Chapter 4: Force and Newton’s Laws of Motion
Physics
Chapter 4
FORCE AND NEWTON’S LAWS OF MOTION
Problems
1. Strategy Determine the forces not acting on the scale.
Solution The scale is in contact with the floor, so a contact force due to the floor is exerted on the scale. The
scale is in contact with the person’s feet, so a contact force due to the person’s feet is exerted on the scale. The
scale is in the proximity of a very large mass (Earth), so the weight of the scale is a force exerted on the scale. The
weight of the person is a force exerted on the person due to the very large mass, so it is not a force exerted on the
scale.
9. Strategy Graph the vectors and their sum. Use the scale of the graph to find the magnitude of the vector sum.
Solution Use graph paper, ruler, and protractor to find the
magnitude and direction of the vector sum of the two forces.
The vector sum points due north. Each side of a grid square
represents 10 N, so the magnitude of the net force on the
sledge is about 120 N north.
21. Strategy Use Newton’s second law for the vertical direction.
Solution Draw a free-body diagram. Find the mass of the potatoes.
T  mg 46.8 N  39.2 N
 Fy  T  mg  ma y , so m 

 4.0 kg .
ay
1.90 m s2
35. Strategy Consider forces acting on the fish suspended by the line.
Solution
One force acting on the fish is an upward force on the fish by the line; its interaction partner is a
downward force on the line by the fish. A second force acting on the fish is the downward gravitational
force on the fish; its interaction partner is the upward gravitational force on the Earth by the fish.
49. Strategy This is the same as asking, “At what altitude is the gravitational field strength half of its value at the
surface of the Earth?” g  GM R 2 , so let the new field strength be g  ng  GM r 2 where n  1 2.
Solution Determine r in terms of R.
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Chapter 4: Force and Newton’s Laws of Motion
Physics
GM
2
g  ng
r 2  R , so r  R .

 n  GM
g
g
n
r2
2
R
Find an expression for the altitude, h.
hrR 
 1

 1

R
 R  R
 1 , so h  (6.371103 km) 
 1  2639 km .
 12 
n
 n



61. Strategy Draw free-body diagrams for each situation. Let the subscripts be the following:
b = book
t = table
e = Earth
h = hand
Solution The diagrams are shown.
(a)
(b)
(c)
(d) Strategy and Solution In cases (a) and (b), the book is accelerating; so in these cases, the net force is not
zero.
(e) Strategy and Solution The normal force on the book is equal to its weight, (0.50 kg)(9.80 m s2 )  4.9 N.
The net force acting on the book in part (b) is equal to the force of kinetic friction. The force of kinetic
friction is opposite the direction of motion. The magnitude is k N  0.40(4.9 N)  2.0 N. Thus, the net force
on the book is 2.0 N opposite the direction of motion.
(f) Strategy and Solution The free-body diagram would look just like the diagram for part (c) and the book
would not slow down because there is no net force on the book (friction is zero).
73. Strategy Use Newton’s laws of motion. The lower cord supports only the lower box, whereas the upper cord
supports both boxes. Draw a diagram.
Solution Find the tension in each cord.
Lower cord
 Fx  Tl  ml g sin   0, so
Tl  ml g sin   (2.0 kg)(9.80 N kg)sin 25  8.3 N .
Upper cord
 Fx  Tu  mu g sin   Tl  0, so
Tu  mu g sin   Tl  (1.0 kg)(9.80 N kg)sin 25  8.3 N  12.4 N .
85. (a) Strategy The force required to start the block moving is that needed to overcome the maximum force of
static friction. Draw a diagram.
2
Chapter 4: Force and Newton’s Laws of Motion
Physics
Solution Find the applied horizontal force.
 Fx  F  fs  0, so F  fs  s N .
 Fy  N  Fg  N  mg  0, so N  mg.
So, the value of the applied force at the instant that the block
starts to slide is F  s mg  0.40(5.0 kg)(9.80 N kg)  20 N .
(b) Strategy The force required to keep the block moving is that needed to overcome kinetic friction. At the
instant the block starts to slide, the net force on the block is the difference between the forces required to
overcome static and kinetic friction.
Solution Calculate the net force.
 F  fs  fk  s mg  k mg  (s  k )mg  (0.40  0.15)(5.0 kg)(9.80 N kg)  12 N
101. Strategy The apparent weight is given by W   m( g  a y ).
Solution
(a) Up is the positive direction. Solve for a y .
 ay
W   m( g  a y )  mg 1 

g


 ay
  W 1 
g



W 
 120 lb 
 1  (9.80 m s2 ) 
 1  1.4 m s2 .
 , so a y  g 
W
140
lb





So, a is 1.4 m s 2 downward .
(b) With a downward acceleration, the elevator could be going up and slowing down, or going down and
speeding up, so the answer is no; one cannot tell whether the elevator is speeding up or slowing down.
113. Strategy Use Newton’s laws of motion.
Solution
(a) Since the airplane is cruising in a horizontal level flight (straight line) at constant velocity, it is in equilibrium
and the net force is zero.
(b) The air pushes upward with a force equal to the weight of the airplane: 2.6  104 N .
129. (a) Strategy Determine the maximum force of static friction and compare it to the force of the push, 5.0 N.
Solution Find the maximum force of static friction.
fs,max  s N  s mg Moon  0.35(2.0 kg)(9.80 m s 2 )  6.9 N  5.0 N
No, the puck does not move, since the maximum force of static friction is greater than the force of the push.
(b) Strategy and Solution Since 7.5 N > 6.9 N, the maximum force of static friction, yes, the puck does move.
(c) Strategy Use Newton’s second law.
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Chapter 4: Force and Newton’s Laws of Motion
Physics
Solution Find the acceleration of the puck.
 F  Fpush  f k  Fpush  k mg  ma, so
a
Fpush
m
 k g 
6.0 N
 0.25(9.80 m s 2 )  0.6 m s 2 .
2.0 kg
(d) Strategy Use Newton’s second law and the fact that gravity is weaker on the Moon.
Solution The acceleration of the puck is
Fpush
a
 k g.
m
Since gravity is weaker on the Moon, the second term on the right side of the equation is smaller than it
would be for Earth. This is the same thing as saying that the force of friction will be less on the Moon.
Therefore, the acceleration of the puck is more on the Moon than on Earth.
141. (a) Strategy Scale A measures the weight of both masses. Scale B only measures the weight of the 4.0-kg mass.
Solution Find the readings of the two scales if the masses of the scales are negligible.
Scale A  (10.0 kg  4.0 kg)(9.80 N kg)  137 N and Scale B  (4.0 kg)(9.80 N kg)  39 N .
(b) Strategy Scale A measures the weight of both masses and scale B. Scale B only measures the weight of
the 4.0-kg mass.
Solution Find the readings if each scale has a mass of 1.0 kg.
Scale A  (10.0 kg  4.0 kg  1.0 kg)(9.80 N kg)  147 N and Scale B  39 N .
153. Strategy Choose the +x-axis to the right and +y-axis up. Use Newton’s second law.
Solution
(a) For m1:
 F1 y  N  W1  N  m1 g  0, so N  m1 g.
 F1x  T  f k  T  k N  T  k m1 g  m1a1x
For m2 :
 F2 x  0 and  F2 y  T  W2  T  m2 g  m2a2 y .
Now, a1x and a2 y must be equal in magnitude, otherwise the cord will
compress or expand. a1x is in the +x-direction and a2 y is in the  y-direction.
So, let a  a1x  a2 y . Then, T  k m1 g  m1a and T  m2 g  m2 a.
Subtract the second equation from the first and solve for a.
 k m1 g  m2 g  m1a  m2 a
g (m2  k m1 )  a(m1  m2 )
a
m2  k m1
g
m1  m2
Find T.
4
Chapter 4: Force and Newton’s Laws of Motion
Physics
T  m2 g  m2 a
T  m2 g  m2
T  (1   k )
m2  k m1
m1  m2
g
m2 (m1  m2 )  m2 (m2  k m1 )
m1  m2
g
m1m2  m22  m22  m1m2 k
m1  m2
g
m1m2
g
m1  m2
(b) For m1  m2 ,
m2  k (0)
g  g and
0  m2
mm
T  (1  k ) 1 2 g  (1  k )m1 g  m2 g , so the tension is negligible compared to the weight of m2 ; .
0  m2
it’s essentially in free fall.
a
If m1  m2 , the force of friction between the table and m1 is so large that m1 will not slide, so a  0 and
T  m2 g  m2 a  0, or T  m2 g .
For m1  m2  m,
a
(c)
m  k m
m(1  k )
1
m2
1
g
g  (1  k ) g and T  (1  k )
g  (1  k )mg .
mm
2m
2
2m
2
a  0 only for m2  0; thus, there is no value at which the two masses slide with constant velocity. For
m2  0, there is no tension in the cord.
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