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CHAPTER 4 FORCES AND NEWTON'S LAWS
OF MOTION
PROBLEMS
1.
REASONING AND SOLUTION According to Newton’s second law, the acceleration is
a = F/m. Since the pilot and the plane have the same acceleration, we can write
 F 
 F 




 m PILOT  m PLANE
F 

 m PLANE
 F PILOT  mPILOT 
or
Therefore, we find
 3.7  104 N 

F

78
kg
 PILOT 
 
  93 N
4
3.1

10
kg


_____________________________________________________________________________
6.
REASONING The time t required for the projectile to come up to a final speed v, starting
from an initial speed v0, can be obtained with the aid of Equation 2.4 of the equations of
kinematics, which is v  v0  at . To use this equation, however, we need a value for the
acceleration a. We can obtain this value with the aid of Newton’s second law.
SOLUTION Solving Equation 2.4 for the time t gives
t
v  v0
a
According to Newton’s second law, the net force is F = ma, which can be solved for the
acceleration to show that
F
a
m
Substituting the expression for a into the expression for t gives
t
v  v0
 F  / m

m  v  v0 
F

 5.0 kg   4.0

 103 m / s   0 m / s  
  0.041 s
5
4.9  10 N
where we have used the fact that the projection starts from rest so that t0 = 0 s.
_____________________________________________________________________________
11. REASONING According to Newton's second law (  F  ma ), the acceleration of the
object is given by a   F / m , where  F is the net force that acts on the object. We must
first find the net force that acts on the object, and then determine the acceleration using
Newton's second law.
SOLUTION The following table gives the x and y components of the two forces that act
on the object. The third row of that table gives the components of the net force.
Force
x-Component
y-Component
F1
40.0 N
0N
F2
(60.0 N) cos 45.0° = 42.4 N
(60.0 N) sin 45.0° = 42.4 N
 F  F1  F2
82.4 N
42.4 N
The magnitude of  F is given by the Pythagorean theorem as
F  (82.4 N)2  (42.4)2  92.7 N
The angle  that  F makes with the +x axis is
 42.4 N 
  27.2
 82.4 N 
  tan 1 
According to Newton's second law, the magnitude of the acceleration of the object is
a
 F 92.7 N

 30.9 m/s2
m
3.00 kg
Since Newton's second law is a vector equation, we know that the direction of the right hand
side must be equal to the direction of the left hand side. In other words, the direction of the
acceleration a is the same as the direction of the net force  F . Therefore, the direction of
the acceleration of the object is 27.2 above the +x axis .
____________________________________________________________________________________________
13.
SSM REASONING To determine the acceleration we will use Newton’s second law
F = ma. Two forces act on the rocket, the thrust T and the rocket’s weight W, which is mg
= (4.50 × 105 kg)(9.80 m/s2) = 4.41 × 106 N. Both of these forces must be considered when
determining the net force F. The direction of the acceleration is the same as the direction
of the net force.
SOLUTION In constructing the free-body diagram for the rocket we choose upward and to
the right as the positive directions. The free-body diagram is as follows:
Fx  T cos55.0
 7.50 106 N cos55.0  4.30 106 N

T
+y
The x component of the net force is
Ty = T sin 55.0º

55.0º
The y component of the net force is
W

+x
Tx = T cos 55.0º

Fy  T sin 55.0  W  7.50 106 N sin 55.0  4.41106 N  1.73 106 N
The magnitudes of the net force and of the acceleration are
F 
a
 Fx 
2
 
 Fy
2
 Fx 
2
 
 Fy
2
 4.30 106 N   1.73106 N 
2

m
2
4.50 10 kg
5
 10.3 m/s2
The direction of the acceleration is the same as the direction of the net force. Thus, it is
directed above the horizontal at an angle of
6
 Fy 
1  1.73  10 N 
  tan 
  21.9
6
 4.30 10 N 
 Fx 
  tan 1 
15. REASONING The acceleration of the sky diver can be obtained directly from Newton’s
second law as the net force divided by the sky diver’s mass. The net force is the vector sum
of the sky diver’s weight and the drag force.
SOLUTION From Newton’s second law, F  ma (Equation 4.1), the sky diver’s
acceleration is
a
F
m
The free-body diagram shows the two forces acting on the sky
diver, his weight W and the drag force f. The net force is
F  f  W . Thus, the acceleration can be written as
f
f W
m
+
The acceleration of the sky diver is

a
W
Free-body diagram
a
f  W 1027 N  915 N

 1.20 m/s2
m
93.4 kg
Note that the acceleration is positive, indicating that it points upward .
____________________________________________________________________________________________
16. REASONING Since there is only one force acting on the man in the horizontal direction, it
is the net force. According to Newton’s second law, Equation 4.1, the man must accelerate
under the action of this force. The factors that determine this acceleration are (1) the
magnitude and (2) the direction of the force exerted on the man, and (3) the mass of the
man.
When the woman exerts a force on the man, the man exerts a force of equal magnitude, but
opposite direction, on the woman (Newton’s third law). It is the only force acting on the
woman in the horizontal direction, so, as is the case with the man, she must accelerate. The
factors that determine her acceleration are (1) the magnitude and (2) the direction of the
force exerted on her, and (3) the her mass.
SOLUTION
a. The acceleration of the man is, according to Equation 4.1, equal to the net force acting on
him divided by his mass.
aman 
F 45 N

 0.55 m / s 2 (due east)
m 82 kg
b. The acceleration of the woman is equal to the net force acting on her divided by her mass.
F 45 N

 0.94 m / s 2 (due west)
m 48 kg
_____________________________________________________________________________
awoman 
20. REASONING The gravitational force acting on each object is specified by Newton’s law of
universal gravitation. The acceleration of each object when released can be determined with
the aid of Newton’s second law. We recognize that the gravitational force is the only force
acting on either object, so that it is the net force to use when applying the second law.
SOLUTION
a. The magnitude of the gravitational force exerted on the rock by the earth is given by
Equation 4.3 as
Frock 

Gmearth mrock

2
rearth
6.67  1011 N  m 2 / kg 2 5.98  10 24 kg 5.0 kg 

 6.38  10 m 
6

2
 49 N
The magnitude of the gravitational force exerted on the pebble by the earth is
Fpebble 

Gmearth mpebble

2
rearth
6.67  1011 N  m 2 / kg 2 5.98  1024 kg 3.0  10 4 kg

 6.38  106 m 

2

2.9  103 N
b. According to the second law, the magnitude of the acceleration of the rock is equal to the
gravitational force exerted on the rock divided by its mass.
arock 
Frock
mrock


Gmearth

2
rearth
6.67  1011 N  m2 / kg 2 5.98  1024 kg

2
 6.38  106 m 

9.80 m /s 2
According to the second law, the magnitude of the acceleration of the pebble is equal to the
gravitational force exerted on the pebble divided by its mass.
apebble 
Fpebble
mpebble


Gmearth

2
rearth
6.67  1011 N  m2 / kg 2 5.98  1024 kg

2
 6.38  106 m 

9.80 m /s 2
_____________________________________________________________________________
22. REASONING The magnitude of the gravitational force that each part exerts on the other is
given by Newton’s law of gravitation as F  Gm1m2 / r 2 . To use this expression, we need
the masses m1 and m2 of the parts, whereas the problem statement gives the weights W1 and
W2. However, the weight is related to the mass by W = mg, so that for each part we know
that m = W/g.
SOLUTION The gravitational force that each part exerts on the other is
F
Gm1m2
r2

G W1 / g W2 / g 
r2
6.67  10 –11 N  m 2 / kg 2  11 000 N  3400 N 



2
2 2
9.80 m/s  12 m 
1.8  10 –7 N
_____________________________________________________________________________
23. REASONING The earth exerts a gravitational force on the raindrop, and simultaneously the
raindrop exerts a gravitational force on the earth. This gravitational force is equal in
magnitude to the gravitational force that the earth exerts on the raindrop. The forces that the
raindrop and the earth exert on each other are Newton’s third law (action–reaction) forces.
Newton’s law of universal gravitation specifies the magnitude of both forces.
SOLUTION
a. The magnitude of the gravitational force exerted on the raindrop by the earth is given by
Equation 4.3:
Fraindrop 

Gmearth mraindrop

2
rearth
6.67  1011 N  m 2 / kg 2 5.98  1024 kg 5.2  10 7 kg

 6.38  10 m 
6

2

5.1  106 N
b. The magnitude of the gravitational force exerted on the earth by the raindrop is
Fearth 

Gmearth mraindrop

2
rearth
6.67  1011 N  m 2 / kg 2 5.98  10 24 kg 5.2  10 7 kg

 6.38  10 m 
6

2

5.1  106 N
_____________________________________________________________________________
24. REASONING Newton’s law of gravitation shows how the weight W of an object of mass m
is related to the mass M and radius r of the planet on which the object is located:
W = GMm/r2. In this expression G is the universal gravitational constant. Using the law of
gravitation, we can express the weight of the object on each planet, set the two weights
equal, and obtain the desired ratio.
SOLUTION According to Newton’s law of gravitation, we have
GM Am
rA2
Weight on planet A

GM Bm
rB2
Weight on planet B
The mass m of the object, being an intrinsic property, is the same on both planets and can be
eliminated algebraically from this equation. The universal gravitational constant can
likewise be eliminated algebraically. As a result, we find that
MA
rA2

rA

rB
27.
MB
rB2
M A rA2

M B rB2
or
MA
 0.60  0.77
MB
SSM REASONING AND SOLUTION According to Equations 4.4 and 4.5, the weight
of an object of mass m at a distance r from the center of the earth is
mg 
GM E m
r2
In a circular orbit that is 3.59 107 m above the surface of the earth ( radius = 6.38 106 m ,
mass  5.98 1024 kg ), the total distance from the center of the earth is
r  3.59 107 m + 6.38 106 m . Thus the acceleration g due to gravity is
g
GM E
r
2

(6.67 1011N  m 2 /kg 2 )(5.98 1024 kg)

(3.59 107 m + 6.38 106m) 2
0.223 m/s 2
____________________________________________________________________________________________
28. REASONING AND SOLUTION
The magnitude of the net force acting on the
moon is found by the Pythagorean theorem to
be
F
SM
Moon
F
Sun
2
2
F  FSM
 FEM
Newton's law of gravitation applied to the sun-moon (the units have been suppressed)
FSM 
GmSmM
2
rSM

 6.67 1011 1.99 1030  7.35 1022 
 4.34 1020 N
2
1.50 1011 
A similar application to the earth-moon gives
FEM 
GmE mM
2
rEM

 6.67 1011  5.98 1024  7.35 1022 
 1.98 1020 N
2
8
 3.85 10 
The net force on the moon is then
F
EM
Earth
F
 4.34  1020 N   1.98  1020 N 
2
2
 4.77  1020 N
____________________________________________________________________________________________
29. REASONING Each particle experiences two gravitational forces, one due to each of the
remaining particles. To get the net gravitational force, we must add the two contributions,
taking into account the directions. The magnitude of the gravitational force that any one
particle exerts on another is given by Newton’s law of gravitation as F  Gm1m2 / r 2 . Thus,
for particle A, we need to apply this law to its interaction with particle B and with particle
C. For particle B, we need to apply the law to its interaction with particle A and with
particle C. Lastly, for particle C, we must apply the law to its interaction with particle A
and with particle B. In considering the directions, we remember that the gravitational force
between two particles is always a force of attraction.
SOLUTION We begin by calculating the magnitude of the gravitational force for each pair
of particles:
FAB 
GmAmB
FBC 
GmB mC
FAC 
GmAmC
r2
r2
r2
6.67  10 –11 N  m 2 / kg 2   363 kg  517 kg 


 5.007  10 –5 N
 0.500 m 2
6.67  10 –11 N  m 2 / kg 2   517 kg 154 kg 


 8.497  10 –5 N
 0.500 m 2
6.67  10 –11 N  m 2 / kg 2   363 kg 154 kg 


 6.629  10 –6 N
 0.500 m 2
In using these magnitudes we take the direction to the right as positive.
a. Both particles B and C attract particle A to the right, the net force being
FA  FAB  FAC  5.007  10 –5 N  6.629  10 –6 N  5.67  10 –5 N, right
b. Particle C attracts particle B to the right, while particle A attracts particle B to the left, the
net force being
FB  FBC – FAB  8.497  10 –5 N – 5.007  10 –5 N  3.49  10 –5 N, right
c. Both particles A and B attract particle C to the left, the net force being
FC  FAC  FBC  6.629  10 –6 N  8.497  10 –5 N  9.16  10 –5 N, left
____________________________________________________________________________________________
30. REASONING The weight of a person on the earth is the gravitational force Fearth that it
exerts on the person. The magnitude of this force is given by Equation 4.3 as
Fearth  G
mearth mperson
2
rearth
where rearth is the distance from the center of the earth to the person. In a similar fashion, the
weight of the person on another planet is
Fplanet  G
mplanet mperson
2
rplanet
We will use these two expressions to obtain the weight of the traveler on the planet.
SOLUTION Dividing Fplanet by Fearth we have
Fplanet
Fearth
G

G
mplanet mperson
2
rplanet
mearth mperson
 mplanet
 
m
 earth
2
rearth
  rearth
 
  rplanet




2
or
Fplanet
Since we are given that
mplanet
mearth
 mplanet
 Fearth 
m
 earth
 3 and
rearth
rplanet

  rearth
 
  rplanet




2
1
, the weight of the space traveler on the
2
planet is
2
1
Fplanet   540.0 N  3   = 405.0 N
2
40. REASONING AND SOLUTION According to Equation 3.3b, the acceleration of the
astronaut is a y  (v y  v0 y ) / t  v y / t , where we have used the fact that v0 y  0 m / s since
the rocket blasts off from rest. The apparent weight and the true weight of the astronaut are
related according to Equation 4.6. Direct substitution gives
vy

FN  mg  ma y  m ( g  a y )  m  g 

t

True
Apparent
weight



weight
45 m/s 

2
 (57 kg)  9.80 m/s 2 
  7.3 10 N
15
s


____________________________________________________________________________________________
42. REASONING AND SOLUTION The apparent weight is
FN = mw(g + a)
We need to find the acceleration a. Let T represent the force applied by the hoisting cable.
Newton's second law applied to the elevator gives
T – (mw + me)g = (mw + me)a
Solving for a gives
a
T
9410 N
g 
 9.80 m/s2  0.954 m/s2
mw  me
60.0 kg  815 kg
Now the apparent weight is
FN = (60.0 kg)(9.80 m/s2 + 0.954 m/s2) = 645 N
____________________________________________________________________________________________
46. REASONING It is the static friction force that accelerates the cup when the plane
accelerates. The maximum possible magnitude of this force will determine the maximum
acceleration, according to Newton’s second law.
SOLUTION According to Newton’s second law and Equation 4.7 for the maximum static
frictional force, we have
F  fsMAX  s FN  s mg  ma
In this result, we have used the fact that the magnitude of the normal force is FN = mg, since
the plane is flying horizontally and the normal force acting on the cup balances the cup’s
weight. Solving for the acceleration a gives


a  s g   0.30 9.80 m/s2  2.9 m/s2
____________________________________________________________________________________________
47. REASONING The magnitude of the kinetic frictional force is given by Equation 4.8 as the
coefficient of kinetic friction times the magnitude of the normal force. Since the slide into
second base is horizontal, the normal force is vertical. It can be evaluated by noting that
there is no acceleration in the vertical direction and, therefore, the normal force must
balance the weight.
To find the player’s initial velocity v0, we will use kinematics. The time interval for the
slide into second base is given as t = 1.6 s. Since the player comes to rest at the end of the
slide, his final velocity is v = 0 m/s. The player’s acceleration a can be obtained from
Newton’s second law, since the net force is the kinetic frictional force, which is known from
part (a), and the mass is given. Since t, v, and a are known and we seek v0, the appropriate
kinematics equation is Equation 2.4 (v = v0 + at).
SOLUTION
a. Since the normal force FN balances the weight mg, we know that FN = mg. Using this
fact and Equation 4.8, we find that the magnitude of the kinetic frictional force is


f k  k FN  k mg   0.49 81 kg  9.8 m/s2  390 N
b. Solving Equation 2.4 (v = v0 + at) for v0 gives v0 = v  at. Taking the direction of the
player’s slide to be the positive direction, we use Newton’s second law and Equation 4.8 for
the kinetic frictional force to write the acceleration a as follows:
a
F k mg

  k g
m
m
The acceleration is negative, because it points opposite to the player’s velocity, since the
player slows down during the slide. Thus, we find for the initial velocity that


v0  v   k g  t  0 m/s    0.49 9.8 m/s2  1.6 s   7.7 m/s


52. REASONING The free-body diagram for the
helicopter is shown in the drawing. Since the velocity
is constant, the acceleration is zero and the helicopter is
at equilibrium. Therefore, according to Newton’s
second law, the net force acting on the helicopter is
zero.
SOLUTION Since the net force is zero, the
components of the net force in the vertical and
horizontal directions are separately zero. Referring to
the free-body diagram, we can see, then, that
L cos 21.0° – W = 0
L sin 21.0° – R = 0
a. Equation (1) gives
L
W
53 800 N

 57 600 N
cos 21.0 cos 21.0
(1)
(2)
b. Equation (2) gives
R  L sin 21.0   57 600 N  sin 21.0  20 600 N
____________________________________________________________________________________________
53.
SSM REASONING In order for the object to move with constant velocity, the net force
on the object must be zero. Therefore, the north/south component of the third force must be
equal in magnitude and opposite in direction to the 80.0 N force, while the east/west
component of the third force must be equal in magnitude and opposite in direction to the
60.0 N force. Therefore, the third force has components: 80.0 N due south and 60.0 N due
east. We can use the Pythagorean theorem and trigonometry to find the magnitude and
direction of this third force.
SOLUTION The magnitude of the third force is
F3  (80.0 N) 2  (60.0 N) 2  1.00 102 N
The direction of F3 is specified by the angle  where
 80.0 N 
  tan –1 
  53.1, south of east
 60.0 N 
____________________________________________________________________________________________
73.
SSM REASONING According to Newton’s second law, the acceleration has the same
direction as the net force and a magnitude given by a = F/m.
SOLUTION Since the two forces are perpendicular, the magnitude of the net force is given
by the Pythagorean theorem as F 
 40.0 N 2   60.0 N 2 . Thus, according to Newton’s
second law, the magnitude of the acceleration is
 40.0 N 2   60.0 N 2
F
a

m
4.00 kg
 18.0 m/s2
The direction of the acceleration vector is given by
 60.0 N 
  56.3 above the +x axis
 40.0 N 
  tan –1 
76. REASONING Newton’s second law can be used in both parts of this problem, because it
applies no matter what value the acceleration. In part (a) the acceleration is non-zero. In
part (b) the acceleration is zero, since the motion occurs at a constant acceleration.
SOLUTION
a. In the vertical direction, Newton’s second law is written as ΣFy = may. Thus, assuming
upward is the positive direction, denoting the cable tension by T, and the man’s weight by
W = mg, we have
T  mg  ma y
Fy
Solving this expression for T reveals that
 ay

 1.10 m / s 2

T  ma y  mg  mg   1   822 N  

1
  914 N
2
 g

9.80
m
/
s




b. The acceleration of the man is zero when his velocity is constant. Using the expression
obtained for the tension in part a, we obtain


T  ma y  mg  m 0 m / s2  mg  mg  822 N
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84. REASONING Newton’s second law, Equation 4.2a, can be used to find the tension in the
coupling between the cars, since the mass and acceleration are known. The tension in the
coupling between the 30th and 31st cars is responsible for providing the acceleration for the
20 cars from the 31st to the 50th car. The tension in the coupling between the 49th and 50th
cars is responsible only for pulling one car, the 50th.
SOLUTION
a. The tension T between the 30th and 31st cars is
Tx  (Mass of 20 cars) ax

(4.2a)


  20 cars  6.8  103 kg/car 8.0  10 2 m / s 2  1.1  10 4 N
b. The tension T between the 49th and 50th cars is
Tx  (Mass of 1 car) ax

(4.2a)


 1 car  6.8  103 kg/car 8.0  102 m / s 2  5.4  10 2 N
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97.
SSM REASONING AND SOLUTION
a. The apparent weight of the person is given by Equation 4.6 as
FN = mg + ma
= (95.0 kg)(9.80 m/s2 + 1.80 m/s2) = 1.10 × 103 N
b.
FN = (95.0 kg)(9.80 m/s2) = 931 N
c.
FN = (95.0 kg)(9.80 m/s2 – 1.30 m/s2) = 808 N
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