Embed Size (px)
Citation preview
Chapter 4 89
67. 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.
____________________________________________________________________________________________________
68. The car has an average acceleration of
a =
v - v 0
t =
1 7 . 0 m / s - 2 7 . 0 m / s
8 . 0 0 s = - 1 . 2 5 m / s
2 .
The magnitude of the net force is then
F = ma = (1380 kg)(1.25 m/s2) =1730 N.
The direction of the net force is opposite the direction of motion. That is, WEST.
____________________________________________________________________________________________________
69. From Newton's second law and the equation: v = v0 + at, we have
F ma m v v 0
t .
a. When the skier accelerates from rest (v0 = 0) to a speed of 11 m/s in
8.0 s, the required net force is
F m v v 0
t ( 73 kg)
(11 m/s) - 0
8.0 s 1.0 10
2 N .
b. When the skier lets go of the tow rope and glides to a halt (v = 0) in 21 s,
the net force acting on the skier is
F m v v 0
t ( 73 kg)
0 - (11 m/s)
21 s-38 N.
The magnitude of the net force is 38 N.
____________________________________________________________________________________________________
90 FORCES AND NEWTON'S LAWS OF MOTION
70. Newton's second law applied
to block 1 (111 N) gives
T = m1a1.
Similarly, for block 2 (258 N)
T - m2g = m2a2.
If the string is not to break or go
slack, both blocks must have
accelerations of the same
magnitude
a. Then a1 = a and a2 = -a.
Object 2
T
W 2
W
T
N 1
1
Object 1
The above equations become
T = m1a (1)
T - m2g = - m2a (2)
a. Substituting (1) into (2) and solving for a yields
a =
m 2 g
m 1 + m
2
= 6 . 8 5 m / s 2 .
b. Using this value in (1) gives
T = m1a = 77.6 N. ____________________________________________________________________________________________________
71 . a. In the vertical direction Fy = may gives
T - mg = may
so
T = may + mg = mg(1 + ay/g)
T = 8 2 2 N 1 + 1 . 1 0 m / s
2
9 . 8 0 m / s 2 = 9 1 4 N .
b. The acceleration of the man is zero if his velocity is constant. From part a
T = mg = 822 N .
Chapter 4 91
____________________________________________________________________________________________________
92 FORCES AND NEWTON'S LAWS OF MOTION
72. The component of the force which accelerates the student down the ramp is
F = mg sin 18°
This force produces an acceleration of
a = g sin 18°.
The speed of the student at the bottom of the ramp is
v = v 0
2 + 2 a s = v
0
2 + 2 g s s i n 1 8 °
v = ( 2 . 6 m / s ) 2 + 2 ( 9 . 8 0 m / s
2 ) ( 6 . 0 m ) s i n 1 8 ° = 6 . 6 m / s .
____________________________________________________________________________________________________
73. From the free body diagram (taking up to be the positive direction)
R - mg = - ma
or R = mg - ma
so R = m (g - a) = (72.0 kg) (9.80 m/s2 - 1.30 m/s2) = 612 N
R
mg
____________________________________________________________________________________________________
74. The boat-trailer has an acceleration
a = v
t =
1 1 m / s
2 8 s = 0 . 3 9 m / s
2
The net force on the boat-trailer is just the tension in the hitch. Newton’s second gives
T = ma = (410 kg)(0.39 m/s2) = 160 N. ____________________________________________________________________________________________________
Chapter 4 93
75. a. A free-body diagram for the crate gives
TB - WC = mCa
TB = WC + mCa
T B
= 1 5 1 0 N +
1 5 1 0 N
9 . 8 0 m / s 2
( 0 . 6 2 0 m / s 2 ) = 1 6 1 0 N .
T
W
B
C
Crate
94 FORCES AND NEWTON'S LAWS OF MOTION
b. An analysis of the free-body diagram for the platform yields
TA - TB - WW = mWa
TA = TB + WW + mWa
T A
= 1 6 1 0 N + 9 6 5 N
9 . 8 0 m / s 2 ( 0 . 6 2 0 m / s
2 ) + 9 6 5 N = 2 6 4 0 N .
T
W
A
W
Man
T B
____________________________________________________________________________________________________
76. Newton’s second law gives
F cos 35.0° - fk = ma
then
a = 7 0 . 0 N
1 5 . 0 k g c o s 3 5 . 0 ° -
3 7 . 8 N
1 5 . 0 k g = 1 . 3 0 m / s
2 .
____________________________________________________________________________________________________
77. The first coupling provides a tension, T1, which pulls the two cars with an acceleration,
a = 0.520 m/s2. Newton's second law gives
T1 = (m1 + m2)a = (51 300 kg + 18 400 kg)(0.520 m/s2)
T1 = 36 200 N.
The second coupling provides a tension, T2, which pulls only the last car. Again, Newton's second
law gives
T2 = m2a = (18 400 kg)(0.520 m/s2)
T2 = 9570 N. ____________________________________________________________________________________________________
78. The acceleration needed so that the craft touches down with zero velocity is
a =
v 2 - v
0
2
2 s =
- ( 1 8 . 0 m / s ) 2
2 ( - 1 6 5 m ) = 0 . 9 8 2 m / s
2 .
Newton's second law applied in the vertical direction gives
Chapter 4 95
F - mg = ma .
96 FORCES AND NEWTON'S LAWS OF MOTION
Then
F = m(a + g) = (11 400 kg)(0.982 m/s2 + 1.60 m/s2)
F = 29 400 N. ____________________________________________________________________________________________________
79 . a. Newton's second law applied to block 1 gives
T - µkFN = m1a1 and FN = m1g.
Similarly, for block 2
T - m2g = m2a2.
If the string is not to break or go slack, both blocks must have the same acceleration, a. Then
a1 = a and a2 = -a.
The above equations become
T - µkm1g = m1a (1)
T - m2g = - m2a (2)
Subtracting (2) from (1) and solving for a yields
a =
( m 2 -
k m
1 ) g
m 1 + m
2
= 5 . 9 7 m / s 2 .
b. Solving (2) for T gives
T = m2(g - a) = 101 N. ___________________________________________________________________________________________________
80. If the +x axis is taken to be parallel to and up the ramp, then Fx = max gives
T - fk - mg sin 30.0° = max
where fk = µkFN . Hence,
T = max + µkFN + mg sin 30.0° (1)
Also, Fy = may gives
FN - mg cos ° = 0
since no acceleration occurs in this direction. Then
FN = mg cos ° . (2)
Substitution of (2) into (1) yields
Chapter 4 97
T = max + µkmg cos 30.0° + mg sin 30.0°
98 FORCES AND NEWTON'S LAWS OF MOTION
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°
T = 2730 N. ___________________________________________________________________________________________________
81. Two cables each with a tension, T, support the mass. Newton's second law applied to the
mass gives
2T - mg = - ma.
Solving
T = m(g - a)/2 =1850 N. ___________________________________________________________________________________________________
82. Three cables each carrying a tension, T, are pulling the cart. Applying Newton's second law
along the incline gives
3T - mg sin 25.0° = ma.
then
a = 3T/m - g sin 25.0° = 0.788 m/s2. ___________________________________________________________________________________________________
83. Friction provides the force which accelerates the book. The maximum force that friction can
provide is
fsmax = µsFN = µsmg.
Newton's second law requires that
fsmax = ma so a = µsg = (0.72)(9.80 m/s2) = 7.1 m/s2
___________________________________________________________________________________________________
84. If the x axis is taken parallel to the slope with +x down the slope, then Fx = max gives
Fw + mg sin - 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 = 0
since there is no acceleration of the sled in this direction. Hence,
FN = mg cos .
Substitution of this into the above result gives
ax = Fw/m + g sin - µkg cos
Chapter 4 99
ax = (105 N)/(65.0 kg) + (9.80 m/s2)sin 30.0° - (0.150)(9.80 m/s2)cos 30.0°
ax = 5.24 m/s2 .
The time required for the sled to travel a distance, s, subject to this acceleration is found from
s = vot +(1/2)axt2 .
Therefore,
t = 2 s
a x
= 2 ( 1 7 5 m )
5 . 2 4 m / s 2
= 8 . 1 7 s .
___________________________________________________________________________________________________
85. 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. ___________________________________________________________________________________________________
86. 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.
mg
P F
N P
F N
mg cos mg sin
a. The acceleration will be a maximum when P = f s max
. From the forces along the line
of motion:
f s max mg sin ma .
100 FORCES AND NEWTON'S LAWS OF MOTION
The force f s max
is equal to sFN. The normal force can be found from the forces
perpendicular to the line of motion.
F N mg cos .
Then s ( mg cos ) mg sin ma
and
a g [s cos sin ] (9.80 m/s2
)[ (0.88) (cos 18 ) - (sin 18 )] = 5.2 m/ s 2
b. When the car is being driven downhill, P (= f s max
) now points down the hill in the
same direction as (mg sin ). Taking the direction of motion as positive, we have
f s max mg sin ma .
Following the same steps as above we obtain
s ( mg cos ) mg sin ma
and
a g [s cos sin ] (9.80 m/s2
)[ (0.88) (cos 18 ) + (sin 18 )] = 11 m/ s2
___________________________________________________________________________________________________
87. The distance required for the truck to stop is found from
s =
v 2 - v
0
2
2 a =
- v 0
2
2 a
The acceleration of the truck is needed. The frictional force decelerates the crate. The
maximum force that friction can supply is
fsmax = µsFN = µsmg .
Newton's second law requires that
fsmax = - ma so a = - µsg.
Now the stopping distance is
s =
v 0
2
2 s g
= ( 2 5 . 0 m / s )
2
2 ( 0 . 6 5 0 ) ( 9 . 8 0 m / s 2 )
= 4 9 . 1 m .
___________________________________________________________________________________________________
Chapter 4 101
88 . a. The rope exerts a tension, T, acting upwardly on each rope. 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
a =
( m 2 - m
1 ) g
m 2 + m
1
= 3 . 6 8 m / s 2 .
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
T - m1g = m1a.
Solving for a gives
a = T/m1 - g = (908 N)/(42.0 kg) - 9.80 m/s2 = 11.8 m/s2.
c. In the first case, the inertia of BOTH blocks affect the acceleration whereas, in the second
case, only the lighter block's inertia remains. ___________________________________________________________________________________________________
89. In order to find the coefficient of kinetic friction, we must first find the acceleration of
the skier down the slope when friction is absent.
If there were no friction, the only forces that act on
the skier are the weight of the skier and the normal
force exerted by the slope. These forces are shown
at the right and have been resolved in directions that
are parallel to and perpendicular to the line of motion.
Newton's second law in the direction of motion (with
the direction of motion taken as positive) is
F mg sin = ma .
This gives
a g sin = (9.80 m/s 2
) sin 35 = 5.6 m/s 2
.
F N
mg cos mg sin
102 FORCES AND NEWTON'S LAWS OF MOTION
Since the skier would reach the bottom in 24 s, the distance traveled along the slope is
x = 1
2 at
2 =
1
2 (5.6 m/s
2 ) (24 s)
2 1600 m .
If it actually takes 45 s for the skier to reach the bottom when friction is present, then,
when friction is present, the acceleration of the skier is found as follows:
x = 1
2 at
2 a =
2 x
t 2
= 2 (1600 m)
(45 s)2
= 1.6 m/s 2 .
When friction is present, we have, in addition to the
forces shown above, the force of kinetic friction which
points up the slope.
Newton's second law in the direction of motion (with
the direction of motion taken as positive) is
F mg sin - f k = ma
where fk = kFN. From the forces perpendicular to
the plane we have
FN = mg cos .
F N
mg cos mg sin
f k
Therefore mg sin - k mg cos = ma.
Solving for k gives
k = g sin - a
g cos =
[(9.80 m/s2
)(sin 35 )] - (1.6 m/s2
)
( 9 . 80 m/s2
)(cos 35 ) = 0.50
__________________________________________________________________________________________________
90. 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)
Chapter 4 103
Solving (1) for a, substituting into (2) and rearranging gives
T =
( 1 / 2 ) m 2 g
1 + ( 1 / 4 ) ( m 2 / m
1 ) = 1 3 . 7 N .
b. Using this result in (1) yields
a = T/m1 = (13.7 N)/(10.0 kg) = 1.37 m/s2.
___________________________________________________________________________________________________
91. 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
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 X 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. ___________________________________________________________________________________________________
92 . a. The acceleration due to gravity at the equator of Saturn is
g s = G
m s
r s
2 = ( 6 . 6 7 x 1 0
- 1 1 N m
2 / k g
2 )
5 . 6 7 x 1 0 2 6
k g
( 6 . 0 0 x 1 0 7 m )
2 = 1 0 . 5 m / s
2 .
b. On Earth the person's weight is W = mgE so
W
S
W =
m g S
m g E
=
g S
g E
= 1 0 . 5 m / s
2
9 . 8 0 m / s 2
= 1 . 0 7 .
104 FORCES AND NEWTON'S LAWS OF MOTION
___________________________________________________________________________________________________
93. a. The mass of the object is, according to Newton's second law,
m = F
a =
5 2 5 N
4 . 2 0 m / s 2
= 1 2 5 k g .
The required force is then
F = ma = (125 kg)(13.7 m/s2) = 1710 N.
b. No net force is needed for motion with a constant velocity.
___________________________________________________________________________________________________
94. Consider a free body diagram for the stunt man 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 ___________________________________________________________________________________________________
95. 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 x 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 X 108 kg)(9.80 m/s2) =1.67 X 109 N.
Chapter 4 105
___________________________________________________________________________________________________
96. Newton's law of gravitation gives
F = G
m 1 m
2
r 2
= ( 6 . 6 7 x 1 0 - 1 1
N m 2 / k g
2 ) ( 0 . 0 0 1 5 0 k g ) ( 0 . 8 7 0 k g )
( 0 . 1 0 0 m ) 2
F = 8 . 7 0 x 1 0 - 1 2
N . ___________________________________________________________________________________________________
97 The net force must be zero for unaccelerated motion. Therefore, the sum of the east-west
components must vanish.
- 60.0 N + FE = 0
so FE = 60.0 N.
Similarly, the sum of the north-south components must vanish so
80.0 N + FN = 0
and FN = - 80.0 N.
106 FORCES AND NEWTON'S LAWS OF MOTION
Now
F = F N
2 + F
E
2 = ( - 8 0 . 0 N )
2 + ( 6 0 . 0 N )
2 = 1 . 0 0 x 1 0
2 N .
The direction is given by the angle
= t a n - 1 - 8 0 . 0 N
6 0 . 0 N = - 5 3 . 1
So the direction is
53.1°, S of E. ___________________________________________________________________________________________________
98. The forces that act on the rock are shown at the right.
Newton's second law (with the direction of motion as positive) is
F mg R ma .
Solving for the acceleration a gives
a mg R
m
(45 kg)(9.80 m/s2
) ( 250 N)
45 kg = 4.2 m/ s
2 .
mg
R
___________________________________________________________________________________________________
99. The apparent weight of the fish cannot be greater than 45 N or the line will break.
a. FN = mg + ma , but a = 0 so
FN = mg = 45 N.
b. Again FN = mg + ma or
FN = mg (1 + a/g) .
The weight of the fish, mg, is then (suppressing some units)
m g =
F N
1 + a / g =
4 5 N
1 + 2 . 0 / 9 . 8 0 = 3 7 N .
___________________________________________________________________________________________________
100. The scale reads the net force, F, on the person. Newton’s second law gives
F - W = ma. Then
a = F - W
m =
6 2 2 N - ( 5 5 k g ) ( 9 . 8 0 m / s 2 )
5 5 k g = 1 . 5 m / s
2 .
___________________________________________________________________________________________________
Chapter 4 107
101. The acceleration due to gravity on the surface is
g = G
m E
r E
2
And at a height, h, above the surface it is
g h = G
m E
( r E
+ h ) 2
Dividing and simplifying yields
h2 + 2rEh - rE2 = 0.
The quadratic formula gives
h = - rE ± 2 rE .
The negative value is rejected so
h = 0.414rE = (0.414)(6.38 X 106m) = 2.64 X 106 m ___________________________________________________________________________________________________
102. The average net force is, according to Newton's second law, F = ma. The acceleration of the
bullet is
a =
v - v 0
t =
7 1 5 m / s - 0 m / s
2 . 5 0 x 1 0 - 3
s
= 2 . 8 6 x 1 0 5 m / s
2 .
F = (0.015 kg)(2.86 X 105 m/s2) = 4290 N ___________________________________________________________________________________________________
103. a. Newton's second law applied to the spacecraft gives
a = F
m =
2 2 4 0 N
3 . 5 0 x 1 0 4 k g
= 0 . 0 6 4 0 m / s 2 .
This is a + acceleration since the thrust is in the direction of motion.
b. The distance traveled is found to be
s v
2 v 0 2
2 a
( 2310 m / s ) 2 ( 1820 m / s )
2
2 ( 0 . 0640 m / s 2
) 1 . 58 10 7
m 1.58 104
km .
___________________________________________________________________________________________________
108 FORCES AND NEWTON'S LAWS OF MOTION
104. The acceleration of the ball is
a =
v 2 - v
0
2
2 s =
( 4 5 m / s ) 2
2 ( 0 . 4 4 m ) = 2 3 0 0 m / s
2 .
The force needed to produce this acceleration is
F = ma = (0.058 kg)(2300 m/s2) = 130 N. ___________________________________________________________________________________________________
105. a. Newton's second law applied to the 45.0 kg mass, m1, gives
T1 -m1g = m1a1 .
A similar application to the 21.0 kg mass, m2, gives
T2 - m2g = m2a2 .
Now a2 = - a1 = a unless the string breaks or goes slack. Also, T1 = T2 for a taut string. The
above equations become for these conditions
T - m1g = -m1a
T - m2g = m2a .
Subtraction of these equations yields
(m2 - m1)g = (m1 + m2) a .
Then
a =
( m 1 - m
2 ) g
m 1 + m
2
= ( 4 5 . 0 k g - 2 1 . 0 k g ) ( 9 . 8 0 m / s
2 )
4 5 . 0 k g + 2 1 . 0 k g = 3 . 5 6 m / s
2 .
b. Solving one of the above equations for the tension gives
T = m1(g - a) = 45.0 kg(9.80 m/s2 - 3.56 m/s2)
T = 281 N. ___________________________________________________________________________________________________
106. The deceleration produced by the frictional force is
a = -
f k
m =
- k m g
m = -
k g
Now the distance traveled by the automobile is
Chapter 4 109
s =
v 2 - v
0
2
- 2 s g
= - ( 1 6 . 1 m / s )
2
- 2 ( 0 . 7 2 0 ) ( 9 . 8 0 m / s 2 )
= 1 8 . 4 m .
___________________________________________________________________________________________________
107. 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 - fk = 0 where fk = µkFN . Then
mg sin = µkFN .
Similarly, the equation for the y direction is Fy = 0 so
FN - mg cos = 0 or
mg cos = FN .
Division of the equations gives tan = µk , hence µk = tan 11.3° = 0.200. ___________________________________________________________________________________________________
108. Let the tension in the left rope be T1 and the tension in the right rope be T2. The sum of the
vertical forces acting on the point where the ropes join must be zero.
T1 cos 47.0° + T2 cos 35.0° - mg = 0. (1)
Similarly, the horizontal forces must add to zero so
- T1 sin 47.0° + T2 sin 35.0° = 0. (2)
Solving (2) for T1, substituting into (1) and rearranging yields
T2 = 317 N.
Using this result in (2) gives
T1 = 249 N. ___________________________________________________________________________________________________
109. The shortest time to pull the person from the cave corresponds to the maximum acceleration,
a, that the rope can withstand. The maximum acceleration of the person without the rope
breaking produces an apparent weight of the person that is given by
FN = mg + ma
which must be equal to the tension in the rope, T. Now
a = T
m - g =
5 6 9 N
( 5 . 2 0 x 1 0 2 N ) / ( 9 . 8 0 m / s
2 )
- 9 . 8 0 m / s 2 = 0 . 9 2 m / s
2 .
The time required to bring the person out of the cave is given by y = (1/2)at2 or
110 FORCES AND NEWTON'S LAWS OF MOTION
t = 2 y
a
=
2 ( 3 5 . 1 m )
0 . 9 2 m / s 2
= 8 . 7 s .
___________________________________________________________________________________________________
Chapter 4 111
110. Combining Newton's second law with v = v0 + at, we have
F ma m v v 0
t = (0.38 kg)
[(-2.0 m/s) - (+2.1 m/s)]
3.3 10 -3
s 470 N
___________________________________________________________________________________________________
111. Take the x-axis in a free body diagram parallel to the incline with +x up the incline. The box
will slide a distance
s =
- v 0
2
2 a before coming to rest.
Newton's second law for the motion in the x direction is
-mg sin - fk = ma .
Also, in the y direction
-mg cos + FN = 0 .
Solving the above equations for, a, with the help of fk = µkFN gives
a = -g( sin + k cos ) = -9.80 m/s2(sin 15.0° + 0.180 cos 15.0°)
a = -4.24 m/s2 .
Now
s = - ( 1 . 5 0 m / s )
2
2 ( - 4 . 2 4 m / s 2 )
= 0 . 2 6 5 m .
___________________________________________________________________________________________________
112. The buoyant force is FB = mig where mi is the initial mass of the balloon and people. After a
mass m is thrown overboard
mig - (mi - m)g = (mi - m)a
m =
m i a
g + a =
( 3 1 0 k g ) ( 0 . 1 5 m / s 2 )
9 . 8 0 m / s 2 + 0 . 1 5 m / s
2 = 4 . 7 k g .
___________________________________________________________________________________________________
113. The frictional force acting on the picture must balance the weight of the picture if it is to
remain stationary. That is,
f s max
= mg .
112 FORCES AND NEWTON'S LAWS OF MOTION
The maximum frictional force is f s max
= µsFN where the normal force, FN, is the applied force, F,
so that
F = (mg)/µs = (1.10 kg)(9.80 m/s2)/0.660 =16.3 N. ___________________________________________________________________________________________________
114. 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 = ma.
The vertical component of the tension in the string supports the weight of the sphere so
T cos = mg.
Eliminating T from the above equations results in
a = g tan .
b. a = (9.80 m/s2) tan 10.0° =1.73 m/s2.
c. Rearranging the result of part a and setting a = 0 m/s2 gives
= tan-1 (a/g) = 0°
___________________________________________________________________________________________________
115 The maximum acceleration that the van can have without having the sofa slide is found
from
f s max ma
where f s max s F N .
Since the sofa rests on a horizontal surface and there are only two vertical forces, namely
the normal force FN and the weight mg, it follows that FN = mg and
s mg ma .
Thus, the maximum acceleration that the van can have without having the sofa slide is
a s g .
Thus, the maximum distance that the van can travel in 5.1 s without having the sofa slide
Chapter 4 113
x 1
2 at
2 1
2
s gt
2 1
2 ( 0 . 30 ) (9.80 m / s
2 )( 5 . 1 s)
2 38 m .
___________________________________________________________________________________________________
114 FORCES AND NEWTON'S LAWS OF MOTION
116. 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 (1) and (2) for T1 and T3, respectively, and substituting into (3) gives
a =
( m 3 - m
1 -
k m
2 ) g
m 1 + m
2 + m
3
.
Hence,
a = 0.60 m/s2.
b. From part a
T1 = m1(g + a) = 104 N.
and
T3 = m3(g - a) = 230 N. ___________________________________________________________________________________________________
117. If the crate is unaccelerated, then the force components along the incline must sum to zero. That
is,
F cos - fs + mg sin = 0
where
fs = µs(mg cos - F sin ).
Combining these and solving for µs results in
s = (F cos + mg sin )
(mg cos - F sin ) = 0.665.
___________________________________________________________________________________________________
