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Unformatted text preview: SSM REASONING AND SOLUTION The free-body diagram is shown at
the right. The forces that act on the picture are the pressing force P, the
normal force FN exerted on the picture by the wall, the weight mg of the
picture, and the force of static friction fsMAX . The maximum magnitude for MAX fS
FN P mg the frictional force is given by Equation 4.7:
= µs FN . The picture is in
equilibrium, and, if we take the directions to the right and up as positive, we
have in the x direction
fsMAX ∑ Fx = P − FN = 0 or P = FN ∑ Fy = fsMAX − mg = 0 or fsMAX = mg and in the y direction Therefore,
fsMAX = µs FN = mg But since FN = P , we have µs P = mg
Solving for P, we have P= mg µs = (1.10 kg)(9.80 m/s 2 )
= 16.3 N
0.660 ____________________________________________________________________________________________ 116. REASONING AND SOLUTION
a. The rope exerts a tension, T, acting upward on each block. Applying Newton's second
law to the lighter block (block 1) gives
T – m1g = m1a
Similarly, for the heavier block (block 2)
T – m2g = – m2a
Subtracting the second equation from the first and rearranging yields m – m1 a= 2
g = 3.68 m/s 2
m +m 2
1 b. The tension in the rope is now 908 N since the tension is the reaction to the applied force
exerted by the hand. Newton's second law applied to the block is Chapter 4 Problems 239 T – m1g = m1a
Solving for a gives
a= ( 908 N ) – 9.80 m/s 2 = 11.8 m/s 2
42.0 kg c. In the first case, the inertia of BOTH blocks affects the acceleration whereas, in the
second case, only the lighter block's inertia remains.
____________________________________________________________________________________________ 117. SSM REASONING AND SOLUTION The penguin comes to a halt on the horizontal
surface because the kinetic frictional force opposes the motion and causes it to slow down.
The time required for the penguin to slide to a halt (v = 0 m/s) after entering the horizontal
patch of ice is, according to Equation 2.4, t= v − v0
ax = −v0
ax We must, therefore, determine the acceleration of the penguin as it slides along the
horizontal patch (see the following drawing). θ fk1 FN1 FN2 mg sin θ fk2 mg cosθ mg Free-body diagram A Free-body diagram B For the penguin sliding on the horizontal patch of ice, we find from free-body diagram B
and Newton's second law in the x direction (motion to the right is taken as positive) that
∑ Fx = – f k2 = ma x or ax = – f k2
m = – µk FN2
m In the y direction in free-body diagram B, we have ∑ Fy = FN2 – mg = 0 , or FN2 = mg .
Therefore, the acceleration of the penguin is 240 FORCES AND NEWTON'S LAWS OF MOTION ax = – µk mg
m = – µk g (1) Equation (1) indicates that, in order to find the acceleration ax, we must find the coefficient
of kinetic friction.
We are told in the problem statement that the coefficient of kinetic friction between the
penguin and the ice is the same for the incline as for the horizontal patch. Therefore, we can
use the motion of the penguin on the incline to determine the coefficient of friction and use
it in Equation (1).
For the penguin sliding down the incline, we find from free-body diagram A (see the
previous drawing) and Newton's second law (taking the direction of motion as positive) that
∑ Fx = mg sin θ – f k1 = max = 0 f k1 = mg sin θ or (2) Here, we have used the fact that the penguin slides down the incline with a constant
velocity, so that it has zero acceleration. From Equation 4.8, we know that f k1 = µ k FN1 .
Applying Newton's second law in the direction perpendicular to the incline, we have
∑ Fy = FN1 – mg cos θ = 0 or FN1 = mg cos θ Therefore, f k1 = µk mg cosθ , so that according to Equation (2), we find
f k1 = µ k mg cos θ = mg sin θ
Solving for the coefficient of kinetic friction, we have µk = sin θ
= tan θ
cos θ Finally, the time required for the penguin to slide to a halt after entering the horizontal patch
of ice is
= 1.2 s
t= 0 =
– µk g g tan θ (9.80 m/s2 ) tan 6.9°
____________________________________________________________________________________________ Chapter 4 Problems 241 118. REASONING The following figure shows the crate on the incline and the free body
diagram for the crate. The diagram at the far right shows all the forces resolved into
components that are parallel and perpendicular to the surface of the incline. We can
analyze the motion of the crate using Newton's second law. The coefficient of friction can
be determined from the resulting equations.
P sin θ y FN x MAX P MAX FN fS fS P θ θ P cos θ θ mg sin θ mg mg cos θ SOLUTION Since the crate is at rest, it is in equilibrium and its acceleration is zero in all
directions. If we take the direction down the incline as positive, Newton's second law
∑ Fx = P cos θ + mg sin θ − fsMAX = 0
According to Equation 4.7, fsMAX = µ s FN . Therefore, we have
P cos θ + mg sin θ − µ s FN = 0 (1) The expression for the normal force can...
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