# Chapter 6 Applications of Newton's Laws Solutions to Problems

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```					                            Chapter 6: Applications of Newton’s Laws
Solutions to Problems
4.   Picture the Problem: The book slides in a straight line across the top of the tabletop.
Strategy: The minimum force required to get the book moving is related to the maximum coefficient of static friction,
and the force required to keep the book sliding at constant speed is equal to the magnitude of the kinetic friction force,
from which k can be determined.

Solution: 1. When the book begins sliding, the applied        F = f s = μs mg
app
force equals the maximum static friction force:
Fapp       2.25 N
μs =      =                     = 0.127
mg (1.80 kg )(9.81 m/s 2 )

2. When the book is sliding at constant speed, the            Fapp = f k = μ k mg
applied force equals the kinetic friction force:
Fapp         1.50 N
μk =        =                    = 0.0849
mg (1.80 kg )(9.81 m/s 2 )
Insight: The coefficient of kinetic friction is usually smaller than the coefficient of static friction. This is the basic idea
behind antilock brakes, which seek to keep the tire of a car rolling so that the friction between the tire and the road
remains in the static regime, where there is a greater force to stop the car and improved handling during the stop.
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43. Picture the Problem: Your car travels along a circular path at constant speed.
Strategy: Static friction between your tires and the road provides the centripetal force required to make the car travel
along a circular path. Set the static friction force equal to the centripetal force and calculate its value.
2
Solution: Set the static friction force equal to                          mv 2 (1200 kg )(15 m/s )
the centripetal force:                                 f s = Fcp = macp =     =                    = 4.7 kN
r          57 m
(          )
Insight: The maximum static friction force is μs mg = (0.88 )(1200 kg ) 9.81 m/s 2 = 10.4 kN which corresponds to a
maximum cornering speed (without skidding) of 22 m/s.
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44. Picture the Problem: The test tube travels along a circular path at constant speed.
Strategy: Solve equation 6-15 for the speed required to attain the desired acceleration.

Solution: Solve equation 6-15 for the speed:       v = racp = r (52, 000 g ) =         (0.075 m )(52, 000 )(9.81 m/s 2 )
= 200 m/s = 0.20 km/s
Insight: This speed corresponds to 25,000 revolutions per minute for the centrifuge, or 415 revolutions per second.
46. Picture the Problem: The forces acting on the car are depicted at
right.
Strategy: Write Newton’s Second Law in the horizontal and
vertical directions and combine them to obtain the radius of the
curve. The procedure is similar to Example 6-9 in the text.

Solution: 1. Write Newton’s                                    mv 2
Fx = N sin = macp =
Second Law in the x direction                                   r
and solve for r:
mv 2
r=
N sin
2. Write Newton’s Second            Fy = N cos         mg = 0
Law
in the y direction and solve              mg
for N:                               N=
cos
2
3. Substitute for N in the         mv 2 cos     v2                     (22.7 m/s )
equation from step 1:           r=          =                  =                               = 85.7 m
sin mg     g tan                (9.81 m/s )tan 31.5°
2

Insight: Note that the car follows the curved path without any help from friction at all. This is because the inward
component of the normal force is sufficient to provide the necessary centripetal force.
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47. Picture the Problem: The free-body diagram for Jill is shown at right.
Strategy: The center of Jill’s circular motion is the pivot point of the vine. There are two
forces acting on Jill, the tension due to the vine upward and gravity downward. These two                 r         r
forces add together to produce her centripetal acceleration in the upward direction. Write                T
Newton’s Second Law for Jill in the vertical direction and solve for T:
Solution: Write Newton’s Second                       Fy = T W = macp                                 r
v         r
Law for Jill in the vertical direction                                            2                             W
and solve for T:                                      T = W + macp = mg + m v r
2
(2.4 m/s )
= (61 kg ) 9.81 m/s + 2

6.5 m
T = 650 N = 0.65 kN
Insight: The tension in the vine will be at its maximum at the bottom of her circular path because it is at that point that
the vine must both support her weight and provide the upward centripetal force. Her speed is a maximum there as well,
making the centripetal force the largest at that point.
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67. Picture the Problem: The force vectors acting on blocks A and B
as well as the rope knot are shown in the diagram at right.
Strategy: Write Newton’s Second Law for blocks A and B, as well
as Newton’s Second Law for the rope knot. In all cases the
r
acceleration is zero. Combine the equations to solve for fs , which
acts on block A and points toward the left. Let the x direction point
to the right for block A and down for block B.
Solution: 1. (a)
Write Newton’s                          Fx =     f s + TA = 0
Second Law for                Block A
block A:
2. Write
Newton’s Second                         Fx = TB + mB g = 0
Law for block B:              Block B

3. Write                             Fx = TA + T3 cos 45° = 0
Newton’s Second               knot
Law for the rope
knot:                                Fy = TB + T3 sin 45° = 0
knot

4. Divide the y
T3 sin 45°                T
equation for the
knot by the x
= tan 45° = 1 = B
equation:                     T3 cos 45°                TA
5. Substitute TA = TB
into the equation             TA = mB g
from step 2:
6. Substitute the
result into the                f s = TA = mB g = (2.25 kg )(9.81 m/s 2 ) = 22.1 N = 22 N
equation from step 1:
7. (b) As long as mass A is heavy enough that
f s, max = μs mA g = (0.320 )(8.50 kg )(9.81 m/s 2 ) = 26.7 N        22.1 N, the friction force is not affected by changes
in mass A. It will stay the same if the mass of block A is doubled.
Insight: The minimum mass of block A that will satisfy the criteria of step (b) is 7.04 kg. The answer to (a) is reported
with only two significant figures because the angle 45° is only given to two significant figures.

71. Picture the Problem: The free-body diagram of the airplane is depicted at right.
Strategy: Let the x axis point horizontally from the airplane towards the center of
its circular motion, and let the y axis point straight upward. Write Newton’s
Second Law in both the horizontal and vertical directions and use the resulting
equations to find and the tension T.

Solution: 1. (a) Write Newton’s         Fx = T sin = macp = m v 2 r
Second Law in the x and y
directions:                             Fy = T cos              mg = 0
2. Solve the y equation for T and        T = mg cos
substitute the result into the x                mg
equation, and solve for :            T sin =              sin = m v 2 r
cos
tan = v 2 rg
2
1         (1.21 m/s )
= tan                                 = 19°
(0.44 m )(9.81 m/s 2 )
3. (c) Calculate the tension from
the equation in step 2:                  mg   (0.075 kg )(9.81 m/s           2
) = 0.78 N
T=     =
cos          cos19°
Insight: This airplane is pretty small. The toy weighs only 0.74 N = 2.6 ounces and flies in a circle of diameter 2.9 ft.
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73. Picture the Problem: The forces that are exerted on each box are depicted at right.
Strategy: Write Newton’s Second Law in the horizontal direction for the top box in
order to show that it is accelerated by the static friction force alone. Find the maximum
acceleration that the static friction force can produce, and then determine the force F that
will produce that acceleration for both boxes.

Solution: 1. (a) Write Newton’s
Second Law for the top box and
Fx = f s, max = μs mtop g = mtop a
solve for the maximum acceleration:                                           a = μs g
2. Write Newton’s Second Law                     Fx = F = (mtop + mbottom )a
for both boxes together, substitute
the maximum acceleration from                     F = (mtop + mbottom )μs g
step 1, and solve for F:
= (2.0 + 5.0 kg )(0.47 )(9.81 m/s 2 )
F = 32 N
3. (b) If mtop is increased, F will increase as well. The maximum acceleration remains the same but a larger force would
be needed to accelerate the larger mass at the same rate.
Insight: Note that the maximum acceleration is independent of the mass of the top block. A more massive top block
will increase the friction force, but it will also have more inertia and therefore require a larger force in order to
accelerate it.
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90. Picture the Problem: The hockey puck travels in a circle with the string
tension providing the necessary centripetal force.
Strategy: The tension in the string will be the same everywhere as long as
the lip of the tabletop is frictionless. From Newton’s Second Law on the
hanging mass M, the tension in the string equals T = Mg . Set the tension
equal to the centripetal force required to keep the hockey puck traveling in
a circle of radius r at constant speed v, then solve for v.
Solution: Set the string tension equal to the     T = Mg = macp = m v 2 r
centripetal force and solve for v:
Mrg
v=
m
Insight: Another way to look at the problem is to ask what mass M would
be required to keep m traveling in a circle of radius r at constant speed v,
and the answer is M = mv 2 rg

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