fisica1-youn-e-freedman-exercicios-resolvidos

5.38: a) There is no net force in the vertical direction, so n + F sin θ − w = 0,
or
n = w − F sinθ
= mg − F sin θ.
The friction force is f
(
k
= µ
kn
= µ
k
mg − F sin θ).
The net
horizontal force is F cosθ
− fk
= F cosθ
− µ
k
( mg − F sinθ)
, and so at constant speed,
µ
kmg
F =
cosθ
+ µ
k
sin θ
b) Using the given values,
2
(0.35)(90 kg)(9.80m s )
F =
= 293 N,
(cos25° + (0.35)sin 25°
)
or 290 N to two figures.
5.39: a)
b) The blocks move with constant speed, so there is no net force on block A; the
tension in the rope connecting A and B must be equal to the frictional force on block A,
µ
k
= (0.35) (25.0 N) = 9 N. c) The weight of block C will be the tension in the rope
connecting B and C; this is found by considering the forces on block B. The components
of force along the ramp are the tension in the first rope (9 N, from part (a)), the
component of the weight along the ramp, the friction on block B and the tension in the
second rope. Thus, the weight of block C is
w C
= 9 N + w B
(sin36.9° + µ
k
cos36.9°
)
= 9 N + (25.0 N)(sin 36.9° + (0.35)cos 36.9°
) = 31.0 N,
or 31 N to two figures. The intermediate calculation of the first tension may be avoided to
obtain the answer in terms of the common weight w of blocks A and B,
w C
= w( µ
k
+ (sinθ
+ µ
k
cosθ)),
giving the same result.
(d) Applying Newton’s Second Law to the remaining masses (B and C) gives:
2
a g(
w − µ w cosθ
− w sin θ)
w + w 1.54m s
( ) .
=
c k B
B
B c
=