fisica1-youn-e-freedman-exercicios-resolvidos

5.92: (a) There is a contact force n between the man (mass M) and the platform (mass m).
The equation of motion for the man is T + n − Mg = Ma,
where T is the tension in the
rope, and for the platform, T − n − mg = ma . Adding to eliminate n, and rearranging,
T = 1 ( M + m)(
a + g ).
2
This result could be found directly by considering the manplatform
combination as a unit, with mass m + M , being pulled upward with a force 2T
due to the two ropes on the combination. The tension T in the rope is the same as the
force that the man applies to the rope. Numerically,
1 2
2
T = (70.0 kg + 25.0 kg)(1.80 m s + 9.80m s ) = 551 N.
2
(b) The end of the rope moves downward 2 m when the platform moves up 1 m, so
2
a = −2a
platform
. Relative to the man, the acceleration of the rope is 3a
= 5.40 m s ,
rope
downward.
5.93: a) The only horizontal force on the two-block combination is the horizontal
component of F r , F cosα.
The blocks will accelerate with a = F cosα ( m 1
+ m2
). b) The
normal force between the blocks is m
1g
+ Fsinα,
for the blocks to move together, the
product of this force and µ
s
must be greater than the horizontal force that the lower block
exerts on the upper block. That horizontal force is one of an action-reaction pair; the
reaction to this force accelerates the lower block. Thus, for the blocks to stay together,
m a ≤ µ m g + F sin ). Using the result of part (a),
( 2 s 1
α
Fcosα
m
2
≤ µ
s(
m1
g + F sinα).
m1
+ m2
Solving the inequality for F gives the desired result.