p<br />
N<br />
KHAU<br />
f{<br />
Fig. 473<br />
U 8<br />
ZHA'O<br />
Fig. 474
362<br />
p<br />
ANSWERS AND SOLUTIONS<br />
612. The torque will reach its<br />
VI<br />
maximum M max = 10 7 X 4Rro dyne-<br />
U<br />
-crn when H = 10 8 X 2AID.<br />
613. As with a series motor, the<br />
power of a shunt-wound motor is<br />
Fig. 475 where R is the resistance of the<br />
armature (see Problem 607). Two<br />
v<strong>al</strong>ues of
ELECTRICITY AND MAGNETISM 363<br />
n<br />
H<br />
<strong>in</strong>g and their area, and H is the <strong>in</strong>tensity<br />
of the magn<strong>et</strong>ic field of the<br />
stator. which is directly proportion<strong>al</strong><br />
to the current.<br />
Upon exclud<strong>in</strong>g cCi from these equations,<br />
we f<strong>in</strong>d that<br />
U 2nMR<br />
n= kH- k2H2<br />
The relation b<strong>et</strong>ween<br />
shown <strong>in</strong> Fig. 476.<br />
nand H is<br />
Fig. 476<br />
2nMR<br />
If H H > H o , the speed <strong>in</strong>creases<br />
when the current grows <strong>in</strong> the w<strong>in</strong>d<strong>in</strong>gs of the stator. and decreases when H>Hm:<br />
If the motor operates without load (M =0). the speed will be n= k~ ,<br />
i.e., it <strong>al</strong>ways decreases with an <strong>in</strong>crease of H.<br />
616. The <strong>in</strong>tensity of the current flow<strong>in</strong>g <strong>in</strong> the w<strong>in</strong>d<strong>in</strong>g of the motor<br />
will be d<strong>et</strong>erm<strong>in</strong>ed by the e.m.I, of the ma<strong>in</strong>s
364 ANSWERS AND SOLUTIONS<br />
and<br />
11+J2+J8= los<strong>in</strong> ( rot+ ~ )+J0 s<strong>in</strong> ( rot+ : :t) =<br />
=2JoS<strong>in</strong>(rot+i-n)cos~ =0<br />
619. The magn<strong>et</strong>ic fields Hi' H 2 and H 3 can be written as follows:<br />
H 1=Hos<strong>in</strong>liJt. H 2=HoS<strong>in</strong>(rot+fn). and H8=Hos<strong>in</strong>(rot+'~ n)<br />
L<strong>et</strong> us select the axes of coord<strong>in</strong>ates x and y as shown <strong>in</strong> Fig. 220 and f<strong>in</strong>d<br />
the sum of the projections of the <strong>in</strong>tensities of these fields on these axes:<br />
H,,=HoS<strong>in</strong>liJt+HoSi~(rot+<br />
~n)cos; n+HoS<strong>in</strong>(rot + fn)cos{ n<br />
Hy=Hos<strong>in</strong> (rot+ ~ n) s<strong>in</strong> ~ n+Hos<strong>in</strong> (rot+ ~ n) s<strong>in</strong> ~ Tl<br />
After simple transformations, we have<br />
H,,=: Hos<strong>in</strong>rot and Hy=: Hocos rot<br />
Such projections are possible only if the vector show<strong>in</strong>g the magn<strong>et</strong>ic field<br />
rotates clockwise with a constant angular velocity (a). .<br />
620. In these conditions the currents <strong>in</strong> coils 1-2 and 3..4 are shifted <strong>in</strong><br />
phase by <strong>al</strong>most n/2. Correspond<strong>in</strong>gly, the magn<strong>et</strong>ic fields created by them<br />
are shifted by the same magnitude. Thus, the space b<strong>et</strong>ween the coils conta<strong>in</strong>s<br />
the fields:<br />
Hi=Hos<strong>in</strong>rot<br />
directed vertic<strong>al</strong>ly and<br />
H2=Hos<strong>in</strong> (6)t+';' )=HoCOS rot<br />
directed horizont<strong>al</strong>ly.<br />
This means (see Problem 619) that a revolv<strong>in</strong>g magn<strong>et</strong>ic field is produced<br />
<strong>in</strong> the space. As the field rotates, it carries <strong>al</strong>ong the cyl<strong>in</strong>der.<br />
This pr<strong>in</strong>ciple underlies the design of s<strong>in</strong>gle-phase <strong>in</strong>duction motors.
CHAPTER 4<br />
OSCILLATIONS<br />
AND WAVES<br />
4-1. Mechanic<strong>al</strong> Oscillations<br />
621. The vertic<strong>al</strong> component of the tension force T is equ<strong>al</strong> to F= t cos ex<br />
(Fig. 477). For the conic<strong>al</strong> pendulum F= mg, s<strong>in</strong>ce the weight has no acceleration<br />
<strong>in</strong> the vertic<strong>al</strong> plane.<br />
When the mathematic<strong>al</strong> pendulum is deflected the maximum from the<br />
position of equilibrium (through the angle a), the result<strong>in</strong>g force is directed<br />
at a tangent to the trajectory of the weight.<br />
Therefore, T = mg cos ct.<br />
When the weight is deflected through the angle a, the tension of the thread<br />
of the conic<strong>al</strong> pendulum will be greater.<br />
622. The period of oscillations of the pendulum on the surface of the Earth<br />
is To= 2n .. / I and at an <strong>al</strong>titude of h above the Earth T I = 2n .. / I<br />
JI g JI it<br />
The number of oscillations a day is N1=24X60x60 ~l ~ :1. Therefore,<br />
at an <strong>al</strong>titude of h above the Earth the clock will be slower by the time<br />
To'<br />
1it 1=N1 (T1-To)=k ( 1-T";)<br />
The ratio b<strong>et</strong>ween the periods is ;: =<br />
as follows from the law of gravitation. Hence.<br />
At kh ~ kh ~ 2 7 d<br />
u 1= R+h - R - . secon s<br />
V ~ = R:h<br />
If the. clock is lowered <strong>in</strong>to a m<strong>in</strong>e. the acceleration<br />
ti . g?, R-h. 4n RS 1 d<br />
ra 101S--g=--r-. smce g="3 p R" an g2=<br />
4n 1<br />
= l' 3 (R - h)3P(R ~ h)t (see Problem 234).<br />
Hence,<br />
;:=V~=VRR h~1-2~<br />
In this case the clock will .be slower by the time<br />
Fig. 477<br />
b-tl=k (<br />
To ) kh.- d<br />
l-r; =2R = 1.35 secon s
366 ANSWERS AND SOLUTIONS<br />
623. Each h<strong>al</strong>f of the rod with a sm<strong>al</strong>l sphere on its end is a mathematic<strong>al</strong><br />
pendulum with a length d/2 that oscillates <strong>in</strong> the field of gravity of the<br />
large sphere. In the fiel d of gravity of the Eart h the period of sma11 oscillations<br />
of a mathematic<strong>al</strong> pendulum is T o=2n V ~. Accord<strong>in</strong>g to the law<br />
of gravitation mg=y m~E , and therefore<br />
.. /fR2<br />
To=2n V yM E<br />
where 1'=6.67X 10- 8 cm 2/g·s2 is the gravity constant, ME is the mass of<br />
the Earth, and R is the distance from the pendulum to the centre of the Earth.<br />
Correspond<strong>in</strong>gly, the period of sm<strong>al</strong>l oscillations of the mathematic<strong>al</strong> pendulum<br />
with a length 1= ~ <strong>in</strong> the field of gravity of the large sphere will be<br />
.. ;([[2<br />
T=2n V 21'M::::: 5.4 hours<br />
624. The period of oscillations of a mathematic<strong>al</strong> pendulum is<br />
T=2n V;,<br />
where g' is the gravity acceleration <strong>in</strong> the correspond<strong>in</strong>g coord<strong>in</strong>ate system.<br />
In our case<br />
g'= JI g2+a 2<br />
where g is the gravity acceleration with respect to the Earth.<br />
Thus,<br />
T=2n, /<br />
V y g2+a 2<br />
~_l__<br />
625. T=2n .. I" 1 • Use the plus sign If the acceleration of the lilt is<br />
V g±a<br />
directed upward and the m<strong>in</strong>us sign if it is directed downward.<br />
626. The oscillations of the block <strong>in</strong> the cup are similar to those ot a<br />
mathematic<strong>al</strong> pendulum, with the only difference that <strong>in</strong>stead of the tension<br />
of the spr<strong>in</strong>g the block is acted upon by the reaction of the support. There..<br />
fore, the sought oscillation period is<br />
T=2n V:<br />
F<br />
627. When M ~ m, the acceleration of the cup is a=Ar-g. Therefore<br />
(see Problem 626),
OSCILLATIONS AND WAVES 367<br />
If F=O, i. e., with free f<strong>al</strong>l<strong>in</strong>g of the cup, T= eo<br />
tions. When F=Mg, we have T=2n { ~ .<br />
and there are no oscilla..<br />
628. The oscillations of the block will periodic<strong>al</strong>ly displace the cup <strong>in</strong> a<br />
horizont<strong>al</strong> plane. Hence, the oscillation period of the block will dim<strong>in</strong>ish,<br />
s<strong>in</strong>ce an addition<strong>al</strong> variable acceleration directed horizont<strong>al</strong>ly will appear <strong>in</strong><br />
the coord<strong>in</strong>ate system related to the cup (see Problem 624).<br />
629. L<strong>et</strong> us compare the motion of the centre of the hoop with that of the<br />
end of a mathematic<strong>al</strong> pendulum with a length R- r, Both po<strong>in</strong>ts describe<br />
the arc of a circle with a radius R - r. L<strong>et</strong> us assume that the hoop and the<br />
pendulum are at rest at the angle
368 ANSWERS AND SOLUTIONS<br />
L<strong>et</strong> us so select l that rot = CJ)2~<br />
For thIs it is necessary that<br />
l= ml~+m2l~<br />
mIll + m 21 2<br />
The angular velocity characterizes the change <strong>in</strong> the angle ~<br />
<strong>in</strong> the course of<br />
time. S<strong>in</strong>ce (1,)1 = (1,)2' the oscillation periods of the two pendulums are the same.<br />
For a mathematic<strong>al</strong> pendulum<br />
T=2n y ~<br />
Therefore, the sought period is<br />
T=2n'" /<br />
m!l~ +m2l: I<br />
V mill +m 2l 2 g<br />
631. This problem can be solved by the same m<strong>et</strong>hod as Problem 630. L<strong>et</strong><br />
the h<strong>al</strong>f-r<strong>in</strong>g be Initi<strong>al</strong>ly deflected from the position of equtlibrlum through<br />
the angle
OSCILLATIONS AND WAVES 369<br />
L<strong>et</strong> us denote the displacement of the weight measured from the position<br />
of equilibrium by x, The velocity of the weight as a function of x can be<br />
found from the law of conservation of energy<br />
k (X O + l)2 _ mv 2 +k (x+ 1)2 mgx<br />
2 mgxo - 2 2<br />
Bear<strong>in</strong>g <strong>in</strong> m<strong>in</strong>d that mg=kl. we f<strong>in</strong>d:<br />
v={k (x~; x 2 )<br />
If M :1= 0, the law of conservation of energy can be written as<br />
It follows that<br />
k (Xo+ 1)2 (m+ M) v 2 +k (x+ l)2<br />
2 -mgxo 2 2 mgx<br />
v= .. / k (X~-X2)<br />
V M+m<br />
Thus, <strong>in</strong> the second case (M ~ 0) the weight<br />
<strong>in</strong>creased by M as compared with the first case.<br />
moves as If Its mass had<br />
Hence, the sought period is<br />
T=2n ym~M<br />
633. When the bottle is displaced from the position of equilibrium by x,<br />
the force act<strong>in</strong>g on the bottle will be equ<strong>al</strong> to F = - yoA x • where Yo is the<br />
specific weight of the water. The m<strong>in</strong>us sign means that the force is directed<br />
aga<strong>in</strong>st the displacement x. Accord<strong>in</strong>g to Newton's second law, the oscillations<br />
of the bottle are d<strong>et</strong>erm<strong>in</strong>ed by the equation ma=-yoAx. This equation is<br />
absolutely similar to the equation for the oscillation of a weight on a spr<strong>in</strong>g:<br />
ma=- kx. S<strong>in</strong>ce for the weight we have CiI=:= 2; = V ~ .then the oscillation<br />
frequency of the bottle will be<br />
v=~=J.. .. / '\'0 A ~ 2 5 _1_<br />
2n 2n V m · second<br />
634. The equation of motion of the mercury has the form<br />
ma=- VA2xwhere<br />
x Is the displacement of the mercury level from the position of equilibrium.<br />
The equation of motion has the same form as <strong>in</strong> the case of oscillations<br />
of a weight on a spr<strong>in</strong>g. Therefore (see Problem 632),<br />
T=2n y 2;A ~ 1.54 seconds<br />
635. The force act<strong>in</strong>g on the body is F= '\' ~ npmr where' is the distance<br />
from the centre of the Earth and ')' is the gravity constant (see Problem 234).<br />
24-2042
370 ANSWERS AND SOLUTIONS<br />
Remember<strong>in</strong>g that<br />
g=y4: pRo this expression can be written as<br />
r<br />
F=mg R<br />
Here R is the radius of the Earth.<br />
the form:<br />
The equation of motion of the body has<br />
ma=- mgr=-k,<br />
R<br />
The force is proportion<strong>al</strong> to the displacement from the position of equilibrium<br />
and directed toward the centre of the Earth. Therefore, the body will perform<br />
harmonic oscillations with a frequency<br />
Hence, the period of oscillations is<br />
.. /_R<br />
T=2n V<br />
g<br />
The body will reach the centre of the Earth dur<strong>in</strong>g<br />
"f=!.="::' 1Il / R ~ 21 m<strong>in</strong>utes<br />
4 2 V g<br />
1t is of <strong>in</strong>terest that the time 't does not depend at <strong>al</strong>l on the distance from<br />
the centre of the Earth at which the body beg<strong>in</strong>s to move, provided this<br />
distance is much greater than the size of the body.<br />
636. The force F act<strong>in</strong>g on the weight deflected from the position of<br />
equilibrium is 21 s<strong>in</strong> qJ (Fig. 478). S<strong>in</strong>ce the angle cp is sm<strong>al</strong>l, it may be<br />
4/x 4/<br />
assumed that F = -1- or F == kx, where k = T .<br />
By us<strong>in</strong>g the formula<br />
we g<strong>et</strong> the follow<strong>in</strong>g expression for the sought quantity:<br />
T=2nI ~;<br />
637. The period of oscillations of the weight on a spr<strong>in</strong>g is<br />
T=2nIi<br />
where k is the coefficient of elasticity of the spr<strong>in</strong>g equ<strong>al</strong> to the ratio b<strong>et</strong>ween<br />
the force that caused the spr<strong>in</strong>g to str<strong>et</strong>ch and its elongation: k = ~ .<br />
x
)SCI LLATIONS AND WAVES<br />
371<br />
Fig. 478<br />
When two identic<strong>al</strong> spr<strong>in</strong>gs str<strong>et</strong>ched by the force F are connected <strong>in</strong> series<br />
F F k<br />
k 1 = Xl =2x ='2<br />
s<strong>in</strong>ce the length of each spr<strong>in</strong>g <strong>in</strong>creases by x. When such spr<strong>in</strong>gs are connected<br />
<strong>in</strong> par<strong>al</strong>lel, the force F) required to <strong>in</strong>crease the<br />
should be two times greater than F.<br />
length of each by x<br />
FI 2F<br />
Hence,<br />
k 2=-=-=2k<br />
x x<br />
With series connection<br />
and with par<strong>al</strong>lel connection<br />
Hence, ~ ~<br />
T) =2n -(~1 =2n -(2:<br />
T2=2n -(2~<br />
=2. The period is h<strong>al</strong>ved.<br />
638. L<strong>et</strong> us deflect both pendulums from the vertic<strong>al</strong> <strong>in</strong> the same direction<br />
and through the same angle. In this case the spr<strong>in</strong>g will not be deformed.<br />
It is easy to see that the pendulums released from this position will oscillate<br />
<strong>in</strong> phase with a frequency of 00= -( ~ • If the pendulums are deflected <strong>in</strong><br />
opposite directions through the same angles they will oscillate <strong>in</strong> antiphase<br />
and the spr<strong>in</strong>g will be deformed. To c<strong>al</strong>culate the frequency of these oscillations,<br />
l<strong>et</strong> us f<strong>in</strong>d the force that r<strong>et</strong>urns the pendulums to the position of<br />
equilibrium. Upon deflection through the angle Q) the force act<strong>in</strong>g on the<br />
mass m from the side of the spr<strong>in</strong>g is 2kl s<strong>in</strong> tp. The sum of the projections<br />
of the forces of gravity and elasticity on a tangent to the circumference, the<br />
so..c<strong>al</strong>led "restor<strong>in</strong>g force" Fr will be equ<strong>al</strong> to<br />
Fr = mg s<strong>in</strong> ~+ 2kl s<strong>in</strong> q> cos cp<br />
(Fig. 479). S<strong>in</strong>ce cos fP ~ 1 at sm<strong>al</strong>l angles.<br />
F,=(mg+2kl!s<strong>in</strong>cp or F,=m(g+2~I)s<strong>in</strong>cp<br />
For a mathematic<strong>al</strong> pendulum the restor<strong>in</strong>g force is mg s<strong>in</strong> cp. The frequency<br />
of oscillations at sm<strong>al</strong>l angles cp can be found from the formula 00= V~·<br />
2kl<br />
In our case the part of g is played by g +- .<br />
m<br />
24 *
372 ANSWERS AND SOLUTIONS<br />
Hence,<br />
0)= Vg~~l<br />
where (0<br />
The period of oscillations is<br />
... I 1<br />
2kl<br />
T=2n V g+m-<br />
639. Yes, it can. For this purpose the<br />
door should be gradu<strong>al</strong>ly swung with a<br />
frequency equ<strong>al</strong> to the natur<strong>al</strong> frequency<br />
of oscillations of the door. Upon resonance,<br />
the amplitude of oscillations may be<br />
very high.<br />
640. From the law of conservation of<br />
.energy:<br />
(02<br />
2 (ml 2 + M,") = Mgra-mgl (I-cos
OSCILLATIONS AND WAVES<br />
373<br />
1<br />
180 Fig. 480<br />
4-2. Electric<strong>al</strong> Oscillations<br />
641. Without a permanent magn<strong>et</strong> the oscillation frequency would be<br />
doubled. When a s<strong>in</strong>usoid<strong>al</strong> current flowed through the coil of the telephone<br />
it would cause the membrane to make two oscillations dur<strong>in</strong>g one period of<br />
current oscillations because the <strong>in</strong>tensity of the magn<strong>et</strong>ic field H created by<br />
this current would have the form shown <strong>in</strong> Fig. 481a, and the force of attraction<br />
of the membrane does not depend on the sign of H.<br />
When there is a permanent magn<strong>et</strong> that generates an <strong>in</strong>tensity of the<br />
magn<strong>et</strong>ic field exceed<strong>in</strong>g the maximum <strong>in</strong>tensity of the current field, the<br />
diagram of the result<strong>in</strong>g <strong>in</strong>tensity has the form shown <strong>in</strong> Fig. 4Blb.<br />
For this reason one oscillation of the current will correspond to one of<br />
the membrane. and the sound will be much less distorted.<br />
642. The frequency of natur<strong>al</strong> oscillations is (0= ..r' if L is <strong>in</strong> henries<br />
f LC<br />
and C <strong>in</strong> farads. As has been shown <strong>in</strong> Problem 591, for a solenoid<br />
L= 10- 9 41t~aAt henries. The capacitance of the capacitor is C= ::d X<br />
1<br />
X 9X 1011<br />
farads.
374<br />
ANSWERS AND SOLUTIONS<br />
H<br />
(a)<br />
H<br />
t'<br />
(b)<br />
Hence,<br />
ro=3X 10 10<br />
t Fig. 481<br />
V- - ld - 1<br />
N2A 1A2<br />
--- =3X 10 8 --d<br />
secon<br />
643. The frequency of natur<strong>al</strong> oscillations of a circuit is d<strong>et</strong>erm<strong>in</strong>ed from<br />
the Thomson formula<br />
I<br />
W=YLC<br />
(a) If the coil conta<strong>in</strong>s the copper core, the periodic changes of the magn<strong>et</strong>ic<br />
field of the coil will produce <strong>in</strong> the core eddy currents whose magn<strong>et</strong>ic<br />
field will weaken the magn<strong>et</strong>ic field of the coil. This will reduce the <strong>in</strong>ductance<br />
of the coil and, consequently, <strong>in</strong>crease the frequency 00.<br />
(b) If the ferrite core is moved <strong>in</strong>to the coil, the magn<strong>et</strong>ic field of the<br />
latter will <strong>in</strong>crease. Accord<strong>in</strong>gly, the <strong>in</strong>ductance L of the coil will grow and<br />
the frequency (r) will decrease.<br />
844. Undamped oscillations will appear <strong>in</strong> the system (if the sm<strong>al</strong>l losses<br />
of energy for the radiation of electromagn<strong>et</strong>ic waves are neglected.) When<br />
the charge is distributed equ<strong>al</strong>ly b<strong>et</strong>ween the capacitors, the energy of the<br />
electrostatic field is m<strong>in</strong>imum, but the current <strong>in</strong>tensity and the energy of the<br />
magn<strong>et</strong>ic field are maximum. The tot<strong>al</strong> energy does not change, but one k<strong>in</strong>d<br />
of energy is converted <strong>in</strong>to another.<br />
645. The displacement of the electron beam by the voltage supplied <strong>al</strong>ong<br />
the vertic<strong>al</strong> can be written as<br />
lL<br />
x=2dV VOl cosrot=a cos rot<br />
(see Problem 543). The displacement of the beam <strong>al</strong>ong the horizont<strong>al</strong> (axis<br />
u) is<br />
lL<br />
1J=2dV V Ot cos (wt-cp)=b cos (oot-q»
OSCILLATIONS AND WAVES<br />
375<br />
!/<br />
Fig. 482<br />
To f<strong>in</strong>d the trajectory, the time should be excluded from the data of the<br />
equations. After simple transformations we have<br />
Xl y2 2xy ..<br />
a 2 +b 2 -lib cos «p=Sln 2
376 ANSWERS AND SOLUTIONS<br />
(curve ab) until the voltage drops to V e . The process is then repeated and<br />
relaxation oscilla tions with a period 't appear.<br />
The charg<strong>in</strong>g and discharg<strong>in</strong>g current of the capacitor is not constant,<br />
because it depends on the voltage across the capacitor, and decreases with a<br />
growth of the voltage. For this reason Oat ab, be, <strong>et</strong>c., are not sections of<br />
straight l<strong>in</strong>es.<br />
647. When the capacitance grows, the time needed to charge the capacitor<br />
to Vs and discharge it to V e <strong>in</strong>creases. The period will therefore be longer.<br />
An <strong>in</strong>crease <strong>in</strong> R will reduce the charg<strong>in</strong>g current of the capacitor and<br />
will <strong>in</strong>crease the period.<br />
648. When the charge across the plates of the capacitor reaches its maximum,<br />
the plates should be moved apart. Here work must be performed to<br />
overcome the forces of attraction b<strong>et</strong>ween the plates. This work is spent to<br />
<strong>in</strong>crease the energy of the circuit. When the charge is zero, the plates should<br />
be moved tog<strong>et</strong>her to their orig<strong>in</strong><strong>al</strong> position. The energy <strong>in</strong> the circuit will<br />
not change.<br />
4-3. Waves<br />
649. The tension of the str<strong>in</strong>g should be <strong>in</strong>creased four times.<br />
650. v = 'Von, where n= 1, 2, 3, 4, ... , and<br />
v _-!.--.iT _2_1_<br />
0- ld V np - second<br />
651. The pipe should accommodate a whole number of h<strong>al</strong>f-waves:<br />
A.<br />
2:k=l (k= 1, 2, 3, ...)<br />
The frequencies of the natura) oscillations<br />
are<br />
c kc I<br />
'Yk=-r= 2l=kX50 second<br />
(c=340 mls is the velocity of sound <strong>in</strong> air).<br />
652. -The sound of the tun<strong>in</strong>g fork will be <strong>in</strong>tensified when the frequency<br />
of the natur<strong>al</strong> oscillations of the air column <strong>in</strong> the vessel co<strong>in</strong>cides with the<br />
frequency of the tun<strong>in</strong>g fork. The natur<strong>al</strong> oscillation frequency of the air<br />
column <strong>in</strong> a tube closed at one end is 'Vk=2k: I • ~ • where I is the length<br />
of the tube and c=340 mls is the velocity of sound. The quantity k takes<br />
v<strong>al</strong>ues of 0, 1, 2, 3, ... Therefore, the possible water levels <strong>in</strong> the vessel<br />
d<strong>et</strong>erm<strong>in</strong>ed by the distance from the surface of the water to the upper edge<br />
of the vessel are<br />
2k+l c<br />
Ik=-4- -:v (k=O, 1, 2, ••.)<br />
When 1= 1 m<strong>et</strong>re, two positions of the water level are possible: 1 0 = 25 em<br />
and 11 =75 em.<br />
653. L<strong>et</strong> us consider a. number of consecutive positions of the bull<strong>et</strong> fly<strong>in</strong>g<br />
<strong>al</strong>ong KA, namely, K. F, E, D, B, and A (Fig. 484). At each po<strong>in</strong>t the
OSCILLATIONS AND WAVES<br />
377<br />
Fig. 484<br />
bull<strong>et</strong> creates before its front a compression that spreads <strong>in</strong> <strong>al</strong>l directions <strong>in</strong><br />
the form of a spheric<strong>al</strong> pulse. S<strong>in</strong>ce the velocity of the bull<strong>et</strong> v is greater<br />
than that of sound c, these pulses appear only beh<strong>in</strong>d the bull<strong>et</strong>. At the<br />
moment when the bull<strong>et</strong> is at po<strong>in</strong>t A, the separate pulses will be as shown<br />
<strong>in</strong> Fig. 484 by circles of various radii. The wave front has the form of a<br />
cone that m.oves forward with the velocity of the bull<strong>et</strong>. The apex angle of<br />
the cone can be d<strong>et</strong>erm<strong>in</strong>ed from the ratio<br />
. BH ct c<br />
S<strong>in</strong> a= AB =vt ='V<br />
654. The sound wave that reaches the man at po<strong>in</strong>t B (Fig. 485) is emit ..<br />
ted when the plane is at a certa<strong>in</strong> po<strong>in</strong>t D (see the solution to Problem 653).<br />
The distance CB=6 km.<br />
The sough t distance is<br />
BC v<br />
AB=-.-=BCs<strong>in</strong><br />
ex. c<br />
where u is the velocity of the plane and c the velocity of sound. Hence,<br />
AB=9 km.<br />
655. Ord<strong>in</strong>arily I the velocity of the w<strong>in</strong>d at a certa<strong>in</strong> <strong>al</strong>ti tude above the<br />
ground is greater than at its surface. For this reason the wave surfaces<br />
which <strong>in</strong> immobile air have the form of spheres with their centre at the<br />
po<strong>in</strong>t of the sound source (dotted l<strong>in</strong>es <strong>in</strong> Fig. 486) change their shape. The<br />
velocity of the waves is higher <strong>in</strong> the direction of the w<strong>in</strong>d than aga<strong>in</strong>st it,<br />
The approximate shapes of the wave surfaces are shown <strong>in</strong> Fig. 486 by solid<br />
l<strong>in</strong>es.<br />
The sound propagates <strong>in</strong> a direction perpendicular at each po<strong>in</strong>t to the<br />
wave surfaces. For this reason the sound propagat<strong>in</strong>g aga<strong>in</strong>st the w<strong>in</strong>d is<br />
deflected upwards (curve AB) and does not reach the man. If the sound propagates<br />
<strong>in</strong> the direction of the w<strong>in</strong>d the sound is deflected towards the ground<br />
(curve AC) and the man hears it.<br />
656. TV stations operate on wavelengths sm<strong>al</strong>ler than 10 m<strong>et</strong>res. For such<br />
waves the ionosphere is "transparent" and does not reflect the waves. Short<br />
waves are practic<strong>al</strong>ly propagated <strong>al</strong>ong a straight l<strong>in</strong>e, s<strong>in</strong>ce they undergo<br />
<strong>al</strong>most no diffraction from such obstacles as houses, <strong>et</strong>c.
A ~-..----4-----r--------r--<br />
Fig. 485<br />
657. To estimate the distance to an object by the position of the reflected<br />
pulse on the screen of a cathode-ray tube, this pulse should arrive not ear ...<br />
I<br />
lier than <strong>in</strong> the time l' and not later than <strong>in</strong> the time T = T after send<strong>in</strong>g<br />
the direct pulse. Hence,<br />
and the maximum distance is<br />
the m<strong>in</strong>imum distance to the object is<br />
C1'<br />
1="2=120 m<br />
L=C~ ~90 km<br />
658. The wave reflected from the roof will reach the aeri<strong>al</strong> with a lag of<br />
AB .<br />
c<br />
,. = - = 10- 6 s. The velocity of the cathode beam <strong>al</strong>ong the screen is<br />
v = ~t ' where M=25~625 s is the time dur<strong>in</strong>g which the beam produces<br />
one l<strong>in</strong>e (neglect the time of reverse travel of the beam).<br />
The shift of the images is ~l =V't' :=!: 7.8 cm.<br />
659. When the vibrator is immersed <strong>in</strong> kerosene its capacitance C <strong>in</strong>creases<br />
e r times. The frequency of the natur<strong>al</strong> oscillations of the circuit is proportion<strong>al</strong><br />
to ";c' Therefore, the oscillation frequency will decrease -ve;<br />
times. The frequency of the natur<strong>al</strong> oscillations of<br />
is v O= 2 c l ' and <strong>in</strong> a dielectric v= -V<br />
21 e r<br />
the vibrator <strong>in</strong> a vacuum
OSCILLATIONS AND WAVES<br />
379<br />
W<strong>in</strong>d<br />
Fig. 486<br />
The wavelength tha t corresponds to this frequency <strong>in</strong> a vacuum is<br />
C ,r-<br />
A=-=21 f er~ 1.4 m<br />
v<br />
In short, this result can be obta<strong>in</strong>ed as follows. The wavelength <strong>in</strong> kero-<br />
sene is A=2l and <strong>in</strong>creases V 8 r times <strong>in</strong> a vacuum. Therefore,<br />
Ao= 2l V £r<br />
660. The hcrizont<strong>al</strong> position of the aeri<strong>al</strong> means that the electric vector<br />
of the wave oscillates ma<strong>in</strong>ly <strong>in</strong> horizont<strong>al</strong> planes. Therefore, the magn<strong>et</strong>ic<br />
vector oscillates <strong>al</strong>ong a vertic<strong>al</strong>.
CHAPTER 5<br />
GEOMETRICAL<br />
OPTICS<br />
5-1. Photom<strong>et</strong>ry<br />
661. The m<strong>in</strong>imum illum<strong>in</strong>ation of a w<strong>al</strong>l (Fig. 487) is<br />
£1=1 cos ex.<br />
,2<br />
The m<strong>in</strong>imum illum<strong>in</strong>ation of the floor is<br />
Accord<strong>in</strong>g to the condition,<br />
£2=1 cos ~<br />
r 2<br />
Hence,<br />
El_~_D_2<br />
E;,- cos ~ -2h-<br />
D<br />
h=-=7.5m<br />
. 4<br />
662. The illum<strong>in</strong>ation of the middle of the table is<br />
E=.!.-!- =.!J...<br />
. H~ H:<br />
where H 2<br />
Is the height of the second lamp above the table.<br />
D<br />
......--- ----~~<br />
Z<br />
h<br />
Fig. 487<br />
Fig. 488
GEOMETRICAL OPTICS<br />
381<br />
Therefore,<br />
( H~ +!-! ~)'/I<br />
£1 /2 4 3<br />
e;= (H~+ ~2r'<br />
The illum<strong>in</strong>ation of the edge of the table will be decreased three times.<br />
663. If a norm<strong>al</strong> to the plate forms an angle a with the direction ASh<br />
the illum<strong>in</strong>a tion of the plate will be<br />
I /<br />
E =(i2 [cos a+cos (gOo-a)) = 'Q2X2 cos 45 0<br />
cos (a-45°)<br />
For this reason the illum<strong>in</strong>ation of the plate will be maximum if it is par<strong>al</strong>lel<br />
to side 8 182 of the triangle. The illum<strong>in</strong>ation is<br />
,r- /<br />
E max = y 2 - a<br />
2<br />
664. When the auxiliary and standard sources are used tog<strong>et</strong>her the illum<strong>in</strong>ations<br />
will be equ<strong>al</strong> if<br />
where /0 is the lum<strong>in</strong>ous <strong>in</strong>tensity of the standard source and /1 that of the<br />
auxiliary source.<br />
In the second case the illum<strong>in</strong>ations are equ<strong>al</strong> when<br />
/ x ':<br />
-r;= r:<br />
where / x is the sought lum<strong>in</strong>ous <strong>in</strong>tensity.<br />
r~.r~<br />
Hence I X = - 2 - 2 =400/ 0 ,<br />
'l,r4<br />
665. The illum<strong>in</strong>ation will be 25 times sm<strong>al</strong>ler.<br />
666. The tot<strong>al</strong> lum<strong>in</strong>ous flux from the lamp is
382 ANSWERS AND SOLUTIONS·<br />
Paper<br />
----Text Fig. 489<br />
669. The illum<strong>in</strong>ation on the edge of the table is<br />
J cos
GEOMETRICAL OPTICS<br />
383<br />
(a) (b) (e) (d) (e) Fig. 490<br />
To understand why the illum<strong>in</strong>ation <strong>in</strong>side the shadow changes as stated<br />
<strong>in</strong> the problem, assume that we look at the Sun <strong>al</strong>ternately from different<br />
sections of the shadow.<br />
The Sun's disk can be seen entirely outside of the shadow. In section A<br />
of the shadow (Fig. 235) the eye is <strong>in</strong> the h<strong>al</strong>f shadow cast by the lower<br />
branch, and only this branch is visible <strong>in</strong> front of the Sun's disk (Fig. 490a).<br />
S<strong>in</strong>ce the branch covers a part of the Sun's disk, the illum<strong>in</strong>ation of this<br />
po<strong>in</strong>t will be weaker. Mov<strong>in</strong>g the eye farther to position B (Fig. 235), we<br />
sh<strong>al</strong>l see that the other branch <strong>al</strong>so partly covers the Sun's disk (Fig. 490b),<br />
and for this reason the illum<strong>in</strong>ation will be still less. Mov<strong>in</strong>g farther, the eye<br />
will occupy position C (Fig. 235) <strong>in</strong> which both branches will be superimposed<br />
(Fig. 490c). Now, the part of the Sun's disk covered by the branches<br />
is sm<strong>al</strong>ler and the illum<strong>in</strong>ation greater. The disk as viewed from D and E is<br />
shown <strong>in</strong> Fig. 490d and e. This expla<strong>in</strong>s why the centr<strong>al</strong> stripe of the shadow<br />
is brighter than the .adjacent parts.<br />
674. As can be seen from. Fig. 491, we have H =L s<strong>in</strong> a, while s<strong>in</strong> a= : '<br />
s<strong>in</strong>ce DE = b is the cross-section<strong>al</strong> diam<strong>et</strong>er of the light cone on the<br />
With the angular dimensions of the Sun's disk ~, we obta<strong>in</strong> L= ~ .<br />
Therefore,<br />
1 b 2<br />
H T a = 9 m<strong>et</strong>res.<br />
ground.<br />
675. If the rays are turned <strong>in</strong> the periscope as shown <strong>in</strong> Fig. 492, the<br />
sought ratio of the widths of the prisms <strong>al</strong>b can be found from the similarity<br />
of the triangles:<br />
a L+l<br />
-r;=-l-<br />
Fig. 49/<br />
676. The height of the mirror<br />
should be equ<strong>al</strong> to h<strong>al</strong>f the height<br />
of the man. The distance from the<br />
.lower edge of the mirror to the floor<br />
should be equ<strong>al</strong> to h<strong>al</strong>f the distance<br />
from the man's eyes to his fe<strong>et</strong><br />
(Fig. 493).<br />
677. L<strong>et</strong> h be the height of the<br />
object and ex. the angle of <strong>in</strong>cidence<br />
of the rays on the mirror (Fig. 494).<br />
If the screen is at a distance of<br />
I ~ h tan (J. from the object, a direct<br />
and an <strong>in</strong>verted shadows with
384 ANSWERS AND SOLUTIONS<br />
DAB<br />
=~-=-·-+-·--3a-·-----·_·-E~<br />
-------- I I I<br />
14; l .I~ l~ Fig. 492'<br />
H<br />
I<br />
Fig. 493<br />
their bases fitted aga<strong>in</strong>st each other will be seen on the screen. The tot<strong>al</strong><br />
length of the shadow is 2h. The shadow is illum<strong>in</strong>ated by the Sun and<br />
contrasts with the other portions of the screen illum<strong>in</strong>ated by both direct<br />
and reflected rays.<br />
If the screen is nearer. the length of the shadow is sm<strong>al</strong>ler than 2h and<br />
will have portions that are illum<strong>in</strong>ated neither by direct nor reflected rays.<br />
678. A po<strong>in</strong>t source of light <strong>al</strong>ways produces a reflection that depends on<br />
the shape of the mirror. The dimensions of the Sun are f<strong>in</strong>ite. Each sm<strong>al</strong>l<br />
section of the lum<strong>in</strong>escent surface produces a bright spot that gives the shape<br />
of the mirror. These spots from various portions of the Sun are superimposed<br />
and produce a more or less diffused pattern.<br />
If the surface on which the reflection is observed is far from the mirror,<br />
the shape of the bright spot will not depend on the shape of the mirror. It<br />
is only at a sm<strong>al</strong>l distance from the mirror that the spot will reproduce the<br />
shape of the mirror, s<strong>in</strong>ce the angles at which the rays from the various<br />
portions of the Sun t<strong>al</strong>l onto the mirror differ very slightly from one another.<br />
Zh<br />
Fig. 494
GEOMETRICAL OPTICS 385<br />
-------,--~-----<br />
" ...... --~_ ,~................ I I<br />
386 ANSWERS AND SOLUTIONS<br />
684. (a) The beam reflected from the first mirror forms an angle 2cx with<br />
the <strong>in</strong>cident beam (a is the angle of <strong>in</strong>cidence). Dur<strong>in</strong>g the time t the mirror<br />
will turn through an angle oot and the new angle of <strong>in</strong>cidence will become<br />
equ<strong>al</strong> to a+ cut, as will the angle of reflection. Therefore, the angle b<strong>et</strong>ween<br />
the <strong>in</strong>cident and reflected beams will <strong>in</strong>crease by 2cut, Le., the reflected beam<br />
will turn through an angle 2mt.<br />
In view of this, the angle of <strong>in</strong>cidence on the second mirror, provided it<br />
does not rotate, would be p+2m! t where p is the angle of <strong>in</strong>cidence with<br />
immobile disks. But the mirror <strong>al</strong>so revolves through the angle rot dur<strong>in</strong>g the<br />
time t, and therefore the angle of <strong>in</strong>cidence becomes P+300t. The angle of<br />
reflection will be the same. Thus, after two reflections the beam will turn<br />
through the angle 3mt from its direction with immobile mirrors. After three<br />
reflections the beam will turn through 5cot and after n reflections through<br />
(2n-l) 2mt. In this way its angular velocity will be Q =(2n-l) 200.<br />
(b) When the mirror moves from the source with a velocity e, the image<br />
will move away from the source with a velocity 2v and from the second<br />
mirror with a velocity 3v. Therefore, the second image moves with a velocity<br />
3t t with respect to the second mirror and with a velocity 4v with respect to<br />
the source. The velocity of the third image with respect to the source will<br />
be 6v and the velocity of the n-th image 2nv.<br />
685. (a) When the first mirror turns through an angle rot the reflected beam<br />
will turn through an angle 2mt (see the solution to Problem 684). 'Hence, the<br />
angle of <strong>in</strong>cidence on the second mirror will <strong>al</strong>so <strong>in</strong>crease by 2m!. and. if the<br />
mirror did not revolve, the angle of reflection would <strong>al</strong>so <strong>in</strong>crease by 2rot.<br />
After two reflections the beam would turn through 2mt as compared with the<br />
case of immobile mirrors.<br />
S<strong>in</strong>ce the second mirror does rotate, however, the angle of the beam <strong>in</strong>cident<br />
on it decreases by rot dur<strong>in</strong>g the time t, The angle of reflection decreases<br />
by the same amount and for this reason the reflected beam will travel <strong>in</strong> the<br />
same direction as with immobile disks.<br />
S<strong>in</strong>ce this l<strong>in</strong>e of reason<strong>in</strong>g may be adopted for any two consecutive reflections,<br />
the angular velocity of rotation of the beam subjected to n reflections<br />
will be Q=O if n is even. and Q=2m if n is odd.<br />
(b) The first image moves away from the source with a velocity 2v and<br />
from the second mirror with a velocity v. Therefore. the second image moves<br />
with respect to the second mirror with a velocity - e, Le., it is immobile<br />
wi th respect to the source.<br />
/1'<br />
~ S<br />
~~~~~---~'--t---<br />
~<br />
~a<br />
~.<br />
o<br />
Fig. 497
GEOMETRICAL OPTICS<br />
381<br />
Fig. 498<br />
Fig. 499<br />
Reason<strong>in</strong>g similarly, we f<strong>in</strong>d that the sought l<strong>in</strong>ear velocity of the n-th<br />
image is zero if n is even, and 2v if n is odd.<br />
686. The beam reflected from mirror ON forms with the <strong>in</strong>cident beam an<br />
angle <br />
degrees), <strong>et</strong>c.<br />
If n is even (n =2k), then image Sk co<strong>in</strong>cides with Sk and will be on one<br />
diam<strong>et</strong>er with the source. Altog<strong>et</strong>her there will be 2k-l =n-1 images.<br />
25 *
388 ANSWERS AND SOLUTIONS<br />
Fig. 500<br />
a<br />
F'<br />
J<br />
I<br />
J<br />
I<br />
I<br />
I<br />
\ I<br />
\ I<br />
\ 0<br />
s<br />
D<br />
If n is odd (n=2i+ 1) it can<br />
easily be seen that the i-th images<br />
lie on the cont<strong>in</strong>uations of<br />
the mirrors and th us co<strong>in</strong>ci de<br />
with the (i + 1) th and <strong>al</strong>l the subsequent<br />
images. For this reason<br />
there will be 2i images, Le.,<br />
n-l as before.<br />
690. Us<strong>in</strong>g the solution of<br />
Problem 689, l<strong>et</strong> us plot consecutively<br />
the first, second, third,<br />
<strong>et</strong>c., images of source S <strong>in</strong> the<br />
mirrors (Fig. 500). All of them<br />
will lie on a circle with a radius<br />
OS and the centre at po<strong>in</strong>to.<br />
If a is an <strong>in</strong>teger, the last i-th<br />
images will either g<strong>et</strong> onto po<strong>in</strong> ts<br />
C and D at which the circle <strong>in</strong>tersects<br />
the cont<strong>in</strong>uations of the<br />
mirrors, or will co<strong>in</strong>cide with<br />
po<strong>in</strong>t<br />
P diam<strong>et</strong>r<strong>al</strong>ly opposite to<br />
the source. In both cases the<br />
. number of images will be a-I.<br />
If a is not an <strong>in</strong>teger, for example a=2i ± ~, where ~ < 1, and i is an<br />
<strong>in</strong>teger, the last i-th images will lie on arc CPD that is beh<strong>in</strong>d both the<br />
first and the second mirrors and there wiII be no more reflections. Thus, the<br />
tot<strong>al</strong> number of images will be 2i.<br />
691. L<strong>et</strong> us plot the image of po<strong>in</strong>t B <strong>in</strong> mirror bd (Fig. 501). L<strong>et</strong> us<br />
then construct image B 1 <strong>in</strong> mirror cd. Also, B 3 is the image of B 2 In mirror<br />
ac and 8 4 is the image of 8 3 <strong>in</strong> mirror abo<br />
L<strong>et</strong> us connect po<strong>in</strong>ts A and B 4 • Po<strong>in</strong>t C is the po<strong>in</strong>t of <strong>in</strong>tersection of ab<br />
with l<strong>in</strong>e AB 4 • L<strong>et</strong> us now draw l<strong>in</strong>e B 3C from B 3 , and connect po<strong>in</strong>t D at<br />
which this l<strong>in</strong>e <strong>in</strong>tersects ac with 8 2 • E with B 1 , and F with B.<br />
It can be stated that broken l<strong>in</strong>e ACDEFB is the sought path of the<br />
beam. Indeed, s<strong>in</strong>ce B 3CB4 is an isosceles triangle, CD is the reflection of<br />
beam AC.
GEOMETRICAL OPTICS<br />
389<br />
Fig. 502<br />
i<br />
Similarly, it is easy to show<br />
that DE is the reflection of CD,<br />
<strong>et</strong>c<br />
Ṫhis solution of the problem<br />
is not unique, s<strong>in</strong>ce the beam<br />
shouId not necessari1y be sent<br />
<strong>in</strong>itiaIJy to mirror abo<br />
692. The coefficient of reflection<br />
of light from the surface of<br />
water dim<strong>in</strong>ishes with a reduction<br />
<strong>in</strong> the angle of <strong>in</strong>cidence.<br />
If the observer looks down,<br />
rays reflected at sm<strong>al</strong>l angles<br />
reach his eyes. The rays reflected<br />
from the sea water at the<br />
, .D horizon reach the eyes at greater<br />
, angles.<br />
" 693. Accord<strong>in</strong>g to the law of<br />
refraction s~n i =n (Fig. 502).<br />
s<strong>in</strong> r<br />
Upon exit from the plate s~n ~ =- . Upon multiply<strong>in</strong>g these expressions, we<br />
s<strong>in</strong> r, n<br />
g<strong>et</strong> s<strong>in</strong> i=s<strong>in</strong>i 1 , i.e., beam CD leav<strong>in</strong>g the plate is par<strong>al</strong>lel to <strong>in</strong>cident beam<br />
AB. A glance at the draw<strong>in</strong>g shows that a,=i-f. The sought displacement<br />
of the beam is x=EC=BC s<strong>in</strong> (i-f).<br />
S<strong>in</strong>ce BC=_d_. then<br />
cosr<br />
x= d s<strong>in</strong> (i - f) d s<strong>in</strong> i (1 _ cos i )<br />
cos, y n2-s<strong>in</strong>2 i<br />
And a maximum displacement of d can be obta<strong>in</strong>ed when i --.. 90°.<br />
694. The angle of <strong>in</strong>cidence of the ray onto AC and BC is 45°. For tot<strong>al</strong><br />
<strong>in</strong>tern<strong>al</strong> reflection it is necessary that s<strong>in</strong> i > J. .<br />
n<br />
Hence, n > y2'~ 1.4.<br />
695. The angle of <strong>in</strong>cidence of the rayon face Be is equ<strong>al</strong> to the sought<br />
angle cx. For the ray to be compl<strong>et</strong>ely reflected from face Be, the angle ex<br />
should exceed the limit<strong>in</strong>g one. .<br />
Therefore, s<strong>in</strong> a > ::' where n z is the refraction <strong>in</strong>dex of water.<br />
Hence, a > 62 030'.<br />
696. This phenomenon is noth<strong>in</strong>g but a mirage frequently observed <strong>in</strong> deserts.<br />
The hot layer of air <strong>in</strong> direct contact with the asph<strong>al</strong>t has a sm<strong>al</strong>ler refraction<br />
<strong>in</strong>dex than the layers above. Tot<strong>al</strong> <strong>in</strong>tern<strong>al</strong> reflection occurs and the<br />
asph<strong>al</strong>t seems to reflect the light just as well as the surface of water.<br />
697. L<strong>et</strong> us divide the plate <strong>in</strong>to many plates so th<strong>in</strong> that their refraction<br />
<strong>in</strong>dex can be assumed constant with<strong>in</strong> the limits of each plate (Fig. 503).<br />
Assume that the beam enters the plate from a medium with a refraction<br />
<strong>in</strong>dex of no and leaves it for a medium with a refraction <strong>in</strong>dex of na·
390<br />
ANSWERS AND SOLUTIONS<br />
Then. accord<strong>in</strong>g to the law of refraction,<br />
s<strong>in</strong> a nt<br />
s<strong>in</strong> ~ = no<br />
s<strong>in</strong> p n'<br />
--=s<strong>in</strong><br />
y nl<br />
s<strong>in</strong> 'V n"<br />
s<strong>in</strong> ~=n'<br />
Fig. 503<br />
s<strong>in</strong> q> n 2<br />
s<strong>in</strong>, = n(n)<br />
s<strong>in</strong> ~ na<br />
s<strong>in</strong>x=n2<br />
Upon multiply<strong>in</strong>g these equations we" g<strong>et</strong><br />
s<strong>in</strong> a ns<br />
SiiiX= no<br />
Hence, the angle at which the beam leaves the plate<br />
x=arcs<strong>in</strong> (:: ~<strong>in</strong> a)<br />
depends only on the angle of <strong>in</strong>cidence of the beam on the plate and on the<br />
refraction <strong>in</strong>dices of the media on both sides of the plate. In particular, if<br />
ng=no. then x=a..
GEOMETRICAL OPTICS<br />
391<br />
Fig. 504<br />
n"<br />
Fig. 505<br />
N<br />
F<br />
Gener<strong>al</strong>ly speak<strong>in</strong>g, the angle O. at which the beam is <strong>in</strong>cl<strong>in</strong>ed to the vertic<strong>al</strong><br />
is related to the refraction <strong>in</strong>dex n at any po<strong>in</strong>t on the plate by the<br />
ratio n s<strong>in</strong> 8=const=no s<strong>in</strong> a.. If the refraction <strong>in</strong>dex reaches a v<strong>al</strong>ue of<br />
n = nos<strong>in</strong> ex anywhere <strong>in</strong>side the plate, full <strong>in</strong>tern<strong>al</strong> reflection will take place.<br />
In this case the beam will leave the plate for the medium at the same angle<br />
ex at which it entered the plate (Fig. 504).<br />
698. The m<strong>in</strong>imum amount of water d<strong>et</strong>erm<strong>in</strong>ed by the level x (Fig. 505)<br />
can be found from the triangle MNF. We have NF=x-b=xtanr. From<br />
the law of refraction<br />
. s<strong>in</strong> i<br />
51nr=-<br />
n<br />
Therefore,<br />
b<br />
x= -t--t-a-n-,<br />
s<strong>in</strong>ce i=45° and 11= ~ .<br />
The amount of water required is V=xa 2 =::: 43.2 litres,<br />
699. The man's eyes are reached by rays com<strong>in</strong>g <strong>in</strong> a narrow beam from<br />
an arbitrary po<strong>in</strong>t C on the bottom. They seem to the eye as issu <strong>in</strong>g from<br />
po<strong>in</strong>t C' (Fig. 506). S<strong>in</strong>ce di and dr are very sm<strong>al</strong>l, we can write:<br />
AD=AC'~'=~ar<br />
cos r<br />
AD'=AC'.~i=~~i<br />
cos ~<br />
By equat<strong>in</strong>g the v<strong>al</strong>ues of AB from triangles ABD and ABD', we have<br />
-!!- ~r= .s.. f1.t<br />
cos 2r cos 2 l
392 ANSWERS AND SOLUTIONS<br />
a<br />
Fig. 506<br />
Us<strong>in</strong>g the law of refraction, we can f<strong>in</strong>d the ratio ~~.<br />
Indeed,<br />
s~n i =n and s<strong>in</strong> (i +ai) =n<br />
s<strong>in</strong>, s<strong>in</strong> (, +ar)<br />
Remember<strong>in</strong>g that ~i and ~r are sm<strong>al</strong>l, we have<br />
s<strong>in</strong> ~i ~ 8i, s<strong>in</strong>~, === ~" and cos L\i === cos ~, =:: t<br />
Therefore, the last equation can be rewritten as<br />
s<strong>in</strong> i+ cos i· Ai=n s<strong>in</strong> ,+n cos r-br<br />
Ai cos r<br />
Hence, T= n--., Uoon <strong>in</strong>sert<strong>in</strong>g this expression <strong>in</strong>to the formula relat<strong>in</strong>g<br />
cos l<br />
a'
GEOMETRICAL OPTICS 393<br />
Fig. 508<br />
Hand h, we f<strong>in</strong>d<br />
h=H coss i =!!.... .<br />
coslj i<br />
n cos 3 , n ( s<strong>in</strong> 2 i ) 3/2<br />
1--- n 2<br />
When i=O, we have h=!!..., i.e., the<br />
n<br />
depth seems reduced by n times. As<br />
i <strong>in</strong>creases, h dim<strong>in</strong>ishes. The approxi..<br />
mate dependence of the seem<strong>in</strong>g depth<br />
on the angle i is shown <strong>in</strong> Fig. 507.<br />
The man's eye is above po<strong>in</strong>t A of the<br />
lake bottom.<br />
700. q>= 120°.<br />
701. The path of the ray <strong>in</strong> the<br />
prism is shown <strong>in</strong> Fig. 508. There is<br />
an obvious relationship b<strong>et</strong>ween the<br />
angles cx and ~, namely. 2ex.+~= 180°,<br />
and a.=2~.<br />
Hence, cx=72°, ~=36°.<br />
702. The path of the ray <strong>in</strong> the prism<br />
is shown <strong>in</strong> Fig. 509. To avoid<br />
full <strong>in</strong>tern<strong>al</strong> reflection on face BN, it is necessary that s<strong>in</strong> ~ ~ ..!... As can<br />
n<br />
be seen from the draw<strong>in</strong>g, ~ = ex.-,. Hence, the greater the v<strong>al</strong>ue of " the<br />
higher is the permissible v<strong>al</strong>ue of cx. The maximum v<strong>al</strong>ue of r is d<strong>et</strong>erm<strong>in</strong>ed<br />
from the condition: s<strong>in</strong> r=...!- (angle of <strong>in</strong>cidence gOO).<br />
n<br />
Therefore, CI max = 2 arc s<strong>in</strong> ~ ~ 83 040'. B<br />
B<br />
Fig. 509 Fig. 5/0
394 ANSWERS AND SOLUTIONS<br />
B<br />
------------t~----aD<br />
Fig. 51/<br />
703. When consider<strong>in</strong>g triangles ABC, AMC and ADC (Fig. 510), it is<br />
easy to see that r+rl=q> and y=ex+P-cp. Accord<strong>in</strong>g to the law of ·refraction,<br />
s<strong>in</strong> ex s<strong>in</strong> rt I<br />
s<strong>in</strong>r=n, and s<strong>in</strong>~=n<br />
Upon solv<strong>in</strong>g this system of equations, we f<strong>in</strong>d that<br />
and<br />
cp=ex+~-y<br />
.. /{ s<strong>in</strong> ex 1}2<br />
n=s<strong>in</strong>~ V s<strong>in</strong>~s<strong>in</strong>(ex+~-y) +tan(a,+~-'V) +1<br />
704. Accord<strong>in</strong>g to the <strong>in</strong>iti<strong>al</strong> condition, the <strong>in</strong>cident beam and the beam<br />
that has passed through the 'prism are mutu<strong>al</strong>ly perpendicular. Therefore,<br />
L q>= L. a, and <strong>al</strong>so L. y= L. Ii (Fig. 511). The sum of the angles of the<br />
quadrangle AKMN is 360°. Therefore, L. K M N = 90° and beam<br />
KM is <strong>in</strong>cident onto face Be at an angle of 45°. If we know the angles of<br />
triangle KBM it is easy to f<strong>in</strong>d that ~=30°. In conformity with the law of<br />
. s<strong>in</strong> ex •<br />
r<strong>et</strong>raction ~=n. Hence,<br />
sm ...<br />
s<strong>in</strong> a=O.5n and a= arc s<strong>in</strong> O.5n<br />
S<strong>in</strong>ce the full <strong>in</strong>tern<strong>al</strong> reflection at an angle of 45 0 is observed only<br />
when n~V~ the angle ex is with<strong>in</strong> 45°~a.EZ;;;90°.<br />
705. Paper parti<strong>al</strong>ly l<strong>et</strong>s through light. But ow<strong>in</strong>g to its fibrous structure<br />
and the great number of pores, the dissipation of light is very high <strong>in</strong> <strong>al</strong>l<br />
directions. For this reason it is impossible to read the text.<br />
When they fill the pores, glue or water reduce the dissipation of light,<br />
s<strong>in</strong>ce their refraction <strong>in</strong>dex is close to that of the paper. The light beg<strong>in</strong>s<br />
to pass through the paper without any appreciable deviation, and the text<br />
can easily be read.
GEOMETRICAL OPTICS<br />
395<br />
6-3. Lenses and Spheric<strong>al</strong> Mirrors<br />
706. n= 1.5.<br />
707. f=2R.<br />
708. The convex surface has a radius of curvature of R 1<br />
=6 cm and the<br />
concave one R 2 = 12 em.<br />
709. In the first case the foc<strong>al</strong> length is d<strong>et</strong>erm<strong>in</strong>ed from the formula<br />
f~ =(:1- 1 ) (~1+~J<br />
S<strong>in</strong>ce <strong>in</strong> a vacuum the foc<strong>al</strong> length of the lens is f, then<br />
1 1 1 (n-l)f<br />
(n-l)f· Hence, ft= n_ =90cm<br />
1<br />
R<br />
1+R2<br />
In the second case the sought foc<strong>al</strong> length is<br />
I, (n-l)f -102cm<br />
~-l<br />
n 2<br />
The lens will be divergent.<br />
710. As shown <strong>in</strong> the solution of Problem 709<br />
-f n 2(n1 - l )<br />
D(n1-n.J<br />
Therefore.<br />
n 2<br />
fDn l ==1.67<br />
fD+ I-n,<br />
711. The image will be m+ 1 times sm<strong>al</strong>ler than the object.<br />
712. The lamp should be moved two m<strong>et</strong>res away from the lens.<br />
713. Obviously, one of the image will be virtu<strong>al</strong>. Therefore, denot<strong>in</strong>g the<br />
distances from the sources of light to the lens by <strong>al</strong> and a2 and from the<br />
lens to the images by b I and b 2 , we obta<strong>in</strong>:<br />
1 1 1 1 1 1<br />
---=- and -+-=<strong>al</strong><br />
b l t : a 2 b 2 t<br />
Accord<strong>in</strong>g to the <strong>in</strong>iti<strong>al</strong> condition, at +a 2 = l and b 1=b2 • Upon solv<strong>in</strong>g<br />
this system of equations, we g<strong>et</strong><br />
1(1± V~)<br />
2<br />
The lens should be placed at a distance of 6 em from one source and 18em<br />
from the other.<br />
714. Apply<strong>in</strong>g the formula of a lens to both cases, we obta<strong>in</strong><br />
1 1 1 1 1 1<br />
-+-=- and -+-=<strong>al</strong><br />
b l t ' <strong>al</strong>a b 2 f<br />
n t
396<br />
ANSWERS AND SOLUTIONS<br />
~-"":;:'-----I--~---<br />
Fig. 512<br />
Accord<strong>in</strong>g to the <strong>in</strong>iti<strong>al</strong> condition,<br />
(magnification <strong>in</strong> the first case);<br />
(magni fication<br />
Hence, t=kk 1k 2k<br />
<strong>in</strong> the second case).<br />
1- 2<br />
[=9 em<br />
715. (1) For this case the path of the rays is shown <strong>in</strong> Fig. 512a. From<br />
the standpo<strong>in</strong>t of reversibility of the light beams, po<strong>in</strong>t B can be regarded<br />
as a source of light and po<strong>in</strong>t A as its image.<br />
Then, accord<strong>in</strong>g to the formula of a lens,<br />
I 1 I<br />
at -7i=-T<br />
<strong>al</strong>b<br />
Hence, f=--b= 20 em.<br />
<strong>al</strong>-<br />
(2) The path of the rays is illustrated <strong>in</strong> Fig. 512b. In this case, both the<br />
image (po<strong>in</strong>t A) and the source (po<strong>in</strong>t B) are virtu<strong>al</strong>. Accord<strong>in</strong>g to the formula<br />
of a lens<br />
lienee, f =a<br />
a2b 2<br />
+ b= 1 em.<br />
2<br />
716. On the basis of the formula of a lens,<br />
1 1 I<br />
a-+d-a =T<br />
where a is the distance b<strong>et</strong>ween the lens and the lamp. Therefore,<br />
a 2-ad+df=O
GEOMETRICAL OPTICS<br />
397<br />
Fig. 513<br />
Upon solv<strong>in</strong>g this equation, we obta<strong>in</strong><br />
yd- 2 <br />
d<br />
a=-<br />
2<br />
±<br />
4<br />
--df<br />
Two positions of the lens are therefore possible: at a distance of <strong>al</strong> = 70 em<br />
and at a distance of a 2 = 30 ern from the lamp.<br />
When I' =26 em, there will be no sharp image on the screen, whatever<br />
the position of the lens, s<strong>in</strong>ce to obta<strong>in</strong> the image it is necessary that d ~ 4'.<br />
717. In the first case hHl = b l , where at and b 1 are the distances from the<br />
a l<br />
object and the image to the lens. In the second case H h 2 = b 2 •<br />
a 2<br />
It follows from the solution of Problem 716 that <strong>al</strong>=b 2<br />
Therefore,<br />
and b 1<br />
=a 2<br />
•<br />
H=Vh lh2<br />
718. On the basis of the formula of a mirror<br />
1 1 1<br />
-a--;;=T<br />
The l<strong>in</strong>ear magnification of the mirror. is<br />
H b<br />
71=(i<br />
Accord<strong>in</strong>g to the <strong>in</strong>iti<strong>al</strong> condition, the angular dimensions of the image on<br />
the concave mirror are 1.5 times greater than those on the fiat mirror: p= 1.5C&<br />
(Fig. 513).<br />
It is obvious that<br />
h<br />
H<br />
tancx=2a and tan p=a+b
398 ANSWERS AND SOLUTIONS<br />
Fig. 514<br />
A'l<br />
F<br />
M<br />
F<br />
Fig. 515<br />
B<br />
M ~<br />
F<br />
F<br />
Fig. 516<br />
S'<br />
Fig. 517
GEOMETRICAL OPTICS<br />
399<br />
When h ~ 2a, ex. and ~ are sm<strong>al</strong>l. For sm<strong>al</strong>l angles<br />
H 01:: 15 h<br />
a+b - · 2a<br />
Upon exclud<strong>in</strong>g the unknown quantities ~<br />
3<br />
that 1=2: a.<br />
and b from the equations, we f<strong>in</strong>d<br />
Hence, R=2f=3a=6 m<strong>et</strong>res.<br />
719. The path of the ray is shown <strong>in</strong> Fig. 514. L<strong>et</strong> us cont<strong>in</strong>ue AB up to<br />
its <strong>in</strong>tersection with the foc<strong>al</strong> plane of lens NN. The beam of par<strong>al</strong>lel<br />
rays after refraction <strong>in</strong> the lens so travels that the cont<strong>in</strong>uations of the rays<br />
should <strong>in</strong>tersect at P'. Ray F'O is not refracted. Thus, ray C A pass<strong>in</strong>g to<br />
po<strong>in</strong>t A is par<strong>al</strong>lel to F'O up to the lens.<br />
720. If A is the source and B is the image, then the lens will be convergent.<br />
The position of the optic<strong>al</strong> centre of the lens 0 and its foci F can be<br />
found by construction as shown <strong>in</strong> Fig. 515.<br />
If B is the source and A is the image, the lens is divergent. The respective<br />
construction is illustrated <strong>in</strong> Fig. 516.<br />
721. The centre of the lens 0 is the po<strong>in</strong>t of <strong>in</strong>tersection of straight l<strong>in</strong>es<br />
SS' and NIN2. The foci can easily be found by construct<strong>in</strong>g the rays par<strong>al</strong>lel<br />
to the major optic<strong>al</strong> axis (Fig. 517).<br />
722. Po<strong>in</strong>t 0, which is the optic<strong>al</strong> centre of the lens, can be found by<br />
dropp<strong>in</strong>g perpendicular 80 onto straight l<strong>in</strong>e N iN2 (Fig. 518). L<strong>et</strong> us draw<br />
an auxiliary optic<strong>al</strong> axis DO par<strong>al</strong>lel to ray AB and extend straight l<strong>in</strong>e BC<br />
until it <strong>in</strong>tersects DO at po<strong>in</strong>t E ly<strong>in</strong>g <strong>in</strong> the foc<strong>al</strong> plane. L<strong>et</strong> us drop a perpendicular<br />
from E onto N IN 2 to f<strong>in</strong>d po<strong>in</strong>t F, one of the ma<strong>in</strong> foci of the<br />
lens. By us<strong>in</strong>g the property of reversibility of the ray. we can f<strong>in</strong>d the other<br />
ma<strong>in</strong> focus Ft.<br />
723. The image S' may be re<strong>al</strong> or virtu<strong>al</strong>. In both cases l<strong>et</strong> us draw an<br />
arbitrary ray ADS' and auxiliary optic<strong>al</strong> axis BOC par<strong>al</strong>lel to it to f<strong>in</strong>d the<br />
position of the source lFig. 519). By connect<strong>in</strong>g the po<strong>in</strong>ts of <strong>in</strong>tersection B<br />
and C (of the auxiliary axis with the foc<strong>al</strong> planes) to po<strong>in</strong>t D by straight<br />
l<strong>in</strong>es, we can f<strong>in</strong>d the position of the source SI (if the image S' is re<strong>al</strong>) and<br />
S2 (if the image is virtu<strong>al</strong>).<br />
724. S<strong>in</strong>ce the ray <strong>in</strong>cident on the mirror at its pole is reflected symm<strong>et</strong>ric<strong>al</strong>ly<br />
with respect to the major optic<strong>al</strong> axis, l<strong>et</strong> us plot po<strong>in</strong>t SI symm<strong>et</strong>ric<strong>al</strong><br />
a<br />
to S' and draw ray SSt until it <strong>in</strong>-<br />
tersects the axis at po<strong>in</strong>t P (Fig. 520).<br />
This po<strong>in</strong>t wilJ be the pole of the mir-<br />
A<br />
ror.<br />
The optic<strong>al</strong> centre C of the mirror<br />
can obviously be found as the po<strong>in</strong>t of<br />
JJ <strong>in</strong>tersection of ray SS' with ax is N N' .<br />
The focus can be found by the usu<strong>al</strong><br />
____..o-.--;;:a~_o--_--_-construction of ray SM par<strong>al</strong>lel to<br />
Ai ~ the axis. The reflected ray must pass<br />
through focus F (ly<strong>in</strong>g on the optic<strong>al</strong><br />
ax is of the mirror) and through 8'.<br />
725 .. (a) L<strong>et</strong> us construct, as <strong>in</strong> the<br />
solution of Problem 724, the ray BAC<br />
Fig. 518 and f<strong>in</strong>d po<strong>in</strong>t C (optic<strong>al</strong> centre of the
8'<br />
Fig. 519<br />
s* IIIIIIII<br />
N<br />
P<br />
H'<br />
Fig. 520<br />
B<br />
Fig. 521
GEOMET~ICAL OPTICS 401<br />
mirror) (Fig. 521a). Pole P can be found by construct<strong>in</strong>g the path of the ray<br />
AP A' reflected <strong>in</strong> the pole with the aid of symm<strong>et</strong>ric<strong>al</strong> po<strong>in</strong>t A'. The position<br />
of the mirror focus F is d<strong>et</strong>erm<strong>in</strong>ed by means of the usu<strong>al</strong> construction<br />
of ray AMF par<strong>al</strong>lel to the axis.<br />
(b) This construction can <strong>al</strong>so be used to f<strong>in</strong>d centre C of the mirror and<br />
pole P (Fig. 52Ib). The reflected ray 8M will pass par<strong>al</strong>lel to the optic<strong>al</strong><br />
axis of the mirror. For this reason, to f<strong>in</strong>d the focus, l<strong>et</strong> us first d<strong>et</strong>erm<strong>in</strong>e<br />
po<strong>in</strong>t M at which straight l<strong>in</strong>e AM, par<strong>al</strong>lel to the optic<strong>al</strong> axis, <strong>in</strong>tersects<br />
the mirror, and then extend 8M to the po<strong>in</strong>t of <strong>in</strong>tersection with the axis at<br />
the focus F.<br />
726. (a) The rays reflected from the flat mirror <strong>in</strong>crease the illum<strong>in</strong>ation at<br />
the centre of the screen. The presence of the mirror is equiv<strong>al</strong>ent to the appearance<br />
of a new source (with the same lum<strong>in</strong>ous <strong>in</strong>tensity) arranged at a<br />
distance from the screen three times greater than that of the first source. For<br />
this reason the illum<strong>in</strong>ation should <strong>in</strong>crease by one-n<strong>in</strong>th of the previous illum<strong>in</strong>ation,<br />
i.e.,<br />
E a=2.51x<br />
(b) The concave mirror is so arranged that the source is <strong>in</strong> its focus. The<br />
rays reflected from the mirror travel <strong>in</strong> a par<strong>al</strong>lel beam. The illum<strong>in</strong>ation<br />
<strong>al</strong>ong the axis of the beam of par<strong>al</strong>lel rays is everywhere the same and equ<strong>al</strong><br />
to the illum<strong>in</strong>ation created by the po<strong>in</strong>t source at the po<strong>in</strong>t of the mirror<br />
closest to it. The tot<strong>al</strong> illum<strong>in</strong>ation at the centre of the screen is equ<strong>al</strong> to the<br />
sum of the illum<strong>in</strong>ations produced by the source at the centre of the screen<br />
and reflected by the rays:<br />
E b=2X2.25lx=4.5lx<br />
(c) The virtu<strong>al</strong> image of the po<strong>in</strong>t source <strong>in</strong> the convex mirror is at a<br />
distance of 2.5r from the screen (r is the distance from the screen to the<br />
source). The lum<strong>in</strong>ous flux
402<br />
ANSWERS AND SOLUTIONS<br />
..... ..........<br />
....................0<br />
d<br />
s<br />
o<br />
.........<br />
................<br />
Fig. 523<br />
727. Each section of the lens produces a full image irrespective of the other<br />
sections. Therefore, there will be no strips on the photograph, and the image<br />
will simply be less bright.<br />
728. Any section of the lens gives an image identic<strong>al</strong> <strong>in</strong> shape to that<br />
produced by the entire lens. The layered lens can be therefore regarded as two<br />
lenses with different foc<strong>al</strong> lengths, but with a common optic<strong>al</strong> centre. Accord<strong>in</strong>gly,<br />
this lens will produce two images: at po<strong>in</strong>t 8 1 and at po<strong>in</strong>t 8 2 (Fig. 523).<br />
The image of the source will be surrounded by a bright h<strong>al</strong>o with a diam<strong>et</strong>er<br />
of ab or, respectively. cd on a screen arranged perpendicular to the optic<strong>al</strong><br />
axis at po<strong>in</strong>t 8 1 or 8 2<br />
•<br />
729. To prove that the visible dimensions of the Sun's disk near the horizon<br />
and high above it are the same, l<strong>et</strong> us project the disk <strong>in</strong> both cases onto<br />
a she<strong>et</strong> of paper with the aid of a long-focus lens. Both the lens and the<br />
she<strong>et</strong> should be perpendicular to the sunrays. A long-focus lens is required<br />
because the dimensions of the image are proportion<strong>al</strong> to the foc<strong>al</strong> length.<br />
If we measure the dimensions of the images, we see that they are equ<strong>al</strong>.<br />
5-4. Optic<strong>al</strong> Systems and Devices<br />
730. The mirror should be placed b<strong>et</strong>ween the focus and a po<strong>in</strong>t ly<strong>in</strong>g on<br />
the double foc<strong>al</strong> length. The path of the rays is shown <strong>in</strong> Fig. 524.<br />
731. The divergent lens should be placed at a distance of 25 em from the<br />
convergent one and the foci of both lenses will co<strong>in</strong>cide. The path of the rays<br />
is shown <strong>in</strong> Fig. 525.<br />
Fig. 524
GEOMETRICAL OPTICS<br />
403<br />
~--Z5cm-"<br />
------:~f2<br />
_-------<br />
I<br />
I<br />
-------~<br />
I<br />
Fig. 525<br />
732. Image A' B' of the object <strong>in</strong> the spheric<strong>al</strong> mirror will be at a distance<br />
b l (Fig. 526) from the mirror d<strong>et</strong>erm<strong>in</strong>ed by the formula of the mirror<br />
1 2<br />
at -b I<br />
=R<br />
Hence, b i =8 em. Distance AA' is 48 ern. Therefore, the plate should be placed<br />
at a distance of 24 cm from object AB.<br />
733. Two cases are possible:<br />
(a) The mirror is at a distance of d = f+R from the lens. The path of the<br />
beam par<strong>al</strong>lel to the optic<strong>al</strong> axis of the system and the image of object AB<br />
are illustrated <strong>in</strong> Fig. 527. Image A'B' (direct and re<strong>al</strong>) is obta<strong>in</strong>ed to full<br />
seaIe with the object <strong>in</strong> any position.<br />
(b) The mirror is at a distance of d==f=R from the lens (Fig. 528).<br />
The image of object A' B', <strong>al</strong>so full-sc<strong>al</strong>e, will be <strong>in</strong>versed and virtu<strong>al</strong><br />
with the .object <strong>in</strong> any position.<br />
734. The path of the rays <strong>in</strong> this optic<strong>al</strong> system is shown <strong>in</strong> Fig. 529.<br />
When the second lens is absent, the first one produces image A' H' that is at<br />
a distance of b i = 60 em from the lens. This distance can be found from the<br />
formula of the lens<br />
~+.!.=.!.<br />
at b I f1<br />
Image A' B' is virtu<strong>al</strong> with respect to<br />
the second lens. Therefore,<br />
_-.!..+J.=..!.<br />
a 2 b2 f2<br />
Fig. 526<br />
where a 2 = b t - d= 30 em.<br />
Hence, b 2=7.5 em.<br />
735. It follows from the solution of the<br />
previous problem that <strong>in</strong> the case of two<br />
convergent lenses at a distance d from each<br />
other the follow<strong>in</strong>g equation is true<br />
1.- +J.. =~+J.. +d (<strong>al</strong>- '1) (12 -b2)<br />
at b 2 II 12 <strong>al</strong>b2f l f 2<br />
26 *
404 ANSWERS AND SOLUTIONS<br />
~---d---~<br />
Fig. 527<br />
In our case the divergent lens is tightly fitted aga<strong>in</strong>st the convergent one<br />
(d=O) and therefore<br />
I<br />
I<br />
I<br />
II<br />
~d<br />
1 1 1 1 1<br />
-+-=--+-=a1<br />
b2 f 1 f 2 f<br />
where f is the sought foc<strong>al</strong> length of the system.<br />
fIf<br />
H t<br />
2<br />
ence, =f--f-'<br />
1- 2<br />
736. The light beam issu<strong>in</strong>g from a po<strong>in</strong>t at a distance of a 2<br />
= 5 em from<br />
the second lens imp<strong>in</strong>ges on it. As follows from the formula of a lens, the<br />
cont<strong>in</strong>uations of the rays refracted by this lens <strong>in</strong>tersect at a distance of<br />
b 2 =4 em from it (Fig. 530). This po<strong>in</strong>t co<strong>in</strong>cides with the focus of the third<br />
lens. For this reason the rays leav<strong>in</strong>g the system will travel <strong>in</strong> a par<strong>al</strong>lel<br />
beam. The given system is telescopic.<br />
737. Figure 531 illustrates the path of the ray through the plate from<br />
po<strong>in</strong>t S of the object. As a result<br />
/<br />
/<br />
of refraction of light by the plate<br />
ray BE seems to be issu<strong>in</strong>g from<br />
I<br />
/<br />
po<strong>in</strong>t S' that is the virtu<strong>al</strong> image<br />
/A<br />
of S <strong>in</strong> the plate. Thus, the distance<br />
/<br />
b<strong>et</strong>ween the image of the object <strong>in</strong><br />
/<br />
/<br />
the plate and the lens is a' =a-SS'.<br />
/ 8<br />
The displacement SS'=AD=<br />
=d-DC.<br />
/<br />
Assum<strong>in</strong>g the angles of <strong>in</strong>cidence<br />
on the plate to be sm<strong>al</strong>l, we have<br />
DC=B.C =dr =~<br />
t r n<br />
Fig. 528<br />
L<br />
s<strong>in</strong>ce - == n.<br />
r
GEOMETRICAL OPTICS<br />
405<br />
B<br />
~1IIr---tt,----.c<br />
~<br />
/I<br />
Fig. 529<br />
Therefore, SS'=d (1- ~ )=4 em. Before the plate was <strong>in</strong>serted, the<br />
screen had been at a distance of b=--!!l-,= 120 em, and after it was tna-<br />
a']<br />
serted at a distance of b' 'f 180 em. The screen should therefore be<br />
a-<br />
shifted by 60 cm.<br />
738. In the absence of the mirror the image A'H' of object AB (Fig. 532)<br />
will be obta<strong>in</strong>ed at a distance of b=-!!:l-, = 180 em from the lens. After<br />
a-<br />
be<strong>in</strong>g reflected from the mirrort the image will occupy position A"B"and<br />
will be at a distance of H'=b-l=80 em from the optic<strong>al</strong> axis.<br />
The layer of water with a thickness of d will shift the image over a distance<br />
of H - H'=d ( 1- ~ ) • where n is the refraction <strong>in</strong>dex of water. This follows<br />
directly from the solution of Problem 737.<br />
Hence. H=H'+d (1- ~ )==85 em.<br />
739. Two cases are possible:<br />
(1) The optic<strong>al</strong> axis of the lens is perpendicular to the front face of the<br />
wedge. The rays reflected from the front face pass through the lens and produce<br />
an image of the po<strong>in</strong>t source that co<strong>in</strong>cides with the source itself. The<br />
Z 9<br />
Fig. 580
406<br />
ANSWERS AND SOLUTIONS<br />
E<br />
a<br />
S<br />
s'<br />
Fig. 531<br />
Fig. 532<br />
rays reflected from the rear face will be deflected by an angle q> (Fig. 533)<br />
d<strong>et</strong>erm<strong>in</strong>ed by the equ<strong>al</strong>ity ~<strong>in</strong>2(j) =n. S<strong>in</strong>ce the angles are sm<strong>al</strong>l cp ~ 2an.<br />
s<strong>in</strong> ex,<br />
The second image of the source will<br />
Fig. 533<br />
be obta<strong>in</strong>ed at a distance d from the<br />
first image, namely d=f(fJ=f2an.<br />
d<br />
Hence, n= 2a.f .<br />
(2) The optic<strong>al</strong> axis of the lens is perpendicular to the rear face of the<br />
wedge. The rays reflected from the front face will be deflected by an angle<br />
2a (Fig. 534) and produce an image at a distance of d 1 =2af from the<br />
source.<br />
The ray s reflected from the rear face will be deflected by an angle e d<strong>et</strong>erm<strong>in</strong>ed<br />
from the equations:<br />
s<strong>in</strong> a d s<strong>in</strong> (a+8)<br />
---n an -n<br />
s<strong>in</strong> ~ - , s<strong>in</strong> (2a-~) -<br />
When the angles are sm<strong>al</strong>l, a= 2a (n - 1). For this reason the second image<br />
will be at a distance d7,=2a.(n-l) f from the source. The tot<strong>al</strong> distance<br />
b<strong>et</strong>ween the images is d=d 1 + d7,= 2anf·<br />
d<br />
Hence, n=2af as <strong>in</strong> the first<br />
case.<br />
740. S<strong>in</strong>ce the image that co<strong>in</strong>cides<br />
with the source is formed<br />
ow<strong>in</strong>g to reflection from the part<br />
of the mirror not covered by the<br />
liquid, the source is obviously arranged<br />
at centre 0 of the hemisphere.<br />
L<strong>et</strong>. us f<strong>in</strong>d the position of the other<br />
image (po<strong>in</strong>t A <strong>in</strong> Fig. 535).<br />
Accord<strong>in</strong>g to the law of refraction<br />
s<strong>in</strong> ex .- a s<strong>in</strong> q>
GEOMETRICAL OPTICS<br />
Fig. 534<br />
As can be seen from the draw<strong>in</strong>g, e=~+2y, where v=a-~ is the angle<br />
of <strong>in</strong>cidence of the refracted rayon the mirror and (R-l-h) tan
408 ANSWERS AND SOLUTIONS<br />
Fig. 536<br />
CD. Ray OBIt passes through the optic<strong>al</strong> centre of the lens after reflection<br />
and is thus not refracted. Ray CD can be plotted as follows. After the first<br />
refraction <strong>in</strong> the lens and reflection, ray BC wiII travel <strong>in</strong> the direction of F,<br />
and will be refracted <strong>in</strong> the lens once more. Its direction after the second<br />
refraction can be found by the m<strong>et</strong>hod described <strong>in</strong> Problem 723: ray OD par<strong>al</strong>lel<br />
to CF 2 is drawn through optic<strong>al</strong> centre 0 until it <strong>in</strong>tersects the foc<strong>al</strong><br />
plane of the lens. The sought ray is obta<strong>in</strong>ed by connect<strong>in</strong>g C and D.<br />
S<strong>in</strong>ce the rays are refracted <strong>in</strong> the lens twice, foc<strong>al</strong> length / of the system<br />
can be found from the follow<strong>in</strong>g ratio (see Problem 735):<br />
I 1 1 1<br />
T=~+ /2 rt:<br />
2=: where f is the foc<strong>al</strong> length of the mirror<br />
f 'lf2 2.5 em<br />
'1 +2ft<br />
Therefore, the distance b to image A'"B'" can be found from the formula<br />
.!.+.!..=..!.<br />
a b f<br />
at<br />
Hence, b=--,=3 em.<br />
a-<br />
742. The foc<strong>al</strong> length of the th<strong>in</strong> lens is f=--!--l t where r is the radius<br />
n-<br />
of the spheric<strong>al</strong> surface.
GEOMETRICAL OPTiCS 409<br />
H<br />
If<br />
Fig. 537<br />
(tt)<br />
PV/;<br />
f//<br />
~/~<br />
(b)<br />
U/ ~<br />
L<strong>et</strong> the rays par<strong>al</strong>lel to the optic<strong>al</strong> axis of the spheric<strong>al</strong> surface f<strong>al</strong>l on<br />
It from the air (Fig. 537). After be<strong>in</strong>g refracted on the surface, ray NK is<br />
deflected by an angle CI- ~ from the optic<strong>al</strong> axis.<br />
As can be seen <strong>in</strong> Fig. 537a, we have OP tan a,=FtP tan (a-p).<br />
Accord<strong>in</strong>g to the law of refraction, s~n ~=n.<br />
s<strong>in</strong> to'<br />
S<strong>in</strong>ce these angles are sm<strong>al</strong>l, ra: ~'1 (a-p) and a=pn.<br />
Therefore,<br />
n<br />
1 = - -1<br />
n-<br />
,=nf<br />
If the par<strong>al</strong>lel rays are <strong>in</strong>cident from the glass (Fig. 537b), then similar<br />
reason<strong>in</strong>g will give the equations:<br />
,<br />
sln e 1<br />
-'-A=- , and r tan a,=fs tan (~-a,)<br />
n<br />
s<strong>in</strong> to'<br />
S<strong>in</strong>ce the angles are sm<strong>al</strong>l, p~na,<br />
Hence, f 2 = --1=,.<br />
and ra,=f" (P-Cl).<br />
n-<br />
743. Two cases are possible: the focus is outside the sphere, and <strong>in</strong>side it.<br />
L<strong>et</strong> us consider the first case. The path of the ray <strong>in</strong>cident on the sphere at<br />
an angle i is shown <strong>in</strong> Fig. 538.<br />
Bear<strong>in</strong>g <strong>in</strong> m<strong>in</strong>d that the angles I and r are sm<strong>al</strong>l, we have, <strong>in</strong> accordance<br />
with the condition of the problem,<br />
BC= R s<strong>in</strong> a=R s<strong>in</strong> (2, -i)~· R (2r-i) e£ Ri (2-n)<br />
n<br />
The focus obviously lies outside the sphere when n < 2 and on the surface<br />
of the sphere if n= 2.
410 ANSWERS AND SOLUTIONS<br />
The distance<br />
Be . 2i (n-I)<br />
CF=BCcot~::!:T'and ~~2(I-T)~ n<br />
8S can easily be found from Fig. 538.<br />
The sought dislance is<br />
Rn<br />
f=R+CF=2(n_l)<br />
Fig. 538<br />
When n > 2, the path of the ray is 8S shown <strong>in</strong> Fig. 539. The sought distance is<br />
As can be seen from Fig. 539,<br />
Hence,<br />
f'=OF~CF-R<br />
CF=ACcot (i-f) ~ .AC :2!!: ~<br />
l-T &-T<br />
f'=-.!L<br />
n-l<br />
744. L<strong>et</strong> us extend ray BF until it <strong>in</strong>tersects the cont<strong>in</strong>uation of the ray<br />
Incident on the sphere par<strong>al</strong>lel to the optic<strong>al</strong> axis (Fig. 538). It can easily<br />
be seen that section DO that connects the po<strong>in</strong>t of <strong>in</strong>tersection with the<br />
centre of the sphere forms a right angle with the direction of the <strong>in</strong>cident<br />
ray. Triangle ODF is a right one, s<strong>in</strong>ce<br />
OFpe!! R _n_ 21 (n-I) Rt (see Problem 743)<br />
2 n-l n<br />
For this reason the ma<strong>in</strong> planes of sphere M N co<strong>in</strong>cide and pass through its<br />
centre.
GEOMETRICAL OPTICS<br />
411<br />
745. The foc<strong>al</strong> length of the sphere is<br />
R n<br />
1=2"" n-l =15 em<br />
Fig. 539<br />
(see <strong>Problems</strong> 743 and 744). By us<strong>in</strong>g the formula of a lens. and this may<br />
be done because the ma<strong>in</strong> planes co<strong>in</strong>cide, l<strong>et</strong> us f<strong>in</strong>d the distance from the<br />
centre of the lens to the image<br />
b= atr:: 15 ern<br />
a-<br />
The image is virtu<strong>al</strong> and is <strong>in</strong> front of the sphere.<br />
746. The th<strong>in</strong> w<strong>al</strong>l of the spheric<strong>al</strong> flask can be regarded as a divergent<br />
lens fith a foc<strong>al</strong> length of<br />
___~1_~~~ R'<br />
11 ( 1 1 ) (n-l) ~R<br />
(n-I) ---<br />
R. R'J,<br />
After pass<strong>in</strong>g through two such lenses at a distance of 2R from each other<br />
(Fig. 540), the rays par<strong>al</strong>lel to the major optic<strong>al</strong> axis (the flask diam<strong>et</strong>er)<br />
will be so refracted that their cont<strong>in</strong>uations will <strong>in</strong>tersect at the focus F of<br />
1'104----41-<br />
I<br />
I<br />
I<br />
I<br />
II<br />
I<br />
lo+-ZH~ Fig. 540
412 ANS\\'ERS AND SOLUTIONS<br />
the system at a distance b from<br />
of a lens,<br />
Hence,<br />
b<br />
the second lens. Accord<strong>in</strong>g to the formula<br />
I I 1<br />
'l+ 2R -1j=-t;<br />
'I ('I+2R)<br />
2 (/1+R)<br />
Po<strong>in</strong>t D of the <strong>in</strong>tersection of AB (cont<strong>in</strong>uation of the <strong>in</strong>cident ray) and CF<br />
(cont<strong>in</strong>uation of the refracted ray) lies on the ma<strong>in</strong> plane of the system at a<br />
distance x from the second lens.<br />
It follows from the similarity of triangles ACB and FICO, and <strong>al</strong>so of<br />
triangles DC8 and FCO that<br />
x 2R<br />
1J=2R+f.<br />
The ma<strong>in</strong> plane lies at the follow<strong>in</strong>g distance from the second lens<br />
2Rb<br />
t.«<br />
x 2R+tl 'I +R<br />
Therefore, the foc<strong>al</strong> length of the system is<br />
f-b - ,~ oc'l _ R2<br />
- -x-2(fl+R)-T- 2(n-l)&R<br />
In view of symm<strong>et</strong>ry of this optic<strong>al</strong> system, the positions of the second<br />
focus and of the other ma<strong>in</strong> plane are obvious.<br />
747. A glance at Fig. 541 shows that the angle of refraction is<br />
r=L. OAB==L. ABO=L. OBC=L. OCB<br />
and L. BAD=L. BCD=i-r<br />
At po<strong>in</strong>t A the beam turns through an angle i-', at po<strong>in</strong>t B through an<br />
angle 1t - 2" and at po<strong>in</strong>t C through i-,. Therefore, the tot<strong>al</strong> angle through<br />
Fig. 541
QEOMETRICAl OPTICS<br />
413<br />
Fig. 542<br />
which the beam is deflected from the <strong>in</strong>iti<strong>al</strong> direction is<br />
e=i-f +.n--2f + i -r=n+2i-4,<br />
. s<strong>in</strong> i<br />
The angle r can be found from the ratio -.-=n.<br />
sm r<br />
748. When a par<strong>al</strong>lel beam of rays is <strong>in</strong>cident on the drop, the ray pass<strong>in</strong>g<br />
<strong>al</strong>ong the diam<strong>et</strong>er has an angle of <strong>in</strong>cidence of i =0°, while the angles of<br />
<strong>in</strong>cidence of the rays above and below it may range from 0 to 90°.<br />
(1) Us<strong>in</strong>g the results of the previous problem and the law of refraction,<br />
we can f<strong>in</strong>d the v<strong>al</strong>ues of 8 for various v<strong>al</strong>ues of i:<br />
Table 3<br />
e<br />
e<br />
0° 180° 55° 138°20'<br />
20° 160°24' 60° 137°56'<br />
40° 144°40' 65° 138°40'<br />
50° 139°40' 70° 140°44'<br />
(2) A diagram of e versus i is shown <strong>in</strong> Fig. 542.<br />
(3) The m<strong>in</strong>imum v<strong>al</strong>ue of the angle of deflection is approximately equ<strong>al</strong><br />
to Sm<strong>in</strong> = 138 0 • The rays leav<strong>in</strong>g the drop are nearly par<strong>al</strong>lel when e= 8mi,..<br />
s<strong>in</strong>ce <strong>in</strong> this case, as can be seen from Table 3 and the diagram, e changes<br />
the slowest when i changes. An approximate path of the rays <strong>in</strong> the drop<br />
Is illustrated <strong>in</strong> Fig 543.
414<br />
J- ------....,.c<br />
8- ----~---<br />
ANSWERS AND SOLUTIONS<br />
749. In accordance with the formula of a lens<br />
The magnification is<br />
l+-l=~<br />
a b f<br />
Fig. 543<br />
b b-f .<br />
k=(i=-f-=24 times<br />
750. The condenser should produce a re<strong>al</strong> image of the source on the lens<br />
hav<strong>in</strong>g the size of the lens. Therefore, us<strong>in</strong>g the formula of a lens and the<br />
expression that d<strong>et</strong>erm<strong>in</strong>es its magnification, we can write two equations:<br />
1 lid x<br />
-X+y=,' and Do =y<br />
Here x is the distance from the source of light to the condenser and y is<br />
the distance from the condenser to the lens. Accord<strong>in</strong>g to the <strong>in</strong>iti<strong>al</strong> condition,<br />
x+y=l.<br />
By cancell<strong>in</strong>g x and y from the expressions obta<strong>in</strong>ed, we can f<strong>in</strong>d the foc<strong>al</strong><br />
length:<br />
diDo<br />
f<br />
7.1 cm<br />
The diam<strong>et</strong>er of the condenser will be m<strong>in</strong>imum if the slide is beh<strong>in</strong>d it.<br />
The m<strong>in</strong>imum permissible diam<strong>et</strong>er is D e:: It cm.<br />
751. Ground glass is needed to fix the plane <strong>in</strong> which the image is obta<strong>in</strong>ed,<br />
and to i ncrease the angle of vision.<br />
•<br />
Transparen t glass is used to exam<strong>in</strong>e the image produced by a lens <strong>in</strong> a<br />
microscope. For this purpose a l<strong>in</strong>e is drawn on a transparent glass to fix<br />
the focuss<strong>in</strong>g plane, and this l<strong>in</strong>e and the adjacent portion of the image produced<br />
by the lens are focussed <strong>in</strong> the microscope. In this case ground glass<br />
cannot be used, s<strong>in</strong>ce the microscope will show <strong>al</strong>l the distortions due to the<br />
structure of the ground surface.<br />
752. (1) The lanterns will appear Equ<strong>al</strong>ly bright, s<strong>in</strong>ce the illum<strong>in</strong>ation of<br />
the r<strong>et</strong><strong>in</strong>a of the eye E= ~~ is the same for both lanterns. (Here L is the
GEOMETRICAL OPTICS 415<br />
lum<strong>in</strong>ance of a lantern, A the area of the pupil, and b the distance from the<br />
crysta Il<strong>in</strong>e lens to the r<strong>et</strong><strong>in</strong>a.)<br />
(2) The image of a far object is closer to the lens than that of a near<br />
object. For this reason a remote lantern produces a higher illum<strong>in</strong>ation of the<br />
film and its image will be brighter on the photograph.<br />
753. The illum<strong>in</strong>ation of the photographic film is<br />
A (a- 1)9<br />
E""",,-I""WS,---<br />
b2 at<br />
where S,= :: is the lens speed, f is the foc<strong>al</strong> length and a is the<br />
distance<br />
from the lens to the object be<strong>in</strong>g photographed (see Problem '152). It is obvious<br />
therefore that the exposure <strong>in</strong> a camera with a short foc<strong>al</strong> length should<br />
be sm<strong>al</strong>ler.<br />
754. The distances b<strong>et</strong>ween the Sun and the Earth and b<strong>et</strong>ween the Sun<br />
and the Moon are practic<strong>al</strong>ly the same. For this reason if the Moon and the<br />
w<strong>al</strong>l had equ<strong>al</strong> coefficients of reflection, their lum<strong>in</strong>ance would seem identic<strong>al</strong>.<br />
It can be assumed, therefore, that the surface of the Moon consists of dark<br />
rock.<br />
755. In air, the extern<strong>al</strong> convex cornea of the eye collects the rays and<br />
produces an image on the r<strong>et</strong><strong>in</strong>a. The cryst<strong>al</strong>l<strong>in</strong>e lens only helps <strong>in</strong> this.<br />
The refraction <strong>in</strong>dex of the liquid <strong>in</strong>side the eye is very close to that of<br />
water. For this reason the cornea refracts <strong>al</strong>most no light and the eye becomes<br />
very far-sighted.<br />
The refractive properties of the cornea are compl<strong>et</strong>ely r<strong>et</strong>a<strong>in</strong>ed when the<br />
swimmer is wear<strong>in</strong>g a mask.<br />
756. When the man looks at remote objects through his spectacles, he sees<br />
them as he would see objects at a distance of <strong>al</strong>=60 em without any spectacles.<br />
Therefore, when the man is wear<strong>in</strong>g spectacles (see the solution to problem 735)<br />
1 I 1 1<br />
(i+7J=7+t;<br />
where a=oo.<br />
When the man is without spectacles<br />
-.!-+-!.=-!.<br />
a 2 b f<br />
Here b is the depth of the eye, +the m<strong>in</strong>imum optic<strong>al</strong> power of the eye<br />
and*the optic<strong>al</strong> power of the spectacles. It is assumed<br />
that the spectacles<br />
are fitted tightly aga<strong>in</strong>st the eye.<br />
Hence, '0=- at·<br />
L<strong>et</strong> us now d<strong>et</strong>erm<strong>in</strong>e the position of the nearest po<strong>in</strong>t of accommodation<br />
of the eye with spectacles<br />
1 I 1 I I 1 1<br />
-+-=-. and -+-=-+a<br />
1 b l fI a 3 b 1 f1 f0
416<br />
ANSWERS AND SOLUTIONS<br />
Therefore.<br />
1 1 1 1 1<br />
-=-+-=---<br />
03 <strong>al</strong> to at at<br />
and aa=15 em<br />
757. The long-sighted man when wear<strong>in</strong>g his friend's spectacles can see<br />
only very far objects. Therefore, the distance a 2 of best vision of the eye of<br />
the long-sighted man can be d<strong>et</strong>erm<strong>in</strong>ed from the equation<br />
1 1<br />
---=D<br />
at at 1<br />
where <strong>al</strong> is a very great distance (a1 -+ 00) and D} is the optic<strong>al</strong> power of<br />
the spectacles of the short-sighted man.<br />
The optic<strong>al</strong> power D 2 of the spectacles that correct the defect of vision of<br />
the long-sighted man can be found from the formula<br />
1 1<br />
---=D ao<br />
2 a2<br />
where a o = O.25 m<strong>et</strong>re is the distance of best vision of the norm<strong>al</strong> eye.<br />
The distance a3 of best vision of a short-sighted eye can be found from the<br />
equation<br />
1 1<br />
---=D<br />
ao aa 1<br />
If the short-sighted man wears the spectacles of his long-sighted friend,<br />
the distance of best vision, Le., the m<strong>in</strong>imum distance a at which the shortsighted<br />
man can easily read a sm<strong>al</strong>l type, can be d<strong>et</strong>erm<strong>in</strong>ed from the formula<br />
1 1<br />
---=D a aa 2<br />
Upon solv<strong>in</strong>g these four equations, we g<strong>et</strong> a= 12.5 ern.<br />
758. When an object with a height of 1 is exam<strong>in</strong>ed from a distance of D,<br />
the angle of vision q>l is d<strong>et</strong>erm<strong>in</strong>ed by the formula<br />
I<br />
QEOMETRICAL OPTICS 417<br />
Fig. 544<br />
Hence,<br />
N- .!!.- b+f_E.- L-r+f<br />
-, L-, L<br />
D<br />
(1) When L=oo, we hhveN=T.<br />
r<br />
(2) When L=D, we have N=T+1-T.<br />
D<br />
71S9. The magnification of a telescope is N= ~: . where II is the foc<strong>al</strong><br />
length of the objective and /s that of the eyepiece. -<br />
S<strong>in</strong>ce <strong>in</strong> a telescope adjusted to <strong>in</strong>f<strong>in</strong>ity the distance b<strong>et</strong>ween the objective<br />
and the eyepiece is 11 +/1' .<br />
D 11 + I,<br />
7=-b-<br />
Here b is the distance from the eyepiece to the image of the diaphragm.<br />
Accord<strong>in</strong>g to the formula of a lens,<br />
_1_+.!.=J..<br />
'.+/s b '2<br />
Upon cancell<strong>in</strong>g b from these equations, we f<strong>in</strong>d that<br />
D<br />
II<br />
7=-,;=N<br />
760. Sharp images of remote objects will be obta<strong>in</strong>ed with the convergent<br />
lens <strong>in</strong> three different positions. The lens can be placed <strong>in</strong> front of the d 1<br />
vergent lens or beh<strong>in</strong>d it.<br />
27-2042
418 ANSWERS AND SOLUTIONS<br />
N<br />
Fig. 545<br />
For the first position the distance d b<strong>et</strong>ween the lenses can be found con..<br />
sider<strong>in</strong>g po<strong>in</strong>t K as the virtu<strong>al</strong> image of po<strong>in</strong>t A <strong>in</strong> the divergent lens (Fig. 545).<br />
1 1 1<br />
-f,-d+T=- t;<br />
Ray MN is par<strong>al</strong>lel to the optic<strong>al</strong> axis of the system.<br />
Hence,<br />
fll<br />
f<br />
d= ,-'l+ l = 3.5 cm<br />
For the second position (the convergent lens is beh<strong>in</strong>d the divergent one)<br />
the path of the rays is shown <strong>in</strong> Fig. 546. Regard<strong>in</strong>g po<strong>in</strong>t A as the image<br />
of K <strong>in</strong> the convergent lens, l<strong>et</strong> us use the formula of a lens .<br />
1 1 1<br />
11 +d+l-d=t;<br />
Hence.<br />
d=l-fl ± l+/1 ../ l-~<br />
2 2 V t+'1<br />
The distance b<strong>et</strong>ween the lenses may be d 2 = 35 em or d a=5 em.<br />
761. L<strong>et</strong> the rays com<strong>in</strong>g from one end of the diam<strong>et</strong>er of the visible disk<br />
of the Moon be directed <strong>al</strong>ong the optic<strong>al</strong> axis of the system. The rays will<br />
produce an image on the optic<strong>al</strong> axis at po<strong>in</strong>t A (Fig. 547) removed by a<br />
distance of l=45 cm from the divergent lens.<br />
IV<br />
---- ..........<br />
K<br />
Fig. 546
GEOMETRICAL OPTICS<br />
419<br />
A.<br />
....---l---~<br />
B<br />
Fig. 547<br />
The rays com<strong>in</strong>g from the other end of the diam<strong>et</strong>er form, accord<strong>in</strong>g to the<br />
condition, an angle cp with the first rays. After pass<strong>in</strong>g through the system,<br />
these rays will produce an image (po<strong>in</strong>t B) ly<strong>in</strong>g <strong>in</strong> a plane perpendicular<br />
to the optic<strong>al</strong> axis and removed by the same distance I from the divergent<br />
lens.<br />
To f<strong>in</strong>d the diam<strong>et</strong>er of the image D 1 = AB, l<strong>et</strong> us consider the path of<br />
the ray pass<strong>in</strong>g through the optic<strong>al</strong> centre of the first lens.<br />
In the first arrangement. the convergent lens is placed <strong>in</strong> front of the<br />
divergent one at a distance of d 1 =3.5 ern. If we consider po<strong>in</strong>t E as the<br />
virtu<strong>al</strong> image of po<strong>in</strong>t 0 we can write<br />
Us<strong>in</strong>g the similarity of triangles ABE and 0lPE. and remember<strong>in</strong>g that<br />
0IP= d 1 tan cp, we obta<strong>in</strong><br />
Upon canceJJ<strong>in</strong>g Xl from these equations, we f<strong>in</strong>d that D 1 ' 0.72 ern,<br />
In the second arrangement of the lenses (d 2=35 em), the path of the rays<br />
is shown <strong>in</strong> Fig. 548. The diam<strong>et</strong>er of the Moon's image D 2 can be found<br />
E<br />
A<br />
~---+----J-----~<br />
Fig. 548
420 ANSWERS AND SOLUTIONS<br />
from the equations:<br />
d 2 tan cp d 2 q><br />
---~-<br />
(x 2 + d 2 )- 1 X 2 - Xz<br />
(consider<strong>in</strong>g triangles EOP, E AB ~nd OPO I ) and<br />
1 1 1<br />
-+-=d<br />
2 .x?, /2<br />
(consider<strong>in</strong>g E as the image of°1 ) . ' Hence DC}. ~ 0.011 em.<br />
Fig. 549<br />
In the third arrangement (d 3=5 em) the path of the rays will be somewhat<br />
different (Fig. 549) than that ShO\\'D <strong>in</strong> Fig. 548.<br />
The equations for 'D 8 can be written similar to the preced<strong>in</strong>g cases as<br />
follows:<br />
Ds ds tan q> e: dscp and ..!.._..!..=..!..<br />
(l- d s ) + Xs Xs Xs ' da Xa fI<br />
Hence, Ds=O.18 em.<br />
762. It follows from the formula of a lens<br />
1 1 1<br />
(i+1J=Fob<br />
that the magnification of the objective is<br />
b' Fob<br />
k1=(i=;-="P o b<br />
where b is the distance from the image to the objective.<br />
The re<strong>al</strong> <strong>in</strong>verse magnification of the object produced by the objective can<br />
be viewed through the eyepiece as through a magnify<strong>in</strong>g glass, the virtu<strong>al</strong><br />
image produced by this magnl fy<strong>in</strong>g glass be<strong>in</strong>g arranged from the eye at the<br />
distance of best vision D=25 em.<br />
Accord<strong>in</strong>g to the formula ~f a magnify<strong>in</strong>g glass<br />
1 1 1<br />
<strong>al</strong> -D = F ey e<br />
where <strong>al</strong> is the distance from the image produced by the objecti ve to the<br />
eyepiece.<br />
1 he magnification of the eyepiece is<br />
D D+Feye<br />
k,,=-=----<br />
at F ey e<br />
The tot<strong>al</strong> magnification of the microscope is<br />
Fob (D+Feye)<br />
k=ktk"<br />
180 times<br />
(a-Fob) Feye
CHAPTER 6<br />
PHYSICAL<br />
OPTICS<br />
6-1. Interference of Light<br />
763. NOt it does not. The presence of Illum<strong>in</strong>ation m<strong>in</strong>ima on the <strong>in</strong>terference<br />
pattern means that no quantity of light enters the given section of<br />
space. /<br />
764. The maximum illum<strong>in</strong>ation will be observed at an arbitrary po<strong>in</strong>t of<br />
the screen C (Fig. 550) if the difference of the paths d l-d1 =kA, where<br />
k=O, 1, 2, ... is an <strong>in</strong>teger.<br />
Accord<strong>in</strong>g to the Pythagorean theorem,<br />
d: = DI+( hk +;r<br />
d~=D'+( hk - ;r<br />
whence<br />
d:-d~=(d,+dl)<br />
(d,-dt)=2hlc'<br />
In accordance with ~he <strong>in</strong>iti<strong>al</strong> conditlon d, +d t ~ 2D. Therefore, d, -d 1 ==<br />
= k). ~ 2:~' . The distance tothe k-th light band from the centre of the screen is<br />
h,,-= k~D • The distance b<strong>et</strong>ween the bands is Ah=hk+l-hk=).f .<br />
785. The distance b<strong>et</strong>ween Interference bands is Ah=~ (see Problem 764).<br />
In our case D==AB es:a+b, and 1=8 1 8, is the distance b<strong>et</strong>ween the images<br />
Sl and 52 of the source <strong>in</strong> the ftat mirrors (Fig. 551). The v<strong>al</strong>ue of I can<br />
r----------------------------<br />
I<br />
IIII<br />
c<br />
4 1 ~~~-----D -------:l~<br />
Fig. 550
422<br />
S,<br />
IT ...........<br />
In --..<br />
U' .,------<br />
I ~----- I<br />
~<br />
----<br />
ANSWERS AND SOLUTIONS<br />
N<br />
f7-=":_-=-1i:;§L~ ----- /I<br />
.---- ()(, r<br />
IV Fig. 551<br />
be found from triangle SlSB:<br />
Hence,<br />
; =2b ~ or 1=2ba<br />
768. The second coherent source is obta<strong>in</strong>ed <strong>in</strong> Lloyd's experiment by<br />
reflection of the rays from mirror AD. In reflection the phase changes by n<br />
(loss of a h<strong>al</strong>f-wave) and the oscillations will be damped at po<strong>in</strong>t 0 where<br />
a bright band should be observed (a' m<strong>in</strong>imum of illum<strong>in</strong>ation). As compared<br />
with Problem 764, the entire pattern will be shifted by the width of a light<br />
(or dark) band.<br />
767. The illum<strong>in</strong>ation on the screen will be <strong>in</strong>tensified when d 2 - d 1 =kA.<br />
The locus of po<strong>in</strong>ts on the screen reached. by rays from both sources with<br />
this difference is a circle with its centre at po<strong>in</strong>t A (Fig. 552). For this reason<br />
the <strong>in</strong>terference bands wi11 ha ve the form of concentric circles.<br />
When l=n'A the illum<strong>in</strong>ation will be <strong>in</strong>tensified at po<strong>in</strong>t A (an <strong>in</strong>terference'<br />
maximum of the n-th order). The nearest bright <strong>in</strong>terference band (circle) of<br />
k-l -~..------p----~/I<br />
Fig. 552
PHYSICAL OPTICS<br />
423<br />
Fig. 653<br />
the (n-I)th order is at a distance from po<strong>in</strong>t A d<strong>et</strong>erm<strong>in</strong>ed from the equation<br />
d 2-d1 = V(nA+Dr~ +h~-l-V D2+h~_1.=(n-l) A.<br />
Bear<strong>in</strong>g <strong>in</strong> m<strong>in</strong>d the conditions of the problem that A ~ D and A. ~ I, we g<strong>et</strong><br />
hn-t~ V 2D
424 ANSWERS AND SOLUTIONS<br />
der<strong>in</strong>g triangle AS t B, we can write<br />
Hence,<br />
I<br />
2+aa~aan<br />
l::r:2aa (n- 1)<br />
Fig. 554<br />
By us<strong>in</strong>g the solution of Problem 764, we f<strong>in</strong>d:<br />
'AD l (a+b)<br />
~h= 2aa(n_l)=O.15cm<br />
L<br />
771. N = Ah' where L i$ the width of the <strong>in</strong>terference pattern.<br />
It cen be seen from Fig.· 256 that L=.! t, where I is the distance b<strong>et</strong>ween<br />
a<br />
the virtu<strong>al</strong> images 8 1 and'S. of the source.<br />
Us<strong>in</strong>g the results ·of the previous problem, w~ g<strong>et</strong><br />
4aba 2 (n -1)2<br />
N= (a+b) A e!5<br />
772. A biprism made of a substance with a refraction <strong>in</strong>dex n 2 deflects<br />
rays by an angle<br />
CPt=( 90"- ~ ) (n,- nl)<br />
where nl is the refraction <strong>in</strong>dex of the medium from which the rays f<strong>al</strong>l. For<br />
a biprism <strong>in</strong> air<br />
If the biprisms are equiv<strong>al</strong>ent, CPt = cp"<br />
Hence,<br />
~=~ n,-n1+ IS0 0 nt-l<br />
n,-I I'lt- I<br />
For the v<strong>al</strong>ues given <strong>in</strong> the <strong>in</strong>iti<strong>al</strong> conditions, 6:!: 179°37'.
PHYSICAL OPTICS<br />
425<br />
~---b---~<br />
Fig. 555<br />
773. The path of the rays <strong>in</strong> the system is illustrated <strong>in</strong> Fig. 554. Here<br />
8 1 and 52 are the images of source S <strong>in</strong> 'the h<strong>al</strong>ves of the lens. Obviously,<br />
b=~ a-I<br />
The distance I b<strong>et</strong>ween 8 1 and 8 2 can be found from the similarity of<br />
triangles SAB and 88 18,:<br />
l=~ a-I<br />
The distance b<strong>et</strong>ween adjacent <strong>in</strong>terference bands on the screen is<br />
~h= A(D-b)=~(Da_DI_af)=lO-2 cm<br />
I ad<br />
(see Problem 764).<br />
The sought number of <strong>in</strong>terference bands is<br />
d(D+a)=25<br />
a~h<br />
174. The distance b<strong>et</strong>ween the virtu<strong>al</strong> sources 8. and S2 can be found by<br />
the m<strong>et</strong>hod described <strong>in</strong> the solut ion of Problem 773 (Fig. 555).<br />
The distance b<strong>et</strong>ween <strong>in</strong>terference bands is<br />
6h<br />
A(0/ - [)a +af)<br />
da<br />
28-2042
426 ANSWERS AND SOLUTIONS<br />
Fig 556<br />
The numberof bands on the screen is<br />
L<br />
N=l1h<br />
where L= ~' is the dimension of the portion of the screen on which <strong>in</strong>terference<br />
bands are observed. Therefore,<br />
0<br />
- NablA 15<br />
- adl +abN'A.-bfNA em<br />
The maximum possible number of bands can be d<strong>et</strong>erm<strong>in</strong>ed (rom the condition<br />
(here 0 --+ 00).<br />
Hence,<br />
adi +Nab'A-bfN"A.=O<br />
adl<br />
N",ax=b/'A._ abA 5<br />
The number of bands is f<strong>in</strong>ite, s<strong>in</strong>ce the distance b<strong>et</strong>ween them <strong>in</strong>creases as<br />
the screenIs moved away, and the dimensions of the portion of the screen<br />
on which the <strong>in</strong>terference pattern appears grow.<br />
775. The distance b<strong>et</strong>ween the <strong>in</strong>terference bands will not depend on the<br />
position of the screen only if the source is arranged <strong>in</strong> .the foc<strong>al</strong> plane of the<br />
lens. This directly follows from the expression • ~<br />
M=a~-(Df-Da+af)<br />
obta<strong>in</strong>ed <strong>in</strong> solv<strong>in</strong>g Problem 774. When.a=f,<br />
Ah=A~=lO-1 em<br />
at any v<strong>al</strong>ue of D.<br />
For this case the path of the rays is shown <strong>in</strong> Fig. 556. A glance at this<br />
illustration will show that the number of <strong>in</strong>terference bands will be maximum<br />
when the screen is <strong>in</strong> position AB. The distance from the screen to the lens
PHYSICAL OPTICS 427<br />
d<br />
can be found Irorn triangle DAB. remember<strong>in</strong>g that the angle a a:T and<br />
AB=R:<br />
Rf<br />
D=7=2 m<strong>et</strong>res<br />
776. The length of a light wave dim<strong>in</strong>ishes n times <strong>in</strong> glass, s<strong>in</strong>ce the<br />
frequency does not change and the velocity decreases n times. This produces<br />
an addition<strong>al</strong> difference <strong>in</strong> the path b<strong>et</strong>ween the coherent waves <strong>in</strong> the beams.<br />
At a distance of d l<br />
the upper beam will accommodate k. =dt wavelengths,<br />
d 2 n d l-d2 •<br />
and the lower beam k 2 = T + -A,- wavelengths at the same distance.<br />
At any po<strong>in</strong>t on the screen the light waves will addition<strong>al</strong>ly be shifted with<br />
respect to each other by k 1-k2 wavelengths. As a result. the entire lnterferenee<br />
pattern will be displaced by k.-k t Cd. ~ d t ) (n-1)= 100 bands.<br />
The process of displacement can be observed at the moment when the pIates<br />
are <strong>in</strong>troduced. After this has been done, the <strong>in</strong>terference pattern on the screen<br />
wi11 be as before.<br />
777. The lens is too thick. Interference occurs only with th<strong>in</strong> films. The air<br />
layer b<strong>et</strong>ween the lens and the glass is th<strong>in</strong>.<br />
778. No, it will not. The difference of the path b<strong>et</strong>ween the waYes me<strong>et</strong><strong>in</strong>g<br />
on the screen and emitted by the sources Sand 51 or 5 and 52 is great.<br />
In these conditions the spectra of various orders that correspond to the spectr<strong>al</strong><br />
<strong>in</strong>terv<strong>al</strong> of the source are superposed <strong>in</strong> the same way as when the waves<br />
are reflected from the boundaries of a thick film.<br />
If the shield is removed, the only result will be superposition of monotonously<br />
chang<strong>in</strong>g illum<strong>in</strong>ation on the <strong>in</strong>terference pattern from sources 8 1<br />
and 5 z .<br />
28 *<br />
Fig. 557
428 ANSWERS AND SOLUTIONS<br />
779. When the r<strong>in</strong>gs are observed <strong>in</strong> reflected light, the <strong>in</strong>tensity ot the<br />
<strong>in</strong>terference beams is about the same.<br />
In transmitted light the <strong>in</strong>tensity of one beam that was not reflected is<br />
much higher than that of another beam that was reflected twice. As a result.<br />
the maxima and m<strong>in</strong>ima will appear aga<strong>in</strong>st the background of uniform illum<strong>in</strong>ation,<br />
the light will not be ext<strong>in</strong>guished compl<strong>et</strong>ely and the entire pattern<br />
wi II be less dist<strong>in</strong>ct than <strong>in</strong> reflected light.<br />
780. In the absence of contact. the radius of the fifth r<strong>in</strong>g is d<strong>et</strong>erm<strong>in</strong>ed<br />
2<br />
by the equation ~l +211=5).. If the dust is removed, the radius of this r<strong>in</strong>g<br />
r: _ ,:-r~_ -t<br />
can be found from the equation7[=5A. Hence, d- - 1.8X IO em.<br />
2r<br />
.. / k'A<br />
781. 'k= V (_1 1_)<br />
R?, R 1<br />
782. To reduce the reflection factor it is necessary that rays J and 2<br />
(Fig. 557) reflected from the extern<strong>al</strong> and <strong>in</strong>tern<strong>al</strong> surfaces of the coat applied<br />
to the optic<strong>al</strong> glass damp each other.<br />
This wi 11 occur if<br />
A<br />
2hn=(2k+ 1)2 (1)<br />
where k = O. I. 2. .••• Hence, the m<strong>in</strong>imum thickness of the coat is<br />
A<br />
hm<strong>in</strong>=-4 .<br />
n<br />
Condition (1) cannot be satisfied for <strong>al</strong>l wavelengths, For this reason. h is<br />
usu<strong>al</strong>ly so selected as to damp the middle part of the spectrum. The thickness<br />
of the coat exceeds hmln by an odd number of times, s<strong>in</strong>ce it is easier to<br />
make thick coats than th<strong>in</strong> ones (a quarter of the wavelength thick).<br />
783. To observe an <strong>in</strong>terference pattern, the maximum of the k-th order<br />
correspond<strong>in</strong>g to the wavelength A should not overlap with the maximum of<br />
the (k+ l)th order correspond<strong>in</strong>gto the wavelength 'A+ l\A, where ~A = 100 A.<br />
This will occur if (A+dA) k ~ A(k+ I).<br />
A<br />
Hence, k E:; ~A •<br />
The maximum permissible thickness of the layer h mtJx satisfies the equation<br />
where k max =;')., . If ').,<br />
2h max={A+l1A) k max<br />
is the wavelength correspond<strong>in</strong>g to the middle 01 the<br />
visible section of the spectrum (A= 5.000A), then<br />
h max e:E 1.3X 10- 3 em<br />
If a th<strong>in</strong> film with a refraction <strong>in</strong>dex n is used <strong>in</strong>stead of the afr layer.<br />
the maximum thickness should be n times sm<strong>al</strong>ler than <strong>in</strong> the air layer.<br />
784. Upon the <strong>in</strong>terference of rays 1 and 2 (Fig. 558) reflected from different<br />
faces of the wedge. the condition of the m<strong>in</strong>imum can be written as<br />
follows: 2hn = kA (k=O. 1, 2).
PHYSICAL OPTICS<br />
429<br />
~---:c -------3~ Fig. 558<br />
S<strong>in</strong>ce the angle a is sm<strong>al</strong>l, h ~ xa.<br />
Therefore, the distance b<strong>et</strong>ween the <strong>in</strong>terference bands on the wedge is<br />
~x=_A_ .<br />
. 2an<br />
Accord<strong>in</strong>g to the lormula lor the magnlficatton of a lens, ~:=1-, where<br />
a is the distance from the screen to the lens and b from the lens to the wedge.<br />
S<strong>in</strong>ce b=d-a, then accord<strong>in</strong>g to the formula of a lens, -!..+-d I =.!-,. Upon<br />
a -a<br />
cancell<strong>in</strong>g a and b from these expressions, we can f<strong>in</strong>d the sought v<strong>al</strong>ue of<br />
the angle a.<br />
A d T Vd<br />
a=---........;.----:--<br />
2-4/d<br />
2nl11 d ± Y d 2-4fd<br />
The solution of .thls problem is not a s<strong>in</strong>gle one, because a sharp image<br />
can be obta<strong>in</strong>ed on the screen with fixed d and f when the lens is <strong>in</strong> one of<br />
two positions.<br />
6-2. Diffraction of Light<br />
785. The radius of the first Fresnel zone can be found from triangles ADE<br />
and DEB (Fig. 559):<br />
r:=<strong>al</strong>-(a-X)I=(b+;r_(b+X)1<br />
S<strong>in</strong>ce the wavelength is sm<strong>al</strong>l, X=:2(ab~b)'<br />
Therefore, r~ = 2ax- Xi.<br />
Neglect<strong>in</strong>g the sm<strong>al</strong>l v<strong>al</strong>ue of Xl,<br />
we f<strong>in</strong><strong>al</strong>Iy obta<strong>in</strong><br />
.. 1""""iiii'A<br />
'1 == JI a+b
430 ANSWERS AND SOLUTIONS<br />
Fig. 559<br />
The same m<strong>et</strong>hod can be used to obta<strong>in</strong> the radii of the follow<strong>in</strong>g Fresnel<br />
zones. For zone k<br />
'1t= 1·/'abk"A<br />
/ a+b<br />
786. A distance from a po<strong>in</strong>t' source to the wave front of a ~ 00 corresponds<br />
to a plane wave.<br />
The sought radii of the zones are<br />
'1t= lim<br />
. {abk"A ..rtix<br />
a<br />
+b =', kb'A<br />
a .... ao<br />
(see the solution to Problem 785).<br />
787. To solve the problem it Is necessary to c<strong>al</strong>culate the number k of<br />
Fresnel zones that can. be accommodated <strong>in</strong> apertures with diam<strong>et</strong>ers of D<br />
and D I .<br />
Us<strong>in</strong>g the results of Problem 785, we have<br />
, / k abA =D<br />
a+b 2<br />
Now it is easy to f<strong>in</strong>d that k=3 (an odd number). When the aperture is<br />
5.2 mm <strong>in</strong> diam<strong>et</strong>er, it can conta<strong>in</strong> aPfroximately four zones (an even number).<br />
Therefore, a greater aperture wil reduce the illum<strong>in</strong>ation at po<strong>in</strong>t B.<br />
788. When the four Fresnel zones are open the dark spot on the axis of<br />
the beam is surrounded by bright and dark r<strong>in</strong>gs. When the aperture<br />
is <strong>in</strong>creased, the tot<strong>al</strong> illum<strong>in</strong>ation of the screen grows <strong>in</strong> magnitude.<br />
but the distribution of the lum<strong>in</strong>ous energy on the screen so changes that<br />
the m<strong>in</strong>imum will be <strong>in</strong> the centre.<br />
•<br />
789. The sought illum<strong>in</strong>ation will be maximum when one Fresnel zone is<br />
accommodated <strong>in</strong> the diaphragm. With a view to the solution of Problem 786,<br />
we have<br />
D=2 YbA=O.2 cm
PHYSICAL OPTICS 431<br />
A<br />
Fig. 560.<br />
790. The diffraction will be noticeable if the aperture accommodates a<br />
sm<strong>al</strong>l number of Fresnel zones, l.e., the radius of the aperture will be of<br />
the same order as (or smaHer than) the radius of the first Fresnel zone:<br />
(a~b)';=:R<br />
where R is the radius of the aperture.<br />
When a=b, we have a).,;:, 2R2.<br />
791. The Fresnel zones plotted <strong>in</strong> Fig. 560 make it possible to d<strong>et</strong>erm<strong>in</strong>e<br />
the lum<strong>in</strong>ous <strong>in</strong>tensity at po<strong>in</strong>t B. The illum<strong>in</strong>ation at this po<strong>in</strong>t Is created<br />
by the first and subsequent Fresnel zones. If the dimensions of the screen do<br />
not exceed considerably the radius of the first centra) zone, found from the<br />
formula <strong>in</strong> Problem 785, a bright spot is sure to appear at po<strong>in</strong>t B with an<br />
illum<strong>in</strong>ation only slightly differ<strong>in</strong>g from that which would have appeared <strong>in</strong><br />
the absence of the screen.<br />
792. Approximately three m<strong>et</strong>res.<br />
793. Here it is more convenient to select the Fresnel zones <strong>in</strong> the form of<br />
bands par<strong>al</strong>lel to the edges of the slit. The m<strong>in</strong>imum will be observed <strong>in</strong> the<br />
direction q> if an even number of zones can be accommodated <strong>in</strong> slit AB<br />
(Fig. 561). This figure shows four Fresnel zones. ·We have b=2kx, where x<br />
is the width of the Fresnel zone, k= I, 2, 3, .•.. Here AK is the difference<br />
<strong>in</strong> the path b<strong>et</strong>ween the extreme rays sent by one zone:<br />
Hence,<br />
. ).<br />
AK=x s<strong>in</strong> q>=2<br />
A.<br />
%=2 s<strong>in</strong> cp<br />
Therefore, the m<strong>in</strong>imum will be observed <strong>in</strong> the direction
432 ANSWERS AND SOLUTIONS<br />
Fig. 561<br />
bright spot would be equ<strong>al</strong> to AB =2, (Fig. 562). In view of diffraction,<br />
the dimensions of the sfot will <strong>in</strong>crease to DC. The distance DC is d<strong>et</strong>erm<strong>in</strong>ed<br />
by the angle q> tha gives the direction to the first m<strong>in</strong>imum (dark r<strong>in</strong>g).<br />
Accord<strong>in</strong>g to the note, 2,s<strong>in</strong>
PHYSICAL OPT ICS<br />
433<br />
~---- d----~-.I<br />
a<br />
s t A<br />
0------------ Zr----------------- 0<br />
Ḅ<br />
D<br />
Fig. 562<br />
and the distances from the centre of' the ditTraction pattern to the maxima correspond<strong>in</strong>g<br />
to various wavelengths, <strong>al</strong>so decrease n times, s<strong>in</strong>ce accord<strong>in</strong>g to<br />
the <strong>in</strong>iti<strong>al</strong> condition the angles q> are sm<strong>al</strong>l and s<strong>in</strong> q> ~ ~ .<br />
800. The spectra of different orders will be <strong>in</strong> contact if-k).,2=(k+ I) "'1·<br />
Therefore,<br />
_ Al _<br />
k-r---;r-5<br />
""-/\'1<br />
For this reason only the spectra of the sixth and seventh orders can be<br />
parti<strong>al</strong>ly superposed. But this grat<strong>in</strong>g (see Problem 796) can give only the<br />
spectrum of the fourth order for this range of wavelengths. Therefore the<br />
spectra will not be superposed <strong>in</strong> our case.<br />
801. When the rays are <strong>in</strong>cident on the grat<strong>in</strong>g at an angle 8 (Fig. 563),<br />
the difference <strong>in</strong> the path b<strong>et</strong>ween the waves issu<strong>in</strong>g from the edges of adja-<br />
BI\<br />
I \<br />
I \<br />
I \ \\\\<br />
,<br />
\,\ To -first<br />
\mazlmum<br />
Fig. 563
434<br />
ANSWERS AND SOLUTIONS<br />
cent slits Is<br />
6=BD-AC=dsln tp-dsln a<br />
These waves add up and Intensify each other when<br />
d (s<strong>in</strong> cp-s<strong>in</strong> 8)=kl<br />
where k= I, 2, 3, ... for the maxima ly<strong>in</strong>g at the right of the centr<strong>al</strong> one<br />
(Jl=O) and k=~I, -2, -3, ••• for those ly<strong>in</strong>g at its left.<br />
The maximum order of the spectrum will be observed when q>=-9()O.<br />
Thus d (-1- ~ ) =kA. Hence, k=-6. The spectrum of the sixth order<br />
may be observed. The m<strong>in</strong>us sign shows that the spectrum lies to the left<br />
of the centr<strong>al</strong> one.<br />
802. As follows from the formula d (s<strong>in</strong> q>-s<strong>in</strong> 8)=kA (see the solution<br />
to Problem 801), the period of the grat<strong>in</strong>g will be m<strong>in</strong>imum with tangenti<strong>al</strong><br />
Incidence of the rays: 9=90°. In this case d d ~ . Therefore, the period of<br />
the grat<strong>in</strong>g should satisfy the <strong>in</strong>equ<strong>al</strong>ity d~ ~ •<br />
803. In the gener<strong>al</strong> case, as shown <strong>in</strong> the solution of Problem 801, the<br />
sought condition will be<br />
d (s<strong>in</strong> cp-s<strong>in</strong> 9)=kA<br />
It may be rewritten as<br />
2dcos CPt 9 s<strong>in</strong> cP 2 9=kA<br />
q>+9 T 9
PHYSICAL OPTICS<br />
435<br />
S~-------+--+------~IF----';;;~~-<br />
Fig. 564<br />
or<br />
Upon solv<strong>in</strong>g this equation, we f<strong>in</strong>d that<br />
s<strong>in</strong> p<br />
-1±Y4n l -<br />
2 Y2<br />
l<br />
The solution with the plus sign has a physic<strong>al</strong> mean<strong>in</strong>g. For red. rays<br />
s<strong>in</strong> ~r e!! 0.26. Therefore, PI' S!: 15°6'. For viol<strong>et</strong> rays s<strong>in</strong> Po ee 0.31 and Pv:::=<br />
~ 18°6'. The sought angle is e=~o-p,. ~ 3°.<br />
Fig. 665
436 ANSWERS AND SOLUTIONS<br />
805. For the red rays the foc<strong>al</strong> length of the lens is<br />
t,.= 2 (n:-l) e: 27 em<br />
Fig. 566<br />
and for the viol<strong>et</strong> rays If} = 25 em.<br />
Accord<strong>in</strong>g to the formula of a lens, the image produced by the red rays<br />
will be at a distance o"f b; = af'f = 58.7 em and that from the viol<strong>et</strong> rays<br />
a- ,.<br />
at bfJ = 50 em.<br />
The image of the source on the screen (Fig. 564) will have the form of<br />
a spot whose edges are coloured red.<br />
The diam<strong>et</strong>er of the spot d can be found from the similarity of triangles<br />
ABE and CDE:<br />
d=Db,-bf/ :=::: 0.15 cm<br />
b,<br />
808. The sunrays f<strong>al</strong>l<strong>in</strong>g on ra<strong>in</strong> drops may be assumed to be par<strong>al</strong>lel.<br />
As they emerge from a drop after be<strong>in</strong>g reflected once on its <strong>in</strong>tern<strong>al</strong> surface,<br />
the rays diverge <strong>in</strong> <strong>al</strong>l directions. Only the rays subjected to m<strong>in</strong>imum defle..<br />
ction are about par<strong>al</strong>lel. When these rays g<strong>et</strong> <strong>in</strong>to the eye, they produce<br />
the maximum visu<strong>al</strong> impression. These rays travel, as can be said, with the<br />
maximum density. The other rays are diffused <strong>in</strong> <strong>al</strong>l directions. As shown<br />
<strong>in</strong> Problem 748, the angle of deflection for par<strong>al</strong>lel rays is 138°. Therefore,<br />
the angle b<strong>et</strong>ween the sunrays and the direction to the ra<strong>in</strong>bow is 42° (for<br />
red light) (Fig. 565).<br />
The eye will receive light from the drops that are <strong>in</strong> the direction form<strong>in</strong>g<br />
an angle of 420 with the l<strong>in</strong>e pass<strong>in</strong>g through the eye and the Sun.<br />
For viol<strong>et</strong> rays this angle is about 40°.<br />
807 The first (primary) ra<strong>in</strong>bow is observed ow<strong>in</strong>g to the rays that were<br />
reflected once <strong>in</strong>side the water drops. Upon refraction, the viol<strong>et</strong> rays undergo<br />
the greatest deflection from the <strong>in</strong>iti<strong>al</strong> direction (see Problem 747) (L a grows<br />
with n, s<strong>in</strong>ce r decreases). For this reason the extern<strong>al</strong> arc will be red and<br />
the <strong>in</strong>tern<strong>al</strong> one viol<strong>et</strong>.<br />
The reflection ra<strong>in</strong>bow is caused by rays that were reflected twice <strong>in</strong>side<br />
the drops. The approximate path of the ray is shown <strong>in</strong> Fig. 566. As can<br />
be shown, the direction to the ra<strong>in</strong>bow is 51° with the l<strong>in</strong>e that connects<br />
the eye and the Sun. With two refractions and two reflections the colours<br />
<strong>al</strong>ternate <strong>in</strong> the reverse order: the extern<strong>al</strong> arc will be viol<strong>et</strong> and the <strong>in</strong>tern<strong>al</strong><br />
one red.
PHYSICAL OPTICS<br />
437<br />
Fig. 567<br />
The lum<strong>in</strong>ous <strong>in</strong>tensity is much weaker after two reflections, for which<br />
reason a reflection ra<strong>in</strong>bow is much less <strong>in</strong>tensive than the primary one.<br />
808. The geographic<strong>al</strong> latitude of Moscow, i.e., the angle b<strong>et</strong>ween the<br />
plane of the equator and a norm<strong>al</strong> to the surface of the Earth, is q>= 56°.<br />
At this moment the Sun is <strong>in</strong> the zenith above the northern tropic (latitude<br />
a=23.5°). Hence, the angle b<strong>et</strong>ween the direction to the Sun and the horizon<br />
(Fig. 567) is<br />
~ = 90°- «p + (X = 57°30'<br />
A .ra<strong>in</strong>bow can be observed only when the <strong>al</strong>titude of the Sun above the<br />
horizon does not exceed 42° (see Fig. 565). Therefore, no ra<strong>in</strong>bow can be<br />
observed.<br />
809. Our eye perceives a colour when its sensitive elements are irritated<br />
by a light wave of a def<strong>in</strong>ite frequency. The frequency of light waves, however,<br />
does not change dur<strong>in</strong>g transition from one medium <strong>in</strong>to another.<br />
810. The green glass should be used. In this case the word will appear<br />
black aga<strong>in</strong>st the green background of the paper, s<strong>in</strong>ce the red colour of the<br />
word "excellent" does not pass through green glass.<br />
If the red glass is used, the word written by the red pencil will not be<br />
seen aga<strong>in</strong>st the red background of the paper.<br />
811. Camera lenses predom<strong>in</strong>antly reflect the extreme parts of the visible<br />
spectrum: red and viol<strong>et</strong> (see Problem 782). A mixture of these colours produces<br />
a lilac t<strong>in</strong>t.<br />
812. The colours of a ra<strong>in</strong>bow are pure spectr<strong>al</strong> colours (see Problem 806)<br />
s<strong>in</strong>ce only a ray of a def<strong>in</strong>ite wavelength is seen <strong>in</strong> a given direction. Conversely,<br />
the colours of th<strong>in</strong> films are produced by ext<strong>in</strong>guish<strong>in</strong>g (tot<strong>al</strong>ly or<br />
parti<strong>al</strong>ly) of the rays of a certa<strong>in</strong> spectr<strong>al</strong> <strong>in</strong>terv<strong>al</strong> due to <strong>in</strong>terference. The<br />
colour of the film will supplement the colour of this spectr<strong>al</strong> <strong>in</strong>terv<strong>al</strong>.<br />
813. Under the force of gravity. the soapy water dra<strong>in</strong>s down onto the<br />
lower portion of the film, which is <strong>al</strong>ways thicker than the upper one. Hence.<br />
the bands that show the locus of po<strong>in</strong>ts of equ<strong>al</strong> thickness should be arranged<br />
horizont<strong>al</strong>ly. The light-blue (blue-green) t<strong>in</strong>t is obta<strong>in</strong>ed when the longwave<br />
(red-orange) part is excluded from the full spectrum (see Problem 812).<br />
When the midd Ie (green) part of the spectrum is ext<strong>in</strong>guished, the rema<strong>in</strong><strong>in</strong>g<br />
rays impart to the film a purple (crimscn) hue, and when the short-wave
438 ANSWERS AND SOLUTIONS<br />
(blue-viol<strong>et</strong>) faTt is excluded from the solid spectrum. the film will appear<br />
as yellow. I the difference <strong>in</strong> the path of the mutu<strong>al</strong>ly ext<strong>in</strong>guish<strong>in</strong>g rays<br />
forms the same number of h<strong>al</strong>f-waves <strong>in</strong> <strong>al</strong>l three cases, the yellow band<br />
should be on top, followed by the purple band and by the light ..blue band<br />
at the bottom. .<br />
814. In the daytime the light-blue light diffused by the sky is added to<br />
the yellowish light of the Moon itself. This mixture of colours is yerceived<br />
by the eye as a white colour. After suns<strong>et</strong> the Hght-blue colour 0 the sky<br />
is attenuated and the Moon acquires a yellowish hue.<br />
815. The smoke is seen aga<strong>in</strong>st a dark background because it diffuses the<br />
sunrays <strong>in</strong>cident from\ above. The particles of the smoke di ffuse blue ll ght<br />
much more <strong>in</strong>tensively than red or yellow light. Therefore, the smoke seems<br />
blue <strong>in</strong> colour.<br />
The smoke is seen <strong>in</strong> transmitted light aga<strong>in</strong>st the background of a bright<br />
sky. The smoke seems yellowish s<strong>in</strong>ce the blue light is diffused <strong>in</strong> <strong>al</strong>l directions<br />
and only the long-wave part of spectrum of white light reaches the eye.<br />
816. A th<strong>in</strong> film of water cover<strong>in</strong>g a moist object reflects the <strong>in</strong>cident<br />
white light <strong>in</strong> cne def<strong>in</strong>ite direction. The surface of the object no longer diffuses<br />
white light <strong>in</strong> <strong>al</strong>l directions, and its own colour becomes predom<strong>in</strong>ant.<br />
The diffused light is not superposed on the light reflected from the object,<br />
and for this reason the colour seems richer.
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