Thomas Calculus 13th [Solutions]

Section 6.5 Work and Fluid Forces 475
7
4
2
9.14
0
2
W ( y 17.5) (8 y) y in oz which is a Riemann sum. The work is the limit of these sums as the
norm of the partition goes to zero:
7
4 2 7
2 3
( 17.5) (8 )
4
2450 26.25 27
0
2 2
9 14 9 14 0
4 7
4
4 y 3 26.25 2 4 7 3 26.25 2
2
y y y
4 2 2
914 0 914 4 2
W y y dy y y y dy
9 2450 9 7 7 2450 7 91.32 in-oz
30. We fill the pipe and the tank. To find the work required to fill the tank note that radius = 10 ft, then
3
V 100 y ft . The force required will be F = 62.4 V = 62.4 100 y = 6240 y lb. The distance
through which F must act is y so the work done lifting the slab is about W1 6240 y y lb ft. The work it
385 385
takes to lift all the water into the tank is: W W 6240 y y lb ft. Taking the limit we end up
with W
1 1
360 360
385 2
y
385
6240 2 2
1 y dy
360
2 2
360
6240 6240 [385 360 ] 182,557,949 ft-lb
To find the work required to fill the pipe, do as above, but take the radius to be 4 in. 1 ft. Then
1 3
V y ft and 62.4
62.4
.
36
36
360
360
360
62.4 62.4 y 62.4 360
2 2 2
0 36 36 2 36 2
0
0
W W W y dy
2 6
F V y Also take different limits of summation and integration:
2 2
352,864 ft-lb
The total work is W W1 W 2 182,557,949 352,864 182,910,813 ft-lb. The time it takes to fill the tank
W 182,910,813
1650 1650
and the pipe is Time 110,855 sec 31 hr
35,780,000 1000 MG
35,780,000
dr
1
35,780,000
6,370,000 r
6,370,000 r
r 6,370,000
4 11 1 1
10
6,370,000 35,780,000
31. Work dr 1000 MG 1000 MG
2 2
(1000) 5.975 10 6.672 10 5.144 10 J
32. (a) Let be the x-coordinate of the second electron. Then
2 2
r ( 1)
0 ( ) 0 29
(23 10 ) 29
23 10 (23 10 29 ) 1 1
11.5 10 29
2
W F d d
1 1 ( 1) 1
1
2
0
(b) W W1 W 2 where W 1 is the work done against the field of the first electron and W 2 is the work done
2 2
against the field of the second electron. Let be the x-coordinate of the third electron. Then r 1 ( 1)
2 2
and r 2 ( 1)
1
5 29 29
5
23 10
5
23 10 29 29 29
23 10
1
( 23 10 )
1 1 23
10 ,
3
2 2
r
1 4 2 4
1
3 ( 1) 3
W2 5 29 29 5
29
23 10
5 29 29
d 23 10 d 23 10 1 ( 23 10 ) 1 1 23 10 (3
3
2 2
r
1 6 4 12
2
3 ( 1) 3
2)
23 29
10 .
12
29 29 29 29
Therefore W W 23 23 23
1 W2
10 10 10 7.67 10 J
4 12 3
W d d and
Copyright
2014 Pearson Education, Inc.