S O L U T I O N M A N U A L CONTENTS
Chapter 12 General Principles
1
Chapter 13 Force Vectors
245
Chapter 14 Equilibrium of a Particle
378
Chapter 15 Force System Resultants
475
Review 1 Kinematics and Kinetics of a Particle
630
Chapter 16 Equilibrium of a Rigid Body
680
Chapter 17 Structural Analysis
833
Chapter 18 Internal Forces
953
Chapter 19 Friction
1023
Review 2 Planar Kinematics and Kinetics of a Rigid Body
1080
Chapter 20 Center of Gravity and Centroid
1131
Chapter 21 Moments of Inertia
1190
Chapter 22 Virtual Work
1270
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12–1.
A baseball is thrown downward from a 50-ft tower with an
initial speed of 18 ft>s. Determine the speed at which it hits
the ground and the time of travel.
SOLUTION
v22 = v21 + 2ac(s2 - s1)
v22 = (18)2 + 2(32.2)(50 - 0)
v2 = 59.532 = 59.5 ft>s
Ans.
v2 = v1 + ac t
59.532 = 18 + 32.2(t)
t = 1.29 s
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12–2.
When a train is traveling along a straight track at 2 m/s, it
begins to accelerate at a = 160 v-42 m>s2, where v is in m/s.
Determine its velocity v and the position 3 s after the
acceleration.
s
SOLUTION
a =
dv
dt
dt =
dv
a
v
3
dt =
L0
3 =
dv
-4
L2 60v
1
(v5 - 32)
300
v = 3.925 m>s = 3.93 m>s
Ans.
ads = vdv
ds =
s
s =
1
60 L2
3.925
v5 dv
1 v6 3.925
a b`
60 6 2
= 9.98 m
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L0
ds =
1 5
vdv
=
v dv
a
60
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v
12–3.
From approximately what floor of a building must a car be
dropped from an at-rest position so that it reaches a speed
of 80.7 ft>s 155 mi>h2 when it hits the ground? Each floor is
12 ft higher than the one below it. (Note: You may want to
remember this when traveling 55 mi>h.)
SOLUTION
( + T)
v 2 = v20 + 2ac(s - s0)
80.72 = 0 + 2(32.2)(s - 0)
s = 101.13 ft
# of floors =
101.13
= 8.43
12
Ans.
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The car must be dropped from the 9th floor.
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*12–4.
Traveling with an initial speed of 70 km>h, a car accelerates
at 6000 km>h2 along a straight road. How long will it take to
reach a speed of 120 km>h? Also, through what distance
does the car travel during this time?
SOLUTION
v = v1 + ac t
120 = 70 + 6000(t)
t = 8.33(10 - 3) hr = 30 s
Ans.
v2 = v21 + 2 ac(s - s1)
(120)2 = 702 + 2(6000)(s - 0)
s = 0.792 km = 792 m
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12–5.
A bus starts from rest with a constant acceleration of 1 m>s2.
Determine the time required for it to attain a speed of 25 m>s
and the distance traveled.
SOLUTION
Kinematics:
v0 = 0, v = 25 m>s, s0 = 0, and ac = 1 m>s2.
+ B
A:
v = v0 + act
25 = 0 + (1)t
t = 25 s
or
v2 = v02 + 2ac(s - s0)
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A:
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252 = 0 + 2(1)(s - 0)
Ans.
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s = 312.5 m
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12–6.
A stone A is dropped from rest down a well, and in 1 s
another stone B is dropped from rest. Determine the
distance between the stones another second later.
SOLUTION
+ T s = s1 + v1 t +
sA = 0 + 0 +
1 2
a t
2 c
1
(32.2)(2)2
2
sA = 64.4 ft
sA = 0 + 0 +
1
(32.2)(1)2
2
sB = 16.1 ft
¢s = 64.4 - 16.1 = 48.3 ft
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12–7.
A bicyclist starts from rest and after traveling along a
straight path a distance of 20 m reaches a speed of 30 km/h.
Determine his acceleration if it is constant. Also, how long
does it take to reach the speed of 30 km/h?
SOLUTION
v2 = 30 km>h = 8.33 m>s
v22 = v21 + 2 ac (s2 - s1)
(8.33)2 = 0 + 2 ac (20 - 0)
ac = 1.74 m>s2
Ans.
v2 = v1 + ac t
8.33 = 0 + 1.74(t)
t = 4.80 s
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*■12–8.
A particle moves along a straight line with an acceleration
of a = 5>(3s1>3 + s5>2) m>s2, where s is in meters.
Determine the particle’s velocity when s = 2 m, if it starts
from rest when s = 1 m. Use Simpson’s rule to evaluate the
integral.
SOLUTION
5
a =
A 3s3 + s2 B
1
5
a ds = v dv
2
v
5 ds
L1 A 3s + s
1
3
0.8351 =
5
2
B
=
L0
v dv
1 2
v
2
v = 1.29 m>s
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12–9.
If it takes 3 s for a ball to strike the ground when it is released
from rest, determine the height in meters of the building
from which it was released. Also, what is the velocity of the
ball when it strikes the ground?
SOLUTION
Kinematics:
v0 = 0, ac = g = 9.81 m>s2, t = 3 s, and s = h.
A+TB
v = v0 + act
= 0 + (9.81)(3)
= 29.4 m>s
or
laws
A+TB
Ans.
1
s = s0 + v0t + act2
2
1
h = 0 + 0 + (9.81)(32)
2
Ans.
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= 44.1 m
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12–10.
The position of a particle along a straight line is given by
s = (1.5t3 - 13.5t2 + 22.5t) ft, where t is in seconds.
Determine the position of the particle when t = 6 s and the
total distance it travels during the 6-s time interval. Hint:
Plot the path to determine the total distance traveled.
SOLUTION
Position: The position of the particle when t = 6 s is
s|t = 6s = 1.5(63) - 13.5(62) + 22.5(6) = -27.0 ft
Ans.
Total DistanceTraveled: The velocity of the particle can be determined by applying
Eq. 12–1.
v =
ds
= 4.50t2 - 27.0t + 22.5
dt
The times when the particle stops are
laws
or
4.50t2 - 27.0t + 22.5 = 0
t = 5s
and
in
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t = 1s
Dissemination
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The position of the particle at t = 0 s, 1 s and 5 s are
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s t = 0 s = 1.5(03) - 13.5(02) + 22.5(0) = 0
and
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is
s t = 1 s = 1.5(13) - 13.5(12) + 22.5(1) = 10.5 ft
the
(including
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student
the
by
s t = 5 s = 1.5(53) - 13.5(52) + 22.5(5) = - 37.5 ft
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stot = 10.5 + 48.0 + 10.5 = 69.0 ft
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From the particle’s path, the total distance is
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12–11.
If a particle has an initial velocity of v0 = 12 ft>s to the
right, at s0 = 0, determine its position when t = 10 s, if
a = 2 ft>s2 to the left.
SOLUTION
+ B
A:
s = s0 + v0t +
1 2
at
2 c
= 0 + 12(10) +
1
(- 2)(10)2
2
Ans.
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= 20 ft
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*12–12.
Determine the time required for a car to travel 1 km along
a road if the car starts from rest, reaches a maximum speed
at some intermediate point, and then stops at the end of
the road. The car can accelerate at 1.5 m>s2 and decelerate
at 2 m>s2.
SOLUTION
Using formulas of constant acceleration:
v2 = 1.5 t1
1
(1.5)(t21)
2
x =
0 = v2 - 2 t2
1
(2)(t22)
2
1000 - x = v2t2 -
or
Combining equations:
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t1 = 1.33 t2; v2 = 2 t2
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x = 1.33 t22
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1000 - 1.33 t22 = 2 t22 - t22
the
t1 = 27.603 s
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is
of
t2 = 20.702 s;
and
t = t1 + t2 = 48.3 s
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12–13.
Tests reveal that a normal driver takes about 0.75 s before
he or she can react to a situation to avoid a collision. It takes
about 3 s for a driver having 0.1% alcohol in his system to
do the same. If such drivers are traveling on a straight road
at 30 mph (44 ft>s) and their cars can decelerate at 2 ft>s2,
determine the shortest stopping distance d for each from
the moment they see the pedestrians. Moral: If you must
drink, please don’t drive!
v1
44 ft/s
d
SOLUTION
Stopping Distance: For normal driver, the car moves a distance of
d¿ = vt = 44(0.75) = 33.0 ft before he or she reacts and decelerates the car. The
stopping distance can be obtained using Eq. 12–6 with s0 = d¿ = 33.0 ft and v = 0.
+ B
A:
v2 = v20 + 2ac (s - s0)
02 = 442 + 2( -2)(d - 33.0)
d = 517 ft
Ans.
laws
or
For a drunk driver, the car moves a distance of d¿ = vt = 44(3) = 132 ft before he
or she reacts and decelerates the car. The stopping distance can be obtained using
Eq. 12–6 with s0 = d¿ = 132 ft and v = 0.
teaching
Web)
v2 = v20 + 2ac (s - s0)
in
+ B
A:
Dissemination
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02 = 442 + 2( -2)(d - 132)
permitted.
d = 616 ft
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12–14.
A car is to be hoisted by elevator to the fourth floor of a
parking garage, which is 48 ft above the ground. If the
elevator can accelerate at 0.6 ft>s2, decelerate at 0.3 ft>s2,
and reach a maximum speed of 8 ft>s, determine the
shortest time to make the lift, starting from rest and ending
at rest.
SOLUTION
+c
v2 = v20 + 2 ac (s - s0)
v2max = 0 + 2(0.6)(y - 0)
0 = v2max + 2( -0.3)(48 - y)
0 = 1.2 y - 0.6(48 - y)
y = 16.0 ft,
+c
vmax = 4.382 ft>s 6 8 ft>s
v = v0 + ac t
4.382 = 0 + 0.6 t1
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or
t1 = 7.303 s
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0 = 4.382 - 0.3 t2
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t = t1 + t2 = 21.9 s
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not
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permitted.
Dissemination
t2 = 14.61 s
12–15.
A train starts from rest at station A and accelerates at
0.5 m>s2 for 60 s. Afterwards it travels with a constant
velocity for 15 min. It then decelerates at 1 m>s2 until it is
brought to rest at station B. Determine the distance
between the stations.
SOLUTION
Kinematics: For stage (1) motion, v0 = 0, s0 = 0, t = 60 s, and ac = 0.5 m>s2. Thus,
+ B
A:
s = s0 + v0t +
s1 = 0 + 0 +
+ B
A:
1 2
at
2 c
1
(0.5)(602) = 900 m
2
v = v0 + act
laws
or
v1 = 0 + 0.5(60) = 30 m>s
in
teaching
Web)
For stage (2) motion, v0 = 30 m>s, s0 = 900 m, ac = 0 and t = 15(60) = 900 s. Thus,
1 2
at
2 c
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s = s0 + v0t +
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A:
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s2 = 900 + 30(900) + 0 = 27 900 m
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v = v0 + act
assessing
is
+ B
A:
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student
the
by
and
For stage (3) motion, v0 = 30 m>s, v = 0, s0 = 27 900 m and ac = - 1 m>s2. Thus,
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0 = 30 + ( - 1)t
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1 2
at
2 c
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s = s0 + v0t +
their
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:
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t = 30 s
s3 = 27 900 + 30(30) +
= 28 350 m = 28.4 km
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1
( - 1)(302)
2
Ans.
*12–16.
A particle travels along a straight line such that in 2 s it
moves from an initial position sA = +0.5 m to a position
sB = - 1.5 m. Then in another 4 s it moves from sB to
sC = +2.5 m. Determine the particle’s average velocity
and average speed during the 6-s time interval.
SOLUTION
¢s = (sC - sA) = 2 m
sT = (0.5 + 1.5 + 1.5 + 2.5) = 6 m
t = (2 + 4) = 6 s
vavg =
2
¢s
= = 0.333 m>s
t
6
sT
6
= = 1 m>s
t
6
Ans.
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This
provided
integrity
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this
assessing
is
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solely
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protected
the
(including
for
work
student
the
by
and
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United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
(vsp)avg =
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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12–17.
The acceleration of a particle as it moves along a straight
line is given by a = 12t - 12 m>s2, where t is in seconds. If
s = 1 m and v = 2 m>s when t = 0, determine the
particle’s velocity and position when t = 6 s. Also,
determine the total distance the particle travels during this
time period.
SOLUTION
v
L2
t
dv =
L0
(2 t - 1) dt
v = t2 - t + 2
s
L1
t
ds =
s =
L0
(t2 - t + 2) dt
1 3
1
t - t2 + 2 t + 1
3
2
When t = 6 s,
Ans.
s = 67 m
Ans.
teaching
Web)
laws
or
v = 32 m>s
Wide
copyright
in
Since v Z 0 then
Ans.
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This
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integrity
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assessing
is
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solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
d = 67 - 1 = 66 m
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–18.
A freight train travels at v = 6011 - e -t2 ft>s, where t is the
elapsed time in seconds. Determine the distance traveled in
three seconds, and the acceleration at this time.
v
s
SOLUTION
v = 60(1 - e - t)
s
L0
3
ds =
L
v dt =
L0
6011 - e - t2dt
s = 60(t + e - t)|30
s = 123 ft
Ans.
dv
= 60(e - t)
dt
a =
or
At t = 3 s
laws
a = 60e - 3 = 2.99 ft>s2
sale
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This
provided
integrity
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assessing
is
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protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–19.
A particle travels to the right along a straight line with a
velocity v = [5>14 + s2] m>s, where s is in meters. Determine
its position when t = 6 s if s = 5 m when t = 0.
SOLUTION
5
ds
=
dt
4 + s
s
L5
(4 + s) ds =
t
L0
5 dt
4 s + 0.5 s2 - 32.5 = 5 t
When t = 6 s,
s2 + 8 s - 125 = 0
laws
or
Solving for the positive root
s = 7.87 m
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This
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assessing
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protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*12–20.
The velocity of a particle traveling along a straight line is
v = (3t2 - 6t) ft>s, where t is in seconds. If s = 4 ft when
t = 0, determine the position of the particle when t = 4 s.
What is the total distance traveled during the time interval
t = 0 to t = 4 s? Also, what is the acceleration when t = 2 s?
SOLUTION
Position: The position of the particle can be determined by integrating the kinematic
equation ds = v dt using the initial condition s = 4 ft when t = 0 s. Thus,
+ B
A:
ds = v dt
s
L4 ft
s2
t
ds =
s
L0
2
A 3t - 6t B dt
= (t 3 - 3t2) 2
4 ft
t
0
s = A t - 3t + 4 B ft
or
2
teaching
Web)
laws
3
Wide
copyright
in
When t = 4 s,
s|4 s = 43 - 3(42) + 4 = 20 ft
by
and
use
United
on
learning.
is
of
the
The velocity of the particle changes direction at the instant when it is momentarily
brought to rest. Thus,
(including
for
work
student
the
v = 3t2 - 6t = 0
assessing
is
work
solely
of
protected
the
t(3t - 6) = 0
provided
integrity
of
and
work
this
t = 0 and t = 2 s
the
part
any
courses
destroy
sale
will
their
s|2 s = 23 - 3 A 22 B + 4 = 0
of
and
s|0 s = 0 - 3 A 02 B + 4 = 4 ft
is
This
The position of the particle at t = 0 and 2 s is
Using the above result, the path of the particle shown in Fig. a is plotted. From this
figure,
sTot = 4 + 20 = 24 ft
Ans.
Acceleration:
+ B
A:
a =
dv
d
=
(3t2 - 6t)
dt
dt
a = 16t - 62 ft>s2
When t = 2 s,
a ƒ t = 2 s = 6122 - 6 = 6 ft>s2 :
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
not
instructors
States
World
permitted.
Dissemination
Ans.
12–21.
If the effects of atmospheric resistance are accounted for, a
falling body has an acceleration defined by the equation
a = 9.81[1 - v2(10-4)] m>s2, where v is in m>s and the
positive direction is downward. If the body is released from
rest at a very high altitude, determine (a) the velocity when
t = 5 s, and (b) the body’s terminal or maximum attainable
velocity (as t : q ).
SOLUTION
Velocity: The velocity of the particle can be related to the time by applying Eq. 12–2.
(+ T)
dt =
t
L0
dv
a
v
dv
2
L0 9.81[1 - (0.01v) ]
dt =
v
t =
v
1
dv
dv
c
+
d
9.81 L0 2(1 + 0.01v)
2(1
0.01v)
L0
1 + 0.01v
b
1 - 0.01v
laws
or
9.81t = 50ln a
100(e0.1962t - 1)
teaching
Web)
(1)
in
e0.1962t + 1
Wide
copyright
v =
instructors
States
World
permitted.
Dissemination
a) When t = 5 s, then, from Eq. (1)
of
the
not
100[e0.1962(5) - 1]
Ans.
and
use
United
on
learning.
is
= 45.5 m>s
e0.1962(5) + 1
work
the
(including
: 1. Then, from Eq. (1)
of
is
work
solely
e0.1962t + 1
for
e0.1962t - 1
protected
b) If t : q ,
student
the
by
v =
Ans.
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this
assessing
vmax = 100 m>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–22.
The position of a particle on a straight line is given by
s = 1t3 - 9t2 + 15t2 ft, where t is in seconds. Determine the
position of the particle when t = 6 s and the total distance it
travels during the 6-s time interval. Hint: Plot the path to
determine the total distance traveled.
SOLUTION
s = t3 - 9t2 + 15t
ds
= 3t2 - 18t + 15
dt
v =
v = 0 when t = 1 s and t = 5 s
t = 0, s = 0
t = 1 s, s = 7 ft
Ans.
sT = 7 + 7 + 25 + (25 - 18) = 46 ft
Ans.
sale
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This
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integrity
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assessing
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protected
the
(including
for
work
student
the
by
and
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United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
t = 6 s, s = - 18 ft
or
t = 5 s, s = - 25 ft
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–23.
Two particles A and B start from rest at the origin s = 0 and
move along a straight line such that a A = (6t - 3) ft>s2 and
aB = (12t2 - 8) ft>s2, where t is in seconds. Determine the
distance between them when t = 4 s and the total distance
each has traveled in t = 4 s.
SOLUTION
Velocity: The velocity of particles A and B can be determined using Eq. 12-2.
dvA = aAdt
vA
t
dvA =
L0
(6t - 3)dt
L0
vA = 3t2 - 3t
dvB = aBdt
vB
t
dvB =
L0
(12t2 - 8)dt
L0
vB = 4t3 - 8t
The times when particle A stops are
or
t = 0 s and = 1 s
laws
3t2 - 3t = 0
in
teaching
Web)
The times when particle B stops are
World
not
instructors
States
Position:The position of particles A and B can be determined using Eq. 12-1.
permitted.
Dissemination
Wide
copyright
t = 0 s and t = 22 s
4t3 - 8t = 0
United
on
learning.
is
of
the
dsA = vAdt
and
use
the
by
(3t2 - 3t)dt
work
(including
L0
student
L0
t
dsA =
for
sA
work
solely
of
protected
the
3 2
t
2
assessing
is
sA = t3 -
integrity
of
and
work
this
dsB = vBdt
provided
part
the
is
This
(4t3 - 8t)dt
any
L0
courses
L0
t
dsB =
and
sB
will
their
destroy
of
sB = t4 - 4t2
sale
The positions of particle A at t = 1 s and 4 s are
sA |t = 1 s = 13 -
3 2
(1 ) = - 0.500 ft
2
sA |t = 4 s = 43 -
3 2
(4 ) = 40.0 ft
2
Particle A has traveled
dA = 2(0.5) + 40.0 = 41.0 ft
Ans.
The positions of particle B at t = 22 s and 4 s are
sB |t = 12 = (22)4 - 4(22)2 = -4 ft
sB |t = 4 = (4)4 - 4(4)2 = 192 ft
Particle B has traveled
dB = 2(4) + 192 = 200 ft
Ans.
At t = 4 s the distance beween A and B is
¢sAB = 192 - 40 = 152 ft
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
*12–24.
A particle is moving along a straight line such that its
velocity is defined as v = ( - 4s2) m>s, where s is in meters.
If s = 2 m when t = 0, determine the velocity and
acceleration as functions of time.
SOLUTION
v = - 4s2
ds
= - 4s2
dt
s
L2
s - 2 ds =
t
L0
-4 dt
Web)
2
8t + 1
laws
s =
teaching
1 -1
(s - 0.5)
4
in
t =
or
-s - 1| s2 = - 4t|t0
2
2
16
b = m>s
8t + 1
(8t + 1)2
Wide
copyright
permitted.
Dissemination
Ans.
States
World
v = -4a
of
the
not
instructors
16(2)(8t + 1)(8)
dv
256
=
=
m>s2
dt
(8t + 1)4
(8t + 1)3
is
on
learning.
United
and
work
the
(including
for
work
solely
of
protected
this
assessing
is
integrity
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and
work
provided
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the
is
This
destroy
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© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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Ans.
use
student
the
by
a =
12–25.
A sphere is fired downwards into a medium with an initial
speed of 27 m>s. If it experiences a deceleration of
a = ( -6t) m>s2, where t is in seconds, determine the
distance traveled before it stops.
SOLUTION
Velocity: v0 = 27 m>s at t0 = 0 s. Applying Eq. 12–2, we have
A+TB
dv = adt
v
L27
t
dv =
L0
-6tdt
v = A 27 - 3t2 B m>s
(1)
At v = 0, from Eq. (1)
0 = 27 - 3t2
t = 3.00 s
teaching
Web)
laws
or
Distance Traveled: s0 = 0 m at t0 = 0 s. Using the result v = 27 - 3t2 and applying
Eq. 12–1, we have
A+TB
Dissemination
Wide
copyright
States
World
permitted.
A 27 - 3t2 B dt
the
not
instructors
is
L0
learning.
L0
t
ds =
of
s
in
ds = vdt
(2)
student
the
by
and
use
United
on
s = A 27t - t3 B m
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This
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this
assessing
is
s = 27(3.00) - 3.003 = 54.0 m
of
protected
the
(including
for
work
At t = 3.00 s, from Eq. (2)
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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Ans.
12–26.
When two cars A and B are next to one another, they are
traveling in the same direction with speeds vA and vB ,
respectively. If B maintains its constant speed, while A
begins to decelerate at aA , determine the distance d
between the cars at the instant A stops.
A
B
d
SOLUTION
Motion of car A:
v = v0 + act
0 = vA - aAt
t =
vA
aA
v2 = v20 + 2ac(s - s0)
0 = v2A + 2( - aA)(sA - 0)
v2A
2aA
laws
or
sA =
teaching
Web)
Motion of car B:
in
vA
vAvB
b =
aA
aA
instructors
States
World
permitted.
Dissemination
Wide
copyright
sB = vBt = vB a
on
learning.
is
of
the
not
The distance between cars A and B is
and
use
United
v2A
vAvB
2vAvB - v2A
` = `
`
aA
2aA
2aA
work
student
the
by
Ans.
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is
This
provided
integrity
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this
assessing
is
work
solely
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the
(including
for
sBA = |sB - sA| = `
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–27.
A particle is moving along a straight line such that when it
is at the origin it has a velocity of 4 m>s. If it begins to
decelerate at the rate of a = 1- 1.5v1>22 m>s2, where v is in
m>s, determine the distance it travels before it stops.
SOLUTION
1
dv
= - 1.5v2
dt
a =
v
L4
1
t
1
v- 2 dv =
v
L0
- 1.5 dt
t
2v2 4 = - 1.5t 0
1
2av2 - 2 b = - 1.5t
v = (2 - 0.75t)2 m>s
t
L0
t
(2 - 0.75t)2 dt =
L0
(4 - 3t + 0.5625t2) dt
or
L0
ds =
laws
s
(1)
teaching
Web)
s = 4t - 1.5t2 + 0.1875t3
Wide
copyright
in
(2)
States
World
permitted.
Dissemination
From Eq. (1), the particle will stop when
United
on
learning.
is
of
the
not
instructors
0 = (2 - 0.75t)2
student
the
by
and
use
t = 2.667 s
work
s|t = 2.667 = 4(2.667) - 1.5(2.667)2 + 0.1875(2.667)3 = 3.56 m
sale
will
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destroy
of
and
any
courses
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the
is
This
provided
integrity
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and
work
this
assessing
is
work
solely
of
protected
the
(including
for
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*12–28.
A particle travels to the right along a straight line with a
velocity v = [5>14 + s2] m>s, where s is in meters.
Determine its deceleration when s = 2 m.
SOLUTION
5
4+s
v =
v dv = a ds
dv =
-5 ds
(4 + s)2
5
- 5 ds
b = a ds
a
(4 + s) (4 + s)2
laws
or
- 25
(4 + s)3
a =
in
teaching
Web)
When s = 2 m
sale
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is
This
provided
integrity
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and
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this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
Ans.
copyright
a = -0.116 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–29.
A particle moves along a straight line with an acceleration
1>2
a = 2v m>s2, where v is in m>s. If s = 0, v = 4 m>s when
t = 0, determine the time for the particle to achieve a
velocity of 20 m>s. Also, find the displacement of particle
when t = 2 s.
SOLUTION
Velocity:
+ B
A:
dt =
dv
a
t
L0
v
dt =
dv
L0 2v1>2
t
v
t ƒ 0 = v1>2 ƒ 4
t = v1>2 - 2
teaching
Web)
laws
or
v = (t + 222
copyright
in
When v = 20 m>s,
instructors
States
World
permitted.
Dissemination
Wide
20 = 1t + 222
Ans.
and
use
United
on
learning.
is
of
the
not
t = 2.47 s
work
student
the
by
Position:
(including
for
+ B
A:
assessing
is
work
solely
of
protected
the
ds = v dt
integrity
of
and
work
(t + 2)2 dt
provided
the
part
L0
is
L0
this
t
ds =
This
s
any
courses
destroy
of
and
sale
will
0
t
1
(t + 2)3 2
3
0
their
s
s2 =
s =
1
[(t + 2)3 - 23]
3
=
1 2
t(t + 6t + 12)
3
When t = 2 s,
s =
1
(2)[(2)2 + 6(2) + 12]
3
= 18.7 m
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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Ans.
12–30.
As a train accelerates uniformly it passes successive
kilometer marks while traveling at velocities of 2 m>s and
then 10 m>s. Determine the train’s velocity when it passes
the next kilometer mark and the time it takes to travel the
2-km distance.
SOLUTION
Kinematics: For the first kilometer of the journey, v0 = 2 m>s, v = 10 m>s, s0 = 0,
and s = 1000 m. Thus,
+ B
A:
v2 = v0 2 + 2ac (s - s0)
102 = 22 + 2ac (1000 - 0)
ac = 0.048 m>s2
For the second kilometer,
ac = 0.048 m>s2. Thus,
s0 = 1000 m,
s = 2000 m,
and
v2 = v0 2 + 2ac (s - s0)
or
+ B
A:
v0 = 10 m>s,
teaching
Web)
laws
v2 = 102 + 2(0.048)(2000 - 1000)
v = 14 m>s
Wide
copyright
in
Ans.
instructors
States
World
permitted.
Dissemination
For the whole journey, v0 = 2 m>s, v = 14 m>s, and ac = 0.048 m>s.2 Thus,
the
not
v = v0 + act
use
United
on
learning.
is
of
+ B
A:
for
work
student
the
by
and
14 = 2 + 0.048t
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is
This
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is
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protected
the
(including
t = 250 s
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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Ans.
12–31.
The acceleration of a particle along a straight line is defined
by a = 12t - 92 m>s2, where t is in seconds. At t = 0,
s = 1 m and v = 10 m>s. When t = 9 s, determine (a) the
particle’s position, (b) the total distance traveled, and
(c) the velocity.
SOLUTION
a = 2t - 9
v
L10
t
dv =
L0
12t - 92 dt
v - 10 = t2 - 9 t
v = t2 - 9 t + 10
s
or
13
t - 4.5 t2 + 10 t
3
in
teaching
Web)
13
t - 4.5 t2 + 10 t + 1
3
Wide
Dissemination
s =
1t2 - 9t + 102 dt
laws
s-1 =
L0
copyright
L1
t
ds =
of
the
not
instructors
States
World
permitted.
Note when v = t2 - 9 t + 10 = 0:
work
s = - 36.63 m
assessing
is
work
solely
of
protected
the
(including
When t = 7.701 s,
student
s = 7.13 m
for
When t = 1.298 s,
the
by
and
use
United
on
learning.
is
t = 1.298 s and t = 7.701 s
integrity
of
and
work
this
s = - 30.50 m
provided
When t = 9 s,
s = - 30.5 m
(b)
sTo t = (7.13 - 1) + 7.13 + 36.63 + (36.63 - 30.50)
Ans.
sTo t = 56.0 m
(c)
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or transmission in any form or by any means, electronic, mechanical,
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v = 10 m>s
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is
This
(a)
Ans.
Ans.
*12–32.
The acceleration of a particle traveling along a straight line
1 1>2
is a = s m>s2, where s is in meters. If v = 0, s = 1 m
4
when t = 0, determine the particle’s velocity at s = 2 m.
SOLUTION
Velocity:
+ B
A:
v dv = a ds
v
L0
s
v dv =
1 1>2
s ds
L1 4
v
s
v2 2
1
= s3>2 `
2 0
6
1
23
1s3>2 - 121>2 m>s
or
1
laws
v =
When s = 2 m, v = 0.781 m>s.
sale
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and
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courses
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the
is
This
provided
integrity
of
and
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this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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12–33.
At t = 0 bullet A is fired vertically with an initial (muzzle)
velocity of 450 m/s. When t = 3 s, bullet B is fired upward
with a muzzle velocity of 600 m/s. Determine the time t, after
A is fired, as to when bullet B passes bullet A. At what
altitude does this occur?
SOLUTION
+ c sA = (sA)0 + (vA)0 t +
sA = 0 + 450 t +
1 2
a t
2 c
1
(- 9.81) t2
2
+ c sB = (sB)0 + (vB)0 t +
sB = 0 + 600(t - 3) +
1
a t2
2 c
1
( -9.81)(t - 3)2
2
laws
or
Require sA = sB
Ans.
Ans.
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is
This
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integrity
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assessing
is
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protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
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World
h = sA = sB = 4.11 km
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
permitted.
Dissemination
copyright
t = 10.3 s
Wide
in
teaching
Web)
450 t - 4.905 t2 = 600 t - 1800 - 4.905 t2 + 29.43 t - 44.145
12–34.
A boy throws a ball straight up from the top of a 12-m high
tower. If the ball falls past him 0.75 s later, determine the
velocity at which it was thrown, the velocity of the ball when
it strikes the ground, and the time of flight.
SOLUTION
Kinematics: When the ball passes the boy, the displacement of the ball in equal to zero.
Thus, s = 0. Also, s0 = 0, v0 = v1, t = 0.75 s, and ac = - 9.81 m>s2.
A+cB
s = s0 + v0t +
1 2
at
2 c
0 = 0 + v110.752 +
1
1-9.81210.7522
2
v1 = 3.679 m>s = 3.68 m>s
or
Ans.
in
teaching
Web)
laws
When the ball strikes the ground, its displacement from the roof top is s = - 12 m.
Also, v0 = v1 = 3.679 m>s, t = t2, v = v2, and ac = - 9.81 m>s2.
1 2
at
2 c
Wide
permitted.
Dissemination
States
World
s = s0 + v0t +
copyright
A+cB
United
on
learning.
is
of
the
not
instructors
1
1- 9.812t22
2
and
use
- 12 = 0 + 3.679t2 +
the
of
protected
3.679 ; 21 - 3.67922 - 414.90521 -122
work
solely
214.9052
and
any
courses
part
the
is
This
Choosing the positive root, we have
provided
integrity
of
and
work
this
assessing
is
t2 =
(including
for
work
student
the
by
4.905t22 - 3.679t2 - 12 = 0
Ans.
sale
will
their
destroy
of
t2 = 1.983 s = 1.98 s
Using this result,
A+cB
v = v0 + act
v2 = 3.679 + 1 -9.81211.9832
= - 15.8 m>s = 15.8 m>s T
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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Ans.
12–35.
When a particle falls through the air, its initial acceleration
a = g diminishes until it is zero, and thereafter it falls at a
constant or terminal velocity vf. If this variation of the
acceleration can be expressed as a = 1g>v2f21v2f - v22,
determine the time needed for the velocity to become
v = vf>2 . Initially the particle falls from rest.
SOLUTION
g
dv
= a = ¢ 2 ≤ A v2f - v2 B
dt
vf
v
dy
L0 v2f - v2¿
=
t
g
v2f L0
dt
vf + v y
g
1
ln ¢
≤` = 2t
2vf
vf - v 0
vf
vf
2g
ln ¢
vf + v
vf - v
≤
vf + vf> 2
vf - vf> 2
≤
or
2g
ln ¢
b
Ans.
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This
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the
(including
for
work
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the
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and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
g
Wide
vf
copyright
t = 0.549 a
in
teaching
Web)
t =
vf
laws
t =
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*12–36.
A particle is moving with a velocity of v0 when s = 0 and
t = 0. If it is subjected to a deceleration of a = - kv3, where
k is a constant, determine its velocity and position as
functions of time.
SOLUTION
a =
dv
= - ky3
dt
v
Lv0
t
v-3 dv =
-k dt
L0
1
- 1v-2 - v0-22 = - kt
2
1
-2
1
v = ¢ 2kt + ¢ 2 ≤ ≤
v0
Ans.
or
ds = v dt
laws
teaching
Web)
1
2
1
≤≤
v20
Wide
¢ 2kt + ¢
in
L0
dt
permitted.
Dissemination
L0
t
ds =
copyright
s
the
not
instructors
States
is
of
and
use
United
on
learning.
4
by
2k
t
student
the
s =
World
1
2
1
2 ¢ 2kt + ¢ 2 ≤ ≤
v0
protected
1
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is
This
provided
integrity
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this
assessing
s =
is
work
solely
2
1
1
1
C ¢ 2kt + ¢ 2 ≤ ≤ S
v0
k
v0
of
the
(including
for
work
0
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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Ans.
12–37.
As a body is projected to a high altitude above the earth’s
surface, the variation of the acceleration of gravity with
respect to altitude y must be taken into account. Neglecting
air resistance, this acceleration is determined from the
formula a = -g0[R2>(R + y)2], where g0 is the constant
gravitational acceleration at sea level, R is the radius of the
earth, and the positive direction is measured upward. If
g0 = 9.81 m>s2 and R = 6356 km, determine the minimum
initial velocity (escape velocity) at which a projectile should
be shot vertically from the earth’s surface so that it does not
fall back to the earth. Hint: This requires that v = 0 as
y : q.
SOLUTION
v dv = a dy
q
0
Ly
v dv = - g0R
2
dy
L0 (R + y)
2
g0 R2 q
v2 2 0
2
=
2 y
R + y 0
teaching
Web)
laws
or
v = 22g0 R
Wide
copyright
in
= 22(9.81)(6356)(10)3
Ans.
sale
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and
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the
is
This
provided
integrity
of
and
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this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
= 11167 m>s = 11.2 km>s
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or transmission in any form or by any means, electronic, mechanical,
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12–38.
Accounting for the variation of gravitational acceleration
a with respect to altitude y (see Prob. 12–37), derive an
equation that relates the velocity of a freely falling particle
to its altitude. Assume that the particle is released from
rest at an altitude y0 from the earth’s surface. With what
velocity does the particle strike the earth if it is released
from rest at an altitude y0 = 500 km? Use the numerical
data in Prob. 12–37.
SOLUTION
From Prob. 12–37,
(+ c )
a = -g0
R2
(R + y)2
Since a dy = v dv
then
v
=
L0
v dv
y
1
v2
d =
R + y y0
2
g0 R2[
1
1
v2
] =
R + y
R + y0
2
Web)
laws
or
g0 R2 c
teaching
Dissemination
Wide
Ly0 (R + y)
2
in
dy
copyright
y
-g0 R2
States
World
permitted.
Thus
United
on
learning.
is
of
the
not
instructors
2g0 (y0 - y)
A (R + y)(R + y0)
and
use
v = -R
by
y = 0,
(including
for
work
student
the
When y0 = 500 km,
the
2(9.81)(500)(103)
work
solely
of
protected
A 6356(6356 + 500)(106)
this
assessing
is
v = - 6356(103)
sale
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is
This
provided
integrity
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and
work
v = - 3016 m>s = 3.02 km>s T
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Ans.
12–39.
A freight train starts from rest and travels with a constant
acceleration of 0.5 ft>s2. After a time t¿ it maintains a
constant speed so that when t = 160 s it has traveled 2000 ft.
Determine the time t¿ and draw the v–t graph for the motion.
SOLUTION
Total Distance Traveled: The distance for part one of the motion can be related to
time t = t¿ by applying Eq. 12–5 with s0 = 0 and v0 = 0.
+ B
A:
1
ac t2
2
s = s0 + v0 t +
s1 = 0 + 0 +
1
(0.5)(t¿)2 = 0.25(t¿)2
2
The velocity at time t can be obtained by applying Eq. 12–4 with v0 = 0.
+ B
A:
v = v0 + act = 0 + 0.5t = 0.5t
(1)
laws
or
The time for the second stage of motion is t2 = 160 - t¿ and the train is traveling at
a constant velocity of v = 0.5t¿ (Eq. (1)).Thus, the distance for this part of motion is
teaching
Web)
s2 = vt2 = 0.5t¿(160 - t¿) = 80t¿ - 0.5(t¿)2
in
+ B
A:
permitted.
Dissemination
Wide
copyright
If the total distance traveled is sTot = 2000, then
is
of
the
not
instructors
States
World
sTot = s1 + s2
by
and
use
United
on
learning.
2000 = 0.25(t¿)2 + 80t¿ - 0.5(t¿)2
the
(including
for
work
student
the
0.25(t¿)2 - 80t¿ + 2000 = 0
assessing
is
work
solely
of
protected
Choose a root that is less than 160 s, then
Ans.
provided
integrity
of
and
work
this
t¿ = 27.34 s = 27.3 s
sale
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destroy
of
and
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courses
part
the
is
This
v–t Graph: The equation for the velocity is given by Eq. (1).When t = t¿ = 27.34 s,
v = 0.5(27.34) = 13.7 ft>s.
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*12–40.
A sports car travels along a straight road with an
acceleration-deceleration described by the graph. If the car
starts from rest, determine the distance s¿ the car travels
until it stops. Construct the v- s graph for 0 … s … s¿ .
a(ft/s2)
6
1000
SOLUTION
v
4
s Graph: For 0 … s 6 1000 ft, the initial condition is v = 0 at s = 0.
+ B
A:
vdv = ads
v
s
vdv =
L0
L0
6ds
v2
= 6s
2
A 212s1>2 B ft>s
v =
or
When s = 1000 ft,
in
teaching
Web)
laws
v = 212(1000)1>2 = 109.54 ft>s = 110 ft>s
Dissemination
World
permitted.
vdv = ads
not
instructors
States
+ B
A:
Wide
copyright
For 1000 ft 6 s … s¿ , the initial condition is v = 109.54 ft>s at s = 1000 ft.
v
is
of
and
use
United
on
learning.
-4ds
by
L1000 ft
the
L109.54 ft>s
the
s
vdv =
student
v
is
work
solely
of
protected
the
(including
for
work
s
v2
2
= - 4s 1000 ft
2 109.54 ft>s
this
assessing
A 220 000 - 8s B ft>s
part
the
is
and
any
courses
When v = 0,
This
provided
integrity
of
and
work
v =
s¿ = 2500 ft
The v–s graph is shown in Fig. a.
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sale
will
their
destroy
of
0 = 220 000 - 8s¿
Ans.
s¿
s(ft)
12–41.
A train starts from station A and for the first kilometer, it
travels with a uniform acceleration. Then, for the next two
kilometers, it travels with a uniform speed. Finally, the train
decelerates uniformly for another kilometer before coming
to rest at station B. If the time for the whole journey is six
minutes, draw the v- t graph and determine the maximum
speed of the train.
SOLUTION
For stage (1) motion,
vmax = 0 + 1ac21t1
vmax = 1ac21t1
(1)
v12 = v02 + 21ac211s1 - s02
vmax 2 = 0 + 21ac2111000 - 02
1ac21 =
vmax 2
2000
(2)
or
+ B
A:
v1 = v0 + 1ac21t
laws
+ B
A:
copyright
in
teaching
Web)
Eliminating 1ac21 from Eqs. (1) and (2), we have
Wide
2000
vmax
World
permitted.
Dissemination
(3)
not
instructors
States
t1 =
the
by
and
use
United
on
learning.
is
of
the
For stage (2) motion, the train travels with the constant velocity of vmax for
t = 1t2 - t12. Thus,
(including
for
work
student
1
1a 2 t2
2 c 2
the
s2 = s1 + v1t +
1000 + 2000 = 1000 + vmax1t2 - t12 + 0
integrity
of
and
provided
(4)
part
the
is
and
any
courses
2000
vmax
This
t2 - t1 =
work
this
assessing
is
work
solely
of
protected
+ B
A:
+ B
A:
v3 = v2 + 1ac23t
sale
+ B
A:
will
their
destroy
of
For stage (3) motion, the train travels for t = 360 - t2. Thus,
0 = vmax - 1ac231360 - t22
vmax = 1ac231360 - t22
(5)
v32 = v22 + 21ac231s3 - s22
0 = vmax 2 + 23 - 1ac23414000 - 30002
1ac23 =
vmax 2
2000
(6)
Eliminating 1ac23 from Eqs. (5) and (6) yields
360 - t2 =
2000
vmax
(7)
Solving Eqs. (3), (4), and (7), we have
t1 = 120 s
t2 = 240 s
vmax = 16.7 m>s
Based on the above results, the v-t graph is shown in Fig. a.
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v = vmax for 2 min < t < 4 min.
Ans.
12–42.
A particle starts from s = 0 and travels along a straight line
with a velocity v = (t2 - 4t + 3) m > s, where t is in
seconds. Construct the v - t and a -t graphs for the time
interval 0 … t … 4 s.
SOLUTION
a–t Graph:
a =
dv
d 2
=
1t - 4t + 32
dt
dt
a = (2t - 4) m>s2
Thus,
a|t = 0 = 2(0) - 4 = - 4 m>s2
a|t = 2 = 0
teaching
Web)
laws
or
a|t = 4 s = 2(4) - 4 = 4 m>s2
permitted.
not
instructors
States
World
dv
= 0. Thus,
dt
v– t Graph: The slope of the v -t graph is zero when a =
Dissemination
Wide
copyright
in
The a -t graph is shown in Fig. a.
the
t = 2s
and
use
United
on
learning.
is
of
a = 2t - 4 = 0
for
work
student
the
by
The velocity of the particle at t = 0 s, 2 s, and 4 s are
work
solely
of
protected
the
(including
v|t = 0 s = 02 - 4(0) + 3 = 3 m>s
any
sale
will
their
destroy
of
and
The v- t graph is shown in Fig. b.
courses
part
integrity
the
is
This
provided
v|t = 4 s = 42 - 4(4) + 3 = 3 m>s
of
and
work
this
assessing
is
v|t = 2 s = 22 - 4(2) + 3 = - 1 m>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–43.
If the position of a particle is defined by
s = [2 sin [(p>5)t] + 4] m, where t is in seconds, construct
the s - t, v- t, and a - t graphs for 0 … t … 10 s.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
SOLUTION
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*12–44.
An airplane starts from rest, travels 5000 ft down a runway,
and after uniform acceleration, takes off with a speed of
162 mi>h. It then climbs in a straight line with a uniform
acceleration of 3 ft>s2 until it reaches a constant speed of
220 mi>h. Draw the s–t, v–t, and a–t graphs that describe
the motion.
SOLUTION
v1 = 0
v2 = 162
mi (1h) 5280 ft
= 237.6 ft>s
h (3600 s)(1 mi)
v22 = v21 + 2 ac(s2 - s1)
(237.6)2 = 02 + 2(ac)(5000 - 0)
ac = 5.64538 ft>s2
v2 = v1 + act
laws
or
237.6 = 0 + 5.64538 t
in
teaching
Web)
t = 42.09 = 42.1 s
mi (1h) 5280 ft
= 322.67 ft>s
h (3600 s)(1 mi)
instructors
States
World
permitted.
Dissemination
Wide
copyright
v3 = 220
United
on
learning.
is
of
the
not
v23 = v22 + 2ac(s3 - s2)
student
the
by
and
use
(322.67)2 = (237.6)2 + 2(3)(s - 5000)
protected
the
(including
for
work
s = 12 943.34 ft
this
assessing
is
work
solely
of
v3 = v2 + act
the
is
This
provided
integrity
of
and
work
322.67 = 237.6 + 3 t
sale
will
their
destroy
of
and
any
courses
part
t = 28.4 s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–45.
The elevator starts from rest at the first floor of the
building. It can accelerate at 5 ft>s2 and then decelerate at
2 ft>s2. Determine the shortest time it takes to reach a floor
40 ft above the ground. The elevator starts from rest and
then stops. Draw the a–t, v–t, and s–t graphs for the motion.
40 ft
SOLUTION
+ c v2 = v1 + act1
vmax = 0 + 5 t1
+ c v3 = v2 + ac t
0 = vmax - 2 t2
Thus
t1 = 0.4 t2
or
1 2
a t1
2 c
laws
+ c s2 = s1 + v1t1 +
in
teaching
Web)
1
(5)(t21) = 2.5 t21
2
Dissemination
Wide
copyright
h = 0 + 0 +
the
not
instructors
States
World
permitted.
1
(2) t22
2
learning.
is
of
+ c 40 - h = 0 + vmaxt2 -
the
by
and
use
United
on
+ c v2 = v21 + 2 ac(s - s1)
the
(including
for
work
student
v2max = 0 + 2(5)(h - 0)
assessing
is
work
solely
of
protected
v2max = 10h
part
the
is
and
any
courses
v2max = 160 - 4h
This
provided
integrity
of
and
work
this
0 = v2max + 2(-2)(40 - h)
sale
will
their
destroy
of
Thus,
10 h = 160 - 4h
h = 11.429 ft
vmax = 10.69 ft>s
t1 = 2.138 s
t2 = 5.345 s
t = t1 + t2 = 7.48 s
When t = 2.145, v = vmax = 10.7 ft>s
and h = 11.4 ft.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
12–46.
The velocity of a car is plotted as shown. Determine the
total distance the car moves until it stops 1t = 80 s2.
Construct the a–t graph.
v (m/s)
10
SOLUTION
t (s)
40
Distance Traveled: The total distance traveled can be obtained by computing the
area under the v - t graph.
s = 10(40) +
1
(10)(80 - 40) = 600 m
2
Ans.
dv
a – t Graph: The acceleration in terms of time t can be obtained by applying a =
.
dt
For time interval 0 s … t 6 40 s,
a =
or
v - 10
0 - 10
1
, v = a - t + 20 b m>s.
=
t - 40
80 - 40
4
teaching
Web)
laws
For time interval 40 s 6 t … 80 s,
dv
= 0
dt
in
dv
1
= - = - 0.250 m s2
dt
4
instructors
States
World
permitted.
Dissemination
Wide
copyright
a =
by
and
use
United
on
learning.
is
of
the
not
For 0 … t 6 40 s, a = 0.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
For 40 s 6 t … 80, a = - 0.250 m s2 .
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
80
12–47.
The v–s graph for a go-cart traveling on a straight road is
shown. Determine the acceleration of the go-cart at
s = 50 m and s = 150 m. Draw the a–s graph.
v (m/s)
8
SOLUTION
For 0 … s 6 100
v = 0.08 s,
100
dv = 0.08 ds
a ds = (0.08 s)(0.08 ds)
a = 6.4(10 - 3) s
At s = 50 m,
a = 0.32 m>s2
Ans.
For 100 6 s 6 200
or
v = - 0.08 s + 16,
teaching
Web)
laws
dv = - 0.08 ds
Wide
copyright
in
a ds = (-0.08 s + 16)(- 0.08 ds)
not
instructors
States
World
permitted.
Dissemination
a = 0.08(0.08 s - 16)
Ans.
the
a = - 0.32 m>s2
by
and
use
United
on
learning.
is
of
At s = 150 m,
for
work
student
the
Also,
is
work
solely
of
protected
the
(including
v dv = a ds
integrity
of
and
work
this
assessing
dv
)
ds
part
the
is
This
provided
a = v(
will
8
) = 0.32 m>s2
100
sale
a = 4(
their
destroy
of
and
any
courses
At s = 50 m,
Ans.
At s = 150 m,
a = 4(
-8
) = -0.32 m>s2
100
At s = 100 m, a changes from amax = 0.64 m>s2
to amin = -0.64 m>s2 .
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
200
s (m)
*12–48.
The v–t graph for a particle moving through an electric field
from one plate to another has the shape shown in the figure.
The acceleration and deceleration that occur are constant
and both have a magnitude of 4 m>s2. If the plates are
spaced 200 mm apart, determine the maximum velocity vmax
and the time t¿ for the particle to travel from one plate to
the other. Also draw the s–t graph. When t = t¿>2 the
particle is at s = 100 mm.
smax
v
s
vmax
SOLUTION
ac = 4 m/s2
t¿/ 2
s
= 100 mm = 0.1 m
2
v2 = v20 + 2 ac(s - s0)
v2max = 0 + 2(4)(0.1 - 0)
vmax = 0.89442 m>s
= 0.894 m>s
Ans.
v = v0 + ac t¿
in
teaching
Web)
laws
or
t¿
0.89442 = 0 + 4( )
2
Ans.
Dissemination
Wide
= 0.447 s
copyright
t¿ = 0.44721 s
the
not
instructors
States
World
permitted.
1
a t2
2 c
United
on
learning.
is
of
s = s0 + v0 t +
work
student
the
by
and
use
1
(4)(t)2
2
(including
for
s = 0 + 0 +
assessing
is
work
solely
of
protected
the
s = 2 t2
provided
integrity
of
and
work
this
0.44721
= 0.2236 = 0.224 s,
2
part
the
is
This
When t =
of
and
any
courses
s = 0.1 m
L0.894
v = - 4 t +1.788
s
t
ds =
L0.1
1 -4t+ 1.7882 dt
L0.2235
s = - 2 t2 + 1.788 t - 0.2
When t = 0.447 s,
s = 0.2 m
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
will
4 dt
L0.2235
sale
t
ds = -
their
destroy
v
t¿
t
12–49.
The v–t graph for a particle moving through an electric field
from one plate to another has the shape shown in the figure,
where t¿ = 0.2 s and vmax = 10 m>s. Draw the s–t and a–t graphs
for the particle. When t = t¿>2 the particle is at s = 0.5 m.
smax
v
s
vmax
SOLUTION
For 0 6 t 6 0.1 s,
t¿/ 2
v = 100 t
dv
= 100
dt
a =
ds = v dt
s
L0
t
ds =
L0
100 t dt
or
s = 50 t 2
in
teaching
Web)
laws
When t = 0.1 s,
Dissemination
Wide
copyright
s = 0.5 m
of
the
not
instructors
States
World
permitted.
For 0.1 s 6 t 6 0.2 s,
by
and
use
United
on
learning.
is
v = -100 t + 20
work
student
the
dv
= - 100
dt
of
protected
the
(including
for
a =
this
assessing
is
work
solely
ds = v dt
t
part
the
is
destroy
of
and
sale
s = - 50 t 2 + 20 t - 1
will
their
s - 0.5 = ( -50 t 2 + 20 t - 1.5)
any
courses
1 -100t + 202dt
L0.1
This
L0.5
ds =
provided
integrity
of
and
work
s
When t = 0.2 s,
s = 1m
When t = 0.1 s, s = 0.5 m and a changes from 100 m/s2
to -100 m/s2. When t = 0.2 s, s = 1 m.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
t¿
t
12–50.
The v -t graph of a car while traveling along a road is
shown. Draw the s -t and a- t graphs for the motion.
v (m/s)
SOLUTION
20
0 … t … 5
20
¢v
=
= 4 m>s2
¢t
5
a =
5 … t … 20
a =
20 … t … 30
a =
5
¢v
20 - 20
=
= 0 m>s2
¢t
20 - 5
¢v
0 - 20
=
= -2 m>s2
¢t
30 - 20
From the v–t graph at t1 = 5 s , t2 = 20 s , and t3 = 30 s,
or
1
(5)(20) = 50 m
2
laws
s1 = A1 =
in
teaching
Web)
s2 = A1 + A2 = 50 + 20(20 - 5) = 350 m
permitted.
Dissemination
Wide
copyright
1
(30 - 20)(20) = 450 m
2
of
the
not
instructors
States
World
s3 = A1 + A2 + A3 = 350 +
and
use
United
on
learning.
is
The equations defining the portions of the s–t graph are
t
ds =
s = 2t2
work
student
4t dt;
L0
(including
the
L0
the
ds = v dt;
for
v = 4t;
by
s
0 … t … 5s
s = 20t - 50
work
solely
of
t
20 dt;
L5
integrity
of
and
this
L50
ds =
assessing
is
ds = v dt;
work
v = 20;
protected
s
5 … t … 20 s
provided
s
ds = v dt;
part
the
is
any
destroy
of
and
will
their
sale
For 0 … t 6 5 s, a = 4 m>s2.
For 20 s 6 t … 30 s, a = - 2 m>s2.
At t = 5 s, s = 50 m. At t = 20 s, s = 350 m.
At t = 30 s, s = 450 m.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
t
ds =
L350
courses
v = 2(30 - t);
This
20 … t … 30 s
L20
2(30 - t) dt;
s = - t2 + 60t - 450
20
30
t (s)
12–51.
The a–t graph of the bullet train is shown. If the train starts
from rest, determine the elapsed time t¿ before it again
comes to rest. What is the total distance traveled during this
time interval? Construct the v–t and s–t graphs.
a(m/s2)
a
0.1t
30
v t Graph: For the time interval 0 … t 6 30 s, the initial condition is v = 0 when
t = 0 s.
dv = adt
v
L0
t
dv =
L0
0.1tdt
v = A 0.05t2 B m>s
When t = 30 s,
v t = 30 s = 0.05 A 302 B = 45 m>s
laws
or
or the time interval 30 s 6 t … t¿ , the initial condition is v = 45 m>s at t = 30 s.
in
teaching
Web)
dv = adt
permitted.
Dissemination
1
t + 5 ≤ dt
15
is
of
the
not
¢-
World
L30 s
L45 m>s
instructors
t
dv =
States
v
Wide
copyright
+ B
A:
by
and
use
United
on
learning.
1 2
t + 5t - 75 ≤ m>s
30
for
work
student
the
v = ¢-
protected
the
(including
Thus, when v = 0,
this
assessing
is
work
solely
of
1 2
t¿ + 5t¿ - 75
30
and
work
0 = -
part
the
is
This
provided
integrity
of
Choosing the root t¿ 7 75 s,
any
courses
Ans.
destroy
of
and
t¿ = 133.09 s = 133 s
sale
will
their
Also, the change in velocity is equal to the area under the a–t graph. Thus,
¢v =
0 =
L
adt
1
1
1
(3)(75) + B ¢ - t¿ + 5 ≤ (t¿ - 75) R
2
2
15
0 = -
1 2
t¿ + 5t¿ - 75
30
This equation is the same as the one obtained previously.
The slope of the v–tgraph is zero when t = 75 s, which is the instant a =
v t = 75 s = -
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
1
A 752 B + 5(75) - 75 = 112.5 m>s
30
dv
= 0.Thus,
dt
( 15 )t
5
t¿
SOLUTION
+ B
A:
1
a
3
75
t(s)
12–51. continued
The v–t graph is shown in Fig. a.
s t Graph: Using the result of v, the equation of the s–t graph can be obtained by
integrating the kinematic equation ds = vdt. For the time interval 0 … t 6 30 s, the
initial condition s = 0 at t = 0 s will be used as the integration limit. Thus,
+ B
A:
ds = vdt
s
L0
t
ds =
s = a
L0
0.05t2 dt
1 3
t bm
60
When t = 30 s,
1
A 303 B = 450 m
60
s t = 30 s =
For the time interval 30 s 6 t … t¿ = 133.09 s, the initial condition is s = 450 m
when t = 30 s.
ds = vdt
1 2
t + 5t - 75 b dt
30
Web)
L30 s
a-
in
L450 m
or
t
ds =
laws
s
teaching
+ B
A:
instructors
States
World
permitted.
Dissemination
Wide
copyright
5
1
s = a - t3 + t2 - 75t + 750b m
90
2
use
United
on
learning.
is
of
the
not
When t = 75 s and t¿ = 133.09 s,
work
student
the
by
and
1
5
A 753 B + A 752 B - 75(75) + 750 = 4500 m
90
2
the
(including
for
s t = 75 s = -
work
solely
of
protected
5
1
A 133.093 B + A 133.092 B - 75(133.09) + 750 = 8857 m
90
2
and
any
courses
part
the
is
This
The s–t graph is shown in Fig. b.
provided
integrity
of
and
work
this
assessing
is
s t = 133.09 s = -
sale
will
their
destroy
of
When t = 30 s,
v = 45 m/s and s = 450 m.
When t = 75 s,
v = vmax = 112.5 m/s and s = 4500 m.
When t = 133 s,
v = 0 and s = 8857 m.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
*12–52.
The snowmobile moves along a straight course according to
the v –t graph. Construct the s–t and a–t graphs for the same
50-s time interval. When t = 0, s = 0.
v (m/s)
12
SOLUTION
s t Graph: The position function in terms of time t can be obtained by applying
12
ds
2
. For time interval 0 s … t 6 30 s, v =
v =
t = a t b m>s.
dt
30
5
t (s)
30
ds = vdt
s
L0
t
ds =
2
tdt
L0 5
1
s = a t2 b m
5
1
A 302 B = 180 m
5
or
s =
laws
At t = 30 s ,
in
teaching
Web)
For time interval 30 s<t ◊ 50 s,
permitted.
Dissemination
Wide
copyright
ds = vdt
instructors
United
on
learning.
is
of
the
not
12dt
L30 s
and
use
L180 m
World
t
ds =
States
s
(including
for
work
student
the
by
s = (12t - 180) m
the
s = 12(50) - 180 = 420 m
assessing
is
work
solely
of
protected
At t = 50 s,
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
a t Graph: The acceleration function in terms of time t can be obtained by applying
dv
2
dv
a =
. For time interval 0 s ◊ t<30 s and 30 s<t ◊ 50 s, a =
=
dt
dt
5
dv
2
= 0.4 m>s and a =
= 0, respectively.
dt
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
50
12–53.
A two-stage missile is fired vertically from rest with the
acceleration shown. In 15 s the first stage A burns out and
the second stage B ignites. Plot the v -t and s-t graphs
which describe the two-stage motion of the missile for
0 … t … 20 s.
a (m/s2)
SOLUTION
25
Since v =
v–t graph.
L
B
18
a dt, the constant lines of the a–t graph become sloping lines for the
t (s)
15
The numerical values for each point are calculated from the total area under the
a–t graph to the point.
At t = 15 s,
v = (18)(15) = 270 m>s
At t = 20 s,
v = 270 + (25)(20 - 15) = 395 m>s
Since s =
v dt, the sloping lines of the v–t graph become parabolic curves for the
laws
or
s–t graph.
L
Wide
copyright
in
teaching
Web)
The numerical values for each point are calculated from the total area under the v–t
graph to the point.
1
(15)(270) = 2025 m
2
s =
At t = 20 s,
s = 2025 + 270(20 - 15) +
is
of
the
not
instructors
States
World
permitted.
Dissemination
At t = 15 s,
for
work
student
the
by
and
use
United
on
learning.
1
(395 - 270)(20 - 15) = 3687.5 m = 3.69 km
2
work
solely
of
protected
the
(including
Also:
and
work
this
assessing
is
0 … t … 15:
any
destroy
of
and
v = v0 + ac t = 0 + 18t
courses
part
the
is
This
provided
integrity
of
a = 18 m>s 2
will
their
1
a t2 = 0 + 0 + 9t2
2 c
sale
s = s0 + v0 t +
When t = 15:
v = 18(15) = 270 m>s
s = 9(15)2 = 2025 m = 2.025 km
15 … t … 20:
a = 25 m>s 2
v = v0 + ac t = 270 + 25(t - 15)
s = s0 + v0 t +
1
1
ac t2 = 2025 + 270(t - 15) + (25)(t - 15)2
2
2
When t = 20:
v = 395 m>s
s = 3687.5 m = 3.69 km
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
A
20
12–54.
The dragster starts from rest and has an acceleration
described by the graph. Determine the time t¿ for it to stop.
Also, what is its maximum speed? Construct the v - t and
s- t graphs for the time interval 0 … t … t¿ .
a (ft/s2)
80
SOLUTION
t¿
v– t Graph: For the time interval 0 … t 6 5 s, the initial condition is v = 0 when
t = 0 s.
A
+
:
5
1
1
B
a⫽⫺t⫹5
dv = adt
v
t
dv =
L0
L0
80dt
v = (80t) ft>s
The maximum speed occurs at the instant when the acceleration changes sign when
t = 5 s. Thus,
vmax = v|t = 5 s = 80(5) = 400 ft>s
Web)
laws
or
Ans.
Wide
World
permitted.
Dissemination
dv = adt
is
of
the
not
instructors
States
+ B
A:
copyright
in
teaching
For the time interval 5 6 t … t¿ , the initial condition is v = 400 ft>s when t = 5 s.
on
learning.
t
and
( -t + 5)dt
for
work
student
L5 s
by
L400 ft > s
the
dv =
use
United
v
work
solely
of
protected
the
(including
t2
+ 5t + 387.5b ft>s
2
assessing
is
v = a-
part
the
is
and
any
courses
t¿ 2
+ 5t¿ + 387.5
2
This
0 = -
provided
integrity
of
and
work
this
Thus when v = 0,
t¿ = 33.28 s = 33.3 s
sale
will
their
destroy
of
Choosing the positive root,
Ans.
Also, the change in velocity is equal to the area under the a- t graph. Thus
¢v =
L
adt
1
0 = 80(5) + e [( - t¿ + 5)(t¿ - 5)] f
2
0 = -
t¿ 2
+ 5t¿ + 387.5
2
This quadratic equation is the same as the one obtained previously. The v-t graph is
shown in Fig. a.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
t (s)
12–54. continued
s–t Graph: For the time interval 0 … t 6 5 s, the initial condition is s = 0 when
t = 0 s.
+ B
A:
ds = vdt
s
t
ds =
80dt
L0
s = (40t2) ft
L0
When t = 5 s,
s|t = 5 s = 40(52) = 1000 ft
For the time interval 5 s 6 t … t¿ = 45 s, the initial condition is s = 1000 ft when
t = 5 s.
+ B
A:
ds = vdt
t
ds =
L1000 ft
L5 s
t2
+ 5t + 387.5b dt
2
t3
5
+ t2 + 387.5t - 979.17bft
6
2
laws
s = a-
a-
or
s
in
teaching
Web)
When t = t¿ = 33.28 s,
33.283
5
+ (33.282) + 387.5(33.28) - 979.17 = 8542 ft
6
2
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
s|t = 33.28 s = -
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
The s –t graph is shown in Fig. b.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–55.
A race car starting from rest travels along a straight
road and for 10 s has the acceleration shown. Construct
the v–t graph that describes the motion and find the
distance traveled in 10 s.
a (m/s2)
6
6
a=
SOLUTION
6
dv = adt
t
L0
v = a
or
1
A 63 B = 12.0 m>s,
18
teaching
Web)
v =
1 3
t b m>s
18
laws
At t = 6 s,
1 2
t dt
L0 6
dv =
Wide
copyright
in
For time interval 6 s 6 t … 10 s,
instructors
States
World
permitted.
Dissemination
dv = adt
the
is
and
use
United
on
learning.
6dt
by
L6s
the
L12.0m>s
not
t
dv =
of
v
the
(including
for
work
student
v = (6t - 24) m>s
work
solely
of
protected
v = 6(10) - 24 = 36.0 m>s
assessing
is
At t = 10 s,
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
ds
Position: The position in terms of time t can be obtained by applying v =
.
dt
For time interval 0 s … t 6 6 s,
L0
t
ds =
s = ¢
sale
will
their
destroy
of
ds = vdt
s
1 3
t dt
L0 18
1 4
t ≤m
72
1
A 64 B = 18.0 m.
72
When t = 6 s, v = 12.0 m>s and s =
For time interval 6 s 6 t … 10 s,
ds = vdt
s
t
dv =
L18.0 m
L6s
(6t - 24)dt
s = A 3t2 - 24t + 54 B m
When t = 10 s, v = 36.0 m>s and s = 3 102 - 24(10) + 54 = 114 m
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
t2
t (s)
v -t Graph: The velocity function in terms of time t can be obtained by applying
dv
formula a =
. For time interval 0 s … t 6 6 s,
dt
v
1
—
6
10
*12–56.
The v–t graph for the motion of a car as it moves along a
straight road is shown. Draw the a–t graph and determine
the maximum acceleration during the 30-s time interval.
The car starts from rest at s = 0.
v (ft/s)
60
v
40
SOLUTION
t
30
0.4t2
v
For t 6 10 s:
t (s)
v = 0.4t2
a =
10
dv
= 0.8t
dt
At t = 10 s:
a = 8 ft>s2
For 10 6 t … 30 s:
laws
in
teaching
Web)
dv
= 1
dt
Dissemination
Wide
copyright
a =
or
v = t + 30
permitted.
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
amax = 8 ft>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
30
12–57.
The v–t graph for the motion of a car as it moves along a
straight road is shown. Draw the s–t graph and determine
the average speed and the distance traveled for the 30-s time
interval. The car starts from rest at s = 0.
v (ft/s)
60
v
40
SOLUTION
t
30
0.4t2
v
For t 6 10 s,
t (s)
v = 0.4t2
10
ds = v dt
s
L0
t
ds =
0.4t2 dt
L0
s = 0.1333t3
At t = 10 s,
laws
or
s = 133.3 ft
in
teaching
Web)
For 10 6 t 6 30 s,
permitted.
Dissemination
Wide
copyright
v = t + 30
is
of
the
not
instructors
States
World
ds = v dt
on
learning.
t
United
s
and
use
1t + 302 dt
for
work
student
L10
by
L133.3
the
ds =
is
work
solely
of
protected
the
(including
s = 0.5t2 + 30t - 216.7
part
the
is
This
s = 1133 ft
provided
integrity
of
and
work
this
assessing
At t = 30 s,
destroy
of
and
any
courses
1133
¢s
=
= 37.8 ft>s
¢t
30
Ans.
sale
will
their
(vsp)Avg =
sT = 1133 ft = 1.13(103) ft
When t = 0 s, s = 133 ft.
When t = 30 s, s = sI = 1.33 (103) ft
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
30
12–58.
The jet-powered boat starts from rest at s = 0 and travels
along a straight line with the speed described by the graph.
Construct the s -t and a -t graph for the time interval
0 … t … 50 s.
v (m/s)
v ⫽ 4.8 (10 ⫺3)t 3
75
v ⫽ ⫺3t ⫹ 150
SOLUTION
t (s)
s–t Graph: The initial condition is s = 0 when t = 0.
+ B
A:
25
ds = vdt
s
L0
t
ds =
L0
4.8(10-3)t3 dt
s = [1.2110-3)t4 ]m
At t = 25 s,
s|t = 25 s = 1.2(10-3)(254) = 468.75 m
ds = vdt
permitted.
Dissemination
Wide
copyright
+ B
A:
in
teaching
Web)
laws
or
For the time interval 25 s 6 t … 50 s, the initial condition s = 468.75 m when
t = 25 s will be used.
instructors
States
the
not
(- 3t + 150) dt
on
learning.
is
of
L25 s
by
and
use
L468.75 m
World
t
ds =
United
s
solely
of
protected
the
(including
for
work
student
the
3
s = a - t2 + 150t - 2343.75b m
2
this
assessing
is
work
When t = 50 s,
any
sale
a–t Graph: For the time interval 0 … t 6 25 s,
will
their
destroy
of
and
The s -t graph is shown in Fig. a.
courses
part
the
is
This
provided
a =
dv
d
= [4.8(10-3)t3] = (0.0144t2) m > s2
dt
dt
When t = 25 s,
a|t = 25 s = 0.0144(252) m > s2 = 9 m > s2
For the time interval 25 s 6 t … 50 s,
a =
dv
d
= (- 3t + 150) = - 3 m > s2
dt
dt
The a -t graph is shown in Fig. b.
When t = 25 s,
a = amax = 9 m > s2 and s = 469 m.
When t = 50 s,
s = 1406 m.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
integrity
of
and
work
3
s|t = 50 s = - (502) + 150(50) - 2343.75 = 1406.25 m
2
50
12–59.
An airplane lands on the straight runway, originally traveling
at 110 ft> s when s = 0. If it is subjected to the decelerations
shown, determine the time t¿ needed to stop the plane and
construct the s–t graph for the motion.
a (ft/s2)
5
–3
SOLUTION
–8
v0 = 110 ft>s
¢ v = 1 a dt
0 - 110 = - 3(15 - 5) - 8(20 - 15) - 3(t¿ - 20)
t¿ = 33.3 s
Ans.
= 550 ft
5s
st=
15s
= 1500 ft
st=
20s
= 1800 ft
st=
33.3s
laws
or
st=
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
= 2067 ft
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
15
20
t'
t (s)
*12–60.
v (m/s)
A car travels along a straight road with the speed shown by
the v -t graph. Plot the a -t graph.
6
1
v⫽—
5 t
1
v ⫽ ⫺—
3 (t ⫺ 48)
SOLUTION
t (s)
a–t Graph: For 0 … t 6 30 s,
v =
1
t
5
a =
dv
1
= = 0.2 m > s2
dt
5
30
For 30 s 6 t … 48 s
teaching
Web)
laws
or
1
v = - (t - 48)
3
1
dv
= - (1) = - 0.333 m > s2
a =
dt
3
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
Using these results, a-t graph shown in Fig. a can be plotted.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
48
12–61.
v (m/s)
A car travels along a straight road with the speed shown by
the v–t graph. Determine the total distance the car travels
until it stops when t = 48 s. Also plot the s–t and a–t graphs.
6
v
1
—
5 t
v
1
—
3 (t
48)
SOLUTION
For 0 … t … 30 s,
t (s)
30
1
v = t
5
a =
dv
1
=
dt
5
ds = v dt
t
L0
1 2
t
10
in
teaching
Web)
s =
1
t dt
L0 5
or
ds =
laws
s
s = 90 m,
Dissemination
Wide
copyright
When t = 30 s,
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
1
v = - (t - 48)
3
work
student
the
by
and
dv
1
= dt
3
the
(including
for
a =
assessing
is
work
solely
of
protected
ds = v dt
this
integrity
of
and
work
1
- 1t - 482dt
3
L30
any
sale
will
their
destroy
of
and
1
s = - t2 + 16t - 240
6
courses
part
the
is
This
L90
t
ds =
provided
s
When t = 48 s,
s = 144 m
Ans.
Also, from the v–t graph
¢s =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
L
v dt
s-0 =
1
1621482 = 144 m
2
Ans.
48
12–62.
A motorcyclist travels along a straight road with the velocity
described by the graph. Construct the s - t and a -t graphs.
v (ft/s)
150
v 20 t 50
50
v 2 t2
SOLUTION
t (s)
5
s–t Graph: For the time interval 0 … t 6 5 s, the initial condition is s = 0 when
t = 0.
+ B
A:
ds = vdt
s
t
ds =
L0
2
L0
2t dt
2
s = a t3 b ft
3
laws
or
When t = 5 s,
in
teaching
Web)
2 3
(5 ) = 83.33 ft = 83.3 ft and a = 20 ft>s2
3
Wide
copyright
s =
instructors
States
United
on
learning.
is
of
the
not
ds = vdt
and
use
+ B
A:
World
permitted.
Dissemination
For the time interval 5 s 6 t … 10 s, the initial condition is s = 83.33 ft when t = 5 s.
by
t
the
s
(20t - 50)dt
work
(including
of
protected
the
L5 s
student
L83.33 ft
for
ds =
work
solely
t
this
= (10t2 - 50t) 2
assessing
s
is
s2
5s
integrity
of
and
work
83.33 ft
any
their
destroy
of
and
When t = 10 s,
courses
part
the
is
This
provided
s = (10t2 - 50t + 83.33) ft
sale
will
s|t = 10 s = 10(102) - 50(10) + 83.33 = 583 ft
The s -t graph is shown in Fig. a.
a–t Graph: For the time interval 0 … t 6 5 s,
+ B
A:
a =
dv
d
(2t2) = (4t) ft>s 2
=
dt
dt
When t = 5 s,
a = 4(5) = 20 ft>s2
For the time interval 5 s 6 t … 10 s,
+ B
A:
a =
d
dv
=
(20t - 50) = 20 ft > s2
dt
dt
The a -t graph is shown in Fig. b.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
10
12–63.
The speed of a train during the first minute has been
recorded as follows:
t (s)
0 20
40
60
0 16
21
24
v (m>s)
Plot the v-t graph, approximating the curve as straight-line
segments between the given points. Determine the total
distance traveled.
SOLUTION
The total distance traveled is equal to the area under the graph.
1
1
1
(20)(16) + (40 - 20)(16 + 21) + (60 - 40)(21 + 24) = 980 m
2
2
2
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
sT =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*12–64.
A man riding upward in a freight elevator accidentally
drops a package off the elevator when it is 100 ft from the
ground. If the elevator maintains a constant upward speed
of 4 ft>s, determine how high the elevator is from the
ground the instant the package hits the ground. Draw the
v–t curve for the package during the time it is in motion.
Assume that the package was released with the same
upward speed as the elevator.
SOLUTION
For package:
(+ c )
v2 = v20 + 2ac(s2 - s0)
v2 = (4)2 + 2( - 32.2)( 0 - 100)
v = 80.35 ft>s T
(+ c ) v = v0 + act
-80.35 = 4 + ( - 32.2)t
laws
or
t = 2.620 s
teaching
Web)
For elevator:
in
s2 = s0 + vt
Wide
copyright
(+ c )
not
instructors
States
World
permitted.
Dissemination
s = 100 + 4(2.620)
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
s = 110 ft
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–65.
Two cars start from rest side by side and travel along a
straight road. Car A accelerates at 4 m>s2 for 10 s and then
maintains a constant speed. Car B accelerates at 5 m>s2
until reaching a constant speed of 25 m/s and then
maintains this speed. Construct the a–t, v–t, and s–t graphs
for each car until t = 15 s. What is the distance between the
two cars when t = 15 s?
SOLUTION
Car A:
v = v0 + ac t
vA = 0 + 4t
At t = 10 s,
vA = 40 m>s
s = s0 + v0t +
1
(4)t2 = 2t2
2
t 7 10 s,
ds = v dt
teaching
sA = 200 m
Dissemination
Wide
copyright
in
At t = 10 s,
Web)
laws
or
sA = 0 + 0 +
1 2
at
2 c
permitted.
not
instructors
States
World
40 dt
the
is
L10
United
on
learning.
L200
t
ds =
of
sA
work
student
the
by
and
use
sA = 40t - 200
the
(including
for
sA = 400 m
is
work
solely
of
protected
At t = 15 s,
work
this
assessing
Car B:
any
destroy
of
and
vB = 0 + 5t
courses
part
the
is
This
provided
integrity
of
and
v = v0 + a c t
their
25
= 5s
5
will
t =
sale
When vB = 25 m/s,
s = s0 + v0t +
sB = 0 + 0 +
1 2
at
2 c
1
(5)t2 = 2.5t2
2
When t = 10 s, vA = (vA)max = 40 m/s and sA = 200 m.
When t = 5 s, sB = 62.5 m.
When t = 15 s, sA = 400 m and sB = 312.5 m.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–65. continued
At t = 5 s,
sB = 62.5 m
t 7 5 s,
ds = v dt
sB
L62.5
t
ds =
L5
25 dt
sB - 62.5 = 25t - 125
sB = 25t - 62.5
When t = 15 s,
sB = 312.5
Distance between the cars is
¢s = sA - sB = 400 - 312.5 = 87.5 m
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Car A is ahead of car B.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–66.
A two-stage rocket is fired vertically from rest at s = 0 with
an acceleration as shown. After 30 s the first stage A burns
out and the second stage B ignites. Plot the v–t and s–t
graphs which describe the motion of the second stage for
0 … t … 60 s.
a (m/s2)
B
A
15
9
SOLUTION
a
0.01t2
For 0 … t … 30 s
v
l
dv =
L0
0.01 t2 dt
L0
30
v = 0.00333t3
When t = 30 s, v = 90 m>s
For 30 s … t … 60 s
L30
15dt
laws
L90
l
dv =
or
v
in
teaching
Web)
v = 15t - 360
v = 540 m>s
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
When t = 60 s,
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
60
t (s)
12–67.
A two-stage rocket is fired vertically from rest at s = 0 with
an acceleration as shown. After 30 s the first stage A burns
out and the second stage B ignites. Plot the s–t graph which
describes the motion of the second stage for 0 … t … 60 s.
a (m/s2)
B
A
15
9
a ⫽ 0.01t 2
SOLUTION
v– t Graph: When t = 0, v = 0. For 0 … t … 30 s,
A+cB
30
dv = a dt
v
L0
v2
t
dv =
v
0.01t2dt
L0
0.01 3 t
t 2
3
0
=
0
laws
or
v = {0.003333t3} m>s
in
teaching
Web)
When t = 30 s, v = 0.003333(303) = 90 m>s
Wide
Dissemination
World
permitted.
dv = a dt
is
of
the
not
instructors
States
A+cB
copyright
For 30 s 6 t … 60 s,
on
learning.
t
use
United
v
and
15 dt
by
for
work
student
L30 s
the
L90 m>s
dv =
the
(including
t
of
work
= 15t 2
solely
v
protected
v2
30 s
this
assessing
is
90 m>s
part
the
is
destroy
will
their
When t = 60 s, v = 15(60) - 360 = 540 m>s
of
and
any
courses
v = {15t - 360} m>s
This
provided
integrity
of
and
work
v - 90 = 15t - 450
sale
s–t Graph: When t = 0, s = 0. For 0 … t … 30 s,
A+cB
ds = vdt
s
t
ds =
L0
L0
0.003333t3dt
s
s 2 = 0.0008333t4 2
0
t
0
4
s = {0.0008333 t } m
When t = 30 s, s = 0.0008333(304) = 675 m
For 30 s 6 t … 60 s,
A+cB
ds = vdt
s
L675 m
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
t
ds =
L30 s
(15t - 360) dt
60
t (s)
12–67. continued
s2
s
675 m
= (7.5t2 - 360t) 2
t
30 s
s - 675 = (7.5t2 - 360t) - [7.5(302) - 360(30)]
s = {7.5t2 - 360t + 4725} m
When t = 60 s, s = 7.5(602) - 360(60) + 4725 = 10 125 m
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Using these results, the s–t graph shown in Fig. a can be plotted.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*12–68.
a (m/s2)
The a–s graph for a jeep traveling along a straight road is
given for the first 300 m of its motion. Construct the v –s
graph. At s = 0, v = 0.
2
SOLUTION
200
a s Graph: The function of acceleration a in terms of s for the interval
0 m … s 6 200 m is
a - 0
2 - 0
=
s - 0
200 - 0
a = (0.01s) m>s2
For the interval 200 m 6 s … 300 m,
a - 2
0 - 2
=
s - 200
300 - 200
a = (-0.02s + 6) m>s2
laws
or
v s Graph: The function of velocity v in terms of s can be obtained by applying
vdv = ads. For the interval 0 m ◊ s<200 m,
in
teaching
Web)
vdv = ads
s
0.01sds
permitted.
Dissemination
L0
instructors
States
World
L0
vdv =
Wide
copyright
v
use
United
on
learning.
is
of
the
not
v = (0.1s) m>s
At s = 200 m,
for
work
student
the
by
and
v = 0.100(200) = 20.0 m>s
is
work
solely
of
protected
the
(including
For the interval 200 m 6 s … 300 m,
work
this
assessing
vdv = ads
integrity
of
provided
( - 0.02s + 6)ds
part
the
is
and
any
L200 m
courses
L20.0 m>s
and
s
vdv =
This
v
destroy
of
A 2 -0.02s2 + 12s - 1200 B m>s
sale
will
their
v =
At s = 300 m,
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
v = 2 -0.02(3002) + 12(300) - 1200 = 24.5 m>s
300
s (m)
12–69.
The v–s graph for the car is given for the first 500 ft of its
motion. Construct the a–s graph for 0 … s … 500 ft. How long
does it take to travel the 500-ft distance? The car starts at s = 0
when t = 0.
v (ft/s)
60
v = 0.1s + 10
SOLUTION
a – s Graph: The acceleration a in terms of s can be obtained by applying vdv = ads.
a = v
10
dv
= (0.1s + 10)(0.1) = (0.01s + 1) ft>s2
ds
500
At s = 0 and s = 500 ft, a = 0.01(0) + 1 = 1.00 ft>s2 and a = 0.01(500) + 1 =
6.00 ft>s2, respectively.
Position: The position s in terms of time t can be obtained by applying v =
ds
.
dt
ds
v
dt =
laws
or
a s = 0 = 100 ft >s2
Wide
copyright
in
teaching
Web)
a s = 500ft = 6.00 ft >s2
World
permitted.
Dissemination
ds
L0 0.1s + 10
is
of
the
not
instructors
L0
s
dt =
States
t
the
by
and
use
United
on
learning.
t = 10ln (0.01s + 1)
work
student
t = 10ln [0.01(500) + 1] = 17.9 s
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
When s = 500 ft,
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
s (ft)
12–70.
The boat travels along a straight line with the speed
described by the graph. Construct the s–t and a -s graphs.
Also, determine the time required for the boat to travel a
distance s = 400 m if s = 0 when t = 0.
v (m/s)
80
SOLUTION
s t Graph: For 0 … s 6 100 m, the initial condition is s = 0 when t = 0 s.
+ B
A:
v
ds
dt =
v
t
v2
20
s (m)
100
s = A t2 B m
When s = 100 m,
100 = t2
t = 10 s
For 100 m 6 s … 400 m, the initial condition is s = 100 m when t = 10 s.
laws
or
ds
v
Web)
ds
teaching
s
dt =
in
dt =
t
L10 s
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
L100 m 0.2s
s
t - 10 = 5ln
100
s
t
- 2 = ln
5
100
s
et>5 - 2 =
100
s
et>5
=
100
e2
provided
integrity
of
and
work
this
s = A 13.53et>5 B m
and
any
courses
part
the
is
This
When s = 400 m,
sale
t = 16.93 s = 16.9 s
will
their
destroy
of
400 = 13.53et>5
The s–t graph is shown in Fig. a.
a s Graph: For 0 m … s 6 100 m,
a = v
dv
= A 2s1>2 B A s - 1>2 B = 2 m>s2
ds
For 100 m 6 s … 400 m,
a = v
dv
= (0.2s)(0.2) = 0.04s
ds
When s = 100 m and 400 m,
a s = 100 m = 0.04(100) = 4 m>s2
a s = 400 m = 0.04(400) = 16 m>s2
The a–s graph is shown in Fig. b.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
4s
s
ds
dt =
L0 2s1>2
L0
t = s1>2
+ B
A:
0.2s
Ans.
400
12–71.
The v -s graph of a cyclist traveling along a straight road is
shown. Construct the a -s graph.
v (ft/s)
15
v ⫽ ⫺0.04 s ⫹ 19
v ⫽ 0.1s ⫹ 5
5
SOLUTION
s (ft)
a–s Graph: For 0 … s 6 100 ft,
+ B
A:
a = v
100
dv
= A 0.1s + 5 B A 0.1 B = A 0.01s + 0.5 B ft>s2
ds
Thus at s = 0 and 100 ft
a ƒ s = 0 = 0.01 A 0 B + 0.5 = 0.5 ft>s2
teaching
Web)
laws
or
a ƒ s = 100 ft = 0.01 A 100 B + 0.5 = 1.5 ft>s2
permitted.
Dissemination
World
A - 0.04s + 19 B A - 0.04 B = A 0.0016s - 0.76 B ft>s2
is
of
the
not
dv
=
ds
instructors
a = v
States
+ B
A:
Wide
copyright
in
For 100 ft 6 s … 350 ft,
by
and
use
United
on
learning.
Thus at s = 100 ft and 350 ft
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
a ƒ s = 350 ft = 0.0016 A 350 B - 0.76 = - 0.2 ft>s2
work
solely
of
protected
the
(including
for
work
student
the
a ƒ s = 100 ft = 0.0016 A 100 B - 0.76 = - 0.6 ft>s2
sale
will
their
destroy
of
and
any
courses
The a -s graph is shown in Fig. a.
Thus at s = 0 and 100 ft
a ƒ s = 0 = 0.01 A 0 B + 0.5 = 0.5 ft>s2
a ƒ s = 100 ft = 0.01 A 100 B + 0.5 = 1.5 ft>s2
At s = 100 ft, a changes from amax = 1.5 ft>s2 to a min = - 0.6 ft>s2.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
350
*■ 12–72.
a (ft/s2)
The a–s graph for a boat moving along a straight path is
given. If the boat starts at s = 0 when v = 0, determine its
speed when it is at s = 75 ft, and 125 ft, respectively. Use
Simpson’s rule with n = 100 to evaluate v at s = 125 ft.
a
5
6( s
10)5/3
5
SOLUTION
Velocity: The velocity v in terms of s can be obtained by applying vdv = ads. For
the interval 0 ft … s 6 100 ft,
100
vdv = ads
v
L0
s
vdv =
5ds
L0
v = 210s = ft>s
At s = 75 ft, v = 210(75) = 27.4 ft>s
Ans.
At s = 100 ft,
Ans.
or
v = 210(100) = 31.62 ft>s
teaching
Web)
laws
For the interval 100 ft 6 s … 125 ft,
Wide
copyright
in
vdv = ads
v
States
World
permitted.
35 + 61 2s - 1025>34ds
the
not
instructors
use
United
on
learning.
is
L100 ft
of
L31.62 ft>s
Dissemination
125 ft
vdv =
student
the
by
and
Evaluating the integral on the right using Simpson’s rule, we have
assessing
is
work
solely
of
protected
the
(including
for
work
v2 v
`
= 201.032
2 31.62 ft/s
Ans.
provided
integrity
of
and
work
this
At s = 125 ft,
sale
will
their
destroy
of
and
any
courses
part
the
is
This
v = 37.4 ft>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
s (ft)
12–73.
The
position
of
a
particle
is
defined
by
r = 55 cos 2t i + 4 sin 2t j6 m, where t is in seconds and the
arguments for the sine and cosine are given in radians.
Determine the magnitudes of the velocity and acceleration
of the particle when t = 1 s. Also, prove that the path of the
particle is elliptical.
SOLUTION
Velocity: The velocity expressed in Cartesian vector form can be obtained by
applying Eq. 12–7.
v =
dr
= {-10 sin 2ti + 8 cos 2tj} m>s
dt
When t = 1 s, v = - 10 sin 2(1)i + 8 cos 2(1)j = { -9.093i - 3.329j} m>s. Thus, the
magnitude of the velocity is
v = 2v2x + v2y = 2(- 9.093)2 + ( -3.329)2 = 9.68 m>s
Ans.
laws
or
Acceleration: The acceleration expressed in Cartesian vector from can be obtained
by applying Eq. 12–9.
in
teaching
Web)
dv
= { -20 cos 2ti - 16 sin 2tj} m>s2
dt
Wide
copyright
a =
on
Ans.
the
by
and
use
United
a = 2a2x + a2y = 28.3232 + ( -14.549)2 = 16.8 m>s2
learning.
is
of
the
not
instructors
States
World
the
(including
for
work
student
Traveling Path: Here, x = 5 cos 2t and y = 4 sin 2t. Then,
(1)
(2)
destroy
of
and
any
courses
part
the
is
This
y2
= sin2 2t
16
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
x2
= cos2 2t
25
sale
will
their
Adding Eqs (1) and (2) yields
y2
x2
+
= cos2 2t + sin2 2t
25
16
However, cos2 2t + sin2 2t = 1. Thus,
y2
x2
+
= 1
25
16
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
(Equation of an Ellipse) (Q.E.D.)
permitted.
Dissemination
When t = 1 s, a = -20 cos 2(1)i - 16 sin 2(1)j = {8.323i - 14.549j} m>s2. Thus, the
magnitude of the acceleration is
12–74.
The velocity of a particle is v = 53i + (6 - 2t)j6 m>s, where t
is in seconds. If r = 0 when t = 0, determine the
displacement of the particle during the time interval
t = 1 s to t = 3 s.
SOLUTION
Position: The position r of the particle can be determined by integrating the
kinematic equation dr = vdt using the initial condition r = 0 at t = 0 as the
integration limit. Thus,
dr = vdt
t
r
L0
dr =
L0
C 3i + (6 - 2t)j D dt
r = c 3ti + A 6t - t2 B j d m
When t = 1 s and 3 s,
laws
or
r t = 1 s = 3(1)i + C 6(1) - 12 D j = [3i + 5j] m>s
in
teaching
Web)
r t = 3 s = 3(3)i + C 6(3) - 32 D j = [9i + 9j] m>s
Dissemination
Wide
copyright
Thus, the displacement of the particle is
is
of
the
not
instructors
States
World
permitted.
¢r = r t = 3 s - r t = 1 s
by
and
use
United
on
learning.
= (9i + 9j) - (3i + 5j)
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
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this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
= {6i + 4j} m
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Ans.
12–75.
A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft),
is subjected to an acceleration of a = 56ti + 12t2k6 ft>s2.
Determine the particle’s position (x, y, z) at t = 1 s.
SOLUTION
Velocity: The velocity expressed in Cartesian vector form can be obtained by
applying Eq. 12–9.
dv = adt
v
L0
t
dv =
16ti + 12t2k2 dt
L0
v = {3t2i + 4t3k} ft/s
Position: The position expressed in Cartesian vector form can be obtained by
applying Eq. 12–7.
laws
or
dr = vdt
Web)
t
teaching
r
in
13t2i + 4t3k2 dt
Wide
L0
copyright
Lr1
dr =
not
instructors
States
World
permitted.
Dissemination
r - (3i + 2j + 5k) = t3i + t4k
and
use
United
on
learning.
is
of
the
r = {(t3 + 3) i + 2j + (t4 + 5)k} ft
(including
for
work
student
the
by
When t = 1 s, r = (13 + 3)i + 2j + (14 + 5)k = {4i + 2j + 6k} ft.
is
work
solely
of
protected
the
The coordinates of the particle are
Ans.
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This
provided
integrity
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this
assessing
(4 ft, 2 ft, 6 ft)
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*12–76.
The
velocity
of
a
particle
is
given
by
v = 516t2i + 4t3j + (5t + 2)k6 m>s, where t is in seconds. If
the particle is at the origin when t = 0, determine the
magnitude of the particle’s acceleration when t = 2 s. Also,
what is the x, y, z coordinate position of the particle at this
instant?
SOLUTION
Acceleration: The acceleration expressed in Cartesian vector form can be obtained
by applying Eq. 12–9.
a =
dv
= {32ti + 12t2j + 5k} m>s2
dt
When t = 2 s, a = 32(2)i + 12 A 22 B j + 5k = {64i + 48j + 5k} m>s2. The magnitude
of the acceleration is
a = 2a2x + a2y + a2z = 2642 + 482 + 52 = 80.2 m>s2
Ans.
or
Position: The position expressed in Cartesian vector form can be obtained by
applying Eq. 12–7.
teaching
Web)
laws
dr = v dt
in
2
3
A 16t i + 4t j + (5t + 2)k B dt
Wide
L0
permitted.
dr =
Dissemination
L0
copyright
t
r
the
not
instructors
States
World
5
16 3
t i + t4j + a t2 + 2tb k d m
3
2
and
use
United
on
learning.
is
of
r = c
(including
for
work
student
the
by
When t = 2 s,
the
16 3
5
A 2 B i + A 24 B j + c A 22 B + 2(2) d k = {42.7i + 16.0j + 14.0k} m.
3
2
work
this
assessing
is
work
solely
of
protected
r =
part
the
is
This
provided
integrity
of
and
Thus, the coordinate of the particle is
Ans.
sale
will
their
destroy
of
and
any
courses
(42.7, 16.0, 14.0) m
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–77.
The car travels from A to B, and then from B to C, as shown
in the figure. Determine the magnitude of the displacement of
the car and the distance traveled.
2 km
A
B
3 km
SOLUTION
Displacement:
¢r = {2i - 3j} km
¢r = 222 + 32 = 3.61 km
Ans.
Distance Traveled:
C
d = 2 + 3 = 5 km
sale
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destroy
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and
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is
This
provided
integrity
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and
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this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–78.
A car travels east 2 km for 5 minutes, then north 3 km for
8 minutes, and then west 4 km for 10 minutes. Determine
the total distance traveled and the magnitude of
displacement of the car. Also, what is the magnitude of the
average velocity and the average speed?
SOLUTION
Total Distance Traveled and Displacement: The total distance traveled is
s = 2 + 3 + 4 = 9 km
Ans.
and the magnitude of the displacement is
¢r = 2(2 - 4)2 + 32 = 3.606 km = 3.61 km
Ans.
Average Velocity and Speed: The total time is ¢t = 5 + 8 + 10 = 23 min = 1380 s
The magnitude of average velocity is
Ans.
laws
or
vavg
3.606 A 103 B
¢r
=
= 2.61 m>s
=
¢t
1380
teaching
in
9 A 103 B
s
=
= 6.52 m>s
¢t
1380
Wide
copyright
permitted.
Dissemination
Ans.
sale
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destroy
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and
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courses
part
the
is
This
provided
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and
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this
assessing
is
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protected
the
(including
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work
student
the
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is
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A vsp B avg =
Web)
and the average speed is
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or transmission in any form or by any means, electronic, mechanical,
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12–79.
A car traveling along the straight portions of the road has
the velocities indicated in the figure when it arrives at
points A, B, and C. If it takes 3 s to go from A to B, and then
5 s to go from B to C, determine the average acceleration
between points A and B and between points A and C.
y
vC
x
vB
B
SOLUTION
vA = 20 i
vA
vB = 21.21 i + 21.21j
20 m/s
A
vC = 40i
aAB =
21.21 i + 21.21 j - 20 i
¢v
=
¢t
3
aAB = { 0.404 i + 7.07 j } m>s2
40 i - 20 i
¢v
=
¢t
8
or
aAC =
Ans.
laws
aAC = { 2.50 i } m>s2
sale
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is
This
provided
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assessing
is
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solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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45
30 m/s
C
40 m/s
*12–80.
A particle travels along the curve from A to B in 2 s. It takes
4 s for it to go from B to C and then 3 s to go from C to D.
Determine its average speed when it goes from A to D.
y
D
B
5m
15 m
C
SOLUTION
10 m
1
1
sT = (2p)(10)) + 15 + (2p(5)) = 38.56
4
4
sT
38.56
= 4.28 m>s
=
tt
2 + 4 + 3
Ans.
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destroy
of
and
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courses
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the
is
This
provided
integrity
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this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
vsP =
x
A
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or transmission in any form or by any means, electronic, mechanical,
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12–81.
The position of a crate sliding down a ramp is given by
x = (0.25t3) m, y = (1.5t2) m, z = (6 - 0.75t5>2) m, where t
is in seconds. Determine the magnitude of the crate’s
velocity and acceleration when t = 2 s.
SOLUTION
Velocity: By taking the time derivative of x, y, and z, we obtain the x, y, and z
components of the crate’s velocity.
d
#
vx = x =
A 0.25t3 B = A 0.75t2 B m>s
dt
d
#
vy = y =
A 1.5t2 B = A 3t B m>s
dt
A - 1.875t3>2 B m>s
or
d
#
vz = z =
A 6 - 0.75t5>2 B =
dt
teaching
Web)
vz = - 1.875 A 2 B 3>2 = - 5.303 m/s
in
vy = 3(2) = 6 m>s
not
instructors
States
World
permitted.
Dissemination
Thus, the magnitude of the crate’s velocity is
Wide
copyright
vx = 0.75 A 22 B = 3 m>s
laws
When t = 2 s,
Ans.
the
by
and
use
United
on
learning.
is
of
the
v = 2vx 2 + vy 2 + vz 2 = 232 + 62 + ( - 5.303)2 = 8.551 ft>s = 8.55 ft
of
protected
the
(including
for
work
student
Acceleration: The x, y, and z components of the crate’s acceleration can be obtained
by taking the time derivative of the results of vx, vy, and vz, respectively.
part
the
is
any
courses
and
A - 2.815t1>2 B m>s2
will
destroy
of
their
d
#
az = vz =
A - 1.875t3>2 B =
dt
sale
d
#
ay = vy =
(3t) = 3 m>s2
dt
This
provided
integrity
of
and
work
this
assessing
is
work
solely
d
#
ax = vx =
A 0.75t2 B = (1.5t) m>s2
dt
When t = 2 s,
ax = 1.5(2) = 3 m>s2
ay = 3 m>s2
az = - 2.8125 A 21>2 B = - 3.977 m>s2
Thus, the magnitude of the crate’s acceleration is
a = 2ax 2 + ay 2 + az 2 = 232 + 32 + ( - 3.977)2 = 5.815 m>s2 = 5.82 m>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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Ans.
12–82.
A rocket is fired from rest at x = 0 and travels along a
parabolic trajectory described by y2 = [120(103)x] m. If the
1
x component of acceleration is ax = a t2 b m>s2, where t is
4
in seconds, determine the magnitude of the rocket’s velocity
and acceleration when t = 10 s.
SOLUTION
Position: The parameter equation of x can be determined by integrating ax twice
with respect to t.
L
dvx =
L
vx
t
dvx =
L0
axdt
1 3
t b m>s
12
or
vx = a
1 2
t dt
L0 4
L
vxdt
teaching
Web)
dx =
laws
L
in
1 3
t dt
L0 12
Wide
Dissemination
L0
t
dx =
copyright
x
the
not
instructors
States
World
permitted.
1 4
t bm
48
use
United
on
learning.
is
of
x = a
work
the
(including
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solely
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protected
this
assessing
is
integrity
of
and
provided
y = A 50t2 B m
1 4
t b
48
work
y2 = 120 A 103 B a
student
the
by
and
Substituting the result of x into the equation of the path,
part
the
is
This
Velocity:
sale
will
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destroy
of
and
any
courses
d
#
vy = y =
A 50t2 B = A 100t B m>s
dt
When t = 10 s,
vx =
1
A 103 B = 83.33 m>s
12
vy = 100(10) = 1000 m>s
Thus, the magnitude of the rocket’s velocity is
v = 2vx 2 + vy2 = 283.332 + 10002 = 1003 m>s
Ans.
Acceleration:
#
d
ay = vy =
(100t) = 100 m>s2
dt
When t = 10 s,
ax =
1
A 102 B = 25 m>s2
4
Thus, the magnitude of the rocket’s acceleration is
a = 2ax 2 + ay2 = 2252 + 1002 = 103 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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Ans.
12–83.
y
The particle travels along the path defined by the parabola
y = 0.5x2. If the component of velocity along the x axis is
vx = 15t2 ft>s, where t is in seconds, determine the particle’s
distance from the origin O and the magnitude of its
acceleration when t = 1 s. When t = 0, x = 0, y = 0.
y
0.5x2
SOLUTION
dx
.
dt
Position: The x position of the particle can be obtained by applying the vx =
O
dx = vx dt
x
L0
t
dx =
L0
5tdt
x = A 2.50t2 B ft
Thus, y = 0.5 A 2.50t2 B 2 = A 3.125t4 B ft. At
t = 1 s, x = 2.5 A 12 B = 2.50 ft
and
y = 3.125 A 14 B = 3.125 ft. The particle’s distance from the origin at this moment is
d = 2(2.50 - 0)2 + (3.125 - 0)2 = 4.00 ft
laws
or
Ans.
copyright
in
teaching
Web)
#
#
Acceleration: Taking the first derivative of the path y = 0.5x2, we have y = xx.
The second derivative of the path gives
Wide
$
#
$
y = x2 + xx
States
World
permitted.
Dissemination
(1)
use
United
on
learning.
is
of
the
not
instructors
#
$
$
However, x = vx, x = ax and y = ay. Thus, Eq. (1) becomes
and
ay = v2x + xax
work
student
the
by
(2)
work
this
assessing
is
work
solely
of
protected
the
(including
for
dvx
= 5 ft>s 2, and x = 2.50 ft . Then, from
When t = 1 s, vx = 5(1) = 5 ft>s ax =
dt
Eq. (2)
part
the
is
This
provided
integrity
of
and
ay = 52 + 2.50(5) = 37.5 ft>s2
destroy
of
and
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courses
Also,
52 + 37.52 = 37.8 ft s2
will
their
a2x + a2y =
sale
a =
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Ans.
x
*12–84.
The motorcycle travels with constant speed v0 along the
path that, for a short distance, takes the form of a sine curve.
Determine the x and y components of its velocity at any
instant on the curve.
y
v0
y
π x)
c sin ( ––
L
x
c
L
SOLUTION
y = c sin a
p
xb
L
p
p
#
#
ca cos xb x
y =
L
L
vy =
p
p
c vx a cos x b
L
L
v20 = v2y + v2x
p 2
p
cb cos2 a xb R
L
L
vx = v0 B 1 + a
-2
p 2
p
cb cos2 a xb R
L
L
laws
or
v20 = v2x B 1 + a
1
Wide
copyright
in
teaching
Web)
Ans.
1
Dissemination
-2
v0 pc
p
p 2
p
vy =
acos xb B 1 + a cb cos2 a x b R
L
L
L
L
World
the
not
instructors
States
on
learning.
is
of
United
and
use
by
work
student
the
the
(including
for
work
solely
of
protected
this
assessing
is
integrity
of
and
work
provided
any
courses
part
the
is
This
destroy
of
and
will
their
sale
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permitted.
Ans.
c
L
12–85.
A particle travels along the curve from A to B in 1 s. If it
takes 3 s for it to go from A to C, determine its average
velocity when it goes from B to C.
y
B
20 m
SOLUTION
(rAC - rAB)
40i - (20i + 20j)
¢r
=
=
= {10i - 10j} m>s
¢t
¢t
2
Ans.
sale
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and
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is
This
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assessing
is
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the
(including
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work
student
the
by
and
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United
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is
of
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Dissemination
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Web)
laws
or
vang =
C
A
Time from B to C is 3 - 1 = 2 s
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x
12–86.
y
When a rocket reaches an altitude of 40 m it begins to travel
along the parabolic path 1y - 4022 = 160x, where the
coordinates are measured in meters. If the component of
velocity in the vertical direction is constant at vy = 180 m>s,
determine the magnitudes of the rocket’s velocity and
acceleration when it reaches an altitude of 80 m.
(y
40)2
160x
SOLUTION
vy = 180 m>s
40 m
(y - 40)2 = 160 x
2(y - 40)vy = 160vx
x
(1)
2(80 - 40)(180) = 160vx
vx = 90 m>s
v = 2902 + 1802 = 201 m>s
dt
= 0
or
d vy
laws
ay =
Ans.
in
teaching
Web)
From Eq. 1,
Dissemination
Wide
copyright
2 v2y + 2(y - 40)ay = 160 ax
is
of
the
not
instructors
States
World
permitted.
2(180)2 + 0 = 160 ax
by
and
use
United
on
learning.
ax = 405 m>s2
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is
This
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assessing
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the
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work
student
the
a = 405 m>s2
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Ans.
12–87.
Pegs A and B are restricted to move in the elliptical slots
due to the motion of the slotted link. If the link moves with
a constant speed of 10 m/s, determine the magnitude of the
velocity and acceleration of peg A when x = 1 m.
y
A
v
Velocity: The x and y components of the peg’s velocity can be related by taking the
first time derivative of the path’s equation.
B
x2
4
x2
+ y2 = 1
4
1
#
#
(2xx) + 2yy = 0
4
1 #
#
xx + 2yy = 0
2
or
1
xv + 2yvy = 0
2 x
(1)
laws
or
At x = 1 m,
teaching
Web)
23
m
2
Dissemination
Wide
copyright
y =
in
(1)2
+ y2 = 1
4
not
instructors
States
World
permitted.
Here, vx = 10 m>s and x = 1. Substituting these values into Eq. (1),
is
of
the
1
23
(1)(10) + 2 ¢
≤ vy = 0
2
2
for
work
student
the
by
and
use
United
on
learning.
vy = -2.887 m>s = 2.887 m>s T
of
protected
the
(including
Thus, the magnitude of the peg’s velocity is
Ans.
provided
integrity
of
and
work
this
assessing
is
work
solely
v = 2vx 2 + vy 2 = 2102 + 2.8872 = 10.4 m>s
and
any
courses
part
the
is
This
Acceleration: The x and y components of the peg’s acceleration can be related by
taking the second time derivative of the path’s equation.
sale
will
their
destroy
of
1 # #
##
##
# #
(xx + xx) + 2(yy + yy) = 0
2
1 #2
##
##
#
A x + xx B + 2 A y 2 + yy B = 0
2
or
1
A v 2 + xax B + 2 A vy 2 + yay B = 0
2 x
Since vx is constant, ax = 0. When x = 1 m, y =
(2)
23
m, vx = 10 m>s, and
2
vy = -2.887 m>s. Substituting these values into Eq. (2),
23
1
a d = 0
A 102 + 0 B + 2 c ( - 2.887)2 +
2
2 y
ay = - 38.49 m>s2 = 38.49 m>s2 T
Thus, the magnitude of the peg’s acceleration is
a = 2ax 2 + ay 2 = 202 + ( -38.49)2 = 38.5 m>s2
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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D
C
SOLUTION
Ans.
y2
1
x
10 m/s
*12–88.
The van travels over the hill described by
y = (- 1.5(10–3) x2 + 15) ft. If it has a constant speed of
75 ft>s, determine the x and y components of the van’s
velocity and acceleration when x = 50 ft.
y
15 ft
y
( 1.5 (10 3) x2
x
SOLUTION
100 ft
Velocity: The x and y components of the van’s velocity can be related by taking the
first time derivative of the path’s equation using the chain rule.
y = -1.5 A 10 - 3 B x2 + 15
#
#
y = -3 A 10 - 3 B xx
or
vy = -3 A 10 - 3 B xvx
When x = 50 ft,
vy = - 3 A 10 - 3 B (50)vx = - 0.15vx
(1)
The magnitude of the van’s velocity is
v = 2vx 2 + vy 2
laws
or
(2)
in
teaching
Web)
Substituting v = 75 ft>s and Eq. (1) into Eq. (2),
permitted.
Dissemination
Wide
copyright
75 = 2vx 2 + ( - 0.15vx)2
on
learning.
is
of
the
instructors
States
Ans.
not
World
vx = 74.2 ft>s ;
the
by
and
use
United
Substituting the result of nx into Eq. (1), we obtain
Ans.
of
protected
the
(including
for
work
student
vy = -0.15(- 74.17) = 11.12 ft>s = 11.1 ft>s c
work
this
assessing
is
work
solely
Acceleration: The x and y components of the van’s acceleration can be related by
taking the second time derivative of the path’s equation using the chain rule.
and
any
courses
part
the
is
This
provided
integrity
of
and
$
# #
##
y = - 3 A 10 - 3 B (xx + xx)
destroy
of
or
sale
will
their
ay = - 3 A 10 - 3 B A vx 2 + xax B
When x = 50 ft, vx = - 74.17 ft>s. Thus,
ay = -3 A 10 - 3 B c ( - 74.17)2 + 50ax d
ay = -(16.504 + 0.15ax)
(3)
Since the van travels with a constant speed along the path,its acceleration along the tangent
of the path is equal to zero. Here, the angle that the tangent makes with the horizontal at
dy
x = 50 ft is u = tan - 1 ¢ ≤ 2
= tan - 1 c -3 A 10 - 3 B x d 2
= tan - 1(- 0.15) = -8.531°.
dx x = 50 ft
x = 50 ft
Thus, from the diagram shown in Fig. a,
ax cos 8.531° - ay sin 8.531° = 0
(4)
Solving Eqs. (3) and (4) yields
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or transmission in any form or by any means, electronic, mechanical,
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ax = -2.42 ft>s = 2.42 ft>s2 ;
Ans.
ay = -16.1 ft>s = 16.1 ft>s2 T
Ans.
15) ft
12–89.
It is observed that the time for the ball to strike the ground
at B is 2.5 s. Determine the speed vA and angle uA at which
the ball was thrown.
vA
uA
A
1.2 m
50 m
SOLUTION
Coordinate System: The x–y coordinate system will be set so that its origin coincides
with point A.
x-Motion: Here, (vA)x = vA cos uA, xA = 0, xB = 50 m, and t = 2.5 s. Thus,
+ B
A:
xB = xA + (vA)xt
50 = 0 + vA cos uA(2.5)
vA cos uA = 20
(1)
y-Motion: Here, (vA)y = vA sin uA,
= - 9.81 m>s2. Thus,
yB = -1.2 m,
ay = -g
and
or
yA = 0 ,
Web)
laws
1
a t2
2 y
teaching
yB = yA + (vA)y t +
in
A+cB
(2)
learning.
is
of
the
not
instructors
States
World
vA sin uA = 11.7825
by
and
use
United
on
Solving Eqs. (1) and (2) yields
work
student
the
vA = 23.2 m>s
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
uA = 30.5°
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Ans.
permitted.
Dissemination
Wide
copyright
1
( -9.81) A 2.52 B
2
- 1.2 = 0 + vA sin uA (2.5) +
B
12–90.
Determine the minimum initial velocity v0 and the
corresponding angle u0 at which the ball must be kicked in
order for it to just cross over the 3-m high fence.
v0
3m
u0
6m
SOLUTION
Coordinate System: The x - y coordinate system will be set so that its origin
coincides with the ball’s initial position.
x-Motion: Here, (v0)x = v0 cos u, x0 = 0, and x = 6 m. Thus,
+ B
A:
x = x0 + (v0)xt
6 = 0 + (v0 cos u)t
t =
6
v0 cos u
(1)
Web)
laws
teaching
Wide
copyright
World
permitted.
Dissemination
1
(-9.81)t2
2
not
of
the
3 = 0 + v0 (sin u) t +
in
1
a t2
2 y
instructors
y = y0 + (v0)y t +
States
A+cB
or
y-Motion: Here, (v0)x = v0 sin u, ay = - g = - 9.81 m > s2, and y0 = 0. Thus,
is
3 = v0 (sin u) t - 4.905t2
the
by
and
use
United
on
learning.
(2)
the
(including
for
work
student
Substituting Eq. (1) into Eq. (2) yields
work
solely
of
protected
58.86
B sin 2u - cos2 u
(3)
integrity
of
and
work
this
assessing
is
v0 =
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
From Eq. (3), we notice that v0 is minimum when f(u) = sin 2u - cos2 u is
df(u)
= 0
maximum. This requires
du
df(u)
= 2 cos 2u + sin 2u = 0
du
tan 2u = - 2
2u = 116.57°
u = 58.28° = 58.3°
Ans.
Substituting the result of u into Eq. (2), we have
(v0)min =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
58.86
= 9.76 m>s
B sin 116.57° - cos2 58.28°
Ans.
12–91.
During a race the dirt bike was observed to leap up off the
small hill at A at an angle of 60° with the horizontal. If the
point of landing is 20 ft away, determine the approximate
speed at which the bike was traveling just before it left the
ground. Neglect the size of the bike for the calculation.
A
20 ft
SOLUTION
+ )s = s + v t
(:
0
0
20 = 0 + vA cos 60° t
( + c ) s = s0 + v0 +
1
a t2
2 c
0 = 0 + vA sin 60° t +
1
( - 32.2) t2
2
Solving
or
t = 1.4668 s
laws
vA = 27.3 ft s
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
Ans.
sale
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
60°
*12–92.
The girl always throws the toys at an angle of 30° from
point A as shown.Determine the time between throws so
that both toys strike the edges of the pool B and C at the
same instant.With what speed must she throw each toy?
A
30°
1m
B
2.5 m
SOLUTION
4m
To strike B:
+ )s = s + v t
(:
0
0
2.5 = 0 + vA cos 30° t
(+ c ) s = s0 + v0 t +
1 2
a t
2 c
0.25 = 1 + vA siv 30° t -
1
(9.81)t2
2
laws
or
Solving
teaching
Web)
t = 0.6687 s
in
(vA)B = 4.32 m>s
Dissemination
Wide
copyright
Ans.
not
instructors
States
World
permitted.
To strike C:
and
use
United
on
learning.
is
of
the
+ )s = s + v t
(:
0
0
for
work
student
the
by
4 = 0 + vA cos 30° t
work
solely
of
protected
the
(including
1 2
a t
2 c
integrity
of
and
provided
part
the
is
and
any
courses
1
(9.81)t2
2
This
0.25 = 1 + vA siv 30° t -
work
this
assessing
is
(+ c ) s = s0 + v0 t +
(vA)C = 5.85 m>s
sale
t = 0.790 s
will
their
destroy
of
Solving
Ans.
Time between throws:
¢t = 0.790 s - 0.6687 s = 0.121 s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
C
0.25 m
12–93.
The player kicks a football with an initial speed of
v0 = 90 ft>s. Determine the time the ball is in the air and
the angle u of the kick.
v0
u
A
126 ft
SOLUTION
Coordinate System: The x – y coordinate system will be set with its origin coinciding
with starting point of the football.
x-motion: Here, x0 = 0, x = 126 ft, and (v0)x = 90 cos u
+ B
A:
x = x0 + (v0)x t
126 = 0 + (90 cos u) t
t =
126
90 cos u
(1)
laws
teaching
Web)
1
a t2
2 y
in
y = y0 + (v0)y t +
Wide
copyright
A+cB
or
y-motion: Here, y0 = y = 0, (v0)y = 90 sin u, and ay = - g = - 32.2 ft. Thus,
World
permitted.
Dissemination
1
( -32.2)t2
2
not
instructors
States
O = 0 + (90 sin u)t +
of
the
O = (90 sin u)t - 16.1t2
and
use
United
on
learning.
is
(2)
for
work
student
the
by
Substitute Eq. (1) into (2) yields
work
solely
of
protected
the
(including
2
126
126
≤ - 16.1 ¢
≤
90 cos u
90 cos u
integrity
of
and
provided
the
part
any
courses
(3)
destroy
of
O = 126 sin u cos u - 31.556
is
This
126 sin u
31.556
cos u
cos2 u
and
O =
work
this
assessing
is
O = 90 sin u ¢
sale
will
their
Using the trigonometry identity sin 2u = 2 sin u cos u, Eq. (3) becomes
63 sin 2u = 31.556
sin 2u = 0.5009
2u = 30.06 or 149.94
u = 15.03° = 15.0° or u = 74.97° = 75.0°
Ans.
If u = 15.03°,
t =
126
= 1.45 s
90 cos 15.03°
Ans.
126
= 5.40 s
90 cos 74.97°
Ans.
If u = 74.97°,
t =
Thus,
u = 15.0°, t = 1.45 s
u = 75.0°, t = 5.40 s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–94.
From a videotape, it was observed that a pro football player
kicked a football 126 ft during a measured time of
3.6 seconds. Determine the initial speed of the ball and the
angle u at which it was kicked.
v0
θ
A
126 ft
SOLUTION
+ B
A:
s = s0 + v0 t
126 = 0 + (v0)x (3.6)
(v0)x = 35 ft>s
A+cB
s = s0 + v0 t +
1 2
a t
2 c
O = 0 + (v0)y (3.6) +
1
(- 32.2)(3.6)2
2
or
(v0)y = 57.96 ft>s
v0 = 2(35)2 + (57.96)2 = 67.7 ft>s
Ans.
Wide
= 58.9°
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
57.96
35
copyright
u = tan - 1
in
teaching
Web)
laws
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–95.
A projectile is given a velocity v0 at an angle f above the
horizontal. Determine the distance d to where it strikes the
sloped ground. The acceleration due to gravity is g.
y
v0
φ
SOLUTION
+ B
A:
d cos u = 0 + v0 (cos f) t
s = s0 + v0 t +
1 2
a t
2 c
d sin u = 0 + v0 (sin f)t +
1
( - g) t2
2
Thus,
or
d cos u
1
d cos u 2
b - ga
b
v0 cos f
2 v0 cos f
teaching
Web)
laws
d sin u = v0 sin fa
in
gd cos2 u
2v20 cos2 f
Dissemination
Wide
copyright
sin u = cos u tan f -
not
instructors
States
World
permitted.
2v20 cos2 f
the
g cos2 u
use
United
on
learning.
is
of
d = (cos u tan f - sin u)
by
and
v 20
sin 2f - 2 tan u cos2 f
g cos u
work
student
the
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
d =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
x
d
s = s0 + v0 t
A+cB
θ
*12–96.
A projectile is given a velocity v0 . Determine the angle f at
which it should be launched so that d is a maximum. The
acceleration due to gravity is g.
y
v0
φ
SOLUTION
+ B
A:
d cos u = 0 + v0 (cos f) t
sy = s0 + v0 t +
1 2
at
2 c
d sin u = 0 + v0 (sin f)t +
1
( - g)t2
2
Thus,
or
d cos u
1
d cos u 2
b - ga
b
v0 cos f
2 v0 cos f
teaching
Web)
laws
d sin u = v0 sin f a
in
gd cos2 u
2v 20 cos2 f
Dissemination
Wide
copyright
sin u = cos u tan f -
not
instructors
States
World
permitted.
2v 20 cos2 f
the
g cos2 u
use
United
on
learning.
is
of
d = (cos u tan f - sin u)
of
protected
the
(including
for
work
student
the
by
and
v 20
A sin 2f - 2 tan u cos2 f B
g cos u
d =
any
sale
sin 2f
tan u + 1 = 0
cos 2f
will
their
destroy
of
and
cos 2f + tan u sin 2f = 0
courses
part
the
is
This
provided
integrity
of
and
work
d(d)
v 20
=
C cos 2f(2) - 2 tan u(2 cos f)(- sin f) D = 0
df
g cos u
this
assessing
is
work
solely
Require:
tan 2f = - ctn u
1
tan - 1 (- ctn u)
2
f =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
x
d
sx = s0 + v0 t
A+cB
θ
Ans.
12–97.
Determine the maximum height on the wall to which the
firefighter can project water from the hose, if the speed of
the water at the nozzle is vC = 48 ft>s.
A
vC = 48 ft/s
SOLUTION
h
θ
C 3 ft
(+ c ) v = v0 + ac t
B
30 ft
0 = 48 sin u - 32.2 t
+ )s = s + v t
(:
0
0
30 = 0 + 48 (cos u)(t)
48 sin u = 32.2
30
48 cos u
sin u cos u = 0.41927
laws
or
sin 2u = 0.83854
in
teaching
Web)
u = 28.5°
Dissemination
Wide
copyright
t = 0.7111 s
the
not
instructors
States
World
permitted.
1
a t2
2 c
use
United
on
learning.
is
of
(+ c ) s = s0 + v0 t +
work
student
the
by
and
1
( - 32.2)(0.7111)2
2
the
(including
for
h - 3 = 0 + 48 sin 28.5° (0.7111) +
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
h = 11.1 ft
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Ans.
■
12–98.
Determine the smallest angle u, measured above the
horizontal, that the hose should be directed so that the
water stream strikes the bottom of the wall at B. The speed
of the water at the nozzle is vC = 48 ft>s.
A
vC
48 ft/s
SOLUTION
+ )
(:
h
u
C 3 ft
s = s0 + v0 t
B
30 ft
30 = 0 + 48 cos u t
30
48 cos u
s = s0 + v0 t +
1
a t2
2 c
0 = 3 + 48 sin u t +
or
2
48 sin u (30)
30
- 16.1 a
b
48 cos u
48 cos u
Web)
in
0 = 3 +
1
( -32.2)t2
2
laws
(+ c)
teaching
t =
Dissemination
Wide
copyright
0 = 3 cos2 u + 30 sin u cos u - 6.2891
is
of
the
not
instructors
States
World
permitted.
3 cos2 u + 15 sin 2u = 6.2891
by
and
use
United
on
learning.
Solving
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
u = 6.41° or 77.9 °
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Ans.
12–99.
Measurements of a shot recorded on a videotape during a
basketball game are shown. The ball passed through the
hoop even though it barely cleared the hands of the player B
who attempted to block it. Neglecting the size of the ball,
determine the magnitude vA of its initial velocity and the
height h of the ball when it passes over player B.
C
30
vA
B
h
A
7 ft
25 ft
SOLUTION
+ )
(:
s = s0 + v0t
30 = 0 + vA cos 30° tAC
(+ c )
s = s0 + v0t +
1 2
at
2 c
10 = 7 + vA sin 30° tAC -
1
(32.2)(t2AC)
2
Solving
vA = 36.73 = 36.7 ft>s
or
Ans.
teaching
Web)
laws
tAC = 0.943 s
+ )
(:
Wide
copyright
in
s = s0 + v0t
not
instructors
States
World
permitted.
Dissemination
25 = 0 + 36.73 cos 30° tAB
United
on
learning.
is
of
the
1 2
act
2
and
use
s = s0 + v0t +
the
by
(+ c )
assessing
is
work
solely
of
protected
the
(including
for
work
student
1
(32.2)(t2AB)
2
h = 7 + 36.73 sin 30° tAB -
courses
part
the
is
This
any
Ans.
sale
will
their
destroy
of
h = 11.5 ft
and
tAB = 0.786 s
provided
integrity
of
and
work
this
Solving
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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5 ft
10 ft
*12–100.
It is observed that the skier leaves the ramp A at an angle
uA = 25° with the horizontal. If he strikes the ground at B,
determine his initial speed vA and the time of flight tAB.
vA
uA
A
4m
3
5
4
SOLUTION
+ B
A:
100 m
s = v0 t
4
100 a b = vA cos 25°tAB
5
A+cB
s = s0 + v0 t +
B
1
ac t2
2
1
3
-4 - 100 a b = 0 + vA sin 25°tAB + ( - 9.81)t2AB
5
2
Ans.
tAB = 4.54 s
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
vA = 19.4 m>s
or
Solving,
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–101.
It is observed that the skier leaves the ramp A at an angle
uA = 25° with the horizontal. If he strikes the ground at B,
determine his initial speed vA and the speed at which he
strikes the ground.
vA
uA
A
4m
3
5
4
100 m
SOLUTION
Coordinate System: x - y coordinate system will be set with its origin to coincide
with point A as shown in Fig. a.
B
4
x-motion: Here, xA = 0, xB = 100 a b = 80 m and (vA)x = vA cos 25°.
5
+ B
A:
xB = xA + (vA)xt
80 = 0 + (vA cos 25°)t
80
t =
vA cos 25°
or
(1)
the
not
instructors
States
1
a t2
2 y
on
learning.
is
of
yB = yA + (vA)y t +
use
United
A+ c B
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
3
y-motion: Here, yA = 0, yB = - [4 + 100 a b ] = - 64 m and (vA)y = vA sin 25°
5
and ay = - g = - 9.81 m>s2.
work
student
the
by
and
1
(- 9.81)t2
2
(including
for
- 64 = 0 + vA sin 25° t +
(2)
assessing
is
work
solely
of
protected
the
4.905t2 - vA sin 25° t = 64
provided
integrity
of
and
work
this
Substitute Eq. (1) into (2) yieldS
2
80
80
≤ = vA sin 25° ¢
≤ = 64
vA cos 25°
yA cos 25°
2
80
≤ = 20.65
vA cos 25°
sale
¢
will
their
destroy
of
and
any
courses
part
the
is
This
4.905 ¢
80
= 4.545
vA cos 25°
vA = 19.42 m>s = 19.4 m>s
Substitute this result into Eq. (1),
t =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
80
= 4.54465
19.42 cos 25°
Ans.
12–101. continued
Using this result,
A+ c B
(vB)y = (vA)y + ay t
= 19.42 sin 25° + ( -9.81)(4.5446)
= - 36.37 m>s = 36.37 m>s T
And
+ B
A:
(vB)x = (vA)x = vA cos 25° = 19.42 cos 25° = 17.60 m>s :
Thus,
vB = 2(vB)2x + (vB)2y
= 236.372 + 17.602
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
= 40.4 m>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–102.
A golf ball is struck with a velocity of 80 ft>s as shown.
Determine the distance d to where it will land.
vA
80 ft/s
B
A
45
10
d
SOLUTION
Horizontal Motion: The horizontal component of velocity is (v0)x = 80 cos 55°
= 45.89 ft>s.The initial and final horizontal positions are (s0)x = 0 and sx = d cos 10°,
respectively.
+ B
A:
sx = (s0)x + (v0)x t
d cos 10° = 0 + 45.89t
(1)
Vertical Motion: The vertical component of initial velocity is (v0)y = 80 sin 55°
= 65.53 ft>s. The initial and final vertical positions are (s0)y = 0 and sy = d sin 10°,
respectively.
1
(a ) t2
2 cy
1
d sin 10° = 0 + 65.53t + ( - 32.2)t2
2
laws
or
sy = (s0)y + (v0)y t +
in
teaching
Web)
(2)
copyright
(+ c )
permitted.
Dissemination
Wide
Solving Eqs. (1) and (2) yields
Ans.
of
the
not
instructors
States
World
d = 166 ft
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
t = 3.568 s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–103.
The ball is thrown from the tower with a velocity of 20 ft/s
as shown. Determine the x and y coordinates to where the
ball strikes the slope. Also, determine the speed at which
the ball hits the ground.
y
20 ft/s
5
4
3
SOLUTION
80 ft
1
Assume ball hits slope.
+ B
A:
2
s = s0 + v0 t
x = 0 +
3
(20)t = 12t
5
x
20 ft
A+cB
s = s0 + v0 t +
y = 80 +
1 2
a t
2 c
4
1
(20)t + ( -32.2)t2 = 80 + 16t - 16.1t2
5
2
teaching
Web)
laws
or
Equation of slope: y - y1 = m(x - x1)
in
1
(x - 20)
2
Dissemination
Wide
copyright
y - 0 =
learning.
is
of
the
not
instructors
States
World
permitted.
y = 0.5x - 10
by
and
use
United
on
Thus,
(including
for
work
student
the
80 + 16t - 16.1t2 = 0.5(12t) - 10
assessing
is
work
solely
of
protected
the
16.1t2 - 10t - 90 = 0
provided
integrity
of
and
work
this
Choosing the positive root:
and
any
courses
part
the
is
This
t = 2.6952 s
Ans.
Since 32.3 ft 7 20 ft, assumption is valid.
sale
will
their
destroy
of
x = 12(2.6952) = 32.3 ft
y = 80 + 16(2.6952) - 16.1(2.6952)2 = 6.17 ft
3
(20) = 12 ft>s
5
+ B
A:
vx = (v0)x =
A+cB
vy = (v0)y + act =
v =
4
(20) + (- 32.2)(2.6952) = -70.785 ft>s
5
(12)2 + (-70.785)2 = 71.8 ft s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
Ans.
*12–104.
The projectile is launched with a velocity v0. Determine the
range R, the maximum height h attained, and the time of
flight. Express the results in terms of the angle u and v0. The
acceleration due to gravity is g.
y
v0
u
h
x
R
SOLUTION
+ B
A:
s = s0 + v0 t
R = 0 + (v0 cos u)t
s = s0 + v0 t +
1
a t2
2 c
0 = 0 + (v0 sin u) t +
1
R
(g) ¢
≤
2
v0 cos u
laws
0 = v0 sin u -
1
( - g)t2
2
or
A+cB
teaching
Web)
v20
sin 2u
g
in
Ans.
Wide
copyright
R =
the
not
instructors
States
World
permitted.
Dissemination
v20 (2 sin u cos u)
R
=
v0 cos u
v0 g cos u
United
on
learning.
is
of
t =
and
use
2v0
sin u
g
by
Ans.
(including
work
solely
of
protected
the
v2 = v20 + 2ac(s - s0)
is
A+cB
for
work
student
the
=
part
the
is
sale
will
their
destroy
of
and
any
courses
v20
sin2 u
2g
This
h =
provided
integrity
of
and
work
this
assessing
0 = (v0 sin u)2 + 2(- g)(h - 0)
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
12–105.
Determine the horizontal velocity vA of a tennis ball at A so
that it just clears the net at B. Also, find the distance s where
the ball strikes the ground.
vA
B
s
21 ft
SOLUTION
Vertical Motion: The vertical component of initial velocity is (v0)y = 0. For the ball
to travel from A to B, the initial and final vertical positions are (s0)y = 7.5 ft and
sy = 3 ft, respectively.
A+cB
sy = (s0)y + (v0)y t +
3 = 7.5 + 0 +
1
(a ) t2
2 cy
1
( -32.2)t 21
2
t1 = 0.5287 s
laws
or
For the ball to travel from A to C, the initial and final vertical positions are
(s0)y = 7.5 ft and sy = 0, respectively.
Web)
teaching
1
( -32.2)t22
2
Wide
permitted.
Dissemination
0 = 7.5 + 0 +
1
(a ) t2
2 cy
in
sy = (s0)y + (v0)y t +
copyright
A+cB
on
learning.
is
of
the
not
instructors
States
World
t2 = 0.6825 s
protected
the
(including
for
work
student
the
by
and
use
United
Horizontal Motion: The horizontal component of velocity is (v0)x = vA. For the ball
to travel from A to B, the initial and final horizontal positions are (s0)x = 0 and
sx = 21 ft, respectively. The time is t = t1 = 0.5287 s.
work
solely
of
sx = (s0)x + (v0)x t
assessing
is
+ B
A;
provided
integrity
of
and
work
this
21 = 0 + vA (0.5287)
Ans.
and
any
courses
part
the
is
This
vA = 39.72 ft>s = 39.7 ft>s
sale
will
their
destroy
of
For the ball to travel from A to C, the initial and final horizontal positions are
(s0)x = 0 and sx = (21 + s) ft, respectively. The time is t = t2 = 0.6825 s.
+ B
A;
sx = (s0)x + (v0)x t
21 + s = 0 + 39.72(0.6825)
s = 6.11 ft
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
7.5 ft
3 ft
C
Ans.
A
12–106.
The ball at A is kicked with a speed vA = 8 ft > s and at an
angle uA = 30°. Determine the point (x, -y) where it strikes
the ground. Assume the ground has the shape of a parabola
as shown.
y
vA
A
θA
x
–y
SOLUTION
B
y = –0.04x 2
(vA)x = 8 cos 30° = 6.928 ft>s
x
(vA)y = 8 sin 30° = 4 ft > s
+ )s = s + v t
(:
0
0
x = 0 + 6.928 t
( + c ) s = s0 + v0 t +
1 2
a t
2 c
1
( -32.2)t2
2
(2)
or
y = 0 + 4t +
(1)
in
teaching
Web)
laws
y = - 0.04 x2
Wide
copyright
From Eqs. (1) and (2):
instructors
States
World
permitted.
Dissemination
y = 0.5774 x - 0.3354 x2
use
United
on
learning.
is
of
the
not
- 0.04 x2 = 0.5774x - 0.3354 x2
for
work
student
the
by
and
0.2954 x2 = 0.5774x
Ans.
is
work
solely
of
protected
the
(including
x = 1.95 ft
and
work
this
assessing
Thus,
integrity
of
y = - 0.04(1.954)2 = - 0.153 ft
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–107.
The ball at A is kicked such that uA = 30°. If it strikes the
ground at B having coordinates x = 15 ft, y = - 9 ft,
determine the speed at which it is kicked and the speed at
which it strikes the ground.
y
vA
A
θA
x
–y
SOLUTION
B
y = –0.04x 2
+ )s = s + v t
(:
0
0
x
15 = 0 + vA cos 30° t
( + c ) s = s0 + v0 t +
1 2
a t
2 c
-9 = 0 + vA sin 30° t +
1
( - 32.2)t2
2
vA = 16.5 ft>s
Ans.
t = 1.047 s
laws
or
+ ) (v ) = 16.54 cos 30° = 14.32 ft>s
(:
B x
in
teaching
Web)
(+ c ) v = v0 + ac t
Dissemination
Wide
copyright
(vB)y = 16.54 sin 30° + ( - 32.2)(1.047)
of
the
not
instructors
States
World
permitted.
= - 25.45 ft>s
Ans.
United
on
learning.
is
(14.32)2 + (- 25.45)2 = 29.2 ft s
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
vB =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*12–108.
The man at A wishes to throw two darts at the target at B so
that they arrive at the same time. If each dart is thrown with
a speed of 10 m>s, determine the angles uC and uD at which
they should be thrown and the time between each throw.
Note that the first dart must be thrown at uC 1 7uD2, then
the second dart is thrown at uD .
5m
uC
uD
A
s = s0 + v0t
5 = 0 + (10 cos u) t
(+ c )
(1)
v = v0 + act
-10 sin u = 10 sin u - 9.81 t
t =
2(10 sin u)
= 2.039 sin u
9.81
From Eq. (1),
laws
or
5 = 20.39 sin u cos u
in
teaching
Web)
sin 2u = 2 sin u cos u
Wide
copyright
Since
States
World
permitted.
Dissemination
sin 2u = 0.4905
instructors
The two roots are uD = 14.7°
United
on
learning.
is
of
the
not
Ans.
Ans.
work
student
the
by
and
use
uC = 75.3°
of
protected
the
(including
for
From Eq. (1): tD = 0.517 s
work
this
assessing
is
work
solely
tC = 1.97 s
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
So that ¢t = tC - tD = 1.45 s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
D
B
SOLUTION
+ )
(:
C
Ans.
12–109.
A boy throws a ball at O in the air with a speed v0 at an
angle u1. If he then throws another ball with the same speed
v0 at an angle u2 6 u1, determine the time between the
throws so that the balls collide in mid air at B.
B
O
u1
u2
y
SOLUTION
Vertical Motion: For the first ball, the vertical component of initial velocity is
(v0)y = v0 sin u1 and the initial and final vertical positions are (s0)y = 0 and sy = y,
respectively.
(+ c)
sy = (s0)y + (v0)y t +
x
1
(a ) t2
2 cy
y = 0 + v0 sin u1t1 +
1
( -g)t21
2
(1)
For the second ball, the vertical component of initial velocity is (v0)y = v0 sin u2 and
the initial and final vertical positions are (s0)y = 0 and sy = y, respectively.
1
(a ) t2
2 cy
y = 0 + v0 sin u2t2 +
1
( -g)t22
2
(2)
or
sy = (s0)y + (v0)y t +
laws
(+ c)
the
not
instructors
States
sx = (s0)x + (v0)x t
learning.
is
of
+ B
A:
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
Horizontal Motion: For the first ball, the horizontal component of initial velocity is
(v0)x = v0 cos u1 and the initial and final horizontal positions are (s0)x = 0 and
sx = x, respectively.
(3)
student
the
by
and
use
United
on
x = 0 + v0 cos u1 t1
is
work
solely
of
protected
the
(including
for
work
For the second ball, the horizontal component of initial velocity is (v0)x = v0 cos u2
and the initial and final horizontal positions are (s0)x = 0 and sx = x, respectively.
this
assessing
sx = (s0)x + (v0)x t
(4)
and
any
courses
part
the
is
This
x = 0 + v0 cos u2 t2
provided
integrity
of
and
work
+ B
A:
cos u1
t
cos u2 1
sale
t2 =
will
their
destroy
of
Equating Eqs. (3) and (4), we have
(5)
Equating Eqs. (1) and (2), we have
v0 t1 sin u1 - v0 t2 sin u2 =
1
g A t21 - t22 B
2
(6)
Solving Eq. [5] into [6] yields
t1 =
t2 =
2v0 cos u2 sin(u1 - u2)
g(cos2 u2 - cos2 u1)
2v0 cos u1 sin(u1 - u2)
g(cos2 u2 - cos2u1)
Thus, the time between the throws is
¢t = t1 - t2 =
=
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
2v0 sin(u1 - u2)(cos u2 - cos u1)
g(cos2 u2 - cos2 u1)
2v0 sin (u1 - u2)
g(cos u2 + cos u1)
Ans.
12–110.
Small packages traveling on the conveyor belt fall off into a
l-m-long loading car. If the conveyor is running at a constant
speed of vC = 2 m>s, determine the smallest and largest
distance R at which the end A of the car may be placed from
the conveyor so that the packages enter the car.
vc
2 m/s
30
3m
A
B
SOLUTION
R
Vertical Motion: The vertical component of initial velocity is (v0)y = 2 sin 30°
= 1.00 m>s. The initial and final vertical positions are (s0)y = 0 and sy = 3 m,
respectively.
A+TB
sy = (s0)y + (v0)y t +
3 = 0 + 1.00(t) +
1
(a ) t2
2 cy
1
(9.81) A t2 B
2
Choose the positive root t = 0.6867 s
or
Horizontal Motion: The horizontal component of velocity is (v0)x = 2 cos 30°
= 1.732 m>s and the initial horizontal position is (s0)x = 0. If sx = R, then
+ B
A:
teaching
Web)
laws
sx = (s0)x + (v0)x t
in
R = 0 + 1.732(0.6867) = 1.19 m
Dissemination
Wide
copyright
Ans.
United
on
learning.
is
of
the
sx = (s0)x + (v0)x t
and
use
+ B
A:
not
instructors
States
World
permitted.
If sx = R + 1, then
for
work
student
the
by
R + 1 = 0 + 1.732(0.6867)
Ans.
assessing
is
work
solely
of
protected
the
(including
R = 0.189 m
Ans.
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and
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is
This
provided
integrity
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and
work
this
Thus, Rmin = 0.189 m, Rmax = 1.19 m
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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1m
12–111.
The fireman wishes to direct the flow of water from his hose
to the fire at B. Determine two possible angles u1 and u2 at
which this can be done. Water flows from the hose at
vA = 80 ft>s.
A
u
vA
20 ft
B
SOLUTION
+ B
A:
s = s0 + v0 t
35 ft
35 = 0 + (80)(cos u )t
A+cB
s = s0 + v0 t +
1 2
act
2
-20 = 0 - 80 (sin u)t +
1
(-32.2)t 2
2
Thus,
or
0.1914
0.4375
t + 16.1 ¢
≤
cos u
cos2 u
laws
20 = 80 sin u
Wide
copyright
in
teaching
Web)
20 cos2 u = 17.5 sin 2u + 3.0816
Dissemination
Solving,
(below the horizontal)
u2 = 85.2°
(above the horizontal)
permitted.
u1 = 24.9°
is
of
the
not
instructors
States
World
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*12–112.
The baseball player A hits the baseball at vA = 40 ft>s and
uA = 60° from the horizontal. When the ball is directly
overhead of player B he begins to run under it. Determine
the constant speed at which B must run and the distance d
in order to make the catch at the same elevation at which
the ball was hit.
vA = 40 ft/s
θA
A
15 ft
SOLUTION
Vertical Motion: The vertical component of initial velocity for the football is
(v0)y = 40 sin 60° = 34.64 ft>s. The initial and final vertical positions are (s0)y = 0
and sy = 0, respectively.
(+ c )
sy = (s0)y + (v0)y t +
0 = 0 + 34.64t +
1
(a ) t2
2 cy
1
( -32.2) t2
2
t = 2.152 s
in
teaching
Web)
laws
or
Horizontal Motion: The horizontal component of velocity for the baseball is
(v0)x = 40 cos 60° = 20.0 ft>s. The initial and final horizontal positions are
(s0)x = 0 and sx = R, respectively.
sx = (s0)x + (v0)x t
Dissemination
Wide
copyright
+ )
(:
is
of
the
not
instructors
States
World
permitted.
R = 0 + 20.0(2.152) = 43.03 ft
by
and
use
United
on
learning.
The distance for which player B must travel in order to catch the baseball is
Ans.
the
(including
for
work
student
the
d = R - 15 = 43.03 - 15 = 28.0 ft
work
this
assessing
is
work
solely
of
protected
Player B is required to run at a same speed as the horizontal component of velocity
of the baseball in order to catch it.
integrity
of
and
vB = 40 cos 60° = 20.0 ft s
sale
will
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destroy
of
and
any
courses
part
the
is
This
provided
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
B
C
vA
d
12–113.
The man stands 60 ft from the wall and throws a ball at it
with a speed v0 = 50 ft>s. Determine the angle u at which
he should release the ball so that it strikes the wall at the
highest point possible. What is this height? The room has a
ceiling height of 20 ft.
v0
50 ft/s
u
h
5 ft
60 ft
SOLUTION
vx = 50 cos u
s = s0 + v0t
x = 0 + 50 cos u t
(+ c)
(1)
v = v0 + act
vy = 50 sin u - 32.2 t
(+ c)
s = s0 + v0t +
(2)
1 2
at
2 c
y = 0 + 50 sin u t - 16.1 t2
(3)
laws
or
+ )
(:
teaching
Web)
v2 = n20 + 2ac(s - s0)
in
(+ c)
(4)
is
of
the
not
instructors
States
World
v2y = 2500 sin2 u - 64.4 s
by
and
use
United
on
learning.
Require vy = 0 at s = 20 - 5 = 15 ft
(including
for
work
student
the
0 = 2500 sin2 u - 64.4 (15)
Ans.
assessing
is
work
solely
of
protected
the
u = 38.433° = 38.4°
part
the
is
destroy
of
and
From Eq. (1)
sale
will
their
t = 0.9652 s
any
courses
0 = 50 sin 38.433° - 32.2 t
This
provided
integrity
of
and
work
this
From Eq. (2)
x = 50 cos 38.433°(0.9652) = 37.8 ft
Time for ball to hit wall
From Eq. (1),
60 = 50(cos 38.433°)t
t = 1.53193 s
From Eq. (3)
y = 50 sin 38.433°(1.53193) - 16.1(1.53193)2
y = 9.830 ft
h = 9.830 + 5 = 14.8 ft
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
permitted.
Dissemination
Wide
copyright
v2y = (50 sin u)2 + 2(-32.2)(s - 0)
20
12–114.
A car is traveling along a circular curve that has a radius of
50 m. If its speed is 16 m>s and is increasing uniformly at
8 m>s2, determine the magnitude of its acceleration at this
instant.
SOLUTION
v = 16 m>s
at = 8 m>s2
r = 50 m
an =
(16)2
v2
=
= 5.12 m>s2
r
50
a = 2(8)2 + (5.12)2 = 9.50 m>s2
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–115.
Determine the maximum constant speed a race car can
have if the acceleration of the car cannot exceed 7.5 m>s2
while rounding a track having a radius of curvature of
200 m.
SOLUTION
Acceleration: Since the speed of the race car is constant, its tangential component of
acceleration is zero, i.e., at = 0. Thus,
a = an =
7.5 =
v2
r
v2
200
v = 38.7 m>s
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*12–116.
A car moves along a circular track of radius 250 ft such that
its speed for a short period of time, 0 … t … 4 s, is
v = 31t + t22 ft>s, where t is in seconds. Determine the
magnitude of its acceleration when t = 3 s. How far has it
traveled in t = 3 s?
SOLUTION
v = 3(t + t 2)
at =
dv
= 3 + 6t
dt
When t = 3 s,
an =
at = 3 + 6(3) = 21 ft>s2
[3(3 + 32)]2
= 5.18 ft>s2
250
a = 2(21)2 + (5.18)2 = 21.6 ft>s2
Ans.
3
or
31t + t22dt
in
teaching
Web)
3
32
t + t3 `
2
0
Wide
Ans.
sale
will
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destroy
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and
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the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
¢s = 40.5 ft
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
permitted.
Dissemination
¢s =
L0
laws
ds =
copyright
L
12–117.
A car travels along a horizontal circular curved road that
has a radius of 600 m. If the speed is uniformly increased at
a rate of 2000 km>h2, determine the magnitude of the
acceleration at the instant the speed of the car is
60 km>h.
SOLUTION
at = ¢
2
1h
2000 km 1000 m
ba
b = 0.1543 m>s2
ba
2
1 km
3600 s
h
v = a
an =
1h
60 km 1000 m
ba
ba
b = 16.67 m>s
h
1 km
3600 s
v2
16.672
= 0.4630 m>s2
=
r
600
a = 2a2t + a2n = 20.15432 + 0.46302 = 0.488 m>s2
sale
will
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destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–118.
The truck travels in a circular path having a radius of 50 m at
a speed of v = 4 m>s. For a short distance from s = 0, its
#
speed is increased by v = 10.05s2 m>s2, where s is in meters.
Determine its speed and the magnitude of its acceleration
when it has moved s = 10 m.
.
v
v
50 m
SOLUTION
v dv = at ds
v
L4
10
v dv =
L0
0.5v 2 - 8 =
0.05s ds
0.05
(10)2
2
v = 4.583 = 4.58 m>s
an =
Ans.
v 2 (4.583)2
= 0.420 m>s2
=
r
50
laws
or
at = 0.05(10) = 0.5 m>s2
Web)
a = 2(0.420)2 + (0.5)2 = 0.653 m>s2
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Ans.
sale
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
(0.05s) m/s2
4 m/s
12–119.
The automobile is originally at rest at s = 0. If its speed is
#
increased by v = 10.05t22 ft>s2, where t is in seconds,
determine the magnitudes of its velocity and acceleration
when t = 18 s.
300 ft
s
240 ft
SOLUTION
at = 0.05t2
v
t
dv =
L0
0.05 t2 dt
L0
v = 0.0167 t3
s
L0
t
ds =
L0
0.0167 t3 dt
s = 4.167(10 - 3) t4
s = 437.4 ft
laws
or
When t = 18 s,
in
teaching
Web)
Therefore the car is on a curved path.
Dissemination
Wide
Ans.
copyright
v = 0.0167(183) = 97.2 ft>s
the
not
instructors
States
World
permitted.
(97.2)2
= 39.37 ft>s2
240
on
learning.
is
of
an =
the
by
and
use
United
at = 0.05(182) = 16.2 ft/s2
the
(including
for
work
student
a = 2(39.37)2 + (16.2)2
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
a = 42.6 ft>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
*12–120.
The automobile is originally at rest s = 0. If it then starts to
#
increase its speed at v = 10.05t22 ft>s2, where t is in seconds,
determine the magnitudes of its velocity and acceleration at
s = 550 ft.
300 ft
s
240 ft
SOLUTION
The car is on the curved path.
at = 0.05 t2
v
t
dv =
L0
0.05 t2 dt
L0
v = 0.0167 t3
s
L0
t
ds =
L0
0.0167 t3 dt
or
s = 4.167(10 - 3) t4
teaching
Web)
laws
550 = 4.167(10-3) t4
Dissemination
Wide
copyright
in
t = 19.06 s
not
instructors
States
World
permitted.
So that
use
United
on
learning.
is
of
the
v = 0.0167(19.06)3 = 115.4
and
v = 115 ft>s
for
work
student
the
by
Ans.
work
solely
of
protected
the
(including
(115.4)2
= 55.51 ft>s2
240
assessing
is
an =
part
the
is
sale
will
their
destroy
of
and
any
courses
a = 2(55.51)2 + (18.17)2 = 58.4 ft>s2
This
provided
integrity
of
and
work
this
at = 0.05(19.06)2 = 18.17 ft>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
12–121.
When the roller coaster is at B, it has a speed of 25 m>s,
which is increasing at at = 3 m>s2. Determine the
magnitude of the acceleration of the roller coaster at this
instant and the direction angle it makes with the x axis.
y
y
1 x2
100
A
s
SOLUTION
B
Radius of Curvature:
x
1 2
y =
x
100
dy
1
=
x
dx
50
1
50
dx2
=
2 3>2
1
xb R
50
5
2 1 2
50
2
= 79.30 m
x = 30 m
in
2
d2y
B1 + a
or
r =
dy 2 3>2
b R
dx
Web)
B1 + a
laws
=
dx2
teaching
d2y
30 m
Dissemination
Wide
copyright
Acceleration:
not
instructors
States
World
permitted.
#
a t = v = 3 m>s2
by
and
use
United
on
learning.
is
of
the
vB 2
252
= 7.881 m>s2
=
r
79.30
the
an =
the
(including
for
work
student
The magnitude of the roller coaster’s acceleration is
Ans.
assessing
is
work
solely
of
protected
a = 2at 2 + an 2 = 232 + 7.8812 = 8.43 m>s2
dy
1
2
≤ = tan-1 c A 30 B d = 30.96°.
dx x = 30 m
50
part
the
is
This
provided
integrity
of
and
work
this
The angle that the tangent at B makes with the x axis is f = tan-1 ¢
and
any
courses
As shown in Fig. a, an is always directed towards the center of curvature of the path. Here,
destroy
of
an
7.881
b = 69.16°. Thus, the angle u that the roller coaster’s acceleration makes
b = tan-1 a
at
3
sale
will
their
a = tan-1 a
with the x axis is
u = a - f = 38.2° b
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
12–122.
If the roller coaster starts from rest at A and its speed
increases at at = (6 – 0.06s) m>s2, determine the magnitude
of its acceleration when it reaches B where sB = 40 m.
y
y
1 x2
100
A
s
SOLUTION
B
Velocity: Using the initial condition v = 0 at s = 0,
x
v dv = at ds
v
L0
s
vdv =
L0
A 6 - 0.06s B ds
30 m
v = a 212s - 0.06s2 b m>s
(1)
Thus,
vB = 412 A 40 B - 0.06 A 40 B 2 = 19.60 m>s
laws
or
Radius of Curvature:
1 2
x
100
dy
1
=
x
dx
50
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
y =
United
on
learning.
is
of
the
1
50
and
use
=
the
dx2
by
d2y
work
student
the
(including
for
of
= 79.30 m
this
assessing
2
5
work
solely
protected
is
2 1 2
50
integrity
of
dx2
=
x = 30 m
and
any
courses
part
the
is
This
provided
2
d2y
2 3>2
1
xb R
50
B1 + a
and
r =
dy 2 3>2
b R
dx
work
B1 + a
destroy
of
Acceleration:
sale
will
their
#
a t = v = 6 - 0.06(40) = 3.600 m>s2
an =
19.602
v2
= 4.842 m>s2
=
r
79.30
The magnitude of the roller coaster’s acceleration at B is
a = 2at 2 + an 2 = 23.6002 + 4.8422 = 6.03 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
12–123.
The speedboat travels at a constant speed of 15 m> s while
making a turn on a circular curve from A to B. If it takes
45 s to make the turn, determine the magnitude of the
boat’s acceleration during the turn.
A
r
SOLUTION
Acceleration: During the turn, the boat travels s = vt = 15(45) = 675 m. Thus, the
675
s
radius of the circular path is r =
=
m. Since the boat has a constant speed,
p
p
at = 0. Thus,
v2
=
r
152
= 1.05 m>s2
675
a
b
p
Ans.
sale
will
their
destroy
of
and
any
courses
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the
is
This
provided
integrity
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and
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this
assessing
is
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solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
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Dissemination
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or
a = an =
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or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
B
*12–124.
The car travels along the circular path such that its speed is
increased by a t = (0.5et) m>s2, where t is in seconds.
Determine the magnitudes of its velocity and acceleration
after the car has traveled s = 18 m starting from rest.
Neglect the size of the car.
s
SOLUTION
v
L0
t
dv =
L0
0.5e t dt
ρ
v = 0.5(e t - 1)
t
18
L0
ds = 0.5
L0
(e t - 1)dt
18 = 0.5(e t - t - 1)
Solving,
t = 3.7064 s
v = 0.5(e 3.7064 - 1) = 19.85 m>s = 19.9 m>s
#
at = v = 0.5e t ƒ t = 3.7064 s = 20.35 m>s2
in
teaching
Web)
laws
or
Ans.
19.852
v2
= 13.14 m>s2
=
r
30
World
permitted.
Dissemination
Wide
copyright
an =
States
a = 2a2t + a2n = 220.352 + 13.142 = 24.2 m>s2
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
30 m
18 m
12–125.
The car passes point A with a speed of 25 m>s after which its
speed is defined by v = (25 - 0.15s) m>s. Determine the
magnitude of the car’s acceleration when it reaches point B,
where s = 51.5 m.
y
y
B
16 m
Velocity: The speed of the car at B is
vB = C 25 - 0.15 A 51.5 B D = 17.28 m>s
Radius of Curvature:
1 2
x
625
dy
= -3.2 A 10-3 B x
dx
= - 3.2 A 10-3 B
dx2
or
laws
4
2 - 3.2 A 10-3 B 2
Web)
= 324.58 m
teaching
=
2
Wide
d2y
2 3>2
c 1 + a -3.2 A 10-3 B x b d
x = 50 m
instructors
States
World
permitted.
2
3>2
in
r =
dy 2
b
dx
Dissemination
B1 + a
copyright
dx2
B
d2y
on
learning.
is
of
the
not
Acceleration:
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
vB 2
17.282
= 0.9194 m>s2
=
r
324.58
dv
= A 25 - 0.15s B A - 0.15 B = A 0.225s - 3.75 B m>s2
at = v
ds
an =
integrity
of
and
work
this
assessing
When the car is at B A s = 51.5 m B
and
any
courses
part
the
is
This
provided
a t = C 0.225 A 51.5 B - 3.75 D = - 2.591 m>s2
their
destroy
of
Thus, the magnitude of the car’s acceleration at B is
sale
will
a = 2a2t + a2n = 2( - 2.591)2 + 0.91942 = 2.75 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
1 2
x
625
s
A x
SOLUTION
y = 16 -
16
12–126.
y
If the car passes point A with a speed of 20 m>s and begins
to increase its speed at a constant rate of at = 0.5 m>s2,
determine the magnitude of the car’s acceleration when
s = 100 m.
y
B
16 m
Velocity: The speed of the car at C is
vC 2 = vA 2 + 2a t (sC - sA)
vC 2 = 202 + 2(0.5)(100 - 0)
vC = 22.361 m>s
Radius of Curvature:
1 2
x
625
laws
teaching
Web)
= -3.2 A 10-3 B
Wide
copyright
dx2
in
d2y
or
dy
= -3.2 A 10-3 B x
dx
permitted.
United
on
learning.
is
of
the
not
instructors
= 312.5 m
use
2
Dissemination
World
States
4
- 3.2 A 10-3 B
x=0
and
dx2
=
for
work
student
2
d2y
2 3>2
c1 + a -3.2 A 10-3 B x b d
by
r =
dy 2 3>2
b R
dx
the
B1 + a
of
protected
the
(including
Acceleration:
work
this
assessing
is
integrity
of
and
work
provided
vC 2
22.3612
=
= 1.60 m>s2
r
312.5
their
destroy
The magnitude of the car’s acceleration at C is
of
and
any
courses
part
the
is
This
an =
solely
#
a t = v = 0.5 m>s
sale
will
a = 2a2t + a2n = 20.52 + 1.602 = 1.68 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
1 2
x
625
s
A x
SOLUTION
y = 16 -
16
12–127.
A train is traveling with a constant speed of 14 m/s along
the curved path. Determine the magnitude of the
acceleration of the front of the train, B, at the instant it
reaches point A 1y = 02.
y (m)
10 m
( —y )
x = 10 e 15
x (m)
SOLUTION
A
vt = 14 m/s
y
x = 10e(15)
y = 15 lna
x
b
10
dy
10
1
15
= 15 a b a b =
x
x
dx
10
d2y
= -
dx2
15
x2
At x = 10,
laws
Web)
= 39.06 m
Wide
|- 0.15|
teaching
=
World
permitted.
dx2
2
in
d2y
3
Dissemination
2
C 1 + (1.5)2 D 2
copyright
r =
dy 2 2
b R
dx
or
3
B1 + a
and
use
United
on
learning.
is
of
the
not
instructors
States
dv
= 0
dt
at =
work
student
the
by
(14)2
v2
=
= 5.02 m s2
r
39.06
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
an = a =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
B
*12–128.
When a car starts to round a curved road with the radius of
curvature of 600 ft, it is traveling at 75 ft>s . If the car’s
#
speed begins to decrease at a rate of v = ( - 0.06t2) ft>s2 ,
determine the magnitude of the acceleration of the car
when it has traveled a distance of s = 700 ft.
s
SOLUTION
Velocity: Using the initial condition v = 75 ft>s when t = 0 s,
L
dt =
L
atdt
v
t
dv =
Lv = 75 ft>s
L0
-0.06t2dt
v = 175 - 0.02t3) ft>s
or
Position: Using the initial condition s = 0 at t = 0 s,
laws
ds = vdt
L0
in
teaching
Web)
175 - 0.02t32dt
Wide
L0
t
ds =
copyright
s
of
the
not
instructors
States
World
permitted.
Dissemination
s = 375t - 0.005t44 ft
and
use
United
on
learning.
is
At s = 700 ft,
for
work
student
the
by
700 = 75t - 0.005t4
solely
of
protected
the
(including
Solving the above equation by trial and error,
and
destroy
of
and
sale
will
v2
552
=
= 5.042 ft>s2
r
600
their
an =
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
t = 10 s and t = 20 s. Pick the first solution.
#
Acceleration:
When
t = 10 s,
at = v = - 0.0611022 = - 6 ft>s2
v = 75 - 0.0211032 = 55 ft>s
Thus, the magnitude of the truck’s acceleration is
a = 2at 2 + an2 = 21-622 + 5.0422 = 7.84 ft>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
r ⫽ 600 ft
12–129.
When the motorcyclist is at A, he increases his speed along
#
the vertical circular path at the rate of v = 10.3t2 ft>s2,
where t is in seconds. If he starts from rest at A, determine
the magnitudes of his velocity and acceleration when he
reaches B.
300 ft
A
60°
300 ft
SOLUTION
v
L0
B
t
dv =
L0
0.3tdt
v = 0.15t2
s
L0
t
ds =
L0
0.15t2 dt
s = 0.05t3
When s = p3 (300) ft,
p
3 (300)
= 0.05t3
t = 18.453 s
v = 0.15(18.453)2 = 51.08 ft>s = 51.1 ft>s
or
Ans.
in
teaching
Web)
laws
#
at = v = 0.3t|t = 18.453 s = 5.536 ft>s2
v2
51.082
=
= 8.696 ft>s2
r
300
States
World
permitted.
Dissemination
Wide
copyright
an =
the
Ans.
is
on
learning.
United
and
use
by
work
student
the
the
(including
for
work
solely
of
protected
this
assessing
is
integrity
of
and
work
provided
any
courses
part
the
is
This
destroy
of
and
will
their
sale
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
not
instructors
(5.536)2 + (8.696)2 = 10.3 ft s2
of
a2t + a2n =
a =
12–130.
When the motorcyclist is at A, he increases his speed along
#
the vertical circular path at the rate of v = (0.04s) ft>s2
where s is in ft. If he starts at vA = 2 ft>s where s = 0 at A,
determine the magnitude of his velocity when he reaches B.
Also, what is his initial acceleration?
300 ft
A
SOLUTION
v dv =
L
at ds
v
s
v dv =
L2
L0
0.04s ds
s
v
v2 2
= 0.0252 2
2 2
0
teaching
Web)
laws
or
v2
- 2 = 0.0252
2
Wide
copyright
in
v2 = 0.0452 + 4 = 0.041s2 + 1002
permitted.
Dissemination
v = 0.2 2s2 + 100
use
United
on
learning.
is
of
the
not
instructors
States
World
At B, s = ru = 300 A p3 B = 100p ft. Thus
and
v2
Ans.
student
the
by
= 0.2 2(100p22 + 100 = 62.9 ft>s
(including
for
work
s = 100p ft
is
work
solely
of
protected
the
Acceleration: At t = 0, s = 0, and v = 2.
integrity
of
and
work
this
assessing
#
at = v = 0.04 s
part
the
is
= 0
This
provided
at 2
an 2
=
s=0
any
destroy
of
and
300
will
their
1222
sale
v2
r
courses
s=0
an =
= 0.01333 ft>s2
a = 21022 + 10.0133322 = 0.0133 ft>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
300 ft
B
#
Velocity: At s = 0, v = 2. Here, ac = v = 0.045. Then
L
60⬚
12–131.
At a given instant the train engine at E has a speed of
20 m>s and an acceleration of 14 m>s2 acting in the
direction shown. Determine the rate of increase in the
train’s speed and the radius of curvature r of the path.
v
75
a
r
SOLUTION
Ans.
at = 14 cos 75° = 3.62 m>s2
an = 14 sin 75°
an =
(20)2
r
Ans.
sale
will
their
destroy
of
and
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courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
r = 29.6 m
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
14 m/s2
E
20 m/s
*12–132.
v
Car B turns such that its speed is increased by
(at)B = (0.5et) m>s2, where t is in seconds. If the car starts
from rest when u = 0°, determine the magnitudes of its
velocity and acceleration when the arm AB rotates u = 30°.
Neglect the size of the car.
B
SOLUTION
Velocity: The speed v in terms of time t can be obtained by applying a =
5m
dv
.
dt
A
dv = adt
v
L0
t
dv =
0.5et dt
L0
v = 0.5 A et - 1 B
(1)
30°
pb = 2.618 m.
180°
The time required for the car to travel this distance can be obtained by applying
When u = 30°, the car has traveled a distance of s = ru = 5a
or
ds
.
dt
laws
v =
teaching
in
Wide
copyright
0.5 A e - 1 B dt
permitted.
Dissemination
t
World
L0
instructors
L0
States
t
ds =
Web)
ds = vdt
2.618 m
and
use
United
on
learning.
is
of
the
not
2.618 = 0.5 A et - t - 1 B
by
t = 2.1234 s
(including
for
work
student
the
Solving by trial and error
is
work
solely
of
protected
the
Substituting t = 2.1234 s into Eq. (1) yields
v = 0.5 A e2.1234 - 1 B = 3.680 m>s = 3.68 m>s
provided
integrity
of
and
work
this
assessing
Ans.
of
and
any
courses
part
the
is
This
Acceleration: The tangential acceleration for the car at t = 2.1234 s is
at = 0.5e2.1234 = 4.180 m>s2. To determine the normal acceleration, apply Eq. 12–20.
will
their
destroy
3.6802
v2
=
= 2.708 m>s2
r
5
sale
an =
The magnitude of the acceleration is
a = 2a2t + a2n = 24.1802 + 2.7082 = 4.98 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
u
12–133.
v
Car B turns such that its speed is increased by
(at)B = (0.5et) m>s2, where t is in seconds. If the car starts
from rest when u = 0°, determine the magnitudes of its
velocity and acceleration when t = 2 s. Neglect the size of
the car.
B
SOLUTION
5m
dv
Velocity: The speed v in terms of time t can be obtained by applying a =
.
dt
A
dv = adt
v
L0
t
dv =
L0
t
0.5e dt
v = 0.5 A et - 1 B
When t = 2 s,
v = 0.5 A e2 - 1 B = 3.195 m>s = 3.19 m>s
Ans.
laws
or
Acceleration: The tangential acceleration of the car at t = 2 s is
at = 0.5e2 = 3.695 m>s2. To determine the normal acceleration, apply Eq. 12–20.
in
teaching
Web)
v2
3.1952
= 2.041 m>s2
=
r
5
Dissemination
Wide
copyright
an =
not
instructors
States
World
permitted.
The magnitude of the acceleration is
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
a = 2a2t + a2n = 23.6952 + 2.0412 = 4.22 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
u
12–134.
A boat is traveling along a circular curve having a radius of
100 ft. If its speed at t = 0 is 15 ft/s and is increasing at
#
v = 10.8t2 ft>s2, determine the magnitude of its acceleration
at the instant t = 5 s.
SOLUTION
v
L15
5
dv =
L0
0.8tdt
v = 25 ft>s
an =
v2
252
=
= 6.25 ft>s2
r
100
At t = 5 s,
#
at = v = 0.8(5) = 4 ft>s2
a2t + a2n =
42 + 6.252 = 7.42 ft s2
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
a =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–135.
A boat is traveling along a circular path having a radius of
20 m. Determine the magnitude of the boat’s acceleration
when the speed is v = 5 m>s and the rate of increase in the
#
speed is v = 2 m>s2.
SOLUTION
at = 2 m>s2
an =
y2
52
=
= 1.25 m>s2
r
20
a = 2a2t + a2n = 222 + 1.252 = 2.36 m>s2
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*■12–136.
Starting from rest, a bicyclist travels around a horizontal circular
path, r = 10 m, at a speed of v = 10.09t2 + 0.1t2 m>s,
where t is in seconds. Determine the magnitudes of his velocity
and acceleration when he has traveled s = 3 m.
SOLUTION
s
L0
t
ds =
L0
10.09t2 + 0.1t2dt
s = 0.03t3 + 0.05t2
When s = 3 m,
3 = 0.03t3 + 0.05t2
Solving,
t = 4.147 s
or
ds
= 0.09t2 + 0.1t
dt
laws
v =
v = 0.09(4.147)2 + 0.1(4.147) = 1.96 m>s
Wide
1.962
v2
= 0.3852 m>s2
=
r
10
permitted.
an =
Dissemination
dv
= 0.18t + 0.1 `
= 0.8465 m>s2
dt
t = 4.147 s
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
at =
copyright
in
teaching
Web)
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
a = 2a2t + a2n = 2(0.8465)2 + (0.3852)2 = 0.930 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
12–137.
A particle travels around a circular path having a radius of
50 m. If it is initially traveling with a speed of 10 m> s and its
#
speed then increases at a rate of v = 10.05 v2 m>s2,
determine the magnitude of the particle’s acceleraton four
seconds later.
SOLUTION
Velocity: Using the initial condition v = 10 m>s at t = 0 s,
dt =
dv
a
t
L0
v
dt =
t = 20 ln
L10 m>s
dv
0.05v
v
10
laws
or
v = (10et>20) m>s
in
teaching
Web)
When t = 4 s,
instructors
States
World
permitted.
Dissemination
Wide
copyright
v = 10e4>20 = 12.214 m>s
and
use
United
on
learning.
is
of
the
not
Acceleration: When v = 12.214 m>s (t = 4 s),
for
work
student
the
by
at = 0.05(12.214) = 0.6107 m>s2
work
solely
of
protected
the
(including
(12.214)2
v2
=
= 2.984 m>s2
r
50
this
assessing
is
an =
the
is
This
provided
integrity
of
and
work
Thus, the magnitude of the particle’s acceleration is
sale
will
their
destroy
of
and
any
courses
part
a = 2at2 + an2 = 20.61072 + 2.9842 = 3.05 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
12–138.
When the bicycle passes point A, it has a speed of 6 m> s,
#
which is increasing at the rate of v = 10.52 m>s2. Determine
the magnitude of its acceleration when it is at point A.
y
y ⫽ 12 ln (
x
)
20
A
x
50 m
SOLUTION
Radius of Curvature:
y = 12 ln a
x
b
20
dy
1
1
12
= 12 a
ba b =
x
dx
x>20 20
`
dx2
=
`-
12
`
x2
4
= 226.59 m
Web)
d2y
x = 50 m
Wide
`
12 2 3>2
≤ R
x
B1 + ¢
World
permitted.
Dissemination
r =
dy 2 3>2
b R
dx
laws
B1 + a
or
12
x2
teaching
= -
in
dx2
copyright
d2y
learning.
is
of
the
not
instructors
States
Acceleration:
the
by
and
use
United
on
at = v = 0.5 m>s2
the
(including
for
work
student
v2
62
=
= 0.1589 m>s2
r
226.59
is
work
solely
of
protected
an =
integrity
of
and
work
this
assessing
The magnitude of the bicycle’s acceleration at A is
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
a = 2at2 + an2 = 20.52 + 0.15892 = 0.525 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
12–139.
The motorcycle is traveling at a constant speed of 60 km> h.
Determine the magnitude of its acceleration when it is at
point A.
y
y2 ⫽ 2x
A
x
25 m
SOLUTION
Radius of Curvature:
y = 22x1>2
dy
1
= 22x - 1>2
dx
2
1
22x - 3>2
4
r =
d2y
`
dx2
2
1
2
B 1 + ¢ 22x - 1>2 ≤ R
=
3>2
or
4
1
` - 22x - 3>2 `
4
= 364.21 m
x = 25 m
Dissemination
Wide
copyright
`
dy 2 3>2
b R
dx
Web)
B1 + a
laws
= -
teaching
dx2
in
d2y
is
of
the
not
instructors
States
World
permitted.
Acceleration: The speed of the motorcycle at a is
by
and
use
United
on
learning.
km 1000 m
1h
≤¢
≤¢
≤ = 16.67 m>s
h
1 km
3600 s
for
work
student
the
v = ¢ 60
work
solely
of
protected
the
(including
v2
16.672
=
= 0.7627 m>s2
r
364.21
this
assessing
is
an =
part
the
is
This
provided
integrity
of
and
work
Since the motorcycle travels with a constant speed, at = 0. Thus, the magnitude of
the motorcycle’s acceleration at A is
sale
will
their
destroy
of
and
any
courses
a = 2at 2 + an2 = 202 + 0.76272 = 0.763 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
*12–140.
The jet plane travels along the vertical parabolic path.
When it is at point A it has a speed of 200 m>s, which is
increasing at the rate of 0.8 m>s2. Determine the magnitude
of acceleration of the plane when it is at point A.
y
y ⫽ 0.4x2
A
SOLUTION
y = 0.4x2
10 km
dy
= 0.8x `
= 4
dx
x = 5 km
d2y
dx2
r =
= 0.8
x
5 km
[1 + (4)2]3/2
= 87.62 km
0.8
at = 0.8 m/s2
or
(0.200)2
= 0.457(10 - 3) km>s2
87.62
teaching
Web)
laws
an =
Wide
copyright
in
an = 0.457 km/s2
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
permitted.
Dissemination
a = 210.822 + 10.45722 = 0.921 m>s2
12–141.
The ball is ejected horizontally from the tube with a speed
of 8 m>s. Find the equation of the path, y = f1x2, and then
find the ball’s velocity and the normal and tangential
components of acceleration when t = 0.25 s.
y
vA
A
SOLUTION
vx = 8 m>s
+ )
(:
s = v0t
x = 8t
A+cB
s = s0 + v0 t +
y = 0 + 0 +
1 2
a t
2 c
1
( - 9.81)t2
2
y = -4.905t2
in
teaching
Web)
laws
or
x 2
y = -4.905 a b
8
Ans.
Dissemination
Wide
(Parabola)
copyright
y = -0.0766x2
of
the
not
instructors
States
World
permitted.
v = v0 + act
the
by
and
use
United
on
learning.
is
vy = 0 - 9.81t
the
(including
for
work
student
When t = 0.25 s,
assessing
is
work
solely
of
protected
vy = - 2.4525 m>s
Ans.
part
the
is
any
courses
will
ay = 9.81 m>s2
sale
ax = 0
their
destroy
of
2.4525
b = 17.04°
8
and
u = tan - 1 a
This
provided
integrity
of
and
work
this
v = 31822 + 12.452522 = 8.37 m>s
an = 9.81 cos 17.04° = 9.38 m>s2
Ans.
at = 9.81 sin 17.04° = 2.88 m>s2
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
8 m/s
x
12–142.
A toboggan is traveling down along a curve which can be
approximated by the parabola y = 0.01x2. Determine the
magnitude of its acceleration when it reaches point A, where
its speed is vA = 10 m>s, and it is increasing at the rate of
#
vA = 3 m>s2.
y
y = 0.01x2
A
36 m
SOLUTION
x
60 m
Acceleration: The radius of curvature of the path at point A must be determined
d2y
dy
first. Here,
= 0.02x and 2 = 0.02, then
dx
dx
r =
[1 + (dy>dx)2]3>2
|d2y>dx2|
=
[1 + (0.02x)2]3>2
2
= 190.57 m
|0.02|
x = 60 m
To determine the normal acceleration, apply Eq. 12–20.
an =
v2
102
=
= 0.5247 m>s2
r
190.57
Web)
laws
or
#
Here, at = vA = 3 m>s. Thus, the magnitude of acceleration is
teaching
32 + 0.52472 = 3.05 m s2
Ans.
in
a2t + a2n =
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
a =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–143.
A particle P moves along the curve y = 1x2 - 42 m with a
constant speed of 5 m>s. Determine the point on the curve
where the maximum magnitude of acceleration occurs and
compute its value.
SOLUTION
y = (x2 - 4)
at =
dv
= 0,
dt
To obtain maximum a = an, r must be a minimum.
This occurs at:
x = 0,
y = -4 m
Ans.
laws
= 2
Web)
dx2
teaching
d2y
in
dy
`
= 2x = 0;
dx x = 0
or
Hence,
Wide
permitted.
Dissemination
World
the
not
instructors
States
is
on
by
and
`
learning.
dx2
of
d2y
United
`
3
[1 + 0]2
1
=
=
020
2
use
rmin =
dy 2 2
b d
dx
copyright
3
c1 + a
the
(including
for
work
student
the
v2
52
= 1 = 50 m>s2
rmin
2
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
(a)max = (an)max =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
*12–144.
The Ferris wheel turns such that the speed of the passengers is
#
increased by v = 14t2 ft>s2, where t is in seconds. If the wheel
starts from rest when u = 0°, determine the magnitudes of
the velocity and acceleration of the passengers when the
wheel turns u = 30°.
40 ft
u
SOLUTION
v
t
dv =
L0
L0
4tdt
v = 2t2
s
t
ds =
2t2 dt
L0
2
s = 3 t3
p
When s = (40) ft,
6
L0
p
2
(40) = t 3
6
3
t = 3.1554 s
v = 2(3.1554)2 = 19.91 ft>s = 19.9 ft>s
or
Ans.
in
teaching
Web)
laws
#
at = v = 4t 0 t = 3.1554 s = 12.62 ft>s2
permitted.
Dissemination
Wide
copyright
19.912
v2
=
= 9.91 ft>s2
r
40
instructors
States
World
an =
not
a = 2at 2 + an2 = 212.622 + 9.912 = 16.0 ft>s2
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–145.
If the speed of the crate at A is 15 ft> s, which is increasing at
#
a rate v = 3 ft>s2, determine the magnitude of the
acceleration of the crate at this instant.
y
y ⫽ 1 x2
16
A
x
SOLUTION
10 ft
Radius of Curvature:
y =
1 2
x
16
dy
1
= x
dx
8
d2y
1
=
8
dx2
Thus,
or
Web)
laws
teaching
4
1
` `
8
in
= 32.82 ft
Wide
`
dx2
=
x = 10 ft
not
instructors
States
World
permitted.
`
d2y
2 3>2
1
8
B1 + ¢ x≤ R
Dissemination
r =
dy 2 3>2
b R
dx
copyright
B1 + a
and
use
United
on
learning.
is
of
the
Acceleration:
for
work
student
the
by
#
at = v = 3ft>s2
work
solely
of
protected
the
(including
v2
152
=
= 6.856 ft>s2
r
32.82
this
assessing
is
an =
the
is
This
provided
integrity
of
and
work
The magnitude of the crate’s acceleration at A is
sale
will
their
destroy
of
and
any
courses
part
a = 2at 2 + an2 = 232 + 6.8562 = 7.48 ft>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
12–146.
The race car has an initial speed vA = 15 m>s at A. If it
increases its speed along the circular track at the rate
a t = 10.4s2 m>s2, where s is in meters, determine the time
needed for the car to travel 20 m. Take r = 150 m.
r
SOLUTION
s
A
n dn
at = 0.4s =
ds
a ds = n dn
n
s
0.4s ds =
L0
L15
n dn
0.4s2 s
n2 n
` =
`
2 0
2 15
laws
or
n2
225
0.4s2
=
2
2
2
in
teaching
Web)
n2 = 0.4s2 + 225
ds
= 20.4s2 + 225
dt
States
World
permitted.
Dissemination
Wide
copyright
n =
not
the
of
on
learning.
is
dt
United
and
L0
work
the
(including
for
of
solely
= 0.632 456t
work
s
assessing
1n (s + 2s2 + 562.5) `
student
= 0.632 456t
protected
ds
L0 2s2 + 562.5
is
s
the
by
L0 20.4s2 + 225
=
instructors
t
ds
use
s
work
this
0
any
will
sale
t = 1.21 s
their
destroy
of
and
At s = 20 m,
courses
part
the
is
This
provided
integrity
of
and
1n (s + 2s2 + 562.5) - 3.166 196 = 0.632 456t
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
12–147.
A boy sits on a merry-go-round so that he is always located
at r = 8 ft from the center of rotation. The merry-go-round
is originally at rest, and then due to rotation the boy’s speed
is increased at 2 ft>s2. Determine the time needed for his
acceleration to become 4 ft>s2.
SOLUTION
a = 2a2n + a2t
at = 2
v = v0 + act
v = 0 + 2t
an =
(2t)2
v2
=
r
8
4 =
(2)2 + a
or
A
(2t)2 2
b
8
teaching
Web)
laws
16 t4
64
in
16 = 4 +
Wide
copyright
t = 2.63 s
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*12–148.
A particle travels along the path y = a + bx + cx 2, where a,
b, c are constants. If the speed of the particle is constant,
v = v0, determine the x and y components of velocity and the
normal component of acceleration when x = 0.
SOLUTION
y = a + bx + cx2
#
#
#
y = bx + 2 c x x
$
$
#
$
y = b x + 2 c (x)2 + 2 c xx
#
#
y = bx
When x = 0,
#
#
v20 + x 2 + b2 x 2
vy =
v0
Ans.
21 + b2
v0 b
Ans.
Dissemination
Wide
copyright
in
teaching
Web)
laws
21 + b2
v20
an =
r
or
#
vx = x =
dy 2 2
C 1 + A dx
B D
World
the
not
instructors
States
United
on
`
and
2
learning.
is
of
dx
use
d2y
the
`
by
r =
permitted.
3
is
this
assessing
= 2c
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
integrity
of
and
part
the
is
any
courses
and
destroy
of
will
2 c v20
(1 + b2)3>2
2c
(1 + b2)3>2
sale
an =
r =
their
At x = 0,
This
provided
dx2
work
d2y
work
solely
of
protected
the
(including
for
work
student
dy
= b + 2c x
dx
Ans.
12–149.
The two particles A and B start at the origin O and travel in
opposite directions along the circular path at constant
speeds vA = 0.7 m>s and vB = 1.5 m>s, respectively.
Determine in t = 2 s, (a) the displacement along the path
of each particle, (b) the position vector to each particle, and
(c) the shortest distance between the particles.
y
SOLUTION
(a)
sA = 0.7(2) = 1.40 m
Ans.
sB = 1.5(2) = 3 m
Ans.
5m
B
1.40
uA =
= 0.280 rad. = 16.04°
5
(b)
uB =
A
vB
1.5 m/s
vA
3
= 0.600 rad. = 34.38°
5
For A
laws
or
x = 5 sin 16.04° = 1.382 = 1.38 m
in
teaching
Web)
y = 5(1 - cos 16.04°) = 0.1947 = 0.195 m
Dissemination
Wide
Ans.
copyright
rA = {1.38i + 0.195j} m
of
the
not
instructors
States
World
permitted.
For B
by
and
use
United
on
learning.
is
x = -5 sin 34.38° = -2.823 = -2.82 m
for
work
student
the
y = 5(1 -cos 34.38°) = 0.8734 = 0.873 m
Ans.
is
work
solely
of
protected
the
(including
rB = { - 2.82i + 0.873j} m
integrity
of
and
work
this
assessing
¢r = rB - rA = {-4.20i + 0.678j} m
(c)
provided
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
¢r = 31 - 4.2022 + 10.67822 = 4.26 m
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
x
O
0.7 m/s
12–150.
The two particles A and B start at the origin O and travel in
opposite directions along the circular path at constant
speeds vA = 0.7 m>s and vB = 1.5 m>s, respectively.
Determine the time when they collide and the magnitude of
the acceleration of B just before this happens.
y
SOLUTION
st = 2p(5) = 31.4159 m
5m
sA = 0.7 t
B
sB = 1.5 t
A
vB
Require
0.7 t + 1.5 t = 31.4159
t = 14.28 s = 14.3 s
Ans.
laws
or
(1.5)2
v2B
= 0.45 m>s2
=
r
5
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
Ans.
sale
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
x
O
vA
sA + sB = 31.4159
aB =
1.5 m/s
0.7 m/s
12–151.
The position of a particle traveling along a curved path is
s = (3t3 - 4t2 + 4) m, where t is in seconds. When t = 2 s,
the particle is at a position on the path where the radius of
curvature is 25 m. Determine the magnitude of the
particle’s acceleration at this instant.
SOLUTION
Velocity:
v =
d 3
A 3t - 4t2 + 4 B = A 9t2 - 8t B m>s
dt
When t = 2 s,
v ƒ t = 2 s = 9 A 22 B - 8122 = 20 m>s
Acceleration:
or
dv
d 2
=
A 9t - 8t B = A 18t - 8 B m>s2
ds
dt
laws
at =
teaching
Web)
at ƒ t = 2 s = 18(2) - 8 = 28 m>s2
in
202
= 16 m>s2
25
Wide
States
World
permitted.
r
=
Dissemination
Av ƒ t = 2 sB2
copyright
an =
on
learning.
is
of
the
not
instructors
Thus,
sale
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destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
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solely
of
protected
the
(including
for
work
student
the
by
and
use
United
a = 2at 2 + an2 = 2282 + 162 = 32.2 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Ans.
*12–152.
If the speed of the box at point A on the track is 30 ft>s
#
which is increasing at the rate of v = 5 ft>s2, determine the
magnitude of the acceleration of the box at this instant.
y
y ⫽ 0.004x2 ⫹10
A
10 ft
x
SOLUTION
50 ft
Radius of Curvature:
y = 0.004x2 + 10
dy
= 0.008x
dx
d2y
dx2
= 0.008
laws
or
Thus,
Web)
teaching
in
4
=
Wide
copyright
= 156.17 ft
permitted.
Dissemination
ƒ 0.008 ƒ
World
`
dx2
3>2
x = 50 ft
on
learning.
is
of
the
not
`
d2y
B 1 + 10.008x22 R
instructors
r =
dy 2 3>2
b R
dx
States
B1 + a
student
the
by
and
use
United
Acceleration:
for
work
v2
302
=
= 5.763 ft>s2
r
156.17
#
at = v = 5 ft>s2
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
an =
part
the
is
This
provided
The magnitude of the box’s acceleration at A is therefore
sale
will
their
destroy
of
and
any
courses
a = 2at 2 + an2 = 252 + 5.7632 = 7.63 ft>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
■
12–153.
A go-cart moves along a circular track of radius 100 ft such
that its speed for a short period of time, 0 … t … 4 s, is
2
v = 6011 - e -t 2 ft>s. Determine the magnitude of its
acceleration when t = 2 s. How far has it traveled in
t = 2 s? Use Simpson’s rule with n = 50 to evaluate the
integral.
SOLUTION
v = 60(1 - e - t )
2
at =
dv
2
2
= 60( -e - t )(-2t) = 120 t e - t
dt
at|t = 2 = 120(2)e - 4 = 4.3958
v|t = 2 = 60(1 - e - 4) = 58.9011
an =
(58.9011)2
= 34.693
100
a = 2(4.3958)2 + (34.693)2 = 35.0 m>s2
6011 - e - t 2 dt
Web)
teaching
L0
laws
2
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
copyright
s = 67.1 ft
Wide
in
L0
2
ds =
or
s
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
■
12–154.
The ball is kicked with an initial speed vA = 8 m>s at an
angle uA = 40° with the horizontal. Find the equation of the
path, y = f(x), and then determine the normal and
tangential components of its acceleration when t = 0.25 s.
y
vA = 8 m/s
uA
40
A
x
x
SOLUTION
Horizontal Motion: The horizontal component of velocity is (v0)x = 8 cos 40°
= 6.128 m>s and the initial horizontal and final positions are (s0)x = 0 and sx = x,
respectively.
+ B
A:
sx = (s0)x + (y0)x t
x = 0 + 6.128t
(1)
or
Vertical Motion: The vertical component of initial velocity is (v0)y = 8 sin 40°
= 5.143 m>s. The initial and final vertical positions are (s0)y = 0 and sy = y,
respectively.
Web)
laws
1
(ac)y t2
2
teaching
sy = (s0)y + (v0)y t +
1
( -9.81) A t2 B
2
(2)
permitted.
Dissemination
copyright
y = 0 + 5.143t +
Wide
in
A+cB
is
of
the
not
instructors
States
World
Eliminate t from Eqs (1) and (2),we have
learning.
y = {0.8391x - 0.1306x2} m = {0.839x - 0.131x2} m
for
work
student
the
by
and
use
United
on
Ans.
destroy
of
and
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courses
part
the
is
This
provided
integrity
of
and
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this
assessing
is
work
solely
of
protected
the
(including
Acceleration: When t = 0.25 s, from Eq. (1), x = 0 + 6.128(0.25) = 1.532 m. Here,
dy
dy
= 0.8391 - 0.2612x. At x = 1.532 m,
= 0.8391 - 0.2612(1.532) = 0.4389
dx
dx
and the tangent of the path makes an angle u = tan-1 0.4389 = 23.70° with the x axis.
The magnitude of the acceleration is a = 9.81 m>s2 and is directed downward. From
the figure, a = 23.70°. Therefore,
Ans.
an = a cos a = 9.81 cos 23.70° = 8.98 m>s2
Ans.
sale
will
their
at = - a sin a = - 9.81 sin 23.70° = - 3.94 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
y
12–155.
The race car travels around the circular track with a speed of
16 m>s. When it reaches point A it increases its speed at
at = (43 v1>4) m>s2, where v is in m>s. Determine the
magnitudes of the velocity and acceleration of the car when
it reaches point B. Also, how much time is required for it to
travel from A to B?
y
A
200 m
SOLUTION
B
x
4 1
at = v4
3
dv = at dt
dv =
4 1
v 4 dt
3
v
L16
3
t
dv
0.75
=
1
v4
L0
dt
v
v4 16 = t
3
laws
or
v4 - 8 = t
4
Wide
copyright
in
teaching
Web)
v = (t + 8)3
permitted.
Dissemination
ds = v dt
World
instructors
States
4
is
of
the
not
(t + 8)3 dt
on
learning.
and
L0
United
L0
t
ds =
use
s
by
t
7
3
(t + 8)3 2
7
0
s =
7
3
(t + 8)3 - 54.86
7
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
s =
7
3
p
(200) = 100p = (t + 8)3 - 54.86
2
7
destroy
of
and
any
courses
part
the
is
This
For s =
will
their
Ans.
sale
t = 10.108 s = 10.1 s
4
v = (10.108 + 8)3 = 47.551 = 47.6 m>s
at =
an =
1
4
(47.551)4 = 3.501 m>s2
3
(47.551)2
v2
= 11.305 m>s2
=
r
200
a = 2(3.501)2 + (11.305)2 = 11.8 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
Ans.
*12–156.
A particle P travels along an elliptical spiral path such
that
its
position
vector
r
is
defined
by
r = 52 cos10.1t2i + 1.5 sin10.1t2j + 12t2k6 m, where t is in
seconds and the arguments for the sine and cosine are given
in radians. When t = 8 s, determine the coordinate
direction angles a, b, and g, which the binormal axis to the
osculating plane makes with the x, y, and z axes. Hint: Solve
for the velocity vP and acceleration a P of the particle in
terms of their i, j, k components. The binormal is parallel to
vP * a P . Why?
z
P
r
SOLUTION
y
rP = 2 cos (0.1t)i + 1.5 sin (0.1t)j + 2tk
#
vP = r = - 0.2 sin (0.1t)i + 0.15 cos (0.1t)j + 2k
x
$
aP = r = -0.02 cos 10.1t2i - 0.015 sin 10.1t2j
When t = 8 s,
or
vP = -0.2 sin (0.8 rad)i + 0.15 cos (0.8 rad)j + 2k = -0.143 47i + 0.104 51j + 2k
in
teaching
Web)
laws
aP = -0.02 cos (0.8 rad)i - 0.015 sin (0.8 rad)j = -0.013 934i - 0.010 76 j
States
World
permitted.
Dissemination
Wide
copyright
Since the binormal vector is perpendicular to the plane containing the n–t axis, and
ap and vp are in this plane, then by the definition of the cross product,
by
and
use
United
on
learning.
is
of
the
not
instructors
j
k
0.104 51 2 3 = 0.021 52i - 0.027 868j + 0.003k
-0.010 76 0
(including
the
of
protected
b = 210.0215222 + 1- 0.02786822 + 10.00322 = 0.035 338
for
work
student
the
i
b = vP * aP = 3 -0.14 347
- 0.013 934
this
assessing
is
work
solely
ub = 0.608 99i - 0.788 62j + 0.085k
Ans.
the
is
This
provided
integrity
of
and
work
a = cos - 1(0.608 99) = 52.5°
part
b = cos - 1(-0.788 62) = 142°
destroy
Ans.
sale
will
their
g = cos - 1(0.085) = 85.1°
of
and
any
courses
Ans.
Note: The direction of the binormal axis may also be specified by the unit vector
ub¿ = - ub, which is obtained from b¿ = ap * vp.
For this case, a = 128°, b = 37.9°, g = 94.9°
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
12–157.
The motion of a particle is defined by the equations
x = (2t + t2) m and y = (t2) m, where t is in seconds.
Determine the normal and tangential components of the
particle’s velocity and acceleration when t = 2 s.
SOLUTION
Velocity: Here, r =
E A 2 t + t 2 B i + t2 j F m.To determine the velocity v, apply Eq. 12–7.
v =
dr
= {(2 + 2t) i + 2tj } m>s
dt
When t = 2 s, v = [2 + 2(2)]i + 2(2)j = {6i + 4j} m>s. Then v = 262 + 42
= 7.21 m>s. Since the velocity is always directed tangent to the path,
vn = 0
and
vt = 7.21 m>s
The velocity v makes an angle u = tan-1
Ans.
4
= 33.69° with the x axis.
6
laws
or
Acceleration: To determine the acceleration a, apply Eq. 12–9.
in
teaching
Web)
dv
= {2i + 2j} m>s2
dt
Wide
copyright
a =
Dissemination
Then
learning.
is
of
the
not
instructors
States
World
permitted.
a = 222 + 22 = 2.828 m>s2
2
= 45.0° with the x axis. From the
2
the
by
and
use
United
on
The acceleration a makes an angle f = tan-1
the
(including
for
work
student
figure, a = 45° - 33.69 = 11.31°. Therefore,
Ans.
assessing
is
work
solely
of
protected
an = a sin a = 2.828 sin 11.31° = 0.555 m>s2
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
at = a cos a = 2.828 cos 11.31° = 2.77 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–158.
The motorcycle travels along the elliptical track at a
constant speed v. Determine the greatest magnitude of the
acceleration if a 7 b.
y
b
x2
a2
x
a
SOLUTION
b
2a2 - x2, we have
a
Acceleration: Differentiating twice the expression y =
dy
bx
= dx
a2a2 - x2
d2y
dx2
= -
ab
(a2 - x2)3>2
The radius of curvature of the path is
dx2
=
2
2-
2 3>2
bx
≤ R
a2a2 - x2
ab
c1 +
=
2
(a2 - x2)3>2
3>2
b2x2
d
a2(a2 - x2)
ab
(1)
(a2 - x2)3>2
in
teaching
Web)
2
d2y
B1 + ¢ -
or
r =
dy 2 3>2
b d
dx
laws
c1 + a
permitted.
Dissemination
Wide
copyright
To have the maximum normal acceleration, the radius of curvature of the path must
be a minimum. By observation, this happens when y = 0 and x = a. When x : a,
of
the
not
instructors
States
World
3>2
3>2
b2x2
b2x2
b3x3
b2x2
.
7 7 1. Then, c1 + 2 2
d :c 2 2
d
= 3 2
2
2
2
2
a (a - x )
a (a - x )
a (a - x )
a (a - x2)3>2
work
student
(including
for
Substituting this value into Eq. [1] yields r =
the
by
b2 3
x . At x = a,
a4
and
use
United
on
learning.
is
2
work
solely
of
protected
the
b2 3
b2
a B =
4 A
a
a
this
assessing
is
r =
part
the
is
This
provided
integrity
of
and
work
To determine the normal acceleration, apply Eq. 12–20.
any
courses
v2
a
v2
= 2 = 2 v2
r
b >a
b
destroy
of
and
(an)max =
sale
will
their
Since the motorcycle is traveling at a constant speed, at = 0. Thus,
amax = (an)max =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
a 2
v
b2
Ans.
y2
b2
1
12–159.
A particle is moving along a circular path having a radius
of 4 in. such that its position as a function of time is given
by u = cos 2t, where u is in radians and t is in seconds.
Determine the magnitude of the acceleration of the particle
when u = 30°.
SOLUTION
When u =
p
6
rad,
p
6
= cos 2t
t = 0.5099 s
#
du
= - 2 sin 2t 2
u =
= - 1.7039 rad>s
dt
t = 0.5099 s
$
d2u
= - 2.0944 rad>s2
u = 2 = - 4 cos 2t 2
dt
t = 0.5099 s
r = 4
#
r = 0
$
r = 0
or
#
$
ar = r - ru2 = 0 - 4(-1.7039)2 = - 11.6135 in.>s2
$
##
au = ru + 2ru = 4( - 2.0944) + 0 = - 8.3776 in.>s2
( -11.6135)2 + ( -8.3776)2 = 14.3 in. s2
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
a2r + a2u =
laws
a =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*12–160.
A particle travels around a limaçon, defined by the
equation r = b - a cos u, where a and b are constants.
Determine the particle’s radial and transverse components
of velocity and acceleration as a function of u and its time
derivatives.
SOLUTION
r = b - a cos u
#
#
r = a sin uu
$
#
#
r = a cos uu2 + a sin uu
#
#
vr = r = a sin uu
Ans.
#
vu = r u = (b - a cos u)u
#
#
$
#
$
ar = r - r u2 = a cos uu2 + a sin uu - (b - a cos u)u2
$
#
= (2a cos u - b) u2 + a sin u u
$
$
# #
# #
au = ru + 2 r u = (b - a cos u)u + 2 a a sin uu bu
Ans.
laws
or
Ans.
teaching
Web)
$
#
= (b - a cos u)u + 2au2 sin u
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–161.
If a particle’s position is described by the polar coordinates
r = 411 + sin t2 m and u = 12e -t2 rad, where t is in seconds
and the argument for the sine is in radians, determine the
radial and transverse components of the particle’s velocity
and acceleration when t = 2 s.
SOLUTION
When t = 2 s,
r = 4(1 +sin t) = 7.637
#
r = 4 cos t = -1.66459
$
r = -4 sin t = -3.6372
u = 2 e-t
#
u = - 2 e - t = - 0.27067
$
u = 2 e-t = 0.270665
#
vr = r = -1.66 m>s
#
vu = ru = 7.637(-0.27067) = - 2.07 m>s
#
$
ar = r - r1u22 = - 3.6372 - 7.6371- 0.2706722 = - 4.20 m>s2
$
##
au = ru + 2ru = 7.63710.2706652 + 21- 1.6645921- 0.270672 = 2.97 m>s2
Web)
laws
or
Ans.
Wide
copyright
in
teaching
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
instructors
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
not
States
World
permitted.
Dissemination
Ans.
12–162.
An airplane is flying in a straight line with a velocity of
200 mi>h and an acceleration of 3 mi>h2. If the propeller
has a diameter of 6 ft and is rotating at an angular rate of
120 rad>s, determine the magnitudes of velocity and
acceleration of a particle located on the tip of the
propeller.
SOLUTION
vPl = ¢
200 mi 5280 ft
1h
≤¢
≤¢
≤ = 293.3 ft>s
h
1 mi
3600 s
aPl = ¢
3 mi 5280 ft
1h 2
≤
¢
≤
¢
≤ = 0.001 22 ft>s2
1 mi
3600 s
h2
vPr = 120(3) = 360 ft>s
v = 2v2Pl + v2Pr = 2(293.3)2 + (360)2 = 464 ft>s
(360)2
v2Pr
=
= 43 200 ft>s2
r
3
a = 2a2Pl + a2Pr = 2(0.001 22)2 + (43 200)2 = 43.2(103) ft>s2
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aPr =
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12–163.
A car is traveling along the circular curve of radius r = 300 ft.
At
the instant shown, its angular rate of
$ rotation is
#
u = 0.4 rad>s, which is increasing at the rate of u = 0.2 rad>s2.
Determine the magnitudes of the car’s velocity and
acceleration at this instant.
A
r = 300 ft
.
θ = 0.4 rad/s
θ = 0.2 rad/s2
..
SOLUTION
θ
Velocity: Applying Eq. 12–25, we have
#
#
vr = r = 0
vu = ru = 300(0.4) = 120 ft>s
Thus, the magnitude of the velocity of the car is
v = 2v2r + v2u = 202 + 1202 = 120 ft>s
Ans.
laws
or
Acceleration: Applying Eq. 12–29, we have
#
$
ar = r - ru2 = 0 - 300 A 0.42 B = - 48.0 ft>s2
$
##
au = ru + 2ru = 300(0.2) + 2(0)(0.4) = 60.0 ft>s2
teaching
Web)
Thus, the magnitude of the acceleration of the car is
in
(-48.0)2 + 60.02 = 76.8 ft s2
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a2r + a2u =
copyright
a =
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*12–164.
A
velocity of
$
# radar gun at O rotates with the angular
2
u = 0.1 rad>s and angular acceleration of u = 0.025 rad>s ,
at the instant u = 45°, as it follows the motion of the car
traveling along the circular road having a radius of
r = 200 m. Determine the magnitudes of velocity and
acceleration of the car at this instant.
r ⫽ 200 m
u
O
SOLUTION
Time Derivatives: Since r is constant,
#
$
r = r = 0
Velocity:
#
vr = r = 0
#
vu = ru = 200(0.1) = 20 m>s
Thus, the magnitude of the car’s velocity is
v = 2vr2 + vu2 = 202 + 202 = 20 m>s
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Acceleration:
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#
#
ar = r - ru2 = 0 - 200(0.12) = - 2 m>s2
$
# #
au = r u + 2ru = 200(0.025) + 0 = 5 m>s2
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United
on
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is
of
the
Thus, the magnitude of the car’s acceleration is
and
a = 2ar2 + au2 = 2(- 2)2 + 52 = 5.39 m>s2
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or transmission in any form or by any means, electronic, mechanical,
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12–165.
If a particle moves along a path such that r = 12 cos t2 ft
and u = 1t>22 rad, where t is in seconds, plot the path
r = f1u2 and determine the particle’s radial and transverse
components of velocity and acceleration.
SOLUTION
r = 2 cos t
t
2
u =
#
r = - 2 sin t
#
1
u =
2
$
r = - 2 cos t
$
u = 0
Ans.
#
1
vu = ru = (2 cos t ) a b = cos t
2
Ans.
#2
1 2
5
$
ar = r - ru = -2 cos t - (2cos t) a b = - cos t
2
2
Ans.
$
1
##
au = ru + 2ru = 2 cos t102 + 21- 2 sin t2 a b = - 2 sin t
2
Ans.
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#
vr = r = - 2 sin t
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or transmission in any form or by any means, electronic, mechanical,
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12–166.
If a particle’s position is described by the polar coordinates
r = 12 sin 2u2 m and u = 14t2 rad, where t is in seconds,
determine the radial and transverse components of its
velocity and acceleration when t = 1 s.
SOLUTION
When t = 1 s,
u = 4t = 4
#
u = 4
$
u = 0
r = 2 sin 2u = 1.9787
#
#
r = 4 cos 2u u = -2.3280
$
#
$
r = -8 sin 2u(u)2 + 8 cos 2u u = -126.638
#
vr = r = -2.33 m>s
#
vu = ru = 1.9787(4) = 7.91 m>s
#
$
ar = r - r(u)2 = -126.638 - (1.9787)(4)2 = - 158 m>s2
$
##
au = ru + 2 ru = 1.9787(0) + 2( - 2.3280)(4) = - 18.6 m>s2
Web)
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or
Ans.
12–167.
The car travels along the circular curve having a radius
r = # 400 ft. At the instant shown, its angular rate of rotation
is
$ u = 0.025 rad>s, which is decreasing at the rate
u = - 0.008 rad>s2. Determine the radial and transverse
components of the car’s velocity and acceleration at this
instant and sketch these components on the curve.
r
.
u
SOLUTION
r = 400
#
u = 0.025
#
r = 0
$
r = 0
u = -0.008
#
vr = r = 0
#
vu = ru = 400(0.025) = 10 ft>s
#
$
ar = r - r u2 = 0 - 400(0.025)2 = - 0.25 ft>s2
$
# #
au = r u + 2 r u = 400(- 0.008) + 0 = -3.20 ft>s2
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photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
400 ft
*12–168.
The car travels along the circular curve of radius r = 400 ft
with a constant speed
of v = 30 ft>s. Determine the angular
#
rate of rotation u of the radial line r and the magnitude of
the car’s acceleration.
r
.
u
SOLUTION
r = 400 ft
#
vr = r = 0
#
r = 0
$
r = 0
#
#
vu = r u = 400 a u b
# 2
v = 6(0)2 + a 400 u b = 30
#
u = 0.075 rad>s
$
u = 0
#
$
ar = r - r u2 = 0 - 400(0.075)2 = - 2.25 ft>s2
$
# #
au = r u + 2 r u = 400(0) + 2(0)(0.075) = 0
laws
or
Ans.
a = 2( -2.25)2 + (0)2 = 2.25 ft>s2
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400 ft
12–169.
The time rate of change of acceleration is referred to as the
jerk, which is often used as a means of measuring passenger
#
discomfort. Calculate this vector, a, in terms of its
cylindrical components, using Eq. 12–32.
SOLUTION
$
$ #
##
$
a = a r -ru 2 bur + a ru + 2ru buu + z uz
#$
# #
$#
$
###
###
$
##
$
#$
$#
#$
## #
$#
a = ar - ru2 - 2ruu b ur + a r - ru2 bu r + a r u + ru + 2ru + 2r u b uu + a ru + 2ru buu + z uz + zu z
#
#
#
But, ur = uuu uu = - uur
#
uz = 0
Substituting and combining terms yields
#$
#
$#
#
###
#$
$#
###
#
a = a r - 3ru 2 - 3ruu b ur + a 3r u + ru + 3ru - ru3 b uu + az buz
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12–170.
A particle is moving along a circular path having a radius of
6 in. such that its position as a function of time is given by
u = sin 3t, where u is in radians, the argument for the sine are
in radians , and t is in seconds. Determine the acceleration of
the particle at u = 30°. The particle starts from rest at
u = 0°.
SOLUTION
r = 6 in.,
#
r = 0,
$
r = 0
u = sin 3t
#
u = 3 cos 3t
$
u = -9 sin 3t
At u = 30°,
30°
p = sin 3t
180°
laws
or
t = 10.525 s
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Thus,
#
u = 2.5559 rad>s
$
u = -4.7124 rad>s2
#
$
ar = r - r u2 = 0 - 6(2.5559)2 = -39.196
$
##
au = r u + 2ru = 6( - 4.7124) + 0 = - 28.274
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a = 2( - 39.196)2 + ( - 28.274)2 = 48.3 in.>s2
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Ans.
12–171.
The slotted link is pinned
# at O, and as a result of the
constant angular velocity u = 3 rad>s it drives the peg P for
a short distance along the spiral guide r = 10.4u2 m, where
u is in radians. Determine the radial and transverse
components of the velocity and acceleration of P at the
instant u = p>3 rad.
0.5 m
P
r
·
u
r
3 rad/s
SOLUTION
#
u = 3 rad>s
At u =
p
,
3
u
r = 0.4 u
#
#
r = 0.4 u
$
$
r = 0.4 u
O
r = 0.4189
#
r = 0.4(3) = 1.20
$
r = 0.4(0) = 0
#
v = r = 1.20 m>s
#
vu = r u = 0.4189(3) = 1.26 m>s
#
$
ar = r - ru2 = 0 - 0.4189(3)2 = - 3.77 m>s2
$
##
au = r u + 2ru = 0 + 2(1.20)(3) = 7.20 m>s2
laws
or
Ans.
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
0.4 u
12–172.
Solve Prob. # 12–171
if the slotted
link has . an angular
#
#
acceleration u = 8 rad>s 2 when u = 3 rad>s at u = p>3 rad.
SOLUTION
#
u = 3 rad/s
u =
0.5 m
r = 0.4 u
#
#
r = 0.4 u
$
$
r = 0.4 u
r
·
u
p
3
O
r = 0.4189
#
r = 1.20
$
r = 0.4(8) = 3.20
laws
or
#
vr = r = 1.20 m>s
#
vu = r u = 0.4189(3) = 1.26 m>s
#
$
ar = r - ru2 = 3.20 - 0.4189(3)2 = - 0.570 m>s2
$
##
au = r u + 2 ru = 0.4189(8) + 2(1.20)(3) = 10.6 m>s2
teaching
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r
3 rad/s
u
#
u = 3
$
u = 8
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
P
0.4 u
12–173.
The slotted link is pinned
# at O, and as a result of the
constant angular velocity u = 3 rad>s it drives the peg P for
a short distance along the spiral guide r = 10.4u2 m, where
u is in radians. Determine the velocity and acceleration of
the particle at the instant it leaves the slot in the link, i.e.,
when r = 0.5 m.
0.5 m
P
r
·
u
SOLUTION
u
r = 0.4 u
#
#
r = 0.4 u
$
$
r = 0.4 u
#
u = 3
$
u = 0
O
At r = 0.5 m,
laws
or
0.5
= 1.25 rad
0.4
u =
in
teaching
Web)
#
r = 1.20
Dissemination
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$
r = 0
permitted.
#
vr = r = 1.20 m>s
#
vu = r u = 0.5(3) = 1.50 m>s
#
$
ar = r - r(u)2 = 0 - 0.5(3)2 = - 4.50 m>s2
$
##
au = ru + 2ru = 0 + 2(1.20)(3) = 7.20 m>s2
World
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r
3 rad/s
Ans.
0.4u
12–174.
A particle moves in the x–y plane such that its position is
defined by r = 52ti + 4t2j6 ft, where t is in seconds.
Determine the radial and transverse components of the
particle’s velocity and acceleration when t = 2 s.
SOLUTION
r = 2ti + 4 t2j|t = 2 = 4i + 16j
v = 2i + 8 tj|t = 2 = 2i + 16j
a = 8j
u = tan - 1 a
16
b = 75.964°
4
v = 2(2)2 + (16)2 = 16.1245 ft>s
16
b = 82.875°
2
or
f = tan - 1 a
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laws
a = 8 ft>s2
Wide
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f - u = 6.9112°
instructors
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vu = 16.1245 sin 6.9112° = 1.94 ft>s
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student
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by
and
d = 90° - u = 14.036°
Ans.
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ar = 8 cos 14.036° = 7.76 ft>s2
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au = 8 sin 14.036° = 1.94 ft>s2
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Ans.
permitted.
Dissemination
vr = 16.1245 cos 6.9112° = 16.0 ft>s
12–175.
A particle P moves along the spiral path r = 110>u2 ft,
where u is in radians. If it maintains a constant speed of
v = 20 ft>s, determine the magnitudes vr and vu as functions
of u and evaluate each at u = 1 rad.
v
P
r
r = 10
––
θ
SOLUTION
10
u
r =
#
10 #
r = - a 2 bu
u
102 # 2
102 #
b u + a 2 b u2
4
u
u
(20)2 = a
#
102
b (1 + u2)u2
u4
laws
(20)2 = a
or
# 2
#
Since v2 = r2 + a ru b
#
Thus, u =
in
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2u2
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21 + u2
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#
2u2
10
20u
vu = ru = a b £
≥ =
2
u
21 + u
21 + u2
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When u = 1 rad
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vr = ¢ -
vu =
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20
2
= 14.1 ft s
Ans.
not
instructors
States
World
10
20
2u2
#
vr = r = - a 2 b £
≥ = 2
u
21 + u
21 + u2
θ
*12–176.
The driver of the car maintains a constant speed of 40 m>s.
Determine the angular velocity of the camera tracking the
car when u = 15°.
r
(100 cos 2u) m
SOLUTION
u
Time Derivatives:
r = 100 cos 2u
#
#
r = (-200 (sin 2 u)u ) m>s
At u = 15°,
r u = 15° = 100 cos 30° = 86.60 m
#
#
#
r u = 15° = -200 sin 30°u = - 100u m>s
Velocity: Referring to Fig. a, vr = - 40 cos f and vu = 40 sin f.
laws
or
#
vr = r
teaching
Web)
#
-40 cos f = - 100u
Dissemination
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(1)
States
World
permitted.
and
on
learning.
is
of
the
not
instructors
#
vu = ru
and
use
United
#
40 sin f = 86.60u
(including
for
work
student
the
by
(2)
is
work
solely
of
protected
the
Solving Eqs. (1) and (2) yields
integrity
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and
work
this
assessing
f = 40.89°
#
u = 0.3024 rad>s = 0.302 rad>s
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This
provided
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–177.
When u = 15°, the car has a speed of 50 m>s which is
increasing at 6 m>s2. Determine the angular velocity of the
camera tracking the car at this instant.
r
(100 cos 2u) m
SOLUTION
u
Time Derivatives:
r = 100 cos 2u
#
#
r = ( - 200 (sin 2u) u ) m>s
$
#
$
r = -200 C (sin 2u) u + 2 (cos 2u) u 2 D m>s2
At u = 15°,
r u = 15° = 100 cos 30° = 86.60 m
#
#
#
r u = 15° = -200 sin 30°u = - 100u m>s
$
#
#
r u = 15° = -200 C sin 30° u + 2 cos 30° u2 D =
$
#
laws
or
A -100u - 346.41u2 B m>s2
Dissemination
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in
teaching
Web)
Velocity: Referring to Fig. a, vr = - 50 cos f and vu = 50 sin f. Thus,
#
vr = r
#
-50 cos f = -100u
learning.
is
of
the
not
instructors
States
World
permitted.
(1)
use
United
on
and
work
student
the
by
and
#
vu = ru
(2)
assessing
is
work
solely
of
protected
the
(including
for
#
50 sin f = 86.60u
part
the
is
This
any
courses
Ans.
sale
will
their
destroy
of
and
f = 40.89°
#
u = 0.378 rad>s
provided
integrity
of
and
work
this
Solving Eqs. (1) and (2) yields
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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12–178.
The small washer slides down the cord OA. When it is at the
midpoint, its speed is 200 mm>s and its acceleration is
10 mm>s2. Express the velocity and acceleration of the
washer at this point in terms of its cylindrical components.
z
A
v, a
700 mm
z
O
SOLUTION
u
2
2
x
OB = 2(400) + (300) = 500 mm
2
vr = (200) a
2
500
b = 116 mm>s
860.23
vu = 0
vz = (200) a
700
b = 163 mm>s
860.23
Thus, v = { -116ur - 163uz} mm>s
Ans.
or
500
b = 5.81
860.23
laws
ar = 10 a
in
teaching
Web)
au = 0
permitted.
Dissemination
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copyright
700
b = 8.14
860.23
States
World
az = 10 a
instructors
Thus, a = { -5.81ur - 8.14uz} mm>s2
United
on
learning.
is
of
the
not
Ans.
and
use
by
work
student
the
the
(including
for
work
solely
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protected
this
assessing
is
integrity
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and
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provided
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the
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This
destroy
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© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
r
y
300 mm
OA = 2(400) + (300) + (700) = 860.23 mm
2
400 mm
12–179.
A block moves outward along the slot in the platform with
#
a speed of r = 14t2 m>s, where t is in seconds. The platform
rotates at a constant rate of 6 rad/s. If the block starts from
rest at the center, determine the magnitudes of its velocity
and acceleration when t = 1 s.
·
θ = 6 rad/s
r
θ
SOLUTION
#
r = 4t|t = 1 = 4
#
$
u = 6
u = 0
1
L0
$
r = 4
1
dr =
L0
4t dt
r = 2t2 D 10 = 2 m
#
#
v = 3 A r B 2 + A ru B 2 = 2 (4)2 + [2(6)]2 = 12.6 m>s
#
$
$
##
a =
r - ru2 2 + ru + 2 ru 2 =
[4 - 2(6)2 ]2 + [0 + 2(4)(6)]2
Ans.
Ans.
= 83.2 m s
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This
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this
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is
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the
(including
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student
the
by
and
use
United
on
learning.
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of
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Dissemination
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Web)
laws
or
2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*12–180.
Pin P is constrained to move along the curve defined by the
lemniscate r = (4 sin 2u) ft. If the slotted arm OA rotates
counterclockwise
with a constant angular velocity of
#
u = 1.5 rad>s, determine the magnitudes of the velocity and
acceleration of peg P when u = 60°.
A
r
P
O
SOLUTION
Time Derivatives:
r = 4 sin 2u
#
u = 1.5 rad>s
$
u = 0
#
#
r = (8(cos 2u)u ) ft>s
#
$
$
r = 8[(cos 2u)u - 2 sin 2u(u )2] ft>s2
When u = 60°,
r|u = 60° = 4 sin 120° = 3.464 ft
laws
or
#
r |u = 60° = 8 cos 120°(1.5) = - 6 ft>s
in
teaching
Web)
$
r|u = 60° = 8[0 - 2 sin 120°(1.52)] = - 31.18 ft>s2
Dissemination
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Velocity:
World
permitted.
#
vu = ru = 3.464(1.5) = 5.196 ft>s
of
the
not
instructors
States
#
vr = r = - 6 ft>s
and
use
United
on
learning.
is
Thus, the magnitude of the peg’s velocity is
by
v = 2vr2 + vu2 = 2( - 6)2 + 5.1962 = 7.94 ft>s
for
work
student
the
Ans.
of
protected
the
(including
Acceleration:
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
#
$
ar = r - ru2 = - 31.18 - 3.464(1.52) = - 38.97 ft>s2
$
##
au = ru + 2ru = 0 + 2( -6)(1.5) = - 18 ft>s2
and
any
courses
Thus, the magnitude of the peg’s acceleration is
sale
will
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destroy
of
a = 2ar2 + au2 = 2( -38.97)2 + ( -18)2 = 42.9 ft>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Ans.
u
r ⫽ (4 sin 2u) ft
12–181.
Pin P is constrained to move along the curve defined by the
lemniscate r = (4 sin 2u) ft. If the angular position of the
slotted arm OA is defined by u = 13t 3>22 rad, determine
the magnitudes of the velocity and acceleration of the pin P
when u = 60°.
A
r
P
O
SOLUTION
Time Derivatives:
r = 4 sin 2u
#
#
r = (8(cos 2u)u ) ft>s
#
#
$
r = 8[(cos 2u)u - 2 (sin 2u)u 2] ft>s2
When u = 60° =
p
rad,
3
p
= 3t 3>2
3
teaching
Web)
laws
or
t = 0.4958 s
States
World
permitted.
Dissemination
Wide
copyright
in
Thus, the angular velocity and angular acceleration of arm OA when
p
u = rad (t = 0.4958 s) are
3
by
and
use
United
on
learning.
is
of
the
not
instructors
#
9
u = t1>2 `
= 3.168 rad>s
2
t = 0.4958 s
of
protected
the
(including
for
work
student
the
$ 9
u = t1>2 `
= 3.196 rad>s2
4
t = 0.4958 s
this
assessing
is
work
solely
Thus,
the
is
This
provided
integrity
of
and
work
r|u = 60° = 4 sin 120° = 3.464 ft
and
any
courses
part
#
r|u = 60° = 8 cos 120°(3.168) = - 12.67 ft>s
sale
will
their
destroy
of
$
r|u = 60° = 8[cos 120°(3.196) - 2 sin 120°(3.1682)] = - 151.89 ft>s2
Velocity:
#
vr = r = - 12.67 ft>s
#
vu = ru = 3.464(3.168) = 10.98 ft>s
Thus, the magnitude of the peg’s velocity is
v = 2vr2 + vu2 = 2(- 12.67)2 + 10.982 = 16.8 ft>s
Ans.
Acceleration:
#
$
ar = r - ru2 = - 151.89 - 3.464(3.1682) = - 186.67 ft>s2
$
##
au = ru + 2ru = 3.464(3.196) + 2( - 12.67)(3.168) = - 69.24 ft>s2
Thus, the magnitude of the peg’s acceleration is
a = 2ar2 + au2 = 2(- 186.67)2 + (- 69.24)2 = 199 ft>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
u
r ⫽ (4 sin 2u) ft
12–182.
A cameraman standing at A is following the movement of a
race car, B, which is traveling around a curved track at
# a
constant speed of 30 m>s. Determine the angular rate u at
which the man must turn in order to keep the camera
directed on the car at the instant u = 30°.
vB
BB
r r
2020
mm
AA
r = 2(20) cos u = 40 cos u
#
#
r = -(40 sin u ) u
#
#
v = r u r + r u uu
#
#
v = 2(r)2 + (r u)2
#
#
(30)2 = ( - 40 sin u)2 (u)2 + (40 cos u)2 (u)2
#
(30)2 = (40)2 [sin2 u + cos2 u](u)2
#
30
u =
= 0.75 rad>s
40
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This
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is
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the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
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in
teaching
Web)
laws
or
Ans.
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or transmission in any form or by any means, electronic, mechanical,
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2020
mm
uu
2020
mm
SOLUTION
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
30 m/s
2020
mm
12–183.
The slotted arm AB drives pin C through the spiral groove
described by
# the equation r = a u. If the angular velocity is
constant at u, determine the radial and transverse components
of velocity and acceleration of the pin.
B
C
r
SOLUTION
#
$
Time Derivatives: Since u is constant, then u = 0.
#
$
#
$
r = au
r = au
r = au = 0
u
A
Velocity: Applying Eq. 12–25, we have
#
#
vr = r = au
#
#
vu = ru = auu
Ans.
Ans.
Acceleration: Applying Eq. 12–29, we have
#
#
#
$
ar = r - ru2 = 0 - auu2 = - auu2
$
# #
#
# #
au = ru + 2r u = 0 + 2(au)(u) = 2au2
or
Ans.
sale
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is
This
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integrity
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this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
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in
teaching
Web)
laws
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*12–184.
The slotted arm AB drives pin C through the spiral groove
described by the equation r = (1.5 u) ft, where u is in
radians. If the arm starts from # rest when u = 60° and is
driven at an angular velocity of u = (4t) rad>s, where t is in
seconds, determine the radial and transverse components of
velocity and acceleration of the pin C when t = 1 s.
B
C
r
SOLUTION
#
$
Time Derivatives: Here, u = 4 t and u = 4 rad>s2.
#
$
$
#
r = 1.5 u = 1.5 (4) = 6 ft>s2
r = 1.5 u
r = 1.5 u = 1.5(4t) = 6t
u
Velocity: Integrate the angular rate,
1
Then, r = b (6t2 + p) r ft.
2
At
t
du =
Lp3
r =
t = 1 s,
A
1
(6t2 + p) rad .
3
4tdt, we have u =
L0
u
1
C 6(12) + p D = 4.571 ft, r = 6(1) = 6.00 ft>s #
2
laws
or
#
and u = 4(1) = 4 rad>s. Applying Eq. 12–25, we have
#
vr = r = 6.00 ft>s
#
vu = ru = 4.571 (4) = 18.3 ft>s
permitted.
Dissemination
copyright
Ans.
Wide
in
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Web)
Ans.
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This
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solely
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the
(including
for
work
student
the
by
and
use
United
#
$
ar = r - ru2 = 6 - 4.571 A 42 B = - 67.1 ft>s2
$
# #
au = r u + 2ru = 4.571(4) + 2(6) (4) = 66.3 ft>s2
is
of
the
not
instructors
States
World
Acceleration: Applying Eq. 12–29, we have
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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Ans.
12–185.
If the slotted arm AB rotates# counterclockwise with a
constant angular velocity of u = 2 rad>s, determine the
magnitudes of the velocity and acceleration of peg P at
u = 30°. The peg is constrained to move in the slots of the
fixed bar CD and rotating bar AB.
D
P
r ⫽ (4 sec u) ft
u
A
C
SOLUTION
4 ft
Time Derivatives:
r = 4 sec u
#
u = 2 rad>s
$
u = 0
#
#
r = (4 secu(tanu)u ) ft>s
#
#
#
$
$
r = 4 [secu(tanu)u + u (sec u (sec2u)u + tan u secu(tan u)u )]
#
#
= 4[secu(tanu)u + u2(sec3u + tan2u secu)] ft>s2
When u = 30°,
laws
or
r|u = 30° = 4 sec 30° = 4.619 ft
in
teaching
Web)
#
r |u = 30° = (4 sec30° tan30°)(2) = 5.333 ft>s
Dissemination
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copyright
$
r |u = 30° = 4[0 + 2 2(sec3 30° + tan2 30° sec 30°)] = 30.79 ft>s2
not
instructors
States
World
permitted.
Velocity:
by
and
use
United
on
learning.
is
of
the
#
vu = ru = 4.619(2) = 9.238 ft>s
#
vr = r = 5.333 ft>s
the
Ans.
work
solely
of
protected
v = 2vr2 + vu2 = 25.3332 + 9.2382 = 10.7 ft>s
(including
for
work
student
the
Thus, the magnitude of the peg’s velocity is
this
assessing
is
Acceleration:
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
#
$
ar = r - ru2 = 30.79 - 4.619(2 2) = 12.32 ft>s2
$
##
au = ru + 2ru = 0 + 2(5.333)(2) = 21.23 ft>s2
sale
will
their
destroy
Thus, the magnitude of the peg’s acceleration is
a = 2ar2 + au2 = 212.32 2 + 21.232 = 24.6 ft>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
12–186.
The peg is constrained to move in the slots of the fixed bar
CD and rotating bar AB. When u = 30°, the angular
velocity
and angular
acceleration of arm AB are
$
#
u = 2 rad>s and u = 3 rad>s2 , respectively. Determine
the magnitudes of the velocity and acceleration of the peg
P at this instant.
D
P
r ⫽ (4 sec u) ft
u
A
C
SOLUTION
4 ft
Time Derivatives:
r = 4 sec u
#
u = 2 rad>s
$
u = 3 rad>s2
#
#
r = (4 secu(tanu)u ) ft>s
#
#
#
$
$
r = 4[secu(tanu)u + u (secusec2uu + tanu secu(tanu)u )]
#
$
= 4[secu(tanu)u + u 2(sec3u° + tan2u°secu°)] ft>s2
When u = 30°,
laws
or
r|u = 30° = 4 sec 30° = 4.619 ft
in
teaching
Web)
#
r|u = 30° = (4 sec 30° tan 30°)(2) = 5.333 ft>s
Dissemination
Wide
copyright
$
r|u = 30° = 4[(sec 30° tan 30°)(3) + 22(sec3 30° + tan2 30° sec 30°)] = 38.79 ft>s2
not
instructors
States
World
permitted.
Velocity:
by
and
use
United
on
learning.
is
of
the
#
vu = ru = 4.619(2) = 9.238 ft>s
#
vr = r = 5.333 ft>s
the
Ans.
work
solely
of
protected
v = 2vr2 + vu2 = 25.3332 + 9.2382 = 10.7 ft>s
(including
for
work
student
the
Thus, the magnitude of the peg’s velocity is
this
assessing
is
Acceleration:
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
#
$
ar = r - ru2 = 38.79 - 4.619(22) = 20.32 ft>s2
$
##
au = ru + 2ru = 4.619(3) + 2(5.333)(2) = 35.19 ft>s2
sale
will
their
destroy
Thus, the magnitude of the peg’s acceleration is
a = 2ar2 + au2 = 220.322 + 35.192 = 40.6 ft>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
12–187.
If the circular plate
# rotates clockwise with a constant
angular velocity of u = 1.5 rad>s, determine the magnitudes
of the velocity and acceleration of the follower rod AB
when u = 2>3p rad.
u
A
B
1/2
r ⫽ (10 ⫹ 50 u
SOLUTION
Time Derivaties:
r = A 10 + 50u1>2 B mm
#
#
r = 25u-1>2u mm>s
#
#
1
$
r = 25 c u-1>2u - u-3>2u2 d mm>s2
2
2p
rad,
3
or
2p 1>2
b d = 82.36 mm
3
Web)
3
= c 10 + 50a
laws
2p
in
r|u =
teaching
When u =
the
on
learning.
is
of
United
and
use
for
work
student
the
by
1 2p -3>2
$ 2p
r|u = = 25 c 0 - a
b
11.522 d = - 9.279 mm>s2
3
2 3
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
2p -1>2
# 2p
r|u = = 25 a b
11.52 = 25.91 mm>s
3
3
is
work
solely
of
protected
the
(including
Velocity: The radial component gives the rod’s velocity.
assessing
#
vr = r = 25.9 mm>s
provided
integrity
of
and
work
this
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
Acceleration: The radial component gives the rod’s acceleration.
#
$
ar = r - ru2 = - 9.279 - 82.36 A 1.52 B = - 195 mm>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Ans.
) mm
r
*12–188.
When u = 2>3p rad, the angular velocity
and angular
#
u
=
1.5
rad>s and
acceleration
of
the
circular
plate
are
$
u = 3 rad>s2, respectively. Determine the magnitudes of the
velocity and acceleration of the rod AB at this instant.
u
A
B
1/2
r ⫽ (10 ⫹ 50 u
SOLUTION
Time Derivatives:
r = (10 + 50u1>2) mm
#
#
r = 25u - 1>2u mm>s
#2
$
1
$
2
r = 25 cu - 1>2 u - u - 3>2u d mm>s
2
or
2p
rad,
3
laws
When u =
teaching
Web)
2p 1>2
b d = 82.36 mm
3
in
Wide
= c 10 + 50a
Dissemination
2p
3
copyright
r|u =
of
protected
the
(including
for
work
student
the
by
2p - 1>2
1 2p - 3>2
$ 2p
(3) - a b
(1.52) d = 42.55 mm>s2
r|u = = 25 c a b
3
3
2 3
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
2p - 1>2
# 2p
r|u = = 25 a b
(1.5) = 25.91 mm>s
3
3
this
assessing
is
work
solely
For the rod,
Ans.
part
the
is
This
provided
integrity
of
and
work
#
v = r = 25.9 mm>s
any
courses
Ans.
sale
will
their
destroy
of
and
$
a = r = 42.5 mm>s2
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
) mm
r
12–189.
The box slides down the helical ramp with a constant speed
of v = 2 m>s. Determine the magnitude of its acceleration.
The ramp descends a vertical distance of 1 m for every full
revolution. The mean radius of the ramp is r = 0.5 m.
0.5 m
SOLUTION
Velocity: The inclination angle of the ramp is f = tan-1
L
1
= tan-1 B
R = 17.66°.
2pr
2p(0.5)
Thus, from Fig. a, vu = 2 cos 17.66° = 1.906 m>s and vz = 2 sin 17.66° = 0.6066 m>s. Thus,
#
vu = ru
#
1.906 = 0.5u
#
u = 3.812 rad>s
Wide
copyright
in
teaching
Web)
laws
or
#
$
#
$
Acceleration: Since r = 0.5 m is constant, r = r = 0. Also, u is constant, then u = 0.
Using the above results,
#
$
ar = r - r u2 = 0 - 0.5(3.812)2 = - 7.264 m>s2
$
##
au = r u + 2ru = 0.5(0) + 2(0)(3.812) = 0
States
World
permitted.
Dissemination
Since vz is constant az = 0. Thus, the magnitude of the box’s acceleration is
instructors
( -7.264)2 + 02 + 02 = 7.26 m>s2
the
not
Ans.
sale
will
their
destroy
of
and
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courses
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the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
is
work
student
the
by
and
use
United
on
learning.
ar 2 + au 2 + az 2 =
of
a =
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–190.
The box slides down the helical ramp which is defined by
r = 0.5 m, u = (0.5t3) rad, and z = (2 – 0.2t2) m, where t is
in seconds. Determine the magnitudes of the velocity and
acceleration of the box at the instant u = 2p rad.
0.5 m
SOLUTION
Time Derivatives:
r = 0.5 m
#
$
r = r = 0
#
u = A 1.5t2 B rad>s
$
u = (3t) rad>s2
z = 2 - 0.2t2
$
z = - 0.4 m>s2
#
z = ( -0.4t) m>s
or
When u = 2p rad,
t = 2.325 s
teaching
Web)
laws
2p = 0.5t3
in
Thus,
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
#
u t = 2.325 s = 1.5(2.325)2 = 8.108 rad>s
$
u t = 2.325 s = 3(2.325) = 6.975 rad>s2
the
by
and
use
United
on
#
z t = 2.325 s = - 0.4(2.325) = - 0.92996 m>s
work
solely
of
protected
the
(including
for
work
student
$
z t = 2.325 s = - 0.4 m>s2
destroy
will
their
sale
#
vz = z = - 0.92996 m>s
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
#
vr = r = 0
#
vu = ru = 0.5(8.108) = 4.05385 m>s
this
assessing
is
Velocity:
Thus, the magnitude of the box’s velocity is
v = 2vr 2 + vu 2 + vz 2 = 202 + 4.053852 + ( -0.92996)2 = 4.16 m>s
Ans.
Acceleration:
#
$
a r = r - r u2 = 0 - 0.5(8.108)2 = - 32.867 m>s2
$
# #
a u = r u + 2r u = 0.5(6.975) + 2(0)(8.108)2 = 3.487 m>s2
$
az = z = - 0.4 m>s2
Thus, the magnitude of the box’s acceleration is
a =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
ar 2 + au 2 + az 2 =
( -32.867)2 + 3.4872 + ( -0.4)2 = 33.1 m>s2 Ans.
12–191.
For a short distance the train travels along a track having
the shape of a spiral, r = 11000>u2 m, where u is in radians.
If it maintains a constant speed v = 20 m>s, determine the
radial and transverse components of its velocity when
u = 19p>42 rad.
r
u
r
SOLUTION
1000
u
r =
1000 #
#
r = - 2 u
u
Since
#
#
v2 = (r)2 + (r u)2
or
#
(1 + u2)(u)2
Web)
u4
laws
(1000)2
in
(20)2 =
(1000)2 # 2
(1000)2 # 2
(u) +
(u)
4
u
u2
teaching
(20)2 =
Dissemination
Wide
copyright
Thus,
permitted.
#
u =
not
instructors
States
World
0.02u2
by
and
use
United
on
learning.
is
of
the
21 + u2
work
student
the
9p
4
(including
for
At u =
integrity
of
and
provided
any
courses
part
the
is
This
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
sale
#
1000
(0.140) = 19.8 m>s
vu = r u =
(9p/4)
will
their
destroy
of
#
vr = r = -2.80 m>s
and
-1000
#
r =
(0.140) = -2.80
(9p/4)2
work
this
assessing
is
work
solely
of
protected
the
#
u = 0.140
Ans.
1000
u
12–192.
For a short distance the train travels along a track having
the shape of a spiral, r = 11000>u2 m, where u is in radians.
#
If the angular rate is constant, u = 0.2 rad>s, determine the
radial and transverse components of its velocity and
acceleration when u = 19p>42 rad.
r
u
r
SOLUTION
#
u = 0.2
$
u = 0
r =
1000
u
#
#
r = -1000(u - 2) u
#
$
$
r = 2000(u - 3)(u)2 - 1000(u - 2) u
9p
4
or
When u =
teaching
Web)
laws
r = 141.477
Wide
copyright
in
#
r = -4.002812
States
World
permitted.
Dissemination
$
r = 0.226513
not
instructors
#
vr = r = -4.00 m>s
#
vu = r u = 141.477(0.2) = 28.3 m>s
#
$
ar = r - r(u)2 = 0.226513 - 141.477(0.2)2 = -5.43 m>s2
$
##
au = r u + 2ru = 0 + 2( -4.002812)(0.2) = - 1.60 m>s2
and
use
United
on
learning.
is
of
the
Ans.
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
1000
u
12–193.
A particle moves along an Archimedean
spiral r = (8u) ft,
#
where u is given in radians. If u = 4 rad>s (constant),
determine the radial and transverse components of the
particle’s velocity and acceleration at the instant
u = p>2 rad. Sketch the curve and show the components on
the curve.
y
r
(8 u) ft
r
u
SOLUTION
#
$
Time Derivatives: Since u is constant, u = 0.
p
r = 8u = 8a b = 4p ft
2
x
#
#
r = 8 u = 8(4) = 32.0 ft>s
$
$
r = 8u = 0
Velocity: Applying Eq. 12–25, we have
#
v r = r = 32.0 ft>s
#
vu = ru = 4p (4) = 50.3 ft>s
Ans.
Ans.
or
Acceleration: Applying Eq. 12–29, we have
#
$
ar = r - ru2 = 0 - 4p A 42 B = - 201 ft>s2
$
# #
au = ru + 2r u = 0 + 2(32.0)(4) = 256 ft>s2
teaching
Web)
laws
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–194.
Solve Prob. $12–193 if the particle
has an
angular
#
acceleration u = 5 rad>s2 when u = 4 rad>s at u = p 2 rad.
y
r
(8 u) ft
r
u
SOLUTION
x
Time Derivatives: Here,
p
r = 8u = 8 a b = 4p ft
2
$
$
r = 8 u = 8(5) = 40 ft>s2
#
#
r = 8 u = 8(4) = 32.0 ft>s
Velocity: Applying Eq. 12–25, we have
#
vr = r = 32.0 ft>s
#
vu = r u = 4p(4) = 50.3 ft>s
Ans.
Ans.
laws
or
Acceleration: Applying Eq. 12–29, we have
#
$
ar = r - r u2 = 40 - 4p A 42 B = - 161 ft>s2
$
# #
au = r u + 2r u = 4p(5) + 2(32.0)(4) = 319 ft>s2
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
copyright
Ans.
Wide
in
teaching
Web)
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–195.
The arm of the robot has a length of r = 3 ft grip
A moves along the path z = 13 sin 4u2 ft, where
u is in radians. If u = 10.5t2 rad, where t is in
seconds, determine the magnitudes of the grip’s
velocity and acceleration when t = 3 s.
A
z
r
u
SOLUTION
u = 0.5 t
#
u = 0.5
$
u = 0
r = 3
z = 3 sin 2t
#
r = 0
#
z = 6 cos 2t
$
r = 0
$
z = - 12 sin 2t
At t = 3 s,
z = -0.8382
#
z = 5.761
or
$
z = 3.353
teaching
Web)
laws
vr = 0
Wide
copyright
in
vu = 3(0.5) = 1.5
Ans.
United
on
learning.
is
of
the
v = 2(0)2 + (1.5)2 + (5.761)2 = 5.95 ft>s
student
the
by
and
use
ar = 0 - 3(0.5)2 = -0.75
protected
the
(including
for
work
au = 0 + 0 = 0
this
assessing
is
work
solely
of
az = 3.353
sale
will
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destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
a = 2(- 0.75)2 + (0)2 + (3.353)2 = 3.44 ft>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
not
instructors
States
World
permitted.
Dissemination
vz = 5.761
*12–196.
For a short time the arm of the robot is extending at a
#
constant rate such that r = 1.5 ft>s when r = 3 ft,
z = 14t22 ft, and u = 0.5t rad, where t is in seconds.
Determine the magnitudes of the velocity and acceleration
of the grip A when t = 3 s.
A
z
r
u
SOLUTION
u = 0.5 t rad
#
u = 0.5 rad>s
$
u = 0
r = 3 ft
z = 4 t2 ft
#
r = 1.5 ft>s
#
z = 8 t ft>s
$
r = 0
$
z = 8 ft>s2
At t = 3 s,
r = 3
z = 36
#
r = 1.5
#
z = 24
$
r = 0
$
z = 8
or
u = 1.5
#
u = 0.5
$
u = 0
teaching
Web)
laws
vr = 1.5
Wide
copyright
in
vu = 3(0.5) = 1.5
instructors
States
World
permitted.
Dissemination
vz = 24
Ans.
United
on
learning.
is
of
the
not
v = 2(1.5)2 + (1.5)2 + (24)2 = 24.1 ft>s
student
the
by
and
use
ar = 0 - 3(0.5)2 = - 0.75
protected
the
(including
for
work
au = 0 + 2(1.5)(0.5) = 1.5
this
assessing
is
work
solely
of
az = 8
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
a = 2( -0.75)2 + (1.5)2 + (8)2 = 8.17 ft>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Ans.
12–197.
The partial surface of the cam is that of a logarithmic spiral
r = (40e0.05u) mm, where u is in #radians. If the cam is rotating
at a constant angular rate of u = 4 rad>s, determine the
magnitudes of the velocity and acceleration of the follower
rod at the instant u = 30°.
u
SOLUTION
r = 40e0.05 u
#
#
r = 2e 0.05u u
·
u
# 2
$
$
r = 0.1e 0.0 5 u a u b + 2e 0.05 u u
p
6
u =
#
u = -4
$
u = 0
r = 40e 0.05A 6 B = 41.0610
or
p
teaching
Web)
laws
p
#
r = 2e 0.05A 6 B (-4) = -8.2122
Wide
copyright
in
p
$
r = 0.1e 0.05A 6 B (-4) 2 + 0 = 1.64244
of
the
not
instructors
States
World
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
permitted.
Dissemination
#
v = r = -8.2122 = 8.21 mm >s
#
$
a = r - r u2 = 1.642 44 - 41.0610(-4) 2 = - 665.33 = -665 mm>s 2
4 rad/s
r
40e0.05u
12–198.
Solve
if the cam has an angular# acceleration
$ Prob. 12–197,
2
of
u
=
2
rad>s
when
its angular velocity is u = 4 rad>s at
#
u = 30°.
u
SOLUTION
·
r = 40e0.05u
#
#
r = 2e0.05uu
u ⫽ 4 rad/s
#
$
$
r = 0.1e0.05u A u B 2 + 2e0.05uu
u =
p
p
$
r = 0.1e0.05A 6 B( -4)2 + 2e0.05A 6 B(- 2) = - 2.4637
#
v = r = 8.2122 = 8.21 mm>s
p
6
#
u = -4
$
u = -2
#
$
a = r - ru 2 = - 2.4637 - 41.0610(- 4)2 = - 659 mm>s2
or
r = 40e0.05A 6 B = 41.0610
teaching
Web)
laws
p
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
p
#
r = 2e0.05AA 6 B ( -4) = - 8.2122
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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r ⫽ 40e0.05u
12–199.
If the end of the cable at A is pulled down with a speed of
2 m>s, determine the speed at which block B rises.
D
C
2 m/s
A
SOLUTION
Position-Coordinate Equation: Datum is established at fixed pulley D. The
position of point A, block B and pulley C with respect to datum are sA, sB, and sC
respectively. Since the system consists of two cords, two position-coordinate
equations can be derived.
(sA - sC) + (sB - sC) + sB = l1
(1)
sB + sC = l2
(2)
B
Eliminating sC from Eqs. (1) and (2) yields
sA + 4sB = l1 = 2l2
laws
or
Time Derivative: Taking the time derivative of the above equation yields
Web)
(3)
Wide
copyright
in
teaching
vA + 4vB = 0
States
World
permitted.
Dissemination
Since vA = 2 m>s, from Eq. (3)
the
not
instructors
2 + 4vB = 0
learning.
is
of
(+ T)
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on
vB = - 0.5 m>s = 0.5 m>s c
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Ans.
*12–200.
The motor at C pulls in the cable with an acceleration
aC = (3t2) m>s2, where t is in seconds. The motor at D draws
in its cable at aD = 5 m>s2. If both motors start at the same
instant from rest when d = 3 m, determine (a) the time
needed for d = 0, and (b) the relative velocity of block A
with respect to block B when this occurs.
D
B
A
d=3m
SOLUTION
For A:
sA + (sA - sC) = l
2vA = vC
2aA = aC = - 3t2
aA = - 1.5t2 = 1.5t2
vA = 0.5t3
:
:
:
laws
or
sA = 0.125 t4
teaching
Web)
For B:
in
Wide
World
permitted.
Dissemination
;
instructors
States
vB = 5t
;
copyright
aB = 5 m>s2
the
not
;
and
use
United
on
learning.
is
of
sB = 2.5t2
for
work
student
the
by
Require sA + sB = d
is
work
solely
of
protected
the
(including
0.125t4 + 2.5t2 = 3
this
assessing
0.125u2 + 2.5u = 3
part
the
is
courses
any
Ans.
sale
will
their
destroy
of
t = 1.0656 = 1.07 s
and
The positive root is u = 1.1355. Thus,
This
provided
integrity
of
and
work
Set u = t2
vA = 0.5(1.0656)3 = 0.6050
vB = 5(1.0656) = 5.3281 m>s
vA = vB + vA>B
0.6050i = - 5.3281i + vA>B i
vA B = 5.93 m s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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:
Ans.
C
12–201.
The crate is being lifted up the inclined plane using the
motor M and the rope and pulley arrangement shown.
Determine the speed at which the cable must be taken up
by the motor in order to move the crate up the plane with a
constant speed of 4 ft>s.
B
A
C
SOLUTION
Position-Coordinate Equation: Datum is established at fixed pulley B. The position
of point P and crate A with respect to datum are sP and sA, respectively.
2sA + (sA - sP) = l
3sA - sP = 0
Time Derivative: Taking the time derivative of the above equation yields
3vA - vP = 0
(1)
Since vA = 4 ft>s, from Eq. [1]
or
3(4) - vP = 0
laws
vP = 12 ft>s
Ans.
sale
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is
This
provided
integrity
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assessing
is
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protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
(+ )
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
M
12–202.
Determine the time needed for the load at B to attain a
speed of 8 m>s, starting from rest, if the cable is drawn into
the motor with an acceleration of 0.2 m>s2.
A
vA
SOLUTION
4 sB + sA = l
B
4 nB = - vA
4 aB = - aA
4 aB = -0.2
aB = -0.05 m>s2
(+T)
vB = (vB)0 + aB t
-8 = 0 - (0.05)(t)
t = 160 s
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This
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the
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the
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and
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is
of
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permitted.
Dissemination
Wide
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in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
vB
12–203.
Determine the displacement of the log if the truck at C
pulls the cable 4 ft to the right.
B
SOLUTION
2sB + (sB - sC) = l
3sB - sC = l
3¢sB - ¢sC = 0
Since ¢sC = -4, then
3¢sB = -4
¢sB = -1.33 ft = 1.33 ft :
sale
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This
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the
(including
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the
by
and
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on
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of
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permitted.
Dissemination
Wide
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Web)
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or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
C
*12–204.
Determine the speed of cylinder A, if the rope is drawn
towards the motor M at a constant rate of 10 m>s.
SOLUTION
Position Coordinates: By referring to Fig. a, the length of the rope written in terms
of the position coordinates sA and sM is
A
3sA + sM = l
h
Time Derivative: Taking the time derivative of the above equation,
A+TB
3vA + vM = 0
Here, vM = 10 m>s. Thus,
3vA + 10 = 0
vA = -3.33 m>s = 3.33 m>s c
sale
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destroy
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and
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courses
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the
is
This
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the
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work
student
the
by
and
use
United
on
learning.
is
of
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States
World
permitted.
Dissemination
Wide
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in
teaching
Web)
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or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
M
12–205.
If the rope is drawn toward the motor M at a speed of
vM = (5t3>2) m>s, where t is in seconds, determine the speed
of cylinder A when t = 1 s.
SOLUTION
Position Coordinates: By referring to Fig. a, the length of the rope written in terms
of the position coordinates sA and sM is
A
3sA + sM = l
M
h
Time Derivative: Taking the time derivative of the above equation,
A+TB
3vA + vM = 0
Here, vM = A 5t3>2 B m>s. Thus,
3vA + 5t3>2 = 0
or
5
5
vA = ¢ - t3>2 ≤ m>s = ¢ t3>2 ≤ m>s 2
= 1.67 m>s
3
3
t=1 s
sale
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This
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the
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the
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and
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is
of
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not
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States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–206.
If the hydraulic cylinder H draws in rod BC at 2 ft>s,
determine the speed of slider A.
A
B
C
H
SOLUTION
2sH + sA = l
2vH = -vA
2(2) = -vA
vA = -4 ft>s = 4 ft>s ;
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
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this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–207.
If block A is moving downward with a speed of 4 ft>s while
C is moving up at 2 ft>s, determine the speed of block B.
SOLUTION
B
C
sA + 2sB + sC = l
A
vA + 2vB + vC = 0
4 + 2vB - 2 = 0
vB = -1 ft>s = 1 ft>s c
sale
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is
This
provided
integrity
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assessing
is
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protected
the
(including
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student
the
by
and
use
United
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is
of
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not
instructors
States
World
permitted.
Dissemination
Wide
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in
teaching
Web)
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or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*12–208.
If block A is moving downward at 6 ft>s while block C is
moving down at 18 ft>s, determine the speed of block B.
SOLUTION
B
C
sA + 2sB + sC = l
A
vA + 2vB + vC = 0
6 + 2vB + 18 = 0
vB = -12 ft/s = 12 ft>s c
sale
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This
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assessing
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protected
the
(including
for
work
student
the
by
and
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United
on
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is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–209.
Determine the displacement of the block B if A is pulled
down 4 ft.
C
SOLUTION
A
2 sA + 2 sC = l1
B
¢sA = - ¢sC
sB - sC + sB = l2
2 ¢sB = ¢sC
Thus,
2 ¢sB = - ¢sA
or
2 ¢sB = -4
¢sB = - 2 ft = 2 ft c
sale
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is
This
provided
integrity
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assessing
is
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solely
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the
(including
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work
student
the
by
and
use
United
on
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is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
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in
teaching
Web)
laws
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–210.
The pulley arrangement shown is designed for hoisting
materials. If BC remains fixed while the plunger P is pushed
downward with a speed of 4 ft>s, determine the speed of the
load at A.
B
C
P
SOLUTION
5 sB + (sB - sA) = l
6 sB - sA = l
6 vB - vA = 0
6(4) = vA
vA = 24 ft>s
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
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permitted.
Dissemination
Wide
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in
teaching
Web)
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or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
4 ft/s
A
12–211.
Determine the speed of block A if the end of the rope is
pulled down with a speed of 4 m>s.
4 m/s
B
SOLUTION
A
Position Coordinates: By referring to Fig. a, the length of the cord written in terms
of the position coordinates sA and sB is
sB + sA + 2(sA - a) = l
sB + 3sA = l + 2a
Time Derivative: Taking the time derivative of the above equation,
vB + 3vA = 0
or
(+ T )
in
teaching
Web)
laws
Here, vB = 4 m>s. Thus,
vA = - 133 m>s = 1.33 m>s c
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
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this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
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is
of
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not
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World
permitted.
Dissemination
Wide
copyright
4 + 3vA = 0
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*12–212.
The cylinder C is being lifted using the cable and pulley
system shown. If point A on the cable is being drawn toward
the drum with a speed of 2 m>s, determine the speed of the
cylinder.
A
vA
SOLUTION
l = sC + (sC - h) + (sC - h - sA)
C
l = 3sC - 2h - sA
s
0 = 3vC - vA
vA
-2
=
= -0.667 m>s = 0.667 m>s c
3
3
Ans.
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is
This
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the
(including
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the
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or
vC =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–213.
The man pulls the boy up to the tree limb C by walking
backward at a constant speed of 1.5 m>s. Determine the
speed at which the boy is being lifted at the instant
xA = 4 m. Neglect the size of the limb. When xA = 0,
yB = 8 m, so that A and B are coincident, i.e., the rope is
16 m long.
C
yB
8m
B
A
SOLUTION
Position-Coordinate Equation: Using the Pythagorean theorem to determine lAC,
we have lAC = 2x2A + 82. Thus,
l = lAC + yB
xA
16 = 2x2A + 82 + yB
yB = 16 - 2x2A + 64
(1)
or
dxA
Time Derivative: Taking the time derivative of Eq. (1) and realizing that yA =
dt
dyB
and yB =
, we have
dt
laws
dyB
xA
dxA
= 2
dt
2xA + 64 dt
xA
yB = yA
2
2xA + 64
(2)
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
yB =
by
(1.5) = - 0.671 m>s = 0.671 m>s c
Ans.
work
24 + 64
student
the
4
the
(including
2
for
yB = -
and
use
United
on
learning.
is
of
the
not
At the instant xA = 4 m, from Eq. [2]
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This
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Note: The negative sign indicates that velocity yB is in the opposite direction to that
of positive yB.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–214.
The man pulls the boy up to the tree limb C by walking
backward. If he starts from rest when xA = 0 and moves
backward with a constant acceleration a A = 0.2 m>s2,
determine the speed of the boy at the instant yB = 4 m.
Neglect the size of the limb.When xA = 0, yB = 8 m, so that A
and B are coincident, i.e., the rope is 16 m long.
C
yB
8m
B
A
SOLUTION
Position-Coordinate Equation: Using the Pythagorean theorem to determine lAC,
we have lAC = 2x2A + 82. Thus,
l = lAC + yB
xA
16 = 2x2A + 82 + yB
yB = 16 - 2x2A + 64
(1)
or
dxA
Time Derivative: Taking the time derivative of Eq. (1) Where vA =
and
dt
dyB
vB =
, we have
dt
laws
dyB
xA
dxA
= 2
dt
2xA + 64 dt
xA
vA
vB = 2
2xA + 64
Wide
copyright
in
teaching
Web)
vB =
not
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Dissemination
(2)
the
by
and
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on
learning.
is
of
the
At the instant yB = 4 m, from Eq. (1), 4 = 16 - 2x2A + 64, xA = 8.944 m. The
velocity of the man at that instant can be obtained.
the
(including
for
work
student
v2A = (v0)2A + 2(ac)A C sA - (s0)A D
assessing
is
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solely
of
protected
v2A = 0 + 2(0.2)(8.944 - 0)
the
is
This
provided
integrity
of
and
work
this
vA = 1.891 m>s
destroy
(1.891) = -1.41 m>s = 1.41 m>s c
28.9442 + 64
Ans.
will
their
8.944
sale
vB = -
of
and
any
courses
part
Substitute the above results into Eq. (2) yields
Note: The negative sign indicates that velocity vB is in the opposite direction to that
of positive yB.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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12–215.
The roller at A is moving upward with a velocity of
vA = 3 ft>s and has an acceleration of aA = 4 ft>s2 when
sA = 4 ft. Determine the velocity and acceleration of block
B at this instant.
A
3 ft
SOLUTION
B
sB + 3 1sA22 + 32 = l
- 12
#
1
#
sB + C (sA)2 + 32 D (2sA) sA = 0
2
#
- 12
#
sB + C s2A + 9 D asA sA b = 0
3
1
1
$
#
#
$
sB - C (sA)2 + 9 D - 2 as2A s2A b + C s2A + 9 D - 2 a s2A b + C s2A + 9 D - 2 asAsA b = 0
#
$
At sA = 4 ft, sA = 3 ft>s, sA = 4 ft>s2
or
1
#
sB + a b (4)(3) = 0
5
laws
vB = -2.4 ft>s = 2.40 ft>s :
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Web)
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Dissemination
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1 3
1
1
##
s B - a b 14221322 + a b 1322 + a b 142142 = 0
5
5
5
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This
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Ans.
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not
States
aB = -3.85 ft>s2 = 3.85 ft>s2 :
sA
4 ft
*12–216.
The girl at C stands near the edge of the pier and pulls in the
rope horizontally at a constant speed of 6 ft>s. Determine
how fast the boat approaches the pier at the instant the
rope length AB is 50 ft.
6 ft/s
xC
C
A
8 ft
B
xB
SOLUTION
The length l of cord is
2(8)2 + x2B + xC = l
Taking the time derivative:
#
#
1
[(8)2 + x2B] - 1/2 2 xBxB + xC = 0
2
(1)
#
xC = 6 ft/s
When AB = 50 ft,
teaching
Web)
laws
or
xB = 2(50)2 - (8)2 = 49.356 ft
Wide
copyright
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From Eq. (1)
is
of
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instructors
States
World
permitted.
Dissemination
#
1
[(8)2 + (49.356)2] - 1/2 2(49.356)(xB) + 6 = 0
2
#
xB = - 6.0783 = 6.08 ft>s ;
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Ans.
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12–217.
The crate C is being lifted by moving the roller at A
downward with a constant speed of vA = 2 m>s along the
guide. Determine the velocity and acceleration of the crate
at the instant s = 1 m. When the roller is at B, the crate
rests on the ground. Neglect the size of the pulley in the
calculation. Hint: Relate the coordinates xC and xA using
the problem geometry, then take the first and second time
derivatives.
4m
B
xA
xC
4m
SOLUTION
s
xC + 2x2A + (4)2 = l
#
#
1
xC + (x2A + 16) - 1/2(2xA)(xA) = 0
2
1
# 2
$
# 2
$
xC - (x2A + 16) - 3/2 (2x2A)(xA ) + (x2A + 16) - 1/2 (xA) + (x2A + 16) - 1/2 (xA)(xA) = 0
2
l = 8 m , and when s = 1 m ,
xC = 3 m
laws
or
xA = 3 m
teaching
Web)
#
vA = xA = 2 m>s
World
permitted.
Dissemination
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$
aA = x A = 0
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States
Thus,
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United
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learning.
vC + [(3)2 + 16] - 1/2 (3)(2) = 0
Ans.
(including
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student
vC = - 1.2 m>s = 1.2 m>s c
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the
aC - [(3)2 + 16] - 3/2 (3)2(2)2 + [(3)2 + 16] - 1/2 (2)2 + 0 = 0
Ans.
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aC = - 0.512 m>s2 = 0.512 m>s2 c
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A
C
12–218.
The man can row the boat in still water with a speed of
5 m>s. If the river is flowing at 2 m>s, determine the speed
of the boat and the angle u he must direct the boat so that it
travels from A to B.
B
vw
5 m/s
u
SOLUTION
A
Solution I
25 m
Vector Analysis: Here, the velocity vb of the boat is directed from A to B. Thus,
f = tan - 1 a
50
b = 63.43°. The magnitude of the boat’s velocity relative to the
25
flowing river is vb>w = 5 m>s. Expressing vb, vw, and vb/w in Cartesian vector form,
we have vb = vb cos 63.43i + vb sin 63.43j = 0.4472vb i + 0.8944vb j, vw = [2i] m>s,
and vb>w = 5 cos ui + 5 sin uj. Applying the relative velocity equation, we have
vb = vw + vb>w
or
0.4472vb i + 0.8944vb j = 2i + 5 cos ui + 5 sin uj
teaching
Web)
laws
0.4472vb i + 0.8944vb j = (2 + 5 cos u)i + 5 sin uj
(1)
instructors
States
World
Dissemination
0.4472vb = 2 + 5 cos u
(2)
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0.8944vb = 5 sin u
for
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student
the
by
Solving Eqs. (1) and (2) yields
the
(including
u = 84.4°
Ans.
assessing
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vb = 5.56 m>s
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Solution II
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52 = 22 + vb 2 - 2(2)(vb) cos 63.43°
of
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This
Scalar Analysis: Referring to the velocity diagram shown in Fig. a and applying the
law of cosines,
vb 2 - 1.789vb - 21 = 0
vb =
-(-1.789) ; 2( -1.789)2 - 4(1)(- 21)
2(1)
Choosing the positive root,
vb = 5.563 m>s = 5.56 m>s
Ans.
Using the result of vb and applying the law of sines,
sin (180° - u) sin 63.43°
=
5.563
5
u = 84.4°
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Ans.
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Wide
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Equating the i and j components, we have
50 m
2 m/s
12–219.
Vertical motion of the load is produced by movement of the
piston at A on the boom. Determine the distance the piston
or pulley at C must move to the left in order to lift the load
2 ft. The cable is attached at B, passes over the pulley at C,
then D, E, F, and again around E, and is attached at G.
A
6 ft/s
B
C
D
G
E
F
SOLUTION
2 sC + 2 s F = l
2 ¢sC = - 2 ¢sF
¢sC = - ¢sF
¢sC = -( - 2 ft) = 2 ft
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This
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Dissemination
Wide
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teaching
Web)
laws
or
Ans.
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12–220.
If block B is moving down with a velocity vB and has an
acceleration a B , determine the velocity and acceleration of
block A in terms of the parameters shown.
sA
A
h
vB, aB
B
SOLUTION
l = sB + 3s2B + h2
1
#
#
0 = sB + (s2A + h2) - 1/2 2sA sA
2
#
- sB(s2A + h2)1/2
#
vA = sA =
sA
vA = - vB (1 + a
h 2 1/2
b )
sA
Ans.
laws
or
h 2
h 2
1
#
#
#
aA = vA = - vB(1 + a b )1/2 - vB a b (1 + a b ) - 1/2(h2)( - 2)(sA) - 3sA
sA
2
sA
h 2 1/2 vAvBh2
h 2 - 1/2
b ) +
(1
+
a
b )
sB
sA
s3A
in
teaching
Web)
Ans.
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Dissemination
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aA = - aB(1 + a
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12–221.
Collars A and B are connected to the cord that passes over
the small pulley at C. When A is located at D, B is 24 ft to the
left of D. If A moves at a constant speed of 2 ft>s to the right,
determine the speed of B when A is 4 ft to the right of D.
C
10 ft
B
A
D
SOLUTION
l = 312422 + 11022 + 10 = 36 ft
311022 + s2B + 3 11022 + s2A = 36
1
1
1
1
#
#
(100 + s2B)- 2 (2s B sB) + (100 + s2A)- 2 (2sAsA) = 0
2
2
1
#
sA sA
100 + s2B 2
#
ba
b
sB = - a
sB
100 + s2A
At sA = 4,
or
311022 + s2B + 3 11022 + 1422 = 36
in
teaching
Web)
laws
sB = 23.163 ft
Dissemination
Wide
copyright
Thus,
1
permitted.
4(2)
100 + (23.163)2 2
#
b = - 0.809 ft>s = 0.809 ft>s :
sB = - a
ba
23.163
100 + 42
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This
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not
instructors
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World
Ans.
2 ft/s
12–222.
Two planes, A and B, are flying at the same altitude. If their
velocities are vA = 600 km>h and vB = 500 km>h such that
the angle between their straight-line courses is u = 75°,
determine the velocity of plane B with respect to plane A.
A
vA
B
u
SOLUTION
vB
vB = vA + vB/A
75°
[500 ; ] = [600 c u ] + vB/A
+ )
(;
500 = -600 cos 75° + (vB/A)x
(vB/A)x = 655.29 ;
(+ c)
0 = -600 sin 75° + (vB/A)y
(vB/A)y = 579.56 c
1vB>A2 = 31655.2922 +1579.5622
vB/A = 875 km/h
Web)
laws
or
Ans.
teaching
579.56
b = 41.5° b
655.29
in
Ans.
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This
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Dissemination
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u = tan-1 a
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or transmission in any form or by any means, electronic, mechanical,
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12–223.
At the instant shown, cars A and B are traveling at speeds
of 55 mi/h and 40 mi/h, respectively. If B is increasing its
speed by 1200 mi>h2, while A maintains a constant speed,
determine the velocity and acceleration of B with respect
to A. Car B moves along a curve having a radius of curvature
of 0.5 mi.
vB = 40 mi/h
B
A
vA = 55 mi/h
SOLUTION
vB = -40 cos 30°i + 40 sin 30°j = {-34.64i + 20j} mi>h
vA = { -55i } mi>h
vB>A = nB - nA
= ( - 34.64i + 20j) - (- 55i) = {20.36i + 20j} mi>h
vB>A = 220.362 + 202 = 28.5 mi>h
Ans.
20
= 44.5°
20.36
Ans.
(aB)t = 1200 mi>h2
or
v2A
402
=
= 3200 mi>h2
r
0.5
teaching
Web)
(aB)n =
a
laws
u = tan - 1
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aB = (3200 cos 60° - 1200 cos 30°)i + (3200 sin 60° + 1200 sin 30°)j
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Dissemination
= {560.77i + 3371.28j} mi>h2
United
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the
not
aA = 0
protected
the
(including
for
= {560.77i + 3371.28j} - 0 = {560.77i + 3371.28j} mi>h2
work
student
the
by
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aB>A = aB - aA
Ans.
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3371.28
= 80.6°
560.77
This
u = tan - 1
work
this
assessing
is
work
solely
of
aB>A = 2(560.77)2 + (3371.28)2 = 3418 mi>h2
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30°
12–224.
At the instant shown, car A travels along the straight
portion of the road with a speed of 25 m>s. At this same
instant car B travels along the circular portion of the road
with a speed of 15 m>s. Determine the velocity of car B
relative to car A.
15
r
A
SOLUTION
B
vA = [25 cos 30° i - 25 sin 30° j] m>s = [21.65i - 12.5j] m>s
vB = [15 cos 15° i - 15 sin 15° j] m>s = [14.49i - 3.882j] m>s
Applying the relative velocity equation,
vB = vA + vB>A
14.49i - 3.882j = 21.65i - 12.5j + vB>A
laws
or
vB>A = [- 7.162i + 8.618j] m>s
teaching
Web)
Thus, the magnitude of vB/A is given by
in
vB>A = 2( -7.162)2 + 8.6182 = 11.2 m>s
of
the
not
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World
permitted.
Dissemination
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Ans.
The direction angle uv of vB/A measured down from the negative x axis, Fig. b is
learning.
is
8.618
b = 50.3° d
7.162
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This
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uv = tan - 1 a
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
C
30
Velocity: Referring to Fig. a, the velocity of cars A and B expressed in Cartesian
vector form are
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
200 m
15
12–225.
An aircraft carrier is traveling forward with a velocity of
50 km>h. At the instant shown, the plane at A has just
taken off and has attained a forward horizontal air speed of
200 km>h, measured from still water. If the plane at B is
traveling along the runway of the carrier at 175 km>h in the
direction shown, determine the velocity of A with respect
to B.
B
15
SOLUTION
50 km/h
vB = vC + vB>C
v B = 50i + 175 cos 15°i + 175 sin 15°j = 219.04i + 45.293j
vA = vB + vA>B
200i = 219.04i + 45.293j + (vA>B)xi + (vA>B)y j
200 = 219.04 + (vA>B)x
0 = 45.293 + (vA>B)y
(vA>B)x = - 19.04
Web)
laws
or
(vA>B)y = -45.293
teaching
vA>B = 2( -19.04)2 + ( -45.293)2 = 49.1 km>h
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Ans.
Dissemination
45.293
b = 67.2° d
19.04
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Ans.
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This
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u = tan - 1 a
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A
12–226.
A car is traveling north along a straight road at 50 km>h.
An instrument in the car indicates that the wind is directed
toward the east. If the car’s speed is 80 km>h, the
instrument indicates that the wind is directed toward the
north-east. Determine the speed and direction of the wind.
SOLUTION
Solution I
Vector Analysis: For the first case, the velocity of the car and the velocity of the wind
relative to the car expressed in Cartesian vector form are vc = [50j] km>h and
vW>C = (vW>C)1 i. Applying the relative velocity equation, we have
vw = vc + vw>c
vw = 50j + (vw>c)1 i
vw = (vw>c)1i + 50j
(1)
laws
or
For the second case, vC = [80j] km>h and vW>C = (vW>C)2 cos 45°i + (vW>C)2 sin 45° j.
Applying the relative velocity equation, we have
Wide
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Web)
vw = vc + vw>c
permitted.
Dissemination
vw = 80j + (vw>c)2 cos 45°i + (vw>c)2 sin 45° j
States
World
vw = (vw>c)2 cos 45° i + C 80 + (vw>c)2 sin 45° D j
United
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(2)
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by
and
use
Equating Eqs. (1) and (2) and then the i and j components,
(3)
the
(including
for
work
student
(vw>c)1 = (vw>c)2 cos 45°
(4)
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50 = 80 + (vw>c)2 sin 45°
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This
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Solving Eqs. (3) and (4) yields
(vw>c)1 = - 30 km>h
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(vw>c)2 = -42.43 km>h
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Substituting the result of (vw>c)1 into Eq. (1),
vw = [ -30i + 50j] km>h
Thus, the magnitude of vW is
vw = 2( - 30)2 + 502 = 58.3 km>h
Ans.
and the directional angle u that vW makes with the x axis is
u = tan - 1 a
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50
b = 59.0° b
30
Ans.
12–227.
Two boats leave the shore at the same time and travel in the
directions shown. If vA = 20 ft>s and vB = 15 ft>s, determine
the velocity of boat A with respect to boat B. How long after
leaving the shore will the boats be 800 ft apart?
vA
A
B
30
O
SOLUTION
vA = vB + vA>B
-20 sin 30°i + 20 cos 30°j = 15 cos 45°i + 15 sin 45°j + vA>B
vA>B = {-20.61i + 6.714j} ft>s
vA>B = 2( -20.61)2 + ( +6.714)2 = 21.7 ft>s
u = tan - 1 (
Ans.
6.714
) = 18.0° b
20.61
Ans.
(800)2 = (20 t)2 + (15 t)2 - 2(20 t)(15 t) cos 75°
t = 36.9 s
or
Ans.
laws
Also
800
800
= 36.9 s
=
vA>B
21.68
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Dissemination
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t =
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45
vB
*12–228.
At the instant shown, the bicyclist at A is traveling at 7 m/s
around the curve on the race track while increasing his speed
at 0.5 m>s2. The bicyclist at B is traveling at 8.5 m/s along the
straight-a-way and increasing his speed at 0.7 m>s2. Determine
the relative velocity and relative acceleration of A with respect
to B at this instant.
vB = 8.5 m/s
B
50 m
vA = vB + vA>B
[7 |R ] = [8.5 : ] + [(vA>B)x : ] + [(vA>B)y T]
40°
7 sin 40° = 8.5 + (vA>B)x
(+ T )
7 cos 40° = (vA>B)y
Thus,
(vA>B)x = 4.00 m>s ;
or
(vA>B)y = 5.36 m>s T
teaching
Web)
laws
(vA>B) = 2(4.00)2 + (5.36)2
vA>B = 6.69 m>s
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copyright
in
Ans.
World
Ans.
the
not
instructors
is
of
by
and
use
United
on
learning.
72
= 0.980 m>s2
50
for
work
student
the
(aA)n =
solely
of
protected
the
(including
aA = aB + aA>B
work
= [0.7 : ] + [(aA>B)x : ] + [(aA>B)y T]
assessing
this
40°
is
[0.980] ud
40° + [0.5] |R
provided
integrity
of
and
work
- 0.980 cos 40° + 0.5 sin 40° = 0.7 + (aA>B)x
any
of
and
(aA>B)x = 1.129 m>s2 ;
courses
part
the
is
This
(+ : )
destroy
0.980 sin 40° + 0.5 cos 40° = (aA>B)y
sale
will
their
(+ T )
(aA>B)y = 1.013 m>s2 T
(aA>B) = 2(1.129)2 + (1.013)2
aA>B = 1.52 m>s2
u = tan - 1
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or transmission in any form or by any means, electronic, mechanical,
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1.013
1.129
Ans.
= 41.9° d
Ans.
permitted.
Dissemination
5.36
b = 53.3° d
4.00
States
u = tan - 1 a
50 m
40°
SOLUTION
+ )
(:
vA = 7 m/s
A
12–229.
Cars A and B are traveling around the circular race track.
At the instant shown, A has a speed of 90 ft>s and is
increasing its speed at the rate of 15 ft>s2, whereas B has a
speed of 105 ft>s and is decreasing its speed at 25 ft>s2.
Determine the relative velocity and relative acceleration of
car A with respect to car B at this instant.
vA
A
B
vB
rA
300 ft
60
rB
SOLUTION
vA = vB + vA>B
- 90i = -105 sin 30° i + 105 cos 30°j + vA>B
vA>B = 5- 37.5i - 90.93j6 ft>s
vA/B = 2( -37.5)2 + (- 90.93)2 = 98.4 ft>s
Ans.
90.93
b = 67.6° d
37.5
Ans.
u = tan - 1 a
aA = aB + aA>B
j = 25 cos 60°i - 25 sin 60°j - 44.1 sin 60°i - 44.1 cos 60°j + aA>B
teaching
Web)
300
or
19022
laws
-15i -
Wide
copyright
in
aA>B = {10.69i + 16.70j} ft>s2
aA>B = 2(10.69)2 + (16.70)2 = 19.8 ft>s2
instructors
States
World
permitted.
Dissemination
Ans.
of
the
not
16.70
b = 57.4° a
10.69
United
on
learning.
is
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
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and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
u = tan - 1 a
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or transmission in any form or by any means, electronic, mechanical,
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250 ft
12–230.
The two cyclists A and B travel at the same constant speed v.
Determine the speed of A with respect to B if A travels along
the circular track, while B travels along the diameter of
the circle.
v
A
r
f
B
v
SOLUTION
vA = v sin ui + v cos uj
vB = vi
vA>B = vA - vB
= (v sin ui + v cos uj) - vi
= (v sin u - v)i + v cos uj
vA>B = 2(v sin u - v)2 + (v cos u)2
laws
or
= 22v2 - 2v2 sin u
= v 22(1 - sin u)
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and
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part
the
is
This
provided
integrity
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this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
Ans.
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or transmission in any form or by any means, electronic, mechanical,
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u
12–231.
At the instant shown, cars A and B travel at speeds of
70 mi>h and 50 mi>h, respectively. If B is increasing its
speed by 1100 mi>h2, while A maintains a constant speed,
determine the velocity and acceleration of B with respect to
A.Car B moves along a curve having a radius of curvature of
0.7 mi.
A
vA
Relative Velocity:
vB = vA + vB>A
50 sin 30°i + 50 cos 30°j = 70j + vB>A
vB>A = {25.0i - 26.70j} mi>h
Thus, the magnitude of the relative velocity vB/A is
yB>A = 225.02 + ( -26.70)2 = 36.6 mi>h
Ans.
laws
or
The direction of the relative velocity is the same as the direction of that for relative
acceleration. Thus
26.70
= 46.9° c
25.0
in
teaching
Web)
Ans.
Wide
copyright
u = tan - 1
is
of
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not
instructors
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World
the
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on
learning.
aB = aA + aB>A
the
(including
for
work
student
(1100 sin 30° + 3571.43 cos 30°)i + (1100 cos 30° - 3571.43 sin 30°)j = 0 + aB>A
assessing
is
work
solely
of
protected
aB>A = {3642.95i - 833.09j} mi>h2
provided
integrity
of
and
work
this
Thus, the magnitude of the relative velocity aB/A is
Ans.
and
any
courses
part
the
is
This
aB>A = 23642.952 + (- 833.09)2 = 3737 mi>h2
their
destroy
of
And its direction is
or transmission in any form or by any means, electronic, mechanical,
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will
833.09
= 12.9° c
3642.95
sale
f = tan - 1
Ans.
permitted.
Dissemination
Relative Acceleration: Since car B is traveling along a curve, its normal
y2B
502
acceleration is (aB)n =
= 3571.43 mi>h2. Applying Eq. 12–35 gives
=
r
0.7
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
vB
30
SOLUTION
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
70 mi/h
B
50 mi/h
12–232.
At the instant shown, cars A and B travel at speeds of 70 mi>h
and 50 mi>h, respectively. If B is decreasing its speed at
1400 mi>h2 while A is increasing its speed at 800 mi>h2,
determine the acceleration of B with respect to A. Car B
moves along a curve having a radius of curvature of 0.7 mi.
A
vA
Relative Acceleration: Since car B is traveling along a curve, its normal acceleration
v2B
502
=
is (aB)n =
= 3571.43 mi>h2. Applying Eq. 12–35 gives
r
0.7
aB = aA + aB>A
(3571.43 cos 30° - 1400 sin 30°)i + ( -1400 cos 30° - 3571.43 sin 30°)j = 800j + aB>A
aB>A = {2392.95i - 3798.15j} mi>h2
Thus, the magnitude of the relative acc. aB/A is
aB>A = 22392.952 + (- 3798.15)2 = 4489 mi>h2
laws
or
Ans.
copyright
in
teaching
Web)
And its direction is
Ans.
World
permitted.
Dissemination
Wide
3798.15
= 57.8° c
2392.95
sale
will
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destroy
of
and
any
courses
part
the
is
This
provided
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of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
f = tan - 1
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
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vB
30
SOLUTION
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
70 mi/h
B
50 mi/h
12–233.
A passenger in an automobile observes that raindrops make
an angle of 30° with the horizontal as the auto travels
forward with a speed of 60 km/h. Compute the terminal
(constant)velocity vr of the rain if it is assumed to fall
vertically.
vr
va = 60 km/h
SOLUTION
vr = va + vr>a
-vr j = - 60i + vr>a cos 30°i - vr>a sin 30°j
+ )
(:
0 = - 60 + vr>a cos 30°
(+ c)
- vr = 0 - vr>a sin 30°
vr>a = 69.3 km>h
vr = 34.6 km h
sale
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destroy
of
and
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courses
part
the
is
This
provided
integrity
of
and
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this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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12–234.
A man can swim at 4 ft/s in still water. He wishes to cross
the 40-ft-wide river to point B, 30 ft downstream. If the river
flows with a velocity of 2 ft/s, determine the speed of the
man and the time needed to make the crossing. Note: While
in the water he must not direct himself toward point B to
reach this point. Why?
30 ft
B
vr = 2 ft/s
A
SOLUTION
Relative Velocity:
vm = vr + vm>r
3
4
n i + vm j = 2i + 4 sin ui + 4 cos uj
5 m
5
(1)
4
v = 4 cos u
5 m
(2)
teaching
Web)
laws
3
v = 2 + 4 sin u
5 m
or
Equating the i and j components, we have
Wide
copyright
in
Solving Eqs. (1) and (2) yields
States
World
permitted.
Dissemination
u = 13.29°
instructors
vm = 4.866 ft>s = 4.87 ft>s
and
use
United
on
learning.
is
of
the
not
Ans.
(including
the
Ans.
work
solely
of
protected
sAB
2402 + 302
=
= 10.3 s
vb
4.866
this
assessing
is
t =
for
work
student
the
by
Thus, the time t required by the boat to travel from points A to B is
sale
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and
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courses
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the
is
This
provided
integrity
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In order for the man to reached point B, the man has to direct himself at an angle
u = 13.3° with y axis.
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40 ft
12–235.
The ship travels at a constant speed of vs = 20 m>s and the
wind is blowing at a speed of vw = 10 m>s, as shown.
Determine the magnitude and direction of the horizontal
component of velocity of the smoke coming from the smoke
stack as it appears to a passenger on the ship.
vs
20 m/s
30
vw
10 m/s
45
y
x
SOLUTION
Solution I
Vector Analysis: The velocity of the smoke as observed from the ship is equal to the
velocity of the wind relative to the ship. Here, the velocity of the ship and wind
expressed in Cartesian vector form are vs = [20 cos 45° i + 20 sin 45° j] m>s
= [14.14i + 14.14j] m>s and vw = [10 cos 30° i - 10 sin 30° j] = [8.660i - 5j] m>s.
Applying the relative velocity equation,
vw = vs + vw>s
8.660i - 5j = 14.14i + 14.14j + vw>s
or
vw>s = [- 5.482i - 19.14j] m>s
Web)
laws
Thus, the magnitude of vw/s is given by
teaching
vw = 2(-5.482)2 + ( -19.14)2 = 19.9 m>s
Wide
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Ans.
instructors
States
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permitted.
Dissemination
and the direction angle u that vw/s makes with the x axis is
of
the
not
19.14
b = 74.0° d
5.482
United
on
learning.
is
Ans.
for
work
student
the
by
and
use
u = tan - 1 a
of
protected
the
(including
Solution II
part
the
is
This
any
courses
Ans.
their
destroy
of
= 19.91 m>s = 19.9 m>s
and
vw>s = 2202 + 102 - 2(20)(10) cos 75°
provided
integrity
of
and
work
this
assessing
is
work
solely
Scalar Analysis: Applying the law of cosines by referring to the velocity diagram
shown in Fig. a,
sale
will
Using the result of vw/s and applying the law of sines,
sin f
sin 75°
=
10
19.91
f = 29.02°
Thus,
u = 45° + f = 74.0° d
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Ans.
*12–236.
Car A travels along a straight road at a speed of 25 m>s
while accelerating at 1.5 m>s2. At this same instant car C is
traveling along the straight road with a speed of 30 m>s
while decelerating at 3 m>s2. Determine the velocity and
acceleration of car A relative to car C.
1.5 m/s
r
100 m
15 m/s
Velocity: The velocity of cars A and C expressed in Cartesian vector form are
vA = [ -25 cos 45°i - 25 sin 45°j] m>s = [- 17.68i - 17.68j] m>s
vC = [-30j] m>s
Applying the relative velocity equation, we have
vA = vC + vA>C
-17.68i - 17.68j = - 30j + vA>C
laws
or
vA>C = [ -17.68i + 12.32j] m>s
teaching
Web)
Thus, the magnitude of vA/C is given by
vA>C = 2( -17.68)2 + 12.322 = 21.5 m>s
Dissemination
Wide
copyright
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Ans.
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instructors
States
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permitted.
and the direction angle uv that vA/C makes with the x axis is
is
learning.
for
work
student
the
by
and
use
United
on
Ans.
of
protected
the
(including
Acceleration: The acceleration of cars A and C expressed in Cartesian vector form are
this
assessing
is
work
solely
aA = [ -1.5 cos 45°i - 1.5 sin 45°j] m>s2 = [ -1.061i - 1.061j] m>s2
any
will
their
sale
aA = aC + aA>C
destroy
of
and
Applying the relative acceleration equation,
courses
part
the
is
This
provided
integrity
of
and
work
aC = [3j] m>s2
-1.061i - 1.061j = 3j + aA>C
aA>C = [ -1.061i - 4.061j] m>s2
Thus, the magnitude of aA/C is given by
aA>C = 2( -1.061)2 + (- 4.061)2 = 4.20 m>s2
Ans.
and the direction angle ua that aA/C makes with the x axis is
ua = tan - 1 a
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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4.061
b = 75.4° d
1.061
3 m/s2
30
SOLUTION
12.32
b = 34.9° b
17.68
45
2
2 m/s2
uv = tan - 1 a
A
25 m/s
Ans.
B
C
30 m/s
12–237.
Car B is traveling along the curved road with a speed of
15 m>s while decreasing its speed at 2 m>s2. At this same
instant car C is traveling along the straight road with a
speed of 30 m>s while decelerating at 3 m>s2. Determine the
velocity and acceleration of car B relative to car C.
r
100 m
15 m/s
Velocity: The velocity of cars B and C expressed in Cartesian vector form are
vB = [15 cos 60° i - 15 sin 60° j] m>s = [7.5i - 12.99j] m>s
vC = [-30j] m>s
Applying the relative velocity equation,
vB = vC + vB>C
7.5i - 12.99j = - 30j + vB>C
or
vB>C = [7.5i + 17.01j] m>s
teaching
Web)
laws
Thus, the magnitude of vB/C is given by
vB>C = 27.52 + 17.012 = 18.6 m>s
Dissemination
Wide
copyright
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Ans.
not
instructors
States
World
permitted.
and the direction angle uv that vB/C makes with the x axis is
is
of
the
17.01
b = 66.2° a
7.5
for
work
student
the
by
and
use
United
on
learning.
Ans.
of
protected
the
(including
Acceleration: The normal component of car B’s acceleration is (aB)n =
vB 2
r
is
work
solely
152
= 2.25 m>s2. Thus, the tangential and normal components of car B’s
100
acceleration and the acceleration of car C expressed in Cartesian vector form are
this
assessing
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the
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This
destroy
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and
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courses
(aB)t = [- 2 cos 60° i + 2 sin 60°j] = [- 1i + 1.732j] m>s2
sale
will
their
(aB)n = [2.25 cos 30° i + 2.25 sin 30° j] = [1.9486i + 1.125j] m>s2
aC = [3j] m>s2
Applying the relative acceleration equation,
aB = aC + aB>C
(- 1i + 1.732j) + (1.9486i + 1.125j) = 3j + aB>C
aB>C = [0.9486i - 0.1429j] m>s2
Thus, the magnitude of aB/C is given by
aB>C = 20.94862 + ( -0.1429)2 = 0.959 m>s2
Ans.
and the direction angle ua that aB/C makes with the x axis is
ua = tan - 1 a
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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0.1429
b = 8.57° c
0.9486
3 m/s2
30
SOLUTION
=
45
1.5 m/s2
2 m/s2
uv = tan - 1 a
A
25 m/s
Ans.
B
C
30 m/s
12–238.
At a given instant the football player at A throws a football C
with a velocity of 20 m/s in the direction shown. Determine
the constant speed at which the player at B must run so that
he can catch the football at the same elevation at which it was
thrown. Also calculate the relative velocity and relative
acceleration of the football with respect to B at the instant the
catch is made. Player B is 15 m away from A when A starts to
throw the football.
A
15 m
Ball:
+ )s = s + v t
(:
0
0
sC = 0 + 20 cos 60° t
v = v0 + ac t
- 20 sin 60° = 20 sin 60° - 9.81 t
t = 3.53 s
laws
or
sC = 35.31 m
teaching
Web)
Player B:
Dissemination
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+ ) s = s + n t
(:
B
B
0
of
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World
permitted.
Require,
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is
35.31 = 15 + vB (3.53)
Ans.
the
(including
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work
student
the
vB = 5.75 m>s
is
work
solely
of
protected
At the time of the catch
the
part
sale
10i - 17.32j = 5.751i + (vC>B)x i + (vC>B)y j
will
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destroy
of
and
any
courses
vC = vB + vC>B
is
(vC)y = 20 sin 60° = 17.32 m>s T
This
provided
integrity
of
and
work
this
assessing
(vC)x = 20 cos 60° = 10 m>s :
+ )
(:
10 = 5.75 + (vC>B)x
(+ c)
- 17.32 = (vC>B)y
(vC>B)x = 4.25 m>s :
(vC>B)y = 17.32 m>s T
vC>B = 2(4.25)2 + (17.32)2 = 17.8 m>s
Ans.
17.32
b = 76.2°
4.25
Ans.
u = tan - 1 a
c
aC = aB + aC>B
- 9.81 j = 0 + aC>B
aC B = 9.81 m s2 T
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60°
B
SOLUTION
(+ c )
C
20 m/s
Ans.
12–239.
Both boats A and B leave the shore at O at the same time. If
A travels at vA and B travels at vB, write a general
expression to determine the velocity of A with respect to B.
A
B
SOLUTION
O
Relative Velocity:
vA = vB + vA>B
vA j = vB sin ui + vB cos uj + vA>B
vA>B = - vB sin ui + (vA - vB cos u)j
Thus, the magnitude of the relative velocity vA>B is
vA>B = 2( - vB sin u)2 + (vA - vB cos u)2
= 2vA2 + vB2 - 2vA vB cos u
laws
or
Ans.
in
teaching
Web)
And its direction is
vA - vB cos u
vB sin u
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Wide
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b
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the
(including
for
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student
the
by
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on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
u = tan-1
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13–1.
z
The 6-lb particle is subjected to the action of
its weight and forces F1 = 52i + 6j - 2tk6 lb, F2 =
5t2i - 4tj - 1k6 lb, and F3 = 5- 2ti6 lb, where t is in
seconds. Determine the distance the ball is from the origin
2 s after being released from rest.
F2
y
F3
x
SOLUTION
©F = ma;
(2i + 6j - 2tk) + (t2i - 4tj - 1k) - 2ti - 6k = ¢
6
≤ (axi + ay j + azk)
32.2
Equating components:
¢
6
≤ a = t2 - 2t + 2
32.2 x
¢
6
≤ a = - 4t + 6
32.2 y
6
≤ a = -2t - 7
32.2 z
¢
Since dv = a dt, integrating from n = 0, t = 0, yields
6
t3
- t2 + 2t
≤ vx =
32.2
3
¢
6
≤ v = - 2t2 + 6t
32.2 y
¢
6
≤ v = - t2 - 7t
32.2 z
laws
or
¢
6
t3
7t2
≤ sz = - 32.2
3
2
¢
Wide
6
2t3
+ 3t2
≤ sy = 32.2
3
instructors
States
World
permitted.
¢
Dissemination
6
t4
t3
+ t2
≤ sx =
32.2
12
3
copyright
¢
in
teaching
Web)
Since ds = v dt, integrating from s = 0, t = 0 yields
not
sz = -89.44 ft
the
sy = 35.78 ft
is
and
use
United
on
learning.
sx = 14.31 ft,
of
When t = 2 s then,
for
work
student
the
by
Thus,
the
(including
(14.31)2 + (35.78)2 + (- 89.44)2 = 97.4 ft
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s =
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Ans.
F1
13–2.
The 10-lb block has an initial velocity of 10 ft>s on the
smooth plane. If a force F = 12.5t2 lb, where t is in seconds,
acts on the block for 3 s, determine the final velocity of the
block and the distance the block travels during this time.
v = 10 ft/s
F = (2.5t) lb
SOLUTION
+ ©F = ma ;
:
x
x
2.5t = ¢
10
≤a
32.2
a = 8.05t
dv = a dt
n
L10
t
dv =
L0
8.05t dt
v = 4.025t2 + 10
laws
or
When t = 3 s,
v = 46.2 ft>s
in
teaching
Web)
Ans.
Dissemination
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copyright
ds = v dt
not
instructors
States
World
(4.025t2 + 10) dt
the
is
L0
United
on
learning.
L0
permitted.
t
ds =
of
s
work
student
the
by
and
use
s = 1.3417t3 + 10t
of
protected
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(including
for
When t = 3 s,
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s = 66.2 ft
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Ans.
13–3.
P
If the coefficient of kinetic friction between the 50-kg crate
and the ground is mk = 0.3, determine the distance the
crate travels and its velocity when t = 3 s. The crate starts
from rest, and P = 200 N.
30⬚
SOLUTION
Free-Body Diagram: The kinetic friction Ff = mkN is directed to the left to oppose
the motion of the crate which is to the right, Fig. a.
Equations of Motion: Here, ay = 0. Thus,
+ c ©Fy = 0;
N - 50(9.81) + 200 sin 30° = 0
N = 390.5 N
+ ©Fx = max;
:
200 cos 30° - 0.3(390.5) = 50a
a = 1.121 m>s2
laws
teaching
Web)
v = v0 + act
in
+ B
A:
or
Kinematics: Since the acceleration a of the crate is constant,
v = 0 + 1.121(3) = 3.36 m>s
Dissemination
Wide
copyright
Ans.
World
instructors
States
not
the
is
of
on
by
and
use
United
1
(1.121) A 32 B = 5.04 m
2
Ans.
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the
(including
s = 0 + 0 +
1 2
at
2 c
learning.
s = s0 + v0t +
for
+ B
A:
permitted.
and
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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*13–4.
P
If the 50-kg crate starts from rest and achieves a velocity of
v = 4 m>s when it travels a distance of 5 m to the right,
determine the magnitude of force P acting on the crate.
The coefficient of kinetic friction between the crate and the
ground is mk = 0.3.
30⬚
SOLUTION
Kinematics: The acceleration a of the crate will be determined first since its motion
is known.
+ )
(:
v2 = v 2 + 2a (s - s )
c
0
0
42 = 02 + 2a(5 - 0)
a = 1.60 m>s2 :
Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.3N is required to be
directed to the left to oppose the motion of the crate which is to the right, Fig. a.
Equations of Motion:
or
N + P sin 30° - 50(9.81) = 50(0)
laws
+ c ©Fy = may;
Wide
copyright
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Web)
N = 490.5 - 0.5P
States
World
permitted.
Dissemination
Using the results of N and a,
the
not
instructors
P cos 30° - 0.3(490.5 - 0.5P) = 50(1.60)
on
learning.
is
of
+ ©F = ma ;
:
x
x
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the
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the
by
and
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United
P = 224 N
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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Ans.
13–5.
The water-park ride consists of an 800-lb sled which slides
from rest down the incline and then into the pool. If the
frictional resistance on the incline is Fr = 30 lb, and in
the pool for a short distance Fr = 80 lb, determine how
fast the sled is traveling when s = 5 ft.
100 ft
SOLUTION
+ b a Fx = max;
800
a
32.2
800 sin 45° - 30 =
a = 21.561 ft>s
s
2
v21 = v20 + 2ac(s - s0)
v21 = 0 + 2(21.561)(10022 - 0))
v1 = = 78.093 ft>s
- 80 =
800
a
32.2
or
+
;
a Fx = ma x;
teaching
Web)
laws
a = - 3.22 ft>s2
Wide
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in
v22 = v21 + 2ac(s2 - s1)
instructors
States
World
permitted.
Dissemination
v22 = (78.093)2 + 2(- 3.22)(5 - 0)
not
v2 = 77.9 ft>s
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the
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of
the
Ans.
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or transmission in any form or by any means, electronic, mechanical,
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100 ft
13–6.
P
If P = 400 N and the coefficient of kinetic friction between
the 50-kg crate and the inclined plane is mk = 0.25,
determine the velocity of the crate after it travels 6 m up the
plane. The crate starts from rest.
30°
SOLUTION
30°
Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.25N is required to be
directed down the plane to oppose the motion of the crate which is assumed to be
directed up the plane. The acceleration a of the crate is also assumed to be directed
up the plane, Fig. a.
Equations of Motion: Here, ay ¿ = 0. Thus,
©Fy ¿ = may ¿;
N + 400 sin 30° - 50(9.81) cos 30° = 50(0)
N = 224.79 N
laws
teaching
Web)
400 cos 30° - 50(9.81) sin 30° - 0.25(224.79) = 50a
in
©Fx ¿ = may ¿;
or
Using the result of N,
States
World
permitted.
Dissemination
Wide
copyright
a = 0.8993 m>s2
learning.
is
of
the
not
instructors
Kinematics: Since the acceleration a of the crate is constant,
the
by
and
use
United
on
v2 = v0 2 + 2ac(s - s0)
the
(including
for
work
student
v2 = 0 + 2(0.8993)(6 - 0)
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v = 3.29 m>s
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Ans.
13–7.
P
If the 50-kg crate starts from rest and travels a distance of 6 m
up the plane in 4 s, determine the magnitude of force P acting
on the crate. The coefficient of kinetic friction between the
crate and the ground is mk = 0.25.
30°
SOLUTION
30°
Kinematics: Here, the acceleration a of the crate will be determined first since its
motion is known.
s = s0 + v0t +
6 = 0 + 0 +
1 2
at
2 c
1
a(42)
2
a = 0.75 m>s2
in
teaching
Web)
laws
or
Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.25N is required to be
directed down the plane to oppose the motion of the crate which is directed up the
plane, Fig. a.
Dissemination
Wide
copyright
Equations of Motion: Here, ay ¿ = 0. Thus,
World
permitted.
N + P sin 30° - 50(9.81) cos 30° = 50(0)
of
the
not
instructors
States
©Fy ¿ = may ¿;
the
by
and
use
United
on
learning.
is
N = 424.79 - 0.5P
(including
for
work
student
Using the results of N and a,
the
P cos 30° - 0.25(424.79 - 0.5P) - 50(9.81) sin 30° = 50(0.75)
assessing
is
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solely
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protected
©Fx ¿ = max ¿;
Ans.
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P = 392 N
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*13–8.
The speed of the 3500-lb sports car is plotted over the 30-s
time period. Plot the variation of the traction force F needed
to cause the motion.
v (ft/s)
F
80
60
SOLUTION
dv
60
Kinematics: For 0 … t 6 10 s. v =
t = {6t} ft>s. Applying equation a =
,
10
dt
we have
dv
= 6 ft>s2
dt
a =
10
80 - 60
v - 60
=
For 10 6 t … 30 s,
, v = {t + 50} ft>s. Applying equation
t - 10
30 - 10
dv
a =
, we have
dt
dv
= 1 ft>s2
dt
teaching
Web)
laws
or
a =
v
Wide
copyright
in
Equation of Motion:
the
is
on
the
by
and
3500
≤ (1) = 109 lb
32.2
learning.
F = ¢
Ans.
of
3500
≤ (6) = 652 lb
32.2
United
F = ¢
use
+
;
a Fx = max ;
not
instructors
States
World
permitted.
Dissemination
For 0 … t 6 10 s
of
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+
;
a Fx = max ;
protected
the
(including
for
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student
For 10 6 t … 30 s
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Ans.
30
t (s)
13–9.
p
The crate has a mass of 80 kg and is being towed by a chain
which is always directed at 20° from the horizontal as
shown. If the magnitude of P is increased until the crate
begins to slide, determine the crate’s initial acceleration if
the coefficient of static friction is ms = 0.5 and the
coefficient of kinetic friction is mk = 0.3.
20
SOLUTION
Equations of Equilibrium: If the crate is on the verge of slipping, Ff = ms N = 0.5N.
From FBD(a),
+ c ©Fy = 0;
N + P sin 20° - 80(9.81) = 0
(1)
+ ©F = 0;
:
x
P cos 20° - 0.5N = 0
(2)
Solving Eqs.(1) and (2) yields
P = 353.29 N
N = 663.97 N
N - 80(9.81) + 353.29 sin 20° = 80(0)
teaching
Web)
+ c ©Fy = may ;
laws
or
Equations of Motion: The friction force developed between the crate and its
contacting surface is Ff = mkN = 0.3N since the crate is moving. From FBD(b),
Wide
copyright
World
permitted.
Dissemination
353.29 cos 20° - 0.3(663.97) = 80a
not
instructors
States
+ ©F = ma ;
:
x
x
in
N = 663.97 N
Ans.
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protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
a = 1.66 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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13–10.
p
The crate has a mass of 80 kg and is being towed by a chain
which is always directed at 20° from the horizontal as
shown. Determine the crate’s acceleration in t = 2 s if the
coefficient of static friction is ms = 0.4, the coefficient of
kinetic friction is mk = 0.3, and the towing force is
P = (90t2) N, where t is in seconds.
20
SOLUTION
Equations of Equilibrium: At t = 2 s, P = 90 A 22 B = 360 N. From FBD(a)
+ c ©Fy = 0;
N + 360 sin 20° - 80(9.81) = 0
+ ©F = 0;
:
x
360 cos 20° - Ff = 0
N = 661.67 N
Ff = 338.29 N
Since Ff 7 (Ff)max = ms N = 0.4(661.67) = 264.67 N, the crate accelerates.
Equations of Motion: The friction force developed between the crate and its
contacting surface is Ff = mkN = 0.3N since the crate is moving. From FBD(b),
+ c ©Fy = may ;
N - 80(9.81) + 360 sin 20° = 80(0)
laws
in
teaching
Web)
360 cos 20° - 0.3(661.67) = 80a
Wide
copyright
+ ©F = ma ;
:
x
x
or
N = 661.67 N
a = 1.75 m>s2
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This
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the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Ans.
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or transmission in any form or by any means, electronic, mechanical,
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13–11.
The safe S has a weight of 200 lb and is supported by the
rope and pulley arrangement shown. If the end of the rope
is given to a boy B of weight 90 lb, determine his
acceleration if in the confusion he doesn’t let go of the rope.
Neglect the mass of the pulleys and rope.
S
SOLUTION
B
Equation of Motion: The tension T developed in the cord is the same throughout
the entire cord since the cord passes over the smooth pulleys.
From FBD(a),
+ c ©Fy = may;
T- 90 = - a
90
ba
32.2 B
(1)
From FBD(b),
2T - 200 = - a
200
ba
32.2 S
(2)
or
+ c ©Fy = may ;
teaching
Web)
laws
Kinematic: Establish the position-coordinate equation, we have
Dissemination
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in
2sS +sB = l
not
instructors
States
World
permitted.
Taking time derivative twice yields
the
2aS + aB = 0
(3)
by
and
use
United
on
learning.
is
of
1+ T 2
for
work
student
the
Solving Eqs.(1),(2), and (3) yields
work
solely
of
protected
the
(including
c
aB = -2.30 ft>s2 = 2.30 ft>s2
T = 96.43 lb
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aS = 1.15 ft>s2 T
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Ans.
*13–12.
The boy having a weight of 80 lb hangs uniformly from the
bar. Determine the force in each of his arms in t = 2 s if the
bar is moving upward with (a) a constant velocity of 3 ft>s,
and (b) a speed of v = 14t22 ft>s, where t is in seconds.
SOLUTION
(a) T = 40 lb
Ans.
(b) v = 4t 2
a = 8t
+ c a Fy = may ;
2T - 80 =
80
18t2
32.2
At t = 2 s.
T = 59.9 lb
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the
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for
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student
the
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of
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not
instructors
States
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permitted.
Dissemination
Wide
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in
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Web)
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or
Ans.
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or transmission in any form or by any means, electronic, mechanical,
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13–13.
The bullet of mass m is given a velocity due to gas pressure
caused by the burning of powder within the chamber of the
gun. Assuming this pressure creates a force of
F = F0 sin 1pt>t02 on the bullet, determine the velocity of
the bullet at any instant it is in the barrel. What is the
bullet’s maximum velocity? Also, determine the position of
the bullet in the barrel as a function of time.
F
F0
t0
SOLUTION
+ ©F = ma ;
:
x
x
F0
pt
dv
= a b sin a b
m
dt
t0
v
dv =
v = a
t
L0
a
F0
pt
b sin a b dt
m
t0
v = -a
F0t0
pt t
b cos a b d
pm
t0 0
F0t0
pt
b a 1 - cos a b b
pm
t0
pt
b = -1, or t = t0.
t0
teaching
Web)
vmax occurs when cos a
Ans.
or
L0
pt
b = ma
t0
laws
a =
F0 sin a
Ans.
permitted.
Dissemination
Wide
in
2F0t0
pm
copyright
vmax =
World
States
the
not
instructors
F0t0
pt
b a 1 - cos a b b dt
pm
t0
is
of
on
work
(including
the
t0
F0t0
pt
b at sin a b b
p
pm
t0
student
s = a
for
t0
F0t0
pt t
b ct sin a b d
pm
p
t0 0
is
work
solely
of
protected
s = a
the
by
and
a
learning.
L0
United
L0
t
ds =
use
s
sale
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This
provided
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assessing
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t
13–14.
The 2-Mg truck is traveling at 15 m> s when the brakes on all its
wheels are applied, causing it to skid for a distance of 10 m
before coming to rest. Determine the constant horizontal force
developed in the coupling C, and the frictional force
developed between the tires of the truck and the road during
this time.The total mass of the boat and trailer is 1 Mg.
C
SOLUTION
Kinematics: Since the motion of the truck and trailer is known, their common
acceleration a will be determined first.
+ b
a:
v2 = v0 2 + 2ac(s - s0)
0 = 152 + 2a(10 - 0)
a = -11.25 m>s2 = 11.25 m>s2 ;
Free-Body Diagram:The free-body diagram of the truck and trailer are shown in
Figs. (a) and (b), respectively. Here, F representes the frictional force developed
when the truck skids, while the force developed in coupling C is represented by T.
laws
in
teaching
Web)
-T = 1000(-11.25)
Wide
copyright
+ ©F = ma ;
:
x
x
or
Equations of Motion:Using the result of a and referrning to Fig. (a),
T = 11 250 N = 11.25 kN
instructors
States
World
permitted.
Dissemination
Ans.
on
learning.
is
of
the
not
Using the results of a and T and referring to Fig. (b),
by
and
use
United
11 250 - F = 2000( -11.25)
the
+ c ©Fx = max ;
sale
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This
provided
integrity
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assessing
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the
(including
for
work
student
F = 33 750 N = 33.75 kN
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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Ans.
13–15.
A freight elevator, including its load, has a mass of 500 kg. It
is prevented from rotating by the track and wheels mounted
along its sides.When t = 2 s, the motor M draws in the cable
with a speed of 6 m> s, measured relative to the elevator. If it
starts from rest, determine the constant acceleration of the
elevator and the tension in the cable. Neglect the mass of
the pulleys, motor, and cables.
M
SOLUTION
3sE + sP = l
3vE = - vP
A+TB
vP = vE + vP>E
-3vE = vE + 6
vE = -
A+cB
6
= -1.5 m>s = 1.5 m>s c
4
v = v0 + ac t
laws
or
1.5 = 0 + aE (2)
aE = 0.75 m>s2 c
in
teaching
Web)
Ans.
4T - 500(9.81) = 500(0.75)
Dissemination
Wide
copyright
+ c ©Fy = may ;
permitted.
T = 1320 N = 1.32 kN
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This
provided
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this
assessing
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the
(including
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student
the
by
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on
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is
of
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Ans.
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*13–16.
The man pushes on the 60-lb crate with a force F. The force
is always directed down at 30° from the horizontal as
shown, and its magnitude is increased until the crate begins
to slide. Determine the crate’s initial acceleration if the
coefficient of static friction is ms = 0.6 and the coefficient of
kinetic friction is mk = 0.3.
F
30⬚
SOLUTION
Force to produce motion:
+ ©F = 0;
:
x
Fcos 30° - 0.6N = 0
+ c ©Fy = 0;
N - 60 - F sin 30° = 0
N = 91.80 lb
F = 63.60 lb
Since N = 91.80 lb,
63.60 cos 30° - 0.3(91.80) = a
60
ba
32.2
a = 14.8 ft>s2
Ans.
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This
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the
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student
the
by
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on
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is
of
the
not
instructors
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World
permitted.
Dissemination
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copyright
in
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laws
or
+ ©F = ma ;
:
x
x
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13–17.
The double inclined plane supports two blocks A and B,
each having a weight of 10 lb. If the coefficient of kinetic
friction between the blocks and the plane is mk = 0.1,
determine the acceleration of each block.
B
A
60
SOLUTION
Equation of Motion: Since blocks A and B are sliding along the plane, the friction
forces developed between the blocks and the plane are (Ff)A = mk NA = 0.1 NA
and (Ff)B = mk NB = 0.1NB. Here, aA = aB = a. Applying Eq. 13–7 to FBD(a),
we have
a + a Fy¿ = may¿;
Q + a Fx¿ = max¿;
10
b (0) NA = 5.00 lb
32.2
NA - 10 cos 60° = a
10
ba
32.2
T + 0.1(5.00) - 10 sin 60° = - a
(1)
laws
or
From FBD(b),
teaching
T - 0.1(8.660) - 10 sin 30° = a
in
a + a Fx¿ = max¿;
Wide
copyright
NB - 10 cos 30° = a
Web)
10
b (0) NB = 8.660 lb
32.2
Q + a Fy¿ = may¿;
(2)
is
of
the
not
instructors
States
World
permitted.
Dissemination
10
ba
32.2
by
and
use
United
on
learning.
Solving Eqs. (1) and (2) yields
Ans.
(including
for
work
student
the
a = 3.69 ft>s2
sale
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This
provided
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assessing
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the
T = 7.013 lb
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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30
13–18.
A 40-lb suitcase slides from rest 20 ft down the smooth
ramp. Determine the point where it strikes the ground at C.
How long does it take to go from A to C?
A
20 ft
30
4 ft
SOLUTION
+ R ©Fx = m ax ;
40 sin 30° =
40
a
32.2
( +R)v2 = v20 + 2 ac(s - s0);
v2B = 0 + 2(16.1)(20)
vB = 25.38 ft>s
( + R) v = v0 + ac t ;
laws
or
25.38 = 0 + 16.1 tAB
in
teaching
Web)
tAB = 1.576 s
permitted.
Dissemination
Wide
copyright
+ )s = (s ) + (v ) t
(:
x
x 0
x 0
learning.
is
of
the
not
instructors
States
World
R = 0 + 25.38 cos 30°(tBC)
for
work
student
the
by
and
use
United
on
1 2
at
2 c
the
(including
1
(32.2)(tBC)2
2
is
work
solely
of
protected
4 = 0 + 25.38 sin 30° tBC +
integrity
of
and
work
this
assessing
tBC = 0.2413 s
provided
Ans.
part
the
is
This
R = 5.30 ft
any
courses
Ans.
sale
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of
and
Total time = tAB + tBC = 1.82 s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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C
R
a = 16.1 ft>s2
( + T) sy = (sy)0 + (vy)0 t +
B
13–19.
Solve Prob. 13–18 if the suitcase has an initial velocity down
the ramp of vA = 10 ft>s and the coefficient of kinetic
friction along AB is mk = 0.2.
A
20 ft
30⬚
4 ft
SOLUTION
+R©Fx = max;
40
40 sin 30° - 6.928 =
a
32.2
a = 10.52 ft>s
v2B = (10)2 + 2(10.52)(20)
vB = 22.82 ft>s
(+R) v = v0 + ac t;
22.82 = 10 + 10.52 tAB
laws
or
tAB = 1.219 s
Wide
copyright
in
teaching
Web)
+ ) s = (s ) + (v ) t
(:
x
x 0
x 0
Dissemination
R = 0 + 22.82 cos 30° (tBC)
permitted.
1
a t2
2 c
1
+ (32.2)(tBC)2
2
the
by
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is
of
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(+ T ) sy = (sy)0 + (vy)0 t +
the
(including
for
work
student
tBC = 0.2572 s
Ans.
this
assessing
is
work
solely
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R = 5.08 ft
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Total time = tAB + tBC = 1.48 s
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C
R
2
( +R) v2 = v20 + 2 ac (s - s0);
4 = 0 + 22.82 sin 30° tBC
B
Ans.
*13–20.
The 400-kg mine car is hoisted up the incline using the cable
and motor M. For a short time, the force in the cable is
F = 13200t22 N, where t is in seconds. If the car has an
initial velocity v1 = 2 m>s when t = 0, determine its
velocity when t = 2 s.
M
v1
SOLUTION
17
8
15
Q+ ©Fx¿ = max¿ ;
3200t2 - 40019.812a
8
b = 400a
17
a = 8t2 - 4.616
dv = adt
v
2
dv =
L2
L0
18t 2 -4.6162 dt
v = 14.1 m>s
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This
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assessing
is
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protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
2 m/s
13–21.
The 400-kg mine car is hoisted up the incline using the cable
and motor M. For a short time, the force in the cable is
F = 13200t22 N, where t is in seconds. If the car has an
initial velocity v1 = 2 m>s at s = 0 and t = 0, determine the
distance it moves up the plane when t = 2 s.
M
v1
SOLUTION
17
8
15
Q+ ©Fx¿ = max¿;
3200t2 - 40019.812 a
8
b = 400a
17
a = 8t2 - 4.616
dv = adt
v
v =
L0
ds
= 2.667t3 - 4.616t + 2
dt
s
2
ds =
L0
(2.667 t3 - 4.616t + 2) dt
laws
L0
18t2 - 4.6162 dt
or
L2
t
dv =
s = 5.43 m
sale
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This
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assessing
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the
(including
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work
student
the
by
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of
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not
instructors
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permitted.
Dissemination
Wide
copyright
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teaching
Web)
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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2 m/s
13–22.
Determine the required mass of block A so that when it is
released from rest it moves the 5-kg block B a distance of
0.75 m up along the smooth inclined plane in t = 2 s.
Neglect the mass of the pulleys and cords.
E
C
D
SOLUTION
B
1
a t2, we have
2 c
Kinematic: Applying equation s = s0 + v0 t +
A
60
(a+ )
1
0.75 = 0 + 0 + aB A 22 B
2
aB = 0.375 m>s2
Establishing the position - coordinate equation, we have
2sA + (sA - sB) = l
3sA - sB = l
Taking time derivative twice yields
3aA - aB = 0
laws
or
(1)
in
teaching
Web)
From Eq.(1),
aA = 0.125 m>s2
permitted.
Dissemination
Wide
copyright
3aA - 0.375 = 0
use
United
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learning.
is
of
the
by
and
T - 5(9.81) sin 60° = 5(0.375)
for
work
student
the
a + ©Fy¿ = may¿ ;
is
work
solely
of
protected
the
(including
T = 44.35 N
provided
integrity
of
3(44.35) - 9.81mA = mA ( -0.125)
part
the
is
This
+ c ©Fy = may ;
and
work
this
assessing
From FBD(a),
any
courses
Ans.
sale
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destroy
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and
mA = 13.7 kg
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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not
instructors
States
World
Equation of Motion: The tension T developed in the cord is the same throughout
the entire cord since the cord passes over the smooth pulleys. From FBD(b),
13–23.
The winding drum D is drawing in the cable at an
accelerated rate of 5 m>s2. Determine the cable tension if
the suspended crate has a mass of 800 kg.
D
SOLUTION
sA + 2 sB = l
aA = - 2 aB
5 = - 2 aB
aB = - 2.5 m>s2 = 2.5 m>s2 c
+ c ©Fy = may ;
2T - 800(9.81) = 800(2.5)
T = 4924 N = 4.92 kN
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This
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the
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work
student
the
by
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on
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is
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Dissemination
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Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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*13–24.
If the motor draws in the cable at a rate of v = (0.05s3>2) m>s,
where s is in meters, determine the tension developed in the
cable when s = 10 m. The crate has a mass of 20 kg, and the
coefficient of kinetic friction between the crate and the ground
is mk = 0.2.
s
v
M
SOLUTION
Kinematics: Since the motion of the create is known, its acceleration a will be
determined first.
a = v
dv
3
= A 0.05s3>2 B c (0.05) a bs1>2 d = 0.00375s2 m>s2
ds
2
When s = 10 m,
a = 0.00375(102) = 0.375 m>s2 :
Free-Body Diagram: The kinetic friction Ff = mkN = 0.2N must act to the left to
oppose the motion of the crate which is to the right, Fig. a.
laws
in
teaching
Web)
N - 20(9.81) = 20(0)
Wide
copyright
+ c ©Fy = may;
or
Equations of Motion: Here, ay = 0. Thus,
instructors
States
World
permitted.
Dissemination
N = 196.2 N
United
on
learning.
is
of
the
not
Using the results of N and a,
by
and
use
T - 0.2(196.2) = 20(0.375)
student
the
+ ©F = ma ;
:
x
x
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This
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the
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T = 46.7 N
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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Ans.
13–25.
If the motor draws in the cable at a rate of v = (0.05t2) m>s,
where t is in seconds, determine the tension developed in the
cable when t = 5 s. The crate has a mass of 20 kg and the
coefficient of kinetic friction between the crate and the ground
is mk = 0.2.
s
v
M
SOLUTION
Kinematics: Since the motion of the crate is known, its acceleration a will be
determined first.
a =
dv
= 0.05(2t) = (0.1t) m>s2
dt
When t = 5 s,
a = 0.1(5) = 0.5 m>s2 :
Free-Body Diagram: The kinetic friction Ff = mkN = 0.2N must act to the left to
oppose the motion of the crate which is to the right, Fig. a.
laws
in
teaching
Web)
N - 20(9.81) = 0
Wide
copyright
+ c ©Fy = may;
or
Equations of Motion: Here, ay = 0. Thus,
instructors
States
World
permitted.
Dissemination
N = 196.2 N
United
on
learning.
is
of
the
not
Using the results of N and a,
by
and
use
T - 0.2(196.2) = 20(0.5)
student
the
+ ©F = ma ;
:
x
x
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This
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T = 49.2 N
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Ans.
13–26.
The 2-kg shaft CA passes through a smooth journal bearing
at B. Initially, the springs, which are coiled loosely around
the shaft, are unstretched when no force is applied to the
shaft. In this position s = s¿ = 250 mm and the shaft is at
rest. If a horizontal force of F = 5 kN is applied, determine
the speed of the shaft at the instant s = 50 mm,
s¿ = 450 mm. The ends of the springs are attached to the
bearing at B and the caps at C and A.
s¿
s
3 kN/m
kCB
SOLUTION
FCB = kCBx = 3000x
+ ©F = ma ;
;
x
x
FAB = kABx = 2000x
5000 - 3000x - 2000x = 2a
2500 - 2500x = a
a dx - v dy
v
0.2
(2500 - 2500x) dx =
v dv
or
2500(0.2)2
v2
≤ =
2
2
teaching
Web)
2500(0.2) - ¢
L0
laws
L0
v = 30 m>s
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This
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A
B
C
kAB
2 kN/m
F
5 kN
13–27.
The 30-lb crate is being hoisted upward with a constant
acceleration of 6 ft>s2. If the uniform beam AB has a weight
of 200 lb, determine the components of reaction at the fixed
support A. Neglect the size and mass of the pulley at B.
Hint: First find the tension in the cable, then analyze the
forces in the beam using statics.
y
5 ft
A
B
x
SOLUTION
6 ft/s2
Crate:
+ c ©Fy = may ;
T - 30 = a
30
b(6)
32.2
T = 35.59 lb
Beam:
+ ©F = 0;
:
x
a + c ©Fy = 0;
Ax = 35.6 lb
Ay - 200 - 35.59 = 0
Ans.
Ay = 236 lb
Ans.
MA = 678 lb # ft
MA - 200(2.5) - (35.59)(5) = 0
Ans.
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not
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permitted.
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Web)
laws
or
+ ©MA = 0;
- Ax + 35.59 = 0
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*13–28.
The driver attempts to tow the crate using a rope that has a
tensile strength of 200 lb. If the crate is originally at rest and
has a weight of 500 lb, determine the greatest acceleration it
can have if the coefficient of static friction between the
crate and the road is ms = 0.4, and the coefficient of kinetic
friction is mk = 0.3.
30
SOLUTION
Equilibrium: In order to slide the crate, the towing force must overcome static
friction.
+ ©F = 0;
:
x
- T cos 30° +0.4N = 0
(1)
+ ©F = 0;
:
N + T sin 30° -500 = 0
(2)
Solving Eqs.(1) and (2) yields:
T = 187.6 lb
N = 406.2 lb
laws
or
Since T 6 200 lb, the cord will not break at the moment the crate slides.
teaching
Web)
After the crate begins to slide, the kinetic friction is used for the calculation.
N = 400 lb
in
N + 200 sin 30° - 500 = 0
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+ c ©Fy = ma y;
World
permitted.
Dissemination
500
a
32.2
learning.
is
of
the
not
instructors
200 cos 30° - 0.3(400) =
States
+ ©F = ma ;
:
x
x
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a = 3.43 ft>s2
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Ans.
13–29.
F (lb)
The force exerted by the motor on the cable is shown in the
graph. Determine the velocity of the 200-lb crate when
t = 2.5 s.
250 lb
M
t (s)
2.5
SOLUTION
Free-Body Diagram: The free-body diagram of the crate is shown in Fig. a.
Equilibrium: For the crate to move, force F must overcome the weight of the crate.
Thus, the time required to move the crate is given by
A
+ c ©Fy = 0;
t = 2s
100t - 200 = 0
250
t = (100t) lb. By referring to Fig. a,
2.5
Equation of Motion: For 2 s 6 t 6 2.5 s, F =
+ c ©Fy = may;
100t - 200 =
200
a
32.2
permitted.
Dissemination
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Kinematics: The velocity of the crate can be obtained by integrating the kinematic
equation, dv = adt. For 2 s … t 6 2.5 s, v = 0 at t = 2 s will be used as the lower
integration limit.Thus,
Web)
laws
or
a = (16.1t - 32.2) ft>s2
World
adt
not
the
on
learning.
is
L
States
dv =
instructors
L
of
(+ c )
United
t
and
use
v
by
(16.1t - 32.2)dt
work
student
the
L2 s
for
L0
dv =
the
(including
t
of
protected
is
2s
work
solely
= A 8.05t2 - 32.2t + 32.2 B ft>s
part
the
is
When t = 2.5 s,
This
provided
integrity
of
and
work
this
assessing
v = A 8.05t2 - 32.2t B 2
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v = 8.05(2.52) - 32.2(2.5) + 32.2 = 2.01 ft>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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Ans.
13–30.
F (N)
The force of the motor M on the cable is shown in the graph.
Determine the velocity of the 400-kg crate A when t = 2 s.
2500
F ⫽ 625 t 2
M
2
SOLUTION
Free-Body Diagram: The free-body diagram of the crate is shown in Fig. a.
Equilibrium: For the crate to move, force 2F must overcome its weight. Thus, the
time required to move the crate is given by
+ c ©Fy = 0;
2(625t2) - 400(9.81) = 0
A
t = 1.772 s
Equations of Motion: F = A 625t2 B N. By referring to Fig. a,
+ c ©Fy = may;
2 A 625t2 B - 400(9.81) = 400a
laws
or
a = (3.125t2 - 9.81) m>s2
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
Kinematics: The velocity of the crate can be obtained by integrating the kinematic
equation, dv = adt. For 1.772 s … t 6 2 s, v = 0 at t = 1.772 s will be used as the
lower integration limit. Thus,
adt
not
the
is
L
instructors
dv =
learning.
L
of
(+ c )
and
use
United
on
A 3.125t2 - 9.81 B dt
work
student
L1.772 s
by
L0
t
dv =
the
v
for
v = A 1.0417t3 - 9.81t B 2
of
protected
the
(including
t
part
the
is
and
any
courses
When t = 2 s,
This
provided
integrity
of
and
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this
assessing
is
= A 1.0417t - 9.81t + 11.587 B m>s
work
solely
1.772 s
3
sale
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destroy
of
v = 1.0417(23) - 9.81(2) + 11.587 = 0.301 m>s
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Ans.
t (s)
13–31.
The tractor is used to lift the 150-kg load B with the 24-mlong rope, boom, and pulley system. If the tractor travels to
the right at a constant speed of 4 m> s, determine the tension
in the rope when sA = 5 m. When sA = 0, sB = 0.
12 m
sB
SOLUTION
B
A
12 - sB + 2s2A + (12)2 = 24
1
#
-sB + A s2A + 144 B - 2 asAsA b = 0
sA
3
1
1
# 2
#
$
$
-sB - A s2A + 144 B - 2 a sAsA b + A s2A + 144 B - 2 a s2A b + A s2A + 144 B - 2 asA sA b = 0
3
2
((5)2 + 144)
-
A s2A + 144 B 2
1
S
or
(4)2 + 0
1
((5)2 + 144)2
S = 1.0487 m>s2
Web)
(5)2(4)2
$
#
s2A + sAsA
laws
A s2A + 144 B
-
3
2
Wide
copyright
in
aB = - C
#
s2As2A
teaching
$
sB = - C
T - 150(9.81) = 150(1.0487)
States
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Dissemination
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Ans.
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T = 1.63 kN
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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*13–32.
The tractor is used to lift the 150-kg load B with the 24-mlong rope, boom, and pulley system. If the tractor travels to
the right with an acceleration of 3 m>s2 and has a velocity of
4 m> s at the instant sA = 5 m, determine the tension in the
rope at this instant. When sA = 0, sB = 0.
12 m
sB
SOLUTION
B
A
12 = sB + 2s2A + (12)2 = 24
sA
3
1
#
#
-sB + A s2A + 144 B - 2 a 2sA sA b = 0
2
3
1
1
# 2
#
$
$
-sB - A s2A + 144 B - 2 a sAsA b + A s2A + 144 B - 2 a s2A b + A s2A + 144 B - 2 asAsA b = 0
(5)2(4)2
3
2
((5)2 + 144)
-
A s2A + 144 B 2
1
S
(4)2 + (5)(3)
1
((5)2 + 144)2
or
3
S = 2.2025 m>s2
Web)
A s2A + 144 B 2
$
#
s2A + sA sA
laws
-
Wide
copyright
in
aB = - C
#
s2A s2A
teaching
$
sB = - C
T - 150(9.81) = 150(2.2025)
States
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Dissemination
+ c ©Fy = may ;
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This
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the
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the
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of
the
Ans.
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not
instructors
T = 1.80 kN
13–33.
Each of the three plates has a mass of 10 kg. If the
coefficients of static and kinetic friction at each surface of
contact are ms = 0.3 and mk = 0.2, respectively, determine
the acceleration of each plate when the three horizontal
forces are applied.
18 N
D
C
15 N
B
A
SOLUTION
Plates B, C and D
+ ©F = 0;
:
x
100 - 15 - 18 - F = 0
F = 67 N
Fmax = 0.3(294.3) = 88.3 N 7 67 N
Plate B will not slip.
aB = 0
Ans.
100 - 18 - F = 0
teaching
Web)
laws
+ ©F = 0;
:
x
or
Plates D and C
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F = 82 N
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Dissemination
Fmax = 0.3(196.2) = 58.86 N 6 82N
by
and
use
United
on
learning.
is
of
the
Slipping between B and C.
for
work
student
the
Assume no slipping between D and C,
the
(including
100 - 39.24 - 18 = 20 ax
is
work
solely
of
protected
+ ©F = ma ;
:
x
x
part
the
is
Check slipping between D and C.
This
provided
integrity
of
and
work
this
assessing
ax = 2.138 m>s2 :
any
courses
F - 18 = 10(2.138)
will
sale
F = 39.38 N
their
destroy
of
and
+ ©F = m a ;
:
x
x
Fmax = 0.3(98.1) = 29.43 N 6 39.38 N
Slipping between D and C.
Plate C:
+ ©F = m a ;
:
x
x
100 - 39.24 - 19.62 = 10 ac
ac = 4.11 m>s2 :
Ans.
Plate D:
+ ©F = m a ;
:
x
x
19.62 - 18 = 10 a D
a D = 0.162m>s2 :
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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Ans.
100 N
13–34.
Each of the two blocks has a mass m. The coefficient of
kinetic friction at all surfaces of contact is m. If a horizontal
force P moves the bottom block, determine the acceleration
of the bottom block in each case.
B
P
A
(a)
B
SOLUTION
P
A
Block A:
+ ©F = ma ;
(a) ;
x
x
(b)
P - 3mmg = m aA
aA =
P
-3mg
m
Ans.
(b) sB + sA = l
aA = - aB
(1)
Block A:
P - T - 3mmg = m aA
or
(2)
laws
+ ©F = m a ;
;
x
x
in
teaching
Web)
Block B:
mmg - T = maB
Wide
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(3)
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Dissemination
+ ©F = ma ;
;
x
x
learning.
is
of
the
not
instructors
Subtract Eq.(3) from Eq.(2):
work
student
the
by
and
use
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P - 4mmg = m (aA - aB)
the
(including
for
P
- 2mg
2m
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aA =
protected
Use Eq.(1);
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Ans.
13–35.
The conveyor belt is moving at 4 m>s. If the coefficient of
static friction between the conveyor and the 10-kg package
B is ms = 0.2, determine the shortest time the belt can stop
so that the package does not slide on the belt.
B
SOLUTION
+ ©F = m a ;
:
x
x
0.2198.12 = 10 a
a = 1.962 m>s2
+ 2v = v + a t
1:
0
c
4 = 0 + 1.962 t
t = 2.04 s
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This
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the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
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in
teaching
Web)
laws
or
Ans.
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or transmission in any form or by any means, electronic, mechanical,
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*13–36.
The 2-lb collar C fits loosely on the smooth shaft. If the
spring is unstretched when s = 0 and the collar is given a
velocity of 15 ft> s, determine the velocity of the collar when
s = 1 ft.
15 ft/s
s
C
1 ft
k
SOLUTION
Fs = kx;
Fs = 4 A 21 + s2 - 1 B
+ ©F = ma ;
:
x
x
1
-
L0
¢ 4s ds -
s
-4 A 21 + s2 - 1 B ¢
4s ds
21 + s
1
- C 2s2 - 4 31 + s2 D 0 =
2
≤ =
v
L15
a
21 + s
2
2
dv
b av b
32.2
ds
≤ = a
2
b v dv
32.2
1
A v2 - 152 B
32.2
v = 14.6 ft>s
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This
provided
integrity
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this
assessing
is
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of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
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permitted.
Dissemination
Wide
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in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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4 lb/ft
13–37.
Cylinder B has a mass m and is hoisted using the cord and
pulley system shown. Determine the magnitude of force F
as a function of the cylinder’s vertical position y so that
when F is applied the cylinder rises with a constant
acceleration aB. Neglect the mass of the cord, pulleys, hook
and chain.
d/2
d/2
F
y
SOLUTION
+ c ©Fy = may ;
2F cos u - mg = maB
2F £
2y +
A d2 B 2
aB
y
A B
d 2
2
2y +
2
≥ - mg = maB
m(aB + g)24y2 + d2
4y
Ans.
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is
This
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integrity
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assessing
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the
(including
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student
the
by
and
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F =
y
2
where cos u =
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or transmission in any form or by any means, electronic, mechanical,
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B
13–38.
vA
The conveyor belt delivers each 12-kg crate to the ramp at
A such that the crate’s speed is vA = 2.5 m>s, directed down
along the ramp. If the coefficient of kinetic friction between
each crate and the ramp is mk = 0.3, determine the speed at
which each crate slides off the ramp at B. Assume that no
tipping occurs. Take u = 30°.
2.5 m/s
A
3m
u
SOLUTION
Q +©Fy = may;
NC - 1219.812 cos 30° = 0
NC = 101.95 N
+ R©Fx = max ;
1219.812 sin 30° - 0.31101.952 = 12 aC
aC = 2.356 m>s2
(+ R)
2
vB2 = vA
+ 2ac1sB - sA2
v2B = 12.522 + 212.356213 - 02
vB = 4.5152 = 4.52 m>s
sale
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This
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integrity
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this
assessing
is
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of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
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in
teaching
Web)
laws
or
Ans.
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B
13–39.
y
An electron of mass m is discharged with an initial
horizontal velocity of v0. If it is subjected to two fields of
force for which Fx = F0 and Fy = 0.3F0, where F0 is
constant, determine the equation of the path, and the speed
of the electron at any time t.
++++++++++++++
v0
+ ©F = ma ;
:
x
x
F0 = max
+ c ©Fy = may;
0.3 F0 = may
Thus,
vx
t
dvx =
Lv0
F0
dt
L0 m
F0
t + v0
m
vy
t
0.3F0
dvy =
dt
L0
L0 m
vx =
0.3F0
t
m
2
F0
0.3F0 2
t + v0 b + a
tb
m
C m
Web)
laws
teaching
permitted.
Dissemination
copyright
World
F0
t + v0 b dt
m
the
not
instructors
is
use
United
on
a
learning.
L0
States
t
dx =
Ans.
of
x
in
1
21.09F20 t2 + 2F0tmv0 + m2v20
m
=
L0
or
a
v =
work
student
the
by
and
F0 t2
+ v0 t
2m
(including
for
x =
t
the
of
work
solely
integrity
of
and
provided
part
the
is
This
any
courses
x =
F0 2m
1
2m
a
by + v0 a
by2
2m 0.3F0
B 0.3F0
x =
y
1
2m
+ v0 a
by 2
0.3
B 0.3F0
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destroy
sale
will
1
2m
by 2
B 0.3F0
of
t = a
and
0.3F0 t2
2m
their
y =
work
this
assessing
L0
0.3F0
t dt
L0 m
protected
dy =
is
y
Wide
vy =
Ans.
++++++++++++++
SOLUTION
x
*13–40.
v
The engine of the van produces a constant driving traction
force F at the wheels as it ascends the slope at a constant
velocity v. Determine the acceleration of the van when it
passes point A and begins to travel on a level road, provided
that it maintains the same traction force.
A
F
u
SOLUTION
Free-Body Diagram: The free-body diagrams of the van up the slope and on the
level road are shown in Figs. a and b, respectively.
Equations of Motion: Since the van is travelling up the slope with a constant
velocity, its acceleration is a = 0. By referring to Fig. a,
©Fx ¿ = max ¿;
F - mg sin u = m(0)
F = mg sin u
Since the van maintains the same tractive force F when it is on level road, from Fig. b,
mg sin u = ma
or
+ ©F = ma ;
:
x
x
a = g sin u
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the
(including
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by
and
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on
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is
of
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permitted.
Dissemination
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copyright
in
teaching
Web)
laws
Ans.
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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13–41.
The 2-kg collar C is free to slide along the smooth shaft AB.
Determine the acceleration of collar C if (a) the shaft is
fixed from moving, (b) collar A, which is fixed to shaft AB,
moves downward at constant velocity along the vertical rod,
and (c) collar A is subjected to a downward acceleration of
2 m>s2. In all cases, the collar moves in the plane.
B
45
A
SOLUTION
(a) + b©Fx¿ = max¿ ;
219.812 sin 45° = 2aC
aC = 6.94 m>s2
(b) From part (a) a C>A = 6.94 m>s2
a C = a A + a C>A
Where a A = 0
= 6.94 m>s2
(c)
aC = aA + aC>A
= 2 + aC>A
b
√T
laws
or
(1)
aC>A = 5.5225 m>s2 b
Web)
219.812 sin 45° = 212 cos 45°+ aC>A2
in
teaching
+ b©Fx¿ = max¿ ;
Dissemination
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copyright
From Eq.(1)
on
learning.
is
of
the
not
instructors
States
World
permitted.
aC = 2 + 5.52 25 = 3.905 + 5.905
;
b
T
T
Ans.
student
the
by
and
use
United
aC = 23.9052 + 5.9052 = 7.08 m>s2
the
(including
for
work
5.905
= 56.5° ud
3.905
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is
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this
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protected
u = tan-1
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or transmission in any form or by any means, electronic, mechanical,
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Ans.
C
13–42.
The 2-kg collar C is free to slide along the smooth shaft AB.
Determine the acceleration of collar C if collar A is
subjected to an upward acceleration of 4 m>s2.
B
45⬚
A
SOLUTION
+ ©F = ma ;
;
x
x
N sin 45° = 2aC>AB sin 45°
N = 2 aC>AB
+ c ©Fy = may ;
N cos 45° - 19.62 = 2142 - 2aC>AB cos 45°
aC>AB = 9.76514
aC = aAB + aC>AB
1aC)x = 0 + 9.76514 sin 45° = 6.905 ;
(aC)y = 4 - 9.76514 cos 45° = 2.905T
or
aC = 216.90522 + 12.90522 = 7.49 m>s2
teaching
Web)
laws
Ans.
2.905
b = 22.8° ud
6.905
in
Ans.
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copyright
u = tan-1 a
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C
13–43.
The coefficient of static friction between the 200-kg crate
and the flat bed of the truck is ms = 0.3. Determine the
shortest time for the truck to reach a speed of 60 km> h,
starting from rest with constant acceleration, so that the
crate does not slip.
SOLUTION
Free-Body Diagram: When the crate accelerates with the truck, the frictional
force Ff develops. Since the crate is required to be on the verge of slipping,
Ff = msN = 0.3N.
Equations of Motion: Here, ay = 0. By referring to Fig. a,
+ c ©Fy = may;
N - 200(9.81) = 200(0)
N = 1962 N
+ ©F = ma ;
:
x
x
-0.3(1962) = 200(- a)
or
a = 2.943 m>s2 ;
km 1000 m
1h
ba
ba
b =
h
1 km
3600 s
16.67 m>s. Since the acceleration of the truck is constant,
Dissemination
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Web)
laws
Kinematics: The final velocity of the truck is v = a 60
World
permitted.
v = v0 + ac t
is
of
the
not
instructors
States
+ )
(;
by
and
use
United
on
learning.
16.67 = 0 + 2.943t
Ans.
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the
(including
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work
student
the
t = 5.66 s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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*13–44.
When the blocks are released, determine their acceleration
and the tension of the cable. Neglect the mass of the pulley.
SOLUTION
Free-Body Diagram: The free-body diagram of blocks A and B are shown in Figs. b
and c, respectively. Here, aA and aB are assumed to be directed downwards so that
they are consistent with the positive sense of position coordinates sA and sB of
blocks A and B, Fig. a. Since the cable passes over the smooth pulleys, the tension in
the cable remains constant throughout.
A
10 kg
B
30 kg
Equations of Motion: By referring to Figs. b and c,
+ c ©Fy = may;
2T - 10(9.81) = - 10aA
(1)
T - 30(9.81) = - 30aB
(2)
and
laws
or
+ c ©Fy = may;
Wide
copyright
in
teaching
Web)
Kinematics: We can express the length of the cable in terms of sA and sB by referring
to Fig. a.
instructors
States
World
permitted.
Dissemination
2sA + sB = l
United
on
learning.
is
of
the
not
The second derivative of the above equation gives
(3)
for
work
student
the
by
and
use
2aA + aB = 0
of
protected
the
(including
Solving Eqs. (1), (2), and (3) yields
work
solely
aB = 7.546 m>s2 = 7.55 m>s2 T
Ans.
this
assessing
is
aA = - 3.773 m>s2 = 3.77 m>s2 c
sale
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is
This
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T = 67.92 N = 67.9 N
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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Ans.
13–45.
If the force exerted on cable AB by the motor is
F = (100t3>2) N, where t is in seconds, determine the 50-kg
crate’s velocity when t = 5 s. The coefficients of static and
kinetic friction between the crate and the ground are ms = 0.4
and mk = 0.3, respectively. Initially the crate is at rest.
A
SOLUTION
Free-Body Diagram: The frictional force Ff is required to act to the left to oppose
the motion of the crate which is to the right.
Equations of Motion: Here, ay = 0. Thus,
N - 50(9.81) = 50(0)
+ c ©Fy = may;
N = 490.5 N
Realizing that Ff = mkN = 0.3(490.5) = 147.15 N,
100t3>2 - 147.15 = 50a
laws
or
a = A 2t3>2 - 2.943 B m>s
permitted.
Dissemination
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copyright
in
teaching
Equilibrium: For the crate to move, force F must overcome the static friction of
Ff = msN = 0.4(490.5) = 196.2 N. Thus, the time required to cause the crate to be
on the verge of moving can be obtained from.
Web)
+ c ©Fx = max;
learning.
is
of
the
not
instructors
States
World
100t3>2 - 196.2 = 0
+ ©F = 0;
:
x
the
by
and
use
United
on
t = 1.567 s
is
work
solely
of
protected
the
(including
for
work
student
Kinematics: Using the result of a and integrating the kinematic equation dv = a dt
with the initial condition v = 0 at t = 1.567 as the lower integration limit,
this
work
provided
integrity
of
and
A 2t3>2 - 2.943 B dt
part
any
t
destroy
v = A 0.8t5>2 - 2.943t B 2
v = A 0.8t5>2 - 2.943t + 2.152 B m>s
sale
1.567 s
will
L1.567 s
the
is
This
t
dv =
assessing
adt
courses
L0
L
of
dv =
v
and
L
their
+ )
(:
When t = 5 s,
v = 0.8(5)5>2 - 2.943(5) + 2.152 = 32.16 ft>s = 32.2 ft>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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Ans.
B
13–46.
Blocks A and B each have a mass m. Determine the largest
horizontal force P which can be applied to B so that A will
not move relative to B. All surfaces are smooth.
A
P
u B
C
SOLUTION
Require
aA = aB = a
Block A:
+ c ©Fy = 0;
+ ©F = ma ;
;
x
x
N cos u - mg = 0
N sin u = ma
a = g tan u
P - N sin u = ma
teaching
Web)
laws
+ ©F = ma ;
;
x
x
or
Block B:
Wide
copyright
in
P - mg tan u = mg tan u
P = 2mg tan u
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the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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13–47.
Blocks A and B each have a mass m. Determine the largest
horizontal force P which can be applied to B so that A will
not slip on B. The coefficient of static friction between A and
B is ms. Neglect any friction between B and C.
A
C
SOLUTION
Require
aA = aB = a
Block A:
N sin u + ms N cos u = ma
N =
mg
cos u - ms sin u
sin u + ms cos u
b
cos u - ms sin u
in
teaching
Web)
a = ga
or
+ ©F = ma ;
;
x
x
N cos u - ms N sin u - mg = 0
laws
+ c ©Fy = 0;
Dissemination
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Block B:
World
permitted.
P - ms N cos u - N sin u = ma
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sin u + ms cos u
b
cos u - ms sin u
protected
P = 2mg a
student
the
by
P - mga
is
of
sin u + ms cos u
sin u + ms cos u
b = mga
b
cos u - ms sin u
cos u - ms sin u
the
not
instructors
States
+ ©F = ma ;
;
x
x
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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P
u B
Ans.
*13–48.
A parachutist having a mass m opens his parachute from an
at-rest position at a very high altitude. If the atmospheric
drag resistance is FD = kv2, where k is a constant,
determine his velocity when he has fallen for a time t. What
is his velocity when he lands on the ground? This velocity is
referred to as the terminal velocity, which is found by letting
the time of fall t : q .
FD
v
SOLUTION
+ T ©Fz = m az ;
mg - kv2 = m
v
m
dv
dt
t
m dv
dt
=
2
L0 1mg - kv 2
L0
v
m
dv
= t
mg
k L0
- v2
k
v
mg
+ v
m
A k
1
T = t
ln D
mg
k ¢ mg
2
- v
A k
A k
0
in
teaching
Web)
laws
or
¢
Wide
copyright
mg
+ v
mg
A k
k
t ¢2
= ln
mg
m A k
- v
A k
and
use
United
on
learning.
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of
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¢
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the
(including
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work
student
the
by
mg
+ v
A k
=
mg
- v
A k
and
work
this
assessing
is
e
2t2mg
k
and
any
courses
part
the
is
This
provided
integrity
of
mg
mg e 2t2 k - v e2t2mgk
mg
=
+ v
A k
A k
When t : q
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vt =
mg
A k
sale
will
their
destroy
of
2t2mg
k
- 1
mg e
v =
C
S
mg
A k
e 2t2 k + 1
Ans.
Ans.
13–49.
The smooth block B of negligible size has a mass m and
rests on the horizontal plane. If the board AC pushes on the
block at an angle u with a constant acceleration a0,
determine the velocity of the block along the board and the
distance s the block moves along the board as a function of
time t. The block starts from rest when s = 0, t = 0.
a0
θ
A
B
s
SOLUTION
Q+ ©Fx = m ax;
0 = m aB sin f
aB = aAC + aB>AC
aB = a0 + aB>AC
Q+
aB sin f = - a0 sin u + aB>AC
Thus,
0 = m( -a0 sin u + aB>AC)
laws
Web)
teaching
in
a0 sin u dt
Wide
L0
Dissemination
L0
t
dvB>AC =
copyright
vB>AC
or
aB>AC = a0 sin u
permitted.
vB>AC = a0 sin u t
of
the
not
instructors
States
World
Ans.
on
learning.
is
t
and
use
United
a0 sin u t dt
by
for
work
student
L0
the
sB>AC = s =
the
(including
1
a sin u t2
2 0
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assessing
is
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s =
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Ans.
C
13–50.
A projectile of mass m is fired into a liquid at an angle u0
with an initial velocity v0 as shown. If the liquid develops a
frictional or drag resistance on the projectile which is
proportional to its velocity, i.e., F = kv, where k is a
constant, determine the x and y components of its position
at any instant. Also, what is the maximum distance xmax that
it travels?
y
v0
θ0
O
SOLUTION
+ ©F = ma ;
:
x
x
- kv cos u = m ax
+ c ©Fy = m ay ;
-m g - k v sin u = m ay
or
-k
d2x
dx
= m 2
dt
dt
-mg - k
dy
d2y
= m 2
dt
dt
laws
or
Integrating yields
Dissemination
Wide
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teaching
Web)
-k
#
t + C1
In x =
m
on
learning.
is
of
the
not
instructors
States
World
permitted.
mg
k
#
t + C2
) =
In (y +
m
k
by
and
use
United
#
y = v0 sin u0
the
#
When t = 0, x = v0 cos u0,
the
is
This
provided
integrity
of
and
work
this
assessing
is
mg
mg -(k>m)t
#
y = + (v0 sin u0 +
)e
k
k
work
solely
of
protected
the
(including
for
work
student
#
x = v0 cos u0 e-(k>m)t
and
any
courses
part
Integrating again,
will
their
destroy
of
-m v0
cos u0 e-(k>m)t + C3
k
sale
x =
y = -
mg
mg m -(k>m)t
t - (v0 sin u0 +
)( )e
k
k k
When t = 0, x = y = 0, thus
x =
y = -
m v0
cos u0(1 - e-(k>m)t)
k
Ans.
mgt
mg
m
+
(v0 sin u0 +
)(1 - e-(k>m)t)
k
k
k
Ans.
m v0 cos u0
k
Ans.
As t : q
xmax =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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x
13–51.
The block A has a mass mA and rests on the pan B, which
has a mass mB . Both are supported by a spring having a
stiffness k that is attached to the bottom of the pan and to
the ground. Determine the distance d the pan should be
pushed down from the equilibrium position and then
released from rest so that separation of the block will take
place from the surface of the pan at the instant the spring
becomes unstretched.
A
B
y d
SOLUTION
For Equilibrium
+ c ©Fy = may;
Fs = (mA + mB)g
yeq =
(mA + mB)g
Fs
=
k
k
Block:
+ c ©Fy = may ;
-mA g + N = mA a
Block and pan
or
- (mA + mB)g + k(yeq + y) = (mA + mB)a
teaching
Web)
laws
+ c ©Fy = may;
Wide
copyright
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Thus,
World
permitted.
Dissemination
-mAg + N
mA + mB
bg + y d = (mA + mB) a
b
mA
k
is
of
the
not
instructors
States
-(mA + mB)g + k c a
the
by
and
use
United
on
learning.
Require y = d, N = 0
the
(including
for
work
student
kd = - (mA + mB)g
provided
integrity
of
and
work
(mA + mB)g
k
sale
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This
d =
this
assessing
is
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protected
Since d is downward,
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Ans.
k
*13–52.
A girl, having a mass of 15 kg, sits motionless relative to the
surface of a horizontal platform at a distance of r = 5 m from
the platform’s center. If the angular motion of the platform is
slowly increased so that the girl’s tangential component of
acceleration can be neglected, determine the maximum speed
which the girl will have before she begins to slip off the
platform. The coefficient of static friction between the girl and
the platform is m = 0.2.
z
5m
SOLUTION
Equation of Motion: Since the girl is on the verge of slipping, Ff = msN = 0.2N.
Applying Eq. 13–8, we have
©Fb = 0 ;
©Fn = man ;
N - 15(9.81) = 0
N = 147.15 N
0.2(147.15) = 15 a
v2
b
5
v = 3.13 m>s
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Web)
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Ans.
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13–53.
The 2-kg block B and 15-kg cylinder A are connected to a
light cord that passes through a hole in the center of the
smooth table. If the block is given a speed of v = 10 m>s,
determine the radius r of the circular path along which it
travels.
r
B
v
SOLUTION
A
Free-Body Diagram: The free-body diagram of block B is shown in Fig. (a). The
tension in the cord is equal to the weight of cylinder A, i.e.,
T = 15(9.81) N = 147.15 N. Here, an must be directed towards the center of the
circular path (positive n axis).
v2
102
and referring to Fig. (a),
=
r
r
Equations of Motion: Realizing that an =
©Fn = man;
147.15 = 2 a
102
b
r
r = 1.36 m
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Ans.
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13–54.
The 2-kg block B and 15-kg cylinder A are connected to a
light cord that passes through a hole in the center of the
smooth table. If the block travels along a circular path of
radius r = 1.5 m, determine the speed of the block.
r
B
v
SOLUTION
A
Free-Body Diagram: The free-body diagram of block B is shown in Fig. (a). The
tension in the cord is equal to the weight of cylinder A, i.e.,
T = 15(9.81) N = 147.15 N. Here, an must be directed towards the center of the
circular path (positive n axis).
v2
v2
and referring to Fig. (a),
=
r
1.5
Equations of Motion: Realizing that an =
©Fn = man;
147.15 = 2 a
2
v
b
1.5
v = 10.5 m>s
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Ans.
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13–55.
The 5-kg collar A is sliding around a smooth vertical guide
rod. At the instant shown, the speed of the collar is
v = 4 m>s, which is increasing at 3 m>s 2. Determine the
normal reaction of the guide rod on the collar, and force P
at this instant.
v ⫽ 4 m/s
30⬚
A
0.5 m
SOLUTION
+ ©F = ma ;
:
t
t
P cos 30° = 5(3)
P = 17.32 N = 17.3 N
+ T ©Fn = man;
Ans.
N + 5 A 9.81 B - 17.32 sin 30° = 5a
42
b
0.5
N = 119.61 N = 120 N T
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Ans.
will
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© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
P
*13–56.
Cartons having a mass of 5 kg are required to move along
the assembly line at a constant speed of 8 m/s. Determine
the smallest radius of curvature, r, for the conveyor so the
cartons do not slip. The coefficients of static and kinetic
friction between a carton and the conveyor are ms = 0.7 and
mk = 0.5, respectively.
8 m/s
ρ
SOLUTION
+ c ©Fb = m ab ;
N - W = 0
N = W
Fx = 0.7W
+ ©F = m a ;
;
n
n
0.7W =
W 82
( )
9.81 r
r = 9.32 m
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13–57.
The block B, having a mass of 0.2 kg, is attached to the
vertex A of the right circular cone using a light cord. If the
block has a speed of 0.5 m>s around the cone, determine
the tension in the cord and the reaction which the cone
exerts on the block and the effect of friction.
z
200 mm
B
400 mm
SOLUTION
r
300
=
;
200
500
+ Q©Fy = may;
300 mm
r = 120 mm = 0.120 m
(0.5)2
4
3
T - 0.2(9.81)a b = B 0.2 ¢
≤Ra b
5
0.120
5
T = 1.82 N
+ a©Fx = max;
Ans.
(0.5)2
3
4
NB - 0.2(9.81)a b = - B 0.2 ¢
≤Ra b
5
0.120
5
NB = 0.844 N
or
Ans.
teaching
Web)
laws
Also,
in
(0.5)2
3
4
b
Ta b - NB a b = 0.2 a
5
5
0.120
permitted.
Dissemination
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copyright
+ ©F = ma ;
:
n
n
the
not
instructors
States
World
4
3
Ta b + NB a b - 0.2(9.81) = 0
5
5
use
United
on
learning.
is
of
+ c ©Fb = 0;
and
T = 1.82 N
for
work
student
the
by
Ans.
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NB = 0.844 N
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Ans.
A
13–58.
The 2-kg spool S fits loosely on the inclined rod for which the
coefficient of static friction is ms = 0.2. If the spool is located
0.25 m from A, determine the minimum constant speed the
spool can have so that it does not slip down the rod.
z
5
4
S
0.25 m
A
SOLUTION
4
r = 0.25a b = 0.2 m
5
+ ©F = m a ;
;
n
n
v2
3
4
b
Ns a b - 0.2Ns a b = 2 a
5
5
0.2
+ c ©Fb = m ab;
4
3
Ns a b + 0.2Ns a b - 2(9.81) = 0
5
5
Ns = 21.3 N
v = 0.969 m>s
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This
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© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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3
13–59.
z
The 2-kg spool S fits loosely on the inclined rod for which
the coefficient of static friction is ms = 0.2. If the spool is
located 0.25 m from A, determine the maximum constant
speed the spool can have so that it does not slip up the rod.
5
S
0.25 m
A
SOLUTION
4
r = 0.25( ) = 0.2 m
5
+ ©F = m a ;
;
n
n
+ c ©Fb = m ab ;
3
4
v2
Ns( ) + 0.2Ns( ) = 2( )
5
5
0.2
4
3
Ns( ) - 0.2Ns( ) - 2(9.81) = 0
5
5
Ns = 28.85 N
v = 1.48 m s
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This
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sale
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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3
4
*13–60.
At the instant u = 60°, the boy’s center of mass G has a
downward speed vG = 15 ft>s. Determine the rate of
increase in his speed and the tension in each of the two
supporting cords of the swing at this instant. The boy has a
weight of 60 lb. Neglect his size and the mass of the seat
and cords.
u
10 ft
SOLUTION
+ R©Ft = mat ;
60 cos 60° =
60
a
32.2 t
2T - 60 sin 60° =
at = 16.1 ft>s2
60 152
a
b
32.2 10
Ans.
T = 46.9 lb
Ans.
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Q+ ©Fn = man ;
G
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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13–61.
At the instant u = 60°, the boy’s center of mass G is
momentarily at rest. Determine his speed and the tension in
each of the two supporting cords of the swing when
u = 90°. The boy has a weight of 60 lb. Neglect his size and
the mass of the seat and cords.
u
10 ft
SOLUTION
G
60
60 cos u =
a
32.2 t
+R© t = mat;
Q+ ©Fn = man;
2T - 60 sin u =
v dn = a ds
v
L0
at = 32.2 cos u
60 v2
a b
32.2 10
(1)
however ds = 10du
90°
v dn =
L60°
322 cos u du
v = 9.289 ft>s
or
Ans.
Ans.
Wide
T = 38.0 lb
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Dissemination
60 9.2892
a
b
32.2
10
copyright
2T - 60 sin 90° =
in
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From Eq. (1)
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13–62.
The 10-lb suitcase slides down the curved ramp for which the
coefficient of kinetic friction is mk = 0.2. If at the instant it
reaches point A it has a speed of 5 ft>s, determine the normal
force on the suitcase and the rate of increase of its speed.
y
1 x2
y = ––
8
6 ft
A
SOLUTION
x
1 2
x
8
n =
dy
1
= tan u = x 2
= - 1.5
dx
4 x= -6
d2y
=
2
dx
u = - 56.31°
1
4
3
or
Web)
(5)2
10
b¢
≤
32.2 23.436
N - 10 cos 56.31° = a
Dissemination
+Q©Fn = man ;
Wide
2
= 23.436 ft
laws
dx2
212
4
teaching
2
d2y
C 1 + (- 1.5)2 D 2
3
=
in
r =
dy 2 2
b R
dx
copyright
B1 + a
N = 5.8783 = 5.88 lb
of
the
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permitted.
Ans.
and
use
United
on
learning.
is
10
a
32.2 t
- 0.2(5.8783) + 10 sin 56.31° =
the
by
+R©Ft = mat;
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at = 23.0 ft s2
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Ans.
13–63.
z
The 150-lb man lies against the cushion for which the
coefficient of static friction is ms = 0.5. Determine the
resultant normal and frictional forces the cushion exerts on
him if, due to rotation about the z axis, he has a constant
speed v = 20 ft>s. Neglect the size of the man.Take u = 60°.
8 ft
G
SOLUTION
+ a a Fy = m1an2y ;
u
150 202
a
b sin 60°
N - 150 cos 60° =
32.2 8
N = 277 lb
+ b a Fx = m1an2x ;
Ans.
150 202
a
b cos 60°
32.2 8
- F + 150 sin 60° =
F = 13.4 lb
Ans.
Note: No slipping occurs
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Since ms N = 138.4 lb 7 13.4 lb
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*13–64.
z
The 150-lb man lies against the cushion for which the
coefficient of static friction is ms = 0.5. If he rotates about
the z axis with a constant speed v = 30 ft>s, determine the
smallest angle u of the cushion at which he will begin to
slip off.
8 ft
G
SOLUTION
+ ©F = ma ;
;
n
n
150 1302
a
b
0.5N cos u + N sin u =
32.2
8
+ c ©Fb = 0;
- 150 + N cos u - 0.5 N sin u = 0
u
2
N =
150
cos u - 0.5 sin u
10.5 cos u + sin u2150
1cos u - 0.5 sin u2
150 1302
a
b
32.2
8
2
=
0.5 cos u + sin u = 3.49378 cos u - 1.74689 sin u
u = 47.5°
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or transmission in any form or by any means, electronic, mechanical,
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13–65.
Determine the constant speed of the passengers on the
amusement-park ride if it is observed that the supporting
cables are directed at u = 30° from the vertical. Each chair
including its passenger has a mass of 80 kg. Also, what are
the components of force in the n, t, and b directions which
the chair exerts on a 50-kg passenger during the motion?
4m
b
6m
u
t
SOLUTION
n
+ ©F = m a ;
;
n
n
v2
)
T sin 30° = 80(
4 + 6 sin 30°
+ c ©Fb = 0;
T cos 30° - 8019.812 = 0
T = 906.2 N
v = 6.30 m>s
16.3022
) = 283 N
Fn = 50(
7
Ans.
©Ft = m at;
Ft = 0
Ans.
©Fb = m ab ;
Fb - 490.5 = 0
or
Ans.
laws
©Fn = m an ;
Fb = 490 N
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This
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Ans.
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13–66.
The man has a mass of 80 kg and sits 3 m from the center of
the rotating platform. Due to the rotation his speed is
#
increased from rest by v = 0.4 m>s2. If the coefficient of
static friction between his clothes and the platform is
ms = 0.3, determine the time required to cause him to slip.
3m
10 m
SOLUTION
©Ft = m at ;
Ft = 80(0.4)
Ft = 32 N
©Fn = m an ;
Fn = (80)
v2
3
F = ms Nm = 2(Ft)2 + (Fn)2
0.3(80)(9.81) =
2
A
(32)2 + ((80) v )2
3
laws
or
v4
55 432 = 1024 + (6400)( )
9
Wide
copyright
in
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Web)
v = 2.9575 m>s
not
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permitted.
Dissemination
dv
= 0.4
dt
at =
v
learning.
is
of
the
t
and
use
United
on
0.4 dt
by
student
L0
the
L0
dv =
protected
the
(including
for
work
v = 0.4 t
this
assessing
is
work
solely
of
2.9575 = 0.4 t
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t = 7.39 s
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Ans.
13–67.
The vehicle is designed to combine the feel of a motorcycle
with the comfort and safety of an automobile. If the vehicle
is traveling at a constant speed of 80 km> h along a circular
curved road of radius 100 m, determine the tilt angle u of
the vehicle so that only a normal force from the seat acts on
the driver. Neglect the size of the driver.
u
SOLUTION
Free-Body Diagram: The free-body diagram of the passenger is shown in Fig. (a).
Here, an must be directed towards the center of the circular path (positive n axis).
1h
km 1000 m
ba
ba
b
h
1 km
3600 s
= 22.22 m>s. Thus, the normal component of the passenger’s acceleration is given by
22.222
v2
= 4.938 m>s2. By referring to Fig. (a),
=
an =
r
100
in
teaching
Web)
laws
or
Equations of Motion: The speed of the passenger is v = a80
9.81m
cos u
N =
Wide
instructors
States
World
permitted.
Dissemination
N cos u - m(9.81) = 0
copyright
+ c ©Fb = 0;
United
on
learning.
is
of
the
not
9.81m
sin u = m(4.938)
cos u
by
and
use
+ ©F = ma ;
;
n
n
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u = 26.7°
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
*13–68.
The 0.8-Mg car travels over the hill having the shape of a
parabola. If the driver maintains a constant speed of 9 m> s,
determine both the resultant normal force and the
resultant frictional force that all the wheels of the car exert
on the road at the instant it reaches point A. Neglect the
size of the car.
y
y
20 (1
x2 )
6400
A
SOLUTION
80 m
dy
d2y
= - 0.00625x and 2 = - 0.00625. The slope angle u at point
Geometry: Here,
dx
dx
A is given by
tan u =
dy
2
= - 0.00625(80)
dx x = 80 m
u = -26.57°
and the radius of curvature at point A is
r =
[1 + (dy>dx)2]3>2
2
2
|d y>dx |
=
[1 + (- 0.00625x)2]3>2
2
= 223.61 m
| -0.00625|
x = 80 m
in
teaching
Web)
laws
or
Equations of Motion: Here, at = 0. Applying Eq. 13–8 with u = 26.57° and
r = 223.61 m, we have
800(9.81) sin 26.57° - Ff = 800(0)
Dissemination
Wide
copyright
©Ft = mat;
permitted.
Ans.
of
the
not
instructors
States
World
Ff = 3509.73 N = 3.51 kN
student
the
and
United
on
learning.
is
92
b
223.61
use
800(9.81) cos 26.57° - N = 800 a
by
©Fn = man;
work
N = 6729.67 N = 6.73 kN
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(including
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Ans.
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
x
13–69.
The 0.8-Mg car travels over the hill having the shape of a
parabola. When the car is at point A, it is traveling at 9 m> s
and increasing its speed at 3 m>s2. Determine both the
resultant normal force and the resultant frictional force that
all the wheels of the car exert on the road at this instant.
Neglect the size of the car.
y
y
20 (1
x2 )
6400
A
SOLUTION
80 m
dy
d2y
Geometry: Here,
= - 0.00625x and 2 = - 0.00625. The slope angle u at point
dx
dx
A is given by
tan u =
dy
2
= - 0.00625(80)
dx x = 80 m
u = -26.57°
and the radius of curvature at point A is
r =
C 1 + (dy>dx)2 D 3>2
2
2
|d y>dx |
=
C 1 + ( -0.00625x)2 D 3>2
2
|- 0.00625|
= 223.61 m
x = 80 m
laws
or
Equation of Motion: Applying Eq. 13–8 with u = 26.57° and r = 223.61 m, we have
in
teaching
Web)
800(9.81) sin 26.57° - Ff = 800(3)
Wide
copyright
©Ft = mat;
Ans.
States
World
permitted.
Dissemination
Ff = 1109.73 N = 1.11 kN
the
not
instructors
92
≤
223.61
by
and
use
on
learning.
is
of
800(9.81) cos 26.57° - N = 800 a
United
©Fn = man;
Ans.
sale
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This
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the
N = 6729.67 N = 6.73 kN
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
x
13–70.
The package has a weight of 5 lb and slides down the chute.
When it reaches the curved portion AB, it is traveling at
8 ft>s 1u = 0°2. If the chute is smooth, determine the speed
of the package when it reaches the intermediate point
C 1u = 30°2 and when it reaches the horizontal plane
1u = 45°2. Also, find the normal force on the package at C.
45°
8 ft/s
θ = 30°
20 ft
A
B
SOLUTION
+ b ©Ft = mat ;
5
a
32.2 t
5 cos f =
at = 32.2 cos f
+ a©Fn = man ;
N - 5 sin f =
5
v2
( )
32.2 20
v dv = at ds
v
Lg
f
v dv =
L45°
32.2 cos f (20 df)
in
teaching
Web)
laws
or
1 2
1
v - (8)2 = 644 (sin f - sin 45°)
2
2
Dissemination
Wide
copyright
At f = 45° + 30° = 75°,
permitted.
vC = 19.933 ft>s = 19.9 ft>s
the
not
instructors
States
World
Ans.
is
of
Ans.
the
by
and
use
United
on
learning.
NC = 7.91 lb
the
(including
for
work
student
At f = 45° + 45° = 90°
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vB = 21.0 ft s
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or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
45°
Ans.
C
13–71.
If the ball has a mass of 30 kg and a speed v = 4 m>s at the
instant it is at its lowest point, u = 0°, determine the tension
in the cord at this instant. Also, determine the angle u to
which the ball swings and momentarily stops. Neglect the
size of the ball.
u
4m
SOLUTION
+ c ©Fn = man;
T - 30(9.81) = 30 a
(4)2
b
4
T = 414 N
+Q©Ft = mat;
Ans.
- 30(9.81) sin u = 30at
at = - 9.81 sin u
at ds = v dv Since ds = 4 du, then
u
L4
v dv
or
L0
0
sin u(4 du) =
laws
-9.81
1
2
Wide
copyright
in
teaching
Web)
C 9.81(4)cos u D u0 = - (4)2
States
World
permitted.
Dissemination
39.24(cos u - 1) = - 8
instructors
u = 37.2°
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assessing
is
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the
(including
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work
student
the
by
and
use
United
on
learning.
is
of
the
not
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*13–72.
The ball has a mass of 30 kg and a speed v = 4 m>s at the
instant it is at its lowest point, u = 0°. Determine the tension
in the cord and the rate at which the ball’s speed is decreasing
at the instant u = 20°. Neglect the size of the ball.
u
4m
SOLUTION
+a©Fn = man;
T - 30(9.81) cos u = 30 a
+Q©Ft = mat;
- 30(9.81) sin u = 30at
v2
b
4
at = - 9.81 sin u
at ds = v dv Since ds = 4 du, then
u
u
9.81(4) cos u 2 =
v dv
1
1
(v)2 - (4)2
2
2
teaching
Web)
0
L4
or
L0
v
sin u (4 du) =
laws
-9.81
in
1 2
v
2
permitted.
Dissemination
Wide
copyright
39.24(cos u - 1) + 8 =
learning.
is
of
the
not
instructors
States
World
At u = 20°
the
by
and
use
United
on
v = 3.357 m>s
Ans.
the
(including
for
work
student
at = -3.36 m>s2 = 3.36 m>s2 b
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T = 361 N
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
13–73.
Determine the maximum speed at which the car with mass
m can pass over the top point A of the vertical curved road
and still maintain contact with the road. If the car maintains
this speed, what is the normal reaction the road exerts on
the car when it passes the lowest point B on the road?
r
A
r
B
r
SOLUTION
Free-Body Diagram: The free-body diagram of the car at the top and bottom of the
vertical curved road are shown in Figs. (a) and (b), respectively. Here, an must be
directed towards the center of curvature of the vertical curved road (positive n axis).
Equations of Motion: When the car is on top of the vertical curved road, it is
required that its tires are about to lose contact with the road surface. Thus, N = 0.
v2
v2
=
Realizing that an =
and referring to Fig. (a),
r
r
+ T ©Fn = man;
mg = m ¢
v2
≤
r
v = 2gr
Ans.
Wide
copyright
in
teaching
Web)
laws
or
Using the result of v, the normal component of car acceleration is
gr
v2
an =
=
= g when it is at the lowest point on the road. By referring to Fig. (b),
r
r
permitted.
Dissemination
N - mg = mg
States
World
+ c ©Fn = man;
instructors
N = 2mg
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student
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of
the
not
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
r
13–74.
If the crest of the hill has a radius of curvature r = 200 ft,
determine the maximum constant speed at which the car
can travel over it without leaving the surface of the road.
Neglect the size of the car in the calculation. The car has a
weight of 3500 lb.
v
r
SOLUTION
T ©Fn = man;
3500 =
3500 v2
a
b
32.2 200
v = 80.2 ft>s
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This
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the
(including
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work
student
the
by
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use
United
on
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is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
200 ft
13–75.
Bobs A and B of mass mA and mB (mA 7 mB) are
connected to an inextensible light string of length l that
passes through the smooth ring at C. If bob B moves as a
conical pendulum such that A is suspended a distance of h
from C, determine the angle u and the speed of bob B.
Neglect the size of both bobs.
C
h
SOLUTION
A
B
Free-Body Diagram: The free-body diagram of bob B is shown in Fig. a. The tension
developed in the string is equal to the weight of bob A, i.e., T = mAg. Here, an must
be directed towards the center of the horizontal circular path (positive n axis).
Equations of Motion: The radius of the horizontal circular path is r = (l - h) sin u.
vB2
v2
Thus, an =
. By referring to Fig. a,
=
r
(l - h) sin u
+ c ©Fb = 0;
mAg cos u - mBg = 0
mB
b
mA
vB 2
d
(l - h) sin u
or
mAg sin u = mB c
teaching
Web)
+ ©F = ma ;
;
n
n
Ans.
laws
u = cos - 1 a
(1)
Dissemination
Wide
in
mAg(l - h)
sin u
mB
B
copyright
vB =
the
not
instructors
States
World
permitted.
2mA2 - mB2
. Substituting this value into Eq. (1),
mA
and
use
United
on
learning.
is
of
From Fig. b, sin u =
work
student
the
by
mAg(l - h) 3mA2 - mB2
£
≥
mB
mA
B
this
integrity
of
and
work
provided
g(l - h)(mA2 - mB2)
mAmB
B
sale
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This
=
assessing
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the
(including
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vB =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
u
Ans.
*13–76.
Prove that if the block is released from rest at point B of a
smooth path of arbitrary shape, the speed it attains when it
reaches point A is equal to the speed it attains when it falls
freely through a distance h; i.e., v = 22gh.
B
h
A
SOLUTION
+ R©Ft = mat;
mg sin u = mat
v dv = at ds = g sin u ds
v
L0
at = g sin u
However dy = ds sin u
h
v dv =
L0
g dy
v2
= gh
2
v = 22gh
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This
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the
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student
the
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United
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of
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instructors
States
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permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Q.E.D.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
13–77.
The skier starts from rest at A(10 m, 0) and descends the
smooth slope, which may be approximated by a parabola.
If she has a mass of 52 kg, determine the normal force the
ground exerts on the skier at the instant she arrives at
point B. Neglect the size of the skier. Hint: Use the result of
Prob. 13–76.
y
10 m
y
5m
B
SOLUTION
Geometry: Here,
dy
d2y
1
1
=
x and 2 =
. The slope angle u at point B is given by
dx
10
10
dx
tan u =
dy
2
= 0
dx x = 0 m
u = 0°
and the radius of curvature at point B is
C 1 + (dy>dx) D
=
|d2y>dx2|
= 10.0 m
or
x=0 m
teaching
Web)
Equations of Motion:
laws
r =
2 3>2
1
xb d
10
4
|1>10|
c1 + a
2 3>2
in
Wide
copyright
World
(1)
not
on
learning.
is
of
United
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
Kinematics: The speed of the skier can be determined using v dv = at ds. Here, at
must be in the direction of positive ds. Also, ds = 21 + (dy>dx)2dx
1 2
= 21 + 100
x dx
this
integrity
of
and
work
provided
the
is
This
part
x
1 2
100 x
≤ A 21 +
any
courses
1 2
100 x dx
B
destroy
¢
L10 m 10 21 +
sale
v2 = 9.81 m2>s2
will
their
L0
0
v dv = - 9.81
.
1 2
100 x
of
v
(+ )
assessing
1
x
x. Then, sin u =
10
10 21 +
and
Here, tan u =
Substituting v2 = 98.1 m2>s2, u = 0°, and r = 10.0 m into Eq.(1) yields
N - 52(9.81) cos 0° = 52 a
98.1
b
10.0
N = 1020.24 N = 1.02 kN
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
permitted.
Dissemination
y2
b
r
N - 52(9.81) cos u = ma
the
+a©Fn = man;
at = -9.81 sin u
instructors
52(9.81) sin u = - 52at
States
+b©Ft = mat;
1 2
––
x
20
5
A
x
13–78.
A spring, having an unstretched length of 2 ft, has one end
attached to the 10-lb ball. Determine the angle u of the
spring if the ball has a speed of 6 ft> s tangent to the
horizontal circular path.
6 in.
A
u
SOLUTION
Free-Body Diagram: The free-body diagram of the bob is shown in Fig. (a). If we
denote the stretched length of the spring as l, then using the springforce formula,
Fsp = ks = 20(l - 2) lb. Here, an must be directed towards the center of the
horizontal circular path (positive n axis).
Equations of Motion: The radius of the horizontal circular path is r = 0.5 + l sin u.
62
v2
Since an =
=
, by referring to Fig. (a),
r
0.5 + l sin u
62
10
a
b
32.2 0.5 + l sin u
(2)
or
20(l - 2) sin u =
(1)
teaching
Web)
+ ©F = ma ;
;
n
n
20(l - 2) cos u - 10 = 0
laws
+ c ©Fb = 0;
Wide
copyright
in
Solving Eqs. (1) and (2) yields
u = 31.26° = 31.3°
Ans.
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l = 2.585 ft
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
not
States
World
permitted.
Dissemination
Ans.
k
20 lb>ft
13–79.
u
The airplane, traveling at a constant speed of 50 m>s, is
executing a horizontal turn. If the plane is banked at
u = 15°, when the pilot experiences only a normal force on
the seat of the plane, determine the radius of curvature r of
the turn. Also, what is the normal force of the seat on the
pilot if he has a mass of 70 kg.
r
SOLUTION
+ c a Fb = mab;
NP sin 15° - 7019.812 = 0
NP = 2.65 kN
+
;
a Fn = man;
NP cos 15° = 70 a
Ans.
502
b
r
r = 68.3 m
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the
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Wide
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Web)
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© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*13–80.
A 5-Mg airplane is flying at a constant speed of
350 km> h along a horizontal circular path of radius
r = 3000 m. Determine the uplift force L acting on the
airplane and the banking angle u. Neglect the size of the
airplane.
L
u
r
SOLUTION
Free-Body Diagram: The free-body diagram of the airplane is shown in Fig. (a).
Here, an must be directed towards the center of curvature (positive n axis).
Equations of Motion: The speed of the airplane is v = ¢ 350
= 97.22 m>s. Realizing that an =
1h
km 1000 m
≤¢
≤¢
≤
h
1 km
3600 s
97.222
v2
= 3.151 m>s2 and referring to Fig. (a),
=
r
3000
+ c ©Fb = 0;
T cos u - 5000(9.81) = 0
(1)
+ ©F = ma ;
;
n
n
T sin u = 5000(3.151)
(2)
laws
or
Solving Eqs. (1) and (2) yields
T = 51517.75 = 51.5 kN
teaching
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student
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in
u = 17.8°
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or transmission in any form or by any means, electronic, mechanical,
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13–81.
A 5-Mg airplane is flying at a constant speed of
350 km> h along a horizontal circular path. If the banking
angle u = 15°, determine the uplift force L acting on the
airplane and the radius r of the circular path. Neglect the
size of the airplane.
L
u
r
SOLUTION
Free-Body Diagram: The free-body diagram of the airplane is shown in Fig. (a).
Here, an must be directed towards the center of curvature (positive n axis).
1h
km 1000 m
ba
ba
b
h
1 km
3600 s
Equations of Motion: The speed of the airplane is v = a350
= 97.22 m>s. Realizing that an =
97.222
v2
and referring to Fig. (a),
=
r
r
+ c ©Fb = 0;
L cos 15° - 5000(9.81) = 0
L = 50780.30 N = 50.8 kN
or
97.222
≤
r
50780.30 sin 15° = 5000 ¢
laws
+ ©F = ma ;
;
n
n
Ans.
teaching
Web)
r = 3595.92 m = 3.60 km
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the
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on
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Dissemination
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in
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
13–82.
The 800-kg motorbike travels with a constant speed of
80 km > h up the hill. Determine the normal force the
surface exerts on its wheels when it reaches point A.
Neglect its size.
y
A
y2
2x
x
100 m
SOLUTION
d2y
dy
22
22
=
and
= - 3>2 . The angle that
2
1>2
dx
dx
2x
4x
the hill slope at A makes with the horizontal is
Geometry: Here, y = 22x1>2. Thus,
u = tan - 1 a
dy
22
b2
= tan - 1 ¢ 1>2 ≤ 2
= 4.045°
dx x = 100 m
2x
x = 100 m
The radius of curvature of the hill at A is given by
dx2
2
= 2849.67 m
22
2-
2
4(1003>2)
x = 100 m
3>2
or
2(100 )
=
b R
Web)
dy
5
2
1>2
in
2
2
22
B1 + a
laws
rA =
dy 2 3>2
b R
dx
teaching
B1 + a
States
World
permitted.
Dissemination
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copyright
Free-Body Diagram: The free-body diagram of the motorcycle is shown in Fig. (a).
Here, an must be directed towards the center of curvature (positive n axis).
work
the
solely
of
protected
Thus, an =
(including
for
v2
22.222
= 0.1733 m>s2. By referring to Fig. (a),
=
rA
2849.67
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
Equations of Motion: The speed of the motorcycle is
1h
km 1000 m
ba
ba
b = 22.22 m>s
v = a 80
h
1 km
3600 s
work
800(9.81)cos 4.045° - N = 800(0.1733)
this
assessing
is
R+ ©Fn = man;
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N = 7689.82 N = 7.69 kN
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
13–83.
The ball has a mass m and is attached to the cord of length l.
The cord is tied at the top to a swivel and the ball is given a
velocity v0. Show that the angle u which the cord makes with
the vertical as the ball travels around the circular path
must satisfy the equation tan u sin u = v20>gl. Neglect air
resistance and the size of the ball.
O
u
l
SOLUTION
+ c ©Fb = 0;
Since r = l sin u
T sin u = m a
v20
b
r
T cos u - mg = 0
T =
a
v0
mv20
l sin2 u
mv20
cos u
b a 2 b = mg
l
sin u
v20
gl
Q.E.D.
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This
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laws
tan u sin u =
or
+ ©F = ma ;
:
n
n
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*13–84.
The 5-lb collar slides on the smooth rod, so that when it is at
A it has a speed of 10 ft> s. If the spring to which it is
attached has an unstretched length of 3 ft and a stiffness of
k = 10 lb>ft, determine the normal force on the collar and
the acceleration of the collar at this instant.
y
10 ft/s
A
1 x2
y 8 ––
2
SOLUTION
y = 8 -
1 2
x
2
O
dy
= tan u = x 2
= 2
dx
x=2
x
u = 63.435°
2 ft
d2y
= -1
dx2
3
2
dx2
2
(1 + (- 2)2)2
= 11.18 ft
| -1|
in
teaching
Web)
1
(2)2 = 6
2
Dissemination
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y = 8 -
d2y
3
=
or
r =
dy 2 2
b R
dx
laws
B1 + a
not
instructors
States
World
permitted.
OA = 2(2)2 + (6)2 = 6.3246
and
use
United
on
learning.
is
of
the
Fs = kx = 10(6.3246 - 3) = 33.246 lb
work
student
the
by
6
; f = 71.565°
2
(10)2
5
ba
b
32.2 11.18
solely
of
protected
the
(including
for
tan f =
work
5 cos 63.435° - N + 33.246 cos 45.0° = a
Ans.
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any
courses
part
the
is
This
N = 24.4 lb
provided
integrity
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and
work
this
assessing
is
+b©Fn = man;
destroy
of
5 sin 63.435° + 33.246 sin 45.0° = a
5
b at
32.2
sale
will
their
+ R©Ft = mat;
at = 180.2 ft>s2
an =
(10)2
v2
= 8.9443 ft>s2
=
r
11.18
a = 2(180.2)2 + (8.9443)2
a = 180 ft>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
13–85.
The spring-held follower AB has a weight of 0.75 lb and
moves back and forth as its end rolls on the contoured
surface of the cam, where r = 0.2 ft and z = 10.1 sin u2 ft. If
the cam is rotating at a constant rate of 6 rad>s, determine
the force at the end A of the follower when u = 90°. In this
position the spring is compressed 0.4 ft. Neglect friction at
the bearing C.
SOLUTION
z = 0.1 sin 2u
#
#
z = 0.2 cos 2uu
#
$
$
z = -0.4 sin 2uu2 + 0.2 cos 2uu
#
u = 6 rad>s
##
u = 0
$
z = -14.4 sin 2u
or
$
FA - 12(z + 0.3) = mz
laws
a Fz = maz;
in
teaching
Web)
0.75
( -14.4 sin 2u)
32.2
Dissemination
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FA - 12(0.1 sin 2u + 0.3) =
is
of
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World
permitted.
For u = 45°,
by
and
use
United
on
learning.
0.75
( -14.4)
32.2
work
student
the
FA - 12(0.4) =
for
FA = 4.46 lb
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the
(including
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
13–86.
Determine the magnitude of the resultant force acting on a
5-kg particle at the instant t = 2 s, if the particle is moving
along a horizontal path defined by the equations
r = (2t + 10) m and u = (1.5t2 - 6t) rad, where t is in
seconds.
SOLUTION
r = 2t + 10|t = 2 s = 14
#
r = 2
$
r = 0
teaching
Web)
laws
or
u = 1.5t2 - 6t
#
u = 3t - 6冷t = 2 s = 0
$
u = 3
#
$
ar = r - ru2 = 0 - 0 = 0
$
##
au = ru + 2ru = 14(3) + 0 = 42
Fr = 5(0) = 0
©Fu = mau;
Fu = 5(42) = 210 N
United
on
learning.
is
of
the
not
instructors
States
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permitted.
©Fr = mar;
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Hence,
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the
by
and
use
F = 2(Fr)2 + (Fu)2 = 210 N
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Ans.
13–87.
The path of motion of a 5-lb particle in the horizontal plane
is described in terms of polar coordinates as r = (2t + 1) ft
and u = (0.5t2 - t) rad, where t is in seconds. Determine
the magnitude of the resultant force acting on the particle
when t = 2 s.
SOLUTION
#
$
r = 2 ft>s
r = 0
#
= 0 rad
u = t - 1|t = 2 s = 1 rad>s
r = 2t + 1|t = 2 s = 5 ft
$
u = 1 rad>s2
u = 0.5t2 - t|t = 2 s
#
$
ar = r - ru2 = 0 - 5(1)2 = - 5 ft>s2
$
##
au = ru + 2ru = 5(1) + 2(2)(1) = 9 ft>s2
Fr =
5
( - 5) = - 0.7764 lb
32.2
©Fu = mau;
Fu =
5
(9) = 1.398 lb
32.2
or
©Fr = mar;
F = 2F2r + F2u = 2(-0.7764)2 + (1.398)2 = 1.60 lb
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work
student
the
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Dissemination
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teaching
Web)
laws
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*13–88.
A particle, having a mass of 1.5 kg, moves along a path
defined by the equations r = 14 + 3t2 m, u = 1t2 + 22 rad,
and z = 16 - t32 m, where t is in seconds. Determine the r,
u, and z components of force which the path exerts on the
particle when t = 2 s.
SOLUTION
u = t2 + 2
#
r = 3 m>s
#
u = 2t| t = 2 s = 4 rad>s
z = 6 - t3
z = - 3t2
r = 4 + 3t| t = 2 s = 10 m
##
$
r = 0
$
u = 2 rad>s2
#
##
z = - 6t| t = 2 s = - 12 m>s2
#
ar = r - r u 2 = 0 -10(4)2 = - 160 m>s2
$
##
au = ru + 2ru = 10(2) + 2(3)(4) = 44 m>s2
##
az = z = - 12 m>s2
Fr = 1.5(- 160) = - 240 N
Ans.
©Fu = mau;
Fu = 1.5(44) = 66 N
Ans.
©Fz = maz;
Fz - 1.5(9.81) = 1.5(- 12)
laws
or
©Fr = mar;
teaching
Web)
Fz = - 3.28 N
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This
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the
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work
student
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Dissemination
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Ans.
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or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
13–89.
Rod OA rotates
# counterclockwise with a constant angular
velocity of u = 5 rad>s. The double collar B is pinconnected together such that one collar slides over the
rotating rod and the other slides over the horizontal curved
rod, of which the shape is described by the equation
r = 1.512 - cos u2 ft. If both collars weigh 0.75 lb,
determine the normal force which the curved rod exerts on
one collar at the instant u = 120°. Neglect friction.
A
B
r
SOLUTION
·
#
$
Kinematic: Here, u = 5 rad>s and u = 0. Taking the required time derivatives at
u = 120°, we have
r = 1.5 (2 – cos ) ft.
r = 1.5(2 - cos u)|u = 120° = 3.75 ft
#
#
r = 1.5 sin uu|u = 120° = 6.495 ft>s
$
#
$
r = 1.5(sin uu + cos uu 2)|u = 120° = - 18.75 ft>s2
teaching
Web)
laws
or
Applying Eqs. 12–29, we have
#
$
ar = r - ru2 = - 18.75 - 3.75(52) = - 112.5 ft>s2
$
# #
au = r u + 2r u = 3.75(0) + 2(6.495)(5) = 64.952 ft>s2
Wide
copyright
in
Equation of Motion: The angle c must be obtained first.
Dissemination
1.5(2 - cos u)
r
2
=
= 2.8867
dr>du
1.5 sin u
u = 120°
World
permitted.
c = 70.89°
on
learning.
is
of
the
not
instructors
States
tan c =
the
by
and
use
United
Applying Eq. 13–9, we have
work
student
0.75
(-112.5)
32.2
the
(including
for
- N cos 19.11° =
is
work
solely
of
protected
a Fr = mar ;
Ans.
integrity
of
and
work
this
assessing
N = 2.773 lb = 2.77 lb
provided
0.75
(64.952)
32.2
FOA + 2.773 sin 19.11° =
destroy
sale
will
their
FOA = 0.605 lb
of
and
any
courses
part
the
is
This
a Fu = mau ;
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
O
= 5 rad/s
13–90.
z
The boy of mass 40 kg is sliding down the spiral slide at a
constant speed such that his position, measured from the
top of the chute, has components r = 1.5 m, u = 10.7t2 rad,
and z = 1 -0.5t2 m, where t is in seconds. Determine the
components of force Fr , Fu , and Fz which the slide exerts on
him at the instant t = 2 s. Neglect the size of the boy.
u
SOLUTION
r = 1.5
#
$
r = r = 0
$
u = 0
u = 0.7t
#
u = 0.7
z
z = - 0.5t
#
z = - 0.5
$
z = 0
#
$
ar = r - r(u)2 = 0 - 1.5(0.7)2 = - 0.735
$
##
au = ru + 2ru = 0
$
az = z = 0
Fr = 40( -0.735) = - 29.4 N
Ans.
©Fu = mau;
Fu = 0
Ans.
©Fz = maz;
Fz - 40(9.81) = 0
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or
©Fr = mar;
Fz = 392 N
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the
(including
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work
student
the
by
and
use
United
on
learning.
is
of
the
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permitted.
Dissemination
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
r 1.5 m
13–91.
The 0.5-lb particle is guided along the circular path using the
slotted
arm guide. If the arm has an$ angular velocity
#
u = 4 rad>s and an angular acceleration u = 8 rad>s2 at the
instant u = 30°, determine the force of the guide on the
particle. Motion occurs in the horizontal plane.
r
u
0.5 ft
SOLUTION
0.5 ft
r = 2(0.5 cos u) = 1 cos u
#
#
r = - sin uu
#
$
r = - cos uu2 - sin uu
#
$
At u = 30°, u = 4 rad>s and u = 8 rad>s2
r = 1 cos 30° = 0.8660 ft
#
r = - sin 30° (4) = - 2 ft>s
..
r = - cos 30° (4)2 - sin 30° (8) = - 17.856 ft>s2
Wide
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Web)
laws
or
#2
$
ar = r - ru = - 17.856 - 0.8660(4)2 = - 31.713 ft>s2
$
##
au = ru + 2ru = 0.8660(8) + 2( -2)(4) = - 9.072 ft>s2
Dissemination
0.5
(- 31.713)
32.2
World
permitted.
N = 0.5686 lb
learning.
is
of
the
not
instructors
- N cos 30° =
States
Q+ ©Fr = mar;
by
and
use
United
on
0.5
( - 9.072)
32.2
for
work
student
F - 0.5686 sin 30° =
the
+a©Fu = mau;
sale
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and
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courses
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is
This
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integrity
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the
(including
F = 0.143 lb
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Ans.
*13–92.
Using a forked rod, a smooth cylinder C having a mass of
0.5 kg is forced to move along the vertical slotted path
r = 10.5u2 m, where u is in radians. If the angular position
of the arm is u = 10.5t22 rad, where t is in seconds,
determine the force of the rod on the cylinder and the
normal force of the slot on the cylinder at the instant
t = 2 s. The cylinder is in contact with only one edge of the
rod and slot at any instant.
C
r
SOLUTION
r = 0.5u
u = 0.5t2
#
r = 0.5u
#
u = t
$
$
r = 0.5u
$
u = 1
At t = 2 s,
#
u = 2 rad>2
u = 2 rad = 114.59°
r = 1m
#
r = 1 m>s
$
u = 1 rad>s2
$
r = 0.5 m>s2
0.5(2)
r
=
c = 63.43°
dr>du
0.5
#
#
ar = r - ru2 = 0.5 - 1(2)2 = - 3.5
$
##
au = ru + 2ru = 1(1) + 2(1)(2) = 5
Wide
copyright
in
teaching
Web)
laws
or
tan c =
NC cos 26.57° - 4.905 cos 24.59° = 0.5( - 3.5)
States
World
permitted.
Dissemination
+ a©Fr = mar;
on
learning.
is
of
the
Ans.
by
and
use
United
F - 3.030 sin 26.57° + 4.905 sin 24.59° = 0.5(5)
the
+ b© Fu = mau;
sale
will
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destroy
of
and
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courses
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the
is
This
provided
integrity
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and
work
this
assessing
is
work
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protected
the
(including
for
work
student
F = 1.81 N
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Ans.
not
instructors
NC = 3.030 = 3.03 N
0.5 u
u
13–93.
If arm OA .rotates with a constant clockwise angular
velocity of u = 1.5 rad>s. determine the force arm OA
exerts on the smooth 4-lb cylinder B when u = 45°.
A
B
r
SOLUTION
u
u
Kinematics: Since the motion of cylinder B is known, ar and au will be determined
4
first. Here, = cos u or r = 4 sec u ft. The value of r and its time derivatives at the
r
instant u = 45° are
O
4 ft
r = 4 sec u |u = 45° = 4 sec 45° = 5.657 ft
#
#
r = 4 sec u(tan u)u|u = 45° = 4 sec 45° tan 45°(1.5) = 8.485 ft>s
$
#
#
#
$
r = 4 C sec u(tan u)u + u A sec u sec2 uu + tan u sec u tan uu B D
#
$
#2
= 4 C sec u(tan u)u + sec3 uu2 + sec u tan2 uu D 2
u = 45°
laws
or
= 4 C sec 45° tan 45°(0) + sec 45°(1.5) + sec 45° tan2 45°(1.5)2 D
Web)
2
teaching
3
Dissemination
Wide
copyright
in
= 38.18 ft>s2
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Using the above time derivatives,
#
$
ar = r - ru 2 = 38.18 - 5.657 A 1.52 B = 25.46 ft>s2
$
##
au = ru - 2ru = 5.657(0) + 2(8.485)(1.5) = 25.46 ft>s2
4
(25.46)
32.2
the
is
This
integrity
of
and
work
N cos 45° - 4 cos 45° =
provided
©Fr = mar;
this
assessing
is
work
solely
of
protected
the
(including
Equations of Motion: By referring to the free-body diagram of the cylinder shown in
Fig. a,
and
any
courses
part
N = 8.472 lb
destroy
of
FOA - 8.472 sin 45° - 4 sin 45° =
4
(25.46)
32.2
sale
will
their
©Fu = mau;
FOA = 12.0 lb
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
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Ans.
13–94.
The collar has a mass of 2 kg and travels along the smooth
horizontal rod defined by the equiangular spiral
r = 1eu2 m, where u is in radians. Determine the tangential
force F and the normal force N acting on the collar when
u = 90°, if the force F maintains a constant angular motion
#
u = 2 rad>s.
F
SOLUTION
r
r = eu
u
#
#
r = euu
#
$
$
r = eu(u)2 + euu
At u = 90°
#
u = 2 rad>s
$
u = 0
r = 4.8105
laws
or
#
r = 9.6210
in
teaching
Web)
$
r = 19.242
and
by
work
student
(including
A B
= eu>eu = 1
the
r
dr
du
for
tan c =
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
#
$
ar = r - r(u)2 = 19.242 - 4.8105(2)2 = 0
$
##
au = ru + 2ru = 0 + 2(9.6210)(2) = 38.4838 m>s2
is
work
solely
of
protected
the
c = 45°
-NC cos 45° + F cos 45° = 2(0)
+
;
a Fu = mau;
F sin 45° + NC sin 45° = 2(38.4838)
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
+ c a Fr = mar;
Ans.
F = 54.4 N
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
sale
will
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destroy
of
NC = 54.4 N
Ans.
r eu
13–95.
The ball has a mass of 2 kg and a negligible size. It is originally
traveling around the horizontal circular path of radius
r# 0 = 0.5 m such that the angular rate of rotation is
u0 = 1 rad>s. If the attached cord ABC is drawn down
through the hole at a constant speed of 0.2 m>s, determine the
tension the cord exerts on the ball at the instant r = 0.25 m.
Also, compute the angular velocity of the ball at this instant.
Neglect the effects of friction between the ball and horizontal
plane. Hint: First show that$ the equation
of motion
# in the u
##
direction yields au = ru + 2ru = 11>r21d1r2u2>dt2 = 0.
#
When integrated, r2u = c, where the constant c is determined
from the problem data.
A
r
F
$
1 d 2#
##
0 = m[ru + 2ru] = mc
(r u) d = 0
r dt
Thus,
or
#
d(r2u) = 0
#
r2u = C
Dissemination
copyright
Ans.
Wide
in
teaching
Web)
laws
#
(0.5)2(1) = C = (0.25)2u
#
u = 4.00 rad>s
World
permitted.
$
r = 0
not
instructors
States
#
Since r = - 0.2 m>s,
the
by
and
use
United
on
learning.
is
of
the
$
#
ar = r - r(u)2 = 0 - 0.25(4.00)2 = - 4 m>s2
work
student
-T = 2( - 4)
(including
for
a Fr = mar;
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This
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the
T = 8N
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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r0
0.2 m/s
C
SOLUTION
a Fu = mau;
u
B
Ans.
·
u0
*13–96.
A
The particle has a mass of 0.5 kg and is confined to move
along the smooth horizontal slot due to the rotation of the
arm OA. Determine the force of the rod on the particle and
the normal force of the slot on the particle when u = 30°.
The
rod is rotating with a constant angular velocity
#
u = 2 rad>s. Assume the particle contacts only one side of
the slot at any instant.
·
u
2 rad/s
0.5 m
u
SOLUTION
0.5
= 0.5 sec u
cos u
#
#
r = 0.5 sec u tan uu
#
# #
$
$
r = 0.5 E C (sec u tan uu) tan u + sec u(sec2 uu D u + sec u tan uu F
#
#
$
= 0.5 C sec u tan2 uu + sec3 uu2 + sec u tan uu D
O
r =
#
$
When u = 30°, u = 2 rad>s and u = 0
r = 0.5 sec 30° = 0.5774 m
laws
or
#
r = 0.5 sec 30° tan 30°(2) = 0.6667 m>s
in
teaching
Web)
$
r = 0.5 C sec 30° tan2 30°(2)2 + sec3 30°(2)2 + sec 30° tan 30°(0) D
Dissemination
Wide
copyright
= 3.849 m>s2
work
the
(including
for
N cos 30° - 0.5(9.81) cos 30° = 0.5(1.540)
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
#
$
ar = r - ru2 = 3.849 - 0.5774(2)2 = 1.540 m>s2
$
##
au = ru + 2ru = 0.5774(0) + 2(0.6667)(2) = 2.667 m>s2
Q+ ©Fr = mar;
Ans.
assessing
is
work
solely
of
protected
N = 5.79 N
this
F + 0.5(9.81) sin 30° - 5.79 sin 30° = 0.5(2.667)
provided
integrity
of
and
work
+ R©Fu = mau;
sale
will
their
destroy
of
and
any
courses
part
the
is
This
F = 1.78 N
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
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r
Ans.
13–97.
Solve
$ Problem 13–96# if the arm has an angular acceleration
of u = 3 rad/s2 and u = 2 rad/s at this instant. Assume the
particle contacts only one side of the slot at any instant.
A
·
u
2 rad/s
r
0.5 m
u
SOLUTION
0.5
= 0.5 sec u
cos u
#
#
r = 0.5 sec u tan uu
r =
O
#
# #
$
$
r = 0.5 E C (sec u tan uu) tan u + sec u(sec2 uu) D u + sec u tan uu F
#
#
$
= 0.5 C sec u tan2 uu2 + sec3 uu2 + sec u tan uu D
#
$
When u = 30°, u = 2 rad>s and u = 3 rad>s2
r = 0.5 sec 30° = 0.5774 m
laws
or
#
r = 0.5 sec 30° tan 30°(2) = 0.6667 m>s
teaching
in
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
= 4.849 m>s2
#
$
ar = r - ru2 = 4.849 - 0.5774(2)2 = 2.5396 m>s2
$
##
au = ru + 2ru = 0.5774(3) + 2(0.6667)(2) = 4.3987 m>s2
Web)
$
r = 0.5 C sec 30° tan2 30°(2)2 + sec3 30°(2)2 + sec 30° tan 30°(3) D
work
student
N cos 30° - 0.5(9.81) cos 30° = 0.5(2.5396)
the
(including
for
Q+ ©Fr = mar;
Ans.
assessing
is
work
solely
of
protected
N = 6.3712 = 6.37 N
this
F + 0.5(9.81) sin 30° - 6.3712 sin 30° = 0.5(4.3987)
provided
integrity
of
and
work
+R©Fu = mau;
sale
will
their
destroy
of
and
any
courses
part
the
is
This
F = 2.93 N
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
13–98.
The collar has a mass of 2 kg and travels along the smooth
horizontal rod defined by the equiangular spiral r = 1eu2 m,
where u is in radians. Determine the tangential force F and
the normal force N acting on the collar when u =# 45°, if the
force F maintains a constant angular motion u = 2 rad>s.
F
r
SOLUTION
r = eu
θ
#
#
r = eu u
#
#
$
r = eu(u)2 + eu u
At u = 45°
#
u = 2 rad>s
$
u = 0
r = 2.1933
laws
or
#
r = 4.38656
in
teaching
Web)
$
r = 8.7731
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
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copyright
#
$
ar = r - r(u)2 = 8.7731 - 2.1933(2)2 = 0
$
# #
au = r u + 2 r u = 0 + 2(4.38656)(2) = 17.5462 m>s2
the
(including
for
work
student
the
by
and
r
= e u>e u = 1
dr
a #b
du
solely
of
protected
tan c =
integrity
of
and
work
this
assessing
is
work
c = u = 45°
provided
- NC cos 45° + F cos 45° = 2(0)
destroy
of
and
any
courses
part
the
is
This
Q+ a Fr = mar ;
will
their
F sin 45° + NC sin 45° = 2(17.5462)
sale
+ a a Fu = mau ;
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
N = 24.8 N
Ans.
F = 24.8 N
Ans.
r = eθ
13–99.
For a short time, the 250-kg roller coaster car is traveling
along the spiral track such that its position measured from
the top of the track has components r = 8 m,
u = 10.1t + 0.52 rad, and z = 1-0.2t2 m, where t is in
seconds. Determine the magnitudes of the components of
force which the track exerts on the car in the r, u, and z
directions at the instant t = 2 s. Neglect the size of the car.
r=8m
SOLUTION
#
$
Kinematic: Here, r = 8 m, r = r = 0. Taking the required time derivatives at
t = 2 s, we have
#
$
u = 0.100 rad>s
u = 0
u = 0.1t + 0.5|t = 2s = 0.700 rad
#
z = - 0.200 m>s
z = - 0.2t|t = 2 s = - 0.400 m
$
z = 0
Applying Eqs. 12–29, we have
$
$
ar = r - ru2 = 0 - 8(0.1002) = - 0.0800 m>s2
$
##
au = r u + 2 ru = 8(0) + 2(0)(0.200) = 0
teaching
Web)
laws
or
$
az = z = 0
Fr = 250( -0.0800) = - 20.0 N
©Fu = mau ;
Fu = 250(0) = 0
©Fz = maz ;
Fz - 250(9.81) = 250(0)
Ans.
States
World
permitted.
©Fr = mar ;
Dissemination
Wide
copyright
in
Equation of Motion:
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is
This
provided
integrity
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assessing
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the
(including
for
Fz = 2452.5 N = 2.45 kN
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Ans.
*13–100.
The 0.5-lb ball is guided along the vertical circular path
r = 2rc cos# u using the arm OA. If the arm has an angular
u = 0.4 rad>s and an angular acceleration
velocity
$
u = 0.8 rad>s2 at the instant u = 30°, determine the force of
the arm on the ball. Neglect friction and the size of the ball.
Set rc = 0.4 ft.
P
r
u
O
SOLUTION
r = 2(0.4) cos u = 0.8 cos u
#
#
r = -0.8 sin uu
#
$
$
r = -0.8 cos uu2 - 0.8 sin uu
#
$
At u = 30°, u = 0.4 rad>s, and u = 0.8 rad>s2
r = 0.8 cos 30° = 0.6928 ft
#
r = - 0.8 sin 30°(0.4) = - 0.16 ft>s
Wide
copyright
in
teaching
Web)
laws
or
$
r = - 0.8 cos 30°(0.4)2 - 0.8 sin 30°(0.8) = - 0.4309 ft>s2
#
$
ar = r - ru2 = -0.4309 - 0.6928(0.4)2 = - 0.5417 ft>s2
$
##
au = ru + 2ru = 0.6928(0.8) + 2( -0.16)(0.4) = 0.4263 ft>s2
World
N = 0.2790 lb
permitted.
Dissemination
0.5
(-0.5417)
32.2
learning.
is
of
the
not
instructors
N cos 30° - 0.5 sin 30° =
States
+Q©Fr = mar;
United
on
0.5
(0.4263)
32.2
by
and
use
FOA + 0.2790 sin 30° - 0.5 cos 30° =
for
work
student
the
a+ ©Fu = mau;
sale
will
their
destroy
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and
any
courses
part
the
is
This
provided
integrity
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and
work
this
assessing
is
work
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protected
the
(including
FOA = 0.300 lb
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
rc
A
13–101.
The ball of mass m is guided along the vertical circular path
r = 2rc cos u using
# the arm OA. If the arm has a constant
angular velocity u0, determine the angle u … 45° at which
the ball starts to leave the surface of the semicylinder.
Neglect friction and the size of the ball.
P
r
u
O
SOLUTION
r = 2rc cos u
#
#
r = -2rc sin uu
#
$
$
r = - 2rc cos uu2 - 2rc sin uu
#
$
Since u is constant, u = 0.
#
#
#
#
$
ar = r - ru2 = -2rc cos uu20 - 2rc cos uu20 = -4rc cos uu20
#
- mg sin u = m(- 4rc cos uu20)
#
#
4rc u20
4rc u20
-1
tan u =
u = tan ¢
≤
g
g
or
Ans.
sale
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This
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assessing
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the
(including
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work
student
the
by
and
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United
on
learning.
is
of
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not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
+ Q©Fr = mar;
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
rc
A
13–102.
Using a forked rod, a smooth cylinder P, having a mass of
0.4 kg, is forced to move along the vertical slotted path
r = 10.6u2 m, where u is in radians. If the cylinder has a
constant speed of vC = 2 m>s, determine the force of the
rod and the normal force of the slot on the cylinder at the
instant u = p rad. Assume the cylinder is in contact with
only one edge of the rod and slot at any instant. Hint: To
obtain the time derivatives necessary to compute the
cylinder’s acceleration components ar and au, take the first
and second time derivatives of r = 0.6u. Then,
for further
#
information, use Eq. 12–26 to determine u. Also, take the
#
time derivative
of Eq. 12–26, noting that vC = 0, to
$
determine u.
P
r
SOLUTION
$
$
#
r = 0.6 u
r = 0.6 u
#
#
#
#
vr = r = 0.6u
vu = ru = 0.6uu
r = 0.6 u
2
#
v2 = r 2 + a rub
2
laws
#
uu2
Web)
$
u = -
teaching
1 + u2
World
= 1.011 rad>s
the
not
instructors
on
learning.
is
2
0.621 + p2
States
#
u =
of
At u = p rad,
permitted.
Dissemination
Wide
#$
#
#$
0 = 0.72u u + 0.36 a 2uu3 + 2u2u u b
or
0.6 21 + u2
in
#
u =
copyright
# 2
# 2
22 = a 0.6u b + a 0.6uu b
and
use
by
work
student
the
the
(including
for
work
solely
of
protected
#
r = 0.6(1.011) = 0.6066 m>s
United
$
(p)(1.011)2
= - 0.2954 rad>s2
u = 1 + p2
r = 0.6(p) = 0.6 p m
c = 72.34°
sale
0.6u
r
=
= u = p
dr>du
0.6
will
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is
This
provided
integrity
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this
assessing
is
$
r = 0.6( -0.2954) = - 0.1772 m>s2
#
$
ar = r - ru 2 = - 0.1772 - 0.6 p(1.011)2 = - 2.104 m>s2
$
##
au = ru + 2ru = 0.6p( - 0.2954) + 2(0.6066)(1.011) = 0.6698 m>s2
tan c =
+ ©F = ma ;
;
r
r
-N cos 17.66° = 0.4( - 2.104)
+ T ©Fu = mau ;
- F + 0.4(9.81) + 0.883 sin 17.66° = 0.4(0.6698)
F = 3.92 N
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
u
r
N = 0.883 N
Ans.
Ans.
0.6u
p
13–103.
A ride in an amusement park consists of a cart which is
supported by small wheels. Initially the cart is traveling in a
circular path# of radius r0 = 16 ft such that the angular rate of
rotation is u0 = 0.2 rad>s. If the attached cable OC is drawn
#
inward at a constant speed of r = - 0.5 ft>s, determine the
tension it exerts on the cart at the instant r = 4 ft. The cart
and its passengers have a total weight of 400 lb. Neglect
the effects of friction. Hint: First show that the equation
of
..
..
motion in# the u direction yields # au = ru + 2ru =
11>r2 d1r2u2>dt = 0. When integrated, r2u = c, where the
constant c is determined from the problem data.
r
u
O
SOLUTION
+Q©Fr = mar ;
-T = ¢
+a©Fu = mau ;
0 = ¢
#
400
$
≤ ¢ r - ru 2 ≤
32.2
(1)
$
400
##
≤ ¢ ru + 2ru ≤
32.2
#
1 d
From Eq. (2), ¢ ≤ ¢ r2u ≤ = 0
r dt
(2)
#
r 2u = c
teaching
Web)
laws
or
#
Since u0 = 0.2 rad>s when r0 = 16 ft, c = 51.2.
Wide
copyright
in
Hence, when r = 4 ft,
is
of
the
not
instructors
States
World
permitted.
Dissemination
#
51.2
u = ¢ 2 ≤ = 3.2 rad>s
(4)
student
the
by
and
use
United
on
learning.
$
Since r = - 0.5 ft>s, r = 0, Eq. (1) becomes
the
(including
for
work
400
a0 - (4)(3.2)2 b
32.2
is
work
solely
of
protected
-T =
assessing
T = 509 lb
this
Ans.
integrity
of
and
work
provided
any
courses
part
the
is
This
destroy
of
and
will
their
sale
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
C
·
u0
*13–104.
#
The
arm is rotating at a rate of u = 5 rad>s when
$
u = 2 rad>s2 and u = 90°. Determine the normal force it
must exert on the 0.5-kg particle if the particle is confined
to move along the slotted path defined by the horizontal
hyperbolic spiral ru = 0.2 m.
0.2
—
u
r
r u·
··
5 rad/s, u
u
SOLUTION
p
= 90°
2
u =
#
u = 5 rad>s
$
u = 2 rad>s2
Wide
copyright
in
teaching
Web)
laws
or
r = 0.2>u = 0.12732 m
#
#
r = -0.2 u - 2 u = -0.40528 m>s
#
$
$
r = -0.2[- 2u - 3 (u)2 + u - 2 u] = 2.41801
#
$
a r = r - r(u)2 = 2.41801 - 0.12732(5)2 = - 0.7651 m>s2
$
##
au = ru + 2 ru = 0.12732(2) + 2( -0.40528)(5) = - 3.7982 m>s2
Dissemination
0.2>u
World
permitted.
-0.2u - 2
not
is
of
the
=
instructors
r
dr
(du
)
States
tan c =
student
the
by
and
use
United
on
learning.
p
c = tan - 1(- ) = -57.5184°
2
Np cos 32.4816° = 0.5( - 0.7651)
+ ©F = ma ;
;
u
u
F + Np sin 32.4816° = 0.5( -3.7982)
this
assessing
is
work
solely
of
protected
the
(including
for
work
+ c ©Fr = m ar ;
part
the
is
This
provided
integrity
of
and
work
NP = - 0.453 N
F = - 1.66 N
sale
will
their
destroy
of
and
any
courses
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
90
2 rad/s2
13–105.
The forked rod is used to move the smooth
2-lb particle around the horizontal path in the
# shape of a
limaçon, r = (2 + cos u) ft. If at all times u = 0.5 rad>s,
determine the force which the rod exerts on the particle at
the instant u = 90°. The fork and path contact the particle
on only one side.
2 ft
u
·
u
SOLUTION
r = 2 + cos u
#
#
r = -sin uu
$
#
$
r = -cos uu2 - sin uu
#
$
At u = 90°, u = 0.5 rad>s, and u = 0
r = 2 + cos 90° = 2 ft
#
r = - sin 90°(0.5) = -0.5 ft>s
Wide
copyright
in
teaching
Web)
laws
or
$
r = - cos 90°(0.5)2 - sin 90°(0) = 0
#
$
ar = r - ru2 = 0 - 2(0.5)2 = - 0.5 ft>s2
$
##
au = ru + 2ru = 2(0) + 2( -0.5)(0.5) = - 0.5 ft>s2
Dissemination
r
2 + cos u 2
=
= -2
dr>du
- sin u u = 90°
World
permitted.
c = - 63.43°
use
United
on
learning.
is
of
the
not
instructors
States
tan c =
and
2
(- 0.5)
32.2
by
N = 0.03472 lb
work
student
the
- N cos 26.57° =
the
(including
for
+ c ©Fr = mar;
2
(-0.5)
32.2
work
solely
of
protected
F - 0.03472 sin 26.57° =
integrity
of
provided
sale
will
their
destroy
of
and
any
courses
part
the
is
This
F = - 0.0155 lb
and
work
this
assessing
is
+ ©F = ma ;
;
u
u
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
r
Ans.
3 ft
13–106.
The forked rod is used to move the smooth 2-lb particle
around the horizontal path in the
# shape of a limaçon,
r = (2 + cos u) ft. If at all times u = 0.5 rad>s, determine
the force which the rod exerts on the particle at the instant
u = 60°. The fork and path contact the particle on only
one side.
2 ft
u
·
u
SOLUTION
r = 2 + cos u
#
#
r = -sin uu
#
$
$
r = -cos uu2 - sin uu
#
$
At u = 60°, u = 0.5 rad>s, and u = 0
r = 2 + cos 60° = 2.5 ft
#
r = - sin 60°(0.5) = -0.4330 ft>s
Wide
copyright
in
teaching
Web)
laws
or
$
r = - cos 60°(0.5)2 - sin 60°(0) = - 0.125 ft>s2
#
$
ar = r - ru2 = - 0.125 - 2.5(0.5)2 = - 0.75 ft>s2
$
##
au = ru + 2ru = 2.5(0) + 2( - 0.4330)(0.5) = - 0.4330 ft>s2
Dissemination
r
2 + cos u 2
=
= - 2.887
dr>du
- sin u u = 60°
World
permitted.
c = -70.89°
use
United
on
learning.
is
of
the
not
instructors
States
tan c =
by
work
student
F - 0.04930 sin 19.11° =
the
+a©Fu
N = 0.04930 lb
the
(including
for
- N cos 19.11° =
and
2
( -0.75)
32.2
+ Q©Fr = mar;
work
solely
of
protected
2
( -0.4330)
32.2
work
this
assessing
is
= mau;
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
F = - 0.0108 lb
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
r
Ans.
3 ft
13–107.
The forked rod is used to move the smooth
2-lb particle around the horizontal path in the shape of a
limaçon, r = (2 + cos u) ft. If u = (0.5t2) rad, where t is in
seconds, determine the force which the rod exerts on the
particle at the instant t = 1 s. The fork and path contact the
particle on only one side.
2 ft
u
·
u
SOLUTION
r = 2 + cos u
#
r = - sin uu
#
$
$
r = - cos uu2 - sin uu
u = 0.5t2
#
u = t
$
u = 1 rad>s2
$
At t = 1 s, u = 0.5 rad, u = 1 rad>s, and u = 1 rad>s2
r = 2 + cos 0.5 = 2.8776 ft
#
r = - sin 0.5(1) = - 0.4974 ft>s2
Wide
copyright
in
teaching
Web)
laws
or
$
r = - cos 0.5(1)2 - sin 0.5(1) = - 1.357 ft>s2
#
$
ar = r - ru2 = - 1.375 - 2.8776(1)2 = - 4.2346 ft>s2
$
##
au = ru + 2ru = 2.8776(1) + 2( - 0.4794)(1) = 1.9187 ft>s2
Dissemination
r
2 + cos u 2
=
= - 6.002
dr>du
- sin u u = 0.5 rad
World
permitted.
c = - 80.54°
use
United
on
learning.
is
of
the
not
instructors
States
tan c =
by
work
student
F - 0.2666 sin 9.46° =
the
+a©Fu = mau;
N = 0.2666 lb
the
(including
for
-N cos 9.46° =
and
2
( -4.2346)
32.2
+ Q©Fr = mar;
work
this
assessing
is
work
solely
of
protected
2
(1.9187)
32.2
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
F = 0.163 lb
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
r
Ans.
3 ft
*13–108.
The collar, which has a weight of 3 lb, slides along the
smooth rod lying in the horizontal plane and having the
shape of a parabola r = 4>11 - cos u2, where u is in radians
and r is #in feet. If the collar’s angular rate is constant and
equals u = 4 rad>s, determine the tangential retarding
force P needed to cause the motion and the normal force
that the collar exerts on the rod at the instant u = 90°.
P
r
u
SOLUTION
4
1 - cos u
r =
#
-4 sin u u
#
r =
(1 - cos u)2
$
- 4 sin u u
$
r =
+
(1 - cos u)2
#
- 4 cos u (u)2
(1 - cos u)2
#
u = 4,
At u = 90°,
+
#
8 sin2 u u2
(1 - cos u)3
$
u = 0
r = 4
laws
or
#
r = - 16
in
teaching
Web)
$
r = 128
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
$
ar = r - r(u)2 = 128 - 4(4)2 = 64
$
##
au = ru + 2 ru = 0 + 2(- 16)(4) = -128
(including
for
work
student
the
by
and
use
4
1 - cos u
r =
integrity
of
and
4
= -1
-4
the
=
part
2
- 4 sin u
°
(1 - cos u)2 u = 90
provided
4
1 - cos u)
and
any
courses
dr
(du
)
=
is
r
This
tan c =
work
this
assessing
is
work
solely
of
protected
the
-4 sin u
dr
=
du
(1 - cos u)2
sale
will
their
destroy
of
c = - 45° = 135°
3
(64)
32.2
+ c ©Fr = m ar ;
P sin 45° - N cos 45° =
+ © F = ma ;
;
u
u
- P cos 45° - N sin 45° =
3
( -128)
32.2
Solving,
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
P = 12.6 lb
Ans.
N = 4.22 lb
Ans.
13–109.
The smooth particle has a mass of 80 g. It is attached to an
elastic cord extending from O to P and due to the slotted
arm guide moves along the horizontal circular path
r = 10.8 sin u2 m. If the cord has a stiffness k = 30 N>m and
an unstretched length of 0.25 m, determine the force of the
guide on the particle
when u = 60°. The guide has a constant
#
angular velocity u = 5 rad>s.
P
r
·
u
0.4 m
SOLUTION
u
r = 0.8 sin u
#
#
r = 0.8 cos u u
O
#
$
$
r = -0.8 sin u (u)2 + 0.8 cos uu
#
$
u = 5,
u = 0
r = 0.6928
At u = 60°,
#
r = 2
laws
or
$
r = - 17.321
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
#
$
ar = r - r(u)2 = -17.321 - 0.6928(5)2 = - 34.641
$
##
au = ru + 2 ru = 0 + 2(2)(5) = 20
World
Fs = 30(0.6928 - 0.25) = 13.284 N
a+ ©Fu = mau;
F - NP sin 30° = 0.08(20)
and
-13.284 + NP cos 30° = 0.08(-34.641)
the
(including
for
work
student
the
by
Q+ ©Fr = m ar;
use
United
on
learning.
is
of
the
not
instructors
States
Fs = ks;
this
assessing
is
work
solely
of
protected
F = 7.67 N
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
NP = 12.1 N
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
5 rad/s
13–110.
The smooth particle has a mass of 80 g. It is attached to an
elastic cord extending from O to P and due to the slotted arm
guide moves along the horizontal circular path r = (0.8 sin u)
m. If the cord has a stiffness k = 30 N>m and an unstretched
length of 0.25$ m, determine
# the force of the guide on the
particle when u = 2 rad>s2, u = 5 rad>s, and u = 60°.
P
r
·
u
0.4 m
SOLUTION
r = 0.8 sin u
u
#
#
r = 0.8 cos u u
O
$
#
$
r = -0.8 sin u (u)2 + 0.8 cos uu
$
#
u = 2
u = 5,
At u = 60°,
r = 0.6928
#
r = 2
or
$
r = -16.521
Dissemination
Wide
copyright
in
teaching
Web)
laws
#
$
ar = r - r(u)2 = -16.521 - 0.6928(5)2 = - 33.841
$
##
au = r u + 2 ru = 0.6925(2) + 2(2)(5) = 21.386
World
permitted.
Fs = 30(0.6928 - 0.25) = 13.284 N
is
of
the
not
instructors
States
Fs = ks;
and
use
United
on
learning.
- 13.284 + NP cos 30° = 0.08( -33.841)
by
Q+ ©Fr = m ar;
work
student
the
F - NP sin 30° = 0.08(21.386)
(including
for
+ a©Fu = mau;
assessing
is
work
solely
of
protected
the
F = 7.82 N
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
NP = 12.2 N
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
5 rad/s
13–111.
A 0.2-kg spool slides down along a smooth rod.
If
# the rod has a constant angular rate of rotation
u = 2 rad>s in the vertical plane, show that the equations of
$
motion for the spool are r - 4r - 9.81 sin u = 0 and
#
0.8r + Ns - 1.962 cos u = 0, where Ns is the magnitude of
the normal force of the rod on the spool. Using the
methods of differential equations, it can be shown that
the solution of the first of these equations is
#
r = C1e -2t + C2e2t - 19.81>82 sin 2t. If r, r, and u are
zero when t = 0, evaluate the constants C1 and C2 to
determine r at the instant u = p>4 rad.
u
u
2 rad/s
SOLUTION
#
#
Kinematic: Here, u. = 2 rad>s and u = 0. Applying Eqs. 12–29, we have
#
$
$
$
ar = r - ru2 = r - r(22) = r - 4r
$
##
#
#
au = ru + 2ru = r(0) + 2r(2) = 4r
Equation of Motion: Applying Eq. 13–9, we have
$
1.962 sin u = 0.2(r - 4r)
or
$
r - 4r - 9.81 sin u = 0
(Q.E.D.)
#
0.8r + Ns - 1.962 cos u = 0
(2)
instructors
States
World
Dissemination
(Q.E.D.)
not
the
of
United
on
learning.
is
2dt, u = 2t. The solution of the differential
and
use
L0
the
(including
for
work
student
the
L0
equation (Eq.(1)) is given by
1
by
#
u =
u
Since u. = 2 rad>s, then
9.81
sin 2t
8
(3)
and
work
this
assessing
is
work
solely
of
protected
r = C1 e - 2 t + C2 e2t -
part
the
is
This
provided
integrity
of
Thus,
(4)
will
their
destroy
of
and
any
courses
9.81
#
cos 2t
r = - 2 C1 e - 2t + 2C2 e2t 4
(5)
9.81
#
At t = 0, r = 0 . From Eq.(4) 0 = - 2 C1 (1) + 2C2 (1) 4
(6)
sale
At t = 0, r = 0 . From Eq.(3) 0 = C1 (1) + C2 (1) - 0
Solving Eqs. (5) and (6) yields
C1 = -
9.81
16
C2 =
9.81
16
Thus,
r = -
9.81 - 2t
9.81 2t
9.81
e
e +
sin 2t
16
16
8
9.81 -e - 2t + e2t
a
- sin 2tb
8
2
9.81
(sin h 2t - sin 2t)
=
8
=
At u = 2t =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
p
,
4
r =
p
p
9.81
asin h - sin 3 b = 0.198 m
8
4
4
Ans.
permitted.
Wide
copyright
in
teaching
Web)
#
1.962 cos u - Ns = 0.2(4r)
©Fu = mau;
(1)
laws
©Fr = mar ;
r
*13–112.
The pilot of an airplane executes a vertical
loop which in part follows the path of a cardioid,
r = 600(1 + cos u) ft. If his speed at A ( u = 0°) is a
constant vP = 80 ft>s , determine the vertical force the
seat belt must exert on him to hold him to his seat when
the plane is upside down at A. He weighs 150 lb.
A
r
u
SOLUTION
r = 600(1 + cos u)|u = 0° = 1200 ft
#
#
r = -600 sin uu u = 0° = 0
$
#
#
$
r = -600 sin uu - 600 cos uu2 u = 0° = -600u2
# 2
#
v2p = r2 + a ru b
# 2
(80)2 = 0 + a1200u b
#
u = 0.06667
#
$
$
#
2vpvp = 2rr + 2 a ru b a ru + ru b
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
$
$
0 = 0 + 0 + 2r2 uu
u = 0
#
$
ar = r - ru2 = -600(0.06667)2 - 1200(0.06667)2 = -8 ft>s2
$
##
au = ru + 2ru = 0 + 0 = 0
of
the
150
b ( -8)
32.2
on
learning.
United
and
work
student
the
the
(including
for
work
solely
of
protected
this
assessing
is
integrity
of
and
work
provided
any
courses
part
the
is
This
destroy
of
and
will
their
sale
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
is
N = 113 lb
use
N - 150 = a
by
+ c ©Fr = mar;
600 (1 + cos u ) ft
13–113.
The earth has an orbit with eccentricity e = 0.0821 around the
d
sun. Knowing that the earth’s minimum distance from the
sun is 151.3(106) km, find the speed at which the earth travels
when it is at this distance. Determine the equation in polar
coordinates which describes the earth’s orbit about the sun.
SOLUTION
e =
Ch2
GMS
e =
GMS
1
¢1 ≤ (r0v0)2
GMS r0
r0 v20
y0 =
=
where C =
GMS
1
¢1 ≤ and h = r0 v0
r0
r0 v20
e = ¢
r0 v20
- 1≤
GMS
r0v20
= e + 1
GMS
GMS (e + 1)
r0
B
66.73(10 - 12)(1.99)(1030)(0.0821 + 1)
= 30818 m>s = 30.8 km>s
B
151.3(109)
Ans.
teaching
Web)
laws
or
GMS
GMS
1
1
=
a1 bcos u + 2 2
2
r
r0
r0 v0
r0v0
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
66.73(10 - 12)(1.99)(1030)
66.73(10 - 12)(1.99)(1030)
1
1
=
b
cos
u
+
a1
r
151.3(109)
151.3(109)(30818)2
C 151.3(109) D 2 (30818)2
of
the
not
1
= 0.502(10 - 12) cos u + 6.11(10 - 12)
r
sale
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of
and
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the
is
This
provided
integrity
of
and
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this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
13–114.
A communications satellite is in a circular orbit above the
earth such that it always remains directly over a point on
the earth’s surface. As a result, the period of the satellite
must equal the rotation of the earth, which is approximately
24 hours. Determine the satellite’s altitude h above the
earth’s surface and its orbital speed.
SOLUTION
The period of the satellite around
r0 = h + re = C h + 6.378(106) D m is given by
T =
circular
orbit
of
radius
2pr0
vs
24(3600) =
vs =
the
2p C h + 6.378(106) D
vs
2p C h + 6.378(106)
(1)
86.4(103)
teaching
Web)
laws
or
The velocity of the satellite orbiting around the circular orbit of radius
r0 = h + re = C h + 6.378(106) D m is given by
in
GMe
C r0
Dissemination
Wide
copyright
yS =
World
instructors
(2)
the
is
on
United
and
use
by
for
work
student
the
Solving Eqs.(1) and (2),
h = 35.87(106) m = 35.9 Mm
sale
will
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destroy
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and
any
courses
part
the
is
This
provided
integrity
of
and
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this
assessing
is
work
solely
of
protected
the
(including
yS = 3072.32 m>s = 3.07 km>s Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
not
States
h + 6.378(106)
learning.
C
permitted.
66.73(10-12)(5.976)(1024)
of
yS =
13–115.
The speed of a satellite launched into a circular orbit
about the earth is given by Eq. 13–25. Determine the
speed of a satellite launched parallel to the surface of
the earth so that it travels in a circular orbit 800 km
from the earth’s surface.
SOLUTION
For a 800-km orbit
v0 =
66.73(10 - 12)(5.976)(1024)
B
(800 + 6378)(103)
= 7453.6 m>s = 7.45 km>s
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*13–116.
A rocket is in circular orbit about the earth at an altitude of
h = 4 Mm. Determine the minimum increment in speed it
must have in order to escape the earth’s gravitational field.
h = 4 Mm
SOLUTION
Circular Orbit:
vC =
66.73(10 - 12)5.976(1024)
GMe
=
= 6198.8 m>s
A r0
B 4000(103) + 6378(103)
Parabolic Orbit:
ve =
2(66.73)(10 - 12)5.976(1024)
2GMe
=
= 8766.4 m>s
A r0
B 4000(103) + 6378(103)
¢v = ve - vC = 8766.4 - 6198.8 = 2567.6 m>s
¢v = 2.57 km>s
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
13–117.
Prove Kepler’s third law of motion. Hint: Use Eqs. 13–19,
13–28, 13–29, and 13–31.
SOLUTION
From Eq. 13–19,
GMs
1
= C cos u +
r
h2
For u = 0° and u = 180°,
GMs
1
= C +
rp
h2
or
GMs
1
= -C +
ra
h2
teaching
Web)
laws
Eliminating C, from Eqs. 13–28 and 13–29,
World
permitted.
Dissemination
Wide
copyright
in
2GMs
2a
=
2
b
h2
on
United
and
use
T2h2
4p2a2
(including
for
work
student
b2 =
by
p
(2a)(b)
h
the
T =
learning.
is
of
the
not
instructors
States
From Eq. 13–31,
any
courses
destroy
of
and
T2 = a
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
sale
4p2
b a3
GMs
will
their
GMs
4p2a3
=
2 2
Th
h2
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
Thus,
Q.E.D.
13–118.
The satellite is moving in an elliptical orbit with an eccentricity
e = 0.25. Determine its speed when it is at its maximum
distance A and minimum distance B from the earth.
2 Mm
A
SOLUTION
Ch2
GMe
e =
GMe
1
¢1 ≤ and h = r0 v0.
r0
r0v20
GMe
1
¢1 ≤ (r0 v0)2
GMe r0
r0 v20
e = ¢
v0 =
B
GMe (e + 1)
r0
teaching
Web)
r0 v20
= e + 1
GMe
r0 v20
- 1≤
GMe
or
e =
laws
where C =
Wide
copyright
permitted.
= 7713 m>s = 7.71 km>s
World
Ans.
States
C
Dissemination
66.73(10 - 12)(5.976)(1024)(0.25 + 1)
the
not
instructors
8.378(106)
by
work
solely
is
(7713) = 4628 m>s = 4.63 km>s
13.96(106)
this
assessing
8.378(106)
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
ra
vB =
and
rp
work
vA =
of
protected
the
(including
for
work
student
the
ra =
and
use
United
8.378(106)
r0
=
= 13.96 A 106 B m
2GMe
2(66.73)(10 - 12)(5.976)(1024)
- 1
- 1
r0 v0
8.378(106)(7713)2
on
learning.
is
of
vB = v0 =
in
where r0 = rp = 2 A 106 B + 6378 A 103 B = 8.378 A 106 B m.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
B
13–119.
The elliptical orbit of a satellite orbiting the earth has an
eccentricity of e = 0.45. If the satellite has an altitude of 6 Mm
at perigee, determine the velocity of the satellite at apogee and
the period.
P
A
SOLUTION
6 Mm
Here, rO = rP = 6(106) + 6378(103) = 12.378(106) m.
h = rPvP
h = 12.378(106)vP
(1)
and
GMe
1
a1 b
rP
rPvP 2
12.378(106)
c1 -
12.378(106)vP 2
d
2.6027
vP 2
(2)
in
teaching
Web)
C = 80.788(10 - 9) -
66.73(10 - 12)(5.976)(1024)
or
1
C =
laws
C =
Wide
World
permitted.
Dissemination
Ch2
GMe
by
and
use
United
on
learning.
is
of
the
not
instructors
States
e =
copyright
Using Eqs. (1) and (2),
66.73(10 - 12)(5.976)(1024)
the
(including
for
work
student
the
2
2.6027
d c 12.378(106)vP d
2
vP
assessing
is
work
solely
of
protected
0.45 =
c 80.788(10 - 9) -
the
is
This
provided
integrity
of
and
work
this
vP = 6834.78 m>s
rP vP 2
any
courses
destroy
of
and
- 1
will
their
rP
2GMe
sale
ra =
part
Using the result of vP,
12.378(106)
=
2(66.73)(10 - 12)(5.976)(1024)
12.378(106)(6834.782)
- 1
= 32.633(106) m
Since h = rPvP = 12.378(106)(6834.78) = 84.601(109) m2>s is constant,
rava = h
32.633(106)va = 84.601(109)
va = 2592.50 m>s = 2.59 km>s
Ans.
Using the result of h,
T =
=
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
p
(r + ra) 2rPra
h P
p
C 12.378(106) + 32.633(106) D 312.378(106)(32.633)(106)
84.601(109)
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
= 33 592.76 s = 9.33 hr
Ans.
*13–120.
Determine the constant speed of satellite S so that it
circles the earth with an orbit of radius r = 15 Mm. Hint:
Use Eq. 13–1.
r ⫽ 15 Mm
SOLUTION
F = G
ms a
y =
ms me
r2
Also
S
y2s
F = ms a b
r
Hence
y20
ms me
b = G
r
r2
me
5.976(1024)
b = 5156 m>s = 5.16 km>s
=
66.73(10 - 12) a
r
B
15(106)
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
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and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
A
G
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
13–121.
The rocket is in free flight along an elliptical trajectory A¿A.
The planet has no atmosphere, and its mass is 0.70 times that
of the earth. If the rocket has an apoapsis and periapsis as
shown in the figure, determine the speed of the rocket when
it is at point A.
r
B
A
6 Mm
SOLUTION
Central-Force Motion: Use ra =
M = 0.70Me, we have
r0
(2 GM>r0 y20 B -1
, with r0 = rp = 6 A 106 B m and
6(10)6
¢
2(66.73) (10-12) (0.7) [5.976(1024)]
6(106)y2P
≤ - 1
or
9 A 106 B =
laws
yA = 7471.89 m>s = 7.47 km>s
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
3 Mm
O
A¿
9 Mm
13–122.
A satellite S travels in a circular orbit around the earth. A
rocket is located at the apogee of its elliptical orbit for
which e = 0.58. Determine the sudden change in speed that
must occur at A so that the rocket can enter the satellite’s
orbit while in free flight along the blue elliptical trajectory.
When it arrives at B, determine the sudden adjustment in
speed that must be given to the rocket in order to maintain
the circular orbit.
120 Mm
B
S
10 Mm
SOLUTION
GMe
1
11 2 [Eq. 13–21] and h = r0 v0
r0
r0 v20
[Eq. 13–20]. Substitute these values into Eq. 13–17 gives
Central-Force Motion: Here, C =
1
GM
A 1 - r2v2e B A r20v20 B
0 0
r0v02
r0
ch 2
e =
=
=
-1
GMe
GMe
GMe
(1)
Rearrange Eq. (1) gives
GMe
1
=
1 + e
r0 v02
(2)
laws
or
Rearrange Eq. (2), we have
teaching
in
D
Web)
11 + e2GMe
Wide
World
, we have
the
not
instructors
on
learning.
is
A 2 GMe >r0 v20 B - 1
permitted.
Dissemination
r0
States
Substitute Eq. (2) into Eq. 13–27, ra =
(3)
copyright
r0
of
v0 =
United
and
use
by
(4)
work
the
(including
B -1
1 - e
bra
1 + e
student
2A
r0 = a
or
the
r0
1
1 + e
for
ra =
assessing
is
work
solely
of
protected
or the first elliptical orbit e = 0.58, from Eq. (4)
integrity
of
and
work
this
1 - 0.58
b C 12011062 D = 31.89911062 m
1 + 0.58
part
the
is
This
provided
1rp21 = r0 = a
their
destroy
of
and
any
courses
Substitute r0 = 1rp21 = 31.89911062 m into Eq. (3) yields
will
11 + 0.582166.732110 - 12215.9762110242
sale
1vp21 =
D
31.89911062
= 4444.34 m>s
Applying Eq. 13–20, we have
1va21 = a
rp
ra
b 1vp21 = c
31.89911062
12011062
d14444.342 = 1181.41 m>s
When the rocket travels along the second elliptical orbit, from Eq. (4), we have
1011062 = a
1 - e
b C 12011062 D
1 + e
e = 0.8462
Substitute r0 = 1rp22 = 1011062 m into Eq. (3) yields
1vp2 =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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11 + 0.84622166.732110 - 12215.9672110242
D
101106)
= 8580.25 m>s
A
13–122. continued
And in Eq. 13–20, we have
(va22 = c
1rp22
1ra22
d 1vp22 = c
1011062
12011062
d 18580.252 = 715.02 m>s
For the rocket to enter into orbit two from orbit one at A, its speed must be
decreased by
¢v = 1va21 - 1va22 = 1184.41 - 715.02 = 466 m>s
Ans.
If the rocket travels in a circular free-flight trajectory, its speed is given by Eq. 13–25.
vc =
66.73110 - 12215.9762110242
GMe
=
= 6314.89 m>s
D r0
D
1011062
The speed for which the rocket must be decreased in order to have a circular orbit is
¢v = 1vp22 - ve = 8580.25 - 6314.89 = 2265.36 m>s = 2.27 km>s
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
13–123.
An asteroid is in an elliptical orbit about the sun such that
its periheliondistance is 9.3011092 km. If the eccentricity of
the orbit is e = 0.073, determine the aphelion distance of
the orbit.
SOLUTION
rp = r0 = 9.3011092 km
e =
GMs
r0v20
1
ch2
= a1 b
a
b
r0
GMs
GMs
r0v20
e = a
r0 v20
- 1b
GMs
r0 v20
= e + 1
GMs
or
laws
Web)
teaching
in
B -1
(2)
Wide
World
permitted.
Dissemination
9.301109211.0732
0.927
learning.
is
11 - e2
=
A
not
r01e + 12
r0
2
e + 1
the
- 1
=
copyright
r0
2GMs
r0 v20
ra =
1
b
e + 1
instructors
ra =
= a
States
r0v20
of
GMs
(1)
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
ra = 10.811092 km
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
*13–124.
An elliptical path of a satellite has an eccentricity
e = 0.130. If it has a speed of 15 Mm>h when it is at perigee,
P, determine its speed when it arrives at apogee, A. Also,
how far is it from the earth’s surface when it is at A?
A
P
SOLUTION
e = 0.130
vp = v0 = 15 Mm>h = 4.167 km>s
e =
GMe
r20 v20
Ch2
1
= ¢1 b
≤a
2
r0
GMe
GMe
r0v0
e = ¢
r0 v20
-1≤
GMe
r0 v20
= e +1
GMe
or
laws
teaching
Web)
v20
in
1.130166.732110 - 12215.9762110242
Wide
permitted.
Dissemination
C 4.16711032 D 2
States
World
=
(e + 1)GMe
copyright
r0 =
United
on
learning.
is
of
the
not
instructors
= 25.96 Mm
and
use
1
e + 1
by
(including
for
work
student
=
r0 v20
the
GMe
solely
of
protected
the
r0
work
B -1
this
A
2
e + 1
assessing
- 1
=
is
r0
2GMe
r0v02
work
rA =
provided
integrity
of
and
r01e + 12
is
This
part
the
1 - e
and
any
courses
rA =
0.870
will
their
destroy
of
25.961106211.1302
sale
=
= 33.7111062 m = 33.7 Mm
vA =
=
v0r0
rA
15125.96211062
33.7111062
= 11.5 Mm>h
Ans.
d = 33.7111062 - 6.37811062
= 27.3 Mm
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
13–125.
A satellite is launched with an initial velocity
v0 = 2500 mi>h parallel to the surface of the earth.
Determine the required altitude (or range of altitudes)
above the earth’s surface for launching if the free-flight
trajectory is to be (a) circular, (b) parabolic, (c) elliptical,
and (d) hyperbolic. Take G = 34.4110-921lb # ft22>slug2,
Me = 409110212 slug, the earth’s radius re = 3960 mi, and
1 mi = 5280 ft.
SOLUTION
v0 = 2500 mi>h = 3.67(103) ft>s
e =
(a)
1 =
C2h
= 0
GMe
or C = 0
GMe
r0 v20
GMe = 34.4(10 - 9)(409)(1021)
= 14.07(1015)
14.07(1015)
[3.67(1013)]2
= 1.046(109) ft
Wide
C2h
= 1
GMe
on
learning.
is
of
United
and
use
by
student
the
(b)
the
not
instructors
States
World
permitted.
e =
Ans.
Dissemination
1.047(109)
- 3960 = 194(103) mi
5280
copyright
r =
in
teaching
Web)
v20
=
or
GMe
laws
r0 =
= 2.09(109) ft = 396(103) mi
this
2(14.07)(1015)
work
integrity
[3.67(103)]2
of
=
and
v20
provided
2GMe
part
the
is
This
r0 =
assessing
is
work
solely
of
protected
the
(including
for
work
GMe
1
1
(r20 v20)a b ¢ 1 ≤ = 1
r0
GMe
r0 v20
Ans.
sale
will
their
destroy
of
and
any
courses
r = 396(103) - 3960 = 392(103) mi
e 6 1
(c)
194(103) mi 6 r 6 392(103) mi
Ans.
e 7 1
(d)
r 7 392(103) mi
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
13–126.
A probe has a circular orbit around a planet of radius R and
mass M. If the radius of the orbit is nR and the explorer is
traveling with a constant speed v0, determine the angle u at
which it lands on the surface of the planet B when its speed
is reduced to kv0, where k 6 1 at point A.
B
u
A
SOLUTION
nR
When the probe is orbiting the planet in a circular orbit of radius rO = nR, its speed
is given by
GM
GM
=
B rO
B nR
vO =
The probe will enter the elliptical trajectory with its apoapsis at point A if its speed is
decreased to va = kvO = k
GM
at this point. When it lands on the surface of the
B nR
planet, r = rB = R.
1
1
GM
GM
=
a1 b cos u + 2 2
r
rP
rPvP2
rP vP
1
1
GM
GM
= a
- 2 2 b cos u + 2 2
rP
R
rP vP
rP vP
laws
or
(1)
GM
b = k 2nGMR is constant,
A nR
Wide
copyright
in
teaching
Web)
Since h = rava = nR ak
States
World
permitted.
Dissemination
rPvP = h
(2)
by
and
use
United
on
learning.
is
of
the
not
instructors
k 2nGMR
rP
vP =
rP
2GM
- 1
rPvP2
rP
nR =
2GM
- 1
rPvP2
2nGMR
vP 2 =
rp(rp + nR)
(including
for
work
student
the
Also,
(3)
sale
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of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
ra =
Solving Eqs.(2) and (3),
rp =
k2n
R
2 - k2
vp =
2 - k2 GM
k B nR
Substituting the result of rp and vp into Eq. (1),
1
2 - k2
1
1
- 2 b cos u + 2
= a 2
R
k nR
k nR
k nR
u = cos - 1 a
k2n - 1
b
1 - k2
Here u was measured from periapsis.When measured from apoapsis, as in the figure then
u = p - cos - 1 a
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
R
k2n - 1
b
1 - k2
Ans.
13–127.
Upon completion of the moon exploration mission, the
command module, which was originally in a circular orbit as
shown, is given a boost so that it escapes from the moon’s
gravitational field. Determine the necessary increase in
velocity so that the command module follows a parabolic
trajectory. The mass of the moon is 0.01230 Me.
A
SOLUTION
3 Mm
When the command module is moving around the circular orbit of radius
r0 = 3(106) m, its velocity is
vc =
66.73(10 - 12)(0.0123)(5.976)(1024)
GMm
=
B r0
B
3(106)
= 1278.67 m>s
The escape velocity of the command module entering into the parabolic trajectory is
ve =
2(66.73)(10 - 12)(0.0123)(5.976)(1024)
2GMm
=
B r0
B
3(106)
or
= 1808.31 m>s
teaching
Web)
laws
Thus, the required increase in the command module is
¢v = ve - vc = 1808.31 - 1278.67 = 529.64 m>s = 530 m>s
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*13–128.
The rocket is traveling in a free-flight elliptical orbit about
the earth such that e = 0.76 and its perigee is
9 Mm as shown. Determine its speed when it is at point B.
Also determine the sudden decrease in speed the rocket
must experience at A in order to travel in a circular orbit
about the earth.
A
B
SOLUTION
9 Mm
GMe
1
a1 b [Eq. 13–21] and h = r0 v 0
r0
r0 v20
[Eq. 13–20] Substitute these values into Eq. 13–17 gives
Central-Force Motion: Here C =
1
e
A 1 - GM
B 1 r20 v022
r20v20
r0v02
r0
ch2
e =
=
=
- 1
GMe
GMe
GMe
(1)
Rearrange Eq.(1) gives
GMe
1
=
1 + e
r0 v02
laws
or
(2)
in
teaching
Web)
Rearrange Eq.(2), we have
Wide
Dissemination
(3)
World
r0
not
instructors
is
of
the
r0
and
use
United
on
learning.
, we have
by
12GMe >r0 v022 - 1
the
Substitute Eq.(2) into Eq. 13–27, ra =
(including
for
work
student
r0
the
B -1
(4)
of
assessing
is
work
solely
2A
1
1 + e
protected
ra =
provided
integrity
of
and
work
this
Rearrange Eq.(4), we have
1 + e
1 + 0.76
br = a
b C 911062 D = 66.011062 m
1 - e 0
1 - 0.76
destroy
of
and
any
courses
part
the
is
This
ra = a
sale
will
their
Substitute r0 = rp = 9 A 106 B m into Eq.(3) yields
vp =
(1 + 0.76)(66.73)(10 - 12)(5.976) (1024)
D
9(106)
= 8830.82 m>s
Applying Eq. 13–20, we have
va = a
rp
ra
b np = B
911062
66.011062
R 18830.822 = 1204.2 m>s = 1.20 km>s
Ans.
If the rocket travels in a circular free-flight trajectory, its speed is given by Eq. 13–25.
ve =
66.73110 - 12215.9762110242
GMe
=
= 6656.48 m>s
D r0
D
911062
The speed for which the rocket must be decreased in order to have a circular orbit is
¢v = v p - v c = 8830.82 - 6656.48 = 2174.34 m>s = 2.17 km>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
permitted.
B
copyright
11 + e2 GMe
States
v0 =
13–129.
A rocket is in circular orbit about the earth at an altitude above
the earth’s surface of h = 4 Mm. Determine the minimum
increment in speed it must have in order to escape the earth’s
gravitational field.
h
SOLUTION
Circular orbit:
vC =
66.73(10 - 12)5.976(1024)
GMe
=
= 6198.8 m>s
B r0
B 4000(103) + 6378(103)
Parabolic orbit:
ve =
2(66.73)(10 - 12)5.976(1024)
2GMe
=
= 8766.4 m>s
B r0
B 4000(103) + 6378(103)
¢v = ve - vC = 8766.4 - 6198.8 = 2567.6 m>s
¢v = 2.57 km>s
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
4 Mm
13–130.
The satellite is in an elliptical orbit having an eccentricity of
e = 0.15. If its velocity at perigee is vP = 15 Mm>h,
determine its velocity at apogee A and the period of the
satellite.
15 Mm/h
P
SOLUTION
Here, vP = c 15(106)
m
1h
da
b = 4166.67 m>s.
h 3600 s
h = rPvP
h = rP (4166.67) = 4166.67rp
(1)
and
GMe
1
¢1 ≤
rP
rP vP 2
C =
66.73(10-12)(5.976)(1024)
1
B1 R
rp
rp(4166.672)
C =
22.97(106)
1
B1 R
rP
rP
e =
Ch2
GMe
or
C =
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
(2)
World
the
not
instructors
States
on
learning.
is
of
United
use
by
and
66.73(10-12)(5.976)(1024)
work
student
the
0.15 =
22.97(106)
1
B1 R (4166.67 rP)2
rP
rP
is
work
solely
of
protected
the
(including
for
rP = 26.415(106) m
integrity
provided
part
the
is
any
courses
- 1
2(66.73)(10-12)(5.976)(1024)
26.415(106)(4166.672)
will
26.415(106)
sale
=
their
destroy
of
rP vP 2
This
rP
2GMe
and
rA =
of
and
work
this
assessing
Using the result of rp
- 1
= 35.738(106) m
Since h = rP vP = 26.415(106)(4166.672) = 110.06(109) m2>s is constant,
rA vA = h
35.738(106)vA = 110.06(109)
vA = 3079.71 m>s = 3.08 km>s
Ans.
Using the results of h, rA, and rP,
T =
=
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
p
A r + rA B 2rP rA
6 P
p
C 26.415(106) + 35.738(106) D 226.415(106)(35.738)(106)
110.06(109)
= 54 508.43 s = 15.1 hr
Ans.
A
13–131.
A rocket is in a free-flight elliptical orbit about the earth
such that the eccentricity of its orbit is e and its perigee is r0.
Determine the minimum increment of speed it should have
in order to escape the earth’s gravitational field when it is at
this point along its orbit.
SOLUTION
To escape the earth’s gravitational field, the rocket has to
make a parabolic trajectory.
ParabolicTrajectory:
ye =
A
2GMe
r0
Elliptical Orbit:
e =
Ch2
GMe
e =
GMe
1
¢1 ≤ (r0 y0)2
GMe r0
r0 y20
GMe
1
¢1 ≤ and h = r0 y0
r0
r0y20
in
teaching
Web)
laws
or
where C =
World
permitted.
Dissemination
Wide
copyright
r0 y20
- 1b
GMe
not
instructors
States
e = a
United
on
learning.
is
of
the
GMe (e + 1)
r0
and
use
B
by
y0 =
the
r0 y20
= e + 1
GMe
of
work
this
assessing
is
integrity
of
and
work
provided
any
courses
part
the
is
This
destroy
of
and
will
their
sale
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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the
(including
for
work
student
GMe (e + 1)
2GMe
GMe
=
a 22 - 21 + eb
r0
r0
A
A r0
solely
A
protected
¢y =
Ans.
*13–132.
The rocket shown is originally in a circular orbit 6 Mm
above the surface of the earth.It is required that it travel
in another circular orbit having an altitude of 14 Mm.To
do this,the rocket is given a short pulse of power at A so
that it travels in free flight along the gray elliptical path
from the first orbit to the second orbit.Determine the
necessary speed it must have at A just after the power
pulse, and at the time required to get to the outer orbit
along the path AA¿. What adjustment in speed must be
made at A¿ to maintain the second circular orbit?
A
O
6 Mm
14 Mm
SOLUTION
r0
Central-Force Motion: Substitute Eq. 13–27, ra =
(2GM>r0v20) - 1
, with
ra = (14 + 6.378)(106) = 20.378(106) m and r0 = rp = (6 + 6.378)(106)
= 12.378(106) m, we have
12.378(106)
or
20.378(106) =
laws
2(66.73)(10 - 12)[5.976(1024)]
teaching
Web)
≤ - 1
12.378(106)v2p
in
¢
States
World
permitted.
Dissemination
Wide
copyright
vp = 6331.27 m>s
United
R (6331.27) = 3845.74 m>s
and
(including
for
work
20.378(106)
use
12.378(106)
student
ra
≤ vp = B
by
rp
the
va = ¢
on
learning.
is
of
the
not
instructors
Applying Eq. 13–20. we have
integrity
of
and
provided
and
any
courses
part
the
is
This
p
(r + ra)2rp ra
h p
T =
work
this
assessing
is
work
solely
of
protected
the
Eq. 13–20 gives h = rp vp = 12.378(106)(6331.27) = 78.368(109) m2>s. Thus, applying
Eq. 13–31, we have
p
[(12.378 + 20.378)(106)]212.378(20.378)(106)
78.368(109)
sale
will
their
destroy
of
=
= 20854.54 s
The time required for the rocket to go from A to A¿ (half the orbit) is given by
t =
T
= 10427.38 s = 2.90 hr
2
Ans.
In order for the satellite to stay in the second circular orbit, it must achieve a speed
of (Eq. 13–25)
vc =
66.73(10 - 12)(5.976)(1024)
GMe
=
= 4423.69 m>s = 4.42 km>s
A r0
A
20.378(106)
Ans.
The speed for which the rocket must be increased in order to enter the second
circular orbit at A¿ is
¢v = vc - va = 4423.69 - 3845.74 = 578 m s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
A'
14–1.
The 20-kg crate is subjected to a force having a constant
direction and a magnitude F = 100 N. When s = 15 m, the
crate is moving to the right with a speed of 8 m/s. Determine
its speed when s = 25 m. The coefficient of kinetic friction
between the crate and the ground is mk = 0.25.
F
30°
SOLUTION
Equation of Motion: Since the crate slides, the friction force developed between the
crate and its contact surface is Ff = mkN = 0.25N. Applying Eq. 13–7, we have
+ c a Fy = may ;
N + 100 sin 30° - 20(9.81) = 20(0)
N = 146.2 N
in
teaching
Web)
laws
or
Principle of Work and Energy: The horizontal component of force F which acts
in the direction of displacement does positive work, whereas the friction force
Ff = 0.25(146.2) = 36.55 N does negative work since it acts in the opposite direction
to that of displacement. The normal reaction N, the vertical component of force F
and the weight of the crate do not displace hence do no work. Applying Eq.14–7,
we have
Dissemination
Wide
copyright
T1 + a U1 - 2 = T2
instructors
States
World
permitted.
25 m
on
learning.
is
of
the
not
100 cos 30° ds
use
United
1
(20)(8 2) +
2
L15 m
by
1
(20)v2
2
work
student
the
(including
the
36.55 ds =
for
L15 m
and
25 m
-
sale
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their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
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and
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this
assessing
is
work
solely
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protected
v = 10.7 m s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
14–2.
F (lb)
For protection, the barrel barrier is placed in front of the
bridge pier. If the relation between the force and deflection
of the barrier is F = (90(103)x1>2) lb, where x is in ft,
determine the car’s maximum penetration in the barrier.
The car has a weight of 4000 lb and it is traveling with a
speed of 75 ft>s just before it hits the barrier.
F ⫽ 90(10)3 x1/2
x (ft)
SOLUTION
Principle of Work and Energy: The speed of the car just before it crashes into the
barrier is v1 = 75 ft>s. The maximum penetration occurs when the car is brought to a
stop, i.e., v2 = 0. Referring to the free-body diagram of the car, Fig. a, W and N do no
work; however, Fb does negative work.
T1 + ©U1 - 2 = T2
1 4000
a
b (752) + c 2 32.2
L0
xmax
90(103)x1>2dx d = 0
xmax = 3.24 ft
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
14–3.
The crate, which has a mass of 100 kg, is subjected to the
action of the two forces. If it is originally at rest, determine
the distance it slides in order to attain a speed of 6 m>s. The
coefficient of kinetic friction between the crate and the
surface is mk = 0.2.
5
30
SOLUTION
Equations of Motion: Since the crate slides, the friction force developed between
the crate and its contact surface is Ff = mk N = 0.2N. Applying Eq. 13–7, we have
+ c ©Fy = may;
3
N + 1000 a b - 800 sin 30° - 100(9.81) = 100(0)
5
N = 781 N
Wide
copyright
in
teaching
Web)
laws
or
Principle of Work and Energy: The horizontal components of force 800 N and
1000 N which act in the direction of displacement do positive work, whereas the
friction force Ff = 0.2(781) = 156.2 N does negative work since it acts in the
opposite direction to that of displacement. The normal reaction N, the vertical
component of 800 N and 1000 N force and the weight of the crate do not displace,
hence they do no work. Since the crate is originally at rest, T1 = 0. Applying
Eq. 14–7, we have
of
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Dissemination
T1 + a U1-2 = T2
the
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learning.
is
4
1
0 + 800 cos 30°(s) + 1000 a bs - 156.2s = (100) A 62 B
5
2
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the
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student
s = 1.35m
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1000 N
800 N
Ans.
3
4
*14–4.
The 2-kg block is subjected to a force having a constant
direction and a magnitude F = 1300>11 + s22 N, where s is
in meters. When s = 4 m, the block is moving to the left
with a speed of 8 m>s. Determine its speed when s = 12 m.
The coefficient of kinetic friction between the block and the
ground is mk = 0.25.
s
SOLUTION
+ c ©Fy = 0;
NB = 2(9.81) +
150
1 + s
T1 + ©U1 - 2 = T2
12
12
300
1
1
150
(2)(8)2 - 0.25[2(9.81)(12 - 4)] - 0.25
ds +
b ds cos 30° = (2)(v22 )
a
2
1
+
s
1
+
s
2
L4
L4
v22 = 24.76 - 37.5 lna
1 + 12
1 + 12
b + 259.81 lna
b
1 + 4
1 + 4
v2 = 15.4 m>s
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sale
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
F
30
v
14–5.
When a 7-kg projectile is fired from a cannon barrel that
has a length of 2 m, the explosive force exerted on the
projectile, while it is in the barrel, varies in the manner
shown. Determine the approximate muzzle velocity of the
projectile at the instant it leaves the barrel. Neglect the
effects of friction inside the barrel and assume the barrel is
horizontal.
F (MN)
15
10
5
SOLUTION
0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
The work done is measured as the area under the force–displacement curve. This
area is approximately 31.5 squares. Since each square has an area of 2.5 A 106 B (0.2),
T1 + ©U1-2 = T2
0 + C (31.5)(2.5) A 106 B (0.2) D =
Ans.
(approx.)
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v2 = 2121 m>s = 2.12 km>s
1
(7)(v2)2
2
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s (m)
14–6.
The spring in the toy gun has an unstretched length of
100 mm. It is compressed and locked in the position shown.
When the trigger is pulled, the spring unstretches 12.5 mm,
and the 20-g ball moves along the barrel. Determine the
speed of the ball when it leaves the gun. Neglect friction.
50 mm
k
Principle of Work and Energy: Referring to the free-body diagram of the
ball bearing shown in Fig. a, notice that Fsp does positive work. The spring
has an initial and final compression of s1 = 0.1 - 0.05 = 0.05 m and
s2 = 0.1 - (0.05 + 0.0125) = 0.0375 m.
T1 + ©U1-2 = T2
1
1
1
0 + B ks1 2 - ks2 2 R = mvA 2
2
2
2
or
1
1
1
0 + B (2000)(0.05)2 - (2000)(0.03752) R = (0.02)vA 2
2
2
2
vA = 10.5 m>s
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Ans.
sale
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150 mm
D
A
B
SOLUTION
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
2 kN/m
14–7.
As indicated by the derivation, the principle of work and
energy is valid for observers in any inertial reference frame.
Show that this is so, by considering the 10-kg block which
rests on the smooth surface and is subjected to a horizontal
force of 6 N. If observer A is in a fixed frame x, determine the
final speed of the block if it has an initial speed of 5 m>s and
travels 10 m, both directed to the right and measured from
the fixed frame. Compare the result with that obtained by an
observer B, attached to the x¿ axis and moving at a constant
velocity of 2 m>s relative to A. Hint: The distance the block
travels will first have to be computed for observer B before
applying the principle of work and energy.
A
B
5 m/s
6N
10 m
Observer A:
T1 + ©U1 - 2 = T2
1
1
(10)(5)2 + 6(10) = (10)v22
2
2
v2 = 6.08 m>s
or
Ans.
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Observer B:
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F = ma
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a = 0.6 m>s2
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1 2
at
2 c
United
the
by
and
s = s0 + v0t +
use
+ B
A:
the
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1
(0.6)t2
2
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10 = 0 + 5t +
work
this
assessing
is
work
t2 + 16.67t - 33.33 = 0
any
sale
Block moves 10 - 3.609 = 6.391 m
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and
At v = 2 m>s, s¿ = 2(1.805) = 3.609 m
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This
provided
integrity
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and
t = 1.805 s
Thus
T1 + ©U1 - 2 = T2
1
1
(10)(3)2 + 6(6.391) = (10)v22
2
2
v2 = 4.08 m>s
Note that this result is 2 m>s less than that observed by A.
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x¿
2 m/s
SOLUTION
6 = 10a
x
Ans.
*14–8.
If the 50-kg crate is subjected to a force of P = 200 N,
determine its speed when it has traveled 15 m starting from
rest. The coefficient of kinetic friction between the crate
and the ground is mk = 0.3.
P
SOLUTION
Free-Body Diagram: Referring to the free-body diagram of the crate, Fig. a,
+ c Fy = may;
N - 50(9.81) = 50(0)
N = 490.5 N
Thus, the frictional force acting on the crate is Ff = mkN = 0.3(490.5) = 147.15 N.
Principle of Work and Energy: Referring to Fig. a, only P and Ff do work.The work of
P will be positive, whereas Ff does negative work.
T1 + g U1 - 2 = T2
0 + 200(15) - 147.15(15) =
1
(50)v2
2
v = 5.63 m>s
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14–9.
If the 50-kg crate starts from rest and attains a speed of
6 m>s when it has traveled a distance of 15 m, determine the
force P acting on the crate. The coefficient of kinetic friction
between the crate and the ground is mk = 0.3.
P
SOLUTION
Free-Body Diagram: Referring to the free-body diagram of the crate, Fig. a,
+ c Fy = may;
N - 50(9.81) = 50(0)
N = 490.5 N
Thus, the frictional force acting on the crate is Ff = mkN = 0.3(490.5) = 147.15 N.
Principle of Work and Energy: Referring to Fig. a, only P and Ff do work.The work of
P will be positive, whereas Ff does negative work.
T1 + gU1 - 2 = T2
0 + P(15) - 147.15(15) =
1
(50)(62)
2
P = 207 N
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14–10.
The 2-Mg car has a velocity of v1 = 100 km>h when the
driver sees an obstacle in front of the car. If it takes 0.75 s for
him to react and lock the brakes, causing the car to skid,
determine the distance the car travels before it stops. The
coefficient of kinetic friction between the tires and the road
is mk = 0.25.
v1
SOLUTION
Free-Body Diagram: The normal reaction N on the car can be determined by
writing the equation of motion along the y axis. By referring to the free-body
diagram of the car, Fig. a,
+ c ©Fy = may;
N - 2000(9.81) = 2000(0)
N = 19 620 N
Since the car skids, the frictional force acting on the car is
Ff = mkN = 0.25(19620) = 4905N.
teaching
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or
Principle of Work and Energy: By referring to Fig. a, notice that only Ff does work,
m
1h
which is negative. The initial speed of the car is v1 = c 100(103) d a
b =
h 3600 s
27.78 m>s. Here, the skidding distance of the car is denoted as s¿ .
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T1 + ©U1-2 = T2
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1
(2000)(27.782) + (-4905s¿) = 0
2
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s¿ = 157.31 m
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(including
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The distance traveled by the car during the reaction time is
s– = v1t = 27.78(0.75) = 20.83 m. Thus, the total distance traveled by the car
before it stops is
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s = s¿ + s– = 157.31 + 20.83 = 178.14 m = 178 m
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Ans.
100 km/h
14–11.
The 2-Mg car has a velocity of v1 = 100 km>h when the
driver sees an obstacle in front of the car. It takes 0.75 s for
him to react and lock the brakes, causing the car to skid. If
the car stops when it has traveled a distance of 175 m,
determine the coefficient of kinetic friction between the
tires and the road.
v1
100 km/h
SOLUTION
Free-Body Diagram: The normal reaction N on the car can be determined by
writing the equation of motion along the y axis and referring to the free-body
diagram of the car, Fig. a,
+ c ©Fy = may;
N - 2000(9.81) = 2000(0)
N = 19 620 N
Since the car skids, the frictional force acting on the car can be computed from
Ff = mkN = mk(19 620).
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or
Principle of Work and Energy: By referring to Fig. a, notice that only Ff does work,
1h
m
which is negative. The initial speed of the car is v1 = c 100(103) d a
b =
h 3600 s
27.78 m>s. Here, the skidding distance of the car is s¿ .
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T1 + ©U1-2 = T2
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1
(2000)(27.782) + C - mk(19 620)s¿ D = 0
2
work
student
the
by
39.327
mk
protected
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s¿ =
the
is
This
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assessing
is
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solely
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The distance traveled by the car during the reaction time is
s– = v1t = 27.78(0.75) = 20.83 m. Thus, the total distance traveled by the car
before it stops is
destroy
will
their
39.327
+ 20.83
mk
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175 =
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s = s¿ + s–
mk = 0.255
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Ans.
*14–12.
Design considerations for the bumper B on the 5-Mg train
car require use of a nonlinear spring having the loaddeflection characteristics shown in the graph. Select the
proper value of k so that the maximum deflection of the
spring is limited to 0.2 m when the car, traveling at 4 m>s,
strikes the rigid stop. Neglect the mass of the car wheels.
F (N)
F
ks2
s (m)
SOLUTION
B
0.2
1
ks2 ds = 0
(5000)(4)2—
2
L0
40 000 - k
(0.2)3
= 0
3
k = 15.0 MN>m2
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14–13.
The 2-lb brick slides down a smooth roof, such that when it
is at A it has a velocity of 5 ft>s. Determine the speed of the
brick just before it leaves the surface at B, the distance d
from the wall to where it strikes the ground, and the speed
at which it hits the ground.
y
A
5 ft/s
15 ft
5
3
4
B
SOLUTION
TA + ©UA-B = TB
30 ft
2
2
1
1
a
b(5)2 + 2(15) = a
b v2B
2 32.2
2 32.2
vB = 31.48 ft>s = 31.5 ft>s
+ b
a:
Ans.
x
d
s = s0 + v0t
4
d = 0 + 31.48 a bt
5
or
1
ac t2
2
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s = s0 + v0t -
laws
A+TB
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1
3
30 = 0 + 31.48 a bt + (32.2)t2
5
2
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16.1t2 + 18.888t - 30 = 0
work
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the
by
and
Solving for the positive root,
solely
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protected
the
(including
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t = 0.89916 s
is
work
4
d = 31.48a b (0.89916) = 22.6 ft
5
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the
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destroy
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2
2
1
1
a
b(5)2 + 2(45) = a
b v2C
2 32.2
2 32.2
of
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TA + ©UA-C = TC
This
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this
assessing
Ans.
vC = 54.1 ft>s
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Ans.
14–14.
If the cord is subjected to a constant force of F = 300 N
and the 15-kg smooth collar starts from rest at A, determine
the velocity of the collar when it reaches point B. Neglect
the size of the pulley.
200 mm
B
C
30⬚
200 mm
F ⫽ 300 N
200 mm
SOLUTION
300 mm
Free-Body Diagram: The free-body diagram of the collar and cord system at an
arbitrary position is shown in Fig. a.
A
Principle of Work and Energy: Referring to Fig. a, only N does no work since it
always acts perpendicular to the motion. When the collar moves from position A to
position B, W displaces vertically upward a distance h = (0.3 + 0.2) m = 0.5 m,
while
force
F
displaces
a
distance
s = AC - BC = 20.72 + 0.42 -
of
20.2 + 0.2 = 0.5234 m . Here, the work of F is positive, whereas W does
negative work.
2
2
TA + g UA - B = TB
1
(15)vB2
2
0 + 300(0.5234) + [- 15(9.81)(0.5)] =
vB = 3.335 m>s = 3.34 m>s
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14–15.
36
Barrier stopping force (kip)
The crash cushion for a highway barrier consists of a nest of
barrels filled with an impact-absorbing material. The barrier
stopping force is measured versus the vehicle penetration
into the barrier. Determine the distance a car having a
weight of 4000 lb will penetrate the barrier if it is originally
traveling at 55 ft>s when it strikes the first barrel.
SOLUTION
T1 + ©U1 - 2 = T2
1 4000
a
b (55)2 - Area = 0
2 32.2
27
18
9
0
Area = 187.89 kip # ft
2(9) + (5 - 2)(18) + x(27) = 187.89
x = 4.29 ft 6 (15 - 5) ft
(O.K!)
Thus
s = 5 ft + 4.29 ft = 9.29 ft
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2
5
10
15
20
25
Vehicle penetration (ft)
*14–16.
Determine the velocity of the 60-lb block A if the two
blocks are released from rest and the 40-lb block B moves
2 ft up the incline. The coefficient of kinetic friction
between both blocks and the inclined planes is mk = 0.10.
SOLUTION
Block A:
+ a©Fy = may;
A
NA - 60 cos 60° = 0
B
NA = 30 lb
60
FA = 0.1(30) = 3 lb
Block B:
+Q©Fy = may;
NB - 40 cos 30° = 0
or
NB = 34.64 lb
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FB = 0.1(34.64) = 3.464 lb
Dissemination
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Use the system of both blocks. NA, NB, T, and R do no work.
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T1 + ©U1-2 = T2
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learning.
is
1 60
1 40
a
b v2A + a
b v2
2 32.2
2 32.2 B
student
the
by
use
United
(0 + 0) + 60 sin 60°|¢sA| - 40 sin 30°|¢sB| - 3|¢sA| -3.464|¢sB| =
the
(including
for
work
2sA + sB = l
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2¢sA = - ¢sB
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2vA = -vB
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When |¢sB| = 2 ft, |¢sA| = 1 ft
Substituting and solving,
vA = 0.771 ft>s
vB = -1.54 ft>s
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Ans.
30
14–17.
If the cord is subjected to a constant force of F = 30 lb and
the smooth 10-lb collar starts from rest at A, determine its
speed when it passes point B. Neglect the size of pulley C.
y ⫽ 1 x2
2
y
C
B
4.5 ft
F ⫽ 30 lb
A
SOLUTION
x
1 ft
Free-Body Diagram: The free-body diagram of the collar and cord system at an
arbitrary position is shown in Fig. a.
Principle of Work and Energy: By referring to Fig. a, only N does no work since it
always acts perpendicular to the motion. When the collar moves from position A to
position B, W displaces upward through a distance h = 4.5 ft, while force F displaces a
distance of s = AC - BC = 262 + 4.52 - 2 = 5.5 ft. The work of F is positive,
whereas W does negative work.
laws
or
TA + gUA - B = TB
in
teaching
Web)
1 10
a
bv 2
2 32.2 B
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copyright
0 + 30(5.5) + [-10(4.5)] =
vB = 27.8 ft>s
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Dissemination
Ans.
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3 ft
2 ft
14–18.
The two blocks A and B have weights WA = 60 lb and
WB = 10 lb. If the kinetic coefficient of friction between the
incline and block A is mk = 0.2, determine the speed of A
after it moves 3 ft down the plane starting from rest. Neglect
the mass of the cord and pulleys.
A
SOLUTION
5
Kinematics: The speed of the block A and B can be related by using position
coordinate equation.
sA + (sA - sB) = l
2¢sA - ¢sB = 0
4
2sA - sB = l
¢sB = 2¢sA = 2(3) = 6 ft
2vA - vB = 0
(1)
Equation of Motion: Applying Eq. 13–7, we have
60
4
(0)
N - 60 a b =
5
32.2
N = 48.0 lb
or
+ ©Fy¿ = may¿ ;
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the
(including
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3
1
1
0 + WA ¢ ¢sA ≤ - Ff ¢sA - WB ¢sB = mAv2A + mB v2B
5
2
2
work
student
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by
and
use
T1 + a U1 - 2 = T2
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this
3
1 60
1 10
60 B (3) R - 9.60(3) - 10(6) = ¢
≤ v2A + ¢
≤ v2B
5
2 32.2
2 32.2
1236.48 = 60v2A + 10v2B
will
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Eqs. (1) and (2) yields
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vA = 3.52 ft>s
vB = 7.033 ft s
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Ans.
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Principle of Work and Energy: By considering the whole system, WA which acts in
the direction of the displacement does positive work. WB and the friction force
Ff = mkN = 0.2(48.0) = 9.60 lb does negative work since they act in the opposite
direction to that of displacement Here, WA is being displaced vertically (downward)
3
¢s and WB is being displaced vertically (upward) ¢sB. Since blocks A and B are
5 A
at rest initially, T1 = 0. Applying Eq. 14–7, we have
3
B
14–19.
y
If the 10-lb block passes point A on the smooth track with a
speed of vA = 5 ft>s, determine the normal reaction on the
block when it reaches point B.
y ⫽ 1 x2
32
vA ⫽ 5 ft/s
A
8 ft
SOLUTION
Free-Body Diagram: The free-body diagram of the block at an arbitrary position is
shown in Fig. a.
x
B
Principle of Work and Energy: Referring to Fig. a, N does no work since it always
acts perpendicular to the motion. When the block slides down the track from
position A to position B, W displaces vertically downward h = 8 ft and does positive
work.
TA + g UA - B = TB
10
1 10
1
a
b A 52 B + 10(8) = a
bv 2
2 32.2
2 32.2 B
laws
or
vB = 23.24 ft>s
in
teaching
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v2
. By referring to Fig. a,
r
Dissemination
Wide
copyright
Equation of Motion: Here, an =
the
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10
v2
ba b
r
32.2
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N - 10 cos u = a
g Fn = man;
use
United
on
10 v2
a b + 10 cos u
32.2 r
by
and
(1)
the
(including
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student
the
N =
solely
of
protected
dy
d2y
1
1
. The slope that the track at position B
=
x and 2 =
dx
16
16
dx
dx
makes with the horizontal is uB = tan - 1 a b 2
= tan - 1(0) = 0°. The radius of
dy x = 0
curvature of the track at position B
rB =
`
dy 2 3>2
b R
dx
d2y
dx2
`
B1 + ¢
=
`
2 3>2
1
x≤ R
16
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Geometry: Here,
1
`
16
4
= 16 ft
x=0
Substituting u = uB = 0°, v = vB = 23.24 ft>s, and r = rB = 16 ft into Eq. (1),
NB =
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10 23.242
c
d + 10 cos 0° = 20.5 lb
32.2 16
Ans.
*14–20.
The steel ingot has a mass of 1800 kg. It travels along the
conveyor at a speed v = 0.5 m>s when it collides with the
“nested” spring assembly. Determine the maximum
deflection in each spring needed to stop the motion of the
ingot. Take kA = 5 kN>m, kB = 3 kN>m.
0.5 m
0.45 m
kB
A
SOLUTION
B
Assume both springs compress
T1 + ©U1 - 2 = T2
1
1
1
(1800)(0.5)2 - (5000)s2 - (3000)(s - 0.05)2 = 0
2
2
2
225 - 2500 s2 - 1500(s2 - 0.1 s + 0.0025) = 0
s2 - 0.0375 s - 0.05531 = 0
s = 0.2547 m 7 0.05 m
sB = 0.205 m
Ans.
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Ans.
or
(O.K!)
sA = 0.255 m
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kA
C
14–21.
The steel ingot has a mass of 1800 kg. It travels along the
conveyor at a speed v = 0.5 m>s when it collides with the
“nested” spring assembly. If the stiffness of the outer spring
is kA = 5 kN>m, determine the required stiffness kB of the
inner spring so that the motion of the ingot is stopped at
the moment the front, C, of the ingot is 0.3 m from the wall.
0.5 m
0.45 m
kA
kB
A
SOLUTION
B
T1 + ©U1 - 2 = T2
1
1
1
(1800)(0.5)2 - (5000)(0.5 - 0.3)2 - (kB)(0.45 - 0.3)2 = 0
2
2
2
kB = 11.1 kN m
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or
Ans.
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or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
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C
14–22.
The 25-lb block has an initial speed of v0 = 10 ft>s when it
is midway between springs A and B. After striking spring B,
it rebounds and slides across the horizontal plane toward
spring A, etc. If the coefficient of kinetic friction between
the plane and the block is mk = 0.4, determine the total
distance traveled by the block before it comes to rest.
2 ft
kA = 10 lb/in.
1 ft
v0 = 10 ft/s
A
B
SOLUTION
Principle of Work and Energy: Here, the friction force Ff = mk N = 0.4(25) =
10.0 lb. Since the friction force is always opposite the motion, it does negative work.
When the block strikes spring B and stops momentarily, the spring force does
negative work since it acts in the opposite direction to that of displacement.
Applying Eq. 14–7, we have
Tl + a U1 - 2 = T2
1
1 25
a
b (10)2 - 10(1 + s1) - (60)s21 = 0
2 32.2
2
s1 = 0.8275 ft
Wide
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laws
or
Assume the block bounces back and stops without striking spring A. The spring
force does positive work since it acts in the direction of displacement. Applying
Eq. 14–7, we have
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Dissemination
T2 + a U2 - 3 = T3
United
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is
of
the
not
1
(60)(0.82752) - 10(0.8275 + s2) = 0
2
by
and
use
0 +
for
work
student
the
s2 = 1.227 ft
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Since s2 = 1.227 ft 6 2 ft, the block stops before it strikes spring A. Therefore, the
above assumption was correct. Thus, the total distance traveled by the block before
it stops is
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the
is
This
provided
sTot = 2s1 + s2 + 1 = 2(0.8275) + 1.227 + 1 = 3.88 ft
Ans.
kB = 60 lb/in.
14–23.
The train car has a mass of 10 Mg and is traveling at 5 m> s
when it reaches A. If the rolling resistance is 1> 100 of the
weight of the car, determine the compression of each spring
when the car is momentarily brought to rest.
300 kN/m
500 kN/m
A
SOLUTION
Free-Body Diagram: The free-body diagram of the train in contact with the spring
1
[10 000(9.81)] = 981 N.
is shown in Fig. a. Here, the rolling resistance is Fr =
100
The compression of springs 1 and 2 at the instant the train is momentarily at rest
will be denoted as s1 and s2. Thus, the force developed in springs 1 and 2 are
A Fsp B 1 = k1s1 = 300(103)s1 and A Fsp B 2 = 500(103)s2. Since action is equal to
reaction,
A Fsp B 1 = A Fsp B 2
300(103)s1 = 500(103)s2
laws
or
s1 = 1.6667s2
Wide
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Principle of Work and Energy: Referring to Fig. a, W and N do no work, and Fsp and
Fr do negative work.
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T1 + g U1 - 2 = T2
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by
1
1
C 300(103) D s12 f + e - C 500(103) D s22 f = 0
2
2
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not
1
(10 000)(52) + [- 981(30 + s1 + s2)] +
2
e-
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150(103)s12 + 250(103)s22 + 981(s1 + s2) - 95570 = 0
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Substituting Eq. (1) into Eq. (2),
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This
666.67(103)s22 + 2616s2 - 95570 = 0
sale
s2 = 0.3767 m = 0.377 m
will
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Solving for the positive root of the above equation,
Substituting the result of s2 into Eq. (1),
s1 = 0.6278 m = 0.628 m
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vA ⫽ 5 m/s
30 m
Ans.
*14–24.
The 0.5-kg ball is fired up the smooth vertical circular track
using the spring plunger. The plunger keeps the spring
compressed 0.08 m when s = 0. Determine how far s it must
be pulled back and released so that the ball will begin to
leave the track when u = 135°.
B
135
SOLUTION
u
s
Equations of Motion:
A
©Fn = man;
y2B
0.5(9.81) cos 45° = 0.5 a
b
1.5
y2B = 10.41 m2>s2
k
Principle of Work and Energy: Here, the weight of the ball is being displaced
vertically by s = 1.5 + 1.5 sin 45° = 2.561 m and so it does negative work. The
spring force, given by Fsp = 500(s + 0.08), does positive work. Since the ball is at
rest initially, T1 = 0. Applying Eq. 14–7, we have
or
TA + a UA-B = TB
in
teaching
Web)
1
(0.5)(10.41)
2
500(s + 0.08) ds - 0.5(9.81)(2.561) =
copyright
L0
laws
s
0 +
Ans.
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s = 0.1789 m = 179 mm
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or transmission in any form or by any means, electronic, mechanical,
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500 N/m
1.5 m
14–25.
The skier starts from rest at A and travels down the ramp. If
friction and air resistance can be neglected, determine his
speed vB when he reaches B. Also, find the distance s to
where he strikes the ground at C, if he makes the jump
traveling horizontally at B. Neglect the skier’s size. He has a
mass of 70 kg.
A
50 m
B
4m
s
C
SOLUTION
30
TA + © UA - B = TB
0 + 70(9.81)(46) =
1
(70)(vB)2
2
vB = 30.04 m>s = 30.0 m>s
+ B
A:
Ans.
s = s0 + v0 t
s cos 30° = 0 + 30.04t
s = s0 + v0 t +
1 2
a t
2 c
laws
or
A+TB
in
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Web)
1
(9.81)t2
2
Wide
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s sin 30° + 4 = 0 + 0 +
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Eliminating t,
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is
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s2 - 122.67s - 981.33 = 0
work
student
the
by
and
Solving for the positive root
for
s = 130 m
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(including
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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14–26.
The catapulting mechanism is used to propel the 10-kg
slider A to the right along the smooth track. The propelling
action is obtained by drawing the pulley attached to rod BC
rapidly to the left by means of a piston P. If the piston
applies a constant force F = 20 kN to rod BC such that it
moves it 0.2 m, determine the speed attained by the slider if
it was originally at rest. Neglect the mass of the pulleys,
cable, piston, and rod BC.
A
P
B
F
SOLUTION
2 sC + sA = l
2 ¢ sC + ¢ sA = 0
2(0.2) = - ¢ sA
-0.4 = ¢ sA
T1 + ©U1 - 2 = T2
1
(10)(vA)2
2
vA = 28.3 m>s
Ans.
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or
0 + (10 000)(0.4) =
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C
14–27.
Block A has a weight of 60 lb and B has a weight of 10 lb.
Determine the distance A must descend from rest before it
obtains a speed of 8 ft>s. Also, what is the tension in the
cord supporting A? Neglect the mass of the cord and
pulleys.
SOLUTION
B
2 sA + sB = l
2 ¢ sA = - ¢ s B
A
2 vA = -vB
For vA = 8 ft/s,
vB = - 16 ft/s
For the system:
T1 + ©U1 - 2 = T2
or
1 60
1 10
a
b(8)2 + a
b ( -16)2
2 32.2
2 32.2
laws
[0 + 0] + [60(sA) - 10(2sA)] =
sA = 2.484 = 2.48 ft
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For block A:
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T1 + ©U1 - 2 = T2
by
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use
1 60
a
b (8)2
2 32.2
(including
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student
the
0 + 60(2.484) - TA(2.484) =
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the
TA = 36.0 lb
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Ans.
*14–28.
The cyclist travels to point A, pedaling until he reaches a
speed vA = 4 m>s. He then coasts freely up the curved
surface. Determine how high he reaches up the surface
before he comes to a stop. Also, what are the resultant normal
force on the surface at this point and his acceleration? The
total mass of the bike and man is 75 kg. Neglect friction, the
mass of the wheels, and the size of the bicycle.
y
C
SOLUTION
1
2
x1/2
y1/2
B y
x
45
A
1
2
x + y = 2
4m
1 -1
1 1 dy
x 2 + y-2
= 0
2
2
dx
1
dy
-x - 2
=
1
dx
y-2
T1 + ©U1 - 2 = T2
1
(75)(4)2 - 75(9.81)(y) = 0
2
y = 0.81549 m = 0.815 m
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x1>2 + (0.81549)1>2 = 2
permitted.
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x = 1.2033 m
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-(1.2033) - 1>2
dy
=
= - 0.82323
dx
(0.81549) - 1>2
and
use
tan u =
for
work
student
the
by
u = -39.46°
the
(including
Nb - 9.81(75) cos 39.46° = 0
is
work
solely
of
protected
Q+ ©Fn = m an ;
Ans.
integrity
of
and
work
this
assessing
Nb = 568 N
provided
75(9.81) sin 39.46° = 75 at
part
the
is
This
+R©Ft = m at ;
any
courses
Ans.
sale
will
their
destroy
of
and
a = at = 6.23 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
2
4m
x
14–29.
The collar has a mass of 20 kg and slides along the smooth
rod. Two springs are attached to it and the ends of the rod as
shown. If each spring has an uncompressed length of 1 m
and the collar has a speed of 2 m>s when s = 0, determine
the maximum compression of each spring due to the backand-forth (oscillating) motion of the collar.
s
kA
50 N/m
kB
1m
1m
0.25 m
SOLUTION
T1 + ©U1 - 2 = T2
1
1
1
(20)(2)2 - (50)(s)2 - (100)(s)2 = 0
2
2
2
s = 0.730 m
sale
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and
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the
is
This
provided
integrity
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and
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this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
100 N/m
14–30.
A
The 30-lb box A is released from rest and slides down along
the smooth ramp and onto the surface of a cart. If the cart
is prevented from moving, determine the distance s from the
end of the cart to where the box stops. The coefficient of
kinetic friction between the cart and the box is mk = 0.6.
10 ft
4 ft
s
C
B
SOLUTION
Principle of Work and Energy: WA which acts in the direction of the vertical
displacement does positive work when the block displaces 4 ft vertically. The friction
force Ff = mk N = 0.6(30) = 18.0 lb does negative work since it acts in the
opposite direction to that of displacement Since the block is at rest initially and is
required to stop, TA = TC = 0. Applying Eq. 14–7, we have
TA + a UA - C = TC
0 + 30(4) - 18.0s¿ = 0
s = 10 - s¿ = 3.33 ft
Ans.
sale
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destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Thus,
s¿ = 6.667 ft
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
14–31.
Marbles having a mass of 5 g are dropped from rest at A through
the smooth glass tube and accumulate in the can at C. Determine
the placement R of the can from the end of the tube and the
speed at which the marbles fall into the can. Neglect the size
of the can.
A
B
3m
2m
SOLUTION
C
TA + © UA-B = TB
R
1
0 + [0.005(9.81)(3 - 2)] = (0.005)v2B
2
vB = 4.429 m>s
A+TB
s = s0 + v0t +
2 = 0 + 0 =
1
a t2
2 c
1
(9.81)t2
2
laws
in
teaching
Web)
s = s0 + v0 t
Wide
copyright
+ b
a:
or
t = 0.6386 s
R = 0 + 4.429(0.6386) = 2.83 m
permitted.
Dissemination
Ans.
on
learning.
is
of
the
not
instructors
States
World
TA + © UA-C = T1
by
and
use
United
1
(0.005)v2C
2
for
work
student
the
0 + [0.005(9.81)(3) =
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
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this
assessing
is
work
solely
of
protected
the
(including
vC = 7.67 m>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
*14–32.
The cyclist travels to point A, pedaling until he reaches a
speed vA = 8 m>s. He then coasts freely up the curved
surface. Determine the normal force he exerts on the
surface when he reaches point B. The total mass of the bike
and man is 75 kg. Neglect friction, the mass of the wheels,
and the size of the bicycle.
y
C
x1/2
y1/2
B y
x
2
4m
SOLUTION
1
2
45
A
1
2
x + y = 2
4m
1 1 dy
1 -1
x 2 + y- 2
= 0
2
2
dx
1
dy
- x- 2
=
1
dx
y- 2
For y = x,
1
2 x2 = 2
laws
or
x = 1, y = 1 (Point B)
in
teaching
Web)
Thus,
permitted.
Dissemination
Wide
copyright
dy
= -1
dx
instructors
States
World
tan u =
use
United
on
learning.
is
of
the
not
u = -45°
the
(including
for
work
student
the
by
and
dy
1
1
= (- x - 2)(y2)
dx
d2y
is
work
solely
of
protected
dy
1
1
1
1 3
1
= y2 a x - 2 b - x - 2 a b ay - 2 b a b
2
2
dx
dx
integrity
provided
the
part
1 1 -3
1 1
y2 x 2 + a b
2
2 x
is
=
of
and
any
courses
dx2
This
d2y
of
and
work
this
assessing
2
sale
will
their
destroy
For x = y = 1
dy
= -1,
dx
r =
d2y
dx2
= 1
[1 + ( -1)2]3>2
= 2.828 m
1
T1 + ©U1 - 2 = T2
1
1
(75)(8)2 - 75(9.81)(1) = (75)(v2B)
2
2
v2B = 44.38
Q+ ©Fn = man ;
NB - 9.81(75) cos 45° = 75 a
NB = 1.70 kN
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
44.38
b
2.828
Ans.
x
14–33.
The man at the window A wishes to throw the 30-kg sack on
the ground. To do this he allows it to swing from rest at B to
point C, when he releases the cord at u = 30°. Determine
the speed at which it strikes the ground and the distance R.
A
B
8m
8m
u
16 m
C
SOLUTION
TB + ©UB - C = TC
D
0 + 30(9.81)8 cos 30° =
R
1
(30)v2C
2
yC = 11.659 m>s
TB + ©UB - D = TD
0 + 30(9.81)(16) =
1
(30) v2D
2
vD = 17.7 m>s
or
Ans.
teaching
Web)
laws
During free flight:
in
1 2
act
2
permitted.
Dissemination
Wide
copyright
( + T) s = s0 + v0t +
the
not
instructors
States
World
1
(9.81)t2
2
use
United
on
learning.
is
of
16 = 8 cos 30° - 11.659 sin 30°t +
for
work
student
the
by
and
t2 - 1.18848 t - 1.8495 = 0
solely
of
protected
the
(including
Solving for the positive root:
this
assessing
is
work
t = 2.0784 s
any
will
sale
s = 24.985 m
their
destroy
of
and
s = 8 sin 30° + 11.659 cos 30°(2.0784)
courses
part
the
is
This
provided
integrity
of
and
work
+ )s = s + v t
(:
0
0
Thus,
R = 8 + 24.985 = 33.0 m
Ans.
Also,
(vD)x = 11.659 cos 30° = 10.097 m>s
(+ T) (vD)x = -11.659 sin 30° + 9.81(2.0784) = 14.559 m>s
nD = 2(10.097)2 + (14.559) 2 = 7.7 m>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
14–34.
The spring bumper is used to arrest the motion of the
4-lb block, which is sliding toward it at v = 9 ft>s. As
shown, the spring is confined by the plate P and wall using
cables so that its length is 1.5 ft. If the stiffness of the
spring is k = 50 lb>ft, determine the required unstretched
length of the spring so that the plate is not displaced more
than 0.2 ft after the block collides into it. Neglect friction,
the mass of the plate and spring, and the energy loss
between the plate and block during the collision.
1.5 ft.
5 ft.
SOLUTION
T1 + ©U1 - 2 = T2
4
1
1
1
a
b (9)2 - c (50)(s - 1.3)2 - (50)(s - 1.5)2 d = 0
2 32.2
2
2
0.20124 = s2 - 2.60 s + 1.69 - (s2 - 3.0 s + 2.25)
0.20124 = 0.4 s - 0.560
s = 1.90 ft
will
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destroy
of
and
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courses
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the
is
This
provided
integrity
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and
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this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
sale
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
vA
P
k
A
14–35.
The collar has a mass of 20 kg and is supported on the
smooth rod. The attached springs are undeformed when
d = 0.5 m. Determine the speed of the collar after the
applied force F = 100 N causes it to be displaced so that
d = 0.3 m. When d = 0.5 m the collar is at rest.
k' = 15 N/m
F = 100 N
60°
SOLUTION
d
T1 + a U1 - 2 = T2
k = 25 N/m
1
(15)(0.5 - 0.3)2
2
0 + 100 sin 60°(0.5 - 0.3) + 196.2(0.5 - 0.3) 1
1
(25)(0.5 - 0.3)2 = (20)v2C
2
2
-
vC = 2.36 m>s
sale
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and
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courses
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is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*14–36.
If the force exerted by the motor M on the cable is 250 N,
determine the speed of the 100-kg crate when it is hoisted
to s = 3 m. The crate is at rest when s = 0.
M
SOLUTION
Kinematics: Expressing the length of the cable in terms of position coordinates sC and
sP referring to Fig. a,
s
3sC + (sC - sP) = l
4sC - sP = l
(1)
Using Eq. (1), the change in position of the crate and point P on the cable can be
written as
(+ T)
4¢sC - ¢sP = 0
Here, ¢sC = - 3 m. Thus,
¢sp = - 12 m = 12 m c
or
4(- 3) - ¢sP = 0
not
instructors
States
World
permitted.
Dissemination
T1 + g U1 - 2 = T2
and
use
United
on
learning.
is
of
the
1
m v2
2 C
the
by
0 + T¢sP + [ - WC ¢sC] =
the
(including
for
work
student
1
(100)v2
2
work
solely
of
protected
0 + 250(12) + [- 100(9.81)(3)] =
Ans.
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and
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courses
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is
This
provided
integrity
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and
work
this
assessing
is
v = 1.07 m>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Wide
copyright
in
teaching
Principle of Work and Energy: Referring to the free-body diagram of the pulley
system, Fig. b, F1 and F2 do no work since it acts at the support; however, T does
positive work and WC does negative work.
Web)
laws
(+ T )
14–37.
If the track is to be designed so that the passengers of the
roller coaster do not experience a normal force equal to
zero or more than 4 times their weight, determine the
limiting heights hA and hC so that this does not occur. The
roller coaster starts from rest at position A. Neglect friction.
A
C
rC
20 m
rB
hC
B
SOLUTION
Free-Body Diagram:The free-body diagram of the passenger at positions B and C
are shown in Figs. a and b, respectively.
Equations of Motion: Here, an =
v2
. The requirement at position B is that
r
NB = 4mg. By referring to Fig. a,
+ c ©Fn = man;
4mg - mg = m ¢
vB 2
≤
15
or
vB 2 = 45g
teaching
Web)
laws
At position C, NC is required to be zero. By referring to Fig. b,
in
vC 2
≤
20
Wide
permitted.
Dissemination
mg - 0 = m ¢
copyright
+ T ©Fn = man;
on
learning.
is
of
the
not
instructors
States
World
vC 2 = 20g
of
protected
the
(including
for
work
student
the
by
and
use
United
Principle of Work and Energy: The normal reaction N does no work since it always
acts perpendicular to the motion. When the rollercoaster moves from position A
to B, W displaces vertically downward h = hA and does positive work.
this
assessing
is
work
solely
We have
part
the
is
This
provided
integrity
of
and
work
TA + ©UA-B = TB
any
courses
1
m(45g)
2
sale
hA = 22.5 m
will
their
destroy
of
and
0 + mghA =
Ans.
When the rollercoaster moves from position A to C, W displaces vertically
downward h = hA - hC = (22.5 - hC) m.
TA + ©UA-B = TB
0 + mg(22.5 - hC) =
hC = 12.5 m
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
1
m(20g)
2
Ans.
15 m
hA
14–38.
The 150-lb skater passes point A with a speed of 6 ft>s.
Determine his speed when he reaches point B and the
normal force exerted on him by the track at this point.
Neglect friction.
y
y2
B
25 ft
Free-Body Diagram:The free-body diagram of the skater at an arbitrary position is
shown in Fig. a.
Principle of Work and Energy: By referring to Fig. a, notice that N does no work since it
always acts perpendicular to the motion. When the skier slides down the track from A
to B, W displaces vertically downward h = yA - yB = 20 - C 2(25)1>2 D = 10 ft and
does positive work.
TA + ©UA-B = TB
or
1 150
1 150
¢
≤ A 62 B + C 150(10) D = ¢
≤v 2
2 32.2
2 32.2 B
vB = 26.08 ft>s = 26.1 ft>s
teaching
Web)
laws
Ans.
in
v2
. By referring to Fig. a,
r
not
the
is
on
150 v2
a b
32.2 r
learning.
N = 150 cos u -
of
150 v2
a b
32.2 r
United
150 cos u - N =
(1)
solely
of
protected
the
(including
for
work
student
the
by
and
use
R+ ©Fn = man ;
instructors
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World
permitted.
Dissemination
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Equations of Motion:Here, an =
is
work
dy
d2y
1
1
= 1>2 , and 2 = . The slope that the
dx
dx
x
2x 3>2
dx
track at position B makes with the horizontal is uB = tan-1 a b 2
dy x = 25 ft
1
= 11.31°. The radius of curvature of the track at position B is
= tan a 1>2 b 2
x
x = 25 ft
given by
sale
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destroy
of
and
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courses
part
the
is
This
provided
integrity
of
and
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this
assessing
Geometry: Here, y = 2x 1>2,
3>2
dy 2 3>2
B1 + a b R
dx
2
d2y
dx2
C1 + ¢
1
x 1>2
=
2
2-
2
≤ S
1
2x 3>2
6
= 265.15 ft
2
x = 25 ft
Substituting u = uB = 11.31°, v = vB = 26.08 ft>s, and r = rB = 265.15 ft into
Eq. (1),
NB = 150 cos 11.31° = 135 lb
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
20 ft
x
SOLUTION
rB =
4x
A
150 26.082
¢
≤
32.2 265.15
Ans.
14–39.
The 8-kg cylinder A and 3-kg cylinder B are released from
rest. Determine the speed of A after it has moved 2 m
starting from rest. Neglect the mass of the cord and pulleys.
SOLUTION
Kinematics: Express the length of cord in terms of position coordinates sA and sB by
referring to Fig. a
2sA + sB = l
A
(1)
B
Thus
(2)
2¢sA + ¢sB = 0
If we assume that cylinder A is moving downward through a distance of ¢sA = 2 m,
Eq. (2) gives
(+ T )
2(2) + ¢sB = 0
¢sB = - 4 m = 4 m c
laws
or
Taking the time derivative of Eq. (1),
2vA + vB = 0
teaching
Web)
(3)
in
(+ T )
Dissemination
Wide
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g T1 + gU1 - 2 = g T2
the
not
instructors
States
World
permitted.
1
1
(8)vA2 + (3)vB2
2
2
learning.
is
of
0 + 8(2)9.81 - 3(4)9.81 =
for
work
student
the
by
and
use
United
on
Positive net work on left means assumption of A moving down is correct. Since
vB = - 2vA,
Ans.
of
protected
the
(including
vA = 1.98 m>s T
sale
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destroy
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and
any
courses
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the
is
This
provided
integrity
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and
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this
assessing
is
work
solely
vB = - 3.96 m>s = 3.96 m>s c
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*14–40.
Cylinder A has a mass of 3 kg and cylinder B has a mass of
8 kg. Determine the speed of A after it moves upwards 2 m
starting from rest. Neglect the mass of the cord and pulleys.
SOLUTION
a T1 + a U1 - 2 = a T2
A
1
1
(3)v2A + (8)v2B
2
2
0 + 2[F1 - 3(9.81)] + 4[8(9.81) - F2] =
B
Also, vB = 2vA, and because the pulleys are massless, F1 = 2F2. The F1 and F2
terms drop out and the work-energy equation reduces to
255.06 = 17.5v2A
vA = 3.82 m>s
sale
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and
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courses
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the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
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teaching
Web)
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or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
14–41.
A 2-lb block rests on the smooth semicylindrical surface. An
elastic cord having a stiffness k = 2 lb>ft is attached to the
block at B and to the base of the semicylinder at point C. If
the block is released from rest at A (u = 0°), determine the
unstretched length of the cord so that the block begins to
leave the semicylinder at the instant u = 45°. Neglect the
size of the block.
k
2 lb/ft
B
1.5 ft
u
C
SOLUTION
+b©Fn = man;
2 sin 45° =
v2
2
a
b
32.2 1.5
v = 5.844 ft>s
T1 + © U1-2 = T2
2
2
3p
2
1
1
1
(2) C p(1.5) - l0 D - (2) c
(1.5) - l0 d - 2(1.5 sin 45°) = a
b (5.844)2
2
2
4
2 32.2
0 +
l0 = 2.77 ft
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This
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the
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student
the
by
and
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United
on
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is
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World
permitted.
Dissemination
Wide
copyright
in
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Web)
laws
or
Ans.
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
A
14–42.
The jeep has a weight of 2500 lb and an engine which
transmits a power of 100 hp to all the wheels. Assuming the
wheels do not slip on the ground, determine the angle u of
the largest incline the jeep can climb at a constant speed
v = 30 ft>s.
θ
SOLUTION
P = FJv
100(550) = 2500 sin u(30)
u = 47.2°
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This
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Dissemination
Wide
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in
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or
Ans.
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14–43.
Determine the power input for a motor necessary to lift 300 lb
at a constant rate of 5 ft> s. The efficiency of the motor is
P = 0.65.
SOLUTION
Power: The power output can be obtained using Eq. 14–10.
P = F # v = 300(5) = 1500 ft # lb>s
Using Eq. 14–11, the required power input for the motor to provide the above
power output is
power input =
1500
= 2307.7 ft # lb>s = 4.20 hp
0.65
Ans.
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and
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courses
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is
This
provided
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the
(including
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the
by
and
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United
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Dissemination
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in
teaching
Web)
laws
or
=
power output
P
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*14–44.
An automobile having a mass of 2 Mg travels up a 7° slope
at a constant speed of v = 100 km>h. If mechanical friction
and wind resistance are neglected, determine the power
developed by the engine if the automobile has an efficiency
P = 0.65.
7⬚
SOLUTION
Equation of Motion: The force F which is required to maintain the car’s constant
speed up the slope must be determined first.
+ ©Fx¿ = max¿;
F - 2(103)(9.81) sin 7° = 2(103)(0)
F = 2391.08 N
100(103) m
1h
b = 27.78 m>s.
R * a
h
3600 s
The power output can be obtained using Eq. 14–10.
Power: Here, the speed of the car is y = B
laws
or
P = F # v = 2391.08(27.78) = 66.418(103) W = 66.418 kW
Wide
copyright
in
teaching
Web)
Using Eq. 14–11, the required power input from the engine to provide the above
power output is
World
permitted.
Dissemination
power output
e
of
the
not
instructors
States
power input =
learning.
is
66.418
= 102 kW
0.65
on
United
and
work
student
the
(including
for
work
solely
of
protected
this
assessing
is
integrity
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and
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provided
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This
destroy
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
use
by
the
=
14–45.
The Milkin Aircraft Co. manufactures a turbojet engine that
is placed in a plane having a weight of 13000 lb. If the engine
develops a constant thrust of 5200 lb, determine the power
output of the plane when it is just ready to take off with a
speed of 600 mi>h.
SOLUTION
At 600 ms>h.
88 ft>s 1
b
= 8.32 (103) hp
60 m>h 550
Ans.
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This
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P = 5200(600)a
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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14–46.
To dramatize the loss of energy in an automobile, consider a
car having a weight of 5000 lb that is traveling at 35 mi>h. If
the car is brought to a stop, determine how long a 100-W light
bulb must burn to expend the same amount of energy.
11 mi = 5280 ft.2
SOLUTION
5280 ft
1h
35 mi
b * a
b * a
b =
h
1 mi
3600 s
Energy: Here, the speed of the car is y = a
51.33 ft>s. Thus, the kinetic energy of the car is
1 2
1 5000
my = a
b A 51.332 B = 204.59 A 103 B ft # lb
2
2 32.2
The power of the bulb is
73.73 ft # lb>s. Thus,
550 ft # lb>s
1 hp
b * a
b =
746 W
1 hp
or
204.59(103)
U
= 2774.98 s = 46.2 min
=
Pbulb
73.73
Ans.
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This
provided
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the
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t =
Pbulb = 100 W * a
laws
U =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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14–47.
The escalator steps move with a constant speed of 0.6 m>s.
If the steps are 125 mm high and 250 mm in length,
determine the power of a motor needed to lift an average
mass of 150 kg per step. There are 32 steps.
250 mm
125 mm
v
0.6 m/s
SOLUTION
Step height: 0.125 m
The number of steps:
4
= 32
0.125
Total load: 32(150)(9.81) = 47 088 N
4
= 2 m, then
2
If load is placed at the center height, h =
laws
or
4
U = 47 088 a b = 94.18 kJ
2
2(32(0.25))2 + 42
teaching
Web)
≤ = 0.2683 m>s
in
4
permitted.
World
94.18
U
=
= 12.6 kW
t
7.454
Ans.
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
P =
instructors
h
2
=
= 7.454 s
ny
0.2683
States
t =
Dissemination
Wide
copyright
ny = n sin u = 0.6 ¢
the
(including
Also,
is
work
solely
of
protected
P = F # v = 47 088(0.2683) = 12.6 kW
sale
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This
provided
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Ans.
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4m
*14–48.
If the escalator in Prob. 14–47 is not moving, determine the
constant speed at which a man having a mass of 80 kg must
walk up the steps to generate 100 W of power—the same
amount that is needed to power a standard light bulb.
250 mm
125 mm
v
0.6 m/s
SOLUTION
P =
(80)(9.81)(4)
U1 - 2
=
= 100
t
t
n =
2(32(0.25))2 + 42
s
=
= 0.285 m>s
t
31.4
t = 31.4 s
sale
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and
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courses
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is
This
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the
(including
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work
student
the
by
and
use
United
on
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is
of
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States
World
permitted.
Dissemination
Wide
copyright
in
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Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
4m
14–49.
The 2-Mg car increases its speed uniformly from rest to
25 m>s in 30 s up the inclined road. Determine the
maximum power that must be supplied by the engine, which
operates with an efficiency of P = 0.8. Also, find the
average power supplied by the engine.
1
10
SOLUTION
Kinematics:The constant acceleration of the car can be determined from
+ B
A:
v = v0 + ac t
25 = 0 + ac (30)
ac = 0.8333 m>s2
Equations of Motion: By referring to the free-body diagram of the car shown in
Fig. a,
©Fx¿ = max¿ ;
F - 2000(9.81) sin 5.711° = 2000(0.8333)
teaching
Web)
laws
or
F = 3618.93N
instructors
States
World
permitted.
Dissemination
Wide
copyright
(Pout)max = F # vmax = 3618.93(25) = 90 473.24 W
in
Power: The maximum power output of the motor can be determined from
by
Pout
90473.24
= 113 091.55 W = 113 kW
=
e
0.8
work
student
the
Ans.
of
work
solely
assessing
is
The average power output can be determined from
protected
the
(including
for
Pin =
and
use
United
on
learning.
is
of
the
not
Thus, the maximum power input is given by
provided
integrity
of
and
work
this
25
b = 45 236.62 W
2
courses
part
the
is
This
(Pout)avg = F # vavg = 3618.93 a
their
destroy
of
and
any
Thus,
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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e
45236.62
= 56 545.78 W = 56.5 kW
0.8
will
(Pout)avg
=
sale
(Pin)avg =
Ans.
14–50.
The crate has a mass of 150 kg and rests on a surface for
which the coefficients of static and kinetic friction are
ms = 0.3 and mk = 0.2, respectively. If the motor M supplies
a cable force of F = (8t2 + 20) N, where t is in seconds,
determine the power output developed by the motor when
t = 5 s.
M
SOLUTION
Equations of Equilibrium: If the crate is on the verge of slipping, Ff = ms N = 0.3N.
From FBD(a),
+ c ©Fy = 0;
N - 150(9.81) = 0
+ ©F = 0;
:
x
0.3(1471.5) - 3 A 8 t2 + 20 B = 0
N = 1471.5 N
t = 3.9867 s
Equations of Motion: Since the crate moves 3.9867 s later, Ff = mk N = 0.2N.
From FBD(b),
+ c ©Fy = may ;
+ ©F = ma ;
:
x
x
N - 150(9.81) = 150 (0)
N = 1471.5 N
or
0.2 (1471.5) - 3 A 8 t2 + 20 B = 150 ( -a)
in
teaching
Web)
laws
a = A 0.160 t2 - 1.562 B m>s2
the
not
instructors
States
World
A 0.160 t2 - 1.562 B dt
is
United
on
L3.9867 s
learning.
L0
5
dy =
of
y
permitted.
Dissemination
Wide
copyright
Kinematics:Applying dy = adt, we have
work
student
the
by
and
use
y = 1.7045 m>s
is
work
solely
of
protected
the
(including
for
Power: At t = 5 s, F = 8 A 52 B + 20 = 220 N. The power can be obtained using
Eq. 14–10.
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This
provided
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this
assessing
P = F # v = 3 (220) (1.7045) = 1124.97 W = 1.12 kW
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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Ans.
14–51.
The 50-kg crate is hoisted up the 30° incline by the pulley
system and motor M. If the crate starts from rest and, by
constant acceleration, attains a speed of 4 m>s after
traveling 8 m along the plane, determine the power that
must be supplied to the motor at the instant the crate has
moved 8 m. Neglect friction along the plane. The motor has
an efficiency of P = 0.74.
M
SOLUTION
30
Kinematics:Applying equation y2 = y20 + 2ac (s - s0), we have
42 = 02 + 2a(8 - 0)
a = 1.00 m>s2
Equations of Motion:
+ ©Fx¿ = max¿ ;
F - 50(9.81) sin 30° = 50(1.00)
F = 295.25 N
Power: The power output at the instant when y = 4 m>s can be obtained using
Eq. 14–10.
or
P = F # v = 295.25 (4) = 1181 W = 1.181 kW
in
teaching
Web)
laws
Using Eq. 14–11, the required power input to the motor in order to provide the
above power output is
permitted.
Dissemination
Wide
copyright
power output
e
not
instructors
States
World
power input =
is
of
the
1.181
= 1.60 kW
0.74
on
learning.
United
and
work
the
(including
for
work
solely
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assessing
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This
destroy
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Ans.
use
student
the
by
=
*14–52.
The 50-lb load is hoisted by the pulley system and motor M.
If the motor exerts a constant force of 30 lb on the cable,
determine the power that must be supplied to the motor if
the load has been hoisted s = 10 ft starting from rest. The
motor has an efficiency of P = 0.76.
M
SOLUTION
+ c ©Fy = m ay ;
B
2(30) - 50 =
50
a
32.2 B
s
aB = 6.44 m>s2
( + c) v2 = v20 + 2ac (s - s0)
v2B = 0 + 2(6.44)(10 - 0)
vB = 11.349 ft>s
2sB + sM = l
laws
or
2nB = - vM
teaching
Web)
vM = -2( 11.349) = 22.698 ft>s
Dissemination
Wide
copyright
in
Po = F # v = 30(22.698) = 680.94 ft # lb>s
the
not
instructors
States
World
permitted.
680.94
= 895.97 ft # lb>s
0.76
United
on
learning.
is
of
Pi =
sale
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This
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student
the
by
and
use
Pi = 1.63 hp
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Ans.
14–53.
The 10-lb collar starts from rest at A and is lifted by
applying a constant vertical force of F = 25 lb to the cord.
If the rod is smooth, determine the power developed by the
force at the instant u = 60°.
3 ft
4 ft
u
F
SOLUTION
A
Work of F
U1 - 2 = 25(5 - 3.464) = 38.40 lb # ft
T1 + ©U1 - 2 = T2s
0 + 38.40 - 10(4 - 1.732) =
1 10 2
(
)v
2 32.2
v = 10.06 ft>s
P = F # v = 25 cos 60°(10.06) = 125.76 ft # lb>s
P = 0.229 hp
sale
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This
provided
integrity
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and
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is
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solely
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the
(including
for
work
student
the
by
and
use
United
on
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is
of
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not
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permitted.
Dissemination
Wide
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Web)
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Ans.
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
14–54.
The 10-lb collar starts from rest at A and is lifted with a
constant speed of 2 ft>s along the smooth rod. Determine the
power developed by the force F at the instant shown.
3 ft
4 ft
u
F
SOLUTION
+ c ©Fy = m ay ;
A
4
F a b - 10 = 0
5
F = 12.5 lb
4
P = F # v = 12.5 a b (2) = 20 lb # ft>s
5
= 0.0364 hp
sale
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destroy
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and
any
courses
part
the
is
This
provided
integrity
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and
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this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
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is
of
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Dissemination
Wide
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
14–55.
The elevator E and its freight have a total mass of 400 kg.
Hoisting is provided by the motor M and the 60-kg block C.
If the motor has an efficiency of P = 0.6, determine the
power that must be supplied to the motor when the elevator
is hoisted upward at a constant speed of vE = 4 m>s.
M
C
SOLUTION
vE
Elevator:
E
Since a = 0,
+ c ©Fy = 0;
60(9.81) + 3T - 400(9.81) = 0
T = 1111.8 N
2sE + (sE - sP) = l
3vE = vP
Since vE = -4 m>s,
or
(1111.8)(12)
F # vP
=
= 22.2 kW
e
0.6
Ans.
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This
provided
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the
(including
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student
the
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is
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Dissemination
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Pi =
vP = - 12 m>s
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*14–56.
The sports car has a mass of 2.3 Mg, and while it is traveling
at 28 m/s the driver causes it to accelerate at 5 m>s2. If the
drag resistance on the car due to the wind is FD = 10.3v22 N,
where v is the velocity in m/s, determine the power supplied
to the engine at this instant. The engine has a running
efficiency of P = 0.68.
FD
SOLUTION
+ ©F = m a ;
:
x
x
F - 0.3v2 = 2.3(103)(5)
F = 0.3v2 + 11.5(103)
At v = 28 m>s
F = 11 735.2 N
PO = (11 735.2)(28) = 328.59 kW
PO
328.59
= 438 kW
=
e
0.68
Ans.
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(including
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the
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Dissemination
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in
teaching
Web)
laws
or
Pi =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
14–57.
The sports car has a mass of 2.3 Mg and accelerates at
6 m>s2, starting from rest. If the drag resistance on the car
due to the wind is FD = 110v2 N, where v is the velocity in
m/s, determine the power supplied to the engine when
t = 5 s. The engine has a running efficiency of P = 0.68.
FD
SOLUTION
+ ©F = m a ;
:
x
x
F - 10v = 2.3(103)(6)
F = 13.8(103) + 10 v
+ )v = v + a t
(:
0
c
v = 0 + 6(5) = 30 m>s
PO = F # v = [13.8(103) + 10(30)](30) = 423.0 kW
PO
423.0
= 622 kW
=
e
0.68
Ans.
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This
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Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Pi =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
14–58.
The block has a mass of 150 kg and rests on a surface for
which the coefficients of static and kinetic friction are
ms = 0.5 and mk = 0.4, respectively. If a force
F = 160t22 N, where t is in seconds, is applied to the cable,
determine the power developed by the force when t = 5 s.
Hint: First determine the time needed for the force to
cause motion.
F
SOLUTION
+ ©F = 0;
:
x
2F - 0.5(150)(9.81) = 0
F = 367.875 = 60t2
t = 2.476 s
+ ©F = m a ;
:
x
x
2(60t2)- 0.4(150)(9.81) = 150ap
ap = 0.8t2 - 3.924
dv = a dt
L2.476
A 0.8t2 - 3.924 B dt
laws
L0
5
dv =
or
v
in
teaching
Web)
5
0.8 3
b t - 3.924t `
= 19.38 m>s
3
2.476
Wide
copyright
v = a
instructors
States
World
permitted.
Dissemination
sP + (sP - sF) = l
United
on
learning.
is
of
the
not
2vP = vF
student
the
by
and
use
vF = 2(19.38) = 38.76 m>s
the
(including
for
work
F = 60(5)2 = 1500 N
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P = F # v = 1500(38.76) = 58.1 kW
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Ans.
14–59.
v
The rocket sled has a mass of 4 Mg and travels from rest
along the horizontal track for which the coefficient of
kinetic friction is mk = 0.20. If the engine provides a
constant thrust T = 150 kN, determine the power output of
the engine as a function of time. Neglect the loss of fuel
mass and air resistance.
T
SOLUTION
+ ©F = ma ;
:
x
x
150(10)3 - 0.2(4)(10)3(9.81) = 4(10)3 a
a = 35.54 m>s2
+ Bv = v + a t
A:
0
c
= 0 + 35.54t = 35.54t
P = T # v = 150(10)3 (35.54t) = 5.33t MW
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This
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permitted.
Dissemination
Wide
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Web)
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or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*14–60.
A loaded truck weighs 16(103) lb and accelerates uniformly
on a level road from 15 ft>s to 30 ft>s during 4 s. If the
frictional resistance to motion is 325 lb, determine the
maximum power that must be delivered to the wheels.
SOLUTION
a =
¢v
30 - 15
=
= 3.75 ft>s2
¢t
4
+ ©F = ma ;
;
x
x
F - 325 = ¢
16(103)
≤ (3.75)
32.2
F = 2188.35 lb
2188.35(30)
= 119 hp
550
Ans.
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This
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laws
or
Pmax = F # vmax =
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
14–61.
If the jet on the dragster supplies a constant thrust of
T = 20 kN, determine the power generated by the jet as a
function of time. Neglect drag and rolling resistance, and
the loss of fuel. The dragster has a mass of 1 Mg and starts
from rest.
T
SOLUTION
Equations of Motion: By referring to the free-body diagram of the dragster shown
in Fig. a,
+ ©F = ma ;
:
x
x
20(103) = 1000(a)
a = 20 m>s2
Kinematics: The velocity of the dragster can be determined from
+ b
a:
v = v0 + ac t
v = 0 + 20 t = (20 t) m>s
Power:
laws
or
P = F # v = 20(103)(20 t)
teaching
Web)
= C 400(103)t D W
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Dissemination
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Ans.
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
14–62.
An athlete pushes against an exercise machine with a force
that varies with time as shown in the first graph. Also, the
velocity of the athlete’s arm acting in the same direction as
the force varies with time as shown in the second graph.
Determine the power applied as a function of time and the
work done in t = 0.3 s.
F (N)
800
t (s)
0.2
0.3
SOLUTION
v (m/s)
For 0 … t … 0.2
F = 800 N
v =
20
t = 66.67t
0.3
P =
F#v
20
t (s)
= 53.3 t kW
0.3
Ans.
For 0.2 … t … 0.3
F = 2400 - 8000 t
laws
or
v = 66.67t
P = F # v = 1160 t - 533t22 kW
in
teaching
Web)
Ans.
Wide
World
permitted.
Dissemination
P dt
instructors
L0
States
u =
copyright
0.3
of
the
not
0.3
2
United
on
learning.
is
1160t - 533t 2 dt
and
use
L0.2
the
L0
53.3t dt +
by
0.2
u =
the
(including
for
work
student
53.3
160
533
(0.2)2 +
[(0.3)2 - (0.2)2] [(0.3)3 - (0.2)3]
2
2
3
is
work
solely
of
protected
=
Ans.
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This
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assessing
= 1.69 kJ
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
14–63.
An athlete pushes against an exercise machine with a
force that varies with time as shown in the first graph.
Also, the velocity of the athlete’s arm acting in the same
direction as the force varies with time as shown in the
second graph. Determine the maximum power developed
during the 0.3-second time period.
F (N)
800
t (s)
0.2
0.3
SOLUTION
v (m/s)
See solution to Prob. 14–62.
P = 160 t - 533 t2
20
dP
= 160 - 1066.6 t = 0
dt
t (s)
0.3
t = 0.15 s 6 0.2 s
Thus maximum occurs at t = 0.2 s
Pmax = 53.3(0.2) = 10.7 kW
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This
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Dissemination
Wide
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Web)
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or
Ans.
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*14–64.
The 500-kg elevator starts from rest and travels upward
with a constant acceleration a c = 2 m>s2. Determine the
power output of the motor M when t = 3 s. Neglect the
mass of the pulleys and cable.
M
SOLUTION
+ c ©Fy = m ay ;
3T - 500(9.81) = 500(2)
E
T = 1968.33 N
3sE - sP = l
3 vE = vP
When t = 3 s,
(+ c ) v0 + ac t
laws
or
vE = 0 + 2(3) = 6 m>s
in
teaching
Web)
vP = 3(6) = 18 m>s
permitted.
Dissemination
Wide
copyright
PO = 1968.33(18)
Ans.
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PO = 35.4 kW
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
14–65.
The 50-lb block rests on the rough surface for which
the coefficient of kinetic friction is m k = 0.2. A force
F = (40 + s 2) lb, where s is in ft, acts on the block in the
direction shown. If the spring is originally unstretched
(s = 0) and the block is at rest, determine the power
developed by the force the instant the block has moved
s = 1.5 ft.
F
k
30
SOLUTION
+ c ©Fy = 0;
NB - A 40 + s2 B sin 30° - 50 = 0
NB = 70 + 0.5s2
T1 + ©U1 - 2 + T2
1.5
0 +
L0
A 40 + s2 B cos 30° ds -
1
2
(20)(1.5) - 0.2
2
L0
1.5
1 50
2
a
bv
2 32.2 2
A 70 + 0.5s2 B ds =
0 + 52.936 - 22.5 - 21.1125 = 0.7764 v22
v2 = 3.465 ft>s
laws
or
When s = 1.5 ft,
in
teaching
Web)
F = 40 + (1.5)2 = 42.25 lb
Dissemination
Wide
copyright
P = F # v = (42.25 cos 30°)(3.465)
permitted.
P = 126.79 ft # lb>s = 0.231 hp
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This
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the
(including
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is
of
the
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
not
instructors
States
World
Ans.
20 lb/ft
14–66.
The girl has a mass of 40 kg and center of mass at G. If she
is swinging to a maximum height defined by u = 60°,
determine the force developed along each of the four
supporting posts such as AB at the instant u = 0°. The
swing is centrally located between the posts.
A
2m
␪
G
SOLUTION
The maximum tension in the cable occurs when u = 0°.
B
T1 + V1 = T2 + V2
0 + 40(9.81)(- 2 cos 60°) =
1
(40)v2 + 40(9.81)(- 2)
2
v = 4.429 m>s
4.4292
b
2
+ c ©Fy = 0;
2(2F) cos 30° - 784.8 = 0
T = 784.8 N
F = 227 N
Ans.
or
T - 40(9.81) = (40)a
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This
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the
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laws
+ c ©Fn = ma n;
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
30⬚
30⬚
14–67.
Two equal-length springs are “nested” together in order to
form a shock absorber. If it is designed to arrest the motion
of a 2-kg mass that is dropped s = 0.5 m above the top of
the springs from an at-rest position, and the maximum
compression of the springs is to be 0.2 m, determine the
required stiffness of the inner spring, kB, if the outer spring
has a stiffness kA = 400 N>m.
s
A
SOLUTION
B
T1 + V1 = T2 + V2
1
1
(400)(0.2)2 + (kB)(0.2)2
2
2
0 + 0 = 0 - 2(9.81)(0.5 + 0.2) +
kB = 287 N>m
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This
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Dissemination
Wide
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Web)
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Ans.
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*14–68.
The collar has a weight of 8 lb. If it is pushed down so as to
compress the spring 2 ft and then released from rest
1h = 02, determine its speed when it is displaced h = 4.5 ft.
The spring is not attached to the collar. Neglect friction.
h
SOLUTION
T1 + V1 = T2 + V2
0 +
k = 30 lb/ft
8
1
1
(30)(2)2 = a
b v22 + 8(4.5)
2
2 32.2
v2 = 13.9 ft s
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This
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the
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the
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United
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Dissemination
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or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
14–69.
The collar has a weight of 8 lb. If it is released from rest at
a height of h = 2 ft from the top of the uncompressed
spring, determine the speed of the collar after it falls and
compresses the spring 0.3 ft.
h
SOLUTION
T1 + V1 = T2 + V2
0 + 0 =
k = 30 lb/ft
8
1
1
a
b v22 - 8(2.3) + (30)(0.3)2
2 32.2
2
v2 = 11.7 ft s
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This
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the
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student
the
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Dissemination
Wide
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Web)
laws
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Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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14–70.
The 2-kg ball of negligible size is fired from point A with
an initial velocity of 10 m/s up the smooth inclined plane.
Determine the distance from point C to where it hits the
horizontal surface at D. Also, what is its velocity when it
strikes the surface?
B
10 m/s
1.5 m
A
C
SOLUTION
2m
Datum at A:
TA + VA = TB + VB
1
1
(2)(10)2 + 0 = (2)(nB)2 + 2(9.81)(1.5)
2
2
vB = 8.401 m>s
+ B s = s + v t
A:
0
0
or
4
d = 0 + 8.401 a b t
5
Web)
laws
1 2
at
2 c
teaching
s = s0 + v0 t +
Wide
copyright
in
(+ c )
not
instructors
States
World
permitted.
Dissemination
1
3
-1.5 = 0 + 8.401 a b t + ( - 9.81)t2
5
2
by
and
use
United
on
learning.
is
of
the
-4.905t2 + 5.040t + 1.5 = 0
for
work
student
the
Solving for the positive root,
is
work
solely
of
protected
the
(including
t = 1.269 s
assessing
4
d = 8.401a b (1.269) = 8.53 m
5
any
will
their
sale
TA + VA = TD + VD
destroy
of
and
Datum at A:
courses
part
the
is
This
provided
integrity
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and
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this
Ans.
1
1
(2)(10)2 + 0 = (2)(nD)2 + 0
2
2
vD = 10 m s
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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Ans.
D
d
14–71.
y
The ride at an amusement park consists of a gondola which
is lifted to a height of 120 ft at A. If it is released from rest
and falls along the parabolic track, determine the speed at
the instant y = 20 ft. Also determine the normal reaction of
the tracks on the gondola at this instant. The gondola and
passenger have a total weight of 500 lb. Neglect the effects
of friction and the mass of the wheels.
A
1 x2
y ⫽ —–
260
SOLUTION
y =
y ⫽ 20 ft
1 2
x
260
dy
1
=
x
dx
130
d2y
1
=
dx2
130
At y = 120 - 100 = 20 ft
x = 72.11 ft
or
u = 29.02°
teaching
Web)
dy
= 0.555 ,
dx
laws
tan u =
in
[1 + (0.555)2]3>2
= 194.40 ft
1
130
States
World
permitted.
Dissemination
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r =
not
instructors
500
v2
a
b
32.2 194.40
the
(1)
on
learning.
is
of
NG - 500 cos 29.02° =
by
and
use
United
+ag Fn = man ;
(including
for
work
student
the
T1 + V1 = T2 + V2
work
solely
of
protected
the
1 500 2
a
b v - 500(100)
2 32.2
this
assessing
is
0 + 0 =
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the
is
This
provided
integrity
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and
work
v2 = 6440
Ans.
will
sale
Substituting into Eq. (1) yields
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v = 80.2 ft>s
NG = 952 lb
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or transmission in any form or by any means, electronic, mechanical,
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Ans.
120 ft
x
*14–72.
The 2-kg collar is attached to a spring that has an
unstretched length of 3 m. If the collar is drawn to point B
and released from rest, determine its speed when it arrives
at point A.
k = 3 N/m
3m
B
A
SOLUTION
4m
Potential Energy: The initial and final elastic potential energy are
1
1
(3) A 232 + 42 - 3 B 2 = 6.00 J and (3)(3 - 3)2 = 0, respectively.The gravitational
2
2
potential energy remains the same since the elevation of collar does not change when it
moves from B to A.
Conservation of Energy:
TB + VB = TA + VA
1
(2) v2A + 0
2
vA = 2.45 m s
Ans.
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This
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(including
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and
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Web)
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or
0 + 6.00 =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
14–73.
The 2-kg collar is attached to a spring that has an unstretched
length of 2 m. If the collar is drawn to point B and released
from rest, determine its speed when it arrives at point A.
k ⫽ 3 N/m
3m
B
A
SOLUTION
4m
Potential Energy: The stretches of the spring when the collar is at B and A are
sB = 232 + 4 2 - 2 = 3 m and sA = 3 - 2 = 1 m, respectively. Thus, the elastic
potential energy of the system at B and A are
(Ve)B =
(Ve)A =
1
1
ks 2 = (3)(32) = 13.5 J
2 B
2
1
1
ks 2 = (3)(12) = 1.5 J
2 A
2
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laws
or
There is no change in gravitational potential energy since the elevation of the collar
does not change during the motion.
Wide
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Conservation of Energy:
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World
permitted.
Dissemination
TB + VB = TA + VA
by
and
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on
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is
of
the
not
instructors
1
1
mvB2 + (Ve)B = mvA2 + (Ve)A
2
2
work
student
the
1
(2)vA2 + 1.5
2
of
protected
the
(including
for
0 + 13.5 =
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solely
vA = 3.46 m>s
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Ans.
14–74.
The 0.5-lb ball is shot from the spring device shown. The
spring has a stiffness k = 10 lb>in. and the four cords C and
plate P keep the spring compressed 2 in. when no load is on
the plate. The plate is pushed back 3 in. from its initial
position. If it is then released from rest, determine the speed
of the ball when it travels 30 in. up the smooth plane.
s
k
C
P
30
SOLUTION
Potential Energy: The datum is set at the lowest point (compressed position).
30
Finally, the ball is
sin 30° = 1.25 ft above the datum and its gravitational
12
potential energy is 0.5(1.25) = 0.625 ft # lb. The initial and final elastic potential
2 + 3 2
2 2
1
1
(120)a
b = 10.42 ft # lb and
(120)a b = 1.667 ft # lb,
energy are
2
12
2
12
respectively.
Conservation of Energy:
laws
or
©T1 + ©V1 = ©T2 + ©V2
in
teaching
Web)
1 0.5
a
b y2 + 0.625 + 1.667
2 32.2
Wide
copyright
0 + 10.42 =
Ans.
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y = 32.3 ft>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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14–75.
The 0.5-lb ball is shot from the spring device shown. Determine
the smallest stiffness k which is required to shoot the ball a
maximum distance of 30 in. up the smooth plane after the
spring is pushed back 3 in. and the ball is released from rest. The
four cords C and plate P keep the spring compressed 2 in. when
no load is on the plate.
s
k
C
P
30
SOLUTION
Potential Energy: The datum is set at the lowest point (compressed position).
30
sin 30° = 1.25 ft above the datum and its gravitational
Finally, the ball is
12
potential energy is 0.5(1.25) = 0.625 ft # lb. The initial and final elastic potential
2 + 3 2
2 2
1
1
b = 0.08681k and (k) a b = 0.01389k, respectively.
energy are (k) a
2
12
2
12
Conservation of Energy:
©T1 + ©V1 = ©T2 + ©V2
laws
or
0 + 0.08681k = 0 + 0.625 + 0.01389k
k = 8.57 lb>ft
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This
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Ans.
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or transmission in any form or by any means, electronic, mechanical,
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*14–76.
The roller coaster car having a mass m is released from rest
at point A. If the track is to be designed so that the car does
not leave it at B, determine the required height h. Also, find
the speed of the car when it reaches point C. Neglect
friction.
A
B
7.5 m
20 m
C
SOLUTION
Equation of Motion: Since it is required that the roller coaster car is about to leave
vB 2
vB 2
=
the track at B, NB = 0. Here, an =
. By referring to the free-body
rB
7.5
diagram of the roller coaster car shown in Fig. a,
©Fn = ma n;
m(9.81) = m ¢
vB 2
≤
7.5
vB 2 = 73.575 m2>s2
or
Potential Energy: With reference to the datum set in Fig. b, the gravitational
potential energy of the rollercoaster car at positions A, B, and C are
A Vg B A = mghA = m(9.81)h = 9.81mh, A Vg B B = mghB = m(9.81)(20) = 196.2 m,
teaching
Web)
laws
and A Vg B C = mghC = m(9.81)(0) = 0.
permitted.
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Conservation of Energy: Using the result of vB 2 and considering the motion of the
car from position A to B,
learning.
is
of
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World
TA + VA = TB + VB
for
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student
the
by
and
use
United
on
1
1
mvA 2 + A Vg B A = mvB 2 + A Vg B B
2
2
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(including
1
m(73.575) + 196.2m
2
this
assessing
is
0 + 9.81mh =
Ans.
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the
is
This
provided
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and
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h = 23.75 m
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TB + VB = TC + VC
of
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Also, considering the motion of the car from position B to C,
1
1
mvB 2 + A Vg B B = mvC 2 + A Vg B C
2
2
1
1
m(73.575) + 196.2m = mvC 2 + 0
2
2
vC = 21.6 m>s
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Ans.
h
14–77.
A 750-mm-long spring is compressed and confined by the
plate P, which can slide freely along the vertical 600-mm-long
rods. The 40-kg block is given a speed of v = 5 m>s when it is
h = 2 m above the plate. Determine how far the plate moves
downwards when the block momentarily stops after striking
it. Neglect the mass of the plate.
v ⫽ 5 m/s
A
h⫽2m
SOLUTION
Potential Energy: With reference to the datum set in Fig. a, the gravitational potential
energy of the block at positions (1) and (2) are A Vg B 1 = mgh1 = 40(9.81)(0) = 0
P
and A Vg B 2 = mgh2 = 40(9.81) C -(2 + y) D = C -392.4(2 + y) D , respectively. The
compression of the spring when the block is at positions (1) and (2) are
s1 = (0.75 - 0.6) = 0.15 m and s2 = s1 + y = (0.15 + y) m. Thus, the initial and
final elastic potential energy of the spring are
k ⫽ 25 kN/m
A Ve B 1 = ks21 = (25)(103)(0.152) = 281.25 J
1
2
1
2
A Ve B 2 = ks22 = (25)(103)(0.15 + y)2
1
2
laws
or
1
2
in
teaching
Web)
Conservation of Energy:
Dissemination
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T1 + V1 = T2 + V2
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the
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student
the
by
1
(40)(52) + (0 + 281.25) = 0 + [ - 392.4(2 + y)] +
2
1
(25)(103)(0.15 + y)2
2
12500y2 + 3357.6y - 1284.8 = 0
and
use
United
on
learning.
is
of
the
not
instructors
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World
permitted.
1
1
mv21 + c A Vg B 1 + A Ve B 1 d = mv22 + c A Vg B 2 + A Ve B 2 d
2
2
part
the
is
This
Solving for the positive root of the above equation,
any
courses
Ans.
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and
y = 0.2133 m = 213 mm
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or transmission in any form or by any means, electronic, mechanical,
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600 mm
14–78.
The 2-lb block is given an initial velocity of 20 ft/s when it is
at A. If the spring has an unstretched length of 2 ft and a
stiffness of k = 100 lb>ft, determine the velocity of the block
when s = 1 ft.
2 ft
B
vA = 20 ft/s
A
s
SOLUTION
k = 100 lb/ft
Potential Energy: Datum is set along AB. The collar is 1 ft below the datum when it
is at C. Thus, its gravitational potential energy at this point is -2(1) = -2.00 ft # lb.
1
(100)(2 - 2)2 = 0 and
The initial and final elastic potential energy are
2
1
(100) A 222 + 12 - 2 B 2 = 2.786 ft # lb, respectively.
2
C
Conservation of Energy:
TA + VA = TC + VC
2
2
1
1
a
b A 202 B + 0 = a
b v2C + 2.786 + ( -2.00)
2 32.2
2 32.2
vC = 19.4 ft s
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14–79.
The block has a weight of 1.5 lb and slides along the smooth
chute AB. It is released from rest at A, which has coordinates
of A(5 ft, 0, 10 ft). Determine the speed at which it slides off
at B, which has coordinates of B(0, 8 ft, 0).
z
5 ft
A
SOLUTION
10 ft
Datum at B:
B
8 ft
TA + VA = TB + VB
0 + 1.5(10) =
x
1 1.5
a
b (vB)2 + 0
2 32.2
vB = 25.4 ft s
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This
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y
*14–80.
Each of the two elastic rubber bands of the slingshot has
an unstretched length of 200 mm. If they are pulled back
to the position shown and released from rest, determine
the speed of the 25-g pellet just after the rubber bands
become unstretched. Neglect the mass of the rubber
bands. Each rubber band has a stiffness of k = 50 N>m.
50 mm
50 mm
240 mm
SOLUTION
T1 + V1 = T2 + V2
1
1
0 + (2) a b(50)[2(0.05)2 + (0.240)2 - 0.2]2 = (0.025)v2
2
2
v = 2.86 m>s
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This
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Ans.
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or transmission in any form or by any means, electronic, mechanical,
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14–81.
Each of the two elastic rubber bands of the slingshot has an
unstretched length of 200 mm. If they are pulled back to the
position shown and released from rest, determine the
maximum height the 25-g pellet will reach if it is fired
vertically upward. Neglect the mass of the rubber bands and
the change in elevation of the pellet while it is constrained
by the rubber bands. Each rubber band has a stiffness
k = 50 N>m.
50 mm
50 mm
240 mm
SOLUTION
T1 + V1 = T2 + V2
1
0 + 2a b (50)[2(0.05)2 + (0.240)2 - 0.2]2 = 0 + 0.025(9.81)h
2
h = 0.416 m = 416 mm
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This
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14–82.
If the mass of the earth is Me, show that the gravitational
potential energy of a body of mass m located a distance r
from the center of the earth is Vg = −GMem>r. Recall that
the gravitational force acting between the earth and the
body is F = G(Mem>r 2), Eq. 13–1. For the calculation, locate
the datum at r : q. Also, prove that F is a conservative
force.
r2
r
SOLUTION
r1
The work is computed by moving F from position r1 to a farther position r2.
Vg = - U = -
L
F dr
r2
= -G Me m
Lr1
= - G Me ma
dr
r2
1
1
- b
r2
r1
or
As r1 : q , let r2 = r1, F2 = F1, then
teaching
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-G Me m
r
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Vg :
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To be conservative, require
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G Me m
0
b
ar
0r
and
use
F = - § Vg = -
work
student
the
by
-G Me m
the
(including
for
r2
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=
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Q.E.D.
14–83.
A rocket of mass m is fired vertically from the surface of the
earth, i.e., at r = r1 . Assuming no mass is lost as it travels
upward, determine the work it must do against gravity to
reach a distance r2 . The force of gravity is F = GMem>r2
(Eq. 13–1), where Me is the mass of the earth and r the
distance between the rocket and the center of the earth.
r2
r
SOLUTION
F = G
r1
Mem
r2
r2
F1-2 =
L
F dr = GMem
2
Lr1 r
1
1
- b
r1
r2
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= GMema
dr
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*14–84.
The firing mechanism of a pinball machine consists of a
plunger P having a mass of 0.25 kg and a spring of stiffness
k = 300 N>m. When s = 0, the spring is compressed
50 mm. If the arm is pulled back such that s = 100 mm and
released, determine the speed of the 0.3-kg pinball B just
before the plunger strikes the stop, i.e., s = 0. Assume all
surfaces of contact to be smooth. The ball moves in the
horizontal plane. Neglect friction, the mass of the spring,
and the rolling motion of the ball.
s
B
P
k
SOLUTION
T1 + V1 = T2 + V2
0 +
1
1
1
1
(300)(0.1 + 0.05)2 = (0.25)(n2)2 + (0.3)(n2)2 + (300)(0.05)2
2
2
2
2
n2 = 3.30 m s
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solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
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World
permitted.
Dissemination
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or
Ans.
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300 N/m
14–85.
A 60-kg satellite travels in free flight along an elliptical orbit
such that at A, where rA = 20 Mm, it has a speed vA = 40 Mm>h.
What is the speed of the satellite when it reaches point B, where
rB = 80 Mm? Hint: See Prob. 14–82, where Me = 5.976(1024) kg
and G = 66.73(10-12) m3>(kg # s2).
B
vB
rB
80 Mm
rA
SOLUTION
yA = 40 Mm>h = 11 111.1 m>s
Since V = -
GMe m
r
T1 + V1 = T2 + V2
66.73(10) - 12(5.976)(10)23(60)
66.73(10) - 12(5.976)(10)24(60)
1
1
2
(60)(11 111.1)2 (60)v
=
B
2
2
20(10)6
80(10)6
vB = 9672 m>s = 34.8 Mm>h
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This
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20 Mm
vA
A
14–86.
Just for fun, two 150-lb engineering students A and B intend
to jump off the bridge from rest using an elastic cord
(bungee cord) having a stiffness k = 80 lb>ft. They wish to
just reach the surface of the river, when A, attached to the
cord, lets go of B at the instant they touch the water.
Determine the proper unstretched length of the cord to do
the stunt, and calculate the maximum acceleration of
student A and the maximum height he reaches above the
water after the rebound. From your results, comment on the
feasibility of doing this stunt.
120 ft
SOLUTION
T1 + V1 = T2 + V2
0 + 2(150)(120) = 0 +
1
(80)(x)2
2
x = 30 ft
Unstretched length of cord.
or
120 = l + 30
l = 90 ft
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Ans.
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When A lets go of B.
not
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T2 + V2 = T3 + V3
and
use
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is
of
the
1
(80)(30)2 = 0 + (150) h
2
the
by
0 +
the
(including
for
work
student
h = 240 ft
part
the
is
any
courses
their
destroy
of
T2 + V2 = T3 + V3
and
Thus, h 7 120 + 90 = 210 ft
This
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This is not possible since the 90 ft cord would have to stretch again, i.e., hmax = 120 +
90 = 210 ft.
will
1
1
(80)(30)2 = 0 + 150 h + (80)[(h - 120) - 90]2
2
2
sale
0 +
36 000 = 150 h + 40(h2 - 420 h + 44 100)
h2 - 416.25 h + 43 200 = 0
Choosing the root 7 210 ft
h = 219 ft
+ c ©Fy = may;
Ans.
800(30) - 150 =
a = 483 ft>s2
150
a
32.2
Ans.
It would not be a good idea to perform the stunt since a = 15 g which is excessive
and A rises 219¿ - 120¿ = 99 ft above the bridge!
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A
B
14–87.
z
The 20-lb collar slides along the smooth rod. If the collar is
released from rest at A, determine its speed when it passes
point B. The spring has an unstretched length of 3 ft.
A
6 ft
k ⫽ 20 lb/ft
SOLUTION
3 ft
rOA = {-2i + 3j + 6k} ft,
rOA = 7 ft
rOB = {4i} ft,
rOB = 4 ft
O
2 ft
B
Put datum at x–y plane
x
TA + VA = TB + VB
0 + (20 lb)(6 ft) +
1
1 20
(20 lb>ft)(7 ft - 3 ft)2 = a
b v2B + 0 +
2
2 32.2
laws
or
1
(20 lb>ft)(4 ft - 3 ft)2
2
vB = 29.5 ft>s
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This
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4 ft
y
*14–88.
Two equal-length springs having a stiffness kA = 300 N>m
and kB = 200 N>m are “nested” together in order to form
a shock absorber. If a 2-kg block is dropped from an at-rest
position 0.6 m above the top of the springs, determine
their deformation when the block momentarily stops.
0.6 m
B
SOLUTION
Datum at initial position:
T1 + V1 = T2 + V2
0 + 0 = 0 - 2(9.81)(0.6 + x) +
1
(300 + 200)(x)2
2
250x2 - 19.62x - 11.772 = 0
Solving for the positive root,
x = 0.260 m
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This
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the
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work
student
the
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Wide
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Web)
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or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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A
14–89.
When the 6-kg box reaches point A it has a speed of
vA = 2 m>s. Determine the angle u at which it leaves the
smooth circular ramp and the distance s to where it falls
into the cart. Neglect friction.
vA = 2 m/s
20°
A
θ
B
1.2 m
SOLUTION
s
At point B:
+b©Fn = man;
6(9.81) cos f = 6 a
n2B
b
1.2
(1)
Datum at bottom of curve:
TA + VA = TB + VB
1
1
(6)(2)2 + 6(9.81)(1.2 cos 20°) = (6)(vB)2 + 6(9.81)(1.2 cos f)
2
2
13.062 = 0.5v2B + 11.772 cos f
laws
or
(2)
in
teaching
Web)
Substitute Eq. (1) into Eq. (2), and solving for vB,
Dissemination
Wide
copyright
vB = 2.951 m>s
World
permitted.
(2.951)2
b = 42.29°
1.2(9.81)
the
not
instructors
States
United
on
learning.
is
f = cos - 1 a
of
Thus,
Ans.
work
s = s0 + v0 t + 12 ac t2
of
protected
the
(including
for
A+cB
student
the
by
and
use
u = f - 20° = 22.3°
solely
1
( -9.81)t2
2
integrity
of
and
work
this
assessing
is
work
-1.2 cos 42.29° = 0 - 2.951(sin 42.29°)t +
the
part
any
courses
will
s = s0 + v0 t
sale
+ b
a:
their
destroy
of
and
Solving for the positive root: t = 0.2687 s
is
This
provided
4.905t2 + 1.9857t - 0.8877 = 0
s = 0 + (2.951 cos 42.29°)(0.2687)
s = 0.587 m
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Ans.
14–90.
The Raptor is an outside loop roller coaster in which riders
are belted into seats resembling ski-lift chairs. Determine
the minimum speed v0 at which the cars should coast down
from the top of the hill, so that passengers can just make the
loop without leaving contact with their seats. Neglect
friction, the size of the car and passenger, and assume each
passenger and car has a mass m.
v0
r
h
SOLUTION
Datum at ground:
T1 + V1 = T2 + V2
1 2
1
mv + mgh = mv21 + mg2r
2 0
2
v1 = 2v20 + 2g(h-2r)
+ T ©Fn = man;
mg = ma
v21
b
r
laws
or
v1 = 2gr
teaching
Web)
Thus,
Wide
copyright
Ans.
the
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is
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and
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student
the
the
(including
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this
assessing
is
integrity
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and
work
provided
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courses
part
the
is
This
destroy
of
and
will
their
sale
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permitted.
Dissemination
g(5r - 2h)
World
v0 =
in
gr = v20 + 2gh - 4gr
14–91.
The Raptor is an outside loop roller coaster in which riders
are belted into seats resembling ski-lift chairs. If the cars
travel at v0 = 4 m>s when they are at the top of the hill,
determine their speed when they are at the top of the loop
and the reaction of the 70-kg passenger on his seat at this
instant. The car has a mass of 50 kg. Take h = 12 m, r = 5 m.
Neglect friction and the size of the car and passenger.
v0
r
h
SOLUTION
Datum at ground:
T1 + V1 = T2 + V2
1
1
(120)(4)2 + 120(9.81)(12) = (120)(v1)2 + 120(9.81)(10)
2
2
v1 = 7.432 m>s
+ T ©Fn = man;
Ans.
(7.432)2
b
5
70(9.81) + N = 70 a
N = 86.7 N
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This
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the
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work
student
the
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permitted.
Dissemination
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Ans.
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*14–92.
The 75-kg man bungee jumps off the bridge at A with an
initial downward speed of 1.5 m>s. Determine the required
unstretched length of the elastic cord to which he is
attached in order that he stops momentarily just above the
surface of the water. The stiffness of the elastic cord is
k = 80 N>m. Neglect the size of the man.
A
150 m
SOLUTION
B
Potential Energy: With reference to the datum set at the surface of the water, the
gravitational potential energy of the man at positions A and B are A Vg B A = mghA =
75(9.81)(150) = 110362.5 J and A Vg B B = mghB = 75(9.81)(0) = 0. When the man
is at position A, the elastic cord is unstretched (sA = 0), whereas the elastic cord
stretches sB = A 150 - l0 B m, where l0 is the unstretched length of the cord.Thus, the
elastic potential energy of the elastic cord when the man is at these two positions are
1
1
1
A Ve B A = ksA 2 = 0 and A Ve B B = ksB 2 = (80)(150 - l0)2 = 40(150 - l0)2.
2
2
2
Conservation of Energy:
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or
TA + VA = TB + VB
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is
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the
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States
1
(75)(1.52) + A 110362.5 + 0 B = 0 + C 0 + 40(150 - l0)2 D
2
permitted.
Dissemination
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1
1
mvA 2 + B a Vg b + A Ve B A R = mvB 2 + B a Vg b + A Ve B B R
2
2
A
B
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the
(including
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work
student
the
by
and
l0 = 97.5 m
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Ans.
14–93.
The 10-kg sphere C is released from rest when u = 0° and
the tension in the spring is 100 N. Determine the speed of
the sphere at the instant u = 90°. Neglect the mass of rod
AB and the size of the sphere.
E
0.4 m
SOLUTION
A
u
Potential Energy: With reference to the datum set in Fig. a, the gravitational potential
energy of the sphere at positions (1) and (2) are A Vg B 1 = mgh1 = 10(9.81)(0.45) =
D
44.145 J and A Vg B 2 = mgh2 = 10(9.81)(0) = 0. When the sphere is at position (1),
100
the spring stretches s1 =
= 0.2 m. Thus, the unstretched length of the spring is
500
or
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1
1
m A v B 2 + c A Vg B 1 + A Ve B 1 d = ms A vs B 2 2 + c A Vg B 2 + A Ve B 2 d
2 s s 1
2
the
(including
for
work
student
the
by
and
1
(10)(vs)22 + (0 + 40)
2
sale
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This
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(vs)2 = 1.68 m>s
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Ans.
C
0.15 m
Conservation of Energy:
T1 + V1 = T2 + V2
B
0.3 m
l0 = 30.32 + 0.42 - 0.2 = 0.3 m, and the elastic potential energy of the spring is
1
1
A Ve B 1 = ks12 = (500)(0.22) = 10 J. When the sphere is at position (2), the spring
2
2
stretches s2 = 0.7 - 0.3 = 0.4 m, and the elastic potential energy of the spring is
1
1
A Ve B 2 = ks22 = (500)(0.42) = 40 J.
2
2
0 + (44.145 + 10) =
k ⫽ 500 N/m
14–94.
v
The double-spring bumper is used to stop the 1500-lb steel
billet in the rolling mill. Determine the maximum displacement
of the plate A if the billet strikes the plate with a speed of 8 ft>s.
Neglect the mass of the springs, rollers and the plates A and B.
Take k1 = 3000 lb>ft, k2 = 45 000 lb>ft.
A
SOLUTION
T1 + V1 = T2 + V2
1
1
2
1 1500
a
b (8)2 + 0 = 0 + (3000)s2L + (4500)s
2 32.2
2
2
2
(1)
Fs = 3000s1 = 4500s2 ;
(2)
s1 = 1.5s2
Solving Eqs. (1) and (2) yields:
s1 = 0.7722 ft
sA = s1 + s2 = 0.7722 + 0.5148 = 1.29 ft
Ans.
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This
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s2 = 0.5148 ft
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8 ft/s
B
k1
k2
14–95.
y
The 2-lb box has a velocity of 5 ft/s when it begins to slide
down the smooth inclined surface at A. Determine the point
C (x, y) where it strikes the lower incline.
5 ft/s
A
15 ft
5
3
4
B
SOLUTION
30 ft
Datum at A:
1
C
TA + VA = TB + VB
2
y
x
2
2
1
1
a
b (5)2 + 0 = a
b v2B - 2(15)
2 32.2
2 32.2
x
vB = 31.48 ft>s
+ B
A:
s = s0 + v0t
4
x = 0 + 31.48 a b t
5
or
(1)
Web)
laws
1 2
at
2 c
teaching
s = s0 + v0t +
Wide
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(+ c )
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permitted.
Dissemination
3
1
y = 30 - 31.48a b t + (- 32.2)t2
5
2
by
and
use
United
on
learning.
Equation of inclined surface:
work
student
the
1
x
2
(2)
the
(including
for
y =
work
solely
of
protected
y
1
= ;
x
2
work
this
assessing
is
Thus,
any
will
sale
Solving for the positive root,
their
destroy
of
and
- 16.1t2 - 31.480t + 30 = 0
courses
part
the
is
This
provided
integrity
of
and
30 - 18.888t - 16.1t2 = 12.592t
t = 0.7014 s
From Eqs. (1) and (2):
4
x = 31.48 a b (0.7014) = 17.66 = 17.7 ft
5
1
(17.664) = 8.832 = 8.83 ft
2
y =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Ans.
Ans.
*14–96.
y
The 2-lb box has a velocity of 5 ft/s when it begins to slide
down the smooth inclined surface at A. Determine its speed
just before hitting the surface at C and the time to travel
from A to C. The coordinates of point C are x = 17.66 ft,
and y = 8.832 ft.
5 ft/s
A
15 ft
5
3
4
B
SOLUTION
30 ft
Datum at A:
1
C
TA + VA = TC + VC
2
y
x
2
1 2
1
a
b (5)2 + 0 = a
b (vC)2 - 2[15 + (30 - 8.832)]
2 32.2
2 32.2
x
vC = 48.5 ft>s
Ans.
2
3
ba
2a b = a
5
32.2 x¿
+R©Fx¿ = max¿;
or
ax¿ = 19.32 ft>s2
teaching
Web)
laws
TA + VA = TB + VB
permitted.
Dissemination
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in
1 2
2
1
a
b (5)2 + 0 = a
b v2B - 2(15)
2 32.2
2 32.2
is
of
the
not
instructors
States
World
vB = 31.48 ft>s
and
use
United
on
learning.
vB = vA + act
by
(+R)
(including
for
work
student
the
31.48 = 5 + 19.32tAB
is
work
solely
of
protected
the
tAB = 1.371 s
this
assessing
s = s0 + n0t
part
the
is
any
courses
(1)
1 2
at
2 c
will
s = s0 + v0 t +
sale
(+ c )
their
destroy
of
and
4
x = 0 + 31.48 a b t
5
This
provided
integrity
of
and
work
+ B
A:
1
3
y = 30 - 31.48 a b t + (- 32.2)t2
5
2
Equation of inclined surface:
y
1
= ;
x
2
y =
1
x
2
(2)
Thus
30 - 18.888t - 16.1t2 = 12.592t
- 16.1t2 - 31.480t + 30 = 0
Solving for the positive root:
t = 0.7014 s
Total time is
t = 1.371 + 0.7014 = 2.07 s
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Ans.
14–97.
A pan of negligible mass is attached to two identical springs of
stiffness k = 250 N>m. If a 10-kg box is dropped from a height
of 0.5 m above the pan, determine the maximum vertical
displacement d. Initially each spring has a tension of 50 N.
1m
k
250 N/m
1m
k
250 N/m
0.5 m
d
SOLUTION
Potential Energy: With reference to the datum set in Fig. a, the gravitational potential
energy of the box at positions (1) and (2) are A Vg B 1 = mgh1 = 10(9.81)(0) = 0 and
A Vg B 2 = mgh2 = 10(9.81) C - A 0.5 + d B D = - 98.1 A 0.5 + d B . Initially, the spring
50
= 0.2 m. Thus, the unstretched length of the spring
250
is l0 = 1 - 0.2 = 0.8 m and the initial elastic potential of each spring is
1
A Ve B 1 = (2) ks1 2 = 2(250 > 2)(0.22) = 10 J . When the box is at position (2), the
2
stretches s1 =
spring stretches s2 = a 2d2 + 12 - 0.8 b m. The elastic potential energy of the
in
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or
springs when the box is at this position is
2
1
2
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A Ve B 2 = (2) ks2 2 = 2(250 > 2) c 2d2 + 1 - 0.8 d = 250a d2 - 1.62d2 + 1 + 1.64 b .
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Conservation of Energy:
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T1 + V1 + T2 + V2
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1
1
mv1 2 + B a Vg b + A Ve B 1 R = mv2 2 + B aVg b + A Ve B 2 R
2
2
1
2
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0 + A 0 + 10 B = 0 + B - 98.1 A 0.5 + d B + 250 ¢ d2 - 1.6 2d2 + 1 + 1.64 ≤ R
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Solving the above equation by trial and error,
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This
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250d2 - 98.1d - 400 2d2 + 1 + 350.95 = 0
Ans.
15–1.
A 2-lb ball is thrown in the direction shown with an initial
speed vA = 18 ft>s. Determine the time needed for it to
reach its highest point B and the speed at which it is
traveling at B. Use the principle of impulse and momentum
for the solution.
B
vA
A
SOLUTION
1 + c2
m1vy21 + ©
L
F dt = m1vy22
2
118 sin 30°2 - 21t2 = 0
32.2
t = 0.2795 = 0.280 s
+ B
A:
m1vx21 + ©
L
Ans.
Fx dt = m1vx22
laws
or
2
2
(v )
(18 cos 30°) + 0 =
32.2
32.2 B
vB = 15.588 = 15.6 ft>s
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18 ft/s
30
15–2.
A 20-lb block slides down a 30° inclined plane with an
initial velocity of 2 ft>s. Determine the velocity of the block
in 3 s if the coefficient of kinetic friction between the block
and the plane is mk = 0.25.
SOLUTION
A +aB
m1vvœ2 + ©
t2
Lt1
Fyœdt = m1vyœ22
0 + N(3) - 20 cos 30°(3) = 0
A +bB
N = 17.32 lb
t2
m1vxœ21 + ©
Lt1
Fxœdt = m1vxœ22
20
20
(2) + 20 sin 30°(3) - 0.25(17.32)(3) =
v
32.2
32.2
v = 29.4 ft>s
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15–3.
A 5-lb block is given an initial velocity of 10 ft>s up a 45°
smooth slope. Determine the time for it to travel up the
slope before it stops.
SOLUTION
A Q+ B
t2
m(vx¿)1 + ©
Lt1
Fx dt = m(vx¿)2
5
(10) + ( -5 sin 45°)t = 0
32.2
t = 0.439 s
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*15–4.
The 180-lb iron worker is secured by a fall-arrest system
consisting of a harness and lanyard AB, which is fixed to the
beam. If the lanyard has a slack of 4 ft, determine the average
impulsive force developed in the lanyard if he happens to fall
4 feet. Neglect his size in the calculation and assume the
impulse takes place in 0.6 seconds.
A
B
SOLUTION
T1 + ©U1 - 2 = T2
0 + 180(4) =
1 180 2
(
)v
2 32.2
y = 16.05 ft>s
(+ T )
mv1 +
L
F dt = mv2
180
(16.05) + 180(0.6) - F(0.6) = 0
32.2
F = 329.5 lb = 330 lb
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15–5.
A man hits the 50-g golf ball such that it leaves the tee at an
angle of 40° with the horizontal and strikes the ground at
the same elevation a distance of 20 m away. Determine the
impulse of the club C on the ball. Neglect the impulse
caused by the ball’s weight while the club is striking the ball.
v2
C
SOLUTION
+ )
(:
sx = (s0)x + (v0)x t
20 = 0 + v cos 40°(t)
(+ c)
s = s0 + v0t +
1 2
at
2 c
0 = 0 + v sin 40°(t) -
1
(9.81)t2
2
t = 1.85 s
laws
or
v = 14.115 m>s
L
teaching
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F dt = mv2
in
mv1 + ©
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Dissemination
F dt = 10.052114.1152
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F dt = 0.706 N # s au 40°
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L
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L
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40⬚
15–6.
A train consists of a 50-Mg engine and three cars, each
having a mass of 30 Mg. If it takes 80 s for the train to
increase its speed uniformly to 40 km>h, starting from rest,
determine the force T developed at the coupling between
the engine E and the first car A. The wheels of the engine
provide a resultant frictional tractive force F which gives
the train forward motion, whereas the car wheels roll freely.
Also, determine F acting on the engine wheels.
v
A
F
SOLUTION
(yx)2 = 40 km>h = 11.11 m>s
Entire train:
+ b
a:
m(vx)1 + ©
L
Fx dt = m(vx)2
0 + F(80) = [50 + 3(30)] A 103 B (11.11)
F = 19.4 kN
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or
Three cars:
in
Fx dt = m(vx)2
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Dissemination
0 + T(80) = 3(30) A 103 B (11.11)
T = 12.5 kN
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Ans.
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m(vx)1 + ©
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+ b
a:
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E
15–7.
A
Crates A and B weigh 100 lb and 50 lb, respectively. If they
start from rest, determine their speed when t = 5 s. Also,
find the force exerted by crate A on crate B during the
motion. The coefficient of kinetic friction between the
crates and the ground is mk = 0.25.
P ⫽ 50 lb
SOLUTION
Free-Body Diagram: The free-body diagram of crates A and B are shown in
Figs. a and b, respectively. The frictional force acting on each crate is
(Ff)A = mkNA = 0.25NA and (Ff)B = mkNB = 0.25NB.
Principle of Impulse and Momentum: Referring to Fig. a,
t2
(+ c )
m(v1)y + ©
Fy dt = m(v2)y
Lt1
100
100
(0) + NA(5) - 100(5) =
(0)
32.2
32.2
NA = 100 lb
t2
Fx dt = m(v2)x
Lt1
or
m(v1)x + ©
laws
+ )
(:
Wide
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100
100
(0) + 50(5) - 0.25(100)(5) - FAB (5) =
v
32.2
32.2
(1)
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Dissemination
v = 40.25 - 1.61FAB
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By considering Fig. b,
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Fy dt = m(v2)y
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Lt1
the
m(v1)y + ©
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t2
(+ c )
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50
50
(0) + NB(5) - 50(5) =
(0)
32.2
32.2
work
this
NB = 50 lb
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Fx dt = m(v2)x
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Lt1
This
m(v1)x + ©
and
t2
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(:
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50
50
(0) + FAB(5) - 0.25(50)(5) =
v
32.2
32.2
v = 3.22 FAB - 40.25
(2)
Solving Eqs. (1) and (2) yields
FAB = 16.67 lb = 16.7 lb
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v = 13.42 ft>s = 13.4 ft>s
Ans.
B
*15–8.
If the jets exert a vertical thrust of T = (500t3>2)N, where t
is in seconds, determine the man’s speed when t = 3 s. The
total mass of the man and the jet suit is 100 kg. Neglect the
loss of mass due to the fuel consumed during the lift which
begins from rest on the ground.
T
SOLUTION
Free-Body Diagram: The thrust T must overcome the weight of the man and jet before
they move. Considering the equilibrium of the free-body diagram of the man and jet
shown in Fig. a,
+ c ©Fy = 0;
500t3>2 - 100(9.81) = 0
t = 1.567 s
Principle of Impulse and Momentum: Only the impulse generated by thrust T after
t = 1.567 s contributes to the motion. Referring to Fig. a,
t2
(+ c )
m(v1)y + ©
Lt1
Fy dt = m(v2)y
3s
L1.567 s
500 t3>2dt - 100(9.81)(3 - 1.567) = 100v
or
100(0) +
laws
3s
teaching
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- 1405.55 = 100v
1.567 s
in
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v = 11.0 m>s
Wide
a 200t5>2 b 2
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15–9.
Under a constant thrust of T = 40 kN, the 1.5-Mg dragster
reaches its maximum speed of 125 m>s in 8 s starting from
rest. Determine the average drag resistance FD during this
period of time.
FD
SOLUTION
Principle of Impulse and Momentum: The final speed of the dragster is v2 = 125 m>s.
Referring to the free-body diagram of the dragster shown in Fig. a,
t2
+ )
(;
Fx dt = m(v2)x
Lt1
1500(0) + 40(103)(8) - (FD)avg (8) = 1500(125)
m(v1)x + ©
(FD)avg = 16 562.5 N = 16.6 kN
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T ⫽ 40 kN
15–10.
P
The 50-kg crate is pulled by the constant force P. If the crate
starts from rest and achieves a speed of 10 m>s in 5 s, determine the magnitude of P. The coefficient of kinetic friction
between the crate and the ground is mk = 0.2.
30⬚
SOLUTION
Impulse and Momentum Diagram: The frictional force acting on the crate is
Ff = mkN = 0.2N.
Principle of Impulse and Momentum:
t2
(+ c )
Fy dt = m(v2)y
Lt1
0 + N(5) + P(5) sin 30° - 50(9.81)(5) = 0
m(v1)y + ©
N = 490.5 - 0.5P
(1)
t2
Fx dt = m(v2)x
Lt1
50(0) + P(5) cos 30° - 0.2N(5) = 50(10)
m(v1)x + ©
laws
or
+ )
(:
4.3301P - N = 500
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(2)
Dissemination
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Solving Eqs. (1) and (2), yields
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N = 387.97 N
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P = 205 N
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15–11.
When the 5-kg block is 6 m from the wall, it is sliding at
v1 = 14 m>s. If the coefficient of kinetic friction between
the block and the horizontal plane is mk = 0.3, determine
the impulse of the wall on the block necessary to stop the
block. Neglect the friction impulse acting on the block
during the collision.
v1
SOLUTION
Equation of Motion: The acceleration of the block must be obtained first before
one can determine the velocity of the block before it strikes the wall.
+ c ©Fy = may ;
N - 5(9.81) = 5(0)
N = 49.05 N
+ ©F = ma ;
:
x
x
-0.3(49.05) = - 5a
a = 2.943 m>s2
Kinematics: Applying the equation v2 = v20 + 2ac (s - s0) yields
+ B
A:
v2 = 142 + 2( -2.943)(6 - 0)
v = 12.68 m>s
Principle of Linear Impulse and Momentum: Applying Eq. 15–4, we have
or
t2
Web)
laws
Fxdt = m(vx)2
in
5(12.68) - I = 5(0)
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Dissemination
I = 63.4 N # s
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permitted.
+ B
A:
Lt1
teaching
m(vx)1 + ©
14 m/s
6m
*15–12.
For a short period of time, the frictional driving force acting
on the wheels of the 2.5-Mg van is FD = (600t2) N, where t
is in seconds. If the van has a speed of 20 km>h when t = 0,
determine its speed when t = 5 s.
FD
SOLUTION
Principle of Impulse and Momentum: The initial speed of the van is v1 = c 20(103)
c
m
d
h
1h
d = 5.556 m>s. Referring to the free-body diagram of the van shown in Fig. a,
3600 s
t2
+ )
(:
m(v1)x + ©
Lt1
Fx dt = m(v2)x
5s
2500(5.556) +
2
L0
600t dt = 2500 v2
v2 = 15.6 m>s
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15–13.
The 2.5-Mg van is traveling with a speed of 100 km>h when
the brakes are applied and all four wheels lock. If the speed
decreases to 40 km>h in 5 s, determine the coefficient of
kinetic friction between the tires and the road.
SOLUTION
Free-Body Diagram: The free-body diagram of the van is shown in Fig. a. The frictional
force is Ff = mkN since all the wheels of the van are locked and will cause the van
to slide.
Principle of Impulse and Momentum: The initial and final speeds of the van are
m
1h
m
1h
v1 = c100(103) d c
d = 27.78 m>s and v2 = c40(103) d c
d = 11.11 m>s.
h 3600 s
h 3600 s
Referring to Fig. a,
t2
(+ c )
m(v1)y + ©
Fy dt = m(v2)y
Lt1
2500(0) + N(5) - 2500(9.81)(5) = 2500(0)
laws
or
N = 24 525 N
Web)
t2
Fx dt = m(v2)x
Lt1
2500(27.78) + [ - mk(24525)(5)] = 2500(11.1)
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teaching
m(v1)x + ©
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mk = 0.340
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15–14.
The force acting on a projectile having a mass m as it passes
horizontally through the barrel of the cannon is
F = C sin (pt>t¿). Determine the projectile’s velocity when
t = t¿ . If the projectile reaches the end of the barrel at this
instant, determine the length s.
s
SOLUTION
+ b
a:
m(vx)1 + ©
L
Fx dt = m(vx)2
t
0 +
- Ca
L0
C sin a
pt
b = mv
t¿
pt t
t¿
b cos a b 2 = mv
p
t¿ 0
v =
Ct¿
pt
a 1 - cos a b b
pm
t¿
laws
or
When t = t¿ ,
teaching
Web)
2C t¿
pm
in
`Ans.
Wide
copyright
v2 =
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ds = v dt
not
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is
of
the
pt
C t¿
b a 1 - cos a b b dt
pm
t¿
and
use
L0
a
the
L0
t
ds =
by
s
the
(including
for
work
student
t¿
pt t¿
C t¿
b c t - sin a b d
pm
p
t¿
0
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s = a
assessing
Ct¿ 2
pm
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Ans.
work
is
This
provided
s =
15–15.
During operation the breaker hammer develops on the
concrete surface a force which is indicated in the graph. To
achieve this the 2-lb spike S is fired from rest into the
surface at 200 ft>s. Determine the speed of the spike just
after rebounding.
F (103) lb
112.5
90
67.5
45
22.5
SOLUTION
S
0
t (ms)
0
1+ T 2
mv1 +
L
F dt = mv2
-2
2
(v)
(200) + 2(0.0004) - Area =
32.2
32.2
Area =
1
19021103210.42110 - 32 = 18 lb # s
2
Thus,
v = 89.8 ft>s
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Ans.
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
0.1
0.2
0.3
0.4
*15–16.
The twitch in a muscle of the arm develops a force which
can be measured as a function of time as shown in the
graph. If the effective contraction of the muscle lasts for a
time t0 , determine the impulse developed by the muscle.
F
F ⫽ F0 ( t )e⫺t/T
T
SOLUTION
t0
t0
I =
L
F dt =
t
F0 a b e t>Tdt
T
L0
F0 t0 - 1t>T2
te
dt
T L0
I = - F0 c Te - t>T a
t0
t
+ 1b d
T
0
I = - F0 c Te - t0>T a
t0
+ 1b - T d
T
t0
bd
T
Ans.
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I = TF0 c 1 - e - t0>T a 1 +
or
I =
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t
15–17.
A hammer head H having a weight of 0.25 lb is moving
vertically downward at 40 ft/s when it strikes the head of a
nail of negligible mass and drives it into a block of wood.
Find the impulse on the nail if it is assumed that the grip at
A is loose, the handle has a negligible mass, and the hammer
stays in contact with the nail while it comes to rest. Neglect
the impulse caused by the weight of the hammer head
during contact with the nail.
H
v = 40 ft/s
A
SOLUTION
(+ T)
m(vy)1 + ©
a
L
Fy dt = m(vy)2
0.25
b (40) F dt = 0
32.2
L
F dt = 0.311 lb # s
Ans.
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L
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
15–18.
The 40-kg slider block is moving to the right with a speed of
1.5 m>s when it is acted upon by the forces F1 and F2. If
these loadings vary in the manner shown on the graph,
determine the speed of the block at t = 6 s. Neglect friction
and the mass of the pulleys and cords.
F2
F1
F (N)
SOLUTION
40
The impulses acting on the block are equal to the areas under the graph.
30
+ b
a:
20
m(vx)1 + ©
L
Fx dt = m(vx)2
0
v2 = 12.0 m>s ( : )
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Dissemination
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Ans.
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F1
10
40(1.5) + 4[(30)4 + 10(6 - 4)] - [10(2) + 20(4 - 2)
+ 40(6 - 4)] = 40v2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
F2
2
4
6
t (s)
15–19.
Determine the velocity of each block 2 s after the blocks are
released from rest. Neglect the mass of the pulleys and cord.
SOLUTION
A
Kinematics: The speed of block A and B can be related by using the position
coordinate equation.
2sA + sB = l
2yA + yB = 0
(1)
Principle of Linear Impulse and Momentum: Applying Eq. 15–4 to block A, we have
Fy dt = m A vy B 2
10
10
b (0) + 2T(2) - 10(2) = - a
b (vA)
32.2
32.2
(2)
teaching
Web)
-a
Lt1
laws
(+ c)
t2
or
m A vy B 1 + ©
Wide
copyright
in
Applying Eq. 15–4 to block B, we have
t2
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Fy dt = m A vy B 2
is
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Lt1
States
m A vy B 1 + ©
United
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learning.
50
50
b(0) + T(2) - 50(2) = - a
b(vB)
32.2
32.2
and
use
-a
(3)
for
work
student
the
by
(+ c)
solely
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the
(including
Solving Eqs. (1),(2) and (3) yields
vA = -27.6 ft>s = 27.6 ft>s c
work
this
assessing
is
work
vB = 55.2 ft>s T
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and
T = 7.143 lb
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Ans.
10 lb
B
50 lb
*15–20.
z
The particle P is acted upon by its weight of 3 lb and forces
F1 and F2 , where t is in seconds. If the particle orginally has
a velocity of v1 = 53i + 1j + 6k6 ft>s, determine its speed
after 2 s.
F1 ⫽ {5i ⫹ 2tj ⫹ tk} lb
SOLUTION
2
mv1 + a
F2 ⫽ {t2i} lb
Fdt = mv2
L0
x
Resolving into scalar components,
2
3
3
132 +
15 + t 22dt =
1v 2
32.2
32.2 x
L0
2
3
3
112 +
2tdt =
1vy2
32.2
32.2
L0
2
laws
or
3
3
162 +
1t - 32dt =
1vz2
32.2
32.2
L0
Web)
vz = - 36.933 ft>s
teaching
vy = 43.933 ft>s
in
vx = 138.96 ft>s
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v = 2(138.96)2 + (43.933)2 + ( -36.933)2 = 150 ft>s
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y
P
15–21.
If it takes 35 s for the 50-Mg tugboat to increase its speed
uniformly to 25 km>h, starting from rest, determine the
force of the rope on the tugboat.The propeller provides the
propulsion force F which gives the tugboat forward motion,
whereas the barge moves freely. Also, determine F acting on
the tugboat. The barge has a mass of 75 Mg.
F
SOLUTION
1000
25a
b = 6.944 m/s
3600
System:
+ 2
1:
my1 + ©
L
F dt = my2
[0 + 0] + F(35) = (50 + 75)(103)(6.944)
F = 24.8 kN
Ans.
laws
or
Barge:
teaching
Web)
F dt = mv2
Wide
L
in
mv1 + ©
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Dissemination
0 + T(35) = (75)(103)(6.944)
Ans.
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T = 14.881 = 14.9 kN
for
work
student
the
by
Also, using this result for T,
this
F dt = mv2
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mv1 + ©
the
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This
+ 2
1:
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Tugboat:
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F = 24.8 kN
of
and
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0 + F1352 - 114.88121352 = 15021103216.9442
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Ans.
15–22.
T (lb)
If the force T exerted on the cable by the motor M is indicated by the graph, determine the speed of the 500-lb crate
when t = 4 s, starting from rest. The coefficients of static and
kinetic friction are ms = 0.3 and mk = 0.25, respectively.
60
30
t (s)
2
SOLUTION
Free-Body Diagram: Here, force 3T must overcome the friction Ff before the crate
T - 30
60 - 30
moves. For 0 … t … 2 s,
or T = A 15t + 30 B lb. Considering the
=
t - 0
2 - 0
free-body diagram of the crate shown in Fig. a, where Ff = mk N = 0.3N,
+ c ©Fy = 0;
N - 500 = 0
N = 500 lb
+
: ©Fx = 0;
3(15t + 30) - 0.3(500) = 0
t = 1.333 s
or
Principle of Impulse and Momentum: Only the impulse of 3T after t = 1.333 s contributes to the motion. The impulse of T is equal to the area under the T vs. t graph.
At t = 1.333 s, T = 50 lb. Thus,
Web)
laws
1
3Tdt = 3 c (50 + 60)(2 - 1.333) + 60(4 - 2) d = 470 lb # s
2
in
L
teaching
I =
Dissemination
Wide
copyright
Since the crate moves, Ff = mkN = 0.25(500) = 125 lb. Referring to Fig. a,
States
World
Fx dt = m(v2)x
the
not
instructors
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Lt1
of
m(v1)x + ©
permitted.
t2
+ )
(:
for
work
student
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and
use
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learning.
500
500
(0) + 470 - 125(4 - 1.333) = a
bv
32.2
32.2
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v = 8.80 ft>s
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Ans.
T
M
15–23.
The 5-kg block is moving downward at v1 = 2 m>s when it
is 8 m from the sandy surface. Determine the impulse of the
sand on the block necessary to stop its motion. Neglect the
distance the block dents into the sand and assume the block
does not rebound. Neglect the weight of the block during
the impact with the sand.
v1
8m
SOLUTION
Just before impact
T1 + ©U1 - 2 = T2
1
1
(5)(2)2 + 8(5)(9.81) = (5)(v2)
2
2
v = 12.687 m>s
F dt = mv2
5112.6872 -
teaching
Web)
F dt = 63.4 N # s
in
Ans.
Wide
L
F dt = 0
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I =
L
or
L
laws
mv1 + ©
copyright
1+ T2
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2 m/s
*15–24.
The 5-kg block is falling downward at v1 = 2 m>s when it is
8 m from the sandy surface. Determine the average
impulsive force acting on the block by the sand if the
motion of the block is stopped in 0.9 s once the block strikes
the sand. Neglect the distance the block dents into the sand
and assume the block does not rebound. Neglect the weight
of the block during the impact with the sand.
v1
8m
SOLUTION
Just before impact
T1 + ©U1 - 2 = T2
1
1
(5)(2)2 + 8(5)(9.81) = (5)(v2)
2
2
v = 12.69 m>s
1+ T2
mv1 + ©
L
F dt = mv2
or
5(12.69) - Fang (0.9) = 0
Fang = 70.5 N
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Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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2 m/s
15–25.
The 0.1-lb golf ball is struck by the club and then travels
along the trajectory shown. Determine the average
impulsive force the club imparts on the ball if the club
maintains contact with the ball for 0.5 ms.
v
30
500 ft
SOLUTION
Kinematics: By considering the x-motion of the golf ball, Fig. a,
+ b
a:
sx = A s0 B + A v0 B x t
500 = 0 + v cos 30° t
t =
500
v cos 30°
Subsequently, using the result of t and considering the y-motion of the golf ball,
sy = A s0 B y + A v0 B y t +
1
ay t2
2
or
a+cb
in
teaching
Web)
laws
2
1
500
500
≤ + ( -32.2) ¢
≤
v cos 30°
2
v cos 30°
Wide
copyright
0 = 0 + v sin 30° ¢
instructors
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Dissemination
v = 136.35 ft>s
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is
of
the
not
Principle of Impulse and Momentum: Here, the impulse generated by the weight of
the golf ball is very small compared to that generated by the force of the impact.
Hence, it can be neglected. By referring to the impulse and momentum diagram
shown in Fig. b,
work
solely
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Fx¿ dt = m A v2 B x¿
work
this
assessing
Lt1
the
t2
is
m A v1 B x¿ + ©
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integrity
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and
0.1
A 136.35 B
32.2
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This
0 + Favg (0.5)(10 - 3) =
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Favg = 847 lb
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Ans.
15–26.
As indicated by the derivation, the principle of impulse and
momentum is valid for observers in any inertial reference
frame. Show that this is so, by considering the 10-kg block
which rests on the smooth surface and is subjected to a
horizontal force of 6 N. If observer A is in a fixed frame x,
determine the final speed of the block in 4 s if it has an initial
speed of 5 m>s measured from the fixed frame. Compare the
result with that obtained by an observer B, attached to the x¿
axis that moves at a constant velocity of 2 m>s relative to A.
L
6N
F dt = m v2
10(5) + 6(4) = 10v
v = 7.40 m>s
Ans.
Observer B:
L
or
F dt = m v2
teaching
Web)
m v1 + a
laws
+ 2
(:
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10(3) + 6(4) = 10v
v = 5.40 m>s
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B
x¿
5 m/s
Observer A:
m v1 + a
x
2 m/s
SOLUTION
+ 2
1:
A
15–27.
The winch delivers a horizontal towing force F to its cable
at A which varies as shown in the graph. Determine the
speed of the 70-kg bucket when t = 18 s. Originally the
bucket is moving upward at v1 = 3 m>s.
A
F
F (N)
600
360
v
B
12
SOLUTION
Principle of Linear Impulse and Momentum: For the time period 12 s … t 6 18 s,
600 - 360
F - 360
=
, F = (20t + 120) N. Applying Eq. 15–4 to bucket B, we have
t - 12
24 - 12
t2
Fy dt = m A vy B 2
Lt1
or
m A vy B 1 + ©
in
teaching
(20t + 120)dt d - 70(9.81)(18) = 70v2
L12 s
Web)
laws
18 s
70(3) + 2 c 360(12) +
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(+ c)
Ans.
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Dissemination
v2 = 21.8 m>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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24
t (s)
*15–28.
The winch delivers a horizontal towing force F to its cable
at A which varies as shown in the graph. Determine the
speed of the 80-kg bucket when t = 24 s. Originally the
bucket is released from rest.
A
F
F (N)
600
360
v
B
12
SOLUTION
Principle of Linear Impulse and Momentum: The total impluse exerted on bucket B
laws
or
can be obtained by evaluating the area under the F–t graph. Thus,
t2
1
I = ©
Fy dt = 2 c 360(12) + (360 + 600)(24 - 12) d = 20160 N # s. Applying
2
Lt1
Eq. 15–4 to the bucket B, we have
t2
in
teaching
Web)
Fy dt = m A vy B 2
Lt1
Wide
copyright
m A vy B 1 + ©
permitted.
Dissemination
80(0) + 20160 - 80(9.81)(24) = 80v2
Ans.
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This
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v2 = 16.6m>s
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not
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World
(+ c)
24
t (s)
15–29.
The train consists of a 30-Mg engine E, and cars A, B, and C,
which have a mass of 15 Mg, 10 Mg, and 8 Mg, respectively.
If the tracks provide a traction force of F = 30 kN on the
engine wheels, determine the speed of the train when
t = 30 s, starting from rest. Also, find the horizontal
coupling force at D between the engine E and car A.
Neglect rolling resistance.
C
Principle of Impulse and Momentum: By referring to the free-body diagram of the
entire train shown in Fig. a, we can write
m A v1 B x + ©
t2
Lt1
Fxdt = m A v2 B x
63 000(0) + 30(103)(30) = 63 000v
v = 14.29 m>s
Ans.
Using this result and referring to the free-body diagram of the train’s car shown in
Fig. b,
t2
or
Fx dt = m A v2 B x
Web)
Lt1
laws
m A v1 B x + ©
teaching
+ b
a:
FD = 15 714.29 N = 15.7 kN
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This
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Ans.
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or transmission in any form or by any means, electronic, mechanical,
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permitted.
Dissemination
Wide
copyright
in
33000(0) + FD(30) = 33 000 A 14.29 B
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
A
E
D
SOLUTION
+ b
a:
B
F
30 kN
15–30.
The crate B and cylinder A have a mass of 200 kg and 75 kg,
respectively. If the system is released from rest, determine
the speed of the crate and cylinder when t = 3 s. Neglect
the mass of the pulleys.
B
SOLUTION
Free-Body Diagram: The free-body diagrams of cylinder A and crate B are shown in
Figs. b and c. vA and vB must be assumed to be directed downward so that they are
consistent with the positive sense of sA and sB shown in Fig. a.
A
Principle of Impulse and Momentum: Referring to Fig. b,
t2
m(v1)y + ©
(+ T )
Fy dt = m(v2)y
Lt1
75(0) + 75(9.81)(3) - T(3) = 75vA
vA = 29.43 - 0.04T
(1)
From Fig. b,
Web)
laws
Fy dt = m(v2)y
teaching
Lt1
in
m(v1)y + ©
or
t2
(+ T )
Wide
copyright
200(0) + 2500(9.81)(3) - 4T(3) = 200vB
(2)
the
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is
of
the
Kinematics: Expressing the length of the cable in terms of sA and sB and referring
to Fig. a,
(3)
protected
the
(including
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work
student
sA + 4sB = l
this
assessing
is
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solely
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Taking the time derivative,
(4)
part
the
is
This
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work
vA + 4vB = 0
destroy
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and
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courses
Solving Eqs. (1), (2), and (4) yields
will
their
vA = 8.409 m>s = 8.41 m>s T Ans.
sale
vB = - 2.102 m>s = 2.10 m>s c
T = 525.54 N
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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not
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permitted.
Dissemination
vB = 29.43 - 0.06T
15–31.
Block A weighs 10 lb and block B weighs 3 lb. If B is
moving downward with a velocity 1vB21 = 3 ft>s at t = 0,
determine the velocity of A when t = 1 s. Assume that the
horizontal plane is smooth. Neglect the mass of the pulleys
and cords.
A
SOLUTION
sA + 2sB = l
(vB)1
vA = -2vB
+ 2
1;
mv1 + ©
-
1+ T2
L
F dt = mv2
10
10
(vA)2
(2)(3) - T(1) =
32.2
32.2
mv1 + ©
L
F dt = mv2
in
teaching
Web)
laws
or
(vA)2
3
3
()
(3) + 3(1) - 2T(1) =
32.2
32.2
2
Dissemination
Wide
copyright
- 32.2T - 10(vA)2 = 60
of
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instructors
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World
permitted.
- 64.4T + 1.5(vA)2 = - 105.6
and
use
United
on
learning.
is
T = 1.40 lb
by
(vA22 = -10.5 ft>s = 10.5 ft>s :
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Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
3 ft/s
B
*15–32.
Block A weighs 10 lb and block B weighs 3 lb. If B is
moving downward with a velocity 1vB21 = 3 ft>s at t = 0,
determine the velocity of A when t = 1 s. The coefficient of
kinetic friction between the horizontal plane and block A is
mA = 0.15.
A
SOLUTION
sA + 2sB = l
(vB)1
vA = -2vB
+ 2
1;
mv1 + ©
-
1+ T 2
L
F dt = mv2
10
10
(v )
(2)(3) - T(1) + 0.15(10) =
32.2
32.2 A 2
mv1 + ©
L
F dt = mv2
in
teaching
Web)
laws
or
3
3 (vA)2
(3) - 3(1) - 2T(1) =
a
b
32.2
32.2 2
Dissemination
Wide
copyright
- 32.2T - 10(vA)2 = 11.70
not
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permitted.
- 64.4T + 1.51vA22 = - 105.6
and
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is
of
the
T = 1.50 lb
by
(vA)2 = - 6.00 ft>s = 6.00 ft>s :
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This
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Ans.
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
3 ft/s
B
15–33.
The log has a mass of 500 kg and rests on the ground for
which the coefficients of static and kinetic friction are
ms = 0.5 and mk = 0.4, respectively. The winch delivers a
horizontal towing force T to its cable at A which varies as
shown in the graph. Determine the speed of the log when
t = 5 s. Originally the tension in the cable is zero. Hint: First
determine the force needed to begin moving the log.
T (N)
1800
T
200 t2
3
t (s)
A T
SOLUTION
+ ©F = 0;
:
x
F - 0.5(500)(9.81) = 0
F = 2452.5 N
Thus,
2T = F
2(200t 2) = 2452.5
t = 2.476 s to start log moving
L
F dt = m v2
or
m v1 + ©
laws
+ )
(:
teaching
Web)
3
in
200t 2 dt + 211800215 - 32 - 0.41500219.81215 - 2.4762 = 500v2
Wide
Dissemination
L2.476
copyright
0 + 2
on
learning.
is
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not
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World
permitted.
t3 3
400( ) `
+ 2247.91 = 500v2
3 2.476
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This
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v2 = 7.65 m>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Ans.
15–34.
The 50-kg block is hoisted up the incline using the cable and
motor arrangement shown. The coefficient of kinetic
friction between the block and the surface is mk = 0.4. If the
block is initially moving up the plane at v0 = 2 m>s, and at
this instant (t = 0) the motor develops a tension in the cord
of T = (300 + 120 2 t) N, where t is in seconds, determine
the velocity of the block when t = 2 s.
v0
2 m/s
30
SOLUTION
+a©Fx = 0;
(+Q)
NB - 50(9.81)cos 30° = 0
m(vx)1 + ©
L
NB = 424.79 N
Fx dt = m(vx)2
2
A 300 + 120 2t B dt - 0.4(424.79)(2)
L0
- 50(9.81)sin 30°(2) = 50v2
50(2) +
v2 = 1.92 m>s
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This
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Dissemination
Wide
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in
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Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
15–35.
The bus B has a weight of 15 000 lb and is traveling to the
right at 5 ft>s. Meanwhile a 3000-lb car A is traveling at
4 ft>s to the left. If the vehicles crash head-on and become
entangled, determine their common velocity just after the
collision. Assume that the vehicles are free to roll during
collision.
vB
5 ft/s
B
A
SOLUTION
+ )
(:
mA(vA)1 + mB(vB)1 = (mA + mB)v
3000
18 000
15 000
(5) (4) =
v
32.2
32.2
32.2
v = 3.5 ft>s :
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This
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permitted.
Dissemination
Wide
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in
teaching
Web)
laws
or
Ans.
sale
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
vA
4 ft/s
*15–36.
The 50-kg boy jumps on the 5-kg skateboard with a horizontal velocity of 5 m>s. Determine the distance s the boy
reaches up the inclined plane before momentarily coming
to rest. Neglect the skateboard’s rolling resistance.
s
30⬚
SOLUTION
Free-Body Diagram: The free-body diagram of the boy and skateboard system is
shown in Fig. a. Here, Wb,Wsb, and N are nonimpulsive forces. The pair of impulsive
forces F resulting from the impact during landing cancel each other out since they are
internal to the system.
Conservation of Linear Momentum: Since the resultant of the impulsive force along
the x axis is zero, the linear momentum of the system is conserved along the x axis.
mb(vb)1 + msb(vsb)1 = (mb + msb)v
+ )
(;
50(5) + 5(0) = (50 + 5)v
laws
or
v = 4.545 m>s
teaching
Web)
Conservation of Energy: With reference to the datum set in Fig. b, the
in
gravitational potential energy of the boy and skateboard at positions A and B are
learning.
is
of
the
not
instructors
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= 269.775s.
permitted.
Dissemination
Wide
copyright
A Vg B A = (mb + msb)ghA = 0 and A Vg B B = (mb + msb)ghB = (50 + 5)(9.81)(s sin 30°)
the
by
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on
TA + VA = TB + VB
is
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1
1
(mb + msb)vA 2 + A Vg B A = (mb + msb)vB 2 + A Vg B B
2
2
part
the
is
This
provided
integrity
of
and
work
this
assessing
1
(50 + 5) A 4.5452 B + 0 = 0 + 269.775s
2
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Ans.
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destroy
of
and
s = 2.11 m
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
15–37.
30 km/h
The 2.5-Mg pickup truck is towing the 1.5-Mg car using a
cable as shown. If the car is initially at rest and the truck is
coasting with a velocity of 30 km>h when the cable is slack,
determine the common velocity of the truck and the car just
after the cable becomes taut. Also, find the loss of energy.
SOLUTION
Free-Body Diagram: The free-body diagram of the truck and car system is shown in
Fig. a. Here, Wt,WC, Nt, and NC are nonimpulsive forces. The pair of impulsive forces
F generated at the instant the cable becomes taut are internal to the system and thus
cancel each other out.
Conservation of Linear Momentum: Since the resultant of the impulsive force is
zero, the linear momentum of the system is conserved along the x axis. The initial
m
1h
speed of the truck is A vt B 1 = c 30(103) d c
d = 8.333 m>s.
h 3600 s
mt A vt B 1 + mC A vC B 1 = A mt + mC B v2
laws
or
2500(8.333) + 0 = (2500 + 1500)v2
v2 = 5.208 m>s = 5.21 m>s ;
Ans.
Wide
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teaching
Web)
+ )
(;
instructors
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permitted.
Dissemination
Kinetic Energy: The initial and final kinetic energy of the system is
United
on
learning.
is
of
the
not
1
1
m (v ) 2 + mC(vC)12
2 t t1
2
by
and
use
T1 =
the
(including
for
work
student
the
1
(2500)(8.3332) + 0
2
solely
of
protected
=
integrity
of
and
work
this
assessing
is
work
= 86 805.56 J
any
will
1
(2500 + 1500)(5.2082)
2
sale
=
their
destroy
of
and
T2 = (mt + mC)v22
courses
part
the
is
This
provided
and
= 54 253.47
Thus, the loss of energy during the impact is
¢E = T1 - T2 = 86 805.56 - 54 253.47 = 32.55(103) J = 32.6 kJ
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Ans.
15–38.
A railroad car having a mass of 15 Mg is coasting at 1.5 m>s
on a horizontal track. At the same time another car having a
mass of 12 Mg is coasting at 0.75 m>s in the opposite
direction. If the cars meet and couple together, determine
the speed of both cars just after the coupling. Find the
difference between the total kinetic energy before and after
coupling has occurred, and explain qualitatively what
happened to this energy.
SOLUTION
+ )
(:
©mv1 = ©mv2
15 000(1.5) - 12 000(0.75) = 27 000(v2)
v2 = 0.5 m>s
Ans.
1
1
(15 000)(1.5)2 + (12 000)(0.75)2 = 20.25 kJ
2
2
T2 =
1
(27 000)(0.5)2 = 3.375 kJ
2
or
T1 =
teaching
Web)
laws
¢T = T1 - T2
= 20.25 - 3.375 = 16.9 kJ
Dissemination
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This
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the
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permitted.
This energy is dissipated as noise, shock, and heat during the coupling.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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15–39.
The car A has a weight of 4500 lb and is traveling to the
right at 3 ft>s. Meanwhile a 3000-lb car B is traveling at
6 ft>s to the left. If the cars crash head-on and become
entangled, determine their common velocity just after the
collision. Assume that the brakes are not applied during
collision.
vA
3 ft/s
A
SOLUTION
+ )
(:
mA (vA)1 + mB(vB)1 = (mA + mB)v 2
3000
7500
4500
(3) (6) =
v
32.2
32.2
32.2 2
v 2 = -0.600 ft>s = 0.600 ft>s ;
destroy
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and
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courses
part
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is
This
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integrity
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the
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is
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Dissemination
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in
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Web)
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will
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© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
vB
B
6 ft/s
*15–40.
The 200-g projectile is fired with a velocity of 900 m>s
towards the center of the 15-kg wooden block, which rests
on a rough surface. If the projectile penetrates and
emerges from the block with a velocity of 300 m>s, determine the velocity of the block just after the projectile
emerges. How long does the block slide on the rough surface, after the projectile emerges, before it comes to rest
again? The coefficient of kinetic friction between the surface and the block is mk = 0.2.
900 m/s
Before
300 m/s
SOLUTION
Free-Body Diagram: The free-body diagram of the projectile and block system is
shown in Fig. a. Here, WB, WP, N, and Ff are nonimpulsive forces. The pair of impulsive forces F resulting from the impact cancel each other out since they are internal to
the system.
After
Conservation of Linear Momentum: Since the resultant of the impulsive force along
the x axis is zero, the linear momentum of the system is conserved along the x axis.
+ B
A:
mP A vP B 1 + mB A vB B 1 = mP A vP B 2 + mB A vB B 2
laws
or
0.2(900) + 15(0) = 0.2(300) + 15(vB)2
teaching
Web)
Ans.
(vB)2 = 8 m>s :
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Principle of Impulse and Momentum: Using the result of A vB B 2, and referring to the
free-body diagram of the block shown in Fig. b,
is
of
the
not
t2
United
on
learning.
Fy dt = m(v2)y
and
use
Lt1
by
m A v1 B y + ©
the
A+cB
the
(including
for
work
student
15(0) + N(t) - 15(9.81)(t) = 15(0)
assessing
is
work
solely
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protected
N = 147.15 N
work
this
t2
provided
integrity
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and
Fxdt = m(v2)x
part
the
is
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sale
t = 4.077 s = 4.08 s
will
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destroy
of
and
15(8) + [ -0.2(147.15)(t)] = 15(0)
any
Lt1
courses
m A v1 B x + ©
This
+ )
(:
Ans.
15–41.
The block has a mass of 50 kg and rests on the surface of the
cart having a mass of 75 kg. If the spring which is attached to
the cart and not the block is compressed 0.2 m and the system
is released from rest, determine the speed of the block
relative to the ground after the spring becomes undeformed.
Neglect the mass of the cart’s wheels and the spring in the
calculation. Also neglect friction. Take k = 300 N>m.
B
C
SOLUTION
T1 + V1 = T2 + V2
(0 + 0) +
1
1
1
(300)(0.2)2 = (50)(vb)2 + (75)(vc)2
2
2
2
12 = 50 v2b + 75 v2c
+ )
(:
©mv1 = ©mv2
0 + 0 = 50 vb - 75 vc
vb = 1.5vc
laws
or
vc = 0.253 m>s ;
vb = 0.379 m s :
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This
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15–42.
The block has a mass of 50 kg and rests on the surface of
the cart having a mass of 75 kg. If the spring which is
attached to the cart and not the block is compressed 0.2 m
and the system is released from rest, determine the speed
of the block with respect to the cart after the spring becomes
undeformed. Neglect the mass of the wheels and the spring
in the calculation. Also neglect friction. Take k = 300 N>m.
B
C
SOLUTION
T1 + V1 = T2 + V2
(0 + 0) +
1
1
1
(300)(0.2)2 = (50)(vb)2 + (75)(vc)2
2
2
2
12 = = 50 v2b + 75 v2c
+ )
(:
©mv1 = ©mv2
0 + 0 = 50 vb - 75 vc
vb = 1.5vc
laws
or
vc = 0.253 m>s ;
in
teaching
Web)
vb = 0.379 m>s :
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vb = vc + vb>c
World
permitted.
0.379 = - 0.253 + vb>c
of
the
not
instructors
States
+ )
(:
is
vb c = 0.632 m s :
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This
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the
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Ans.
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15–43.
20 km/h
The three freight cars A, B, and C have masses of 10 Mg, 5 Mg,
and 20 Mg, respectively. They are traveling along the track
with the velocities shown. Car A collides with car B first, followed by car C. If the three cars couple together after collision, determine the common velocity of the cars after the two
collisions have taken place.
5 km/h
A
B
SOLUTION
Free-Body Diagram: The free-body diagram of the system of cars A and B when they
collide is shown in Fig. a. The pair of impulsive forces F1 generated during the collision cancel each other since they are internal to the system. The free-body diagram
of the coupled system composed of cars A and B and car C when they collide is
shown in Fig. b. Again, the internal pair of impulsive forces F2 generated during the
collision cancel each other.
Conservation of Linear Momentum: When A collides with B, and then the coupled
cars A and B collide with car C, the resultant impulsive force along the x axis is zero.
Thus, the linear momentum of the system is conserved along the x axis. The initial
speed of the cars A, B, and C are
or
in
teaching
Web)
m
1h
da
b = 1.389 m>s,
h 3600 s
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A vB B 1 = c5(103)
m
1h
da
b = 5.556 m>s
h 3600 s
laws
A vA B 1 = c20(103)
the
not
instructors
States
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permitted.
m
1h
da
b = 6.944 m>s
h 3600 s
the
by
and
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is
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and A vC B 1 = c 25(103)
protected
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mA(vA)1 + mB(vB)1 = (mA + mB)v2
this
assessing
is
+ B
A:
the
(including
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For the first case,
the
is
This
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integrity
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and
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10000(5.556) + 5000(1.389) = (10000 + 5000)vAB
of
and
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courses
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vAB = 4.167 m>s :
sale
+ B
A:
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Using the result of vAB and considering the second case,
(mA + mB)vAB + mC(vC)1 = (mA + mB + mC)vABC
(10000 + 5000)(4.167) + [ -20000(6.944)] = (10000 + 5000 + 20000)vABC
vABC = - 2.183 m>s = 2.18 m>s ;
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Ans.
25 km/h
C
*15–44.
Two men A and B, each having a weight of 160 lb, stand on
the 200-lb cart. Each runs with a speed of 3 ft>s measured
relative to the cart. Determine the final speed of the cart if
(a) A runs and jumps off, then B runs and jumps off the
same end, and (b) both run at the same time and jump off at
the same time. Neglect the mass of the wheels and assume
the jumps are made horizontally.
A
SOLUTION
(a) A jumps first.
+ ) 0 + 0 = m v - (m + m )vœ
(;
A A
C
B C
0 =
However, vA = -vCœ + 3
160
360 œ
( - vCœ + 3) vC
32.2
32.2
vCœ = 0.9231 ft>s :
And then B jumps
0 + (mC + mB) vCœ = mB vB - mCvC
However, vB = - vC + 3
teaching
Web)
laws
or
160
200
360
( - 0.9231) =
( -vC + 3) vC
32.2
32.2
32.2
in
vC = 2.26 ft>s :
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Ans.
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(b) Both men jump at the same time
United
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is
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However, v = -vC + 3
by
and
use
+ B 0 + 0 = (m + m )v - m v
A;
A
B
C C
work
student
the
160
160
200
+
b ( - vC + 3) v
32.2
32.2
32.2 C
the
(including
for
0 = a
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This
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vC = 1.85 ft>s :
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Ans.
B
15–45.
The block of mass m is traveling at v1 in the direction u1
shown at the top of the smooth slope. Determine its speed
v2 and its direction u2 when it reaches the bottom.
z
v1
y
x
u1
h
SOLUTION
There are no impulses in the v direction:
v2
mv1 sin u1 = mv2 sin u2
u2
T1 + V1 = T2 + V2
1
1
mv12 + mgh = mv22 + 0
2
2
v2 = 3v21 + 2gh
v1 sin u1
or
3v21 + 2gh
in
teaching
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laws
sin u2 =
Ans.
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3v21 + 2gh
copyright
v1 sin u1
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u2 = sin - 1 £
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15–46.
The barge B weighs 30 000 lb and supports an automobile
weighing 3000 lb. If the barge is not tied to the pier P and
someone drives the automobile to the other side of the barge
for unloading, determine how far the barge moves away from
the pier. Neglect the resistance of the water.
200 ft
P
B
SOLUTION
Relative Velocity: The relative velocity of the car with respect to the barge is y c/b.
Thus, the velocity of the car is
+ B
A:
vc = -vb + vc>b
(1)
Conservation of Linear Momentum: If we consider the car and the barge as a
system, then the impulsive force caused by the traction of the tires is internal to the
system. Therefore, it will cancel out. As the result, the linear momentum is
conserved along the x axis.
0 = mc vc + mb vb
3000
30 000
b vc - a
b vb
32.2
32.2
or
(2)
teaching
Web)
0 + 0 = a
laws
+ B
A:
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Substituting Eq. (1) into (2) yields
(3)
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Dissemination
11vb - vc>b = 0
use
by
and
11sb - sc>b = 0
(4)
for
work
student
the
+ B
A:
United
on
learning.
is
of
the
not
Integrating Eq. (3) becomes
solely
of
protected
the
(including
Here, s c/b = 200 ft. Then, from Eq. (4)
work
sb = 18.2 ft
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This
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11sb - 200 = 0
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Ans.
15–47.
20 km/h
The 30-Mg freight car A and 15-Mg freight car B are moving
towards each other with the velocities shown. Determine the
maximum compression of the spring mounted on car A.
Neglect rolling resistance.
SOLUTION
Conservation of Linear Momentum: Referring to the free-body diagram of the freight
cars A and B shown in Fig. a, notice that the linear momentum of the system is conserved along the x axis. The initial speed of freight cars A and B are
m
1h
m
1h
(vA)1 = c20(103) d a
b = 5.556 m>s and (vB)1 = c10(103) d a
b
h 3600 s
h 3600 s
= 2.778 m>s. At this instant, the spring is compressed to its maximum, and no relative
motion occurs between freight cars A and B and they move with a common speed.
+ )
(:
mA(vA)1 + mB(vB)1 = (mA + mB)v2
30(103)(5.556) + c -15(103)(2.778) d = c 30(103) + 15(103) dv2
teaching
Web)
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or
v2 = 2.778 m>s :
is
of
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by
and
use
United
on
learning.
©T1 + ©V1 = ©T2 + ©V2
provided
integrity
of
and
work
this
assessing
is
1
1
(30) A 103 B (5.5562) + (15) A 103 B A 2.7782 B + 0
2
2
work
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the
(including
for
work
student
the
1
1
1
c mA(vA)12 + mB(vB)12 d + (Ve)1 = (mA + mB)v22 + (Ve)2
2
2
2
any
courses
part
the
is
This
1
c30 A 103 B + 15 A 103 B d A 2.7782 B + 1.5 A 106 B smax2
2
smax = 0.4811 m = 481 mm
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=
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Conservation of Energy: The initial and final elastic potential energy of the spring
1
1
1
is (Ve)1 = ks12 = 0 and (Ve)2 = ks22 = (3)(106)smax2 = 1.5(106)smax2.
2
2
2
10 km/h
k ⫽ 3 MN/m
A
B
*15–48.
The barge weighs 45 000 lb and supports two automobiles A
and B, which weigh 4000 lb and 3000 lb, respectively. If the
automobiles start from rest and drive towards each other,
accelerating at aA = 4 ft>s2 and aB = 8 ft>s2 until they reach
a constant speed of 6 ft/s relative to the barge, determine the
speed of the barge just before the automobiles collide. How
much time does this take? Originally the barge is at rest.
Neglect water resistance.
30 ft
A
B
C
SOLUTION
+ B
A;
vA = vC + vA>C = vC - 6
+ B
A;
vB = vC + vB>C = vC + 6
+ B
A;
©mv1 = ©mv2
0 = mA(vC - 6) + mB(vC + 6) + mCvC
0 = a
3000
45 000
4000
b(vC - 6) + a
b (vC + 6) + a
bvC
32.2
32.2
32.2
vC = 0.1154 ft>s = 0.115 ft>s
laws
or
Ans.
in
teaching
Web)
For A:
v = v0 + act
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A:
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6 = 0 + 4tA
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is
tA = 1.5 s
work
student
the
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protected
1
(4)(1.5)2 = 4.5 ft
2
this
assessing
integrity
of
and
work
s = 0 + 0 +
1 2
at
2 c
for
s = s0 + v0t +
is
+ B
A:
part
the
is
This
provided
For B:
any
courses
v = v0 + ac t
destroy
of
and
+ B
A;
sale
will
their
6 = 0 + 8tB
tB = 0.75 s
+ B
A;
s = s0 + v0 t +
s = 0 + 0 +
1 2
at
2 c
1
(8)(0.75)2 = 2.25 ft
2
For the remaining (1.5 - 0.75) s = 0.75 s
s = vt = 6(0.75) = 4.5 ft
Thus,
s = 30 - 4.5 - 4.5 - 2.25 = 18.75 ft
t¿ =
18.75>2
s>2
=
= 1.5625
v
6
t = 1.5 + 1.5625 = 3.06 s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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Ans.
15–49.
The man M weighs 150 lb and jumps onto the boat B which
has a weight of 200 lb. If he has a horizontal component of
velocity relative to the boat of 3 ft>s, just before he enters
the boat, and the boat is traveling vB = 2 ft>s away from the
pier when he makes the jump, determine the resulting
velocity of the man and boat.
M
B
SOLUTION
+ )
(:
vM = vB + vM>B
vM = 2 + 3
vM = 5 ft>s
+ )
(:
©m v1 = ©m v2
200
350
150
152 +
122 =
1vB22
32.2
32.2
32.2
1vB22 = 3.29 ft>s
sale
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This
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the
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the
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of
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
15–50.
The man M weighs 150 lb and jumps onto the boat B which
is originally at rest. If he has a horizontal component of
velocity of 3 ft>s just before he enters the boat, determine
the weight of the boat if it has a velocity of 2 ft>s once the
man enters it.
M
B
SOLUTION
+ )
(:
vM = vB + vM/B
vM = 0 + 3
vM = 3 ft>s
+ )
(:
©m1v12 = ©m1v22
1WB + 1502
WB
150
132 +
102 =
122
32.2
32.2
32.2
WB = 75 lb
sale
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This
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the
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or transmission in any form or by any means, electronic, mechanical,
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15–51.
1.5 m/s
The 20-kg package has a speed of 1.5 m>s when it is delivered to the smooth ramp. After sliding down the ramp it
lands onto a 10-kg cart as shown. Determine the speed of the
cart and package after the package stops sliding on the cart.
2m
SOLUTION
Conservation of Energy: With reference to the datum set in Fig. a, the gravitational
potential energy of the package at positions (1) and (2) are
A Vg B 1 = mgh1 = 20(9.81)(2) = 392.4 J and A Vg B 2 = mgh2 = 0.
T1 + V1 = T2 + V2
1
1
m(vP)12 + (Vg)1 = m(vP)22 + (Vg)2
2
2
laws
or
1
1
(20)(1.52) + 392.4 = (20)(vP)22 + 0
2
2
in
teaching
Web)
(vP)2 2 = 6.441 m>s ;
use
United
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is
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by
and
mP(vP)2 + mC(vC)1 = (mP + mC)v
work
student
the
+ )
(;
of
protected
the
(including
for
20(6.441) + 0 = (20 + 10)v
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This
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v = 4.29 m>s ;
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Conservation of Linear Momentum: Referring to the free-body diagram of the
package and cart system shown in Fig. b, the linear momentum is conserved along
the x axis since no impulsive force acts along it. The package stops sliding on the cart
when they move with a common speed. At this instant,
*15–52.
The free-rolling ramp has a mass of 40 kg. A 10-kg crate is
released from rest at A and slides down 3.5 m to point B. If
the surface of the ramp is smooth, determine the ramp’s
speed when the crate reaches B. Also, what is the velocity of
the crate?
3.5 m
A
30
SOLUTION
Conservation of Energy: The datum is set at lowest point B. When the crate is at
point A, it is 3.5 sin 30° = 1.75 m above the datum. Its gravitational potential energy
is 1019.81211.752 = 171.675 N # m. Applying Eq. 14–21, we have
T1 + V1 = T2 + V2
0 + 171.675 =
1
1
1102v2C + 1402v2R
2
2
171.675 = 5 v2C + 20 v2R
(1)
laws
or
Relative Velocity: The velocity of the crate is given by
in
teaching
Web)
vC = vR + vC>R
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= -vRi + 1vC>R cos 30°i - vC>R sin 30°j2
permitted.
(2)
learning.
is
of
the
not
instructors
States
World
= 10.8660 vC>R - vR2i - 0.5 vC>Rj
by
and
use
United
on
The magnitude of vC is
(including
for
work
student
the
vC = 2(0.8660 vC>R - vR22 + 1-0.5 vC>R22
(3)
assessing
is
work
solely
of
protected
the
= 2v2C>R + v2R - 1.732 vR vC>R
destroy
of
and
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the
is
This
provided
integrity
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this
Conservation of Linear Momentum: If we consider the crate and the ramp as a
system, from the FBD, one realizes that the normal reaction NC (impulsive force) is
internal to the system and will cancel each other. As the result, the linear momentum
is conserved along the x axis.
sale
will
their
0 = mC 1vC2x + mR vR
+ )
(:
0 = 1010.8660 vC>R - vR2 + 401-vR2
0 = 8.660 vC>R - 50 vR
(4)
Solving Eqs. (1), (3), and (4) yields
vR = 1.101 m>s = 1.10 m>s vC = 5.43 m>s
Ans.
vC>R = 6.356 m>s
From Eq. (2)
vC = 30.866016.3562 - 1.1014i - 0.516.3562j = 54.403i - 3.178j6 m>s
Thus, the directional angle f of vC is
f = tan - 1
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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3.178
= 35.8°
4.403
cf
Ans.
B
15–53.
20 ft
The 80-lb boy and 60-lb girl walk towards each other with a
constant speed on the 300-lb cart. If their velocities, measured relative to the cart, are 3 ft>s to the right and 2 ft>s to
the left, respectively, determine the velocities of the boy and
girl during the motion. Also, find the distance the cart has
traveled at the instant the boy and girl meet.
A
SOLUTION
Conservation of Linear Momentum: From the free-body diagram of the boy, girl, and
cart shown in Fig. a, the pairs of impulsive forces F1 and F2 generated during the walk
cancel each other since they are internal to the system. Thus, the resultant of the
impulsive forces along the x axis is zero, and the linear momentum of the system is
conserved along the x axis.
+ )
(:
©mv1 = ©mv2
0 + 0 + 0 =
60
300
80
v (v ) v
32.2 b
32.2 g
32.2 c
(1)
80vb - 60vg - 300vc = 0
laws
or
Kinematics: Applying the relative velocity equation and considering the motion of
the boy,
vb = - vc + 3
(2)
permitted.
Dissemination
copyright
+ )
(:
Wide
in
teaching
Web)
vb = vc + vb>c
learning.
is
of
the
not
instructors
States
World
For the girl,
the
by
and
use
United
on
vg = vc + vg>c
work
student
-vg = - vc - 2
the
(including
for
+ )
(:
(3)
assessing
is
work
solely
of
protected
vg = vc + 2
provided
integrity
of
and
work
this
Solving Eqs. (1), (2), and (3), yields
Ans.
and
any
courses
part
the
is
This
vb = 2.727 ft>s = 2.73 ft>s :
Ans.
vc = 0.2727 ft>s ;
sale
will
their
destroy
of
vg = 2.273 ft>s = 2.27 ft>s ;
The velocity of the girl relative to the boy can be determined from
vg = vb + vg>b
+ )
(:
- 2.273 = 2.727 + vg>b
vg>b = - 5 ft>s = 5 ft>s ;
Here, sg>b = 20 ft and vg>b = 5 ft>s is constant. Thus,
+ )
(;
sg>b = (sg>b)0 + vg>b t
20 = 0 + 5t
t = 4s
Thus, the distance the cart travels is given by
+ )
(;
sc = (sc)0 + vc t
= 0 + 0.2727(4)
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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= 1.09 ft
Ans.
A
15–54.
20 ft
The 80-lb boy and 60-lb girl walk towards each other with
constant speed on the 300-lb cart. If their velocities measured relative to the cart are 3 ft>s to the right and 2 ft>s to
the left, respectively, determine the velocity of the cart while
they are walking.
A
SOLUTION
Conservation of Linear Momentum: From the free-body diagram of the body, girl,
and cart shown in Fig. a, the pairs of impulsive forces F1 and F2 generated during the
walk cancel each other since they are internal to the system. Thus, the resultant of the
impulsive forces along the x axis is zero, and the linear momentum of the system is
conserved along the x axis.
+ )
(:
©mv1 = ©mv2
0 + 0 + 0 =
60
300
80
v (v ) v
32.2 b
32.2 g
32.2 c
(1)
80vb - 60vg - 300vc = 0
laws
or
Kinematics: Applying the relative velocity equation and considering the motion of
the boy,
in
teaching
Web)
vb = vc + vb>c
permitted.
Dissemination
Wide
(2)
copyright
vb = - vc + 3
+ )
(:
learning.
is
of
the
not
instructors
States
World
For the girl,
the
by
and
use
United
on
vg = vc + vg>c
work
student
-vg = - vc - 2
the
(including
for
+ )
(:
(3)
assessing
is
work
solely
of
protected
vg = vc + 2
part
the
is
and
any
courses
vb = 2.727 ft>s :
This
provided
integrity
of
and
work
this
Solving Eqs. (1), (2), and (3), yields
vc = 0.2727 ft>s = 0.273 ft>s ;
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sale
will
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destroy
of
vg = 2.273 ft>s ;
Ans.
A
15–55.
A tugboat T having a mass of 19 Mg is tied to a barge B
having a mass of 75 Mg. If the rope is “elastic” such that it
has a stiffness k = 600 kN>m, determine the maximum
stretch in the rope during the initial towing. Originally both the
tugboat and barge are moving in the same direction with
speeds 1vT21 = 15 km>h and 1vB21 = 10 km>h, respectively.
Neglect the resistance of the water.
(vB)1
B
SOLUTION
(vT)1 = 15 km>h = 4.167 m>s
(vB)1 = 10 km>h = 2.778 m>s
When the rope is stretched to its maximum, both the tug and barge have a common
velocity. Hence,
+ )
(:
©mv1 = ©mv2
19 000(4.167) + 75 000(2.778) = (19 000 + 75 000)v2
v2 = 3.059 m>s
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
1
(19 000 + 75 000)(3.059)2 = 439.661 kJ
2
teaching
T2 =
in
1
1
(19 000)(4.167)2 + (75 000)(2.778)2 = 454.282 kJ
2
2
copyright
T1 =
Web)
laws
or
T1 + V1 = T2 + V2
work
student
the
by
and
Hence,
is
work
solely
of
protected
the
(including
for
1
(600)(103)x 2
2
454.282(103) + 0 = 439.661(103) +
assessing
x = 0.221 m
this
Ans.
integrity
of
and
work
provided
any
courses
part
the
is
This
destroy
of
and
will
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sale
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
(vT)1
T
*15–56.
Two boxes A and B, each having a weight of 160 lb, sit on
the 500-lb conveyor which is free to roll on the ground. If
the belt starts from rest and begins to run with a speed of
3 ft>s, determine the final speed of the conveyor if (a) the
boxes are not stacked and A falls off then B falls off, and (b)
A is stacked on top of B and both fall off together.
A
B
SOLUTION
a) Let vb be the velocity of A and B.
+ b
a:
©mv1 = ©mv2
0 = a
+ b
a:
320
500
b (vb) - a
b(vc)
32.2
32.2
vb = vc + vb>c
vb = -vc + 3
Thus, vb = 1.83 ft>s :
vc = 1.17 ft>s ;
in
teaching
Web)
laws
or
When a box falls off, it exerts no impulse on the conveyor, and so does not alter the
momentum of the conveyor. Thus,
Dissemination
Wide
Ans.
copyright
a) vc = 1.17 ft>s ;
b) vc = 1.17 ft>s ;
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and
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is
This
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this
assessing
is
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the
(including
for
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student
the
by
and
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on
learning.
is
of
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not
instructors
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permitted.
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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15–57.
The 10-kg block is held at rest on the smooth inclined plane
by the stop block at A. If the 10-g bullet is traveling at
300 m>s when it becomes embedded in the 10-kg block,
determine the distance the block will slide up along the
plane before momentarily stopping.
300 m/s
A
30
SOLUTION
Conservation of Linear Momentum: If we consider the block and the bullet as a
system, then from the FBD, the impulsive force F caused by the impact is internal
to the system. Therefore, it will cancel out. Also, the weight of the bullet and the
block are nonimpulsive forces. As the result, linear momentum is conserved along
the x œ axis.
mb(vb)x¿ = (mb + mB) vx œ
0.01(300 cos 30°) = (0.01 + 10) v
v = 0.2595 m>s
Wide
copyright
in
teaching
Web)
laws
or
Conservation of Energy: The datum is set at the blocks initial position. When the
block and the embedded bullet is at their highest point they are h above the datum.
Their gravitational potential energy is (10 + 0.01)(9.81)h = 98.1981h. Applying
Eq. 14–21, we have
States
World
permitted.
Dissemination
T1 + V1 = T2 + V2
United
on
learning.
is
of
the
not
instructors
1
(10 + 0.01) A 0.25952 B = 0 + 98.1981h
2
by
and
use
0 +
(including
for
work
student
the
h = 0.003433 m = 3.43 mm
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This
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protected
the
d = 3.43 > sin 30° = 6.87 mm
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Ans.
15–58.
A ball having a mass of 200 g is released from rest at a
height of 400 mm above a very large fixed metal surface. If
the ball rebounds to a height of 325 mm above the surface,
determine the coefficient of restitution between the ball
and the surface.
SOLUTION
Before impact
T1 + V1 = T2 + V2
0 + 0.2(9.81)(0.4) =
1
(0.2)v21 + 0
2
v1 = 2.801 m>s
After the impact
or
1
(0.2)v22 = 0 + 0.2(9.81)(0.325)
2
teaching
Web)
laws
v2 = 2.525 m>s
Wide
copyright
in
Coefficient of restitution:
Dissemination
(vA)2 - (vB)2
(vB)1 - (vA)1
0 - ( - 2.525)
=
2.801 - 0
by
and
use
United
on
learning.
is
of
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not
instructors
States
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permitted.
e =
the
(+ T )
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and
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courses
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is
This
provided
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this
assessing
is
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protected
the
(including
for
work
student
= 0.901
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
15–59.
The 5-Mg truck and 2-Mg car are traveling with the freerolling velocities shown just before they collide. After the
collision, the car moves with a velocity of 15 km>h to the
right relative to the truck. Determine the coefficient of
restitution between the truck and car and the loss of energy
due to the collision.
30 km/h
10 km/h
SOLUTION
Conservation of Linear Momentum: The linear momentum of the system is
conserved along the x axis (line of impact).
1h
m
da
b = 8.333 m>s
h 3600 s
The initial speeds of the truck and car are (vt)1 = c30 A 103 B
and (vc)1 = c10 A 103 B
m
1h
da
b = 2.778 m>s.
h 3600 s
By referring to Fig. a,
mt A vt B 1 + mc A vc B 1 = mt A vt B 2 + mc A vc B 2
or
+ b
a:
teaching
Web)
laws
5000(8.333) + 2000(2.778) = 5000 A vt B 2 + 2000 A vc B 2
(1)
1h
m
da
b = 4.167 m>s : .
h 3600 s
of
the
not
instructors
States
World
Coefficient of Restitution: Here, (vc>t) = c15 A 103 B
the
by
and
use
United
on
learning.
is
Applying the relative velocity equation,
(vc)2 = (vt)2 + 4.167
assessing
is
work
solely
of
protected
+ B
A:
the
(including
for
work
student
(vc)2 = (vt)2 + (vc>t)2
this
(2)
provided
integrity
of
and
work
(vc)2 - (vt)2 = 4.167
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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(vc)2 - (vt)2
8.333 - 2.778
destroy
e =
will
(vc)2 - (vt)2
(vt)1 - (vc)1
their
e =
sale
+ B
A:
of
and
any
courses
part
the
is
This
Applying the coefficient of restitution equation,
(3)
permitted.
Dissemination
Wide
copyright
in
5 A vt B 2 + 2 A vc B 2 = 47.22
15–59. continued
Substituting Eq. (2) into Eq. (3),
e =
4.167
= 0.75
8.333 - 2.778
Ans.
Solving Eqs. (1) and (2) yields
(vt)2 = 5.556 m>s
(vc)2 = 9.722 m>s
Kinetic Energy: The kinetic energy of the system just before and just after the
collision are
T1 =
=
1
1
mt(vt)1 2 + mc(vc)1 2
2
2
1
1
(5000)(8.3332) + (2000)(2.7782)
2
2
= 181.33 A 103 B J
or
1
1
m (v ) 2 + mc(vc)2 2
2 t t2
2
laws
T2 =
in
teaching
Web)
1
1
(5000)(5.5562) + (2000)(9.7222)
2
2
Wide
copyright
=
of
the
not
instructors
States
World
permitted.
Dissemination
= 171.68 A 103 B J
and
use
United
on
learning.
is
Thus,
for
work
student
the
by
¢E = T1 - T2 = 181.33 A 103 B - 171.68 A 103 B
is
work
solely
of
protected
the
(including
= 9.645 A 103 B J
Ans.
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destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
= 9.65 kJ
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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*15–60.
Disk A has a mass of 2 kg and is sliding forward on the
smooth surface with a velocity 1vA21 = 5 m/s when it strikes
the 4-kg disk B, which is sliding towards A at 1vB21 = 2 m/s,
with direct central impact. If the coefficient of restitution
between the disks is e = 0.4, compute the velocities of A
and B just after collision.
SOLUTION
Conservation of Momentum :
mA (vA)1 + mB (vB)1 = mA (vA)2 + mB (vB)2
+ B
A:
2(5) + 4(- 2) = 2(vA)2 + 4(vB)2
(1)
Coefficient of Restitution :
(vB)2 - (vA)2
(vA)1 - (vB)1
0.4 =
(vB)2 - (vA)2
5 - ( - 2)
(2)
laws
or
+ B
A:
e =
(vB)2 = 1.27m s :
Ans.
sale
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of
and
any
courses
part
the
is
This
provided
integrity
of
and
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this
assessing
is
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of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
copyright
(vA)2 = - 1.53 m s = 1.53 m s ;
Wide
in
teaching
Web)
Solving Eqs. (1) and (2) yields
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
(vA)1 = 5 m/s
(vB)1 = 2 m/s
A
B
15–61.
Block A has a mass of 3 kg and is sliding on a rough horizontal
surface with a velocity 1vA21 = 2 m>s when it makes a direct
collision with block B, which has a mass of 2 kg and is
originally at rest. If the collision is perfectly elastic 1e = 12,
determine the velocity of each block just after collision and
the distance between the blocks when they stop sliding. The
coefficient of kinetic friction between the blocks and the
plane is mk = 0.3.
(vA)1
A
SOLUTION
+ )
(:
a mv1 = a mv2
3122 + 0 = 31vA22 + 21vB22
+ )
(:
e =
1 =
1vB22 - 1vA222
1vA21 - 1vB21
1vB22 - 1vA22
2 - 0
Ans.
1vB22 = 2.40 m>s :
Ans.
Wide
copyright
in
teaching
Web)
laws
1vA22 = 0.400 m>s :
or
Solving
permitted.
Dissemination
Block A:
learning.
is
of
the
not
instructors
States
World
T1 + a U1 - 2 = T2
for
work
student
the
by
and
use
United
on
1
13210.40022 - 319.81210.32dA = 0
2
is
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solely
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protected
the
(including
dA = 0.0272 m
part
the
is
This
T1 + a U1 - 2 = T2
provided
integrity
of
and
work
this
assessing
Block B:
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1
12212.4022 - 219.81210.32dB = 0
2
dB = 0.9786 m
d = dB - dA = 0.951 m
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Ans.
B
15–62.
If two disks A and B have the same mass and are subjected
to direct central impact such that the collision is perfectly
elastic 1e = 12, prove that the kinetic energy before collision
equals the kinetic energy after collision. The surface upon
which they slide is smooth.
SOLUTION
+ B
A:
©m v1 = ©m v2
mA (vA)1 + mB (vB)1 = mA (vA)2 + mB (vB)2
mA C (vA)1 - (vA)2 D = mB C (vB)2 - (vB)1 D
+ B
A:
e =
(1)
(vB)2 - (vA)2
= 1
(vA)1 - (vB)1
(vB)2 - (vA)2 = (vA)1 - (vB)1
(2)
Combining Eqs. (1) and (2):
teaching
Web)
laws
or
mA C (vA)1 - (vA)2 D C (vA)1 + (vA)2 D = mB C (vB)2 - (vB)1 D C (vB)2 + (vB)1 D
Dissemination
Wide
copyright
in
1
Expand and multiply by :
2
States
World
permitted.
1
1
1
1
m (v )21 + mB (vB)21 = mA(vA)22 + mB (vB)22
2 A A
2
2
2
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This
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the
(including
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student
the
by
and
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is
of
the
not
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Q.E.D.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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15–63.
Each ball has a mass m and the coefficient of restitution
between the balls is e. If they are moving towards one another
with a velocity v, determine their speeds after collision. Also,
determine their common velocity when they reach the state of
maximum deformation. Neglect the size of each ball.
SOLUTION
+ B
A:
©m v1 = ©m v2
mv - mv = mvA + mvB
vA = - vB
+ B
A:
e =
(vB)2 - (vA)2
vB - vA
=
(vA)1 - (vB)1
v - (-v)
Ans.
vA = - ve = ve ;
Ans.
laws
vB = ve :
or
2ve = 2vB
teaching
in
©m v1 = ©m v2
permitted.
Dissemination
Wide
copyright
+ B
A:
Web)
At maximum deformation vA = vB = v¿ .
is
of
the
not
instructors
States
World
mv - mv = (2m) v¿
learning.
v¿ = 0
on
United
and
use
by
work
student
the
the
(including
for
work
solely
of
protected
this
assessing
is
integrity
of
and
work
provided
any
courses
part
the
is
This
destroy
of
and
will
their
sale
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
v
v
A
B
*15–64.
The three balls each have a mass m. If A has a speed v just
before a direct collision with B, determine the speed of C
after collision. The coefficient of restitution between each
pair of balls e. Neglect the size of each ball.
v
B
A
SOLUTION
Conservation of Momentum: When ball A strikes ball B, we have
mA (vA)1 + mB (vB)1 = mA (vA)2 + mB (vB)2
+ B
A:
my + 0 = m(vA)2 + m(vB)2
(1)
Coefficient of Restitution:
(vB)2 - (vA)2
(vA)1 - (vB)1
e =
(vB)2 - (vA)2
v - 0
(2)
laws
or
+ B
A:
e =
in
teaching
Web)
Solving Eqs. (1) and (2) yields
y(1 + e)
2
Wide
not
instructors
States
World
permitted.
(yB)2 =
Dissemination
y(1 - e)
2
copyright
(yA)2 =
use
United
on
learning.
is
of
the
Conservation of Momentum:When ball B strikes ball C, we have
work
student
the
by
and
mB (vB)2 + mC (vC)1 = mB (vB)3 + mC (vC)2
the
(including
for
v(1 + e)
d + 0 = m(vB)3 + m(vC)2
2
(3)
of
this
assessing
is
work
solely
mc
protected
+ B
A:
part
the
is
any
courses
e =
will
(vC)2 - (vB)3
v(1 + e)
- 0
2
sale
+ B
A:
their
destroy
of
(vC)2 - (vB)3
(vB)2 - (vC)1
and
e =
This
provided
integrity
of
and
work
Coefficient of Restitution:
(4)
Solving Eqs. (3) and (4) yields
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
(vC)2 =
v(1 + e)2
4
(vB)3 =
v(1 - e2)
4
Ans.
C
15–65.
A 1-lb ball A is traveling horizontally at 20 ft>s when it
strikes a 10-lb block B that is at rest. If the coefficient of
restitution between A and B is e = 0.6, and the coefficient
of kinetic friction between the plane and the block is
mk = 0.4, determine the time for the block B to stop sliding.
SOLUTION
+ b
a:
©m1 v1 = ©m2 v2
a
1
10
1
b (20) + 0 = a
b(vA)2 + a
b(vB)2
32.2
32.2
32.2
(vA)2 + 10(vB)2 = 20
+ b
a:
e =
(vB)2 - (vA)2
(vA)1 - (vB)1
0.6 =
(vB)2 - (vA)2
20 - 0
teaching
Web)
laws
or
(vB)2 - (vA)2 = 12
Wide
copyright
in
Thus,
States
World
permitted.
Dissemination
(vB)2 = 2.909 ft>s :
use
United
on
learning.
is
of
the
not
instructors
(vA)2 = -9.091 ft>s = 9.091 ft>s ;
for
work
student
the
by
and
Block B:
the
(including
F dt = m v2
of
work
solely
assessing
L
protected
m v1 + ©
is
+ b
a:
provided
integrity
of
and
work
this
10
b(2.909) - 4t = 0
32.2
part
the
is
This
a
any
courses
Ans.
sale
will
their
destroy
of
and
t = 0.226 s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
15–66.
If the girl throws the ball with a horizontal velocity of 8 ft>s,
determine the distance d so that the ball bounces once on
the smooth surface and then lands in the cup at C. Take
e = 0.8.
A
vA
3 ft
C
B
d
SOLUTION
(+ T )
v2 = v20 + 2ac1s - s02
(v122y = 0 + 2132.22132
1v12y = 13.90 T
(+ T)
s = s0 + v0 t +
3 = 0 + 0 +
1 2
ac t
2
1
132.221tAB22
2
or
tAB = 0.43167 s
laws
1v22y
teaching
in
1v22y
Wide
Dissemination
World
permitted.
13.90
of
the
not
instructors
0.8 =
Web)
1v12y
copyright
e =
States
(+ T )
by
and
use
United
on
learning.
is
1v22y = 11.1197 c
v = v0 + ac t
for
work
student
the
(+ T)
is
work
solely
of
protected
the
(including
11.1197 = -11.1197 + 32.21tBC2
and
any
courses
part
the
is
This
Total time is tAC = 1.1224 s
provided
integrity
of
and
work
this
assessing
tBC = 0.6907 s
sale
d = vA1tAC2
will
their
destroy
of
Since the x component of momentum is conserved
d = 811.12242
d = 8.98 ft
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
15–67.
The three balls each weigh 0.5 lb and have a coefficient of
restitution of e = 0.85. If ball A is released from rest and
strikes ball B and then ball B strikes ball C, determine the
velocity of each ball after the second collision has occurred.
The balls slide without friction.
A
r
3 ft
B
SOLUTION
Ball A:
Datum at lowest point.
T1 + V1 = T2 + V2
0 + 10.52132 =
1 0.5
1
21vA221 + 0
2 32.2
1vA21 = 13.90 ft>s
Balls A and B:
+ B
A:
laws
in
teaching
Web)
0.5
0.5
0.5
2113.902 + 0 = 1
21vA22 + 1
21vB22
32.2
32.2
32.2
Dissemination
Wide
copyright
1
or
©mv1 = ©mv 2
World
permitted.
1vB22 - 1vA222
States
the
not
instructors
1vA21 - 1vB21
is
on
learning.
e =
of
+ B
A:
by
and
use
United
1vB22 - 1vA22
work
student
the
13.90 - 0
the
(including
for
0.85 =
assessing
is
work
solely
of
protected
Solving:
this
Ans.
provided
integrity
of
and
work
1vA22 = 1.04 ft>s
and
any
courses
part
the
is
This
1vB22 = 12.86 ft>s
their
destroy
of
Balls B and C:
will
+ B
A:
1
+ B
A:
sale
©mv 2 = ©mv 3
0.5
0.5
0.5
2112.862 + 0 = 1
21v 2 + 1
21v 2
32.2
32.2 B 3
32.2 C 3
e =
0.85 =
1vC23 - 1vB23
1vB22 - 1vC22
1vC23 - 1vB23
12.86 - 0
Solving:
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
1vB23 = 0.964 ft>s
Ans.
1vC23 = 11.9 ft>s
Ans.
C
*15–68.
The girl throws the ball with a horizontal velocity of
v1 = 8 ft>s. If the coefficient of restitution between the ball
and the ground is e = 0.8, determine (a) the velocity of the
ball just after it rebounds from the ground and (b) the
maximum height to which the ball rises after the first
bounce.
v1
3 ft
h
SOLUTION
Kinematics: By considering the vertical motion of the falling ball, we have
( + T)
(v1)2y = (v0)2y + 2ac C sy - (s0)y D
(v1)2y = 02 + 2(32.2)(3 - 0)
(v1)y = 13.90 ft>s
Coefficient of Restitution (y):
(v1)y - A vg B 1
0 - (v2)y
or
0.8 =
-13.90 - 0
teaching
Web)
(+ c)
A vg B 2 - (v2)y
laws
e =
World
not
instructors
States
Conservation of “x” Momentum: The momentum is conserved along the x axis.
permitted.
Dissemination
Wide
copyright
in
(v2)y = 11.12 ft>s
on
learning.
is
of
the
(vx)2 = 8 ft>s :
United
m(vx)1 = m(vx)2 ;
by
and
use
+ B
A:
for
work
student
the
The magnitude and the direction of the rebounding velocity for the ball is
Ans.
this
Ans.
part
the
is
This
provided
integrity
of
and
work
11.12
b = 54.3°
8
assessing
is
work
solely
of
protected
the
(including
v2 = 2(vx)22 + A vy B 22 = 282 + 11.122 = 13.7 ft>s
u = tan - 1 a
sale
(v)2y = (v2)2y + 2ac C sy - (s2)y D
will
their
destroy
of
and
any
courses
Kinematics: By considering the vertical motion of tfie ball after it rebounds from
the ground, we have
(+ c)
0 = 11.122 + 2( - 32.2)(h - 0)
h = 1.92 ft
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
8 ft/s
Ans.
15–69.
A 300-g ball is kicked with a velocity of vA = 25 m>s at point A
as shown. If the coefficient of restitution between the ball and
the field is e = 0.4, determine the magnitude and direction u
of the velocity of the rebounding ball at B.
vA
v¿B
30
u
B
A
SOLUTION
Kinematics: The parabolic trajectory of the football is shown in Fig. a. Due to the
symmetrical properties of the trajectory, vB = vA = 25 m>s and f = 30°.
Conservation of Linear Momentum: Since no impulsive force acts on the football
along the x axis, the linear momentum of the football is conserved along the x axis.
+ b
a;
m A vB B x = m A v Bœ B x
0.3 A 25 cos 30° B = 0.3 Av Bœ B x
A v Bœ B x = 21.65 m>s ;
laws
or
Coefficient of Restitution: Since the ground does not move during the impact, the
coefficient of restitution can be written as
Web)
teaching
in
copyright
permitted.
Dissemination
World
not
instructors
the
- 25 sin 30°
is
on
learning.
and
use
United
0.4 =
Wide
A vB B y - 0
- A v Bœ B y
States
e =
0 - A v Bœ B y
of
A+cB
for
work
student
the
by
A v Bœ B y = 5 m>s c
is
work
solely
of
protected
the
(including
Thus, the magnitude of vBœ is
Ans.
integrity
of
and
work
this
assessing
v Bœ = 2 A v Bœ B x + A vBœ B y = 221.652 + 52 = 22.2 m>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
any
5
≤ = 13.0°
21.65
destroy
of
and
S = tan - 1 ¢
will
B
courses
part
the
is
This
A
v Bœ x
their
A v Bœ B y
sale
u = tan - 1 C
provided
and the angle of vBœ is
Ans.
25 m>s
15–70.
Two smooth spheres A and B each have a mass m. If A is
given a velocity of v0, while sphere B is at rest, determine
the velocity of B just after it strikes the wall. The coefficient
of restitution for any collision is e.
v0
A
SOLUTION
Impact: The first impact occurs when sphere A strikes sphere B. When this occurs, the
linear momentum of the system is conserved along the x axis (line of impact).
Referring to Fig. a,
+ )
(:
mAvA + mBvB = mA(vA)1 + mB(vB)1
mv0 + 0 = m(vA)1 + m(vB)1
(vA)1 + (vB)1 = v0
(vB)1 - (vA)1
vA - vB
e =
(vB)1 - (vA)1
v0 - 0
or
e =
laws
+ )
(:
(1)
(vB)1 - (vA)1 = ev0
in
teaching
Web)
(2)
Dissemination
Wide
copyright
Solving Eqs. (1) and (2) yields
permitted.
1 - e
b v0 :
2
World
(vA)1 = a
the
not
instructors
States
is
and
use
United
on
learning.
1 + e
b v0 :
2
of
(vB)1 = a
for
work
student
the
by
The second impact occurs when sphere B strikes the wall, Fig. b. Since the wall does
not move during the impact, the coefficient of restitution can be written as
work
solely
of
protected
the
(including
0 - C - (vB)2 D
(vB)1 - 0
this
assessing
e =
is
+ )
(:
integrity
of
and
provided
part
the
is
any
courses
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
will
e(1 + e)
v0
2
sale
(vB)2 =
their
destroy
of
1 + e
d v0 - 0
2
This
c
work
0 + (vB)2
and
e =
Ans.
B
15–71.
It was observed that a tennis ball when served horizontally
7.5 ft above the ground strikes the smooth ground at B 20 ft
away. Determine the initial velocity vA of the ball and the
velocity vB (and u) of the ball just after it strikes the court at
B. Take e = 0.7.
vA
A
7.5 ft
vB
u
20 ft
SOLUTION
+ )
(:
s = s0 + v0t
20 = 0 + vA t
(+ T)
s = s0 + v0 t +
1 2
a t
2 c
7.5 = 0 + 0 +
1
(32.2)t 2
2
t = 0.682524
vA = 29.303 = 29.3 ft>s
or
Ans.
teaching
Web)
laws
vB x 1 = 29.303 ft>s
Wide
copyright
in
(+ T) v = v0 + ac t
the
not
mv1 = mv 2
use
United
on
learning.
is
of
+ )
(:
instructors
States
World
permitted.
Dissemination
vBy 1 = 0 + 32.2(0.68252) = 21.977 ft>s
work
student
the
by
and
vB2x = vB 1x = 29.303 ft>s :
the
(including
for
vBy 2
work
solely
of
protected
vBy 1
21.977
vBy 2 = 15.384 ft>s c
integrity
of
and
work
,
provided
vBy 2
part
the
is
This
0.7 =
this
assessing
is
e =
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
will
15.384
= 27.7°
29.303
au
sale
u = tan - 1
their
destroy
of
and
any
courses
vB 2 = 2(29.303)2 + (15.384)2 = 33.1 ft>s
Ans.
B
*15–72.
The tennis ball is struck with a horizontal velocity vA ,
strikes the smooth ground at B, and bounces upward at
u = 30°. Determine the initial velocity vA , the final velocity
vB , and the coefficient of restitution between the ball and
the ground.
vA
A
7.5 ft
vB
u
20 ft
SOLUTION
(+ T)
v2 = v20 + 2 ac (s - s0)
(vBy)21 = 0 + 2(32.2)(7.5 - 0)
vBy1 = 21.9773 m>s
(+ T)
v = v0 + ac t
21.9773 = 0 + 32.2 t
t = 0.68252 s
( + T)
s = s0 + v0 t
Web)
laws
or
20 = 0 + vA (0.68252)
in
Ans.
Wide
copyright
mv1 = mv 2
States
World
permitted.
Dissemination
+ )
(:
teaching
vA = 29.303 = 29.3 ft>s
on
learning.
is
of
the
not
instructors
vB x 2 = vB x 1 = vA = 29.303
Ans.
the
by
and
use
United
vB2 = 29.303>cos 30° = 33.8 ft>s
the
(including
for
work
student
vBy 2 = 29.303 tan 30° = 16.918 ft>s
work
solely
of
protected
16.918
= 0.770
21.9773
this
=
assessing
vBy 1
is
vBy 2
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
e =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
B
15–73.
The 1 lb ball is dropped from rest and falls a distance of 4 ft
before striking the smooth plane at A. If e = 0.8, determine
the distance d to where it again strikes the plane at B.
4 ft
A
d
SOLUTION
T1 + V1 = T2 + V2
B
3
5
4
0 + 0 =
1
1m21vA221 - m132.22142
2
1vA21 = 22132.22142 = 16.05 ft>s
R+ 1vA22x =
3
116.052 = 9.63 ft>s
5
4
Q+ 1vA22y = 0.8 a b116.052 = 10.27 ft>s
5
Web)
laws
or
1vA22 = 2(9.6322 + 110.2722 = 14.08 ft>s
in
teaching
10.27
b = 46.85°
9.63
Dissemination
Wide
copyright
u = tan - 1 a
on
learning.
is
of
the
not
instructors
States
World
permitted.
3
f = 46.85° - tan - 1 a b = 9.977°
4
work
student
the
by
and
use
United
+ ) s = s + v t
(:
0
0
assessing
is
work
solely
of
protected
the
(including
for
4
da b = 0 + 14.08 cos 9.977° 1t2
5
provided
integrity
of
and
work
this
1 2
at
2 c
part
the
is
This
( + T) s = s0 + v0 t +
t = 0.798 s
d = 13.8 ft
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
sale
will
their
destroy
of
and
any
courses
1
3
da b = 0 - 14.08 sin 9.977° 1t2 + 132.22t2
5
2
Ans.
15–74.
The 1 lb ball is dropped from rest and falls a distance of 4 ft
before striking the smooth plane at A. If it rebounds and in
t = 0.5 s again strikes the plane at B, determine the
coefficient of restitution e between the ball and the plane.
Also, what is the distance d?
4 ft
A
d
SOLUTION
T1 + V1 = T2 + V2
B
3
5
4
0 + 0 =
1
1m21vA221 - m132.22142
2
1vA21 = 22132.22142 = 16.05 ft>s
3
116.052 = 9.63 ft>s
5
+R
1vA22x¿ =
Q+
4
1vA22y¿ = ea b 116.052 = 12.84e ft>s
5
or
s = s0 + v0 t
teaching
Web)
laws
+ )
(:
permitted.
Dissemination
Wide
copyright
in
4
1d2 = 0 + vA2x10.52
5
the
not
instructors
States
World
1 2
a t
2 c
is
and
use
United
on
learning.
s = s0 + v0 t +
of
(+ T)
the
(including
for
work
student
the
by
3
1
1d2 = 0 - vA2y 10.52 + 132.2210.522
5
2
3
4
4
0.5 c9.63 a b + 12.84e a b d = d
5
5
5
(+ c)
4
3
3
0.5 c -9.63 a b + 12.84e a b d = 4.025 - d
5
5
5
sale
will
Solving,
their
destroy
of
and
any
courses
part
the
is
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provided
integrity
of
and
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this
assessing
is
work
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of
protected
+ )
(:
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
e = 0.502
Ans.
d = 7.23 ft
Ans.
15–75.
The 1-kg ball is dropped from rest at point A, 2 m above the
smooth plane. If the coefficient of restitution between the
ball and the plane is e = 0.6, determine the distance d
where the ball again strikes the plane.
A
2m
B
SOLUTION
30⬚
Conservation of Energy: By considering the ball’s fall from position (1) to position (2)
as shown in Fig. a,
d
TA + VA = TB + VB
1
1
2
2
m v + (Vg)A = mBvB + (Vg)B
2 A A
2
1
2
0 + 1(9.81)(2) = (1)vB + 0
2
vB = 6.264 m>s T
laws
or
Conservation of Linear Momentum: Since no impulsive force acts on the ball along
the inclined plane (x¿ axis) during the impact, linear momentum of the ball is conserved along the x¿ axis. Referring to Fig. b,
in
teaching
Web)
mB(vB)x¿ = mB(v¿B)x¿
Dissemination
Wide
copyright
1(6.264) sin 30° = 1(v¿B) cos u
v¿B cos u = 3.1321
the
by
and
use
United
on
learning.
Coefficient of Restitution: Since the inclined plane does not move during the
impact,
is
of
the
not
instructors
States
World
permitted.
(1)
the
(including
for
work
student
0 - (v¿B)y¿
(v¿B)y¿ - 0
assessing
is
work
solely
of
protected
e =
provided
integrity
of
and
work
this
0 - v¿B sin u
- 6.264 cos 30° - 0
part
the
is
This
0.6 =
(2)
their
destroy
of
and
any
courses
v¿B sin u = 3.2550
u = 46.10°
sale
will
Solving Eqs. (1) and (2) yields
v¿B = 4.517 m>s
Kinematics: By considering the x and y motion of the ball after the impact, Fig. c,
+ )
(:
sx = (s0)x + (v¿B)x t
d cos 30° = 0 + 4.517 cos 16.10°t
t = 0.1995d
(+ c )
sy = (s0)y + (v¿B)y t +
(3)
1 2
at
2 y
- d sin 30° = 0 + 4.517 sin 16.10°t +
4.905t2 - 1.2528t - 0.5d = 0
1
( -9.81)t2
2
(4)
Solving Eqs. (3) and (4) yields
d = 3.84 m
t = 0.7663 s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Ans.
C
*15–76.
A ball of mass m is dropped vertically from a height h0
above the ground. If it rebounds to a height of h1, determine
the coefficient of restitution between the ball and the
ground.
h0
h1
SOLUTION
Conservation of Energy: First, consider the ball’s fall from position A to position B.
Referring to Fig. a,
TA + VA = TB + VB
1
1
2
2
mvA + (Vg)A = mvB + (Vg)B
2
2
1
2
0 + mg(h0) = m(vB)1 + 0
2
Subsequently, the ball’s return from position B to position C will be considered.
TB + VB = TC + VC
laws
or
1
1
2
2
mvB + (Vg)B = mvC + (Vg)C
2
2
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
1
2
m(vB)2 + 0 = 0 + mgh1
2
(vB)2 = 22gh1 c
and
use
United
on
learning.
is
of
the
Coefficient of Restitution: Since the ground does not move,
the
by
(vB)2
(vB)1
work
student
(including
solely
of
protected
the
e = -
for
(+ c )
is
work
22gh1
assessing
this
integrity
of
and
part
the
is
any
courses
destroy
of
and
will
their
sale
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
work
- 22gh0
h1
A h0
provided
=
This
e = -
15–77.
The cue ball A is given an initial velocity (vA)1 = 5 m>s. If
it makes a direct collision with ball B (e = 0.8), determine
the velocity of B and the angle u just after it rebounds from
the cushion at C (e¿ = 0.6). Each ball has a mass of 0.4 kg.
Neglect the size of each ball.
(vA)1 ⫽ 5 m/s
A
B
30⬚
C
u
SOLUTION
Conservation of Momentum: When ball A strikes ball B, we have
mA(vA)1 + mB(vB)1 = mA(vA)2 + mB(vB)2
0.4(5) + 0 = 0.4(vA)2 + 0.4(vB)2
(1)
Coefficient of Restitution:
e =
0.8 =
(vB)2 - (vA)2
5 - 0
(2)
laws
or
+)
(;
(vB)2 - (vA)2
(vA)1 - (vB)1
Wide
copyright
in
teaching
Web)
Solving Eqs. (1) and (2) yields
(vB)2 = 4.50 m>s
not
instructors
States
World
permitted.
Dissemination
(vA)2 = 0.500 m>s
the
by
and
use
United
on
learning.
is
of
the
Conservation of “y” Momentum: When ball B strikes the cushion at C, we have
of
protected
the
(including
for
work
student
mB(vBy)2 = mB(vBy)3
work
solely
0.4(4.50 sin 30°) = 0.4(vB)3 sin u
this
assessing
is
(+ T )
(3)
part
the
is
This
provided
integrity
of
and
work
(vB)3 sin u = 2.25
and
any
courses
Coefficient of Restitution (x):
+ )
(;
(vBx)2 - (vC)1
0.6 =
0 - [- (vB)3 cos u]
4.50 cos 30° - 0
will
their
destroy
of
(vC)2 - (vBx)3
sale
e =
(4)
Solving Eqs. (1) and (2) yields
(vB)3 = 3.24 m>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
u = 43.9°
Ans.
15–78.
Using a slingshot, the boy fires the 0.2-lb marble at the
concrete wall, striking it at B. If the coefficient of restitution
between the marble and the wall is e = 0.5, determine the
speed of the marble after it rebounds from the wall.
vA
45
A
5 ft
SOLUTION
A sB B x = A sA B x + A vA B x t
100 = 0 + 75 cos 45° t
t = 1.886 s
and
1
a t2
2 y
1
( -32.2)(1.8862)
2
teaching
Web)
A sB B y = 0 + 75 sin 45°(1.886) +
or
A sB B y = A sA B y + A vA B y t +
laws
a+cb
Dissemination
Wide
copyright
in
= 42.76 ft
not
instructors
States
World
permitted.
and
United
on
learning.
is
of
the
A vB B y = A vA B y + ay t
and
use
a+cb
for
work
student
the
by
A vB B y = 75 sin 45° + ( -32.2)(1.886) = -7.684 ft>s = 7.684 ft>s T
is
work
solely
of
protected
the
(including
Since A vB B x = A vA B x = 75 cos 45° = 53.03 ft>s, the magnitude of vB is
part
the
is
any
destroy
of
and
7.684
≤ = 8.244°
53.03
will
A vB B x
S = tan - 1 ¢
their
A vB B y
sale
u = tan - 1 C
courses
and the direction angle of vB is
This
provided
integrity
of
and
work
this
assessing
vB = 2 A vB B x 2 + A vB B y 2 = 253.032 + 7.6842 = 53.59 ft>s
Conservation of Linear Momentum: Since no impulsive force acts on the marble
along the inclined surface of the concrete wall (x¿ axis) during the impact, the linear
momentum of the marble is conserved along the x¿ axis. Referring to Fig. b,
A +Q B
mB A v Bœ B x¿ = mB A v Bœ B x¿
0.2
0.2
A 53.59 sin 21.756° B =
A v Bœ cos f B
32.2
32.2
v Bœ cos f = 19.862
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
C
100 ft
Kinematics: By considering the x and y motion of the marble from A to B, Fig. a,
+ b
a:
B
75 ft>s
(1)
60
15–78. continued
Coefficient of Restitution: Since the concrete wall does not move during the impact,
the coefficient of restitution can be written as
A +aB
e =
0 - A v Bœ B y¿
A v Bœ B y¿ - 0
0.5 =
-v Bœ sin f
- 53.59 cos 21.756°
v Bœ sin f = 24.885
(2)
Solving Eqs. (1) and (2) yields
v Bœ = 31.8 ft>s
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
15–79.
The sphere of mass m falls and strikes the triangular block
with a vertical velocity v. If the block rests on a smooth
surface and has a mass 3 m, determine its velocity just after
the collision. The coefficient of restitution is e.
SOLUTION
Conservation of “x¿ ” Momentum:
m (v)1 = m (v)2
( R +)
45°
m( v sin 45°) = m(vsx¿)2
(vsx¿)2 =
22
v
2
Coefficient of Restitution (y¿ ):
or
laws
vb cos 45° - [- A vsy¿ B 2]
Web)
v cos 45° - 0
teaching
e =
A vsy¿ B 1 - (vb)1
Wide
copyright
(+ b )
(vb)2 - A vs y¿ B 2
in
e =
World
permitted.
Dissemination
22
(ev - vb)
2
is
of
the
not
instructors
States
A vsy¿ B 2 =
by
and
use
United
on
learning.
Conservation of “ x ” Momentum:
work
student
the
(including
for
0 + 0 = 3mvb - m A vsy¿ B 2 cos 45° - m(vsx¿)2 cos 45°
this
integrity
of
and
work
22
22
22 22
(ev - vb)
v
= 0
2
2
2
2
any
destroy
of
will
sale
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
and
1 + e
v
7
their
vb =
courses
part
the
is
This
provided
3vb -
assessing
is
work
solely
of
protected
+ B
A;
the
0 = ms (vs)x + mb vb
*15–80.
Block A, having a mass m, is released from rest, falls a
distance h and strikes the plate B having a mass 2m. If the
coefficient of restitution between A and B is e, determine
the velocity of the plate just after collision. The spring has
a stiffness k.
A
h
B
k
SOLUTION
Just before impact, the velocity of A is
T1 + V1 = T2 + V2
0 + 0 =
1
mv 2A - mgh
2
vA = 22gh
(+ T )
e =
(vB)2 - (vA)2
22gh
e 22gh = (vB)2 - (vA)2
or
(1)
©mv1 = ©mv 2
teaching
Web)
laws
(+ T )
(2)
Dissemination
Wide
copyright
in
m(vA) + 0 = m(vA)2 + 2m(vB)2
of
is
2gh(1 + e)
on
learning.
United
and
work
student
the
the
(including
for
work
solely
of
protected
this
assessing
is
integrity
of
and
work
provided
any
courses
part
the
is
This
destroy
of
and
will
their
sale
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
use
1
3
by
(vB)2 =
the
not
instructors
States
World
permitted.
Solving Eqs. (1) and (2) for (vB)2 yields;
15–81.
The girl throws the 0.5-kg ball toward the wall with an
initial velocity vA = 10 m>s. Determine (a) the velocity at
which it strikes the wall at B, (b) the velocity at which it
rebounds from the wall if the coefficient of restitution
e = 0.5, and (c) the distance s from the wall to where it
strikes the ground at C.
B
vA
A
10 m/s
30
1.5 m
C
SOLUTION
3m
Kinematics: By considering the horizontal motion of the ball before the impact,
we have
+ B
A:
s
sx = (s0)x + vx t
3 = 0 + 10 cos 30°t
t = 0.3464 s
By considering the vertical motion of the ball before the impact, we have
(+ c)
vy = (v0)y + (ac)y t
= 10 sin 30° + ( - 9.81)(0.3464)
laws
or
= 1.602 m>s
copyright
in
teaching
Web)
The vertical position of point B above the ground is given by
permitted.
Dissemination
Wide
1
(a ) t2
2 cy
World
sy = (s0)y + (v0)y t +
on
learning.
work
student
the
by
and
use
United
(sB)y = 1.5 + 10 sin 30°(0.3464) +
is
of
1
( -9.81) A 0.34642 B = 2.643 m
2
the
not
instructors
States
(+ c)
Ans.
this
assessing
is
work
solely
(vb)1 = 2(10 cos 30°)2 + 1.6022 = 8.807 m>s = 8.81 m>s
of
protected
the
(including
for
Thus, the magnitude of the velocity and its directional angle are
integrity
of
and
work
1.602
= 10.48° = 10.5°
10 cos 30°
provided
Ans.
and
any
courses
part
the
is
This
u = tan - 1
sale
will
their
destroy
of
Conservation of “y” Momentum: When the ball strikes the wall with a speed of
(vb)1 = 8.807 m>s, it rebounds with a speed of (vb)2.
mb A vby B 1 = mb A vby B 2
+ B
A;
mb (1.602) = mb C (vb)2 sin f D
(vb)2 sin f = 1.602
(1)
Coefficient of Restitution (x):
e =
+ B
A:
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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0.5 =
(vw)2 - A vbx B 2
A vbx B 1 - (vw)1
0 - C -(vb)2 cos f D
10 cos 30° - 0
(2)
15–81. continued
Solving Eqs. (1) and (2) yields
f = 20.30° = 20.3°
(vb)2 = 4.617 m>s = 4.62 m>s
Ans.
Kinematics: By considering the vertical motion of the ball after the impact, we have
(+ c)
sy = (s0)y + (v0)y t +
1
(a ) t2
2 cy
1
( - 9.81)t21
2
- 2.643 = 0 + 4.617 sin 20.30°t1 +
t1 = 0.9153 s
By considering the horizontal motion of the ball after the impact, we have
+ B
A;
sx = (s0)x + vx t
s = 0 + 4.617 cos 20.30°(0.9153) = 3.96 m
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
15–82.
The 20-lb box slides on the surface for which mk = 0.3. The
box has a velocity v = 15 ft>s when it is 2 ft from the plate.
If it strikes the smooth plate, which has a weight of 10 lb and
is held in position by an unstretched spring of stiffness
k = 400 lb>ft, determine the maximum compression
imparted to the spring. Take e = 0.8 between the box and
the plate. Assume that the plate slides smoothly.
v
k
2 ft
SOLUTION
T1 + a U1 - 2 = T2
1 20
1 20
a
b (15)2 - (0.3)(20)(2) = a
b (v2)2
2 32.2
2 32.2
v2 = 13.65 ft>s
+ )
(:
a mv1 = a mv2
20
20
10
b (13.65) = a
b vA +
vB
32.2
32.2
32.2
laws
teaching
Web)
vP - vA
13.65
Wide
permitted.
Dissemination
0.8 =
or
(vB)2 - (vA)2
(vA)1 - (vB)1
in
e =
copyright
a
is
of
the
not
instructors
States
World
Solving,
by
and
use
United
on
learning.
vP = 16.38 ft>s, vA = 5.46 ft>s
(including
for
work
student
the
T1 + V1 = T2 + V2
this
assessing
is
work
solely
of
protected
the
1
1 10
a
b (16.38)2 + 0 = 0 + (400)(s)2
2 32.2
2
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
s = 0.456 ft
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
15 ft/s
Ans.
15–83.
Before a cranberry can make it to your dinner plate, it must
pass a bouncing test which rates its quality. If cranberries
having an e Ú 0.8 are to be accepted, determine the
dimensions d and h for the barrier so that when a cranberry
falls from rest at A it strikes the plate at B and bounces over
the barrier at C.
A
SOLUTION
T1 + V1 = T2 + V2
0 + 3.5W =
1 W
a
b (vc)21 + 0
2 32.2
(vc)1 = 15.01 ft>s
or
Conservation of “x¿ ” Momentum: When the cranberry strikes the plate with a
speed of (vc)1 = 15.01 ft>s, it rebounds with a speed of (vc)2.
teaching
Web)
laws
mc A vcx¿ B 1 = mc A vcx¿ B 2
permitted.
Dissemination
Wide
copyright
in
3
mc (15.01) a b = mc C (vc)2 cos f D
5
learning.
is
of
the
instructors
States
(1)
United
and
use
by
(including
for
work
student
the
(vP)2 - A vcy¿ B 2
the
A vcy¿ B 1 - (vP)1
is
work
solely
of
protected
e =
on
Coefficient of Restitution (y¿ ):
assessing
0 - (vc)2 sin f
integrity
of
and
work
this
(2)
provided
4
-15.01 a b - 0
5
and
any
courses
part
the
is
This
0.8 =
will
their
destroy
of
Solving Eqs. (1) and (2) yields
(vc)2 = 13.17 ft>s
sale
f = 46.85°
Kinematics: By considering the vertical motion of the cranberry after the impact,
we have
vy = (v0)y + ac t
0 = 13.17 sin 9.978° + ( - 32.2) t
(+ c)
sy = (s0)y + (v0)y t +
t = 0.07087 s
1
(a ) t2
2 cy
= 0 + 13.17 sin 9.978° (0.07087) +
= 0.080864 ft
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
1
(- 32.2) A 0.070872 B
2
not
World
(vc)2 cos f = 9.008
(+ c)
B
d
Conservation of Energy: The datum is set at point B. When the cranberry falls from
a height of 3.5 ft above the datum, its initial gravitational potential energy is
W(3.5) = 3.5 W. Applying Eq. 14–21, we have
( a +)
3.5 ft
4
h
(+ b )
5
C
3
15–83. continued
By considering the horizontal motion of the cranberry after the impact, we have
+ B
A;
sx = (s0)x + vx t
4
d = 0 + 13.17 cos 9.978° (0.07087)
5
d = 1.149 ft = 1.15 ft
Ans.
Thus,
3
3
d = 0.080864 + (1.149) = 0.770 ft
5
5
Ans.
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Dissemination
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in
teaching
Web)
laws
or
h = sy +
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or transmission in any form or by any means, electronic, mechanical,
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*15–84.
A ball is thrown onto a rough floor at an angle u. If it rebounds
at an angle f and the coefficient of kinetic friction
is m, determine the coefficient of restitution e. Neglect the size
of the ball. Hint: Show that during impact, the average
impulses in the x and y directions are related by Ix = mIy .
Since the time of impact is the same, Fx ¢t = mFy ¢t or
Fx = mFy .
y
u
x
SOLUTION
0 - 3 - v2 sin f4
( + T)
e =
+ )
(:
m1vx21 +
v1 sin u - 0
e =
v2 sin f
v1 sin u
(1)
t2
Fx dx = m1vx22
Lt1
mv1 cos u - Fx ¢t = mv2 cos f
m1vy21 +
(2)
t2
Lt1
Fy dx = m1vy22
laws
(+ T)
mv1 cos u - mv2 cos f
¢t
or
Fx =
in
teaching
Web)
mv1 sin u - Fy ¢t = - mv2 sin f
mv1 sin u + mv2 sin f
¢t
(3)
instructors
States
World
permitted.
Dissemination
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Fy =
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on
learning.
is
of
the
not
Since Fx = mFy, from Eqs. (2) and (3)
the
(including
for
work
student
the
by
and
m1mv1 sin u + mv2 sin f)
mv1 cos u - mv2 cos f
=
¢t
¢t
(4)
part
the
is
Substituting Eq. (4) into (1) yields:
This
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this
assessing
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protected
cos u - m sin u
v2
=
v1
m sin f + cos f
will
sale
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destroy
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and
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courses
sin f cos u - m sin u
a
b
sin u m sin f + cos f
their
e =
f
Ans.
15–85.
A ball is thrown onto a rough floor at an angle of u = 45°. If
it rebounds at the same angle f = 45°, determine the
coefficient of kinetic friction between the floor and the ball.
The coefficient of restitution is e = 0.6. Hint: Show that
during impact, the average impulses in the x and y
directions are related by Ix = mIy . Since the time of impact
is the same, Fx ¢t = mFy ¢t or Fx = mFy .
y
u
x
SOLUTION
(+ T )
+ B
A:
e =
0 - [-v2 sin f]
v1 sin u - 0
v2 sin f
v1 sin u
e =
(1)
t2
m(vx)1 +
Fx dx = m(vx)2
Lt1
mv1 cos u - Fx ¢t = mv2 cos f
m A vy B 1 +
(2)
t2
Lt1
Fydx = m A yy B 2
laws
(+ c )
mv1 cos u - mv2 cos f
¢t
or
Fx =
in
teaching
Web)
mv1 sin u - Fy ¢t = -mv2 sin f
mv1 sin u + mv2 sin f
¢t
Wide
copyright
permitted.
Dissemination
(3)
instructors
States
World
Fy =
the
(including
for
work
student
the
by
m(mv1 sin u + mv2 sin f)
mv1 cos u - mv2 cos f
=
¢t
¢t
and
use
United
on
learning.
is
of
the
not
Since Fx = mFy, from Eqs. (2) and (3)
(4)
part
the
is
Substituting Eq. (4) into (1) yields:
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cos u - m sin u
v2
=
v1
m sin f + cos f
destroy
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and
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courses
sin f cos u - m sin u
a
b
sin u m sin f + cos f
sale
will
their
e =
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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0.6 =
sin 45° cos 45° - m sin 45°
a
b
sin 45° m sin 45° + cos 45°
0.6 =
1-m
1+m
m = 0.25
f
Ans.
15–86.
The “stone” A used in the sport of curling slides over the ice
track and strikes another “stone” B as shown. If each
“stone” is smooth and has a weight of 47 lb, and the
coefficient of restitution between the “stones” is e = 0.8,
determine their speeds just after collision. Initially A has a
velocity of 8 ft>s and B is at rest. Neglect friction.
3 ft
y
B
A
SOLUTION
(vA)1
Line of impact (x-axis):
©mv1 = ©mv2
47
47
47
(8) cos 30° =
(v ) +
(v )
32.2
32.2 B 2x
32.2 A 2x
( + a)
0 +
( + a)
e = 0.8 =
(vB)2x - (vA)2x
8 cos 30° - 0
Solving:
or
(vA)2x = 0.6928 ft>s
in
teaching
Web)
laws
(vB)2x = 6.235 ft>s
permitted.
Dissemination
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copyright
Plane of impact (y-axis):
of
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not
instructors
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World
Stone A:
by
and
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is
mv1 = mv2
of
protected
the
(including
for
work
student
the
47
47
(8) sin 30° =
(v )
32.2
32.2 A 2y
(Q+)
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and
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this
assessing
is
work
solely
(vA)2y = 4
part
the
is
This
provided
integrity
Stone B:
47
(vB)2y
32.2
will
0 =
sale
(Q+ )
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mv1 = mv2
(vB)2y = 0
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(vA)2 = 2(0.6928)2 + (4)2 = 4.06 ft>s
Ans.
(vB)2 = 2(0)2 + (6.235)2 = 6.235 = 6.24 ft>s
Ans.
30
8 ft/s x
15–87.
Two smooth disks A and B each have a mass of 0.5 kg. If
both disks are moving with the velocities shown when they
collide, determine their final velocities just after collision.
The coefficient of restitution is e = 0.75.
y
(vA)1
x
B
SOLUTION
+ )
(:
(vB)1
©mv1 = ©mv2
3
0.5(4)( ) - 0.5(6) = 0.5(yB)2x + 0.5(yA)2x
5
+ )
(:
e =
(vA)2 - (vB)2
(vB)1 - (vA)1
0.75 =
(vA)2x - (vB)2x
4 A 35 B - ( - 6)
(vA)2x = 1.35 m>s :
laws
or
(vB)2x = 4.95 m>s ;
in
teaching
Web)
mv1 = mv2
Wide
copyright
(+ c )
not
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permitted.
Dissemination
4
0.5( )(4) = 0.5(vB)2y
5
and
use
United
on
learning.
is
of
the
(vB)2y = 3.20 m>s c
by
vA = 1.35 m>s :
work
student
the
Ans.
for
Ans.
is
work
solely
of
protected
the
(including
vB = 2(4.59)2 + (3.20)2 = 5.89 m>s
assessing
3.20
= 32.9˚ ub
4.95
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
work
is
This
provided
u = tan-1
6 m/s
5 4
3
4 m/s
A
*15–88.
Two smooth disks A and B each have a mass of 0.5 kg. If
both disks are moving with the velocities shown when they
collide, determine the coefficient of restitution between the
disks if after collision B travels along a line, 30°
counterclockwise from the y axis.
y
(vA)1
x
B
SOLUTION
(vB)1
©mv1 = ©mv2
+ )
(:
3
0.5(4)( ) - 0.5(6) = - 0.5(vB)2x + 0.5(vA)2x
5
- 3.60 = -(vB)2x + (vA)2x
(+ c )
4
0.5(4)( ) = 0.5(vB)2y
5
(vB)2y = 3.20 m>s c
laws
or
(vB)2x = 3.20 tan 30° = 1.8475 m>s ;
in
teaching
Web)
(vA)2x = - 1.752 m>s = 1.752 m>s ;
(vA)2 - (vB)2
(vB)1 - (vA)1
Wide
permitted.
Dissemination
not
of
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= 0.0113
on
learning.
is
Ans.
United
and
4(35) -( -6)
use
- 1.752- ( -1.8475)
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This
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the
(including
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the
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e =
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e =
copyright
+ )
(:
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or transmission in any form or by any means, electronic, mechanical,
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6 m/s
5 4
3
4 m/s
A
15–89.
Two smooth disks A and B have the initial velocities shown
just before they collide at O. If they have masses mA = 8 kg
and mB = 6 kg, determine their speeds just after impact.
The coefficient of restitution is e = 0.5.
y
13 12
5
A
vA
O
SOLUTION
vB
+b©mv1 = ©mv2
-6(3 cos 67.38°) + 8(7 cos 67.38°) = 6(vB)x¿ + 8(vA)xœ
(+ b)
0.5 =
e =
(vB)2 - (vA)2
(vA)1 - (vB )1
(vB)xœ - (vA)xœ
7 cos 67.38° + 3 cos 67.38°
Solving,
laws
or
(vB)xœ = 2.14 m>s
in
teaching
Web)
(vA)xœ = 0.220 m>s
Dissemination
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copyright
(vB)yœ = 3 sin 67.38° = 2.769 m>s
not
instructors
States
World
permitted.
(vA)y¿ = -7 sin 67.38° = -6.462 m>s
of
the
vB = 2(2.14)2 + (2.769)2 = 3.50 m>s
and
use
United
on
learning.
is
Ans.
Ans.
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vA = 2(0.220)2 + (6.462)2 = 6.47 m>s
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
3 m/s
7 m/s
x
B
15–90.
If disk A is sliding along the tangent to disk B and strikes
B with a velocity v, determine the velocity of B after the
collision and compute the loss of kinetic energy during
the collision. Neglect friction. Disk B is originally at rest.
The coefficient of restitution is e, and each disk has the
same size and mass m.
B
v
A
SOLUTION
Impact: This problem involves oblique impact where the line of impact lies along x¿
axis (line jointing the mass center of the two impact bodies). From the geometry
r
u = sin - 1 a b = 30°. The x¿ and y¿ components of velocity for disk A just before
2r
impact are
A vAx¿ B 1 = - v cos 30° = - 0.8660v
A vAy¿ B 1 = - v sin 30° = - 0.5 v
Conservation of “x¿ ” Momentum:
or
mA A vAx¿ B 1 + mB A vBx¿ B 1 = mA A vAx¿ B 2 + mB A vBx¿ B 2
m( -0.8660v) + 0 = m A vAx B 2 + m A vBx¿ B 2
laws
(1)
in
teaching
Web)
(R +)
Wide
permitted.
Dissemination
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not
A vAx¿ B 1 - A vBx¿ B 1
A vBx¿ B 2 - A vAx¿ B 2
(2)
by
- 0.8660v - 0
work
student
the
(including
for
e =
A vBx¿ B 2 - A vAx¿ B 2
the
e =
copyright
Coefficient of Restitution (x¿ ):
assessing
is
work
solely
of
protected
Solving Eqs. (1) and (2) yields
23
(e - 1) v
4
integrity
of
and
work
this
A vAx¿ B 2 =
provided
23
(1 + e) v
4
part
the
is
This
A vBx¿ B 2 = -
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Conservation of “y¿ ” Momentum: The momentum is conserved along y¿ axis for
both disks A and B.
(+Q)
(+Q)
mB A vBy¿ B 1 = mB A vBy¿ B 2;
mA A vAy¿ B 1 = mA A vAy¿ B 2;
A vBy¿ B 2 = 0
A vAy¿ B 2 = - 0.5 v
Thus, the magnitude the velocity for disk B just after the impact is
(vB)2 = 2(vBx¿)22 + (vBy¿)22
=
D
a-
2
23
23
(1 + e) vb + 0 =
(1 + e) v
4
4
and directed toward negative x¿ axis.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
Ans.
15–90. continued
The magnitude of the velocity for disk A just after the impact is
(vA)2 = 2( vAx¿)22 + (vAy¿)22
=
2
23
(e - 1) v d + ( -0.5 v)2
D 4
=
v2 (3e2 - 6e + 7)
A 16
c
Loss of Kinetic Energy: Kinetic energy of the system before the impact is
Uk =
1
mv2
2
Kinetic energy of the system after the impact is
Uk œ =
mv2
A 6e2 + 10 B
32
or
=
2
2
2
23
1
1
m B v (3e2 - 6e + 7) R + m B
(1 + e) v R
2
A 16
2
4
Web)
laws
Thus, the kinetic energy loss is
in
teaching
1
m v2
mv2 (6e2 + 10)
2
32
Dissemination
Wide
copyright
¢Uk = Uk - Uk œ =
States
World
permitted.
3mv2
(1 - e2)
16
the
not
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the
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of
=
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
15–91.
Two disks A and B weigh 2 lb and 5 lb, respectively. If they
are sliding on the smooth horizontal plane with the velocities
shown, determine their velocities just after impact. The
coefficient of restitution between the disks is e = 0.6.
y
SOLUTION
30
mA A vA B n + mB A vB B n = mA A vAœ B n + mB A vBœ B n
-
5
2 œ
2
5 œ
(5) cos 45° +
(10) cos 30° =
vA cos uA +
vB cos uB
32.2
32.2
32.2
32.2
œ
2vA
cos uA + 5vBœ cos uB = 36.23
(1)
Also, we notice that the linear momentum a of disks A and B are conserved along
the t axis (tangent to the plane of impact). Thus,
a+Tb
œ
mA A vA B t = mA A vA
Bt
Web)
laws
or
2
2
œ
sin uA
(5) sin 45° =
vA
32.2
32.2
(2)
Wide
copyright
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teaching
œ
vA
sin uA = 3.5355
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mB A vB B t = mB A v Bœ B t
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a+Tb
(including
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work
student
the
by
and
2
2
(10) sin 30° = =
v Bœ sin uB
32.2
32.2
(3)
assessing
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the
v Bœ sin uB = 5
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the
is
This
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this
Coefficient of Restitution: The coefficient of restitution equation written along the
n axis (line of impact) gives
any
courses
A vAœ B n - A vBœ B n
and
A vB B n - A vA B n
destroy
of
0.6 =
œ
vA
œ
vA
cos uA - v Bœ cos uB
10 cos 30° -
cos uA -
v Bœ
sale
will
e =
their
+ b
a:
A -5 cos 45° B
cos uB = 7.317
(4)
Solving Eqs. (1), (2), (3), and (4), yields
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
œ
= 11.0 ft>s
vA
uA = 18.8°
Ans.
v Bœ = 5.88 ft>s
uB = 58.3°
Ans.
B
A
45
x
Conservation of Linear Momentum:By referring to the impulse and momentum of
the system of disks shown in Fig. a, notice that the linear momentum of the system is
conserved along the n axis (line of impact). Thus,
+ b
a:
5 ft/s
10 ft/s
*15–92.
Two smooth coins A and B, each having the same mass,
slide on a smooth surface with the motion shown.
Determine the speed of each coin after collision if they
move off along the blue paths. Hint: Since the line of impact
has not been defined, apply the conservation of momentum
along the x and y axes, respectively.
y
(vA)2
(vA)1
A
0.5 ft/s
45
5
A
3
4
30
SOLUTION
30
©mv1 = mv2
+ )
(:
(vB)2
4
-m(0.8) sin 30° - m(0.5)( ) = -m(vA)2 sin 45° - m(vB)2 cos 30°
5
3
m(0.8) cos 30° - m(0.5)( ) = m(vA)2 cos 45° - m(vB)2 sin 30°
5
-0.3928 = - 0.707(vA)2 + 0.5(vB)2
laws
or
Solving,
Ans.
(vA)2 = 0.766 ft>s
Ans.
will
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is
This
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the
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student
the
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is
of
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permitted.
Dissemination
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Web)
(vB)2 = 0.298 ft>s
sale
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(vB)1
B
0.8 = 0.707(vA)2 + 0.866(vB)2
(+ c)
x
O
B
0.8 ft/s
15–93.
Disks A and B have a mass of 15 kg and 10 kg, respectively.
If they are sliding on a smooth horizontal plane with the
velocities shown, determine their speeds just after impact.
The coefficient of restitution between them is e = 0.8.
y
5
4
Line of
impact
3
A
10 m/s
x
SOLUTION
8 m/s
Conservation of Linear Momentum: By referring to the impulse and momentum of
the system of disks shown in Fig. a, notice that the linear momentum of the system is
conserved along the n axis (line of impact). Thus,
¿
+ Q mA A vA B n + mB A vB B n = mA A vA
B n + mB A vB¿ B n
3
3
¿
15(10)a b - 10(8)a b = 15vA
cos fA + 10vB¿ cos fB
5
5
¿
15vA
cos fA + 10vB¿ cos fB = 42
(1)
or
Also, we notice that the linear momentum of disks A and B are conserved along the
t axis (tangent to? plane of impact). Thus,
teaching
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¿
+ a mA A vA B t = mA A vA
Bt
Dissemination
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4
¿
15(10)a b = 15vA
sin fA
5
permitted.
¿
vA
sin fA = 8
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the
+a mB A vB B t = mB A vB¿ B t
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10(8)a b = 10 vB¿ sin fB
5
(3)
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This
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this
vB¿ sin fB = 6.4
+Q e =
0.8 =
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¿
(vB¿ )n - (vA
)n
(vA)n - (vB)n
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Coefficient of Restitution: The coefficient of restitution equation written along the n
axis (line of impact) gives
¿
cos fA
vB¿ cos fB - vA
3
3
10 a b - c -8a b d
5
5
¿
vB¿ cos fB - vA
cos fA = 8.64
(4)
Solving Eqs. (1), (2), (3), and (4), yeilds
¿
= 8.19 m>s
vA
Ans.
fA = 102.52°
vB¿ = 9.38 m>s
fB = 42.99°
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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Ans.
B
15–94.
z
Determine the angular momentum H O of the particle about
point O.
1.5 kg
A
6 m/s
4m
SOLUTION
vA = 6 ¢
rOB = { - 7j} m
4m
2i - 4j - 4k
2(22 + ( -4)2 + (- 4)2
j
-7
1.5( -4)
x
k
3 = {42i + 21k} kg # m2>s
0
1.5(- 4)
Ans.
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i
0
1.5(2)
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3m
≤ = {2i - 4j - 4k} m>s
HO = rOB * mvA
HO = 3
2m
B
O
y
15–95.
Determine the angular momentum H O of the particle about
point O.
z
A
10 lb
14 ft/s
P
6 ft
O
5 ft
SOLUTION
rOB = {8i + 9j}ft
HO = 4
10
a
b(12)
32.2
vA = 14 ¢
12i + 4j - 6k
2122 + 42 + (- 6)2
j
9
10
a
b(4)
32.2
y
8 ft
≤ = {12i + 4j - 6k} ft>s
9 ft
x
B
k
0
4
10
a
b ( -6)
32.2
ft2>s
Ans.
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or
= {-16.8i + 14.9j - 23.6k} slug
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
4 ft
2 ft
HO = rOB * mvA
i
8
5 ft
3 ft
*15–96.
Determine the angular momentum H P of the particle about
point P.
z
A
10 lb
14 ft/s
P
6 ft
O
5 ft
5 ft
4 ft
3 ft
SOLUTION
2 ft
y
rPB = {5i + 11j - 5k} ft
vA = 14 a
8 ft
12i + 4j - 6k
212)2 + (4)2 + ( - 6)2
i
5
H P = rPB * mvA = 4
a
9 ft
x
B
b = {12i + 4j - 6k} ft>s
10
b (12)
32.2
j
11
10
a
b (4)
32.2
k
-5
4
10
a
b (- 6)
32.2
= {-14.3i - 9.32j - 34.8k} slug # ft2>s
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15–97.
z
Determine the total angular momentum H O for the system
of three particles about point O. All the particles are
moving in the x–y plane.
3 kg
6 m/s
C
O
900 mm
SOLUTION
4 m/s
A 1.5 kg
HO = ©r * mv
i
= 3 0.9
0
j
0
- 1.5(4)
x
i
k
0 3 + 3 0.6
-2.5(2)
0
i
k
0 3 + 3 -0.8
0
0
j
0.7
0
j
-0.2
3(-6)
k
03
0
= {12.5k} kg # m2>s
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
700 mm
200 mm
800 mm
y
B
2 m/s 600 mm
2.5 kg
15–98.
Determine the angular momentum H O of each of the two
particles about point O. Use a scalar solution.
y
5m
P
15 m/s
5
A
3
4m
4
2 kg
1.5 m
O
x
2m
SOLUTION
4m
4
3
a +(HA)O = -2(15)a b (1.5) - 2(15)a b (2)
5
5
30⬚
1.5 kg
10 m/s
= -72.0 kg # m2>s = 72.0 kg # m2>s b
Ans.
a + (HB)O = -1.5(10)(cos 30°)(4) -1.5(10)(sin 30°)(1)
= -59.5 kg # m2>s = 59.5 kg # m2>s b
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© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
B
1m
15–99.
Determine the angular momentum H P of each of the two
particles about point P. Use a scalar solution.
y
5m
P
15 m/s
5
A
3
4m
4
2 kg
1.5 m
O
x
2m
SOLUTION
4m
4
3
a +(HA)P = 2(15) a b (2.5) - 2(15) a b (7)
5
5
30⬚
1.5 kg
10 m/s
= -66.0 kg # m2>s = 66.0 kg # m2>s b
Ans.
a + (HB)P = - 1.5(10)(cos 30°)(8) + 1.5(10)(sin 30°)(4)
= -73.9 kg # m2>s b
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© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
B
1m
*15–100.
The small cylinder C has a mass of 10 kg and is attached to
the end of a rod whose mass may be neglected. If the frame
is subjected to a couple M = 18t2 + 52 N # m, where t is in
seconds, and the cylinder is subjected to a force of 60 N,
which is always directed as shown, determine the speed of
the cylinder when t = 2 s. The cylinder has a speed
v0 = 2 m>s when t = 0.
z
60 N
5 4
0.75 m
(Hz)1 + ©
L
Mz dt = (Hz)2
2
8
69 + [ t3 + 5t]20 = 7.5v
3
v = 13.4 m>s
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
C
M
3
(8t 2 + 5)dt = 10v(0.75)
(10)(2)(0.75) + 60(2)( )(0.75) +
5
L0
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
v
x
SOLUTION
3
(8t2
y
5) N m
15–101.
The 10-lb block rests on a surface for which mk = 0.5. It is
acted upon by a radial force of 2 lb and a horizontal force of
7 lb, always directed at 30° from the tangent to the path as
shown. If the block is initially moving in a circular path with
a speed v1 = 2 ft>s at the instant the forces are applied,
determine the time required before the tension in cord AB
becomes 20 lb. Neglect the size of the block for the
calculation.
A
4 ft
B
7 lb
SOLUTION
©Fn = man ;
20 - 7 sin 30° - 2 =
10 v 2
( )
32.2 4
v = 13.67 ft>s
(HA) t + ©
MA dt = (HA)2
teaching
Web)
laws
or
10
10
)(2)(4) + (7 cos 30°)(4)(t) - 0.5(10)(4) t =
(13.67)(4)
32.2
32.2
in
(
L
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copyright
t = 3.41 s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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30
2 lb
15–102.
The 10-lb block is originally at rest on the smooth surface. It
is acted upon by a radial force of 2 lb and a horizontal force
of 7 lb, always directed at 30° from the tangent to the path
as shown. Determine the time required to break the cord,
which requires a tension T = 30 lb. What is the speed of the
block when this occurs? Neglect the size of the block for the
calculation.
A
4 ft
B
7 lb
SOLUTION
©Fn = man ;
30 - 7 sin 30° - 2 =
10 v2
( )
32.2 4
v = 17.764 ft>s
L
MA dt = (HA)2
0 + (7 cos 30°)(4)(t) =
10
(17.764)(4)
32.2
t = 0.910 s
Ans.
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(HA)1 + ©
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
30
2 lb
15–103.
A 4-lb ball B is traveling around in a circle of radius r1 = 3 ft
with a speed 1vB21 = 6 ft>s. If the attached cord is pulled down
through the hole with a constant speed vr = 2 ft>s, determine
the ball’s speed at the instant r2 = 2 ft. How much work has
to be done to pull down the cord? Neglect friction and the size
of the ball.
B
r1 = 3 ft
(v B )1 = 6 ft>s
SOLUTION
H1 = H2
v r = 2 ft>s
4
4
v (2)
(6)(3) =
32.2
32.2 u
vu = 9 ft>s
v2 = 292 + 22 = 9.22 ft>s
Ans.
T1 + ©U1 - 2 = T2
1 4
1 4
(
)(6)2 + ©U1 - 2 = (
)(9.22)2
2 32.2
2 32.2
or
©U1 - 2 = 3.04 ft # lb
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or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*15–104.
A 4-lb ball B is traveling around in a circle of radius r1 = 3 ft
with a speed 1vB21 = 6 ft>s. If the attached cord is pulled
down through the hole with a constant speed vr = 2 ft>s,
determine how much time is required for the ball to reach
a speed of 12 ft/s. How far r2 is the ball from the hole when
this occurs? Neglect friction and the size of the ball.
B
r1 = 3 ft
(v B )1 = 6 ft>s
SOLUTION
v = 2(vu)2 + (2)2
v r = 2 ft>s
12 = 2(vu)2 + (2)2
vu = 11.832 ft>s
H1 = H2
4
4
(6)(3) =
(11.832)(r2)
32.2
32.2
r2 = 1.5213 = 1.52 ft
Ans.
laws
or
¢r = v rt
Ans.
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t = 0.739 s
Wide
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Web)
(3 - 1.5213) = 2t
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
15–105.
The four 5-lb spheres are rigidly attached to the crossbar frame having
a negligible weight. If a couple moment M = (0.5t + 0.8) lb # ft, where t
is in seconds, is applied asshown, determine the speed of each of the
spheres in 4 seconds starting from rest. Neglect the size of the spheres.
M
SOLUTION
(Hz)1 + ©
L
Mz dt = (Hz)2
4
0 +
L0
0.6 ft
(0.5 t + 0.8) dt = 4 B a
5
b (0.6 v 2) R
32.2
7.2 = 0.37267 v2
v2 = 19.3 ft>s
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
(0.5t
0.8) lb · ft
15–106.
A small particle having a mass m is placed inside the
semicircular tube. The particle is placed at the position
shown and released. Apply the principle
of angular
#
momentum about point O 1©MO = HO2, and show that
the motion
$ of the particle is governed by the differential
equation u + 1g>R2 sin u = 0.
O
u
SOLUTION
c + ©MO =
dHO
;
dt
-Rmg sin u =
d
(mvR)
dt
dv
d 2s
= - 2
dt
dt
g sin u = But, s = R u
$
Thus, g sin u = -Ru
$
g
or, u + a b sin u = 0
R
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R
15–107.
The ball B has a weight of 5 lb and is originally rotating in a
circle. As shown, the cord AB has a length of 3 ft and passes
through the hole A, which is 2 ft above the plane of motion. If
1.5 ft of cord is pulled through the hole, determine the speed
of the ball when it moves in a circular path at C.
1.5 ft
C
2 ft
3 ft
SOLUTION
B
vB
Equation of Motion: When the ball is travelling around the first circular path,
2
u = sin - 1 = 41.81° and r1 = 3 cos 41.81° = 2.236. Applying Eq. 13–8, we have
3
©Fb = 0;
©Fn = man;
2
T1 a b -5 = 0
3
T1 = 7.50 lb
7.50 cos 41.81° =
v2t
5
a
b
32.2 2.236
v1 = 8.972 ft>s
laws
or
When the ball is traveling around the second circular path, r2 = 1.5 cos f. Applying
Eq. 13–8, we have
Web)
T2 sin f - 5 = 0
in
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(1)
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©Fb = 0;
permitted.
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Dissemination
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v22
5
a
b
32.2 1.5 cos f
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T2 cos f =
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©Fn = man;
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on
learning.
Conservation of Angular Momentum: Since no force acts on the ball along the
tangent of the circular, path the angular momentum is conserved about z axis.
Applying Eq. 15–23, we have
provided
integrity
of
and
work
r1mv1 = r2mv2
this
assessing
is
work
solely
of
protected
the
(Ho)1 = (Ho)2
5
5
b (8.972) = 1.5 cos f a
b v2
32.2
32.2
the
(3)
will
sale
Solving Eqs. (1),(2) and (3) yields
their
destroy
of
and
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courses
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This
2.236 a
f = 13.8678°
T2 = 20.85 lb
v2 = 13.8 ft>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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T
A
Ans.
vC
*15–108.
A child having a mass of 50 kg holds her legs up as shown as
she swings downward from rest at u1 = 30°. Her center of
mass is located at point G1 . When she is at the bottom
position u = 0°, she suddenly lets her legs come down,
shifting her center of mass to position G2 . Determine her
speed in the upswing due to this sudden movement and the
angle u2 to which she swings before momentarily coming to
rest. Treat the child’s body as a particle.
3m
G2
First before u = 30°;
T1 + V1 = T2 + V2
1
(50)(v1)2 + 0
2
v1 = 2.713 m>s
H1 = H2
50(2.713)(2.80) = 50(v2)(3)
v2 = 2.532 = 2.53 m>s
laws
or
Ans.
Wide
copyright
in
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Web)
Just after u = 0°;
States
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permitted.
Dissemination
T2 + V2 = T3 + V3
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is
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the
not
instructors
1
(50)(2.532)2 + 0 = 0 + 50(9.81)(3)(1 - cos u2)
2
(including
for
work
student
the
0.1089 = 1 - cos u2
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the
u2 = 27.0°
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Ans.
30°
G1
G2
0 + 2.80(1 - cos 30°)(50)(9.81) =
2.80 m
u1
SOLUTION
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
u2
15–109.
The 150-lb car of an amusement park ride is connected to a
rotating telescopic boom. When r = 15 ft, the car is moving
on a horizontal circular path with a speed of 30 ft>s. If the
boom is shortened at a rate of 3 ft>s, determine the speed of
the car when r = 10 ft. Also, find the work done by the axial
force F along the boom. Neglect the size of the car and the
mass of the boom.
F
SOLUTION
Conservation of Angular Momentum: By referring to Fig. a, we notice that the
angular momentum of the car is conserved about an axis perpendicular to the
page passing through point O, since no angular impulse acts on the car about this
axis. Thus,
A HO B 1 = A HO B 2
r1mv1 = r2m A v2 B u
Av2Bu =
15(30)
r1 v1
= 45 ft>s
=
r2
10
or
Thus, the magnitude of v2 is
laws
v 2 = 4A v 2 B r 2 - A v2 B u 2 = 232 + 452 = 45.10 ft>s = 45.1 ft>s
in
teaching
Web)
Ans.
Dissemination
Wide
copyright
Principle of Work and Energy: Using the result of v2,
of
the
not
instructors
States
World
permitted.
T1 + ©U1 - 2 = T2
(including
the
solely
of
protected
1 150
1 150
a
b (302) + UF = a
b (45.102)
2 32.2
2 32.2
for
work
student
the
by
and
use
United
on
learning.
is
1
1
mv1 2 + UF = mv 2 2
2
2
assessing
is
work
UF = 2641 ft # lb
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This
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this
Ans.
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
r
15–110.
An amusement park ride consists of a car which is attached
to the cable OA. The car rotates in a horizontal circular path
and is brought to a speed v1 = 4 ft>s when r = 12 ft. The
cable is then pulled in at the constant rate of 0.5 ft>s.
Determine the speed of the car in 3 s.
O
A
r
SOLUTION
Conservation of Angular Momentum: Cable OA is shorten by
¢r = 0.5(3) = 1.50 ft. Thus, at this instant r2 = 12 - 1.50 = 10.5 ft. Since no force
acts on the car along the tangent of the moving path, the angular momentum is
conserved about point O. Applying Eq. 15–23, we have
(HO)1 = (HO)2
r1 mv1 = r2 mv¿
12(m)(4) = 10.5(m) v¿
v¿ = 4.571 ft>s
laws
or
The speed of car after 3 s is
v2 = 20.52 + 4.5712 = 4.60 ft>s
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This
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the
(including
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student
the
by
and
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United
on
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is
of
the
not
instructors
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permitted.
Dissemination
Wide
copyright
in
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Web)
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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15–111.
The 800-lb roller-coaster car starts from rest on the track
having the shape of a cylindrical helix. If the helix descends
8 ft for every one revolution, determine the speed of the car
in t = 4 s. Also, how far has the car descended in this time?
Neglect friction and the size of the car.
r = 8 ft
8 ft
SOLUTION
u = tan-1(
8
) = 9.043°
2p(8)
©Fy = 0;
N - 800 cos 9.043° = 0
N = 790.1 lb
H1 +
L
M dt = H2
4
0 +
L0
8(790.1 sin 9.043°) dt =
800
(8)vt
32.2
laws
or
vt = 20.0 ft>s
teaching
Web)
20
= 20.2 ft>s
cos 9.043°
in
Ans.
Dissemination
Wide
copyright
v =
of
the
not
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permitted.
T1 + ©U1 - 2 = T2
by
and
use
United
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learning.
is
1 800
(
)(20.2)2
2 32.2
the
0 + 800h =
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This
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the
(including
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student
h = 6.36 ft
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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Ans.
*15–112.
A small block having a mass of 0.1 kg is given a horizontal
velocity of v1 = 0.4 m>s when r1 = 500 mm. It slides along
the smooth conical surface. Determine the distance h it must
descend for it to reach a speed of v2 = 2 m>s. Also, what is
the angle of descent u, that is, the angle measured from the
horizontal to the tangent of the path?
r1 = 500 mm
υ 1 = 0.4 m>s
h
r2
θ
v2
SOLUTION
T1 + V1 = T2 + V2
30°
1
1
(0.1)(0.4)2 + 0.1(9.81)(h) = (0.1)(2)2
2
2
h = 0.1957 m = 196 mm
Ans.
From similar triangles
r2 =
(0.8660 - 0.1957)
(0.5) = 0.3870 m
0.8660
laws
or
(H0)1 = (H0)2
in
teaching
Web)
0.5(0.1)(0.4) = 0.3870(0.1)(v 2 ¿)
Dissemination
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copyright
v 2 ¿ = 0.5168 m>s
not
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World
permitted.
v 2 = cos u = v 2 ¿
and
use
United
on
learning.
is
of
the
2 cos u = 0.5168
Ans.
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the
by
u = 75.0°
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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15–113.
An earth satellite of mass 700 kg is launched into a freeflight trajectory about the earth with an initial speed of
vA = 10 km>s when the distance from the center of the
earth is rA = 15 Mm. If the launch angle at this position is
fA = 70°, determine the speed vB of the satellite and its
closest distance rB from the center of the earth. The earth
has a mass Me = 5.976110242 kg. Hint: Under these
conditions, the satellite is subjected only to the earth’s
gravitational force, F = GMems>r2, Eq. 13–1. For part of
the solution, use the conservation of energy.
vB
rA
vA
SOLUTION
(HO)1 = (HO)2
ms (vA sin fA)rA = ms (vB)rB
700[10(103) sin 70°](15)(106) = 700(vB)(rB)
(1)
TA + VA = TB + VB
laws
or
GMe ms
GMems
1
1
m (v )2 = ms (vB)2 rA
rB
2 s A
2
Wide
copyright
in
teaching
Web)
66.73(10-12)(5.976)(1024)(700)
1
1
= (700)(vB)2
(700)[10(103)]2 6
2
2
[15(10 )]
Dissemination
66.73(10-12)(5.976)(1024)(700)
rB
World
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on
learning.
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of
United
the
by
and
use
Solving,
Ans.
the
(including
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student
vB = 10.2 km>s
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rB = 13.8 Mm
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permitted.
(2)
States
-
rB
fA
Ans.
15–114.
The fire boat discharges two streams of seawater, each at a
flow of 0.25 m3>s and with a nozzle velocity of 50 m> s.
Determine the tension developed in the anchor chain,
needed to secure the boat. The density of seawater is
rsw = 1020 kg>m3.
30⬚
45⬚
60⬚
SOLUTION
Steady Flow Equation: Here, the mass flow rate of the sea water at nozzles A and
dmA
dmB
B are
=
= rsw Q = 1020(0.25) = 225 kg>s. Since the sea water is coldt
dt
lected from the larger reservoir (the sea), the velocity of the sea water entering the
control volume can be considered zero. By referring to the free-body diagram of the
control volume (the boat),
+ ©F = dmA A v B + dmB A v B ;
;
x
A x
B x
dt
dt
T cos 60° = 225(50 cos 30°) + 225(50 cos 45°)
T = 40 114.87 N = 40.1 kN
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This
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the
(including
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laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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15–115.
A jet of water having a cross-sectional area of 4 in2 strikes
the fixed blade with a speed of 25 ft>s. Determine the
horizontal and vertical components of force which the
blade exerts on the water, gw = 62.4 lb>ft3.
130
25 ft/s
SOLUTION
Q = Av = a
4
b(25) = 0.6944 ft3>s
144
62.4
dm
= rQ = a
b (0.6944) = 1.3458 slug>s
dt
32.2
vAx = 25 ft>s
vAy = 0
vBx = -25 cos 50°ft>s
vBy = 25 sin 50°ft>s
+ ©F = dm (v - v );
:
x
A
dt Bx
- Fx = 1.3458[- 25 cos 50° - 25]
or
Fx = 55.3 lb
teaching
Web)
laws
Ans.
in
Fy = 1.3458(25 sin 50° - 0)
Wide
Dissemination
dm
A vBy - vAy);
dt
copyright
+ c ©Fy =
permitted.
Ans.
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Fy = 25.8 lb
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*15–116.
The 200-kg boat is powered by the fan which develops a
slipstream having a diameter of 0.75 m. If the fan ejects air
with a speed of 14 m/s, measured relative to the boat,
determine the initial acceleration of the boat if it is initially at
rest.Assume that air has a constant density of ra = 1.22 kg/m3
and that the entering air is essentially at rest. Neglect the drag
resistance of the water.
0.75 m
SOLUTION
Equations of Steady Flow: Initially, the boat is at rest hence vB = va>b
p
dm
= 14 m>s. Then, Q = vBA = 14 c A 0.752 B d = 6.185 m3>s and
= raQ
4
dt
= 1.22(6.185) = 7.546 kg>s. Applying Eq. 15–26, we have
©Fx =
dm
(v - vAx);
dt Bx
-F = 7.546( - 14 - 0)
F = 105.64 N
Equation of Motion :
105.64 = 200a
a = 0.528 m>s2
Ans.
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This
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the
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laws
or
+ ©F = ma ;
:
x
x
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15–117.
The chute is used to divert the flow of water, Q = 0.6 m3>s.
If the water has a cross-sectional area of 0.05 m2, determine
the force components at the pin D and roller C necessary
for equilibrium. Neglect the weight of the chute and weight
of the water on the chute, rw = 1 Mg>m3 .
A
0.12 m
C
SOLUTION
2m
Equations of Steady Flow: Here, the flow rate Q = 0.6 m2>s. Then,
Q
0.6
dm
=
= 12.0 m>s. Also,
= rw Q = 1000 (0.6) = 600 kg>s. Applying
y =
A
0.05
dt
Eqs. 15–26 and 15–28, we have
D
1.5 m
dm
(dDB yB - dDA yA);
dt
- Cx (2) = 600 [0 - 1.38(12.0)]
a + ©MA =
Cx = 4968 N = 4.97 kN
Dx = 2232N = 2.23 kN
Ans.
laws
or
+ ©F = dm A y - y );
:
x
Bx
Ax
dt
Dx + 4968 = 600 (12.0 - 0)
Ans.
Web)
teaching
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Ans.
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Dissemination
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Dy = 7200 N = 7.20 kN
Wide
dm
A youty - yiny B ;
dt
Dy = 600[0 - ( -12.0)]
+ c ©Fy = ©
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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B
15–118.
The buckets on the Pelton wheel are subjected to a 2-indiameter jet of water, which has a velocity of 150 ft>s. If
each bucket is traveling at 95 ft>s when the water strikes
it, determine the power developed by the bucket.
gw = 62.4 lb>ft3.
20
150 ft/s
SOLUTION
vA = 150 - 95 = 55 ft>s :
+ )(v ) = -55 cos 20° + 95 = 43.317 ft>s :
(:
B x
+ ©F = dm (v - v )
;
x
Ax
dt Bx
1
62.4
b(p)a b 2[(55)(55 cos 20° - ( - 55)]
Fx = a
32.2
12
Fx = 248.07 lb
or
P = 248.07(95) = 23,567 ft # lb>s
laws
P = 42.8 hp
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This
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20
95 ft/s
15–119.
The blade divides the jet of water having a diameter of 3 in.
If one-fourth of the water flows downward while the other
three-fourths flows upwards, and the total flow is
Q = 0.5 ft3>s, determine the horizontal and vertical
components of force exerted on the blade by the jet,
gw = 62.4 lb>ft3.
3 in.
SOLUTION
Equations of Steady Flow: Here, the flow rate Q = 0.5 ft2>s. Then,
Q
0.5
dm
62.4
y =
=
= rw Q =
(0.5) = 0.9689 slug>s.
= 10.19 ft>s. Also,
p 3 2
A
dt
32.2
A B
4
12
Applying Eq. 15–25 we have
©Fx = ©
dm
A youts - yins B ; - Fx = 0 - 0.9689 (10.19)
dt
©Fy = ©
1
dm
3
A youty - yiny B ; Fy = (0.9689)(10.19) + (0.9689)(-10.19)
dt
4
4
Fx = 9.87 lb
Ans.
Fy = 4.93 lb
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This
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*15–120.
The fan draws air through a vent with a speed of 12 ft>s. If
the cross-sectional area of the vent is 2 ft2, determine the
horizontal thrust on the blade. The specific weight of the air
is ga = 0.076 lb>ft3.
v
SOLUTION
dm
= rvA
dt
=
0.076
(12)(2)
32.2
= 0.05665 slug>s
©F =
dm
(v - vA)
dt B
T = 0.05665(12 - 0) = 0.680 lb
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Ans.
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
12 ft/s
15–121.
The gauge pressure of water at C is 40 lb>in2 . If water
flows out of the pipe at A and B with velocities vA = 12 ft>s
and vB = 25 ft>s, determine the horizontal and vertical
components of force exerted on the elbow necessary
tohold the pipe assembly in equilibrium. Neglect the
weight of water within the pipe and the weight of the pipe.
The pipe has a diameter of 0.75 in. at C, and at A and B the
diameter is 0.5 in. gw = 62.4 lb>ft3.
vA
3
B
A
dmA
62.4
0.25 2
=
(12)(p)a
b = 0.03171 slug>s
dt
32.2
12
vC
C
dmB
0.25 2
62.4
(25)(p) a
=
b = 0.06606 slug>s
dt
32.2
12
dm C
= 0.03171 + 0.06606 = 0.09777 slug>s
dt
vC AC = vA AA + vB AB
in
teaching
Web)
laws
or
0.375 2
0.25 2
0.25 2
b = 12(p)a
b + 25(p)a
b
12
12
12
Dissemination
Wide
copyright
vC = 16.44 ft>s
(including
for
work
student
the
by
3
40(p)(0.375)2 - Fx = 0 - 0.03171(12) a b - 0.09777(16.44)
5
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
+ ©F = dmB v + dmA v - dm C v
:
x
dt B s
dt As
dt Cs
Ans.
assessing
is
work
solely
of
protected
the
Fx = 19.5 lb
provided
integrity
of
and
work
this
dmB
dmA
dm C
v +
v v
dt By
dt Ay
dt Cy
any
destroy
of
and
Fy = 1.9559 = 1.96 lb
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
sale
will
their
4
Fy = 0.06606(25) + 0.03171 a b(12) - 0
5
courses
part
the
is
This
+ c ©Fy =
vB
5
4
SOLUTION
vC(p)a
12 ft/s
Ans.
25 ft/s
15–122.
A speedboat is powered by the jet drive shown. Seawater is
drawn into the pump housing at the rate of 20 ft3>s through a
6-in.-diameter intake A. An impeller accelerates the water
flow and forces it out horizontally through a 4-in.- diameter
nozzle B. Determine the horizontal and vertical components
of thrust exerted on the speedboat. The specific weight of
seawater is gsw = 64.3 lb>ft3.
B
SOLUTION
45
Steady Flow Equation: The speed of the sea water at the hull bottom A and B are
Q
Q
20
20
vA =
=
= 101.86 ft>s and vB =
=
= 229.18 ft>s and
AA
AB
p 6 2
p 4 2
a b
a b
4 12
4 12
the mass flow rate at the hull bottom A and nozle B are the same, i.e.,
dmA
dmB
64.3
dm
=
=
= rsw Q = a
b(20) = 39.94 slug>s. By referring to the
dt
dt
dt
32.2
free-body diagram of the control volume shown in Fig. a,
C A vB B x - A vA B x D ;
Fx = 39.94 A 229.18 - 101.86 cos 45° B
laws
or
+ b ©F = dm
a:
x
dt
= 6276.55 lb = 6.28 kip
in
teaching
Web)
Ans.
C A vB B y - A vA B y D ;
Fy = 39.94 A 101.86 sin 45° - 0 B
Wide
States
World
permitted.
Dissemination
dm
dt
copyright
a + c b ©Fy =
instructors
= 2876.53 lb = 2.28 kip
sale
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destroy
of
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any
courses
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is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
A
15–123.
A plow located on the front of a locomotive scoops up snow
at the rate of 10 ft3>s and stores it in the train. If the
locomotive is traveling at a constant speed of 12 ft>s,
determine the resistance to motion caused by the shoveling.
The specific weight of snow is gs = 6 lb>ft3.
SOLUTION
©Fx = m
dm t
dv
+ vD>t
dt
dt
F = 0 + (12 - 0) a
10(6)
b
32.2
F = 22.4 lb
sale
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and
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courses
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is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*15–124.
The boat has a mass of 180 kg and is traveling forward on a
river with a constant velocity of 70 km>h, measured relative
to the river. The river is flowing in the opposite direction at
5 km>h. If a tube is placed in the water, as shown, and it
collects 40 kg of water in the boat in 80 s, determine the
horizontal thrust T on the tube that is required to overcome
the resistance due to the water collection and yet maintain
the constant speed of the boat. rw = 1 Mg>m3.
T
SOLUTION
40
dm
=
= 0.5 kg>s
dt
80
vD>t = (70)a
dm i
dv
+ vD>i
dt
dt
T = 0 + 19.444(0.5) = 9.72 N
Ans.
sale
will
their
destroy
of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
©Fs = m
1000
b = 19.444 m>s
3600
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
vR
5 km/h
15–125.
Water is discharged from a nozzle with a velocity of 12 m>s
and strikes the blade mounted on the 20-kg cart. Determine
the tension developed in the cord, needed to hold the cart
stationary, and the normal reaction of the wheels on the
cart. The nozzle has a diameter of 50 mm and the density of
water is rw = 1000 kg>mg3.
45⬚
A
SOLUTION
Steady Flow Equation: Here, the mass flow rate at sections A and B of the control
p
dm
= rWQ = rWAv = 1000 c (0.052) d (12) = 7.5p kg>s
volume is
dt
4
Referring to the free-body diagram of the control volume shown in Fig. a,
dm
+
: ©Fx = dt [(vB)x - (vA)x];
- Fx = 7.5p(12 cos 45° - 12)
Fx = 82.81 N
dm
+ c ©Fy =
[(vB)y - (vA)y];
dt
Fy = 7.5p(12 sin 45° - 0)
or
Fy = 199.93 N
in
teaching
Web)
laws
Equilibrium: Using the results of Fx and Fy and referring to the free-body diagram
of the cart shown in Fig. b,
+ c ©Fy = 0;
N - 20(9.81) - 199.93 = 0
N = 396 N
Ans.
Wide
T = 82.8 N
Dissemination
82.81 - T = 0
copyright
+ ©F = 0;
:
x
sale
will
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destroy
of
and
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courses
part
the
is
This
provided
integrity
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and
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this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
B
15–126.
Water is flowing from the 150-mm-diameter fire hydrant with
a velocity vB = 15 m>s. Determine the horizontal and vertical
components of force and the moment developed at the base
joint A, if the static (gauge) pressure at A is 50 kPa. The
diameter of the fire hydrant at A is 200 mm. rw = 1 Mg>m3.
vB
15 m/s
B
500 mm
SOLUTION
dm
= rvAA B = 1000(15)(p)(0.075)2
dt
A
dm
= 265.07 kg>s
dt
vA = (
dm 1
265.07
)
=
dt rA A
1000(p)(0.1)2
vA = 8.4375 m>s
laws
or
+ ©F = dm (v - v )
;
x
Ax
dt Bx
A x = 265.07(15 - 0) = 3.98 kN
in
teaching
Web)
Ans.
dm
(vBy - vAy)
dt
instructors
States
World
permitted.
Dissemination
Wide
copyright
+ c ©Fy =
United
on
learning.
is
of
the
not
-A y + 50(103)(p)(0.1)2 = 265.07(0- 8.4375)
Ans.
work
student
the
by
and
use
Ay = 3.81 kN
the
(including
for
dm
(dAB vB - dAA vA)
dt
assessing
is
work
solely
of
protected
a + ©MA =
integrity
provided
sale
will
their
destroy
of
and
any
courses
part
the
is
This
M = 1.99 kN # m
of
and
work
this
M = 265.07(0.5(15) - 0)
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
15–127.
A coil of heavy open chain is used to reduce the stopping
distance of a sled that has a mass M and travels at a speed of
v0. Determine the required mass per unit length of the
chain needed to slow down the sled to (1>2)v0 within a
distance x = s if the sled is hooked to the chain at x = 0.
Neglect friction between the chain and the ground.
SOLUTION
Observing the free-body diagram of the system shown in Fig. a, notice that the pair
of forces F, which develop due to the change in momentum of the chain, cancel each
other since they are internal to the system. Here, vD>s = v since the chain on the
dms
ground is at rest. The rate at which the system gains mass is
= m¿v and the
dt
mass of the system is m = m¿x + M. Referring to Fig. a,
dv
+ v A m¿v B
dt
0 = A m¿x + M B
dv
+ m¿v2
dt
(1)
in
teaching
Web)
laws
0 = A m¿x + M)
or
+ b ©F = m dv + v dms ;
a:
s
D>s
dt
dt
dx
dx
= v or dt =
,
dt
v
instructors
States
World
permitted.
Dissemination
Wide
copyright
Since
United
on
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is
of
the
not
dv
+ m¿v2 = 0
dx
the
by
and
use
A m¿x + M B v
(2)
is
work
solely
of
protected
the
(including
for
work
student
dv
m¿
= -¢
≤ dx
v
m¿x + M
this
assessing
1
v at x = s,
2 0
provided
integrity
of
and
work
Integrating using the limit v = v0 at x = 0 and v =
v0
= -1n A m¿x + M B
the
is
s
part
any
courses
destroy
of
their
1
2 v0
will
1n v
dv
m¿
= b dx
a
v
L0 m¿x + M
and
Lv0
s
This
v0
sale
1
2
0
1
M
=
2
m¿s + M
m¿ =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
M
s
Ans.
*15–128.
The car is used to scoop up water that is lying in a trough at
the tracks. Determine the force needed to pull the car
forward at constant velocity v for each of the three cases.
The scoop has a cross-sectional area A and the density of
water is rw .
v
v
F1
F2
(a)
(b)
v
SOLUTION
F3
The system consists of the car and the scoop. In all cases
©Ft = m
dme
dv
- vD>e
dt
dt
(c)
F = 0 - v(r)(A) v
F = v2 r A
sale
will
their
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of
and
any
courses
part
the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
instructors
States
World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
15–129.
250 mm
The water flow enters below the hydrant at C at the rate of
0.75 m3>s. It is then divided equally between the two outlets at A and B. If the gauge pressure at C is 300 kPa, determine the horizontal and vertical force reactions and the
moment reaction on the fixed support at C. The diameter
of the two outlets at A and B is 75 mm, and the diameter of
the inlet pipe at C is 150 mm. The density of water is
rw = 1000 kg>m3. Neglect the mass of the contained water
and the hydrant.
30⬚
A
B
650 mm
600 mm
SOLUTION
Free-Body Diagram: The free-body diagram of the control volume is shown in
Fig. a. The force exerted on section A due to the water pressure is FC = pCAC =
p
300(103)c A 0.152 B d = 5301.44 N. The mass flow rate at sections A, B, and C, are
4
Q
dmC
dmA
dmB
0.75
=
= rW a b = 1000a
b = 375 kg>s
= rWQ =
and
dt
dt
2
2
dt
1000(0.75) = 750 kg>s.
C
Web)
laws
Q
0.75
=
= 42.44 m>s.
p
AC
(0.152)
4
in
vC =
teaching
Q>2
0.75>2
=
= 84.88 m>s
p
AA
(0.0752)
4
Wide
copyright
vA = vB =
or
The speed of the water at sections A, B, and C are
and
use
United
on
learning.
is
of
the
for
work
student
the
by
Cx = - 375(84.88 cos 30°) + 375(84.88) - 0
Ans.
and
work
this
assessing
is
work
solely
of
protected
the
(including
Cx = 4264.54 N = 4.26 kN
dmA
dmB
dmC
(vA)y +
(vB)y (vC)y;
+ c ©Fy =
dt
dt
dt
part
the
is
This
provided
integrity
of
- Cy + 5301.44 = 375(84.88 sin 30°) + 0 - 750(42.44)
any
courses
Ans.
destroy
of
and
Cy = 21 216.93 N = 2.12 kN
sale
will
their
Writing the steady flow equation about point C,
dmA
dmB
dmC
dvA +
dvB dvC;
+ ©MC =
dt
dt
dt
-MC = 375(0.65)(84.88 cos 30°) - 375(0.25)(84.88 sin 30°)
+ [- 375(0.6)(84.88)] - 0
MC = 5159.28 N # m = 5.16 kN # m
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
not
instructors
States
World
permitted.
Dissemination
Steady Flow Equation: Writing the force steady flow equations along the x and
y axes,
+ ©F = dmA (v ) + dmB (v ) - dm C (v ) ;
:
x
A x
B x
C x
dt
dt
dt
15–130.
The mini hovercraft is designed so that air is drawn in at a
constant rate of 20 m3>s by the-fan blade and channeled to
provide a vertical thrust F, just sufficient to lift the hovercraft
off the water, while the remaining air is channeled to produce
a horizontal thrust T on the hovercraft. If the air is discharged horizontally at 200 m>s and vertically at 800 m>s,
determine the thrust T produced. The hovercraft and its passenger have a total mass of 1.5 Mg, and the density of the air
is ra = 1.20 kg>m3.
T
F
SOLUTION
Steady Flow Equation: The free-body diagram of the control volume A is shown in
dmA
Fig. a. The mass flow rate through the control volume is
= raQA = 1.20QA.
dt
Since the air intakes from a large reservoir (atmosphere), the velocity of the air
entering the control volume can be considered zero; i.e., (vA)in = 0. Also, the force
acting on the control volume is required to equal the weight of the hovercraft. Thus,
F = 1500(9.81) N.
dmA
c (vA)out - (vA)in d ; 1500(9.81) = 1.20QA (800 - 0)
dt
laws
or
QA = 15.328 m3>s
and
use
United
on
learning.
is
of
the
not
instructors
States
World
sale
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destroy
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and
any
courses
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is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
+ )©Fx = dmB c(vB)out - (vB)in d ; T = 5.606(200 - 0) = 1121.25 N = 1.12 kN
(;
dt
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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permitted.
Dissemination
Wide
copyright
in
teaching
The flow rate through the control volume B, Fig. b, is QB = 20 - QA =
20 - 15.328 = 4.672 m3>s. Thus, the mass flow rate of this control volume is
dmB
= raQB = 1.20(4.672) = 5.606 kg>s. Again, the intake velocity of the control
dt
volume B is equal to zero; i.e., (vB)in = 0. Referring to the free-body diagram of this
control volume, Fig. b,
Web)
( + T )©Fy =
15–131.
B
Sand is discharged from the silo at A at a rate of 50 kg>s with
a vertical velocity of 10 m>s onto the conveyor belt, which is
moving with a constant velocity of 1.5 m>s. If the conveyor
system and the sand on it have a total mass of 750 kg and
center of mass at point G, determine the horizontal and vertical components of reaction at the pin support B roller support A. Neglect the thickness of the conveyor.
1.5 m/s
G
30⬚
A
SOLUTION
10 m/s
4m
Steady Flow Equation: The moment steady flow equation will be written about
point B to eliminate Bx and By. Referring to the free-body diagram of the control
volume shown in Fig. a,
+ ©MB =
dm
(dvB - dvA);
dt
750(9.81)(4) - Ay(8) = 50[0 - 8(5)]
Ay = 4178.5 N = 4.18 kN
Ans.
Writing the force steady flow equation along the x and y axes,
-Bx = 50(1.5 cos 30° - 0)
Bx = | -64.95 N| = 65.0 N :
Ans.
laws
or
+ ©F = dm [(v ) - (v ) ];
:
x
B x
A x
dt
teaching
Web)
By + 4178.5 - 750(9.81)
in
dm
[(vB)y - (vA)y];
dt
Wide
copyright
+ c ©Fy =
not
instructors
States
World
permitted.
Dissemination
= 50[1.5 sin 30° - (-10)]
Ans.
sale
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and
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This
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integrity
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this
assessing
is
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solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
By = 3716.25 N = 3.72 kN c
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
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4m
*15–132.
The fan blows air at 6000 ft3>min. If the fan has a weight of
30 lb and a center of gravity at G, determine the smallest
diameter d of its base so that it will not tip over. The specific
weight of air is g = 0.076 lb>ft3.
1.5 ft
G
0.5 ft
SOLUTION
4 ft
1 min
6000 ft3
b * a
b = 100 ft3>s. Then,
Equations of Steady Flow: Here Q = a
min
60 s
Q
dm
100
0.076
y =
= p
= ra Q =
(100) = 0.2360 slug>s.
= 56.59 ft>s. Also,
2
A
dt
32.2
(1.5
)
4
Applying Eq. 15–26 we have
a + ©MO =
dm
d
(dOB yB - dOA yA B ; 30 a 0.5 + b = 0.2360 [4(56.59) - 0]
dt
2
d
d = 2.56 ft
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This
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(including
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the
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Dissemination
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Web)
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
15–133.
The bend is connected to the pipe at flanges A and B as
shown. If the diameter of the pipe is 1 ft and it carries a
discharge of 50 ft3>s, determine the horizontal and vertical
components of force reaction and the moment reaction
exerted at the fixed base D of the support. The total weight
of the bend and the water within it is 500 lb, with a mass
center at point G. The gauge pressure of the water at the
flanges at A and B are 15 psi and 12 psi, respectively.
Assume that no force is transferred to the flanges at A and
B. The specific weight of water is gw = 62.4 lb>ft3.
G
A
45
4 ft
D
SOLUTION
teaching
Web)
laws
or
Free-Body Diagram: The free-body of the control volume is shown in Fig. a. The
force exerted on sections A and B due to the water pressure is
p
p
FA = PA AA = 15c A 122 B d = 1696.46 lb
and
FB = PB AB = 12c A 122 B d
4
4
= 1357.17 lb. The speed of the water at, sections A and B are
Q
50
=
= 63.66 ft>s. Also, the mass flow rate at these two sections
vA = vB =
p 2
A
A1 B
4
dm A
dm B
62.4
are
=
= rW Q = a
b (50) = 96.894 slug>s.
dt
dt
32.2
Dissemination
Wide
copyright
in
Steady Flow Equation: The moment steady flow equation will be wr titen about
point D to eliminate Dx and Dy.
the
not
instructors
States
World
permitted.
dm B
dm A
dvB dvA ;
dt
dt
use
United
on
learning.
is
of
a + ©MD =
for
work
student
the
by
and
MD + 1357.17 cos 45°(4) - 500 A 1.5 cos 45° B - 1696.46(4)
work
solely
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protected
provided
integrity
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and
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assessing
is
Ans.
any
courses
part
the
is
This
destroy
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and
will
their
sale
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
the
(including
= -96.894(4) A 63.66 cos 45° B - [-96.894(4)(63.66)]
MD = 10 704.35 lb # ft = 10.7 kip # ft
B
1.5 ft
15–133. continued
Writing the force steady flow equation along the x and y axes,
A + c B ©Fy =
dm
c A vB B y - A vA B y d ;
dt
Dy - 500 - 1357.17 sin 45° = 96.894(63.66 sin 45° - 0)
Dy = 5821.44 lb = 5.82 kip
+ b ©F = dm
a:
x
dt
Ans.
C A vB B x - A vA B x D ;
1696.46 - 1357.17 cos 45° - Dx = 96.894[63.66 cos 45° - 63.66 D
Dx = 2543.51 lb = 2.54 kip
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This
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copyright
in
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Web)
laws
or
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
15–134.
Each of the two stages A and B of the rocket has a mass of
2 Mg when their fuel tanks are empty. They each carry
500 kg of fuel and are capable of consuming it at a rate of
50 kg>sand eject it with a constant velocity of 2500 m> s,
measured with respect to the rocket. The rocket is launched
vertically from rest by first igniting stage B. Then stage A is
ignited immediately after all the fuel in B is consumed and
A has separated from B. Determine the maximum velocity
of stage A. Neglect drag resistance and the variation of the
rocket’s weight with altitude.
A
B
SOLUTION
The mass of the rocket at any instant t is m = (M + m0) - qt. Thus, its weight at the
same instant is W = mg = [(M + m0) - qt]g.
+ c ©Fs = m
dme
dv
dv
; -[(M + m0) - qt]g = [(M + m0) - qt]
- vD>e
- vD>eq
dt
dt
dt
vD>eq
dv
=
- g
dt
(M + m0) - qt
M = 4000 kg,
q = 50 kg>s ,
and
or
m0 = 1000 kg ,
laws
During the first stage,
vD>e = 2500 m>s. Thus,
Dissemination
Wide
copyright
in
teaching
Web)
2500(50)
dv
=
- 9.81
dt
(4000 + 1000) - 50t
by
and
use
United
on
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is
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the
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instructors
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World
permitted.
2500
dv
= a
- 9.81 b m>s2
dt
100 - t
is
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solely
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protected
the
(including
for
work
student
the
The time that it take to complete the first stage is equal to the time for all the fuel
500
in the rocket to be consumed, i.e., t =
= 10 s. Integrating,
50
this
assessing
2500
- 9.81 b dt
100 - t
integrity
of
and
work
L0
a
the
is
This
L0
10 s
dv =
provided
v1
part
10 s
their
destroy
0
of
and
any
courses
v1 = [- 2500 ln(100 - t) - 9.81t] 2
sale
will
= 165.30 m>s
During the second stage of launching, M = 2000 kg, m0 = 500 kg, q = 50 kg>s, and
vD>e = 2500 m>s. Thus, Eq. (1) becomes
2500(50)
dv
=
- 9.81
dt
(2000 + 500) - 50t
2500
dv
= a
- 9.81 b m>s2
dt
50 - t
The maximum velocity of rocket A occurs when it has consumed all the fuel. Thus,
500
the time taken is given by t =
= 10 s. Integrating with the initial condition
50
v = v1 = 165.30 m>s when t = 0 s,
vmax
10 s
dv =
L165.30 m>s
L0
a
2500
- 9.81b dt
50 - t
vmax - 165.30 = [ - 2500 ln(50 - t) - 9.81t] 2
10 s
0
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
vmax = 625 m>s
Ans.
15–135.
A power lawn mower hovers very close over the ground.
This is done by drawing air in at a speed of 6 m>s through
an intake unit A, which has a cross-sectional area of
A A = 0.25 m2, and then discharging it at the ground, B,
where the cross-sectional area is A B = 0.35 m2. If air at A
is subjected only to atmospheric pressure, determine the
air pressure which the lawn mower exerts on the ground
when the weight of the mower is freely supported and no
load is placed on the handle. The mower has a mass of
15 kg with center of mass at G. Assume that air has a
constant density of ra = 1.22 kg>m3.
vA
A
G
B
SOLUTION
dm
= r A A vA = 1.22(0.25)(6) = 1.83 kg>s
dt
+ c ©Fy =
dm
((vB)y - (vA)y)
dt
P = (0.35) - 15(9.81) = 1.83(0 - ( -6))
P = 452 Pa
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This
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(including
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work
student
the
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and
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on
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of
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in
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Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
*15–136.
The 12-ft-long open-link chain has 2 ft of its end suspended
from the hole as shown. If a force of P = 10 lb is applied to
its end and the chain is released from rest, determine the
velocity of the chain at the instant the entire chain has been
extended. The chain has a weight of 2 lb> ft.
2 ft
SOLUTION
P ⫽ 10 lb
From the free-body diagram of the system shown in Fig. a, F cancels since it is internal to the system. Here, vD>s = v since the chain on the horizontal surface is at
dms
2
rest. The rate at which the chain gains mass is
= a
vb slug>s, and the mass
dt
32.2
2y
of the chain is m = a
b slug.
32.2
Referring to Fig. a,
+ T ©Fs = m
dms
dv
;
+ vD>s
dt
dt
10 + 2y = a
2y dv
2
b
+ va
vb
32.2 dt
32.2
dv
+ 2v2 = 322 + 64.4y
dt
laws
or
2y
in
teaching
Web)
dy
dy
, then
= v or dt =
v
dt
Wide
copyright
Since
World
permitted.
Dissemination
dv
+ 2v2 = 322 + 64.4y
dy
United
on
learning.
is
of
the
not
instructors
States
2yv
the
by
and
use
Multiplying by y dy,
the
(including
for
work
student
dv
+ 2yv2 b dy = A 322y + 64.4y2 B dy
dy
dy
assessing
is
this
dv
+ 2yv2, then
dy
integrity
of
and
= 2vy2
work
d A v2y2 B
provided
Since
work
solely
of
protected
a2vy2
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destroy
of
and
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courses
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the
is
This
d A v2y2 B = A 322y + 64.4y2 B dy
sale
will
Integrating,
v2y2 = 161y2 + 21.467y3 + C
Substituting the initial condition v = 0 at y = 2 ft,
0 = 161 A 22 B + 21.467 A 23 B + C
C = - 815.73
ft4
s2
Thus,
v2y2 = 161y2 + 21.467y3 - 815.73
At the instant the entire chain is in motion y = 12 ft.
v2 A 122 B = 161 A 122 B + 21.467 A 123 B - 815.73
v = 20.3 ft>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Ans.
15–137.
P
A chain of mass m0 per unit length is loosely coiled on the
floor. If one end is subjected to a constant force P when
y = 0, determine the velocity of the chain as a function of y.
y
SOLUTION
From the free-body diagram of the system shown in Fig. a, F cancels since it is
internal to the system. Here, vD>s = v since the chain on the horizontal surface is at
dms
rest. The rate at which the chain gains mass is
= m0v, and the mass of the
dt
system is m = m0y. Referring to Fig. a,
P - m0gy = m0y
dv
+ v(m0v)
dt
P - m0gy = m0y
dv
+ m0v2
dt
dv
P
+ v2 =
- gy
m0
dt
in
teaching
Web)
y
or
dms
dv
;
+ vD>s
dt
dt
laws
+ c ©Fs = m
Wide
copyright
dy
dy
,
= v or dt =
v
dt
instructors
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permitted.
Dissemination
Since
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is
of
the
not
dv
P
+ v2 =
- gy
m0
dy
vy
the
(including
for
work
student
Multiplying by 2y dy,
work
solely
of
protected
dv
2P
+ 2v2y b dy = a y - 2gy2 bdy
m0
dy
(1)
integrity
of
and
provided
dv
+ 2yv2, then Eq. (1) can be written as
dy
part
the
is
= 2vy2
destroy
will
their
2P
y - 2gy2 b dy
m0
sale
d A v2y2 B = a
of
and
any
dy
courses
d A v2y2 B
This
Since
work
this
assessing
is
a2vy2
Integrating,
L
d A v2y2 B =
v2y2 =
2P
a y - 2gy2 b dy
L m0
P 2
2
y - gy3 + C
m0
3
Substituting v = 0 at y = 0,
0 = 0 - 0 + C
C = 0
Thus,
v2y2 =
v =
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
P 2
2
y - gy3
m0
3
2
P
- gy
3
B m0
Ans.
15–138.
The second stage of a two-stage rocket weighs 2000 lb
(empty) and is launched from the first stage with a velocity
of 3000 mi>h. The fuel in the second stage weighs 1000 lb. If
it is consumed at the rate of 50 lb>s and ejected with a
relative velocity of 8000 ft>s, determine the acceleration of
the second stage just after the engine is fired. What is the
rocket’s acceleration just before all the fuel is consumed?
Neglect the effect of gravitation.
SOLUTION
Initially,
©Fs = m
0 =
dme
dv
- v D>e a
b
dt
dt
50
3000
a - 8000 a
b
32.2
32.2
a = 133 ft>s2
Ans.
Finally,
or
50
2000
a - 8000 a
b
32.2
32.2
teaching
Web)
laws
0 =
a = 200 ft>s2
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This
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the
(including
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student
the
by
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on
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is
of
the
not
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World
permitted.
Dissemination
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copyright
in
Ans.
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
15–139.
The missile weighs 40 000 lb. The constant thrust provided
by the turbojet engine is T = 15 000 lb. Additional thrust is
provided by two rocket boosters B. The propellant in each
booster is burned at a constant rate of 150 lb/s, with a
relative exhaust velocity of 3000 ft>s. If the mass of the
propellant lost by the turbojet engine can be neglected,
determine the velocity of the missile after the 4-s burn time
of the boosters.The initial velocity of the missile is 300 mi/h.
T
B
SOLUTION
+ ©F = m dv - v dme
:
s
D>e
dt
dt
At a time t, m = m0 - ct, where c =
T + cvD>e
m0 - ct
c
b1n a
m0
b + y0
m0 - ct
or
T + cvD>e
b dt
(1)
in
v = a
L0
a
Web)
Lv0
t
dv =
laws
v
dv
- vD>e c
dt
teaching
T = (m0 - ct)
dme
.
dt
40 000
150
= 1242.24 slug, c = 2 a
b = 9.3168 slug>s, vD>e = 3000 ft>s,
32.2
32.2
the
on
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of
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and
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student
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the
is
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vmax = 580 ft s
This
provided
integrity
of
and
work
vmax = a
assessing
is
work
solely
15 000 + 9.3168(3000)
1242.24
b 1n a
b + 440
9.3168
1242.24 - 9.3168(4)
of
protected
the
(including
Substitute the numerical values into Eq. (1):
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Ans.
permitted.
Dissemination
World
not
instructors
States
300(5280)
= 440 ft>s.
3600
the
t = 4 s, v0 =
Wide
copyright
Here, m0 =
*15–140.
The 10-Mg helicopter carries a bucket containing 500 kg of
water, which is used to fight fires. If it hovers over the land in
a fixed position and then releases 50 kg>s of water at 10 m>s,
measured relative to the helicopter, determine the initial
upward acceleration the helicopter experiences as the water
is being released.
a
SOLUTION
+ c ©Ft = m
dm e
dy
- yD>e
dt
dt
Initially, the bucket is full of water, hence m = 10(103) + 0.5(103) = 10.5(103) kg
0 = 10.5(103)a - (10)(50)
a = 0.0476 m>s2
sale
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and
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courses
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the
is
This
provided
integrity
of
and
work
this
assessing
is
work
solely
of
protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
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permitted.
Dissemination
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Web)
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or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
15–141.
The earthmover initially carries 10 m3 of sand having a
density of 1520 kg>m3. The sand is unloaded horizontally
through a 2.5 m 2 dumping port P at a rate of 900 kg>s
measured relative to the port. Determine the resultant
tractive force F at its front wheels if the acceleration of the
earthmover is 0.1 m>s2 when half the sand is dumped. When
empty, the earthmover has a mass of 30 Mg. Neglect any
resistance to forward motion and the mass of the wheels. The
rear wheels are free to roll.
P
F
SOLUTION
When half the sand remains,
m = 30 000 +
1
(10)(1520) = 37 600 kg
2
dm
= 900 kg>s = r vD>e A
dt
900 = 1520(vD>e)(2.5)
laws
or
vD>e = 0.237 m>s
in
teaching
Web)
dv
= 0.1
dt
Wide
copyright
a =
of
the
not
instructors
States
World
permitted.
Dissemination
+ ©F = m dv - dm v
;
s
dt
dt
by
and
use
United
on
learning.
is
F = 37 600(0.1) - 900(0.237)
Ans.
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This
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integrity
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assessing
is
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protected
the
(including
for
work
student
the
F = 3.55 kN
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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15–142.
The 12-Mg jet airplane has a constant speed of 950 km>h
when it is flying along a horizontal straight line. Air enters
the intake scoops S at the rate of 50 m3>s. If the engine burns
fuel at the rate of 0.4 kg>s and the gas (air and fuel) is
exhausted relative to the plane with a speed of 450 m>s,
determine the resultant drag force exerted on the plane by
air resistance. Assume that air has a constant density of
1.22 kg>m3. Hint: Since mass both enters and exits the plane,
Eqs. 15–28 and 15–29 must be combined to yield
dme
dmi
dv
©Fs = m
- vD>e
+ vD>i
.
dt
dt
dt
v
950 km/h
S
SOLUTION
©Fs = m
dm e
dm i
dv
(vD>E) +
(v )
dt
dt
dt D>i
v = 950 km>h = 0.2639 km>s,
(1)
dv
= 0
dt
or
vD>E = 0.45 km>s
teaching
Web)
laws
vD>t = 0.2639 km>s
States
World
permitted.
Dissemination
Wide
copyright
in
dm t
= 50(1.22) = 61.0 kg>s
dt
the
by
and
use
United
on
learning.
is
of
the
not
instructors
dm e
= 0.4 + 61.0 = 61.4 kg>s
dt
for
work
student
Forces T and R are incorporated into Eq. (1) as the last two terms in the equation.
is
work
solely
of
protected
the
(including
+ ) - F = 0 - (0.45)(61.4) + (0.2639)(61)
(;
D
assessing
FD = 11.5 kN
sale
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destroy
of
and
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is
This
provided
integrity
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this
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
15–143.
The jet is traveling at a speed of 500 mi>h, 30° with the
horizontal. If the fuel is being spent at 3 lb>s,and the engine
takes in air at 400 lb>s, whereas the exhaust gas (air and
fuel) has a relative speed of 32 800 ft>s, determine the
acceleration of the plane at this instant. The drag resistance
of the air is FD = 10.7v22 lb, where the speed is measured
in ft>s. The jet has a weight of 15 000 lb. Hint: Since mass
both enters and exits the plane, Eqs. 15-28 and 15-29 must
be combined to yield
©Fx = m
500 mi/ h
30
dme
dmi
dv
.
- vD>e
+ vD>i
dt
dt
dt
SOLUTION
dmi
400
=
= 12.42 slug>s
dt
32.2
dme
403
=
= 12.52 slug>s
dt
32.2
v = vD>i = 500 mi>h = 733.3 ft>s
teaching
Web)
laws
or
dme
dmi
dv
- vD>e
+ vD>i
dt
dt
dt
in
a + ©Fs = m
15 000 dv
- 32 800(12.52) + 733.3(12.42)
32.2 dt
instructors
States
World
permitted.
Dissemination
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-(15 000) sin 30° - 0.7(733.3)2 =
of
the
not
dv
= 37.5 ft>s2
dt
United
on
learning.
is
Ans.
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a =
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*15–144.
A four-engine commercial jumbo jet is cruising at a
constant speed of 800 km>h in level flight when all four
engines are in operation. Each of the engines is capable of
discharging combustion gases with a velocity of 775 m>s
relative to the plane. If during a test two of the engines, one
on each side of the plane, are shut off, determine the new
cruising speed of the jet. Assume that air resistance (drag) is
proportional to the square of the speed, that is, FD = cv2,
where c is a constant to be determined. Neglect the loss of
mass due to fuel consumption.
SOLUTION
Steady Flow Equation: Since the air is collected from a large source (the
atmosphere), its entrance speed into the engine is negligible. The exit speed of the
air from the engine is
+ b
a:
ve + vp + ve>p
ve = - 222.22 + 775 = 552.78 m>s :
in
teaching
Web)
+ b
a:
laws
or
When the four engines are in operation, the airplane has a constant speed of
m
1h
b = 222.22 m>s. Thus,
vp = c 800(103) d a
h 3600 s
Dissemination
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Referring to the free-body diagram of the airplane shown in Fig. a,
the
not
instructors
States
World
permitted.
dm
(552.78 - 0)
dt
C(222.222) = 4
dm
dt
work
student
the
(including
for
the
by
C = 0.044775
and
use
United
on
learning.
is
of
+ ©F = dm C A v B - A v B D ;
:
x
B x
A x
dt
integrity
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and
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this
ve = - vp + 775
the
is
This
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a:
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When only two engines are in operation, the exit speed of the air is
and
any
courses
part
Using the result for C,
dm
dm
≤ A vp 2 B = 2 C - vp + 775 B - 0 D
dt
dt
will
destroy
of
their
C A vB B x - A vA B x D ; ¢ 0.044775
sale
+ ©F = dm
:
x
dt
0.044775vp 2 + 2vp - 1550 = 0
Solving for the positive root,
vp = 165.06 m s = 594 km h
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Ans.
15–145.
v
The car has a mass m0 and is used to tow the smooth chain
having a total length l and a mass per unit of length m¿. If
the chain is originally piled up, determine the tractive force
F that must be supplied by the rear wheels of the car,
necessary to maintain a constant speed v while the chain is
being drawn out.
F
SOLUTION
+ ©F = m dv + v dmi
:
s
D>i
dt
dt
At a time t, m = m0 + ct, where c =
Here, vD>i = v,
dmi
m¿dx
=
= m¿v.
dt
dt
dv
= 0.
dt
F = (m0 - m¿v )(0) + v(m¿v) = m¿v2
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or transmission in any form or by any means, electronic, mechanical,
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15–146.
A rocket has an empty weight of 500 lb and carries 300 lb
of fuel. If the fuel is burned at the rate of 1.5 lb / s and ejected
with a velocity of 4400 ft/ s relative to the rocket, determine
the maximum speed attained by the rocket starting from rest.
Neglect the effect of gravitation on the rocket.
SOLUTION
dme
dv
- vD>e
dt
dt
At a time t, m = m0 - ct, where c =
0 = (m0 - ct)
t
dv =
L0
L0
cvD>e
m0 - ct
≤ dt
m0
≤
m0 - ct
(1)
in
teaching
Web)
v = vD> e ln ¢
¢
dv
- vD>e c
dt
or
v
dme
. In space the weight of the rocket is zero.
dt
laws
+ c ©Fs =
the
= 0.4658 slug>s, vD>e = 4400 ft>s.
not
15
32.2
is
and
use
United
on
= 24.8447 slug, c =
learning.
500 + 300
32.2
of
Here, m0 =
instructors
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The maximum speed occurs when all the fuel is consumed, that is, when
t = 300
15 = 20 s.
for
work
student
the
by
Substitute the numerical into Eq. (1):
work
solely
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protected
the
(including
24.8447
b
24.8447 - (0.4658(20))
assessing
is
vmax = 4400 1na
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Ans.
this
vmax = 2068 ft>s
15–147.
Determine the magnitude of force F as a function of time,
which must be applied to the end of the cord at A to raise
the hook H with a constant speed v = 0.4 m>s. Initially the
chain is at rest on the ground. Neglect the mass of the cord
and the hook. The chain has a mass of 2 kg>m.
A
H
v
0.4 m/s
SOLUTION
dv
= 0,
dt
y = vt
mi = my = mvt
dmi
= mv
dt
+ c ©Fs = m
dm i
dv
+ vD>i (
)
dt
dt
F - mgvt = 0 + v(mv)
laws
or
F = m(gvt + v2)
Wide
copyright
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teaching
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= 2[9.81(0.4)t + (0.4)2]
Ans.
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Dissemination
F = (7.85t + 0.320) N
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*15–148.
The truck has a mass of 50 Mg when empty.When it is unloading 5 m3 of sand at a constant rate of 0.8 m3>s, the sand flows
out the back at a speed of 7 m> s, measured relative to the
truck, in the direction shown. If the truck is free to roll, determine its initial acceleration just as the load begins to empty.
Neglect the mass of the wheels and any frictional resistance to
motion. The density of sand is rs = 1520 kg>m3.
a
45⬚
7 m/s
SOLUTION
A System That Loses Mass: Initially, the total mass of the truck is
dme
and
m = 50(103) + 5(1520) = 57.6(103) kg
= 0.8(1520) = 1216 kg>s .
dt
Applying Eq. 15–29, we have
+ ©F = m dv - v dme ;
:
s
D>e
dt
dt
0 = 57.6(103)a - (0.8 cos 45°)(1216)
a = 0.104 m>s2
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This
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Ans.
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or transmission in any form or by any means, electronic, mechanical,
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15–149.
The chain has a total length L 6 d and a mass per unit
length of m¿ . If a portion h of the chain is suspended over
the table and released, determine the velocity of its end A
as a function of its position y. Neglect friction.
h
y
d
A
SOLUTION
©Fs = m
dm e
dv
+ vD>e
dt
dt
m¿gy = m¿y
dv
+ v(m¿v)
dt
m¿gy = m¿ ay
Since dt =
dy
, we have
v
or
dv
+ v2
dy
in
teaching
Web)
laws
gy = vy
dv
+ v2 b
dt
Dissemination
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Multiply by 2y and integrate:
the
not
instructors
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permitted.
dv
+ 2yv2 b dy
dy
is
use
United
on
learning.
L
a 2vy2
of
L
2gy2 dy =
the
(including
for
work
student
the
by
and
2 3 3
g y + C = v2y2
3
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y3 - h3
2
ga
b
B3
y2
their
v =
2 y3 - h3
ga
b
3
y2
sale
Thus, v2 =
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2
when v = 0, y = h, so that C = - gh3
3
Ans.
16–1.
The angular velocity of the disk is defined by
v = 15t2 + 22 rad>s, where t is in seconds. Determine the
magnitudes of the velocity and acceleration of point A on
the disk when t = 0.5 s.
A
0.8 m
SOLUTION
v = (5 t2 + 2) rad>s
a =
dv
= 10 t
dt
t = 0.5 s
v = 3.25 rad>s
a = 5 rad>s2
vA = vr = 3.25(0.8) = 2.60 m>s
Ans.
laws
or
a z = ar = 5(0.8) = 4 m>s2
in
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a n = v2r = (3.25)2(0.8) = 8.45 m>s2
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copyright
a A = 2(4)2 + (8.45)2 = 9.35 m>s2
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16–2.
A flywheel has its angular speed increased uniformly from
15 rad>s to 60 rad>s in 80 s. If the diameter of the wheel is
2 ft, determine the magnitudes of the normal and tangential
components of acceleration of a point on the rim of the
wheel when t = 80 s, and the total distance the point travels
during the time period.
SOLUTION
v = v0 + ac t
60 = 15 + ac(80)
ac = 0.5625 rad/s2
a t = ar = (0.5625)(1) = 0.562 ft/s2
Ans.
a n = v2r = (60)2(1) = 3600 ft/s2
Ans.
v2 = v20 + 2ac (u - u0)
(60)2 = (15)2 + 2(0.5625)(u-0)
laws
or
u = 3000 rad
teaching
Web)
s = ur = 3000(1) = 3000 ft
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This
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Ans.
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
16–3.
The disk is originally rotating at v0 = 8 rad>s. If it is
subjected to a constant angular acceleration of a = 6 rad>s2,
determine the magnitudes of the velocity and the n and t
components of acceleration of point A at the instant
t = 0.5 s.
V0
B
1.5 ft
2 ft
SOLUTION
v = v0 + ac t
v = 8 + 6(0.5) = 11 rad>s
v = rv;
vA = 2(11) = 22 ft>s
Ans.
at = ra;
(aA)t = 2(6) = 12.0 ft>s2
Ans.
(aA)n = (11)2(2) = 242 ft>s2
Ans.
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an = v2r;
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8 rad/s
A
*16–4.
The disk is originally rotating at v0 = 8 rad>s. If it is
subjected to a constant angular acceleration of a = 6 rad>s2,
determine the magnitudes of the velocity and the n and t
components of acceleration of point B just after the wheel
undergoes 2 revolutions.
V0
B
1.5 ft
2 ft
SOLUTION
v2 = v20 + 2ac (u - u0)
v2 = (8)2 + 2(6)[2(2p) - 0]
v = 14.66 rad>s
Ans.
(aB)t = ar = 6(1.5) = 9.00 ft>s2
Ans.
(aB)n = v2r = (14.66)2(1.5) = 322 ft>s2
Ans.
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vB = vr = 14.66(1.5) = 22.0 ft>s
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8 rad/s
A
16–5.
Initially the motor on the circular saw turns its drive shaft at
v = 120t2>32 rad>s, where t is in seconds. If the radii of gears
A and B are 0.25 in. and 1 in., respectively, determine the
magnitudes of the velocity and acceleration of a tooth C on
the saw blade after the drive shaft rotates u = 5 rad starting
from rest.
SOLUTION
v = 20 t2>3
dv
40 -1>3
=
t
dt
3
a =
du = v dt
t
u
L0
du =
L0
20 t2>3 dt
3
u = 20a b t5>3
5
teaching
Web)
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or
When u = 5 rad
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t = 0.59139 s
States
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permitted.
Dissemination
a = 15.885 rad>s2
on
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is
of
the
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v = 14.091 rad>s
the
by
and
use
United
v A r A = vB r B
the
(including
for
work
student
14.091(0.25) = vB(1)
assessing
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vB = 3.523 rad>s
Ans.
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the
is
and
any
courses
aA rA = aB rB
This
provided
integrity
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and
work
this
vC = vB r = 3.523(2.5) = 8.81 in.>s
aB = 3.9712 rad>s2
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of
15.885(0.25) = aB(1)
(aC)t = aB r = 3.9712(2.5) = 9.928 in.>s2
(aC)n = v2B r = (3.523)2 (2.5) = 31.025 in.>s2
aC = 2(9.928)2 + (31.025)2
= 32.6 in.>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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Ans.
16–6.
A wheel has an initial clockwise angular velocity of 10 rad>s
and a constant angular acceleration of 3 rad>s2. Determine
the number of revolutions it must undergo to acquire a
clockwise angular velocity of 15 rad>s. What time is
required?
SOLUTION
v2 = v20 + 2ac(u - u0)
(15)2 = (10)2 + 2(3)(u- 0)
1
≤ = 3.32 rev.
2p
u = 20.83 rad = 20.83 ¢
Ans.
v = v0 + ac t
15 = 10 + 3t
t = 1.67 s
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This
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16–7.
If gear A rotates with a constant angular acceleration of
aA = 90 rad>s2, starting from rest, determine the time
required for gear D to attain an angular velocity of 600 rpm.
Also, find the number of revolutions of gear D to attain this
angular velocity. Gears A, B, C, and D have radii of 15 mm,
50 mm, 25 mm, and 75 mm, respectively.
D
F
SOLUTION
aB rB = aA rA
rA
15
aB = a baA = a b (90) = 27 rad>s2
rB
50
Since gears C and B share the same shaft, aC = aB = 27 rad>s2. Also, gear D is in
mesh with gear C. Thus,
or
aD rD = aC rC
rC
25
aD = a b aC = a b (27) = 9 rad>s2
rD
75
600 rev 2p rad 1 min
ba
ba
b =
min
1 rev
60 s
20p rad>s. Applying the constant acceleration equation,
Dissemination
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The final angular velocity of gear D is vD = a
of
the
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vD = (vD)0 + aD t
and
use
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is
20p = 0 + 9t
Ans.
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t = 6.98 s
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is
This
(20p)2 = 02 + 2(9)(uD - 0)
provided
integrity
of
and
work
this
assessing
vD2 = (vD)0 2 + 2aD [uD - (uD)0]
uD = (219.32 rad) a
sale
will
= 34.9 rev
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B
C
Gear B is in mesh with gear A. Thus,
Ans.
A
*16–8.
If gear A rotates with an angular velocity of vA =
(uA + 1) rad>s, where uA is the angular displacement of
gear A, measured in radians, determine the angular
acceleration of gear D when uA = 3 rad, starting from rest.
Gears A, B, C, and D have radii of 15 mm, 50 mm, 25 mm,
and 75 mm, respectively.
D
F
SOLUTION
aA duA = vA dvA
aA duA = (uA + 1) d(uA + 1)
aA duA = (uA + 1) duA
aA = (uA + 1)
At uA = 3 rad,
laws
or
aA = 3 + 1 = 4 rad>s2
in
teaching
Web)
Motion of Gear D: Gear A is in mesh with gear B. Thus,
of
the
not
instructors
States
World
permitted.
Dissemination
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copyright
aB rB = aA rA
rA
15
aB = a baA = a b (4) = 1.20 rad>s2
rB
50
the
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protected
aD rD = aC rC
rC
25
aD = a b aC = a b (1.20) = 0.4 rad>s2
rD
75
work
student
the
by
and
use
United
on
learning.
is
Since gears C and B share the same shaft aC = aB = 1.20 rad>s2. Also, gear D is in
mesh with gear C. Thus,
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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B
C
Motion of Gear A:
Ans.
A
16–9.
The vertical-axis windmill consists of two blades that have a
parabolic shape. If the blades are originally at rest and
begin to turn with a constant angular acceleration of
ac = 0.5 rad>s2, determine the magnitude of the velocity
and acceleration of points A and B on the blade after the
blade has rotated through two revolutions.
ac
0.5 rad/s2
B
SOLUTION
10 ft
Angular Motion: The angular velocity of the blade after the blade has rotated
2(2p) = 4p rad can be obtained by applying Eq. 16–7.
20 ft
v2 = v20 + 2ac (u - u0)
v2 = 02 + 2(0.5)(4p - 0)
v = 3.545 rad>s
Motion of A and B: The magnitude of the velocity of point A and B on the blade can
be determined using Eq. 16–8.
Ans.
vB = vrB = 3.545(10) = 35.4 ft>s
Ans.
in
teaching
Web)
laws
or
vA = vrA = 3.545(20) = 70.9 ft>s
States
World
permitted.
Dissemination
Wide
copyright
The tangential and normal components of the acceleration of point A and B can be
determined using Eqs. 16–11 and 16–12 respectively.
on
learning.
is
of
the
not
instructors
(at)A = arA = 0.5(20) = 10.0 ft>s2
student
the
by
and
use
United
(an)A = v2 rA = A 3.5452 B (20) = 251.33 ft>s2
the
(including
for
work
(at)B = arB = 0.5(10) = 5.00 ft>s2
work
this
assessing
is
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solely
of
protected
(an)B = v2 rB = A 3.5452 B (10) = 125.66 ft>s2
part
the
is
This
provided
integrity
of
and
The magnitude of the acceleration of points A and B are
Ans.
and
any
courses
(a)A = 2(at)2A + (an)2A = 210.02 + 251.332 = 252 ft>s2
5.002 + 125.662 = 126 ft s2
destroy
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(at)2B + (an)2B =
their
(a)B = =
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Ans.
A
16–10.
The vertical-axis windmill consists of two blades that have a
parabolic shape. If the blades are originally at rest and begin
to turn with a constant angular acceleration of ac = 0.5 rad>s2,
determine the magnitude of the velocity and acceleration of
points A and B on the blade when t=4 s.
ac
0.5 rad/s2
B
SOLUTION
10 ft
Angular Motion: The angular velocity of the blade at t = 4 s can be obtained by
applying Eq. 16–5.
20 ft
v = v0 + ac t = 0 + 0.5(4) = 2.00 rad>s
Motion of A and B: The magnitude of the velocity of points A and B on the blade
can be determined using Eq. 16–8.
vA = vrA = 2.00(20) = 40.0 ft>s
Ans.
vB = vrB = 2.00(10) = 20.0 ft>s
Ans.
teaching
Web)
laws
or
The tangential and normal components of the acceleration of points A and B can be
determined using Eqs. 16–11 and 16–12 respectively.
Wide
copyright
in
(at)A = arA = 0.5(20) = 10.0 ft>s2
instructors
States
World
permitted.
Dissemination
(an)A = v2 rA = A 2.002 B (20) = 80.0 ft>s2
on
learning.
is
of
the
not
(at)B = arB = 0.5(10) = 5.00 ft>s2
for
work
student
the
by
and
use
United
(an)B = v2 rB = A 2.002 B (10) = 40.0 ft>s2
of
protected
the
(including
The magnitude of the acceleration of points A and B are
Ans.
this
assessing
is
work
solely
(a)A = 2(at)2A + (an)2A = 210.02 + 80.02 = 80.6 ft>s2
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This
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(a)B = 2(at)2B + (an)2B = 25.002 + 40.02 = 40.3 ft>s2
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Ans.
A
16–11.
If the angular velocity of the drum is increased uniformly
from 6 rad>s when t = 0 to 12 rad>s when t = 5 s,
determine the magnitudes of the velocity and acceleration
of points A and B on the belt when t = 1 s. At this instant
the points are located as shown.
45°
B
SOLUTION
v = v0 + act
A
a = 1.2 rad>s2
12 = 6 + a(5)
At t = 1 s,
v = 6 + 1.2(1) = 7.2 rad>s
vA = vB = v r = 7.2 ¢
Ans.
4
≤ = 0.4 ft>s2
12
Ans.
laws
or
aA = ar = 1.2 ¢
4
≤ = 2.4 ft>s
12
in
teaching
Web)
4
≤ = 0.4 ft>s2
12
Wide
copyright
(aB)t = ar = 1.2 ¢
World
permitted.
Dissemination
= 17.28 ft>s2
is
of
the
not
instructors
4
12
States
(aB)n = v2r = (7.2)2
Ans.
on
learning.
0.42 + 17.282 = 17.3 ft s2
United
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This
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work
student
the
by
and
(aB)2t + (aB)2n =
use
aB =
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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4 in.
*16–12.
A motor gives gear A an angular acceleration of
aA = 10.25u3 + 0.52 rad>s2, where u is in radians. If this
gear is initially turning at 1vA20 = 20 rad>s, determine the
angular velocity of gear B after A undergoes an angular
displacement of 10 rev.
(vA)0 ⫽ 20 rad/s
B
0.05 m
A
AA
SOLUTION
aA = 0.25 u3 + 0.5
a du = v dv
vA
20p
L0
(0.25 u3 + 0.5)duA =
20p
(0.0625 u4 + 0.5 u)冷 0
=
vA dvA
L20
vA
1
(vA)2冷 20
2
vA = 1395.94 rad>s
vA rA = vB rB
laws
or
1395.94(0.05) = vB (0.15)
vB = 465 rad>s
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This
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assessing
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protected
the
(including
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student
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on
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Web)
Ans.
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0.15 m
16–13.
A motor gives gear A an angular acceleration of
aA = 14t32 rad>s2, where t is in seconds. If this gear is
initially turning at 1vA20 = 20 rad>s, determine the angular
velocity of gear B when t = 2 s.
(vA)0 ⫽ 20 rad/s
B
0.05 m
A
0.15 m
AA
SOLUTION
a A = 4 t3
dv = a dt
t
vA
L20
dvA =
L0
t
aA dt =
4 t3dt
L0
vA = t4 + 20
When t = 2 s,
vA = 36 rad>s
laws
or
vA rA = vB rB
in
teaching
Web)
36(0.05) = vB (0.15)
Wide
copyright
vB = 12 rad>s
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This
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protected
the
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and
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United
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is
of
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States
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permitted.
Dissemination
Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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16–14.
The disk starts from rest and is given an angular acceleration
a = (2t 2) rad>s2, where t is in seconds. Determine the
angular velocity of the disk and its angular displacement
when t = 4 s.
P
0.4 m
SOLUTION
dv
= 2 t2
dt
a =
t
v
2
dv =
L0
L0
v =
2 32t
t
3 0
v =
23
t
3
2 t dt
When t = 4 s,
2 3
(4) = 42.7 rad>s
3
in
teaching
Web)
2 3
t dt
L0 3
permitted.
Dissemination
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copyright
1 4
t
6
States
World
u =
du =
laws
t
u
L0
Ans.
or
v =
learning.
is
of
the
not
instructors
When t = 4 s,
by
and
use
United
on
1 4
(4) = 42.7 rad
6
sale
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This
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the
(including
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student
the
u =
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or transmission in any form or by any means, electronic, mechanical,
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Ans.
16–15.
The disk starts from rest and is given an angular acceleration
a = (5t1>2) rad>s2, where t is in seconds. Determine the
magnitudes of the normal and tangential components of
acceleration of a point P on the rim of the disk when t = 2 s.
P
0.4 m
SOLUTION
Motion of the Disk: Here, when t = 0, v = 0.
dv = adt
t
v
L0
v2
dv =
v
=
0
v = e
L0
1
5t2dt
10 3 2 t
t2
3
0
10 3
t2 f rad>s
3
teaching
Web)
laws
10 3
A 2 2 B = 9.428 rad>s
3
in
v =
or
When t = 2 s,
Dissemination
Wide
copyright
When t = 2 s,
permitted.
a = 5 A 2 2 B = 7.071 rad>s2
United
on
learning.
is
of
the
not
instructors
States
World
1
for
work
student
the
by
and
use
Motion of point P: The tangential and normal components of the acceleration of
point P when t = 2 s are
Ans.
is
work
solely
of
protected
the
(including
at = ar = 7.071(0.4) = 2.83 m>s2
Ans.
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This
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this
assessing
a n = v2r = 9.4282(0.4) = 35.6 m>s2
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*16–16.
The disk starts at v0 = 1 rad>s when u = 0, and is given an
angular acceleration a = (0.3u) rad>s2, where u is in radians.
Determine the magnitudes of the normal and tangential
components of acceleration of a point P on the rim of the
disk when u = 1 rev.
P
0.4 m
SOLUTION
a = 0.3u
v
L1
vdv =
u
L0
0.3udu
u
1 22v
v
= 0.15u2 2
2
1
0
v2
- 0.5 = 0.15u2
2
laws
or
v = 20.3u2 + 1
in
teaching
Web)
At u = 1 rev = 2p rad
Dissemination
Wide
copyright
v = 20.3(2p)2 + 1
learning.
is
of
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not
instructors
States
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permitted.
v = 3.584 rad>s
Ans.
the
by
and
use
United
on
a t = ar = 0.3(2p) rad>s 2 (0.4 m) = 0.7540 m>s2
protected
the
(including
for
work
student
a n = v2r = (3.584 rad>s)2(0.4 m) = 5.137 m>s2
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This
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assessing
is
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solely
of
a p = 2(0.7540)2 + (5.137)2 = 5.19 m>s2
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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Ans.
16–17.
Starting at (vA)0 = 3 rad>s, when u = 0, s = 0, pulley A is
given an angular acceleration a = (0.6u) rad>s2, where u is
in radians. Determine the speed of block B when it has
risen s = 0.5 m. The pulley has an inner hub D which is
fixed to C and turns with it.
150 mm
C
D
A
50 mm
75 mm
SOLUTION
aa = 0.6uA
uC =
B
0.5
= 6.667 rad
0.075
s
uA(0.05) = (6.667)(0.15)
uA = 20 rad
adu = vdv
vA
0.6uAduA =
L3
vAdvA
laws
L0
or
20
teaching
Web)
1 2 vA
v 2
2 A 3
Wide
States
World
permitted.
Dissemination
0
=
in
20
copyright
0.3u2A 2
United
on
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is
of
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instructors
1 2
v - 4.5
2 A
by
and
use
120 =
the
(including
for
work
student
the
vA = 15.780 rad>s
assessing
is
work
solely
of
protected
15.780(0.05) = vC (0.15)
the
is
This
provided
integrity
of
and
work
this
vC = 5.260 rad>s
part
vB = 5.260(0.075) = 0.394 m>s
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Ans.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
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or transmission in any form or by any means, electronic, mechanical,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
16–18.
Starting from rest when s = 0, pulley A is given a constant
angular acceleration ac = 6 rad>s2. Determine the speed of
block B when it has risen s = 6 m. The pulley has an inner
hub D which is fixed to C and turns with it.
150 mm
C
D
A
50 mm
75 mm
SOLUTION
aA r A = aC r C
6(50) = aC(150)
B
aC = 2 rad>s2
s
aB = aC rB = 2(0.075) = 0.15 m>s2
(+ c)
v2 = v20 + 2 ac (s - s0)
v2 = 0 + 2(0.15)(6 - 0)
v = 1.34 m>s
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the
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the
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laws
or
Ans.
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Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
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16–19.
The vacuum cleaner’s armature shaft S rotates with an
angular acceleration of a = 4v3>4 rad>s2, where v is in
rad>s. Determine the brush’s angular velocity when t = 4 s,
starting from v0 = 1 rad>s, at u = 0. The radii of the shaft
and the brush are 0.25 in. and 1 in., respectively. Neglect the
thickness of the drive belt.
SOLUTION
Motion of the Shaft: The angular velocity of the shaft can be determined from
dt =
L
t
L0
2
t
dt =
A
dvS
L aS
vs
dvS
L1 4vS 3>4
2
vs
t 0 = vS 1>4 1
t = vS 1>4 – 1
laws
or
vS = (t+1) 4
in
teaching
Web)
When t = 4 s
States
World
permitted.
Dissemination
Wide
copyright
vs = 54 = 625 rad>s
and
use
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work
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the
by
vB rB = vs rs
work
solely
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protected
the
(including
rs
0.25
b(625) = 156 rad>s
≤v = a
rB s
1
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vB = ¢
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Ans.
not
instructors
Motion of the Beater Brush: Since the brush is connected to the shaft by a non-slip
belt, then
S
A
S
*16–20.
The operation of “reverse” for a three-speed automotive
transmission is illustrated schematically in the figure. If the
crank shaft G is turning with an angular speed of 60 rad>s,
determine
the
angular
speed
of
the
drive
shaft H. Each of the gears rotates about a fixed axis. Note
that gears A and B, C and D, E and F are in mesh. The radii
of each of these gears are reported in the figure.
A
vH
H
G
F
C
B
D
E
rA ⫽ 90 mm
rB ⫽ rC ⫽ 30 mm
rD ⫽ 50 mm
rE ⫽ 70 mm
rF ⫽ 60 mm
SOLUTION
60(90) = vBC (30)
vBC = 180 rad>s
180(30) = 50(vDE)
vDE = 108 rad>s
108(70) = (60) (vH)
vH = 126 rad>s
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This
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the
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Dissemination
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
vG ⫽ 60 rad/s
16–21.
V0 ⫽ 6 rad/s
A motor gives disk A an angular acceleration of
aA = 10.6t2 + 0.752 rad>s2, where t is in seconds. If the
initial angular velocity of the disk is v0 = 6 rad>s,
determine the magnitudes of the velocity and acceleration
of block B when t = 2 s.
A
0.15 m
SOLUTION
dv = a dt
v
L6
2
dv =
L0
(0.6 t2 + 0.75) dt
B
v - 6 = (0.2 t3 + 0.75 t)|20
v = 9.10 rad>s
Ans.
aB = at = ar = [0.6(2)2 + 0.75](0.15) = 0.472 m>s2
Ans.
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destroy
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and
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is
This
provided
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and
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this
assessing
is
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solely
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protected
the
(including
for
work
student
the
by
and
use
United
on
learning.
is
of
the
not
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World
permitted.
Dissemination
Wide
copyright
in
teaching
Web)
laws
or
vB = vr = 9.10(0.15) = 1.37 m>s
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
16–22.
For a short time the motor turns gear A with an angular
acceleration of aA = (30t1>2) rad>s2, where t is in seconds.
Determine the angular velocity of gear D when t = 5 s,
starting from rest. Gear A is initially at rest. The radii of
gears A, B, C, and D are rA = 25 mm, rB = 100 mm,
rC = 40 mm, and rD = 100 mm, respectively.
A
C
SOLUTION
Motion of the Gear A: The angular velocity of gear A can be determined from
dvA =
L
adt
L
t
vA
dvA =
L0
vA
vA
0
30t1>2dt
L0
t
= 20t3>2 2
0
vA = A 20t
3>2
B rad>s
laws
or
When t = 5 s
Wide
copyright
in
teaching
Web)
vA = 20 A 53>2 B = 223.61 rad>s
of
the
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instructors
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World
and
use
United
on
learning.
is
vB rB = vA rA
the
(including
for
work
student
the
by
rA
25
b(223.61) = 55.90 rad>s
≤v = a
rB A
100
work
solely
of
protected
vC = v B = ¢
work
this
assessing
is
Also, gear D is in mesh with gear C. Then
part
the
is
This
provided
integrity
of
and
vD rD = vC rC
any
courses
rC
40
b (55.90) = 22.4 rad>s
≤v = a
rD C
100
sale
will
their
destroy
of
and
vD = ¢
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to:
Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Ans.
permitted.
Dissemination
Motion of Gears B, C, and D: Gears B and C which are mounted on the same axle
will have the same angular velocity. Since gear B is in mesh with gear A, then
B
D
16–23.
The motor turns gear A so that its angular velocity increases
uniformly from zero to 3000 rev>min after the shaft turns
200 rev. Determine the angular velocity of gear D when
t = 3 s. The radii of gears A, B, C, and D are rA = 25 mm,
rB = 100 mm, rC = 40 mm, and rD = 100 mm, respectively.
A
C
SOLUTION
Motion of Wheel A: Here, vA = a 3000
1 min 2p rad
rev
ba
ba
b = 100p rad>s
min
60 s
1rev
2p rad
b = 400p rad. Since the angular acceleration of gear
1 rev
A is constant, it can be determined from
when uA = (200 rev) a
vA 2 = (vA)0 2 + 2aA C uA - (uA)0 D
(100p)2 = 02 + 2aA (400p - 0)
laws
or
aA = 39.27 rad>s2
teaching
Web)
Thus, the angular velocity of gear A when t = 3 s is
Dissemination
Wide
copyright
in
vA = A v A B 0 + a A t
not
instructors
States
World
permitted.
= 0 + 39.27(3)
by
and
use
United
on
learning.
is
of
the
= 117.81 rad>s
the
(including
for
work
student
the
Motion of Gears B, C, and D: Gears B and C which are mounted on the same axle
will have the same angular velocity. Since gear B is in mesh with gear A, then
assessing
is
work
solely
of
protected
vB rB = vB rA
provided
integrity
of
and
work
this
rA
25