Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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CHAPTER 12 PROBLEMS PLUS 0 149
Equ!lting components, we have
JJ-,n- mg sin(}+ hm1n cosO= 0 ,=? hmin cos(}+ J-L •• n = mgsin(} (1)
n- mg cos()- hrnin sinfJ = 0 =? hmin sin(}+ mgcos(} = n (2)
(c) Since (2) is solved for n, we substitute into (1):
hmin cos() + JJ-~ (hmin sin(}+ mg cos 9) = mg sin 9 =?
hmin cos fJ + hmin/1-s sin fJ = mg sin fJ - mgJJ-. cos(}
'*
1
(sin8 - JJ-.• cos'(} ) . ( tau8 - /1-., )
tmin = mg cos(}+ 1-t. sin(} = mg 1 + JJ-. tau(}
() kn (} th
. b h ( tan8 - tanfJ8 ) d . . . . .
From part a we ow JJ-~ = tan ., so IS ecomes min = mg +tan (}, tau(} an usmg a tngonomctnc tdentlty,
1
this is mg tau(8- 8. ) as desired.
Note for fJ = 8., hmin = mgtau 0 = 0, which makes sense since the block is at rest for 8. , thus no additional force H
is necessary to prevent it from moving. As(} increases, the factor tau(8- 8.), and hence the value of hmin, increases
slowly for small values ofB- 8. but much more rapidly as 8 - e. becomes significant. This seems reasonable, as the
steeper the inclined ~lane, the less the hori zont~l components of the various forces affect the movement of the block, so we
would need a much larger magnitude of horizontal force to keep the block motionless. If we allow f) --+ 90°, corresponding
to the inclined plane being placed vertically, the value of hmin is quite large; this is to be expected, as it takes a great
amount of horizontal force to keep an object from moving vertically. In fact, without friction (so (} 8 = 0), we would have
(} --+ 90° :::? hmin --+ oo, and it would be impossible to keep the block from slipping.
(d) Since hmo.x is the largest value of h that keeps the block from slipping, the force of friction is keeping the block from ·
moving up the inclined plane; thus, F is directed down the plane. Our system of forces is similar to that in part {b), then,
except that we have F = -(JJ-.n) i. (Note that IFI is again maximal.) Following our procedure in parts (b) and (c), we
equate components:
- JJ-,n- mgsin(} + hmux cos(} = 0 :::? hme.x cos(}- /)- 8
n = mgsin8
n - mg cos (} - hm:l.X sin (} = 0
'* hmax sin (} + mg cos e = n
Then substituting,
hmax cos 8 - /1-s ( hmax sin 8 + mg cos B) = mg sin 8 '*
hmax cos(} -
hmn.x/1-s sin(} = mg sin(} + mgJJ-. cos e '*
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