Download Chapter 04 Solutions - Mosinee School District

Chapter 4
10.
While the engines operate, their total upward thrust exceeds the weight of the rocket, and the rocket experiences a net
upward force. This net force causes the upward velocity of the rocket to increase in magnitude (speed). The upward
thrust of the engines is constant, but the remaining mass of the rocket (and hence, the downward gravitational force or
weight) decreases as the rocket consumes its fuel. Thus, there is an increasing net upward force acting on a
diminishing mass. This yields an acceleration that increases in time.
12.
→
→
T
R
→
R
→
R
→
f
→
mg
→
→
mg
(a)
(b)
n
→
mg
(c)
ur
ur
In the free-body diagrams give above, R represents a force due to air resistance, T is a force due to the thrust of the
r
r
r
rocket engine, n is a normal force, f is a friction force, and the forces labeled mg are gravitational forces.
14.
If the brakes lock, the car will travel farther than it would travel if the wheels continued to roll, because the coefficient
of kinetic friction is less than that of static friction. Hence, the force of kinetic friction is less than the maximum force
of static friction.
PROBLEM SOLUTIONS
4.1
⎛ 2000 lbs ⎞ ⎛ 4.448 N ⎞
4
w = 2 tons ⎜
⎜
⎟ = 2 × 10 N
⎝ 1 ton ⎟⎠ ⎝ 1 lb ⎠
4.2
From v = v0 + a t , the acceleration given to the football is
(
)
aav =
v − v0
10 m s − 0
=
= 50 m s2
t
0.20 s
Then, from Newton’s 2 nd law, we find
(
)
Fav = m aav = ( 0.50 kg ) 50 m s2 = 25 N
Page 4.38
Chapter 4
4.3
4.4
(
)
(a)
ΣFx = ma x = ( 6.0 kg ) 2.0 m s2 = 12 N
(b)
ax =
(a)
Action: The hand exerts a force to the right on the spring. Reaction: The spring exerts an equal magnitude
Σ Fx
12 N
=
= 3.0 m s2
m
4.0 kg
force to the left on the hand. Action: The wall exerts a force to the left on the spring. Reaction: The spring
exerts an equal magnitude force to the right on the wall. Action: Earth exerts an downward gravitational force
on the spring. Reaction: The spring exerts an equal magnitude gravitational force upward on the Earth.
(b)
Action: The handle exerts a force upward to the right on the wagon. Reaction: The wagon exerts an equal
magnitude force downward to the left on the handle. Action: Earth exerts an upward contact force on the
wagon. Reaction: The wagon exerts an equal magnitude downward contact force on the Earth. Action: Earth
exerts an downward gravitational force on the wagon. Reaction: The wagon exerts an equal magnitude
gravitational force upward on the Earth.
(c)
Action: The player exerts a force upward to the left on the ball. Reaction: The ball exerts an equal magnitude
force downward to the right on the player. Action: Earth exerts an downward gravitational force on the ball.
Reaction: The ball exerts an equal magnitude gravitational force upward on the Earth.
(d)
Action: M exerts a gravitational force to the right on m. Reaction: m exerts an equal magnitude gravitational
force to the left on M.
(e)
Action: The charge +Q exerts an electrostatic force to the right on the charge − q . Reaction: The charge − q
exerts an equal magnitude electrostatic force to the left on the charge +Q.
(f)
Action: The magnet exerts a force to the right on the iron. Reaction: The iron exerts an equal magnitude force
to the left on the magnet.
4.5
The weight of the bag of sugar on Earth is
⎛ 4.448 N ⎞
wE = mgE = ( 5.00 lbs ) ⎜
= 22 .2 N
⎝ 1 lb ⎟⎠
If gM is the free-fall acceleration on the surface of the Moon, the ratio of the weight of an object on the Moon to its
weight when on Earth is
wM
mgM
g
=
= M
wE
mgE
gE
so
Page 4.39
Chapter 4
⎛g ⎞
w M = wE ⎜ M ⎟ .
⎝ gE ⎠
Hence, the weight of the bag of sugar on the Moon is
⎛ 1⎞
wM = ( 22.2 N ) ⎜ ⎟ = 3.71 N .
⎝ 6⎠
On Jupiter, its weight would be
⎛g ⎞
wJ = wE ⎜ J ⎟ = ( 22.2 N ) ( 2.64 ) = 58.7 N
⎝ gE ⎠
The mass is the same at all three locations, and is given by
m =
( 5.00 lb ) ( 4.448 N lb ) = 2.27 kg
wE
=
9.80 m s2
gE
a =
4.6
ΣF
7.5 × 105 N
=
= 5.0 × 10 −2 m s2
m
1.5 × 107 kg
and v = v0 + at gives
t =
4.7
v − v0
80 km h − 0 ⎛ 0.278 m s ⎞ ⎛ 1 min ⎞
=
= 7.4 min
a
5.0 × 10 −2 m s2 ⎜⎝ 1 km h ⎟⎠ ⎜⎝ 60 s ⎟⎠
Summing the forces on the plane shown gives
a 2.0 m/s2
4.8
⇒
From which,
f = 9.6 N
(a)
(
10 N − f = ( 0.20 kg ) 2.0 m s2
 ΣFx = ma x
)
→
f
F 10 N
The sphere has a larger mass than the feather. Hence, the sphere experiences a larger gravitational force
Fg = mg than does the feather.
(b)
The time of fall is less for the sphere than for the feather. This is because air resistance affects the motion of
the feather more than that of the sphere.
(c)
In a vacuum, the time of fall is the same for the sphere and the feather. In the absence of air resistance, both
objects have the free-fall acceleration g.
(d)
In a vacuum, the total force on the sphere is greater than that on the feather. In the absence of air resistance,
the total force is just the gravitational force, and the sphere weighs more than the feather.
Page 4.40
Chapter 4
4.9 The vertical acceleration of the salmon as it goes from v0 y = 3.0 m s (underwater) to v y = 6.0 m s (after moving
upward 1.0 m or 2/3 of its body length) is
ay =
v y2 − v02 y
2 Δy
( 6.0
=
m s ) − ( 3.0 m s )
2
2
2 ( 1.00 m )
= 13.5 m s2
Applying Newton’s second law to the vertical leap of this salmon having a mass of 61 kg , we find
ΣFy = ma y ⇒ F − mg = ma y
or
(
)
m
m⎞
⎛
F = m a y + g = ( 61 kg ) ⎜ 13.5 2 + 9.8 2 ⎟ = 1.4 × 103 N
⎝
s
s ⎠
4.10
The acceleration of the bullet is given by
v 2 − v02
( 320 m s ) − 0
a =
=
2 ( Δx )
2 ( 0.82 m )
2
Then,
⎡ ( 320 m s )2
ΣF = ma = 5.0 × 10 −3 kg ⎢
⎢ 2 ( 0.82 m )
⎣
(
4.11
(a)
)
⎤
⎥ = 3.1 × 102 N
⎥
⎦
From the second law, the acceleration of the
+y
a =
(b)
ur
F
ΣF
2 000 N − 1 800 N
=
= 0.200 m s2
m
1 000 kg
ur
f
The distance moved is
## Δx = v0 t +
f = 1 800 N
F = 2 000 N
+x
boat is
1 2
1
at = 0 +
0.200 m s2
2
2
(c) The final velocity is##
(
) (10.0 s )2
(
= 10.0 m
v = v0 + at = 0 + 0.200 m s2
Page 4.41
) (10.0 s ) =
2.00 m s .
Chapter 4
4.12
(a)
Choose the positive y-axis in the forward direction. We
y
resolve
the forces into their components as
200 N
346 N
450 N
–78.1 N
443 N
Resultant
ΣFx = 122 N
ΣFy = 790 N
30°
00
N
400 N
10°
4
y-component
F1
x-component
0N
F 2 45
Force
x
The magnitude and direction of the resultant force is
( ΣFx )2
FR =
( )
+ ΣFy
2
⎛ Σ Fx ⎞
⎟ = 8.77° to right of y-axis
⎝ Σ Fy ⎠
θ = tan −1 ⎜
= 799 N
r
Thus, FR = 799 N at 8.77 ° to the right of the forward direction
(b)
r
The acceleration is in the same direction as FR and has magnitude
a =
4.13
(a)
At terminal speed, the magnitude of the air resistance force is equal to the weight of the skydiver
(F
drag
k =
(b)
FR
799 N
=
= 0.266 m s2
m
3 000 kg
)
= mg . Therefore,
Fdrag
v2
At the speed v =
=
1
2
mg
2
vterminal
=
( 65.0 kg ) ( 9.80 m s2 )
( 55.0 m s )2
= 0.211 kg m
vterminal , the upward force due to air resistance is less than the downward gravitational
force acting on the skydiver, leaving a net downward force of magnitude Fnet = mg − Fdrag . Thus, the
skydiver will have a downward acceleration of magnitude
a =
Fnet
mg − kv 2
⎛ k⎞
=
= g−⎜ ⎟
⎝ m⎠
m
m
⎛ vterminal ⎞
⎜⎝
2 ⎟⎠
2
⎛ 0.211 kg m ⎞ ⎛ 55.0 m s ⎞
= 9.80 m s2 − ⎜
⎜
⎟⎠
2
⎝ 65.0 kg ⎟⎠ ⎝
4.14
(a)
2
= 7.35 m s2
With the wind force being horizontal, the only vertical force acting on the object is its own weight, mg. This
gives the object a downward acceleration of
Page 4.42
Chapter 4
ay =
Σ Fy
m
=
− mg
= −g
m
The time required to undergo a vertical displacement Δy = −h , starting with initial vertical velocity v0 y = 0 ,
is found from Δ y = v0 y t +
−h = 0 −
(b)
g 2
t
2
Σ Fx
F
=
m
m
With v0 x = 0 , the horizontal displacement the object undergoes while falling a vertical distance h is given by
1
2
a x t 2 as
1 ⎛ F ⎞ ⎛ 2h ⎞
Δx = 0 + ⎜ ⎟ ⎜
2 ⎝ m ⎠ ⎝ g ⎟⎠
2
=
Fh
mg
The total acceleration of this object while it is falling will be
a =
4.15
2h
g
t =
The only horizontal force acting on the object is that due to the wind, so ΣFx = F and the horizontal
Δx = vox t +
(d)
a y t 2 as
or
acceleration will be a x =
(c)
1
2
a x2 + a y2 =
( F m )2 + ( − g )2
=
( F m )2
+ g2
Starting with v0 y = 0 and falling 30 m to the ground, the velocity of the ball just before it hits is
(
v1 = − v02y + 2a y Δy = − 0 + 2 −9.80 m s2
) ( −30 m )
= −24 m s
On the rebound, the ball has v y = 0 after a displacement Δ y = + 20 m . Its velocity as it left the ground must have
been
(
v2 = + vy2 − 2 a y Δy = + 0 − 2 −9.80 m s2
) ( 20 m )
= +20 m s
Thus, the average acceleration of the ball during the 2.0-ms contact with the ground was
aav =
+20 m s − ( −24 m s )
v2 − v1
=
= +2.2 × 10 4 m s2
Δt
2.0 × 10 −3 s
upward
The average resultant force acting on the ball during this time interval must have been
(
)
Fnet = maav = ( 0.50 kg ) +2.2 × 10 4 m s2 = +1.1 × 10 4 N
or
ur
F net = 1.1 × 10 4 N upward
Page 4.43
Chapter 4
4.16
Since the two forces are perpendicular to each other, their resultant is:
( 180 N )2 + ( 390 N )2
FR =
⎛ 390 N ⎞
= 430 N , at θ = tan −1 ⎜
= 65.2° N of E
⎝ 180 N ⎟⎠
Thus,
a =
FR
430 N
=
= 1.59 m s2
m
270 kg
or
r
a = 1.59 m s2 at 65.2° N of E .
4.17
(a)
Since the burglar is held in equilibrium, the tension in the
y
vertical cable equals the burglar’s weight of 600 N Now,
consider the junction in the three cables:##
Σ Fy = 0 , giving T2 sin 37.0 ° − 600 N = 0
→
T2
→
T1
37.0°
x
w600 N
or
T2 =
600 N
= 997 N in the inclined cable
sin 37.0 °
Also, ΣFx = 0 which yields T2 cos 37.0° − T1 = 0 , or
T1 = ( 997 N ) cos 37.0 ° = 796 N in the horizontal cable
(b) If the left end of the originally horizontal cable was attached to a point higher up the wall,
the tension in this cable would then have an upward component. This upward component
would support part of the weight of the cat burglar, thus.
decreasing the tension in the cable on the right
Page 4.44
Chapter 4
4.18
Using the reference axis shown in the sketch at the right, we see that
ΣFx = T cos14.0° − T cos14.0° = 0
and
ΣFy = −T sin 14.0 ° − T sin 14.0° = −2T sin 14.0 °
y
Thus, the magnitude of the resultant force exerted on the tooth by the wire brace is
R =
( ΣFx )2
( )
+ Σ Fy
2
=
0 + ( −2T sin 14.0 ° )
2
x
14°
T
14°
= 2T sin 14.0 °
T
or
R = 2 ( 18.0 N ) sin 14.0 ° = 8.71 N
4.19
From ΣFx = 0 , T1 cos 30.0° − T2 cos 60.0° = 0
or
T2 = ( 1.73 ) T1
[1]
Then Σ Fy = 0 becomes
T1 sin 30.0 ° + (1.73 T1 ) sin 60.0 ° − 150 N = 0
y
→
T2
which gives T1 = 75.0 N in the right side cable .
Finally, Equation [1] above gives T2 = 130 N in the left side cable .
→
T1
60.0°
30.0°
x
150 N
4.20
If the hip exerts no force on the leg, the system must be
in equilibrium with the three forces shown in the free-body diagram.
y
Thus Σ Fx = 0 becomes
w2 cos α = ( 110 N ) cos 40 °
11
[1]
40°
From Σ Fy = 0 , we find
w2 sin α = 220 N − ( 110 N ) sin 40 °
0N
→
w2
α
220 N
[2]
Dividing Equation [2] by Equation [1] yields
Page 4.45
x
Chapter 4
⎛ 220 N − ( 110 N ) sin 40 ° ⎞
= 61°
(110 N ) cos 40° ⎟⎠
⎝
α = tan −1 ⎜
Then, from either Equation [1] or [2], w2 = 1.7 × 102 N .
4.21
(a) Free-body diagrams of the two blocks are shown at the
right.
Note that each block experiences a downward gravitational
force
(
)
Fg = ( 3.50 kg ) 9.80 m s2 = 34.3 N
Also, each has the same upward acceleration as the elevator,
in this case a y = +1.60 m s2 .
Applying Newton’s second law to the lower block:
ΣFy = ma y
⇒
T2 − Fg = ma y
(
)
T2 = Fg + ma y = 34.3 N + ( 3.50 kg ) 1.60 m s2 = 39.9 N
or
Next, applying Newton’s second law to the upper block:
ΣFy = ma y
⇒
T1 − T2 − Fg = ma y
or
(
)
T1 = T2 + Fg + ma y = 39.9 N + 34.3 N + ( 3.50 kg ) 1.60 m s2 = 79.8 N
(b)
Note that the tension is greater in the upper string, and this string will break first as the acceleration of the
system increases. Thus, we wish to find the value of a y when T1 = 85.0 N . Making use of the general
relationships derived in (a) above gives:
(
)
T1 = T2 + Fg + ma y = Fg + ma y + Fg + ma y = 2 Fg + 2ma y
or
ay =
T1 − 2 Fg
2m
=
85.0 N − 2 ( 34.3 N )
2 ( 3.50 kg )
= 2.34 m s2
Page 4.46
Chapter 4
4.22
(a)
Free-body diagrams of the two blocks are shown at the right.
Note that each block experiences a downward gravitational
force Fg = mg .
Also, each has the same upward acceleration as the elevator,
ay = + a .
Applying Newton’s second law to the lower block:
ΣFy = ma y
⇒
T2 − Fg = ma y
or
T2 = mg + ma = m ( g + a )
Next, applying Newton’s second law to the upper block:
ΣFy = ma y
⇒
T1 − T2 − Fg = ma y
or
T1 = T2 + Fg + ma y = ( mg + ma ) + mg + ma = 2 ( mg + ma ) = 2T1
(b)
Note that T1 = 2T2 , so the upper string breaks first as the acceleration of the system increases.
(c)
When the upper string breaks, both blocks will be in free-fall with a = − g . Then, using the results of part (a),
T2 = m ( g + a ) = m ( g − g ) = 0 and T1 = 2T2 = 0 .
4.23
m = 1.00 kg and mg = 9.80 N
⎛ 0.200 m ⎞
= 0.458°
⎝ 25.0 m ⎟⎠
α = tan −1 ⎜
Since a y = 0 , require that ΣFy = T sin α + T sin α − mg = 0 ,
giving 2 T s in α = mg ,
25.0 m
a
or
→
T
9.80 N
= 613 N
T =
2 sinα
4.24
0.200 m
→
mg
25.0 m
a
→
T
The resultant force exerted on the boat by the people is
2 ⎡⎣ ( 600 N ) cos 30.0° ⎤⎦ = 1.04 × 103 N in the forward direction
If the boat moves with constant velocity, the total force acting on it must be zero. Hence, the resistive force exerted
on the boat by the water must be
Page 4.47
Chapter 4
r
f = 1.04 × 103 N in the rearward direction
4.25
The forces on the bucket are the tension in the rope and the weight of
(
)
the bucket, mg = ( 5.0 kg ) 9.80 m s2 = 49 N . Choose the positive
→
T
direction upward and use Newton’s second law:
ΣFy = ma y
(
T − 49 N = ( 5.0 kg ) 3.0 m s2
)
→
a
T = 64 N
→
mg
4.26
(a) From Newton’s second law, we find the acceleration as
ax =
Σ Fx
10 N
=
= 0.33 m s2
m
30 kg
F 10 N
To find the distance moved, we use
Δ x = v0 x t +
(b)
1
1
a t2 = 0 +
0.33 m s2
2 x
2
(
) ( 3.0 s )2
= 1.5 m
If the shopper places her 30 N ( 3.1 kg ) child in the cart, the new acceleration will be
ax =
ΣFx
10 N
=
= 0.30 m s2
mtotal
33 kg
and the new distance traveled in 3.0 s will be
Δx ′ = 0 +
1
0.30 m s2
2
(
) ( 3.0 s )2
= 1.4 m
Page 4.48
Chapter 4
4.27
(a)
The average acceleration is given by
aav =
v − v0
5.00 m s − 20.0 m s
=
= −3.75 m s2
Δt
4.00 s
The average force is found from Newton’s second law as
(
)
Fav = maav = ( 2000 kg ) −3.75 m s2 = − 7.50 × 103 N
(b)
The distance traveled is:
⎛ 5.00 m s + 20.0 m s ⎞
x = vav ( Δt ) = ⎜
⎟⎠ ( 4.00 s ) = 50.0 m
⎝
2
4.28
Let m1 = 10.0 kg , m2 = 5.00 kg , and θ = 40.0° .
→
Applying the second law to each object gives
m1a = m1 g − T
a
→
[1]
and
→
f
m2 a = T − m2 g sin θ
→
n m1g
m1
→
→
m1g
[2]
T
→
T
→
m2
a
→
m2g
Adding these equations yields
m1a + m2 a = m1 g − T + T − m2 g
or
⎛ m − m2 sin θ ⎞
a = ⎜ 1
g
⎝ m1 + m2 ⎟⎠
so
⎛ 10.0 kg − ( 5.00 kg ) sin 40.0 ° ⎞
2
2
a = ⎜
⎟ 9.80 m s = 4.43 m s
15.0 kg
⎝
⎠
(
)
Then, Equation [1] yields
T = m1 ( g − a ) = ( 10.0 kg ) ⎡⎣ ( 9.80 − 4.43 ) m s2 ⎤⎦ = 53.7 N
4.29
(a)
The resultant external force acting on this system, consisting of all three blocks having a total mass of 6.0 kg,
is 42 N directed horizontally toward the right. Thus, the acceleration produced is
a =
(b)
ΣF
42 N
=
= 7.0 m s2 horizontally to the right
6.0 kg
m
Draw a free body diagram of the 3.0-kg block and apply Newton’s second law to the horizontal forces acting
on this block:
Page 4.49
Chapter 4
ΣFx = ma x ⇒
(c)
(
)
42 N − T = ( 3.0 kg ) 7.0 m s2 , and therefore T = 21 N
The force accelerating the 2.0-kg block is the force exerted on it by the 1.0-kg block. Therefore, this force is
given by:
(
)
F = ma = ( 2.0 kg ) 7.0 m s2 , or
r
F = 14 N horizontally to the right .
4.30
The acceleration of the mass down the incline is given by
Δ x = v0 t +
1 2
at ,
2
or
0.80 m = 0 +
1
2
a ( 0.50 s )
2
This gives
a =
2 ( 0.80 N )
( 0.50 s )2
= 6.4 m s2
(
)
Thus, the net force directed down the incline is F = ma = ( 2.0 kg ) 6.4 m s2 = 13 N .
4.31
(a) Assuming frictionless pulleys, the tension is uniform through
the entire length of the rope. Thus, the tension at the point
where the rope attaches to the leg is the same as that at the
8.00-kg block. Part (a) of the sketch at the right gives a freebody diagram of the suspended block. Recognizing that the
block has zero acceleration, Newton’s second law gives
ΣFy = T − mg = 0
or
(
)
T = mg = ( 8.00 kg ) 9.80 m s2 = 78.4 N
(b) Part (b) of the sketch above gives a free-body diagram of the pulley near the foot. Here, F is the magnitude of
the force the foot exerts on the pulley. By Newton’s third law, this is the same as the magnitude of the force
the pulley exerts on the foot. Applying the second law gives:
ΣFx = T + T cos 70.0° − F = ma x = 0
or
F = T ( 1 + cos 70.0° ) = ( 78.4 N ) ( 1 + cos 70.0° ) = 105 N
Page 4.50
Chapter 4
4.32
(a)
(b)
Note that the blocks move on a horizontal surface with a y = 0 . Thus, the net vertical force acting on each
block and on the combined system of both blocks is zero. The net horizontal force acting on the combined
system consisting of both m1 and m2 is ΣFx = F − P + P = F .
(c)
Looking at just m1 , ΣFy = 0 as explained above, while ΣFx = F − P .
(d)
Looking at just m2 , we again have ΣFy = 0 , while ΣFx = + P
(e)
For m1 :
ΣFx = ma x
⇒
F − P = m1a .
For m2 :
ΣFx = ma x
⇒
P = m2 a .
(f)
Substituting the second of the equations found in (e) above into the first gives the following:
m1a = F − P = F − m2 a
( m1 + m2 ) a
or
= F
and
a = F ( m1 + m2 )
Then substituting this result into the second equation from (e), we have
⎛
⎞
F
P = m2 a = m2 ⎜
⎟
+
m
m
⎝ 1
2⎠
(g)
or
⎛ m2 ⎞
P = ⎜
F
⎝ m1 + m2 ⎟⎠
Realize that applying the force to m2 rather than m1 would have the effect of interchanging the roles of
m1 and m2 . We may easily find the results for that case by simply interchanging the labels m1 and m2 in the
results found in (f) above. This gives a = F ( m2 + m1 ) (the same result as in the first case) and
⎛
m1 ⎞
P = ⎜
F
⎝ m2 + m1 ⎟⎠
We see that
the contact force, P, is larger in this case because m1 > m2 .
Page 4.51
Chapter 4
4.33
Taking the downward direction as positive, applying Newton’s second law to the falling person yields
ΣFy = mg − f = ma y , or
ay = g −
⎛ 100 N ⎞
f
= 9.80 m s2 − ⎜
= 8.6 m s2
m
⎝ 80 kg ⎟⎠
Then, v 2y = v02y + 2 a y ( Δy ) gives the velocity just before hitting the net as
vy =
44.34
(a)
v02y + 2 a y ( Δ y ) =
(
0 + 2 8.6 m s2
) ( 30 m )
= 23 m s
First, consider a system consisting of the two blocks combined,
with mass m1 + m2 . For this system, the only external horizontal
force is the tension in cord A pulling to the right. The tension in
cord B is a force one part of our system exerts on another part of
our system, and is therefore an internal force.
Applying Newton’s second law to this system (including only external forces, as we should) gives
ΣFx = ma
⇒
TA = ( m1 + m2 ) a
[1]
Now, consider a system consisting of only m2 . For this system, the tension in cord B is an external force since
it is a force exerted on block 2 by block 1 (which is not part of this system). Applying Newton’s second law to
this system gives
ΣFx = ma
⇒
TB = m2 a
[2]
Comparing equations [1] and [2], and realizing that the acceleration is the same in both cases (see Part (b)
below), it is clear that:
Cord A exerts a larger force on block 1 than cord B exerts on block 2
(b)
Since cord B connecting the two blocks is taut and unstretchable, the two blocks stay a fixed distance apart,
and the velocities of the two blocks must be equal at all times. Thus, the rates at which the velocities of the
two blocks change in time are equal, or the the two blocks must have equal accelerations .
(c)
Yes. Block 1 exerts a forward force on Cord B, so Newton’s third law tells us that Cord B exerts a
force of equal magnitude in the backward direction on Block 1.
Page 4.52
Chapter 4
4.35
(a)
When the acceleration is upward, the total upward force T
(
must exceed the total downward force
)
w = mg = ( 1 500 kg ) 9.80 m s2 = 1.47 × 10 4 N
(b)
When the velocity is constant, the acceleration is zero. The total upward force T and the total downward force
w must be equal in magnitude .
(c)
If the acceleration is directed downward, the total downward force w must exceed the total upward force T.
(d)
ΣFy = ma y
⇒
(
)
T = mg + ma y = ( 1 500 kg ) 9.80 m s2 + 2.50 m s2 = 1.85 × 10 4 N
Yes , T > w .
(e)
ΣFy = ma y
⇒
(
)
T = mg + ma y = ( 1 500 kg ) 9.80 m s2 + 0 = 1.47 × 10 4 N
Yes , T = w .
(f)
ΣFy = ma y
⇒
(
)
T = mg + ma y = ( 1 500 kg ) 9.80 m s2 − 1.50 m s2 = 1.25 × 10 4 N
Yes , T < w .
4.36
Note that if the cord connecting the two blocks has a fixed length, the accelerations of the blocks must have equal
magnitudes, even though they differ in directions. Also, observe from the diagrams, we choose the positive
direction for each block to be in its direction of motion.
(First consider the block moving along the horizontal. The only force in
x
→
T
the direction of movement is T. Thus,
ΣFx = ma x ⇒
→
T = ( 5.00 kg ) a
[1]
5.00 kg
(a)
ΣFy = ma y ⇒ 98.0 N − T = ( 10.0 kg ) a
[2]
Equations [1] and [2] can be solved simultaneously to give:
98.0 N − ( 5.00 kg ) a = ( 10.0 kg ) a
or
and
Page 4.53
a =
y
10.0 kg
mg 98.0 N
Next consider the block which moves vertically. The forces on it are the
tension T and its weight, 98.0 N.
T
98.0 N
= 6.53 m s2
15.0 kg
(b)
Chapter 4
(
)
T = ( 5.00 kg ) 6.53 m s2 = 32.7 N
4.37
Trailer
300 kg
→
T
→
nT
Car
1 000 kg
→
→
→
wT
→
→
nc
→
Rcar
F
T
nc
q
→
wc
→
F
Choose the +x direction to be horizontal and forward with the +y vertical and upward. The common acceleration of
the car and trailer then has components of a x = +2 .15 m s2 and a y = 0 .
(a)
The net force on the car is horizontal and given by
( ΣFx )car
(b)
(
The net force on the trailer is also horizontal and given by
( ΣFx )trailer
(c)
)
= F − T = mcar a x = ( 1000 kg ) 2.15 m s2 = 2.15 × 103 N forward
(
)
= +T = mtrailer a x = ( 300 kg ) 2.15 m s2 = 645 N forward
Consider the free-body diagrams of the car and trailer. The only horizontal force acting on the trailer is
T = 645 N forward , and this is exerted on the trailer by the car. Newton’s third law then states that the force
the trailer exerts on the car is 645 N toward the rear .
(d)
The road exerts two forces on the car. These are F and nc shown in the free-body diagram of the car.
From part (a),
F = T + 2 .15 × 103 N = 645 N + 2 .15 × 103 N = + 2 .80 × 103 N
( )
Also, ΣFy
car
= nc − wc = mcar a y = 0 , so nc = wc = mcar g = 9.80 × 103 N
The resultant force exerted on the car by the road is then
Rcar =
F 2 + nc2 =
( 2 .80 × 103 N )2 + ( 9.80 × 103 N )2
= 1.02 × 10 4 N
⎛n ⎞
at θ = tan −1 ⎜ c ⎟ = tan −1 ( 3.51) = 74.1 ° above the horizontal and forward. Newton’s third law then
⎝ F⎠
states that the resultant force exerted on the road by the car is
1.02 × 10 4 N at 74.1° below the horizontal and rearward
Page 4.54
Chapter 4
44.38
First, consider the 3.00-kg rising mass. The forces on it are the tension, T,
→
law becomes
T − 29.4 N = ( 3.00 kg ) a
→
T
and its weight, 29.4 N. With the upward direction as positive, the second
T
Rising
Mass
m1 3.00 kg
[1]
The forces on the falling 5.00-kg mass are its weight and T, and its
Falling
Mass
m2 5.00 kg
w1 29.4 N
w2 49.0 N
acceleration has the same magnitude as that of the rising mass. Choosing
the positive direction down for this mass, gives
49.0 N − T = ( 5.00 kg ) a
(a)
[2]
Solving Equation (2) for a and substituting into [1] gives
⎛ 3.00 kg ⎞
T − 29.4 N = ⎜
( 49.0 N − T )
⎝ 5.00 kg ⎟⎠
T = 36.8 N
and the tension is
(b)
Equation (2) then gives the acceleration as
a =
(c)
49.0 N − 36.8 N
= 2.44 m s2
5.00 kg
Consider the 3.00-kg mass. We have
Δ y = v0 y t +
4.39
1.60 T = 58.8 N
or
1
1
2
2.44 m s2 ( 1.00 s ) = 1.22 m
ay t 2 = 0 +
2
2
(
)
When the block is on the verge of moving, the static friction force has a magnitude fs =
( fs )max
= μs n .
Since equilibrium still exists and the applied force is 75 N, we have
ΣFx = 75 N − fs = 0
or
( fs )max
= 75 N
In this case, the normal force is just the weight of the crate, or n = mg . Thus, the coefficient of static friction is
μs =
( fs )max
n
=
( fs )max
mg
75 N
=
(
( 20 kg ) 9.80 m s2
)
= 0.38
After motion exists, the friction force is that of kinetic friction, fk = μk n .
Since the crate moves with constant velocity when the applied force is 60 N, we find that ΣFx = 60 N − fk = 0 or
fk = 60 N . Therefore, the coefficient of kinetic friction is
μk =
fk
f
60 N
= k =
n
mg
( 20 kg ) 9.80 m s2
(
)
= 0.31
Page 4.55
Chapter 4
4.40
(a)
The static friction force attempting to prevent motion may reach a maximum value of
( fs )max
(
)
= μs n1 = μs m1 g = ( 0.50 ) ( 10 kg ) 9.80 m s2 = 49 N
This exceeds the force attempting to move the system, w2 = m2 g = 39 N . Hence, the system remains at rest
and the acceleration is a = 0
(b)
Once motion begins, the friction force retarding the motion is
(
)
fk = μk n1 = μk m1 g = ( 0.30 ) ( 10 kg ) 9.80 m s2 = 29 N
This is less than the force trying to move the system, w2 = m2 g . Hence, the system gains speed at the rate
a =
4.41
(a)
(
⎡ 4.0 kg − 0.30 ( 10 kg ) ⎤⎦ 9.80 m s2
Fnet
m g − μk m1 g
= 2
= ⎣
4.0 kg + 10 kg
mtotal
m1 + m2
)
= 0.70 m s2
Since the crate has constant velocity, a x = a y = 0 .
Applying Newton’s second law:
ΣFx = F cos 20.0° − fk = ma x = 0
fk = ( 300 N ) cos 20.0 ° = 282 N
or
and
ΣFy = n − F sin 20.0 ° − w = 0
or
n = ( 300 N ) sin 20.0° + 1 000 N = 1.10 × 103 N
The coefficient of friction is then
μk =
(b)
fk
282 N
=
= 0.256
n
1.10 × 103 N
In this case, ΣFy = n + F sin 20.0 ° − w = 0 , so n = w − F sin 20.0 ° = 897 N
The friction force now becomes fk = μ k n = ( 0.256 ) ( 897 N ) = 230 N
⎛ w⎞
Therefore, ΣFx = F cos 20.0° − fk = ma x = ⎜ ⎟ a x and the acceleration is
⎝ g⎠
a =
4.42
(a)
ax =
( F cos 20.0° − fk ) g
w
(
⎡ ( 300 N ) cos 20.0 ° − 230 N ⎤⎦ 9.80 m s2
= ⎣
1 000 N
vx − v0 x
6.00 m s − 12.0 m s
=
= − 1.20 m s2
5.00 s
t
Page 4.56
)
= 0.509 m s2
Chapter 4
(b)
From Newton’s second law, ΣFx = − fk = ma x , or fk = − ma x .
The normal force exerted on the puck by the ice is n = mg , so the coefficient of friction is
μk =
)
)
= 0.122
⎛ v + v0 x ⎞
⎛ 6.00 m s + 12.0 m s ⎞
Δx = ( vx )av t = ⎜ x
⎟⎠ ( 5.00 s ) = 45.0 m
⎟⎠ t = ⎜⎝
⎝
2
2
(c)
4.43
(
(
− m −1.20 m s2
fk
=
n
m 9.80 m s2
When the load on the verge of sliding forward on the
mload g→
→
bed of the slowing truck, the rearward directed static
fs
friction force has its maximum value
( fs )max =μs n =
→
n
μs mload g
Since slipping is not yet occurring, this single horizontal force must give the load an acceleration equal to that the
truck.
Thus, ΣFx = ma x
(a)
− μ s mload g = mload atruck
⇒
or
atruck = − μs g .
If slipping is to be avoided, the maximum allowable rearward acceleration of the truck is seen to be
atruck = − μ s g , and v2x = v02 x + 2 a x ( Δx ) gives the minimum stopping distance as
( Δx )min
=
v2
0 − v02x
= 0x
2 μs g
2 ( atruck )max
( Δx )min =
If v0 x = 12 m s and μs = 0.500 , then
(b)
4.44
( 12.0 m s )2
2 ( 0.500 ) ( 9.80 m s2 )
Examining the calculation of Part (a) shows that neither mass is necessary
(a) The free-body diagram of the crate is shown at the right. Since the crate has no
(
)
vertical acceleration a y = 0 , we see that
ΣFy = ma y
⇒
n − mg = 0
or
n = mg
The only horizontal force present is the friction force exerted on the crate by
the truck bed. Thus,
ΣFx = ma x
⇒
f = ma
Page 4.57
= 14.7 m
Chapter 4
If the crate is not to slip (i.e., the static case is to prevail), it is necessary that the required friction force not
exceed the maximum possible static friction force, ( fs ) max = μ s n . From this, we find the maximum
allowable acceleration as:
amax =
(b)
fmax
=
m
( fs )max
m
=
μs n
m
=
μs ( mg )
m
(
)
= μs g = ( 0.350 ) 9.80 m s2 = 3.43 m s2
Once slipping has started, the kinetic friction case prevails and f = fk = μk n . The acceleration of the crate
in this case will be
a =
4.45
(a)
μ ( mg )
μ n
f
= k = k
= μ k g = ( 0.320 ) 9.80 m s2 = 3.14 m s2
m
m
m
(
)
The acceleration of the system is found from
Δy = v0 y t +
1 2
1
2
at , or 1.00 m = 0 + a ( 1.20 s )
2
2
→
a
→
→
f
Using the free body diagram of m2 , the second law gives
( 5.00 kg ) ( 9.80
m
s2
→
n m1g
which gives a = 1.39 m s2
) − T = ( 5.00 kg ) (1.39
m
s2
m1
→
→
m1g
) or
T
→
T
→
m2
a
→
m2g
T = 42 .1 N
Then applying the second law to the horizontal motion of m1
(
)
42 .1 N − f = ( 10.0 kg ) 1.39 m s2 or f = 28.2 N
Since n = m1 g = 98.0 N , we have
μk =
4.46
(a)
f
28.2 N
=
= 0.287
n
98.0 N
Since the puck is on a horizontal surface, the normal force is vertical.
With a y = 0 , we see that
ΣFy = ma y
⇒
n − mg = 0
or
n = mg
Once the puck leaves the stick, the only horizontal force is a friction force in the negative x-direction (to
oppose the motion of the puck). The acceleration of the puck is
ax =
− μ k ( mg )
Σ Fx
− fk
− μk n
=
=
=
= − μk g
m
m
m
m
Page 4.58
Chapter 4
(b)
Then v2x = v02 x + 2 a x ( Δx ) gives the horizontal displacement of the puck before coming to rest as
Δx =
4.47
(a)
v2x − v02x
0 − v02
v02
=
=
2a x
2 μk g
2 ( − μk g )
The crate does not accelerate perpendicular to the incline. Thus,
ΣF⊥ = ma⊥ = 0
⇒
n = F + mg cos θ
→
F
→
n
→
The net force tending to move the crate down the incline is
fs
ΣF = mg sin θ − fs , where fs is the force of static friction
q 35.0°
between the crate and the incline. If the crate is in equilibrium,
then mg sin θ − fs = 0
so
fs = Fg sin θ
→
mg
But, we also know fs ≤ μ s n = μ s ( F + mg cos θ )
Therefore, we may write
mg sin θ ≤ μ s ( F + mg cos θ ) or,
⎛ sin θ
⎞
⎛ sin 35.0°
⎞
F ≥ mg ⎜
− cos θ ⎟ = ( 3.00 kg ) 9.80 m s2 ⎜
− cos 35.0° ⎟ = 32.1 N
⎝ 0.300
⎠
⎝ μs
⎠
(
(a)
r
Find the normal force n on the 25.0 kg box:
y
ΣFy = n + ( 80.0 N ) sin 25.0° − 245 N = 0
→
n
or n = 211 N
F
→
f
0N
80.
Now find the friction force, f, as
25.0°
x
mg 245 N
f = μ k n = 0.300 ( 211 N ) = 63.4 N
From the second law, we have ΣFx = ma , or
( 80.0 N ) cos 25.0° − 63.4 N= ( 25.0 kg ) a
which yields a = 0.366 m s2 .
(b)
When the box is on the incline,
y
ΣFy = n + ( 80.0 N ) sin 25.0° − ( 245 N ) cos 10.0° = 0
→
n
→
giving n = 207 N
f
The friction force is f = μ k n = 0.300 ( 207 N ) = 62.2 N
The net force parallel to the incline is then
ΣFx = ( 80.0 N ) cos 25.0° − ( 245 N ) sin 10.0° − 62 .2 N=-32 .3 N
Page 4.59
245 N
4.48
)
10.0°
N
.0
80
F 25.0°
x
Chapter 4
Thus,
a =
4.49
(a)
ΣFx
−32 .3 N
=
= − 1.29 m s2 or 1.29 m s2 down the incline
25.0 kg
m
The object will fall so that ma = mg − bv, or a =
( mg − bv )
→
where the downward direction is taken as positive.
v = vterminal =
(b)
mg
=
b
m s2
15 kg s
)
→
m
Equilibrium ( a = 0 ) is reached when
( 50 kg ) ( 9.80
→
f bv
m
v
→
mg
= 33 m s
If the initial velocity is less than 33 m/s, then a ≥0 and 33 m/s is the largest velocity attained by the object. On
the other hand, if the initial velocity is greater than 33 m/s, then a ≤ 0 and 33 m/s is the smallest velocity
attained by the object. Note also that if the initial velocity is 33 m/s, then a = 0 and the object continues falling
with a constant speed of 33 m/s.
4.50
(a)
The force of friction is found as f = μ k n = μ k ( mg )
Choose the positive direction of the x-axis in the direction of motion and apply Newton’s second law. We
have
ΣFx = − f = ma x
ax =
or
−f
= − μk g
m
From v2 = v02 + 2 a ( Δx ) , with v = 0, v0 = 50.0 km h = 13.9 m s , we find
0 = ( 13.9 m s ) + 2 ( − μ k g ) ( Δx )
2
or
With μk = 0.100 , this gives
Δx =
(b)
(13.9 m s )2
2 ( 0.100 ) ( 9.80 m s2 )
= 98.6 m
With μk = 0.600 , Equation (1) above gives
Δx =
( 13.9 m s )2
2 ( 0.600 ) ( 9.80 m s2 )
= 16.4 m
Page 4.60
Δx =
( 13.9
m s)
2 μk g
2
[1]
Chapter 4
4.51
(a)
Δ x = v0 t +
2 ( Δx )
ax =
(b)
t2
1
1
a x t 2 = 0 + a x t 2 gives:
2
2
=
2 ( 2.00 m )
(1.50 s )
2
= 1.78 m s2
y
Considering forces parallel to the incline, Newton’s second law yields
(
ΣFx = ( 29.4 N ) sin 30.0 ° − fk = ( 3.00 kg ) 1.78 m s2
)
→
or fk = 9.37 N
or
kn
x
Perpendicular to the plane, we have equilibrium, so
ΣFy = n − ( 29.4 N ) cos 30.0° = 0
m
f k
n
30.0°
n = 25.5 N
30.0°
w mg 29.4 N
Then,
μk =
fk
9.37 N
=
= 0.368
n
25.5 N
(c)
From part (b) above, fk = 9.37 N
(d)
Finally, v2 = v02 + 2 a x ( Δx ) gives
(
)
v = v02 + 2ax ( Δx ) = 0 + 2 1.78 m s 2 ( 2.00 m ) = 2.67 m s
4.52
r
(b) When the minimum force F is used, the block tends to slide down
r
the incline so the friction force, f s is directed up the incline.
y
→
While the block is in equilibrium, we have
n
f k
ΣFx = F cos 60.0° + fs − ( 19.6 N )sin 60.0° = 0
m kn
x
[1]
30.0°
and
ΣFy = n − F sin 60.0 ° − ( 19.6 N ) cos 60.0° = 0
( fs )max
30.0°
[2]
= μs n = ( 0.300 ) n
For minimum F (impending motion),
fs =
Equation [2] gives
n = 0.866 F + 9.80 N
Page 4.61
w mg 29.4 N
[3]
[4]
Chapter 4
(a)
Equation [1] gives
4.53
fs = 0.300 ( 0.866 F + 9.80 N ) = 0.260 F + 2.94 N , so
Equation [3] becomes:
0.500 F + 0.260 F + 2.94 N − 17.0 N = 0
F = 18.5 N .
or
n = 0.866 (18.5 N ) + 9.80 N= 25.8 N .
(b)
Finally, Equation (4) gives the normal force
b)
First, taking downward as positive, apply Newton’s second law to the
12.0 kg block:
ΣFy = ( 12.0 kg ) g − T = ( 12 .0 kg ) a
or
x
→
(
T = ( 12.0 kg ) 9.80 m s2 − a
)
[1]
⇒
n
→
T
12.0 kg
For the 7.00 kg block, we have
ΣF⊥ = 0
→
T
7.00 kg
y
→
n = ( 68.6 N ) cos 37.0° = 54.8 N
f
w 118 N
37.0°
w 68.6 N
and
f = μ k n = ( 0.250 ) ( 54.8 N ) = 13.7 N
Taking up the incline as the positive direction and applying Newton’s second law to the 7.00-kg block gives
ΣFx = T − f − ( 68.6 N )sin 37.0° = ( 7.00 kg ) a , or
( 7.00 kg ) a
= T − 13.7 N − 41.3 N
(2)
Substituting Equation [1] into [2] yields
( 7.00 kg+12.0 kg ) a
4.54
(a)
= 62.7 N or a = 3.30 m s2
Both objects start from rest and have accelerations of the same magnitude, a. This magnitude can be
determined by applying Δy = v0 y t +
a =
2 ( Δy )
t2
=
2 ( 1.00 m )
( 4.00s )
2
1
2
a y t 2 to the motion of m1 :
= 0.125 m s2
Page 4.62
Chapter 4
(b) Consider the free-body diagram of m1 and apply Newton’s 2nd law:
ΣFy = ma y
→
T
⇒ T − m1 g = m1 ( + a )
m1
→
or
m1g
(
)
T = m1 ( g + a ) = ( 4.00 kg ) 9.80 m s2 + 0.125 m s2 = 39.7 N
(c) Considering the free-body diagram of m2 :
ΣFy = ma y
so
→
⇒ n − m2 g cos θ = 0 or n = m2 g cosθ ,
(
)
n
2
fk
m2 g sin θ − T − fk = m2 ( + a )
⇒
→
m
→
n = ( 9.00 kg ) 9.80 m s2 cos 40.0 ° = 67.6 N .
ΣFx = ma x
y
T
q
x
→
m2g
Then fk = m2 ( g sin θ − a ) − T ,
or
(
)
fk = ( 9.00 kg ) ⎡⎣ 9.80 m s2 sin 40.0° − 0.125 m s2 ⎤⎦ − 39.7 N = 15.9 N
The coefficient of kinetic friction is
μk =
fk
15.9 N
=
= 0.235
n
67.6 N
4.55
y
y
nground w/2 85.0 lb
22.0° 22.0°
→
→
F2
ntip
→
F1
→
f
F
Free-Body Diagram of Person
x
45.8
lb
x
w 170 lb
22.0°
Free-Body Diagram of Crutch Tip
From the free-body diagram of the person, ΣFx = F1 sin ( 22 .0 ° ) − F2 sin ( 22 .0 ° ) = 0 , which gives or
F1 = F2 = F
Then, ΣFy = 2 F cos 22 .0 ° + 85.0 lbs − 170 lbs = 0 yields F = 45.8 lb
(a)
Now consider the free-body diagram of a crutch tip.
Σ Fx = f − ( 45.8 lb ) sin 22 .0 ° = 0 ,
Page 4.63
Chapter 4
or
f = 17.2 lb
ΣFy = ntip − ( 45.8 lb ) cos 22 .0° = 0 ,
which gives ntip = 42 .5 lb
For minimum coefficient of friction, the crutch tip will be on the verge of slipping, so
f =
( fs )max
and μs =
(b)
= μs ntip
f
ntip
=
17.2 lb
= 0.404
42.5 lb
As found above, the compression force in each crutch is
F1 = F2 = F = 45.8 lb
4.56
(b)
The acceleration of the ball is found from
( 20.0 m s ) − 0 = 133 m s2
v 2 − v02
a =
=
2 ( Δy )
2 ( 1.50 m )
2
→
→
f bv
From the second law, ΣFy = F − w = ma y , so
→
m
(
)
F = w + ma y = 1.47 N + ( 0.150 kg ) 133 m s2 = 21.5 N
v
→
mg
4.57 (a)
(b)
Note that the suspended block on the left, m4 , is heavier than that on the right, m2 . Thus, if the system
overcomes friction and moves, the center block will move right to left with each block’s acceleration being
in the directions shown above .
Page 4.64
Chapter 4
First, consider the center block, m1 , which has no vertical acceleration. Then,
Σ Fy = n − m1 g = 0
(
)
n = m1 g = ( 1.00 kg ) 9.80 m s2 = 9.80 N
or
This means the friction force is:
f = μ k n = ( 0.350 ) ( 9.80 N ) = 3.43 N
Assuming the cords do not stretch, the speeds of the three blocks must always be equal. Thus, the magnitudes
of the blocks’ accelerations must have a common value, a.
uur
uur
uur
a 4 = a1 = a 2 = a
Taking the indicated direction of the acceleration as the positive direction of motion for each block, we apply
Newton’s second law to each block as follows
For m4 :
m4 g − T1 = m4 a
or
T1 = m4 ( g − a ) = ( 4.00 kg ) ( g − a )
[1]
For m1 :
T1 − T2 − f = m1a
or
T1 − T2 = ( 1.00 kg ) a + 3.43 N
[2]
For m2 :
T2 − m2 g = m2 a
or
T2 = m2 ( g + a ) = ( 2.00 kg ) ( g + a )
[3]
Substituting equations [1] and [3] into equation [2], and solving for a yields the following:
( 4.00 kg ) ( g − a ) − ( 2.00 kg ) ( g + a ) = ( 1.00 kg ) a + 3.43 N
a =
(c)
( 4.00 kg − 2.00 kg ) ( 9.80
)
m s2 − 3.43 N
4.00 kg + 2.00 kg + 1.00 kg
= 2.31 m s2
Using this result in equations [1] and [3] gives the tensions in the two cords as:
T1 = ( 4.00 kg ) ( g − a ) = ( 4.00 kg ) ( 9.80 − 2.31) m s2 = 30.0 N
and
T2 = ( 2.00 kg ) ( g + a ) = ( 2.00 kg ) ( 9.80 + 2.31) m s2 = 24.2 N
(e) From the final calculation in part (b), observe that if the friction force had a value of zero (rather than 3.53 N),
the acceleration of the system would increase in magnitude. Then, observe from equations [1] and [3] that this
would mean T1 would decrease while T2 would increase
Page 4.65
Chapter 4
4.58
The sketch at the right gives an edge view of the sail (heavy
r
line) as seen from above. The velocity of the wind, v wind , is
north
→
Fsail
directed to the east and the force the wind exerts on the sail
is perpendicular to the sail. The magnitude of this force is
30°
30°
⎛
N ⎞ r
Fsail = ⎜ 550
v wind
⊥
m s ⎟⎠
⎝
r
where v wind is the component of the wind velocity perpendicular to
→
vwind
east
⊥
the sail.
When the sail is oriented at 30° from the north-south line and the wind speed is vwind = 17 knots ,
we have
⎛
N ⎞ r
Fsail = ⎜ 550
v wind
m s ⎟⎠
⎝
⊥
⎛
N ⎞
= ⎜ 550
m s ⎟⎠
⎝
⎡
⎤
⎛ 0.514 m s ⎞
cos 30° ⎥ = 4.2 × 103 N
⎢ 17 knots ⎜
⎟
⎝ 1 knot ⎠
⎢⎣
⎥⎦
(
)
The eastward component of this force will be counterbalanced by the force of the water on the keel of the boat.
Before the sailboat has significant speed (that is, before the drag force develops), its acceleration is provided by the
ur
northward component of F sail . Thus, the initial acceleration is
ur
F sail
4.2 × 103 N sin 30°
north
a =
= 2.6 m s2
=
800 kg
m
(
4.59
)
(a) The horizontal component of the resultant force exerted on the light by the
cables is
Rx = Σ Fx = ( 60.0 N ) cos 45.0 ° − ( 60.0 N ) cos 45.0 ° = 0
y
60
.0
The resultant y component is:
45.0°
N
.0
N
60
45.0°
Ry = Σ Fy = ( 60.0 N ) sin 45.0° + ( 60.0 N ) sin 45.0° = 84.9 N
Hence, the resultant force is 84.9 N vertically upward .
(b)
The forces on the traffic light are the weight, directed downward, and the 84.9 N vertically upward force
exerted by the cables. Since the light is in equilibrium, the resultant of these forces must be zero. Thus,
w = 84.9 N .
Page 4.66
x
Chapter 4
4.60
(a)
For the suspended block, ΣFy = T − 50.0 N = 0 , so the tension in the rope is T = 50.0 N . Then,
considering the horizontal forces on the 100-N block, we find
ΣFx = T − fs = 0 , or fs = T = 50.0 N
(b)
If the system is on the verge of slipping, fs =
( fs )max
= μs n . Therefore,
the minimum acceptable coefficient of friction is
μs =
(c)
fs
50.0 N
=
= 0.500
n
100 N
If μk = 0.250 , then the friction force acting on the 100-N block is
fk = μ k n = ( 0.250 ) ( 100 N ) = 25.0 N
Since the system is to move with constant velocity, the net horizontal force on the 100-N block must be zero,
or ΣFx = T − fk = T − 25.0 N = 0 . The required tension in the rope is T = 25.0 N . Now, considering
the forces acting on the suspended block when it moves with constant velocity, ΣFy = T − w = 0 , giving the
required weight of this block as w = T = 25.0 N .
4.61
On the level surface, the normal force exerted on the sled by the ice equals the total weight, or n = 600 N. Thus, the
friction force is
f = μ k n = ( 0.050 ) ( 600 N ) = 30 N
Hence, Newton’s second law yields ΣFx = − f = ma x , or
ax =
(
− ( 30 N ) 9.80 m s2
−f
−f
=
=
600 N
m
w g
)
= − 0.49 m s2
The distance the sled travels on the level surface before coming to rest is
0 − ( 7.0 m s )
v2 − v02 x
Δx = x
=
2 ax
2 − 0.49 m s2
2
(
4.62
)
= 50 m
Consider the vertical forces acting on the block:
ΣFy = ( 85.0 N ) sin 55.0° − 39.2 N − n = ma y = 0 ,
so the normal force is
n = 69.6 N − 39.2 N = 30.4 N
→
n
→
Now, consider the horizontal forces:
fk
(
ΣFx = ( 85.0 N ) cos 55.0° − fk = ma x = ( 4.00 kg ) 6.00 m s2
or
)
4.00 kg
55.0°
fk = ( 85.0 N ) cos 55.0 ° − 24.0 N = 24.8 N
Page 4.67
85.0 N
mg 39.2 N
Chapter 4
The coefficient of kinetic friction is then μk =
4.63
fk
24.8 N
=
= 0.814 .
n
30.4 N
(a)
The force that accelerates the box is the friction force between the box and the truck bed.
(b)
The maximum acceleration the truck can have before the box slides is found by considering the maximum
static friction force the truck bed can exert on the box:
( fs )max
= μs n = μs ( mg )
Thus, from Newton’s second law,
amax =
4.64
( f s ) max
m
=
μ s ( mg )
m
(
)
= μ s g = ( 0.300) 9.80 m s 2 = 2.94 m s 2
Let m1 = 5.00 kg, m2 = 4.00 kg, and m3 = 3.00 kg . Let T1 be the tension in the string between m1 and m2 , and
T2 the tension in the string between m2 and m3 .
(a)
We may apply Newton’s second law to each of the masses.
for m1:
m1a = T1 − m1 g
[1]
for m2 :
m2 a = T2 + m2 g − T1
[2]
for m3 :
m3 a = m3 g − T2
[3]
Adding these equations yields ( m1 + m2 + m3 ) a = ( − m1 + m2 + m3 ) g , so
⎛ − m1 + m2 + m3 ⎞
⎛ 2.00 kg ⎞
a = ⎜
g = ⎜
9.80 m s2 = 1.63 m s2
⎟
⎝ 12.0 kg ⎟⎠
⎝ m1 + m2 + m3 ⎠
(
(b)
)
(
)
(
)
From Equation [1], T1 = m1 ( a + g ) = ( 5.00 kg ) 11.4 m s2 = 57.2 N , and
from Equation [3], T2 = m3 ( g − a ) = ( 3.00 kg ) 8.17 m s2 = 24.5 N
Page 4.68
Chapter 4
4.65
When an object of mass m is on this frictionless incline, the only force acting parallel to the incline is the parallel
component of weight, mg sin θ directed down the incline. The acceleration is then
a =
F
mg sin θ
=
= g sin θ = 9.80 m s2 sin 35.0° = 5.62 m s2 (directed down the incline)
m
m
(a)
Taking up the incline as positive, the time for the sled projected up the incline to come to rest is given by
(
t =
)
v − v0
0 − 5.00 m s
=
= 0.890 s
a
− 5.62 m s2
The distance the sled travels up the incline in this time is
⎛ v + v0 ⎞
⎛ 0 + 5.00 m s ⎞
Δs = vav t = ⎜
t = ⎜
⎟⎠ ( 0.890 s ) = 2.22 m
⎝
⎝ 2 ⎟⎠
2
(b)
The time required for the first sled to return to the bottom of the incline is the same as the time needed to go
up, that is, t = 0.890 s . In this time, the second sled must travel down the entire 10.0 m length of the incline.
The needed initial velocity is found from Δs = v0t +
v0 =
) ( 0.890 s )
= −8.74 m s
8.74 m s down the incline .
or
4.66
(
−5.62 m s2
Δs
at
−10.0 m
−
=
−
t
2
0.890 s
2
1 2
at as
2
Before she enters the water, the diver is in free-fall with an acceleration of 9.80 m s2 downward. Taking
downward as the positive direction, her velocity when she reaches the water is given by
v =
v02 + 2 a ( Δy ) =
(
0 + 2 9.80 m s2
) (10.0 m )
= 14.0 m s
This is also her initial velocity for the 2.00 s after hitting the water. Her average acceleration during this 2.00 s
interval is
aav =
0 − ( 14.0 m s )
v − v0
=
= − 7.00 m s2
2 .00 s
t
Continuing to take downward as the positive direction, the average upward force by the water is found as
ΣFy = Fav + mg = maav , or
(
)
Fav = m ( aav − g ) = ( 70.0 kg ) ⎡⎣ −7.00 m s2 − 9.80 m s2 ⎤⎦ = − 1.18 × 103 N
or
Fav = 1.18 × 103 N upward
Page 4.69
Chapter 4
4.67
(a)
Free-body diagrams for the two blocks are given at the right.
→
→
a
The coefficient of kinetic friction for aluminum on steel is
n1
→
μ1 = 0.47 while that for copper on steel is μ2 = 0.36 . Since
T
Aluminum
2.00 kg
a y = 0 for each block,
→
f1
n1 = w1 and n2 = w2 cos 30.0° .
Thus,
→
f1 = μ1n1 = 0.47 ( 19.6 N ) = 9.21 N
w1 19.6 N
→
a
T
→
and
→
f2
f2 = μ2 n2 = 0.36 ( 58.8 N ) cos 30.0° = 18.3 N
n2
Co
6.0 pper
0k
g
x
For the aluminum block:
ΣFx = ma x
⇒
T = 9.21 N + ( 2.00 kg ) a
giving
30°
T − f1 = m ( + a ) or T = f1 + ma
w2 58.8 N
[1]
For the copper block:
ΣFx = ma x
or
⇒
( 58.8 N ) sin 30.0° − T
− 18.3 N = ( 6.00 kg ) a
11.1 N − T = ( 6.00 kg ) a
[2]
Substituting Equation [1] into Equation [2] gives
11.1 N − 9.21 N − ( 2.00 kg ) a = ( 6.00 kg ) a
or
4.68
a =
1.86 N
= 0.232 m s2
8.00 kg
(b)
From Equation [1] above,
(
)
T = 9.21 N + ( 2.00 kg ) 0.232 m s2 = 9.68 N .
In the vertical direction, we have
mg
Σ Fy = T cos 4.0 ° − mg = 0 , or T =
cos 4.0°
In the horizontal direction, the second law becomes:
4.0°
y
→
T
ΣFx = T sin 4.0° = ma , so
a =
T sin 4.0 °
= g tan 4.0 ° = 0.69 m s2
m
Page 4.70
x
→
mg
Chapter 4
4.69
Figure 1 is a free-body diagram for the system consisting of
both blocks. The friction forces are
f1 = μk n1 = μk ( m1 g ) and f2 = μk ( m2 g )
→
a
For this system, the tension in the connecting rope is an
→
The second law gives
→
n1
internal force and is not included in second law calculations.
n2
m1
→
→
m1g
f1
→
f2
ΣFx = 50 N − f1 − f2 = ( m1 + m2 ) a
which reduces to
a =
50 N
− μk g
m1 + m2
[1]
Figure 2 gives a free-body diagram of m1 alone. For this
system, the tension is an external force and must be included in
the second law. We find: ΣF = T − f = m a , or
x
1
1
T = m1 ( a + μk g )
[2]
If the surface is frictionless, μk = 0 . Then, Equation [1] gives
a =
50 N
50 N
−0 =
= 1.7 m s2
m1 + m2
30 kg
(
(b) If μk = 0.10 , Equation [1] gives the acceleration as
a =
50 N
− ( 0.10 ) 9.80 m s2 = 0.69 m s2
30 kg
(
)
while Equation (2) gives the tension as
(
)
T = ( 10 kg ) ⎡⎣ 0.69 m s2 + ( 0.10 ) 9.80 m s2 ⎤⎦ = 17 N
4.70
)
T = ( 10 kg ) 1.7 m s2 + 0 = 17 N .
and Equation [2] yields
(a)
Page 4.71
50 N
m2
→
m 2g
Chapter 4
(b)
No. In general, the static friction force is less than the maximum value of ( fs ) max = μ s n . It is equal to this
maximum value only when the coin is on the verge of slipping, or at the critical angle θc .
For θ ≤ θc , f ≤ ( fs ) max = μ s n .
(c)
Recognize that when the y axis is chosen perpendicular to the incline as shown above, a y = 0 and we find
ΣFy = n − mg cos θ = ma y = 0
n = mg cos θ
or
Also, when static conditions still prevail, but the coin is on the verge of slipping, we have
a x = 0, θ = θc , and f =
( fs )max
= μs n = μs mg cos θc . Then, Newton’s second law becomes
ΣFx = mg sin θ c − μ s mg cos θ c = max = 0
and
μs mg cos θc = mg sin θc yielding μs =
(d)
sin θ c
= tan θ c
cos θ c
Once the coin starts to slide, kinetic conditions prevail and the friction force is
f = f k = μk n = μk mg cos θ
At θ = θ c′ < θ c , the coin slides with constant velocity, and a x = 0 again. Under these conditions, Newton’s
second law gives
ΣFx = mg sin θ c′ − μk mg cos θc′ = ma x = 0
and
μk mg cos θc′ = mg sin θc′
yielding
μk =
sin θ c′
= tan θ c′
cos θ c′
Page 4.72
Chapter 4
4.71
(a)
When the pole exerts a force downward and toward the rear on the lakebed, the
lakebed exerts an oppositely directed force of equal magnitude, F, on the end of the
pole.
As the boat floats on the surface of the lake, its vertical acceleration is a y = 0 .
Thus, Newton’s second law gives the magnitude of the buoyant force, FB , as
ΣFy = FB + F cos θ − mg = 0
and, with θ = 35.0° ,
(
)
FB = mg − F cos θ = ( 370 kg ) 9.80 m s2 − ( 240 N ) cos 35.0°
or
FB = 3.43 × 103 N = 3.43 kN
(b)
Applying Newton’s second law to the horizontal motion of the boat gives
ur
ΣFx = F sin θ − F drag = ma x
ax =
or
( 240 N ) sin 35.0° − 47.5 N
370 kg
= 0.244 m s2
After an elapsed time t = 0.450 s , vx = v0 x + a x t gives the velocity of the boat as
(
vx = 0.857 m s + 0.244 m s2
(c)
) ( 0.450 s ) =
0.967 m s
ur
If angle θ increased while the magnitude of F remained constant, the vertical component of this force would
decrease. The buoyant force would have to increase to support more of the weight of the boat and its
ur
contents. At the same time, the horizontal component of F would increase, which would
increase the acceleration of the boat.
4.72
(a)
(b)
Applying Newton’s second law to the rope yields
ΣFx = ma x
⇒
F − T = mr a
or
T = F − mr a
Then, applying Newton’s second law to the block, we find
ΣFx = ma x
⇒ T = mb a
or
F − mr a = mb a
Page 4.73
[1]
Chapter 4
which gives
a =
(c)
F
mb + mr
Substituting the acceleration found above back into equation [1] gives the tension at the left end of the rope as
⎛ mb + mr − mr ⎞
⎛
⎞
F
T = F − mr a = F − mr ⎜
= F⎜
⎟
⎟
mb + mr
⎝ mb + mr ⎠
⎝
⎠
or
⎛
⎞
mb
T = ⎜
⎟F
m
m
+
⎝ b
r ⎠
(d)
From the result of (c) above, we see that as mr approaches zero, T approaches F. Thus,
the tension in a cord of negligible mass is constant along its length .
4.73
Choose the positive x axis to be down the incline and the y
y
axis perpendicular to this as shown in the free-body
→
diagram of the toy. The acceleration of the toy then has
T
→
components of
a
a y = 0, and a x =
Δv x
+ 30.0 m s
=
= + 5.00 m s2
Δt
6.00 s
Applying the second law to the toy gives:
(a)
ΣFx = mg sin θ = ma x
sin θ = ma x mg = a x g ,
⇒
and
⎛ ax ⎞
⎛ 5.00 m s2 ⎞
= sin −1 ⎜
= 30.7°
⎟
⎝ g ⎠
⎝ 9.80 m s2 ⎟⎠
θ = sin −1 ⎜
(b)
ΣFy = T − mg cos θ = ma y = 0 , or
(
)
T = mg cos θ = ( 0.100 kg ) 9.80 m s2 cos 30.7° = 0.843 N
Page 4.74
x
q
m
→
mg q
Chapter 4
4.74
The sketch at the right gives the free-body diagram of the person. The
scale simply reads the magnitude of the normal force exerted on the
student by the seat. From Newton’s second law, we obtain
ΣFy = ma y =0
⇒
n − mg cos 30.0 ° = 0
or
n = mg cos θ = ( 200 lb ) cos 30.0° = 173 lb
4.75
The acceleration the car has as it is coming to a stop is
a =
v 2 − v02
2 ( Δx )
=
0 − ( 35 m s )
2
= − 0.61 m s2
2 ( 1 000 m )
Thus, the magnitude of the total retarding force acting on the car is
⎛ w⎞
⎛ 8 820 N ⎞
F = m a = ⎜ ⎟ a = ⎜
0.61 m s2 = 5.5 × 102 N
⎝ g⎠
⎝ 9.80 m s2 ⎟⎠
(
4.766
(a)
)
In the vertical direction, we have
8 000 N
ΣFy = ( 8 000 N ) sin 65.0° − w = ma y = 0
so
65.0°
w = ( 8 000 N ) sin 65.0° = 7.25 × 103 N
→
(b)
⎛ w⎞
ΣFx = ( 8 000 N ) cos 65.0° = ma x = ⎜ ⎟ a x
⎝ g⎠
or
ax =
g ⎣⎡ ( 8 000 N ) cos 65.0° ⎦⎤
w
=
( 9.80
m s2
) ( 8 000 N ) cos 65.0°
7.25 × 103 N
Page 4.75
→
w mg
Along the horizontal, Newton’s second law yields
=
Chapter 4
4.77
Since the board is in equilibrium, ΣFx = 0 and we see that
the normal forces must have the same magnitudes on both sides
of the board. Also, if the minimum normal forces (compression
→
forces) are being applied, the board is on the verge of slipping
and the friction force on each side is f =
→
f
f
→
( fs )max = μs n
→
n
n
The board is also in equilibrium in the vertical direction,
so
ΣFy = 2 f − w = 0, or f =
w
2
w 95.5 N
The minimum compression force needed is then
f
n =
4.78
μs
=
w
2 μs
=
95.5 N
= 72.0 N
2 ( 0.663 )
Consider the two free-body
diagrams, one of the penguin
alone
and one of the combined system
→
→
n2
n1
consisting of penguin
plus sled.
The normal force exerted on the
→
→
f1
penguin by the sled is
F
→
f2
w1 70.0 N
n1 = w1 = m1 g
w total 130 N
and the normal force exerted on the combined system by the ground is
n2 = wtotal = mtotal g = 130 N
The penguin is accelerated forward by the static friction force exerted on it by the sled. When the penguin is on the
verge of slipping, this acceleration is
amax =
( f1 )max
m1
=
μ s n1
m1
=
(
μs m1 g
m1
) = μ g = ( 0.700 ) ( 9.80 m s ) = 6.86 m s
s
Page 4.76
2
2
Chapter 4
Since the penguin does not slip on the sled, the combined system must have the same acceleration as the penguin.
Applying Newton’s second law to this system gives
ΣFx = F − f2 = mtotal amax
⎛w
⎞
F = f2 + mtotal amax = μ k ( wtotal ) + ⎜ total ⎟ amax
⎝ g ⎠
or
which yields
⎛ 130 N ⎞
F = ( 0.100 ) ( 130 N ) + ⎜
6.86 m s2 = 104 N
⎝ 9.80 m s2 ⎟⎠
(
4.79
)
First, we will compute the needed accelerations:
(1)
Before it starts to move:
ay = 0
(2)
During the first 0.80 s:
ay =
(3)
v y − v0 y
t
=
1.2 m s − 0
= 1.5 m s2
0.80 s
While moving at constant velocity:
ay = 0
(4)
During the last 1.5 s:
ay =
v y − v0 y
t
=
0 − 1.2 m s
= −0.80 m s2
1.5 s
The spring scale reads the normal force the scale exerts on the man. Applying Newton’s second law to the vertical
motion of the man gives:
ΣFy = n − mg = ma y
(a)
(
n = m g + ay
or
)
When a y = 0 ,
(
)
n = ( 72 kg ) 9.80 m s2 + 0 = 7.1 × 10 2 N
(b)
When a y = 1.5 m s2 ,
(
)
n = ( 72 kg ) 9.80 m s2 + 1.5 m s2 = 8.1 × 10 2 N
(c)
When a y = 0 ,
(
)
n = ( 72 kg ) 9.80 m s2 + 0 = 7.1 × 10 2 N
Page 4.77
Chapter 4
(d)
When a y = −0.80 m s2 ,
(
)
n = ( 72 kg ) 9.80 m s2 − 0.80 m s2 = 6.5 × 102 N
4.80
The friction force exerted on the mug by the moving tablecloth is the only horizontal force the mug experiences
during this process. Thus, the horizontal acceleration of the mug will be
amug =
fk
mmug
=
0.100 N
= 0.500 m s2
0.200 kg
The cloth and the mug both start from rest ( v0 x = 0 ) at time t = 0 . Then, at time t > 0 , the horizontal
displacements of the mug and cloth are given by Δ x = v0 x t +
Δ xmug = 0 +
1
2
( 0.500
)
(
1
2
a x t 2 as:
)
m s2 t 2 = 0.250 m s2 t 2
and
Δxcloth = 0 +
1
2
( 3.00
)
(
)
m s2 t 2 = 1.50 m s2 t 2
In order for the edge of the cloth to slip under the mug, it is necessary that Δ xcloth = Δxmug + 0.300 m , or
(1.50
)
(
)
m s2 t 2 = 0.250 m s2 t 2 + 0.300 m
The elapsed time when this occurs is
t =
0.300 m
= 0.490s
(1.50 − 0.250 ) m s2
At this time, the mug has moved a distance of
(
Δ xmug = 0.250 m s2
) ( 0.490 s )2
= 6.00 × 10 −2 m = 6.00 cm
Page 4.78
Chapter 4
4. 81
(a)
Consider the first free-body diagram
in which Chris and the
chair treated as a combined system. The
weight of this system is
wtotal = 480 N , and its mass is
mtotal =
→
T 250 N
wtotal
= 49.0 kg
g
T 250 N
n
T 250 N
Taking upward as positive, the
→
a
acceleration of this system is found
from Newton’s second law as
ΣFy = 2T − wtotal = mtotal a y
Thus
ay =
2 ( 250 N ) − 480 N
49.0 kg
wtotal 320 N 160 N
wChris 320 N
= + 0.408 m s2
or
0.408 m s2 upward
(b)
The downward force that Chris exerts on the chair has the same magnitude as the upward normal force exerted
on Chris by the chair. This is found from the free-body diagram of Chris alone as
ΣFy = T + n − wChris = mChris a y
so
n = mChris a y + wChris − T
Hence,
⎛ 320 N ⎞
n = ⎜
0.408 m s2 + 320 N − 250 N = 83.3 N
⎝ 9.80 m s2 ⎟⎠
(
4.82
)
ur
Let R represent the horizontal force of air resistance. Since the
helicopter and bucket move at constant velocity, a x = a y = 0 .
The second law then gives:
ΣFy = T cos 40.0° − mg = 0
y
or
mg
T=
cos 40.0°
or
R = T sin 40.0°
Also,
ΣFx = T sin 40.0° − R = 0
40.0°
→
T
x
R
→
→
w mg
Page 4.79
Chapter 4
Thus,
⎛ mg ⎞
R=⎜
sin 40.0° = ( 620 kg ) 9.80 m s 2 tan 40.0° = 5.10 × 103 N
⎟
⎝ cos 40.0° ⎠
(
)
Page 4.80