Download CHAPTER 4 FORCES AND NEWTON`S LAWS OF MOTION

CHAPTER 4 FORCES AND NEWTON'S LAWS
OF MOTION
PROBLEMS
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1.
REASONING AND SOLUTION According to Newton’s second law, the acceleration is
a = ΣF/m. Since the pilot and the plane have the same acceleration, we can write
 ΣF 
 ΣF 
=
 m 


 PILOT  m  PLANE
or
ΣF 

 m  PLANE
( ΣF ) PILOT = mPILOT 
Therefore, we find
 3.7 × 104 N 
( ΣF ) PILOT = ( 78kg ) 
 = 93N
4
 3.1 × 10 kg 
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2.
REASONING AND SOLUTION The acceleration of the bicycle/rider combination is
ΣF
9.78 N
=
= 0.103 m/s 2
m 13.1 kg + 81.7 kg
______________________________________________________________________________
a=
3.
SSM REASONING According to Newton's second law of motion, the net force applied to
the fist is equal to the mass of the fist multiplied by its acceleration. The data in the problem gives
the final velocity of the fist and the time it takes to acquire that velocity. The average acceleration
can be obtained directly from these data using the definition of average acceleration given in
Equation 2.4.
SOLUTION The magnitude of the average net force applied to the fist is, therefore,
∆v
8.0 m/s–0m/s 
∑ F = ma = m   = ( 0.70 kg ) 
 = 37 N
0.15 s
 ∆t 


______________________________________________________________________________
4.
REASONING AND SOLUTION Using Equation 2.4 and assuming that t 0 = 0 s, we have for
the required time that
v – v0
t=
a
Since ΣF = ma, it follows that
114 FORCES AND NEWTON'S LAWS OF MOTION
t=
v – v0
)= (5.0 kg )[(4.0 × 10 3
)
]=
m/s – (0 m/s )
4.1 × 10 –2 s
5
F
/
m
F
Σ
Σ
4.9 × 10 N
______________________________________________________________________________
5.
=
m (v – v 0
SSM REASONING The net force acting on the ball can be calculated using Newton's
second law. Before we can use Newton's second law, however, we must use Equation 2.9 from
the equations of kinematics to determine the acceleration of the ball.
SOLUTION According to Equation 2.9, the acceleration of the ball is given by
a=
v 2 − v20
2x
Thus, the magnitude of the net force on the ball is given by
 v 2 − v02 
 (45 m/s) 2 – (0 m/s) 2 
∑ F = ma = m
 = (0.058 kg) 
 = 130 N
2(0.44 m)


 2x 
______________________________________________________________________________
6.
REASONING AND SOLUTION The acceleration is obtained from
1 2
at
2
x = v 0t +
where v 0 = 0 m/s. So
a = 2x/t 2
Newton’s second law gives
 2 (18 m ) 
 2x 
ΣF = ma = m 2  = (72 kg )
2  = 2900 N
t 
 ( 0.95 s ) 
______________________________________________________________________________
7.
REASONING According to Newton’s second law, Equation 4.1, the average net force Σ F is
equal to the product of the object’s mass m and the average acceleration a . The average
acceleration is equal to the change in velocity divided by the elapsed time (Equation 2.4), where
the change in velocity is the final velocity v minus the initial velocity v 0.
SOLUTION The average net force exerted on the car and riders is
(
)
v − v0
45m/s − 0 m/s
= 5.5 × 103 kg
= 3.5 × 104 N
t − t0
7.0s
______________________________________________________________________________
∑ F = ma = m
Chapter 4 Problems
8.
115
REASONING AND SOLUTION From Equation 2.9,
v 2 = v 20 + 2ax
Since the arrow starts from rest, v0 = 0 m/s. In both cases, x is the same so
v12 2a1 x a1
=
=
v 22 2a 2 x a 2
v1
=
v2
or
a1
a2
Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and
v1
F
= 1
v2
F2
or
v 2 = v1
F2
F1
= v1
2 F1
F1
= ( 25.0 m/s ) 2 = 35.4 m/s
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9.
SSM WWW
REASONING Let due east be chosen as the positive direction. Then,
when both forces point due east, Newton's second law gives
FA + FB = ma1
1
424
3
(1)
ΣF
2
where a1 = 0.50 m/s . When FA points due east and FB points due west, Newton's second law
gives
FA – FB = ma2
1
424
3
(2)
ΣF
2
where a2 = 0.40 m/s . These two equations can be used to find the magnitude of each force.
SOLUTION
a. Adding Equations 1 and 2 gives
FA =
m ( a1 + a 2 )
2
=
(8.0kg ) (0.50m/s 2 + 0.40m/s 2 )
b. Subtracting Equation 2 from Equation 1 gives
2
= 3.6 N
116 FORCES AND NEWTON'S LAWS OF MOTION
FB =
m ( a1 − a 2 )
2
=
(8.0kg ) ( 0.50m/s 2 − 0 . 4 0 m / s2 )
2
= 0.40N
______________________________________________________________________________
10. REASONING AND SOLUTION
FE = F cos θ = (720 N) cos 38° = 570 N
FN = F sin θ = (720 N) sin 38° = 440 N
______________________________________________________________________________
11. REASONING According to Newton’s second law, the acceleration has the same direction as
the net force and a magnitude given by a = ΣF/m.
SOLUTION Since the two forces are perpendicular, the magnitude of the net force is given by
the Pythagorean theorem as Σ F = ( 40.0 N ) 2 + (60.0 N ) 2 . Thus, according to Newton’s
second law, the magnitude of the acceleration is
ΣF
a=
=
m
( 40.0 N ) 2 + (60.0 N ) 2
4.00 kg
= 18.0 m/s 2
The direction of the acceleration vector is given by
 60.0 N 
θ = tan –1 
= 56.3° above the + x axis
 40.0 N 
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12. REASONING AND SOLUTION The free
body diagram is shown at the right, where
F1 = 45.0 N
F2 = 25.0 N
θ = 65.0°
When F1 is replaced by its x and y
components, we obtain the free body
diagram in the second drawing.
FN
F2
θ
W
F1
Chapter 4 Problems
If we choose the right as the positive
direction, then
FN
F1 cos θ – F2 = max
ax =
or
ax =
5.00 kg
F2
F1 cos θ − F2
m
( 45.0 N ) cos65.0° – (25.0 N )
117
F1 cos θ
W
F1 sin θ
= –1.20 m/s 2
The minus sign indicates that the acceleration
is to the left .
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___
13.
SSM 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 θ that ∑ F makes with the +x axis is
 42.4 N  27.2
θ = tan −1 
=
°
82.4 N 
ΣF
θ
82.4 N
According to Newton's second law, the magnitude of the acceleration of the object is
42.4 N
118 FORCES AND NEWTON'S LAWS OF MOTION
a=
∑ F 92.7 N
=
= 30.9 m/s 2
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 .
_________________________________________________________________________________________
___
1
1




14. REASONING Equations 3.5a  x = v 0 x t + a x t 2  and 3.5b  y = v 0 y t + a y t 2  give the




2
2
displacements of an object under the influence of constant accelerations ax and ay. We can add
these displacements as vectors to find the magnitude and direction of the resultant displacement.
To use Equations 3.5a and 3.5b, however, we must have values for ax and ay. We can obtain
these values from Newton’s second law, provided that we combine the given forces to calculate
the x and y components of the net force acting on the duck, and it is here that our solution begins.
SOLUTION Let the directions due east and due north, respectively, be the +x and +y directions.
Then, the components of the net force are
Σ Fx = 0.10 N + (0.20 N)cos 52 ° = 0.2231 N
Σ Fy = – ( 0.20 N ) sin 52 ° = –0.1576 N
According to Newton’s second law, the components of the acceleration are
ax =
ay =
ΣFx 0.2231 N
=
= 0.08924 m/s 2
m
2.5 kg
ΣFy
m
=
–0.1576 N
= –0.06304 m/s 2
2.5 kg
From Equations 3.5a and 3.5b, we now obtain the displacements in the x and y directions:
x = v 0 x t + a x t 2 = ( 0.11 m/s )( 3.0 s ) +
1
2
y = v 0 y t + 1 a y t 2 = (0 m/s )( 3.0 s ) +
2
0.08924 m/s 2 )(3.0 s )2 = 0.7316 m
(
2
1
–0.06304 m/s 2 )(3.0 s )2 = –0.2837 m
(
2
1
The magnitude of the resultant displacement is
Chapter 4 Problems
r=
119
2
2
x 2 + y 2 = ( 0.7316 m ) + ( –0.2837 m ) = 0.78 m
The direction of the resultant displacement is
 0.2837 m 
θ = tan –1 
= 21° south of east
 0.7316 m 
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___
15. REASONING AND SOLUTION Combining Newton's second law with v = v 0 + at, we have
 ( –2.0 m/s ) – ( +2.1 m/s ) 
 v – v0 
Σ F = ma = m
 = ( 0.38 kg )
 = –470 N
 t 


3.3 × 10 –3 s
The answer is negative, indicating that the force is directed away from the cushion.
_________________________________________________________________________________________
___
1


16. REASONING For both the tug and the asteroid, Equation 2.8  x = v 0 t + at 2  applies with


2
v 0 = 0 m/s, since both are initially at rest. In applying this equation, we must be careful and use the
proper acceleration for each object. Newton’s second law indicates that the acceleration is given
by a = ΣF/m. In this expression, we note that the magnitudes of the net forces acting on the tug
and the asteroid are the same, according to Newton’s action-reaction law. The masses of the tug
and the asteroid are different, however. Thus, the distance traveled for either object is given by,
where we use for ΣF only the magnitude of the pulling force
1
1 ΣF  2
x = v 0 t + at 2 = 
t
2
2 m 
SOLUTION Let L be the initial distance between the tug and the asteroid. When the two
objects meet, the distances that each has traveled must add up to equal L. Therefore,
1
1
2
2
L = x T + x A = aT t 2 + a A t 2
 1
1  ΣF  2
1  ΣF  2
1
1  2
L= 
+
t + 
 t = ΣF 
t
2  mT 
2  mA 
2
 mT m A 
Solving for the time t gives
120 FORCES AND NEWTON'S LAWS OF MOTION
t=
2L
 1
1 
ΣF
+

 mT m A 
=
2 ( 450 m )


1
1
+
( 490 N ) 

 3500 kg 6200 kg 
= 64 s
_________________________________________________________________________________________
___
17.
SSM WWW REASONING We first determine the acceleration of the boat. Then, using
Newton's second law, we can find the net force ∑ F that acts on the boat. Since two of the three
forces are known, we can solve for the unknown force FW once the net force ∑ F is known.
SOLUTION Let the direction due east be the positive x direction and the direction due north be
the positive y direction. The x and y components of the initial velocity of the boat are then
v 0 x = ( 2.00 m/s) cos 15.0 ° = 1.93 m/s
v 0 y = (2.00 m/s) sin 15.0 ° = 0.518 m/s
Thirty seconds later, the x and y velocity components of the boat are
v x = ( 4.00 m/s) cos 35.0 ° = 3.28 m/s
v y = ( 4.00 m/s) sin 35.0 ° = 2.29 m/s
Therefore, according to Equations 3.3a and 3.3b, the x and y components of the acceleration of
the boat are
v −v
3.28 m/s – 1.93 m/s
a x = x 0x =
= 4.50 × 10–2 m/s 2
t
30.0 s
v − v0 y 2.29 m/s – 0.518 m/s
ay = y
=
= 5.91 × 10–2 m/s 2
t
30.0 s
Thus, the x and y components of the net force that act on the boat are
∑ Fx = max = (325 kg) (4.50 × 10 –2 m/s 2 ) = 14.6 N
∑ Fy = may = (325 kg) (5.91 ×10–2 m/s 2 ) =19.2 N
The following table gives the x and y components of the net force ∑ F and the two known forces
that act on the boat. The fourth row of that table gives the components of the unknown force
FW .
Chapter 4 Problems
121
Force
x-Component
y-Component
∑F
14.6 N
19.2 N
F1
(31.0 N) cos 15.0° = 29.9 N
(31.0 N) sin 15.0° = 8.02 N
F2
–(23.0 N ) cos 15.0° = –22.2 N
–(23.0 N) sin 15.0° = –5.95 N
FW = ∑ F − F1 − F2
14.6 N – 29.9 N + 22.2 N = 6.9 N
19.2 N – 8.02 N + 5.95 N = 17.1 N
The magnitude of FW is given by the Pythagorean theorem as
FW = (6.9 N) 2 + (17.1 N) 2 = 18.4 N
The angle θ that FW makes with the x axis is
 17.1 N
θ = tan −1 
= 68°
6.9 N 
17.1 N
θ
Therefore, the direction of FW is 68°, north of east .
6.9 N
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___
18. REASONING AND SOLUTION Newton's law of gravitation gives
(
)(
)(
)
6.67 × 10 –11 N ⋅ m 2 / kg 2 1.50 × 10−3 kg 870.0 × 10− 3 kg
G m1 m2
F=
=
= 8.70 × 10 –12 N
2
r2
( 0.100 m )
_________________________________________________________________________________________
___
19.
SSM REASONING AND SOLUTION
a. According to Equation 4.5, the weight of the space traveler of mass m = 115 kg on earth
is
W = mg = (115 kg) (9.80 m/s 2 ) = 1.13 × 10 3 N
b. In interplanetary space where there are no nearby planetary objects, the gravitational force
exerted on the space traveler is zero and g = 0 m/s2. Therefore, the weight is W = 0 N . Since
the mass of an object is an intrinsic property of the object and is independent of its location in the
universe, the mass of the space traveler is still m =115 kg .
122 FORCES AND NEWTON'S LAWS OF MOTION
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___
20. REASONING AND SOLUTION The forces that act on the rock are shown
at the right. Newton's second law (with the direction of motion as positive) is
R
Σ F = mg – R = ma
Solving for the acceleration a gives
(
)
2
mg – R ( 45kg ) 9.80 m/s – ( 250 N)
a=
=
= 4.2 m/s 2
m
45kg
mg
_________________________________________________________________________________________
___
21. REASONING The magnitude of the gravitational force that each part exerts on the other is given
by Newton’s law of gravitation as F = G m1m2 / 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=
G m1 m2 G ( W1 / g ) ( W2 / g )
=
r2
r2
( 6.67 × 10
–11
=
)
N ⋅ m2 / kg 2 (11000 N) ( 3400 N)
( 9.80 m/s ) (12 m )
2 2
2
= 1.8 × 10–7 N
_________________________________________________________________________________________
___
22. 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 = G m1m2 / 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:
Chapter 4 Problems
FAB =
Gm A m B
FBC =
Gm B mC
FAC =
Gm A mC
123
6.67 × 10 –11 N ⋅ m 2 / kg 2 )(363 kg )(517 kg )
(
=
= 5.007 × 10 –5 N
( 0.500 m ) 2
r2
6.67 × 10 –11 N ⋅ m 2 / kg 2 )(517 kg )(154 kg )
(
=
= 8.497 × 10 –5 N
(0.500 m ) 2
r2
6.67 × 10 –11 N ⋅ m 2 / kg 2 )(363 kg )(154 kg )
(
=
= 6.629 × 10 –6 N
(0.500 m )2
r2
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
_________________________________________________________________________________________
___
23.
SSM REASONING AND SOLUTION
a. Combining Equations 4.4 and 4.5, we see that the acceleration due to gravity on the surface of
Saturn can be calculated as follows:
g Saturn = G
M Saturn
2
r Saturn
(
= 6.67 × 10
–11
N ⋅ m /kg
2
2
(5.67 × 10 26 kg ) =
)(6.00 × 10 7 m) 2
10.5 m/s 2
b. The ratio of the person’s weight on Saturn to that on earth is
WSaturn mg Saturn g Saturn 10.5 m/s 2
=
=
=
= 1.07
Wearth
mg earth
gearth
9.80 m/s 2
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___
124 FORCES AND NEWTON'S LAWS OF MOTION
24. 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
F
Sun
2
F=
EM
2
F SM + FEM
Earth
Newton's law of gravitation applied to the sun-moon (the units have been suppressed)
m m
F
SM
= G
S M
r2
SM
30
-11
=(6.67 x 10
)
22
(1.99 x 10 )(7.35 x 10 )
11
(1.50 x 10
m)
2
20
= 4.34 x 10 N.
A similar application to the earth-moon gives
m m
F
= G
EM
E
r
M
2
-11
= (6.67 x 10
24
)
22
(5.98 x 10 )(7.35 x 10 )
8
2
20
= 1.98 x 10 N.
(3.85 x 10 m)
EM
The net force on the moon is then
F=
(4.34 × 10 20 N )2 + (1.98 × 10 20 N )2 =
4.77 × 10 20 N
_________________________________________________________________________________________
___
25. REASONING According to Equation 4.4, the weights of an object of mass m on the surfaces of
planet A (mass = MA, radius = R ) and planet B (mass = MB , radius = R ) are
WA =
GM A m
and
R2
WB =
GM B m
R2
The difference between these weights is given in the problem.
SOLUTION The difference in weights is
W A – WB =
Rearranging this result, we find
GM A m
R2
–
GM B m
R2
=
Gm
(M A – M B )
R2
Chapter 4 Problems
MA – M B
W A – WB )R 2
(
=
=
Gm
(3620 N )(1.33 × 10 7 m)
125
2
(6.67 × 10
−11
N ⋅ m /kg ) (5450 kg )
2
2
= 1.76 × 10 24 kg
_________________________________________________________________________________________
___
26. REASONING The acceleration due to gravity at the surface of a planet is given by Equations
4.4 and 4.5 as g Planet = GM / r 2 , where M is the mass of the planet and r is its radius. By applying
this relation to the unknown planet and to the earth, and noting the ratios of the masses and radii,
we will be able to find the acceleration due to gravity on the surface of the planet.
SOLUTION Taking the ratio of gPlanet to gEarth, and noting that MPlanet/MEarth = 0.10 and
rEarth/rPlanet = 1/0.50, we have that
GM Planet
2
2
2
 M Planet   rEarth 
rPlanet
1 

=
= 
 = ( 0.10) 

 = 0.40
GM Earth
 0.50 
 M Earth   rPlanet 
2
rEarth
gPlanet
gEarth
Therefore, the acceleration due to gravity on the planet is
(
g Planet = ( 0.40 ) g Earth = ( 0.40 ) 9.80m/s 2
)
=
3 . 9 m / s2
_________________________________________________________________________________________
___
27.
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 =
GME m
r
2
In a circular orbit that is 3.59 × 10 7 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 × 10 7 m + 6.38 × 106 m . Thus the acceleration g due to gravity is
g=
GM E
r
2
=
(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(5.98 × 10 24 kg)
(3.59 ×
10 7 m
+ 6.38 ×
10 6
m)
2
=
0.223 m/s 2
_________________________________________________________________________________________
___
126 FORCES AND NEWTON'S LAWS OF MOTION
28. REASONING AND SOLUTION The figure at the right
shows the three spheres with sphere 3 being the sphere of
unknown mass. Sphere 3 feels a force F31 due to the
presence of sphere 1, and a force F32 due to the presence
of sphere 2. The net force on sphere 3 is the resultant of
F31 and F32.
Note that since the spheres form an equilateral triangle,
each interior angle is 60°. Therefore, both F31 and F32
make a 30° angle with the vertical line as shown.
3
F
F
31
32
1.20 m
30°
2
1
Furthermore, F31 and F32 have the same magnitude
given by
F=
GMm 3
r
2
where M is the mass of either sphere 1 or 2 and m3 is the mass of sphere 3. The components of
the two forces are shown in the following drawings:
F
31
F cos θ
30.0°
F cos θ
F sin θ
F
32
30.0°
F sin θ
Clearly, the horizontal components of the two forces add to zero. Thus, the net force on sphere 3
is the resultant of the vertical components of F31 and F32:
F3 = 2F cos θ = 2
GMm 3
r
2
cos θ.
The acceleration of sphere 3 is given by Newton's second law:
F3
(6.67 × 10 −11 N ⋅ m 2 /kg 2 ) (2.80 kg )
GM
a3 =
= 2 2 cosθ = 2
cos 30.0°
m3
r
( 1.20 m )2
= 2.25 × 10 –10 m/s 2
_________________________________________________________________________________________
___
Chapter 4 Problems
127
29. REASONING According to Equation 4.4, the weights of an object of mass m on the surface of
a planet (mass = M, radius = R ) and at a height H above the surface are
GMm
W=
2
142R
43
On surface
and
GMm
WH =
(R + H )2
1442443
At height H above surface
The fact that W is one percent less than WH tells us that the WH/W = 0.9900, which is the starting
point for our solution.
SOLUTION The ratio WH/W is
GMm
WH
(R + H )2
R2
1
=
=
=
= 0.9900
2
GMm
W
( R + H ) (1 + H / R ) 2
R2
Solving for H/R gives
H
1
H
1+ =
or
= 0.0050
R
0.9900
R
_________________________________________________________________________________________
___
30. REASONING The gravitational force that the sun exerts on a person standing on the earth is
2
given by Equation 4.3 as Fsun = GM sun m / rsun-earth
, where Msun is the mass of the sun, m is the
mass of the person, and rsun-earth is the distance from the sun to the earth. Likewise, the
gravitational force that the moon exerts on a person standing on the earth is given by
2
Fmoon = GM moon m / rmoon-earth
, where Mmoon is the mass of the moon and rmoon-earth is the
distance from the moon to the earth. These relations will allow us to determine whether the sun or
the moon exerts the greater gravitational force on the person.
SOLUTION Taking the ratio of Fsun to Fmoon, and using the mass and distance data found on
the inside of the front cover of the text, we have that
Fsun
Fmoon
GM sun m
2
 M
rsun-earth
=
=  sun
GM moon m
 M moon
2
rmoon-earth
2
  rmoon-earth 


  rsun-earth 
 1.99 × 1030 kg  3.85 × 108 m 
= 


22
11
 7.35 × 10 kg  1.50 × 10 m 
2
= 178
128 FORCES AND NEWTON'S LAWS OF MOTION
Therefore, the sun exerts the greater gravitational force.
_________________________________________________________________________________________
___
31.
SSM WWW
REASONING AND SOLUTION There are two forces that act on the
balloon; they are, the combined weight of the balloon and its load, Mg, and the upward buoyant
force FB . If we take upward as the positive direction, then, initially when the balloon is
motionless, Newton's second law gives FB − Mg = 0 . If an amount of mass m is dropped
overboard so that the balloon has an upward acceleration, Newton's second law for this situation
is
FB − (M − m)g = ( M − m)a
But FB = mg , so that
Mg – ( M − m ) g = mg = ( M – m )a
Solving for the mass m that should be dropped overboard, we obtain
m=
Ma
(310 kg )(0.15 m/s 2 )
=
= 4.7 kg
g + a 9.80 m/s 2 + 0.15 m/s 2
_________________________________________________________________________________________
___
32. REASONING AND SOLUTION The acceleration due to gravity at the surface of the neutron
star is
a=
Gm (6.67 ×10 −11 N ⋅ m 2 /kg 2 )(2.0 ×1030 kg)
=
= 5.3 ×1012 m/s2
2
3
2
r
(5.0 × 10 m)
Since the gravitational force is assumed to be constant, the acceleration will be constant and the
speed of the object can be calculated from v 2 = v02 + 2ay , with v0 = 0m/s since the object falls
from rest. Solving for v yields
(
)
v = 2 ay = 2 5.3 × 10 12 m/s 2 (0.010 m ) = 3.3 × 10 5 m/s
_________________________________________________________________________________________
___
33. REASONING We place the third particle (mass = m3) as shown in the following drawing:
L
D
m
m3
2m
Chapter 4 Problems
129
The magnitude of the gravitational force that one particle exerts on another is given by Newton’s
law of gravitation as F = Gm1m2/r2. Before the third particle is in place, this law indicates that the
force on each particle has a magnitude Fbefore = Gm2m/L2. After the third particle is in place,
each of the first two particles experiences a greater net force, because the third particle also exerts
a gravitational force on them.
SOLUTION For the particle of mass m, we have
Gmm3
Fafter
=
Fbefore
Gm 2m
2
D
L2 = L m3 + 1
Gm2 m
2mD 2
2
L
2
+
For the particle of mass 2m, we have
G2 mm3
+
Gm2m
L
Fafter
( L – D)
=
Gm2 m
Fbefore
L2
2
2
=
L2 m3
m( L – D)
2
+1
Since Fafter/Fbefore = 2 for both particles, we have
L2 m 3
2mD
2
+1 =
L2 m3
m (L – D )
2
+1
or
2
2 D2 = ( L – D)
Expanding and rearranging this result gives D2 + 2 LD − L2 = 0 , which can be solved for D using
the quadratic formula:
D=
–2 L ±
( 2 L )2 – 4(1) ( – L2 )
=
2 (1)
0.414 L or –2.414 L
The negative solution is discarded because the third particle lies on the +x axis between m and 2m.
Thus, D = 0.414 L .
_________________________________________________________________________________________
___
34. REASONING
In each case the object is in equilibrium. According to Equation 4.9b,
ΣFy = 0, the net force acting in the y (vertical) direction must be zero. The net force is composed
of the weight of the object(s) and the normal force exerted on them.
130 FORCES AND NEWTON'S LAWS OF MOTION
SOLUTION
a. There are three vertical forces acting on the crate: an upward normal force +FN that the floor
exerts, the weight –m1g of the crate, and the weight –m2g of the person standing on the crate.
Since the weights act downward, they are assigned negative numbers. Setting the sum of these
forces equal to zero gives
FN + ( −m1g ) + ( −m2 g ) = 0
14444
4244444
3
Σ Fy
The magnitude of the normal force is
FN = m1g + m2g = (35 kg + 65 kg)(9.80 m/s2) = 9 8 0 N
b. There are only two vertical forces acting on the person: an upward normal force +FN that the
crate exerts and the weight –m2g of the person. Setting the sum of these forces equal to zero gives
FN + ( −m 2 g ) = 0
144244
3
Σ Fy
The magnitude of the normal force is
FN = m2g = (65 kg)(9.80 m/s2) = 6 4 0 N
_________________________________________________________________________________________
___
35.
SSM REASONING AND SOLUTION According to Equation 3.3b, the acceleration of
the astronaut is a y = (vy − v0y ) / t = vy /t . The apparent weight and the true weight of the
astronaut are related according to Equation 4.6. Direct substitution gives
vy 

F
=
mg
+
ma
=
m
(
g
+
a
)
=
m
g
+


N
y
y
{ {
t 

Apparent
True
weight
weight
45 m/s 

2
= (57 kg)  9.80 m/s 2 +
 = 7.3 × 10 N
15
s


_________________________________________________________________________________________
___
36. REASONING AND SOLUTION
a. The apparent weight of the person is
FN = mg + ma
= (95.0 kg)(9.80 m/s2 + 1.80 m/s2) = 1.10 X 103 N
Chapter 4 Problems
FN = (95.0 kg)(9.80 m/s2) = 931 N
b.
c.
131
FN = (95.0 kg)(9.80 m/s2 – 1.30 m/s2) = 808 N
_________________________________________________________________________________________
___
37. REASONING AND SOLUTION The block will move only if the applied force is greater than
the maximum static frictional force acting on the block. That is, if
F > µsFN = µsmg = (0.650)(45.0 N) = 29.2 N
The applied force is given to be F = 36.0 N which is greater than the maximum static frictional
force, so the block will move .
The block's acceleration is found from Newton's second law.
a=
ΣF F – f k F – µk mg
=
=
= 3.72 m/s 2
m
m
m
_________________________________________________________________________________________
___
38. REASONING It is the static friction force that accelerates the cup when the plane accelerates.
The maximum magnitude that this force can have will determine the maximum acceleration,
according to Newton’s second law.
SOLUTION According to Newton’s second law, we have
ΣF = f sMAX = µs FN = µs 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 gives
(
)
a = µs g = ( 0.30 ) 9.80 m/s 2 = 2.9 m/s 2
_________________________________________________________________________________________
___
39.
SSM REASONING In order to start the crate moving, an external agent must supply a force
that is at least as large as the maximum value fsMAX = µs FN , where µs is the coefficient of static
friction (see Equation 4.7). Once the crate is moving, the magnitude of the frictional force is very
nearly constant at the value f k = µk FN , where µk is the coefficient of kinetic friction (see
132 FORCES AND NEWTON'S LAWS OF MOTION
Equation 4.8). In both cases described in the problem statement, there are only two vertical forces
that act on the crate; they are the upward normal force FN , and the downward pull of gravity (the
weight) mg. Furthermore, the crate has no vertical acceleration in either case. Therefore, if we
take upward as the positive direction, Newton's second law in the vertical direction gives
FN − mg = 0 , and we see that, in both cases, the magnitude of the normal force is FN = mg .
SOLUTION
a. Therefore, the applied force needed to start the crate moving is
fsMAX = µs mg = (0.760)(60.0 kg)(9.80 m/s 2 ) = 447 N
b. When the crate moves in a straight line at constant speed, its velocity does not change, and it
has zero acceleration. Thus, Newton's second law in the horizontal direction becomes P – fk = 0 ,
where P is the required pushing force. Thus, the applied force required to keep the crate sliding
across the dock at a constant speed is
P = fk = µk mg = (0.410)(60.0 kg)(9.80 m/s 2 ) = 241 N
_________________________________________________________________________________________
___
40. REASONING AND SOLUTION
a.
fk = µkmg = (0.61)(92 kg)(9.80 m/s2) = 550 N
b. From kinematics
 f 
 –550 N 
v 0 = – at = –  k  t = – 
 (1.2 s ) = 7.2 m/s
 m
 92 kg 
_________________________________________________________________________________________
___
41. REASONING In each of the three cases under consideration the kinetic frictional force is given
by f k = µk FN. However, the normal force FN varies from case to case. To determine the normal
force, we use Equation 4.6 (FN = mg + ma) and thereby take into account the acceleration of the
elevator. The normal force is greatest when the elevator accelerates upward (a positive) and
smallest when the elevator accelerates downward (a negative).
SOLUTION
a. When the elevator is stationary, its acceleration is a = 0 m/s2. Using Equation 4.6, we can
express the kinetic frictional force as
Chapter 4 Problems
133
f k = µk FN = µk (mg + ma) = µk m (g + a )
[(
) (
= (0.360 )(6.00 kg ) 9.80 m/s 2 + 0 m/s 2
)]=
21.2 N
b. When the elevator accelerates upward, a = +1.20 m/s2. Then,
f k = µk FN = µk (mg + ma) = µk m (g + a )
[(
) (
= (0.360 )(6.00 kg ) 9.80 m/s 2 + 1.20 m/s 2
)]=
23.8 N
c. When the elevator accelerates downward, a = –1.20 m/s2. Then,
f k = µk FN = µk (mg + ma) = µk m (g + a )
[(
) (
= (0.360 )(6.00 kg ) 9.80 m/s 2 + –1.20 m/s 2
)]=
18.6 N
_________________________________________________________________________________________
___
42. REASONING The initial and final velocities of the car are known, so we may use Equation 2.9
( v 2x = v02 x + 2a x x ) of the equations of kinematics to determine the stopping displacement x,
provided the acceleration ax can be found. The acceleration is related to the net force ΣFx acting
on the car by Equation 4.2, ∑ Fx = max , where m is its mass. The only force acting on the car in
the x-direction is the static frictional force that acts on the tires. The minimum stopping distance
occurs when the static frictional force is a maximum, so the net force can be expressed with the aid
of Equation 4.7 as ∑ Fx = − fs MAX = − µs FN , where µs is the coefficient of static friction and FN is
the magnitude of the normal force that acts on the tires. The minus sign indicates that the force
points to the left, along the –x axis. By combining these relations, will be able to find the
displacement of the car.
SOLUTION Solving Equation 2.9 for x and substituting Newton’s second law ( a x = ∑ Fx /m )
into this equation yields
x=
vx2 − v02x
v 2 − v02x
= x
2a x
2 ( Σ Fx / m )
The net force acting on the car is equal to the maximum static frictional force. The maximum static
frictional force is given by − f sMAX = − µs FN . With this substitution, the displacement becomes:
x=
v 2x − v02 x
v 2x − v02 x
=
2 ( ΣFx / m ) 2 ( −µsFN / m )
134 FORCES AND NEWTON'S LAWS OF MOTION
There are two forces acting on the car in the vertical, or y, direction, the downward-acting weight
–mg and the upward-acting normal force +FN. Since the car is traveling on a horizontal road and
does not leave it, the net force in the vertical direction must be zero, so that FN – mg = 0, or FN =
mg. Thus, the displacement can be written as (notice that the mass m algebraically cancels out of
this expression)
v 2x − v02x
v 2x − v02x
x=
=
2 ( − µs FN / m) −2 µs g
The stopping displacements with and without antilock brakes are
v 2 − v02x ( 0 m/s ) − (17.9m/s )
= x
=
−2µs g
−2 ( 0.764) 9.80m/s2
2
xwith
2
(
v 2 − v02 x ( 0 m/s ) − (17.9m/s )
= x
=
−2 µs g 2 ( −0.615 ) 9 . 8 0 m / s2
2
xwithout
= 21.4m
)
2
(
)
= 26.6m
The difference between these two displacements is 26.6 m – 21.4 m = 5 . 2 m .
_________________________________________________________________________________________
___
43.
SSM REASONING
If we assume that kinetic friction is the only horizontal force that
acts on the skater, then, since kinetic friction is a resistive force, it acts opposite to the direction of
motion and the skater slows down. According to Newton's second law ( ∑ F = ma ), the
magnitude of the deceleration is a = f k / m .
The magnitude of the frictional force that acts on the skater is, according to Equation 4.8,
f k = µk FN where µk is the coefficient of kinetic friction between the ice and the skate blades.
There are only two vertical forces that act on the skater; they are the upward normal force FN and
the downward pull of gravity (the weight) mg. Since the skater has no vertical acceleration,
Newton's second law in the vertical direction gives (if we take upward as the positive direction)
FN − mg = 0 . Therefore, the magnitude of the normal force is FN = mg and the magnitude of the
deceleration is given by
f
µF
µ mg
a= k = k N = k
= µk g
m
m
m
SOLUTION
a. Direct substitution into the previous expression gives
Chapter 4 Problems
135
2
2
a = µk g = (0.100)(9.80 m/s ) = 0.980 m/s
Since the skater is slowing down, the
direction of the acceleration must be opposite to the direction of motion
.
b. The displacement through which the skater will slide before he comes to rest can be obtained
from Equation 2.9 (v 2 − v02 = 2 ax ). Since the skater comes to rest, v = 0 m/s. If we take the
direction of motion of the skater as the positive direction, then, solving for x, we obtain
–v 2
–(7.60 m/s) 2
x= 0 =
= 29.5 m
2a
2(–0.980 m/s 2 )
_________________________________________________________________________________________
___
44. REASONING The free-body diagrams for the large cube (mass = M) and the small cube
(mass = m) are shown in the following drawings. In the case of the large cube, we have omitted
the weight and the normal force from the surface, since the play no role in the solution (although
they do balance).
fs MAX
–FN
P
FN
mg
In these diagrams, note that the two blocks exert a normal force on each other; the large block
exerts the force FN on the smaller block, while the smaller block exerts the force –FN on the larger
block. In accord with Newton’s third law these forces have opposite directions and equal
magnitudes FN . Under the influence of the forces shown, the two blocks have the same
acceleration a. We begin our solution by applying Newton’s second law to each one.
SOLUTION According to Newton’s second law, we have
Σ F = P – FN = Ma
144
42444
3
FN = ma
1
424
3
Large block
Small block
Substituting FN = ma into the large-block expression and solving for P gives
P = (M + m) a
For the smaller block to remain in place against the larger block, the static frictional force must
balance the weight of the smaller block, so that f s M A X = mg. But f s M A X is given by
136 FORCES AND NEWTON'S LAWS OF MOTION
f s M A X = µsFN , where, from the Newton’s second law, we know that FN = ma. Thus, we have
µsma = mg or a = g/µs . Using this result in the expression for P gives
P = ( M + m )a =
( M + m )g
µs
=
(25 kg + 4.0 kg )(9.80 m/s 2 )
0.71
= 4.0 × 10 2 N
_________________________________________________________________________________________
___
45. REASONING The free-body diagram for the box is shown in the following drawing on the left.
On the right the same drawing is repeated, except that the pushing force P is resolved into its
horizontal and vertical components.
P
P sin θ
FN
θ
FN
P cos θ
fk
fk
mg
mg
Since the block is moving at a constant velocity, it has no acceleration, and Newton’s second law
indicates that the net vertical and net horizontal forces must separately be zero.
SOLUTION Taking upward and to the right as the positive directions, we write the zero net
vertical and horizontal forces as follows:
FN – mg – Psin θ = 0
144424443
P cos θ – f k = 0
1442443
Vertical
Horizontal
From the equation for the horizontal forces, we have P cos θ = f k . But the kinetic frictional force
is f k = µk FN . Furthermore, from the equation for the vertical forces, we have FN = mg + P sin θ
. With these substitutions, we obtain
Pcos θ = f k = µ k FN = µk (mg + P sinθ )
Solving for P gives
P=
µk mg
cosθ − µk sin θ
Chapter 4 Problems
137
The necessary pushing force becomes infinitely large when the denominator in this expression is
zero. Hence, we find that cos θ – µk sin θ = 0 , which can be rearranged to show that
sin θ
1
1 
= tan θ =
or θ = tan –1 
 = 68°
 0.41 
cos θ
µk
_________________________________________________________________________________________
___
46. REASONING AND SOLUTION
a. In the horizontal direction the thrust, F, is balanced by the resistive force, fr, of the water. That
is,
Fx = 0
or
fr = F = 7.40 × 105 N
b. In the vertical direction, the weight, mg, is balanced by the buoyant force, Fb. So
Fy = 0
gives
Fb = mg = (1.70 × 108 kg)(9.80 m/s2) = 1.67 × 109 N
_________________________________________________________________________________________
___
47.
SSM REASONING AND SOLUTION There are three vertical forces that act on the
lantern. The two upward forces of tension exerted by the wires, and the downward pull of gravity
(the weight). If we let upward be the positive direction, and let T represent the tension in one of
the wires, then Equation 4.9b gives
∑ Fy = 2T − mg = 0
Solving for T gives
T=
mg (12.0 kg )(9.80 m/s 2 )
=
= 58.8 N
2
2
_________________________________________________________________________________________
___
48. REASONING AND SOLUTION The free body diagram for the plane is shown below to the
left. The figure at the right shows the forces resolved into components parallel to and
perpendicular to the line of motion of the plane.
L
L
T
R
T
R
W
θ
W sin θ
W cos θ
138 FORCES AND NEWTON'S LAWS OF MOTION
If the plane is to continue at constant velocity, the resultant force must still be zero after the fuel is
jettisoned. Therefore (using the directions of T and L to define the positive directions),
T – R – W(sin θ) = 0
(1)
L – W (cos θ) = 0
(2)
From Example 13, before the fuel is jettisoned, the weight of the plane is 86 500 N, the thrust is
103 000 N, and the lift is 74 900 N. The force of air resistance is the same before and after the
fuel is jettisoned and is given in Example 13 as R = 59 800 N.
After the fuel is jettisoned, W = 86 500 N – 2800 N = 83 700 N
From Equation (1) above, the thrust after the fuel is jettisoned is
T = R + W (sin θ) = [(59 800 N) + (83 700 N)(sin 30.0°)] = 101 600 N
From Equation (2), the lift after the fuel is jettisoned is
L = W (cos θ) = (83 700 N)(cos 30.0°) = 72 500 N
a. The pilot must, therefore, reduce the thrust by
103 000 N – 101 600 N = 1400 N
b. The pilot must reduce the lift by
74 900 N – 72 500 N = 2400 N
_________________________________________________________________________________________
___
Chapter 4 Problems
49.
SSM REASONING AND SOLUTION The figure at
the right shows the forces that act on the wine bottle.
Newton’s second law applied in the horizontal and vertical
directions gives
139
y
45.0° 45.0°
F1
F2
x
∑ Fy = F1 cos 45.0° + F2 cos 45.0° – W = 0
(1)
∑ Fx = F2 sin 45.0° − F1 sin 45.0° = 0
(2)
W
From Equation (2), we see that F1 = F2 . According to Equation (1), we have
F1 =
W
mg
=
2 cos 45.0° 2 cos 45.0 °
Therefore,
F1 = F2 =
(1.40 kg) (9.80 m/s 2 )
= 9.70 N
2 cos 45.0 °
_________________________________________________________________________________________
___
50. REASONING The drawing shows the I-beam and the three
forces that act on it, its weight W and the tension T in each of the
cables. Since the I-beam is moving upward at a constant velocity,
its acceleration is zero and it is in vertical equilibrium. According to
Equation 4.9b, ∑ Fy = 0 , the net force in the vertical (or y)
direction must be zero. This relation will allow us to find the
magnitude of the tension.
SOLUTION
becomes
Taking up as the +y direction, Equation 4.9b
T
T
70.0°
70.0°
+ T sin70.0° + T sin70.0 °−8.00×103 N = 0
14444444
4244444444
3
ΣF y
Solving this equation for the tension gives T = 4 2 6 0 N .
W = –8.00 × 103 N
_________________________________________________________________________________________
___
140 FORCES AND NEWTON'S LAWS OF MOTION
51. REASONING The book is kept from falling as long as the total static frictional force balances
the weight of the book. The forces that act on the book are shown in the following free-body
diagram, where P is the pressing force applied by each hand.
fsMAX
f sMAX
P
P
W
In this diagram, note that there are two pressing forces, one from each hand. Each hand also
applies a static frictional force, and, therefore, two static frictional forces are shown. The
maximum static frictional force is related in the usual way to a normal force FN, but in this problem
the normal force is provided by the pressing force, so that FN = P.
SOLUTION Since the frictional forces balance the weight, we have
(
) (
)
2 f sMAX = 2 µ s FN = 2 µ s P = W
Solving for P, we find that
P=
W
31 N
=
= 39 N
2µs 2( 0.40 )
_________________________________________________________________________________________
___
52. REASONING AND SOLUTION Newton’s second law applied in the vertical and horizontal
directions gives
L cos 21.0° – W = 0
L sin 21.0° – R = 0
(1)
(2)
Chapter 4 Problems
141
a. Equation (1) gives
2 1.0 °
W
53 800 N
L=
=
= 57 600 N
cos 21.0° cos 21.0°
L
b. Equation (2) gives
R
R = L sin 21.0° = ( 57 600 N) sin 21.0° = 20 600 N
W
_________________________________________________________________________________________
___
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
N
F3 = (80.0 N)2 + (60.0 N) 2 = 1.00 × 102 N
The direction of F3 is specified by the angle θ where
 80.0 N 
θ = tan –1 
= 53.1° , south of east
60.0 N 
E
θ
80.0 N
θ
F3
60.0 N
_________________________________________________________________________________________
___
142 FORCES AND NEWTON'S LAWS OF MOTION
54. REASONING The free-body diagram in the drawing at the
FN
right shows the forces that act on the clown (weight = W). In
this drawing, note that P denotes the pulling force. Since the
rope passes around three pulleys, forces of magnitude P are P
applied both to the clown’s hands and his feet. The normal
force due to the floor is FN , and the maximum static frictional
force is f sMAX . At the instant just before the clown’s feet move,
P
f sMAX
W
the net vertical and net horizontal forces are zero, according to
Newton’s second law, since there is no acceleration at this
instant.
SOLUTION According to Newton’s second law, with upward and to the right chosen as the
positive directions, we have
FN + P – W = 0
1442443
f sMAX – P = 0
1
4
4244
3
and
Vertical forces
Horizontal forces
From the horizontal-force equation we find P = f sMAX . But f sMAX = µsFN . From the verticalforce equation, the normal force is FN = W – P . With these substitutions, it follows that
P = f sMAX = µs FN = µs ( W – P )
Solving for P gives
P=
µs W
1 + µs
=
( 0.53 )( 890 N )
1 + 0.53
= 310 N
_________________________________________________________________________________________
___
55. REASONING The drawing shows the bicycle
(represented as a circle) moving down the hill. Since
the bicycle is moving at a constant velocity, its
acceleration is zero and it is in equilibrium. Choosing
the x axis to be parallel to the hill, Equation 4.9a states
that ∑ Fx = 0 , so the net force along the x axis is
zero. This relation will allow us to find the value of the
numerical constant c that appears in the expression for
fair.
+x
fair
–x
8.0°
mg
SOLUTION Taking up the hill as the +x direction, the x-component of the weight is
–mg sin 8.0°, and the force due to air resistance is +f air = +cv, where the plus sign indicates that
this force points opposite to the motion of the bicycle, or up the hill. Equation 4.9a can be written
as
Chapter 4 Problems
143
+ cv −4
mg
sin8.03
° =0
144
2444
ΣFx
Solving this equation for the constant c gives
(
)
2
mg sin8.0° ( 85kg ) 9.80m/s sin8.0°
c=
=
= 13kg/s
v
8.9m/s
_________________________________________________________________________________________
___
56. REASONING AND SOLUTION If the +x axis is taken in the direction
of motion, Fx = 0 gives
F – fk – mg sin θ = 0
F
N
fk
where
F
fk = µkFN
W
Then
F – µkFN – mg sin θ = 0
Also,
25.0°
(1)
F y = 0 gives
FN – mg cos θ = 0
so
FN = mg cos θ
(2)
Substituting Equation (2) into Equation (1) and solving for F yields
F = mg( sin θ + µk cos θ )
F = (55.0 kg)(9.80 m/s2)[sin 25.0° + (0.120)cos 25.0°] = 286 N
_________________________________________________________________________________________
___
57.
SSM REASONING There are four forces that act on the chandelier; they are the forces of
tension T in each of the three wires, and the downward force of gravity mg. Under the influence
of these forces, the chandelier is at rest and, therefore, in equilibrium. Consequently, the sum of
the x components as well as the sum of the y components of the forces must each be zero. The
figure below shows the free-body diagram for the chandelier and the force components for a
suitable system of x, y axes. Note that the free-body diagram only shows one of the forces of
tension; the second and third tension forces are not shown. The triangle at the right shows the
geometry of one of the cords, where l is the length of the cord, and d is the distance from the
ceiling.
144 FORCES AND NEWTON'S LAWS OF MOTION
y
θ
θ
T
θ
d
l
x
mg
We can use the forces in the y direction to find the magnitude T of the tension in any one wire.
SOLUTION
diagram that
Remembering that there are three tension forces, we see from the free-body
3T sin θ = mg
or
T=
mg
mg
mgl
=
=
3 sin θ 3(d / l)
3d
Therefore, the magnitude of the tension in any one of the cords is
T=
(44 kg)(9.80 m/s 2 )(2.0 m)
= 1.9 × 102 N
3(1.5 m)
_________________________________________________________________________________________
___
58. REASONING AND SOLUTION Let the tension in wire 1 be T1 and the tension in wire 2 be
T2. The sum of the vertical forces acting on the point where the wires join must be zero.
T1 sin 43.0° + T2 sin 55.0° – mg = 0
(1)
Similarly, the horizontal forces must add to zero so
– T1 cos 43.0° + T2 cos 55.0° = 0
(2)
Solving Equation (2) for T1, substituting into Equation (1), and rearranging yields
T2 = 317 N
Using this result in Equation (2) gives T1 = 249 N
_________________________________________________________________________________________
___
Chapter 4 Problems
145
59. REASONING The tree limb is in equilibrium, so we will apply Newton’s second law to it. This
law will allow us to determine the tension in both
segments of the wire.
0.800 m
3.20 m
SOLUTION Let T1 be the tension in the shorter
segment of the wire, T2 be the tension in the longer
segment, and W be the weight of the limb (see the
drawing). In the horizontal direction, Fx = 0
(Equation 4.9a) gives
−T1 cos θ1 + T2 cos θ2 = 0
144424443
T1
0.200 m
θ1
T2
θ2
W
Σ Fx
(1)
 0.200m 
 0.200m
where θ1 = tan− 1 
= 14.0° and θ2 = tan − 1 

 = 3.58° . In the vertical
 0.800m 
 3.20m 
direction, Fy = 0 (Equation 4.9b) gives
+T1 sin θ1 + T2 sin θ2 − W = 0
14444244443
ΣFy
(2)
a. Solving Equation (1) for T2 in terms of T1 , and substituting the result into Equation (2) yields
T1 =
W
151N
=
= 499 N
sin θ1 + ( sin θ 2 ) ( cos θ1 ) / ( cos θ2 ) sin14.0°+ ( sin 3.58°) ( cos14.0°) / ( cos3.58° )
b. Substituting T1 = 499 N into Equation (1) and solving for T2 gives
T2 = T1 ( cos θ1 ) / ( cos θ2 ) = ( 499 N) ( cos14.0° ) / ( cos3.58° ) = 4 8 5 N
_________________________________________________________________________________________
___
60. REASONING AND SOLUTION
a. If the block is not to slide down the wall, then the vertical forces acting on the block must sum
to zero.
F cos 40.0° – mg + µsFN = 0
Additionally, the horizontal forces must sum to zero.
F sin 40.0° – FN = 0
Eliminating FN gives
F=
mg
= 79.0 N
cos40.0° + µs sin 40.0°
146 FORCES AND NEWTON'S LAWS OF MOTION
b. The above analysis applies to the case where the block is starting to slide up the wall except,
that the frictional force will be in the opposite direction. Hence,
mg
F=
= 219 N
cos40.0° – µs sin 40.0°
______________________________________________________________________________
61.
SSM 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
FN
fS
P
P
θ
θ
MAX
fS
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 indicates
that
∑ Fx = P cos θ + mgsin θ − fsMAX = 0
According to Equation 4.7, fsMAX = µs FN . Therefore, we have
Pcos θ+ mgsin θ − µsFN = 0
(1)
The expression for the normal force can be found from analyzing the forces that are perpendicular
to the incline. Taking up to be positive, we have
∑ Fy = Psin θ+ FN – mgcos θ = 0
or
FN = mg cos θ – Psin θ
Equation (1) then becomes
Pcos θ+ mgsin θ − µs (mgcosθ – P sin θ) = 0
Solving for the coefficient of static friction, we find that
Chapter 4 Problems
µs =
147
P cos θ + mg sin θ (535 N) cos 20.0° + (225 kg)(9.80 m/s 2 ) sin 20.0°
=
= 0.665
mg cos θ – P sin θ (225 kg)(9.80 m/s 2 ) cos 20.0°– (535 N) sin 20.0°
_________________________________________________________________________________________
___
62. REASONING The weight of the part of the washcloth off the table is moff g. At the instant just
before the washcloth begins to slide, this weight is supported by a force that has magnitude equal
to f sMAX , which is the static frictional force that the table surface applies to the part of the
washcloth on the table. This force is transmitted “around the bend” in the washcloth hanging over
the edge by the tension forces between the molecules of the washcloth, in much the same way that
a force applied to one end of a rope is transmitted along the rope as it passes around a pulley.
SOLUTION Since the static frictional supports the weight of the washcloth off the table, we have
f sMAX = moff g. The static frictional force is f sMAX = µsFN . The normal force FN is applied by
the table to the part of the washcloth on the table and has a magnitude equal to the weight of that
part of the washcloth. This is so, because the table is assumed to be horizontal and the part of the
washcloth on it does not accelerate in the vertical direction. Thus, we have
f sMAX = µs FN = µs mon g = moff g
The magnitude g of the acceleration due to gravity can be eliminated algebraically from this result,
giving µsmon = moff . Dividing both sides by mon + moff gives
 mon
µs 
 mon + moff


moff
 =
 mon + moff
µs f on = f off
or
where we have used f on and f off to denote the fractions of the washcloth on and off the table,
respectively. Since f on + f off = 1, we can write the above equation on the left as
(
)
µs 1 – f off = f off
or
f off =
µs
1 + µs
=
0.40
= 0.29
1 + 0.40
_________________________________________________________________________________________
___
63. REASONING AND SOLUTION Newton's second law, F = ma, implies that the
acceleration a and the net force are in the same direction. This is 64° N of E . The magnitude of
the net force is
F = ma = (350 kg)(0.62 m/s2) = 220 N
_________________________________________________________________________________________
___
148 FORCES AND NEWTON'S LAWS OF MOTION
64. REASONING Suppose the bobsled is moving along the +x direction. There are two forces
acting on it that are parallel to its motion; a force +Fx propelling it forward and a force of
–450 N that is resisting its motion. The net force is the sum of these two forces. According to
Newton’s second law, Equation 4.2a, the net force is equal to the mass of the bobsled times its
acceleration. Since the mass and acceleration are known, we can use the second law to determine
the magnitude of the propelling force.
SOLUTION
a. Newton’s second law states that
+ Fx − 4 5 0 N = max
144244
3
(4.2a)
ΣFx
Solving this equation for Fx gives
(
)
Fx = max + 4 5 0 N = ( 270kg ) 2.4m/s 2 + 4 5 0 N = 1 1 0 0 N
b. The magnitude of the net force that acts on the bobsled is
(
)
Σ Fx = max = ( 270kg ) 2.4m/s 2 = 6 5 0 N
(4.2a)
_________________________________________________________________________________________
___
65.
SSM REASONING If we assume that the acceleration is constant, we can use Equation 2.4
( v = v0 + at ) to find the acceleration of the car. Once the acceleration is known, Newton's
second law (∑ F = ma ) can be used to find the magnitude and direction of the net force that
produces the deceleration of the car.
SOLUTION The average acceleration of the car is, according to Equation 2.4,
a=
v − v0 17.0m/s − 27.0m/s
=
= − 1.25m/s 2
t
8.00s
where the minus sign indicates that the direction of the acceleration is opposite to the direction of
motion; therefore, the acceleration points due west.
According to Newton's Second law, the net force on the car is
∑ F = ma = (1380 kg)(–1.25 m/s 2 ) = –1730 N
Chapter 4 Problems
149
The magnitude of the net force is 1 7 3 0 N . From Newton's second law, we know that the
direction of the force is the same as the direction of the acceleration, so the force also points
due west .
_________________________________________________________________________________________
___
66. REASONING AND SOLUTION From Newton's second law and the equation: v = v 0 + at,
we have
v − v0
F = ma = m
t
a. When the skier accelerates from rest (v 0 = 0 m/s) to a speed of 11 m/s in 8.0 s, the required
net force is
v - v0
(11 m/s) − 0m/s
F=m
= (73 kg)
= 1.0 × 10 2 N
t
8.0 s
b. When the skier lets go of the tow rope and glides to a halt (v = 0 m/s) in 21 s, the net force
acting on the skier is
v − v0
0 m/s − (11m/s)
F=m
= (73kg)
= −3 8 N
t
21 s
The magnitude of the net force is 38 N .
_________________________________________________________________________________________
___
67. REASONING AND SOLUTION
Newton's second law applied to
block 1 (422 N) gives
Object 1
Object 2
N1
T = m1a1
T
T
Similarly, for block 2 (185 N)
W1
T – m2g = m2a2
W2
If the string is not to break or go slack,
both blocks must have accelerations of
the same magnitude.
Then a1 = a and a2 = –a. The above equations become
T = m1a
(1)
150 FORCES AND NEWTON'S LAWS OF MOTION
T – m2g = – m2a
(2)
a. Substituting Equation (1) into Equation (2) and solving for a yields
a=
m2 g
m1 + m2
= 2.99 m/s 2
b. Using this value in Equation (1) gives
T = m1 a = 129 N
_________________________________________________________________________________________
___
68. REASONING Let’s assume that the rocket is moving in the +y direction. There are three forces
acting on it: the upward thrust of +T, the force –f air of air resistance, and the weight
–mg of the rocket. The net force is the sum of these three forces. According to Newton’s second
law, Equation 4.2b, the net force is equal to the mass of the rocket times its acceleration. We can
use the second law to determine the acceleration.
SOLUTION
Newton’s second law states that
T − mg − f air = may
144244
3
Σ Fy
(4.2b)
Solving this equation for ay gives
ay =
=
+T − mg − f air
m
(
)(
)
7.5 × 10 6 N − 4.5 × 105 kg 9.80m/s 2 − 2.4 × 10 6 N
4.5 × 10 kg
5
= +1 . 5 m / s 2
_________________________________________________________________________________________
___
69.
SSM
WWW
REASONING
The speed of the
skateboarder at the bottom of the ramp can be found by solving
Equation 2.9 (v 2 = v02 + 2 ax where x is the distance that the
skater moves down the ramp) for v. The figure at the right
shows the free-body diagram for the skateboarder. The net
force ∑ F , which accelerates the skateboarder down the ramp,
F
mgsin θ
N
mgcos θ
Chapter 4 Problems
151
is the component of the weight that is parallel to the incline: ∑ F = mg sin θ . Therefore, we know
from Newton's second law that the acceleration of the skateboarder down the ramp is
a=
∑ F mg sin θ
=
= g sin θ
m
m
SOLUTION Thus, the speed of the skateboarder at the bottom of the ramp is
2
2
2
2
v = v 0 + 2ax = v0 + 2gx sin θ = (2.6 m/s) + 2(9.80 m/s )(6.0 m) sin18 ° = 6.6 m/s
_________________________________________________________________________________________
___
Crate
70. REASONING AND SOLUTION
a. A free-body diagram for the crate gives
TB
TB – WC = mCa
TB = WC + mCa
W
C
1510 N 
TB = 1510 N + 
0.620 m/s 2 = 1610 N
2 
 9.80 m/s 
(
)
b. An analysis of the free-body diagram for the platform yields
TA – TB – WW = mW a
Man
TA
TA = TB + WW + mW a
 965 N 
2
TA = 1610 N + 
+ 965 N
2  0.620 m/s
9.80
m/s


(
)
W
W
T
B
= 2640 N
_________________________________________________________________________________________
___
71.
SSM REASONING We can use the appropriate equation of kinematics to find the
acceleration of the bullet. Then Newton's second law can be used to find the average net force on
the bullet.
SOLUTION According to Equation 2.4, the acceleration of the bullet is
152 FORCES AND NEWTON'S LAWS OF MOTION
a=
v − v0 715 m/s − 0 m/s
=
= 2.86 × 105 m/s2
–3
t
2.50 ×10 s
Therefore, the net average force on the bullet is
∑ F = ma = (15 × 10 −3 kg)(2.86 × 10 5 m/s 2 ) = 4290 N
_________________________________________________________________________________________
___
72. REASONING AND SOLUTION Newton’s second law gives
F cos 35.0° – fk = ma
Then
 70.0 N 
37.8 N
a= 
= 1.30 m/s 2
 cos 35.0° –
 15.0 kg 
15.0 kg
_________________________________________________________________________________________
___
73. REASONING AND SOLUTION The acceleration needed so that the craft touches down with
zero velocity is
v2 - v2 - (18.0 m/s) 2
0
2
a=
=
= 0.982 m/s .
2s
2(- 165 m)
Newton's second law applied in the vertical direction gives
F – mg = ma
Then
F = m(a + g) = (1.14 × 104 kg)(0.982 m/s2 + 1.60 m/s2) = 29 400 N
_________________________________________________________________________________________
___
74. REASONING The free-body diagrams for Robin
(mass = m) and for the chandelier (mass = M) are given
at the right. The tension T in the rope applies an upward
force to both. Robin accelerates upward, while the
chandelier accelerates downward, each acceleration
having the same magnitude. Our solution is based on
separate applications of Newton’s second law to Robin
and the chandelier.
T
Robin
Robin Hood
and
Chandelier
mg
Mg
SOLUTION Applying Newton’s second law, we find
=4
T1–4
mg
ma
4
24
3
T
T
Ma
1–4Mg
42=4–4
3
Chandelier
Chapter 4 Problems
153
In these applications we have taken upward as the positive direction, so that Robin’s acceleration
is a, while the chandelier’s acceleration is –a. Solving the Robin-Hood equation for T gives
T = mg + ma
Substituting this expression for T into the Chandelier equation gives
mg + ma – Mg = – Ma
 M – m 
a=
g
 M + m
or
a. Robin’s acceleration is
 (220 kg ) – (82 kg ) 
M – m
2
2
a = 
g=
 9.80 m/s = 4.5 m/s
 M + m
 (220 kg ) + (82 kg )
(
)
b. Substituting the value of a into the expression for T gives
(
)
T = mg + ma = (82 kg ) 9.80 m/s 2 + 4.5 m/s 2 = 1200 N
_________________________________________________________________________________________
___
75.
SSM REASONING The shortest time to pull the person from the cave corresponds to the
maximum acceleration, a y , that the rope can withstand. We first determine this acceleration and
then use kinematic Equation 3.5b ( y = v0 y t + 2 ayt 2 ) to find the time t.
1
SOLUTION As the person is being pulled from the cave, there are two forces that act on him;
they are the tension T in the rope that points vertically upward, and the weight of the person mg
that points vertically downward. Thus, if we take upward as the positive direction, Newton's
second law gives ∑ Fy = T – mg = may . Solving for a y , we have
ay =
T
T
569 N
2
2
–g=
–g=
2
2 − 9.80 m/s = 0.92 m/s
m
W /g
(5.20 ×10 N)/(9.80 m/s )
1
Therefore, from Equation 3.5b with v0y = 0 m/s, we have y = 2 ay t 2 . Solving for t, we find
t=
2y
=
ay
2(35.1 m)
= 8.7 s
0.92 m/s 2
_________________________________________________________________________________________
___
76. REASONING AND SOLUTION If the +x axis is taken to be parallel to and up the ramp, then
Fx = max gives
154 FORCES AND NEWTON'S LAWS OF MOTION
T – fk – mg sin 30.0° = max
where fk = µkFN . Hence,
T = max + µkFN + mg sin 30.0°
Also, F y = may gives
(1)
FN – mg cos 30.0° = 0
since no acceleration occurs in this direction. Then
FN = mg cos 30.0°
(2)
Substitution of Equation (2) into Equation (1) yields
T = max + µkmg cos 30.0° + mg sin 30.0°
T = (205 kg)(0.800 m/s2) + (0.900)(205 kg)(9.80 m/s2)cos 30.0°
+ (205 kg)(9.80 m/s2)sin 30.0° = 2730 N
_________________________________________________________________________________________
___
77. REASONING AND SOLUTION The static frictional force accelerates the crate so that it does
not slip against the bed of the truck. The maximum force that friction can supply is
fsmax = µsFN = µsmg
Newton's second law gives fsmax = ma, so
a = µsg = (0.40)(9.80 m/s2) = 3.9 m/s2
_________________________________________________________________________________________
___
78. 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
Chapter 4 Problems
Tx = (Mass of 20cars)a x
155
(4.2a)
(
)(
)
= ( 20cars ) 6.8 × 103 kg/car 8.0 × 10 −2 m / s 2 = 1.1 × 104 N
b. The tension T between the 49th and 50th cars is
Tx = (Massof1car) a x
(
(4.2a)
)(
)
= (1car ) 6.8 × 103 kg/car 8.0 × 10 −2 m / s 2 = 5.4 × 102 N
_________________________________________________________________________________________
___
79.
SSM REASONING The box comes to a halt because the
kinetic frictional force and the component of its weight parallel to
the incline oppose the motion and cause the box to slow down.
The distance that the box travels up the incline can be can be
found by solving Equation 2.9 (v 2 = v02 + 2 ax ) for x. Before we
use this approach, however, we must first determine the
acceleration of the box as it travels along the incline.
FN
mg sin θ
mg cos θ
fk
SOLUTION The figure above shows the free-body diagram for the box. It shows the resolved
components of the forces that act on the box. If we take the direction up the incline as the positive
x direction, then, Newton's second law gives
∑ Fx = – mgsin θ – fk = max
– mgsin θ – µk FN = max
or
where we have used Equation 4.8, f k = µk FN . In the y direction we have
∑ Fy = FN – mgcos θ = 0
or
FN = mg cosθ
since there is no acceleration in the y direction. Therefore, the equation for the motion in the x
direction becomes
–mgsin θ – µk mgcosθ = max
or
ax = – g(sin θ+ µk cos θ)
According to Equation 2.9, with this value for the acceleration and the fact that v = 0 m/s, the
distance that the box slides up the incline is
v20
v02
(1.50 m/s) 2
x=–
=
=
= 0.265 m
2a 2g(sin θ+ µk cosθ ) 2(9.80 m/s 2 )[sin 15.0 ° + (0.180)cos 15.0°]
_________________________________________________________________________________________
___
156 FORCES AND NEWTON'S LAWS OF MOTION
80. REASONING AND SOLUTION If the x axis is taken parallel to the slope with +x down the
slope, then F x = max gives
Fw + mg sin θ – fk = max
where fk = µkFN and Fw is the force exerted by the wind on the person and sled.
Fy = may gives
FN – mg cos θ = 0
since there is no acceleration of the sled in this direction. Hence,
FN = mg cos θ
Substitution of this into the above result gives
ax = Fw/m + g sin θ – µkg cos θ
ax = (105 N)/(65.0 kg) + (9.80 m/s2)sin 30.0° – (0.150)(9.80 m/s2)cos 30.0° = 5.24 m/s2
The time required for the sled to travel a distance, x, subject to this acceleration is found from
x = vot +(1/2)axt2
Therefore, and using the fact that v 0 = 0 m/s, the time is
t=
2x
=
ax
2(175 m )
5.24 m/s 2
= 8.17 s
_________________________________________________________________________________________
___
81. REASONING AND SOLUTION The figure to the left below shows the forces that act on the
sports car as it accelerates up the hill. The figure to the right below shows these forces resolved
into components parallel to and perpendicular to the line of motion. Forces pointing up the hill will
be taken as positive.
FN
FN
P
P
mg sin θ
mg cos θ
mg
Chapter 4 Problems
157
a. The acceleration will be a maximum when P = f smax . From the forces along the line of motion:
f smax − mg sin θ = ma
max
The force f s
is equal to µs FN. The normal force can be found from the forces perpendicular to
the line of motion:
FN = mg cos θ
Then
and
µ s ( mg cos θ) − mg sin θ = ma
(
) (
)
a = g µs cosθ – sin θ = 9.80 m/s 2 (0.88 cos18° – sin 18° ) = 5.2 m/s 2
b. When the car is being driven downhill, P (= f smax ) now points down the hill in the same
direction as (mg sin θ). Taking the direction of motion as positive, we have
f smax + mg sin θ = ma
Following the same steps as above we obtain
µ s ( mg cos θ) + mg sin θ = ma
and
(
) (
)
a = g µs cosθ + sin θ = 9.80 m/s 2 (0.88 cos18° + sin 18° ) = 11 m/s 2
_________________________________________________________________________________________
___
82. REASONING AND SOLUTION The distance required for the truck to stop is found from
2
v 2 – v02 ( 0 m/s ) – v0
x=
=
2a
2a
2
The acceleration of the truck is needed. The frictional force decelerates the crate. The maximum
force that friction can supply is
f smax = µsFN = µsmg
Newton's second law requires that
f smax = – ma so a = – µsg
Now the stopping distance is
158 FORCES AND NEWTON'S LAWS OF MOTION
( 25 m/s )2
=
= 49.1 m
x=
2 µs g 2 ( 0.650 )(9.80 m/s 2 )
v 02
_________________________________________________________________________________________
___
83.
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 −v0
=
ax
ax
θ
fk1
FN1
FN2
fk2
mg sin θ
mg cosθ
mg
Free-body diagram A
Free-body diagram B
We must, therefore, determine the acceleration of the penguin as it slides along the horizontal
patch.
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 = – fk2 = max
or
ax =
– f k2 –µk FN2
=
m
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
ax =
– µk mg
= – µk g
m
(1)
Equation (1) indicates that, in order to find the acceleration a x , 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).
Chapter 4 Problems
159
For the penguin sliding down the incline, we find from free-body diagram A and Newton's second
law (taking the direction of motion as positive) that
∑ Fx = mgsin θ – fk1 = max = 0
f k1 = mgsin 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 = µk FN1 . Applying
Newton's second law in the direction perpendicular to the incline, we have
∑ Fy = FN1 – mgcos θ = 0
or
FN1 = mg cosθ
Therefore, f k1 = µk mg cosθ, so that according to Equation (2), we find
f k1 = µk mg cos θ = mgsin θ
Solving for the coefficient of kinetic friction, we have
µk =
sin θ
= tan θ
cosθ
Finally, the time required for the penguin to slide to a halt after entering the horizontal patch of ice is
−v
–v0
v0
1.4 m/s
t= 0 =
=
=
= 1.2 s
ax
– µk g g tanθ (9.80 m/s 2 ) tan 6.9°
_________________________________________________________________________________________
___
84. 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
 m2 – m1 
a= 
 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
160 FORCES AND NEWTON'S LAWS OF MOTION
T – m1g = m1a
Solving for a gives
a=
T
(908 N )
–g=
– 9.80 m/s 2 = 11.8 m/s 2
m1
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.
_________________________________________________________________________________________
___
85. REASONING AND SOLUTION
a. The left mass (mass 1) has a tension T1 pulling it up. Newton's second law gives
T1 – m1g = m1a
(1)
The right mass (mass 3) has a different tension, T3, trying to pull it up. Newton's second for it is
T3 – m3g = – m3a
(2)
The middle mass (mass 2) has both tensions acting on it along with friction. Newton's second law
for its horizontal motion is
T3 – T1 – µkm2g = m2a
(3)
Solving Equation (1) and Equation (2) for T1 and T3, respectively, and substituting into Equation
(3) gives
(m - m - µ m )g
a = m3 + 1m +k m2 .
1
2
3
Hence,
a = 0.60 m/s2
b. From part a
T1 = m1(g + a) = 104 N
and
T3 = m3(g - a) = 230 N
_________________________________________________________________________________________
___
86. REASONING AND SOLUTION
a. The static frictional force is responsible for accelerating the top block so that it does not slip
against the bottom one. The maximum force that can be supplied by friction is
Chapter 4 Problems
161
fsmax = µsFN = µsm1g
Newton's second law requires that fsmax = m1a, so
a = µsg
The force necessary to cause BOTH blocks to have this acceleration is
F = (m1 + m2)a = (m1 + m2)µsg
F = (5.00 kg + 12.0 kg)(0.600)(9.80 m/s2) = 1.00 × 102 N
b.
The maximum acceleration that the two block combination can have before slipping
occurs is
a = F/(17.0 kg)
Newton's second law applied to the 5.00 kg block is
F – µsm1g = m1a = (5.00 kg)(F)/(17.0 kg)
Hence
F = 41.6 N
_________________________________________________________________________________________
___
87.
SSM REASONING AND SOLUTION
a. According to Equation 4.4, the weight of an object of mass m on the surface of Mars would
be given by
GM Mm
W=
RM2
where MM is the mass of Mars and RM is the radius of Mars. On the surface of Mars, the weight
of the object can be given as W = mg (see Equation 4.5), so
mg =
GM M m
RM2
or
g=
GM M
RM2
Substituting values, we have
g=
(6.67 ×10−11 N ⋅ m 2 /kg2 )(6.46 ×1023 kg)
= 3.75 m/s2
6
2
(3.39 ×10 m)
b. According to Equation 4.5,
162 FORCES AND NEWTON'S LAWS OF MOTION
W = mg = (65 kg)(3.75 m/s2) = 2.4 ×10 2 N
_________________________________________________________________________________________
___
88. REASONING AND SOLUTION Since the skier is pulled at
constant velocity, the net force on the skier is zero. The free body
diagram is shown at the right, where P is the horizontal pulling force
and R is the resistive force exerted on the skier by the water. The
positive directions are to the right and up in the diagram.
F
W
R
P
mg
a.
Σ Fx = P – R = 0
b.
Σ Fy = FW – mg = 0
or
or
R = P = 350 N
(
)
FW = mg = (92kg ) 9.80 m/s 2 = 9.0 × 102 N
_________________________________________________________________________________________
___
89. REASONING AND SOLUTION Consider a free body diagram for the stuntman with the x-axis
parallel to the ground and the +y-axis vertically upward. The motion is along the +x-axis.
Newton's second law written for no motion along the y-axis is Fy = 0 or
FN – mg = 0
This gives the normal force to be
FN = mg = (109 kg)(9.80 m/s2)
Newton's second law for uniform motion in the x direction is Fx = 0 or
T – fk = 0
Then
T = fk = µkFN = (0.870)FN = 929 N
_________________________________________________________________________________________
___
90. REASONING The magnitude of the gravitational force exerted on the satellite by the earth is
given by Equation 4.3 as F = Gmsatellite mearth / r 2 , where r is the distance between the satellite and
the center of the earth. This expression also gives the magnitude of the gravitational force exerted
on the earth by the satellite. According to Newton’s second law, the magnitude of the earth’s
acceleration is equal to the magnitude of the gravitational force exerted on it divided by its mass.
Chapter 4 Problems
163
Similarly, the magnitude of the satellite’s acceleration is equal to the magnitude of the gravitational
force exerted on it divided by its mass.
SOLUTION
a. The magnitude of the gravitational force exerted on the satellite when it is a distance of two
earth radii from the center of the earth is
Gmsatellite mearth
F=
=
r2
( 6.67 × 10
−11
) ( 425kg ) ( 5.98 × 10
( 2 ) ( 6.38 × 10 m ) 


N ⋅ m2 / k g2
24
6
kg
)
2
=
1.04 × 103 N
b. The magnitude of the gravitational force exerted on the earth when it is a distance of two earth
radii from the center of the satellite is
Gmsatellite mearth
F=
=
r2
( 6.67 × 10
−11
) ( 425kg ) ( 5.98 × 10
( 2 ) ( 6.38 × 10 m ) 


N ⋅ m2 / k g2
24
6
kg
2
)
=
1.04 × 103 N
c. The acceleration of the satellite can be obtained from Newton’s second law.
asatellite =
F
msatellite
1.04 × 103 N
=
=
425kg
2.45m/s2
d. The acceleration of the earth can also be obtained from Newton’s second law.
aearth =
F
1.04 × 103 N
=
= 1.74 × 10−22 m / s2
24
mearth 5.98 × 10 kg
_________________________________________________________________________________________
___
91.
SSM REASONING AND SOLUTION
Four forces act on the sled. They are the pulling
force P, the force of kinetic friction f k , the weight mg of the sled, and the normal force FN exerted
on the sled by the surface on which it slides. The following figures show free-body diagrams for
the sled. In the diagram on the right, the forces have been resolved into their x and y components.
164 FORCES AND NEWTON'S LAWS OF MOTION
y
P sin θ
y
FN
FN
P
fk
P cos θ
fk
θ
x
mg
x
mg
Since the sled is pulled at constant velocity, its acceleration is zero, and Newton's second law in
the direction of motion is (with right chosen as the positive direction)
∑ Fx = P cos θ − f k = ma x = 0
From Equation 4.8, we know that f k = µk FN , so that the above expression becomes
P cos θ − µk FN = 0
(1)
∑ Fy = P sinθ + FN − mg = ma y = 0
(2)
In the vertical direction,
Solving Equation (2) for the normal force, and substituting into Equation (1), we obtain
P cos θ − µk ( mg − P sin θ ) = 0
Solving for µk , the coefficient of kinetic friction, we find
µk =
P cosθ
(80.0 N) cos 30.0°
=
= 0.444
mg − P sin θ (20.0 kg) (9.80 m/s 2 ) − (80.0N) sin 30.0°
_________________________________________________________________________________________
___
92. REASONING The skydiver is falling along the –y direction. There are two forces acting on him:
the upward-acting force +f air of air resistance, and his weight, –mg. The net force is the sum of
these forces. According to Newton’s second law, Equation 4.2b, the net force is equal to the mass
of the skydiver times his acceleration. We can use the second law to determine the acceleration.
SOLUTION
Chapter 4 Problems
165
Newton’s second law states that
+ f air − mg = may
14243
Σ Fy
(4.2b)
Solving this equation for ay and noting that f air = 13 mg , we have
ay =
− mg + 13 mg
= − 23 g = − 6 . 5 m / s2
m
The magnitude of the skydiver’s acceleration is 6 . 5 m / s2 .
b. When the skydiver falls at a constant velocity, his acceleration is zero, and he is in equilibrium.
According to Equation 4.9b, ΣFy = 0, and the net force acting on him must be zero.
f air − mg = 0
14
243
ΣFy
The force of air resistance is
(
f air = mg = (110kg ) 9 . 8 m / s 2
)
= 1100 N,upward
_________________________________________________________________________________________
___
93. REASONING AND SOLUTION
a. In the vertical direction Fy = may gives
T – mg = may
so
T = may + mg = mg(1 + ay/g)

1.10 m/s 2 
T = ( 822 N )  1 +
 = 914 N

9.80 m/s 2 
b. The acceleration of the man is zero if his velocity is constant. From part a
T = mg = 822 N
_________________________________________________________________________________________
___
166 FORCES AND NEWTON'S LAWS OF MOTION
94. REASONING AND SOLUTION The apparent weight is
FN = mw(g + a)
We need to find the acceleration, a. Newton's second law applied to the elevator gives
T – (mw + me)g = (mw + me)a
Solving for a gives
9410 N
a = m T+ m - g =
- 9.80 m/s 2 = 0.954 m/s2 .
60.0 kg + 815 kg
w
e
Now the apparent weight is
FN = 60.0 kg(9.80 m/s2 + 0.954 m/s2) = 645 N
_________________________________________________________________________________________
___
95.
SSM REASONING AND SOLUTION
a. Each cart has the same mass and acceleration; therefore, the net force acting on any one of the
carts is, according to Newton's second law
∑ F = ma = (26 kg)(0.050 m/s 2 ) = 1.3 N
b. The fifth cart must essentially push the sixth, seventh, eight, ninth and tenth cart. In other
words, it must exert on the sixth cart a total force of
∑ F = ma = 5(26 kg)(0.050 m/s 2 ) = 6.5 N
_________________________________________________________________________________________
___
96. REASONING AND SOLUTION The acceleration required is
v2 - v2
a=
0
2x
2
- (15.0 m/s)
2
=
= - 2.25 m/s
2(50.0 m)
Newton's second law then gives the magnitude of the net force as
F = ma = (1580 kg)(2.25 m/s2) = 3560 N
Chapter 4 Problems
167
_________________________________________________________________________________________
___
97. REASONING AND SOLUTION
a. Newton's second law applied to the spacecraft gives
a=
ΣF
2240 N
=
= 0.0640 m/s 2
4
m
3.50 × 10 kg
This is a + acceleration since the thrust is in the direction of motion.
b. The distance traveled is found to be
x=
v 2 – v 02
2a
=
(2310 m/s ) 2 – (1820 m/s ) 2
(
2 0.0640 m/s
2
)
= 1.58 × 10 7 m = 1.58 × 10 4 km
_________________________________________________________________________________________
___
98. REASONING As discussed in Conceptual Example 7, the same net force is required on the
moon as on the earth. This net force is given by Newton’s second law as ΣF = ma, where the
mass m is the same in both places. Thus, from the given mass and acceleration, we can calculate
the net force. On the moon, the net force comes about due to the drive force and the opposing
frictional force. Since the drive force is given, we can find the frictional force.
SOLUTION Newton’s second law, with the direction of motion taken as positive, gives
ΣF = ma
or
(1430 N ) – f = (5.90 × 10 3 kg )(0.220 m/s 2 )
Solving for the frictional force f , we find
(
)(
)
f = (1430 N ) – 5.90 × 10 3 kg 0.220 m/s 2 = 130N
_________________________________________________________________________________________
___
y
99.
SSM REASONING AND SOLUTION
The system is shown in the drawing. We will
let m1 = 21.0 kg, and m2 = 45.0 kg . Then,
m1 will move upward, and m 2 will move
downward. There are two forces that act on
each object; they are the tension T in the
cord and the weight mg of the object. The
forces are shown in the free-body diagrams
at the far right.
y
T
T
x
m1
m2
m2 g
m1 g
168 FORCES AND NEWTON'S LAWS OF MOTION
We will take up as the positive direction. If the acceleration of m1 is a, then the acceleration of
m 2 must be –a.
From Newton's second law, we have for m1
∑ Fy
and for m 2
= T − m1g = m1a
(1)
∑ Fy = T − m2g = – m2a
(2)
a. Eliminating T between these two equations, we obtain
a=
 45.0 kg – 21.0 kg 
m2 – m1
g=
(9.80 m/s 2 ) = 3.56 m/s 2
 45.0 kg + 21.0 kg 
m2 + m1
b. Eliminating a between Equations (1) and (2), we find
T=
2 m1 m2
 2(21.0 kg)(45.0 kg) 
g= 
(9.80 m/s 2 ) = 281 N

m1 + m2
 21.0 kg + 45.0 kg 
_________________________________________________________________________________________
___
100. REASONING AND SOLUTION The deceleration produced by the frictional force is
a= –
fk
m
=
– µk mg
m
= – µk g
The time it takes for the car to come to a halt is given by Equation 2.4 as
t=
v − v0 v − v0
0m/s − 16.1m/s
=
=
= 2.28s
a
− µk g − ( 0.720) 9.80m/s2
(
)
_________________________________________________________________________________________
___
101. REASONING AND SOLUTION Take the x-axis of a free body diagram for the block to be
parallel to the incline with +x down the incline. Newton's second law for uniform motion in the +x
direction is Fx = 0, so
mg sin θ – fk = 0
where fk = µkFN . Then
mg sin θ = µkFN
Similarly, the equation for the y direction is Fy = 0, so
Chapter 4 Problems
FN – mg cos θ = 0
or
169
mg cos θ = FN
Division of the equations gives tan θ = µk , hence µk = tan 11.3° = 0.200
_________________________________________________________________________________________
___
102. REASONING Since the boxes are at rest, they are in equilibrium. According to Equation 4.9b,
the net force in the vertical, or y, direction is zero, ΣFy = 0. There are two unknowns in this
problem, the normal force that the table exerts on box 1 and the tension in the rope that connects
boxes 2 and 3. To determine these unknowns we will apply the relation ΣFy = 0 twice, once to the
boxes on the left of the pulley and once to the box on the right.
SOLUTION There are four forces acting on the two blocks on the
left. The blocks are in equilibrium, so that the net force must be zero.
Choosing the +y direction as being the upward direction, we have that
− W1 − W2 + FN + T = 0
14442444
3
T
ΣF y
where W1 and W2 are the magnitudes of the weights of the boxes, FN
is the magnitude of the normal force that the table exerts on box 1,
and T is the magnitude of the tension in the rope. We know the
weights, but do not know the tension. To find the tension, we note
that the block on the right is also in equilibrium, so that the net force
acting on it must be zero.
− W3 + T = 0
1424
3
sothat
T = W3
3
T
FN
W3
2
1
Σ Fy
Substituting this expression for T back into the first equation and
solving for the normal force gives
FN =W 1 +W2 − W3 = 5 5 N + 3 5 N − 2 8 N = 6 2 N
W2
W1
_________________________________________________________________________________________
___
103. SSM REASONING AND SOLUTION Three forces act on the man. They
are two upward forces of tension of equal magnitude T, and the force of gravity mg.
Therefore, if we take up as the positive direction, Newton's second law gives
∑ F = 2T − mg = ma
Solving for the acceleration a, we find
a=
2T − mg 2T
2(358 N)
=
−g=
− 9.80 m/s2 = 0.14 m/s 2
m
m
72.0 kg
T
T
mg
170 FORCES AND NEWTON'S LAWS OF MOTION
_________________________________________________________________________________________
___
104. REASONING According to Newton’s second law, the acceleration of the probe is a = ΣF/m.
Using this value for the acceleration in Equation 2.8 and noting that the probe starts from rest
(v 0 = 0 m/s), we can write the distance traveled by the probe as
1
1  ΣF  2
t
x = v 0 t + at 2 = 
2
2 m 
This equation is the basis for our solution.
SOLUTION Since each engine produces the same amount of force or thrust T, the net force is
ΣF = 2T when the engines apply their forces in the same direction and Σ F = T 2 + T 2 = 2T
when they apply their forces perpendicularly. Thus, we write the distances traveled in the two
situations as follows:
1 2T  2
x = 
t
 m
2
144244
3
and
Engines fired in
same direction
1  2T  2
x= 
 t⊥
2 m 
1442443
Engines fired
perpendicularly
Since the distances are the same, we have
2T  2 1  2 T  2

t = 
 t⊥
2 m 
2 m 
1
or
2 t 2 = t 2⊥
The firing time when the engines apply their forces perpendicularly is, then,
t⊥ =
(4 2 )t = (4 2 )(28 s ) =
33 s
_________________________________________________________________________________________
___
105. REASONING The tension in each coupling bar is responsible for accelerating the objects behind
it. The masses of the cars are m1, m2, and m3. We can use Newton’s second law to express the
tension in each coupling bar, since friction is negligible:
T A = (m1 + m2 + m3 )a
1444
424444
3
TB = (m 2 + m3 )a
144
42444
3
TC = m 3 a
14243
Coupling bar A
Coupling bar B
Coupling bar C
In these expressions a = 0.12 m/s2 remains constant. Consequently, the tension in a given bar will
change only if the total mass of the objects accelerated by that bar changes as a result of the
Chapter 4 Problems
171
luggage transfer. Using ∆ (Greek capital delta) to denote a change in the usual fashion, we can
express the changes in the above tensions as follows:
[(
)]
[(
)]
(
)
∆TA = ∆ m1 + m2 + m 3 a
14444
4244444
3
∆TB = ∆ m 2 + m 3 a
1444
424444
3
∆TC = ∆m3 a
1442443
Coupling bar A
Coupling bar B
Coupling bar C
SOLUTION
a. Moving luggage from car 2 to car 1 does not change the total mass m1 + m2 + m3, so
∆(m1 + m2 + m3) = 0 kg and ∆TA = 0 N .
The transfer from car 2 to car 1 causes the total mass m2 + m3 to decrease by 39 kg, so
∆(m2 + m3) = –39 kg and
[
]
(
)
∆ TB = ∆ (m2 + m3 ) a = (–39 kg ) 0.12 m/s 2 = –4.7 N
The transfer from car 2 to car 1 does not change the mass m3, so ∆m3 = 0 kg and ∆TC = 0 N .
b. Moving luggage from car 2 to car 3 does not change the total mass m1 + m2 + m3, so
∆(m1 + m2 + m3) = 0 kg and ∆TA = 0 N .
The transfer from car 2 to car 3 does not change the total mass m2 + m3, so ∆(m2 + m3) = 0 kg
and ∆TB = 0 N .
The transfer from car 2 to car 3 causes the mass m3 to increase by 39 kg, so ∆m3 = +39 kg and
(
)
∆TC = (∆m 3 )a = ( +39 kg ) 0.12 m/s 2 = +4.7 N
_________________________________________________________________________________________
___
106. REASONING AND SOLUTION
acceleration, it follows that
Since both motions are characterized by constant
1
yJ
yE
=
2
1
2
a J t 2J
a E t 2E
where the subscripts designate those quantities that pertain to Jupiter and Earth. Since both
objects fall the same distance, the above ratio is equal to unity. Solving for the ratio of the times
yields
172 FORCES AND NEWTON'S LAWS OF MOTION
tJ
tE
=
aE
aJ
GM E / RE2
=
=
GM J / RJ2
RJ
ME
RE
MJ
= (11.2 )
1
= 0.628
318
_________________________________________________________________________________________
___
107. 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
MAX
fS
FN
the picture, and the force of static friction f sMAX . The maximum magnitude
for the frictional force is given by Equation 4.7: fsMAX = µs FN . The picture
P
mg
is in equilibrium, and, if we take the directions to the right and up as positive,
we have in the x direction
∑ Fx = P − FN = 0
or
P = FN
∑ Fy = fsMAX − mg = 0
or
fsMAX = mg
and in the y direction
Therefore,
MAX
fs
But since FN = P , we have
= µs FN = mg
µs P = mg
Solving for P, we have
P=
mg (1.10 kg)(9.80 m/s 2 )
=
= 16.3 N
µs
0.660
_________________________________________________________________________________________
___
108. REASONING AND SOLUTION
a. The force acting on the sphere which accelerates it is the horizontal component of the tension in
the string. Newton's second law for the horizontal motion of the sphere gives
T sin θ = ma
The vertical component of the tension in the string supports the weight of the sphere so
T cos θ = mg
Eliminating T from the above equations results in a = g tan θ .
Chapter 4 Problems
b.
(
173
)
a = g tan θ = 9.80 m/s 2 tan 10.0° = 1.73 m/s 2
c. Rearranging the result of part a and setting a = 0 m/s2 gives
θ = tan –1 (a / g ) = 0°
_________________________________________________________________________________________
___
109. REASONING AND SOLUTION
a. Newton's second law for block 1 (10.0 kg) is
T = m1a
(1)
Block 2 (3.00 kg) has two ropes attached each carrying a tension T. Also, block 2 only travels
half the distance that block 1 travels in the same amount of time so its acceleration is only half of
block 1's acceleration. Newton's second law for block 2 is then
2T – m2g = – (1/2)m2a
(2)
Solving Equation (1) for a, substituting into Equation (2), and rearranging gives
T=
1
m g
2 2
1 + 1 (m 2 / m1 )
= 13.7 N
4
b. Using this result in Equation (1) yields
a=
T
13.7 N
=
= 1.37 m/s 2
m1 10.0 kg
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___
110. REASONING The diagram at the right shows the two applied forces
that act on the crate. These two forces, plus the kinetic frictional force fk
constitute the net force that acts on the crate. Once the net force has
been determined, Newtons’ second law, ΣF = ma (Equation 4.1) can
be used to find the acceleration of the crate.
+y
F1
55.0°
SOLUTION The sum of the two applied forces is F = F1 + F2.
The x-component of this sum is Fx = F1 cos 55.0° + F2
= (88.0 N) cos 55.0° + 54.0 N = 104 N. The y-component of F is Fy = F1 sin 55.0° =
(88.0 N) sin 55.0° = 72.1 N. The magnitude of F is
+x
F2
174 FORCES AND NEWTON'S LAWS OF MOTION
F = F x2 + F y2 =
(104N )
2
+ ( 72.1N) = 127 N
2
Since the crate starts from rest, it moves along the direction of F. The kinetic frictional force fk
opposes the motion, so it points opposite to F. The net force acting on the crate is
the sum of F and fk. The magnitude a of the crate’s acceleration is equal to the magnitude ΣF of
the net force divided by the mass m of the crate
a=
ΣF − f k + F
=
m
m
(4.1)
According to Equation 4.8, the magnitude f k of the kinetic frictional force is given by f k = µk FN ,
where FN is the magnitude of the normal force. In this situation, FN is equal to the magnitude of the
crate’s weight, so FN = mg. Thus, the x-component of the acceleration is
(
)
2
− µk mg + F − ( 0.35 ) ( 25.0kg ) 9.80m/s + 1 2 7 N
a=
=
= 1.65m/s 2
m
25.0kg
The crate moves along the direction of F, whose x and y components have been determined
previously. Therefore, the acceleration is also along F. The angle φ that F makes with the x-axis
can be found using the inverse tangent function.
F 
 F1 sin55.0° 
ϕ = tan− 1  y  = tan − 1 

 F1 cos55.0 ° + F2 
 Fx 


( 88.0N) sin55.0°
= tan− 1 
 = 34.6° abovethe x axis
 ( 88.0N) cos55.0° + 5 4 . 0 N
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111. SSM REASONING When the bicycle is coasting straight down the hill, the forces that act on
it are the normal force FN exerted by the surface of the hill, the force of gravity mg, and the force
of air resistance R. When the bicycle climbs the hill, there is one additional force; it is the applied
force that is required for the bicyclist to climb the hill at constant speed. We can use our
knowledge of the motion of the bicycle down the hill to find R. Once R is known, we can analyze
the motion of the bicycle as it climbs the hill.
Chapter 4 Problems
175
SOLUTION The figure to the left below shows the free-body diagram for the forces during the
downhill motion. The hill is inclined at an angle θ above the horizontal. The figure to the right
shows these forces resolved into components parallel to and perpendicular to the line of motion.
+y
+x
FN
FN
R
R
mg sin θ
mg
mg cos θ
θ
Since the bicyclist is traveling at a constant velocity, his acceleration is zero. Therefore, according
to Newton's second law, we have ∑ Fx = 0 and ∑ Fy = 0 . Taking the direction up the hill as
positive, we have ∑ Fx = R − mg sin θ = 0 , or
R = mg sin θ = (80.0 kg)(9.80 m/s 2 ) sin 15.0 ° = 203 N
When the bicyclist climbs the same hill at
constant speed, an applied force P must push the
system up the hill, and the force of air resistance
will oppose the motion by pointing down the hill.
FN
P
mg sin θ
The figure at the right shows the resolved forces
that act on the system during the uphill motion.
mg cosθ
R
Using the same sign convention as above, we have ∑ Fx = P − mg sin θ – R = 0 , or
P = R + mg sin θ = 203 N + 203 N =
406 N
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___
112. CONCEPT QUESTIONS
a. Yes. 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.
b. Yes. 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).
SOLUTION
176 FORCES AND NEWTON'S LAWS OF MOTION
a. The acceleration of the man is, according to Equation 4.1, equal to the net force acting on him
divided by his mass.
ΣF 4 5 N
=
= 0.55m/s 2 (dueeast)
m 82kg
b. The acceleration of the woman is equal to the net force acting on her divided by her mass.
aman =
a woman =
ΣF 4 5 N
=
= 0.94m/s 2 (duewest)
m 48kg
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___
113. CONCEPT QUESTIONS
a. The other force is the static frictional force exerted on the refrigerator by the floor.
b. Since the refrigerator does not move, the static frictional force must be equal in magnitude, but
opposite in direction, to the horizontal pushing force that the person exerts on the refrigerator.
c. The magnitude of the maximum static frictional force is given by Equation 4.7 as f s MAX = µs FN .
This is also the largest possible force that the person can exert on the refrigerator before it begins
to move. Thus, the factors that determine this force magnitude are the coefficient of static friction
µs and the magnitude FN of the normal force (which is equal to the weight of the refrigerator in this
case).
SOLUTION
a. Since the refrigerator does not move, it is in equilibrium, and the magnitude of the static
frictional force must be equal to the magnitude of the horizontal pushing force. Thus, the magnitude
of the static frictional force is 267 N . The direction of this force must be opposite to that of the
pushing force, so the static frictional force is in the +x direction .
b. The magnitude of the largest pushing force is given by Equation 4.7 as
f sMAX = µs FN = µsmg = (0.65)(57 kg)(9.80 m/s 2 ) = 360 N
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___
114. CONCEPT QUESTIONS
a. The gravitational force exerted on the rock is greater than that on the pebble, because the rock
has the greater mass.
b. The accelerations are equal, because all objects, regardless of their mass, fall to the earth with
the same acceleration due to gravity.
Chapter 4 Problems
177
SOLUTION
a. The magnitude of the gravitational force exerted on the rock by the earth is given by Equation
4.3 as
Gmearth mrock
2
rearth
Frock =
( 6.67 × 10
=
−11
)(
)
N ⋅ m 2 /kg 2 5.98 × 10 24 kg (5.0kg )
( 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 / k g 2 5.98 × 10 24 kg 3.0 × 10 −4 kg
( 6.38 × 10 m)
6
2
)=
2.9 × 10−3 N
b. 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 /kg2
( 6.38 × 10
6
)( 5.98 × 10
m)
24
2
kg
)=
9.80m/s2 , downward
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 ⋅ m 2 /kg 2 5.98 × 1024 kg
( 6.38 × 10 m)
6
2
)=
9.80m/s 2, downward
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115. CONCEPT QUESTION Yes, 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.
178 FORCES AND NEWTON'S LAWS OF MOTION
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 × 1024 kg 5.2 × 10 −7 kg
( 6.38 × 10 m)
6
2
)=
5.1 × 10−6 N
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116. CONCEPT QUESTIONS
a. F2 can be positive and can have any magnitude. F2 can also be negative, provided that its
magnitude is less than the magnitude of F1.
b. F2 must be negative and have a magnitude that is greater than that of F1.
c. F2 must be negative and have a magnitude that is equal to that of F1.
SOLUTION
a. We may use Newton’s second law, ΣFx = max, to find the force F2. Taking the positive x
direction to be to the right, we have
F1 + F2 = max
1424
3
so
F2 = max − F1
ΣFx
F2 = (3.0 kg)(+5.0 m/s2) − (+9.0 N) = +6 N
b. Applying Newton’s second law again gives
F2 = max − F1 = (3.0 kg)(−5.0 m/s2) − (+9.0 N) = −24 N
Chapter 4 Problems
179
c. An application of Newton’s second law gives
F2 = max − F1 = (3.0 kg)(0 m/s2) − (+9.0 N) = −9.0 N
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117. CONCEPT QUESTIONS
a. No, the magnitude of the normal force is not equal to the
weight of the car. As the drawing shows, the normal force FN
points perpendicular to the hill, while the weight W points
vertically down. Since the car does not leave the surface of the
hill, the magnitude of the perpendicular component of the weight
W cos θ must equal the magnitude of the normal force, so FN =
W cos θ . Thus, the magnitude of the normal force is less than the
magnitude of the weight.
FN
θ
θ
W cos θ
W
b. As the hill becomes steeper, θ increases, and cos θ decreases. Consequently, the normal force
decreases as the hill becomes steeper.
c. The magnitude of the normal force does not depend on whether the car is traveling up or down
the hill.
SOLUTION
a. From part a of the Concept Questions, we have that FN = W cos θ . The ratio of the magnitude
of the normal force to the magnitude W of the weight is
FN W cos θ
=
= cos15° = 0.97
W
W
b. When the angle is 35°, the ratio is
FN W cos θ
=
= cos35° = 0.82
W
W
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