fisica1-youn-e-freedman-exercicios-resolvidos

5.24: a) If there is no applied horizontal force, no friction force is needed to keep the
box in equilibrium. b) The maximum static friction force is, from Eq. (5.6),
µ
s
n = µ
sw = (0.40) (40.0 N) = 16.0 N, so the box will not move and the friction force
balances the applied force of 6.0 N. c) The maximum friction force found in part (b),
16.0 N. d) From Eq. (5.5), µ
k
n = (0.20)(40.0 N) = 8.0 N e) The applied force is enough
to either start the box moving or to keep it moving. The answer to part (d), from
Eq. (5.5), is independent of speed (as long as the box is moving), so the friction force is
2
8.0 N. The acceleration is ( F − f ) m 2.45 m s .
k
=
5.25: a) At constant speed, the net force is zero, and the magnitude of the applied force
must equal the magnitude of the kinetic friction force,
2
F r
= fk = µ
kn
= µ
kmg
= (0.12) (6.00 kg) (9.80 m s ) = 7 N.
b) F r
− f k
= ma,
so
F r
= ma + f = ma = µ mg = m(
a + µ g)
c) Replacing
k
= (6.00 kg)(0.180 m
2
g = 9.80 m s with
k
s
2
k
2
+ (0.12)9.80 m s ) = 8 N.
2
1 .62 m s gives 1.2 N and 2.2 N.
f
5.26: The coefficient of kinetic friction is the ratio
k
n
, and the normal force has
a
magnitude 85 N + 25 N = 110 N. The friction force, from F
H
− fk
= ma = w
g
is
2
a
⎛ − 0.9 m s ⎞
f
k
= FH
− w = 20 N − 85 N
⎜
=
2
9.80 m s
⎟
g
⎝ ⎠
28 N
(note that the acceleration is negative), and so µ = 0.25.
k
=
110 N
28 N
5.27: As in Example 5.17, the friction force is µ
kn = µ
kwcosα
and the component of the
weight down the skids is w sin α.
In this case, the angle α is arcsin( 2.00 20.0) = 5.7°
.
µ k cosα
µ k 0.25
The ratio of the forces is = = > 1,
so the friction force holds the safe back,
sin α tan α 0.10
and another force is needed to move the safe down the skids.
b) The difference between the downward component of gravity and the kinetic
friction force is
2
w (sin α − µ cos α)
= (260 kg) (9.80 m s ) (sin 5.7° − (0.25) cos 5.7°
) = −381
N.
k