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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–9.
The acceleration of a particle as it moves along a straight
line is given by a = (2t - 1) 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
a = 2t - 1
dv = a dt
L2
v
t
L0
dv =
(2t - 1)dt
v = t2 - t + 2
dx = v dt
Lt
s
ds =
s =
L0
t
(t2 - t + 2)dt
1 3
1
t - t 2 + 2t + 1
3
2
When t = 6 s
v = 32 m>s
Ans.
s = 67 m
Ans.
Since v ≠ 0 for 0 … t … 6 s, then
Ans.
d = 67 - 1 = 66 m
Ans:
v = 32 m>s
s = 67 m
d = 66 m
9
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–10.
A particle moves along a straight line with an acceleration
of a = 5>(3s1>3 + s 5>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 a numerical method to evaluate
the integral.
SOLUTION
a =
5
1
3
5
A 3s + s2 B
a ds = v dv
2
v
5 ds
1
3
5
2
L1 A 3s + s B
0.8351 =
=
L0
v dv
1 2
v
2
Ans.
v = 1.29 m>s
Ans:
v = 1.29 m>s
10
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–16.
A particle is moving along a straight line with an initial
velocity of 6 m>s when it is subjected to a deceleration of
a = (- 1.5v1>2) m>s2, where v is in m>s. Determine how far it
travels before it stops. How much time does this take?
SOLUTION
Distance Traveled: The distance traveled by the particle can be determined by
applying Eq. 12–3.
ds =
vdv
a
s
L0
v
ds =
v
1
L6 m>s - 1.5v2
v
s =
dv
1
L6 m>s
- 0.6667 v2 dv
3
= a -0.4444v2 + 6.532 b m
When v = 0,
3
Ans.
s = - 0.4444a 0 2 b + 6.532 = 6.53 m
Time: The time required for the particle to stop can be determined by applying
Eq. 12–2.
dt =
dv
a
t
L0
v
dt = 1
dv
1
L6 m>s 1.5v 2
v
1
t = - 1.333a v2 b 6 m>s = a3.266 - 1.333v 2 b s
When v = 0,
1
Ans.
t = 3.266 - 1.333a 0 2 b = 3.27 s
Ans:
s = 6.53 m
t = 3.27 s
16
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–19.
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
v1 = 0 + 0.5(60) = 30 m>s
For stage (2) motion, v0 = 30 m>s, s0 = 900 m, ac = 0 and t = 15(60) = 900 s. Thus,
+ B
A:
s = s0 + v0t +
1 2
at
2 c
s2 = 900 + 30(900) + 0 = 27 900 m
For stage (3) motion, v0 = 30 m>s, v = 0, s0 = 27 900 m and ac = - 1 m>s2. Thus,
+ B
A:
v = v0 + act
0 = 30 + ( -1)t
t = 30 s
+
:
s = s0 + v0t +
1 2
at
2 c
s3 = 27 900 + 30(30) +
1
( - 1)(302)
2
Ans.
= 28 350 m = 28.4 km
Ans:
s = 28.4 km
19
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–36.
The s–t graph for a train has been experimentally
determined. From the data, construct the v–t and a–t graphs
for the motion; 0 … t … 40 s. For 0 … t … 30 s, the curve is
s = (0.4t2) m, and then it becomes straight for t Ú 30 s.
s (m)
600
360
Solution
0 … t … 30:
s = 0.4t 2
30
v =
ds
= 0.8t
dt
a =
dv
= 0.8
dt
t (s)
40
30 … t … 40:
s - 360 = a
600 - 360
b(t - 30)
40 - 30
v =
ds
= 24
dt
a =
dv
= 0
dt
s = 24(t - 30) + 360
Ans:
s = 0.4t 2
ds
v =
= 0.8t
dt
dv
= 0.8
a =
dt
s = 24(t - 30) + 360
ds
v =
= 24
dt
dv
a =
= 0
dt
36
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–42.
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
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
40
80
t (s)
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 =
For time interval 40 s 6 t … 80 s,
a =
dv
= 0
dt
v - 10
0 - 10
1
, v = a - t + 20 b m>s.
=
t - 40
80 - 40
4
dv
1
= - = - 0.250 m s2
dt
4
For 0 … t 6 40 s, a = 0.
For 40 s 6 t … 80, a = - 0.250 m s2 .
Ans:
s = 600 m. For 0 … t 6 40 s,
a = 0. For 40 s 6 t … 80 s,
a = - 0.250 m>s2
42
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12–43.
The motion of a jet plane just after landing on a runway
is described by the a–t graph. Determine the time t′ when
the jet plane stops. Construct the v–t and s–t graphs for the
motion. Here s = 0, and v = 300 ft>s when t = 0.
a (m/s2)
10
10
20
Solution
v–t Graph. The v–t function can be determined by integrating dv = a dt.
For 0 … t 6 10 s, a = 0. Using the initial condition v = 300 ft>s at t = 0,
v
L300 ft>s
L0
dv =
t
0 dt
v - 300 = 0
v = 300 ft>s
Ans.
a - ( - 20)
For 10 s 6 t 6 20 s,
t - 10
=
- 10 - ( -20)
20 - 10
, a = (t - 30) ft>s2. Using the
initial condition v = 300 ft>s at t = 10 s,
t
v
(t - 30) dt
L10 s
L300 ft>s
dv =
t
1
v - 300 = a t 2 - 30tb `
2
10 s
1
v = e t 2 - 30t + 550 f ft>s
2
Ans.
At t = 20 s,
v`
=
t = 20 s
1
( 202 ) - 30(20) + 550 = 150 ft>s
2
For 20 s 6 t 6 t′, a = -10 ft>s. Using the initial condition v = 150 ft>s at t = 20 s,
t
v
L20 s
L150 ft>s
dv =
- 10 dt
v - 150 = ( - 10t) `
t
20 s
Ans.
v = ( - 10t + 350) ft>s
It is required that at t = t′, v = 0. Thus
0 = - 10 t′ + 350
Ans.
t′ = 35 s
Using these results, the v9t graph shown in Fig. a can be plotted s-t Graph.
The s9t function can be determined by integrating ds = v dt. For 0 … t 6 10 s, the
initial condition is s = 0 at t = 0.
s
t
ds =
300 dt
L0
L0
s = {300 t} ft
Ans.
At = 10 s,
s 0 t = 10 s = 300(10) = 3000 ft
43
20
t¿
t (s)
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–70.
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
r
L0
t
dr =
L0
C 3i + (6 - 2t)j D dt
r = c3ti + A 6t - t2 B j d m
When t = 1 s and 3 s,
r t = 1 s = 3(1)i + C 6(1) - 12 D j = [3i + 5j] m>s
r t = 3 s = 3(3)i + C 6(3) - 32 D j = [9i + 9j] m>s
Thus, the displacement of the particle is
¢r = r t = 3 s - r t = 1 s
= (9i + 9j) - (3i + 5j)
Ans.
= {6i + 4j} m
75
Ans:
∆r = 5 6i + 4j 6 m
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–72.
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.
Position: The position expressed in Cartesian vector form can be obtained by
applying Eq. 12–7.
dr = v dt
r
L0
dr =
r = c
t
L0
2
3
A 16t i + 4t j + (5t + 2)k B dt
5
16 3
t i + t4j + a t2 + 2tb k d m
3
2
When t = 2 s,
r =
5
16 3
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
Thus, the coordinate of the particle is
Ans.
(42.7, 16.0, 14.0) m
Ans:
(42.7, 16.0, 14.0) m
77
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12–90.
Show that the girl at A can throw the ball to the boy at B
by launching it at equal angles measured up or down
from a 45° inclination. If vA = 10 m>s, determine the range
R if this value is 15°, i.e., u1 = 45° − 15° = 30° and u2 = 45° +
15° = 60°. Assume the ball is caught at the same elevation
from which it is thrown.
vA 10 m/s
u
A
B
R
Solution
+2 s = s + vt
1S
0
0
R = 0 + (10 cos u)t
1 + c 2 v = v0 + act
- 10 sin u = 10 sin u - 9.81t
t =
20
sin u
9.81
200
sin u cos u
9.81
Thus,
R =
R =
100
sin 2u
9.81
(1)
Since the function y = sin 2u is symmetric with respect to u = 45° as indicated,
Eq. (1) will be satisfied if | f 1 | = | f 2 |
Choosing f = 15° or u1 = 45° - 15° = 30°
substituting into Eq. (1) yields
and
u2 = 45° + 15° = 60°,
and
Ans.
R = 8.83 m
Ans:
R = 8.83 m
95
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–91.
y
The ball at A is kicked with a speed vA = 80 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.
vA
A
uA
x
y
B
Solution
x
y 0.04x2
(vA)x = 80 cos 30° = 69.28 ft>s
(vA)y = 80 sin 30° = 40 ft>s
+ 2s = s + v t
1S
0
0
(1)
x = 0 + 69.28t
1 2
at
2 c
1
- y = 0 + 40t + ( - 32.2)t 2
2
y = - 0.04x2
1 + c 2 s = s0 + v0t +
(2)
From Eqs. (1) and (2):
- y = 0.5774x - 0.003354x2
0.04x2 = 0.5774x - 0.003354x2
0.04335x2 = 0.5774x
Ans.
x = 13.3 ft
Thus
y = - 0.04 (13.3)2 = - 7.09 ft
Ans.
Ans:
(13.3 ft, - 7.09 ft)
96
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*12–112.
A boat has an initial speed of 16 ft>s. If it then increases its
speed along a circular path of radius r = 80 ft at the rate of
#
v = (1.5s) ft>s, where s is in feet, determine the time needed
for the boat to travel s = 50 ft.
Solution
at = 1.5s
L0
s
v
1.5s ds =
L16
v dv
0.75 s2 = 0.5 v2 - 128
v =
s
ds
= 2256 + 1.5 s2
dt
L0 1s2 + 170.7
ds
=
L0
t
1.225 dt
s
ln 1 s + 2s2 + 170.7 2 0 0 = 1.225t
ln 1 s + 2s2 + 170.7 2 - 2.570 = 1.225t
At s = 50 ft,
Ans.
t = 1.68 s
Ans:
t = 1.68 s
120
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–119.
The satellite S travels around the earth in a circular path
with a constant speed of 20 Mm>h. If the acceleration is
2.5 m>s2, determine the altitude h. Assume the earth’s
diameter to be 12 713 km.
S
h
SOLUTION
n = 20 Mm>h =
Since at =
dn
= 0, then,
dt
a = an = 2.5 =
r =
20(106)
= 5.56(103) m>s
3600
n2
r
(5.56(103))2
= 12.35(106) m
2.5
The radius of the earth is
12 713(103)
= 6.36(106) m
2
Hence,
h = 12.35(106) - 6.36(106) = 5.99(106) m = 5.99 Mm
Ans.
Ans:
h = 5.99 Mm
127
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12–153.
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
Ans.
vt = 7.21 m>s
The velocity v makes an angle u = tan-1
4
= 33.69° with the x axis.
6
Acceleration: To determine the acceleration a, apply Eq. 12–9.
a =
dv
= {2i + 2j} m>s2
dt
Then
a = 222 + 22 = 2.828 m>s2
The acceleration a makes an angle f = tan-1
figure, a = 45° - 33.69 = 11.31°. Therefore,
2
= 45.0° with the x axis. From the
2
an = a sin a = 2.828 sin 11.31° = 0.555 m>s2
Ans.
at = a cos a = 2.828 cos 11.31° = 2.77 m>s2
Ans.
Ans:
vn = 0
vt = 7.21 m>s
an = 0.555 m>s2
at = 2.77 m>s2
162
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12–145.
Particles A and B are traveling counter-clockwise around a
circular track at a constant speed of 8 m>s. If at
the instant shown the speed of A begins to increase by
(at)A = (0.4sA) m>s2, where sA is in meters, determine the
distance measured counterclockwise along the track from B
to A when t = 1 s. What is the magnitude of the
acceleration of each particle at this instant?
A
sA
u 120
sB
B
SOLUTION
r5m
Distance Traveled: Initially the distance between the particles is
d0 = rdu = 5 a
120°
b p = 10.47 m
180°
When t = 1 s, B travels a distance of
dB = 8(1) = 8 m
The distance traveled by particle A is determined as follows:
vdv = ads
v
L8 m>s
s
vdv =
L0
0.4 sds
v = 0.6325 2s2 + 160
(1)
ds
dt =
v
t
L0
s
dt =
ds
L0 0.6325 2s2 + 160
2s2 + 160 + s
1
S≥
£ In C
0.6325
2160
1 =
s = 8.544 m
Thus the distance between the two cyclists after t = 1 s is
Ans.
d = 10.47 + 8.544 - 8 = 11.0 m
Acceleration:
For A, when t = 1 s,
A a t B A = vA = 0.4 A 8.544 B = 3.4176 m>s2
#
vA = 0.6325 28.5442 + 160 = 9.655 m>s
(a n)A =
v2A
9.6552
=
= 18.64 m>s2
r
5
The magnitude of the A’s acceleration is
aA = 23.41762 + 18.642 = 19.0 m>s2
Ans.
For B, when t = 1 s,
A at B B = vA = 0
#
(an)B =
v2B
82
=
= 12.80 m>s2
r
5
Ans:
The magnitude of the B’s acceleration is
aB = 202 + 12.802 = 12.8 m>s2
Ans.
154
d = 11.0 m
aA = 19.0 m>s2
aB = 12.8 m>s2
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–155.
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 = p6 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
#
$
ar = r - ru2 = 0 - 4(- 1.7039)2 = - 11.6135 in.>s2
$
##
au = ru + 2ru = 4(- 2.0944) + 0 = - 8.3776 in.>s2
a =
a2r + a2u =
( -11.6135)2 + ( - 8.3776)2 = 14.3 in. s2
Ans.
Ans:
a = 14.3 in.>s2
164
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*12–160.
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
Ans.
Acceleration:
#
#
ar = r - ru2 = 0 - 200(0.12) = - 2 m>s2
$
# #
au = r u + 2ru = 200(0.025) + 0 = 5 m>s2
Thus, the magnitude of the car’s acceleration is
a = 2ar2 + au2 = 2(-2)2 + 52 = 5.39 m>s2
Ans.
Ans:
v = 20 m>s
a = 5.39 m>s2
169
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–207.
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
Position Coordinates: By referring to Fig. a, the length of the cord written in terms
of the position coordinates sA and sB is
A
sB + sA + 2(sA - a) = l
sB + 3sA = l + 2a
Time Derivative: Taking the time derivative of the above equation,
(+ T)
vB + 3vA = 0
Here, vB = 4 m>s. Thus,
4 + 3vA = 0
vA = - 133 m>s = 1.33 m>s c
Ans.
Ans:
vA = 1.33 m>s
216
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–209.
The cord is attached to the pin at C and passes over the two
pulleys at A and D. The pulley at A is attached to the
smooth collar that travels along the vertical rod. Determine
the velocity and acceleration of the end of the cord at B if at
the instant sA = 4 ft the collar is moving upwards at 5 ft>s,
which is decreasing at 2 ft>s2.
3 ft
3 ft
C
D
sB
sA
SOLUTION
B
A
2 2s2A + 32 + sB = l
1
1
#
#
2a b A s2A + 9 B - 2 a 2sA sA b + s B = 0
2
#
2sA sA
#
sB = 1
A s2A + 9 B 2
1
1
3
##
#
1
#
##
#
s B = -2sA2 A s2A + 9 B - 2 - a2sA sA b A s2A + 9 B - 2 - a2sA sA b c a - b A s2A + 9 B - 2 a2sA sA b d
2
#
$
#
2 A sA + sA sA B
2 A sA sA B 2
$
sB = +
1
3
A s2A + 9 B 2
A s2A + 9 B 2
At sA = 4 ft,
2(4)(-5)
#
vB = sB = 1 = 8 ft>s T
(42 + 9)2
$
aB = sB = -
2 c 1 -522 + 142122 d
14 + 92
2
1
2
+
Ans.
231421- 5242
14 + 92
2
3
2
= - 6.80 ft>s2 = 6.80 ft>s2 c Ans.
Ans:
vB = 8 ft>s T
aB = 6.80 ft>s2 c
218
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*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
A
B
C
Motion of Gear A:
aA duA = vA dvA
aA duA = (uA + 1) d(uA + 1)
aA duA = (uA + 1) duA
aA = (uA + 1)
At uA = 3 rad,
aA = 3 + 1 = 4 rad>s2
Motion of Gear D: Gear A is in mesh with gear B. Thus,
aB rB = aA rA
rA
15
aB = a b aA = a b (4) = 1.20 rad>s2
rB
50
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,
aD rD = aC rC
rC
25
aD = a b aC = a b (1.20) = 0.4 rad>s2
rD
75
Ans.
Ans:
aD = 0.4 rad>s2
637
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–9.
At the instant vA = 5 rad>s, pulley A is given an angular
acceleration a = (0.8u) rad>s2, where u is in radians.
Determine the magnitude of acceleration of point B on
pulley C when A rotates 3 revolutions. Pulley C has an inner
hub which is fixed to its outer one and turns with it.
A
vA
aA
50 mm
Solution
40 mm
C
Angular Motion. The angular velocity of pulley A can be determined by integrating
v dv = a du with the initial condition vA = 5 rad>s at uA = 0.
vA
60 mm
uA
L5 rad>s
L0
v2 vA
`
2 5 rad>s
= ( 0.4u 2 ) `
v dv =
B
0.8udu
uA
0
v2A
52
= 0.4u 2A
2
2
vA = e 20.8u 2A + 25 f rad>s
At uA = 3(2p) = 6p rad,
vA = 20.8(6p)2 + 25 = 17.585 rad>s
aA = 0.8(6p) = 4.8p rad>s2
Since pulleys A and C are connected by a non-slip belt,
vCrC = vArA;
vC(40) = 17.585(50)
vC = 21.982 rad>s
aCrC = aArA;
aC(40) = (4.8p)(50)
aC = 6p rad>s2
Motion of Point B. The tangential and normal components of acceleration of
point B can be determined from
(aB)t = aCrB = 6p(0.06) = 1.1310 m>s2
(aB)n = v2CrB = ( 21.9822 ) (0.06) = 28.9917 m>s2
Thus, the magnitude of aB is
aB = 2(aB)2t + (aB)2n = 21.13102 + 28.99172
= 29.01 m>s2 = 29.0 m>s2
Ans.
Ans:
aB = 29.0 m>s2
638
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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
v = 20.3u2 + 1
At u = 1 rev = 2p rad
v = 20.3(2p)2 + 1
v = 3.584 rad>s
a t = ar = 0.3(2p) rad>s 2 (0.4 m) = 0.7540 m>s2
Ans.
a n = v2r = (3.584 rad>s)2(0.4 m) = 5.137 m>s2
Ans.
a p = 2(0.7540)2 + (5.137)2 = 5.19 m>s2
Ans:
at = 0.7540 m>s2
an = 5.137 m>s2
645
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–17.
A motor gives gear A an angular acceleration of
aA = (2 + 0.006 u 2) rad>s2, where u is in radians. If this
gear is initially turning at vA = 15 rad>s, determine the
angular velocity of gear B after A undergoes an angular
displacement of 10 rev.
B
A
175 mm
100 mm
aB
aA
vA
Solution
Angular Motion. The angular velocity of the gear A can be determined by
integrating v dv = a du with initial condition vA = 15 rad>s at uA = 0.
vA
L15 rad>s
v dv =
L0
uA
( 2 + 0.006 u 2 ) du
uA
v2 vA
`
= ( 2u + 0.002 u 3 ) `
2 15 rad>s
0
v2A
152
= 2uA + 0.002 u 3A
2
2
vA = 20.004 u 3A + 4 u + 225 rad>s
At uA = 10(2p) = 20p rad,
vA = 20.004(20p)3 + 4(20p) + 225
= 38.3214 rad>s
Since gear B is meshed with gear A,
vBrB = vArA ; vB(175) = 38.3214(100)
vB = 21.8979 rad>s
Ans.
= 21.9 rad>s d
Ans:
vB = 21.9 rad>s d
646
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–51.
The pins at A and B are confined to move in the vertical
and horizontal tracks. If the slotted arm is causing A to move
downward at vA, determine the velocity of B at the instant
shown.
d
θ
y
90°
A
h
vA
SOLUTION
B
x
Position coordinate equation:
tan u =
d
h
=
x
y
h
x = ¢ ≤y
d
Time derivatives:
#
x =
h #
y
d
vB =
h
v
d A
Ans.
Ans:
h
vB = ¢ ≤vA
d
681
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*16–56.
At the instant shown, the disk is rotating with an angular
velocity of V and has an angular acceleration of A.
Determine the velocity and acceleration of cylinder B at
this instant. Neglect the size of the pulley at C.
A
3 ft
u
V, A
C
5 ft
SOLUTION
B
s = 232 + 52 - 2(3)(5) cos u
#
1
1
#
vB = s = (34 - 30 cos u)- 2(30 sin u)u
2
vB =
15 v sin u
#
aB = s =
=
Ans.
1
(34 - 30 cos u) 2
#
#
15 v cos uu + 15v sin u
234 - 30 cos u
15 (v2 cos u + a sin u)
(34 - 30 cos u)
1
2
-
+
#
1
a - b (15v sin u) a 30 sin uu b
2
3
(34 - 30 cos u) 2
225 v2 sin2 u
Ans.
3
(34 - 30 cos u) 2
Ans:
vB =
aB =
686
15 v sin u
1
(34 - 30 cos u)2
15 (v2 cos u + a sin u)
1
(34 - 30 cos u)2
-
225 v2 sin2 u
3
(34 - 30 cos u)2
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–73.
If the slider block A is moving downward at vA = 4 m>s,
determine the velocity of point E at the instant shown.
B
4
250 mm
400 mm
E
300 mm
3
5
D
vA 4 m/s
300 mm
30
C
Solution
A
See solution to Prob. 16–87.
vE = vD + vE>D
S
vE = 4T + 2.727 + (5.249)(0.3)
3
4
Q5 f 30°
3
5
4
( + T ) (vE)y = 4 - 2.727 a b + 5.249(0.3)(cos 30°)
5
+
(S
) (vE)x = 0 + 2.727 a b + 5.249(0.3)(sin 30°)
(vE)x = 2.424 m>s S
(vE)y = 3.182 m>s T
vE = 2(2.424)2 + (3.182)2 = 4.00 m>s
Ans.
u = tan-1 a
Ans.
Also:
3.182
b = 52.7°
2.424
See solution to Prob. 16–87.
vE = vD + vCE * rE>D
vE = (1.636i - 1.818j) + ( - 5.25k) * {cos 30°(0.3)i - 0.4 sin 30°(0.3)j}
vE = 52.424i - 3.182j 6 m>s
vE = 2(2.424)2 + (3.182)2 = 4.00 m>s
Ans.
u = tan-1 a
Ans.
3.182
b = 52.7°
2.424
Ans:
vE = 4.00 m>s
u = 52.7° c
705
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–77.
The planetary gear system is used in an automatic
transmission for an automobile. By locking or releasing
certain gears, it has the advantage of operating the car at
different speeds. Consider the case where the ring gear R is
held fixed, vR = 0, and the sun gear S is rotating at
vS = 5 rad>s. Determine the angular velocity of each of the
planet gears P and shaft A.
40 mm
vR
P
R
vS
S
SOLUTION
A
80 mm
vA = 5(80) = 400 mm>s ;
vB = 0
vB = vA + v * rB>A
0 = -400i + (vp k) * (80j)
40 mm
0 = -400i - 80vp i
Ans.
vP = -5 rad>s = 5 rad>s
vC = vB + v * rC>B
vC = 0 + ( - 5k) * ( - 40j) = -200i
vA =
200
= 1.67 rad>s
120
Ans.
Ans:
vP = 5 rad>s
vA = 1.67 rad>s
709
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–90.
10 ft/s B
vB
Due to slipping, points A and B on the rim of the disk have
the velocities shown. Determine the velocities of the center
point C and point D at this instant.
D
E
45
C
A
0.8 ft
30
F
vA
5 ft/s
SOLUTION
x
1.6 - x
=
5
10
5x = 16 - 10x
x = 1.06667 ft
v =
10
= 9.375 rad>s
1.06667
rIC-D = 2(0.2667)2 + (0.8)2 - 2(0.2667)(0.8) cos 135° = 1.006 ft
sin f
sin 135°
=
0.2667
1.006
f = 10.80°
vC = 0.2667(9.375) = 2.50 ft>s :
Ans.
vD = 1.006(9.375) = 9.43 ft>s
Ans.
u = 45° + 10.80° = 55.8° h
Ans:
vC = 2.50 ft>s d
vD = 9.43 ft>s
u = 55.8° h
723
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–5.
If blocks A and B of mass 10 kg and 6 kg, respectively, are
placed on the inclined plane and released, determine the
force developed in the link. The coefficients of kinetic
friction between the blocks and the inclined plane are
mA = 0.1 and mB = 0.3. Neglect the mass of the link.
A
B
SOLUTION
30
Free-Body Diagram: Here, the kinetic friction (Ff)A = mANA = 0.1NA and
(Ff)B = mB NB = 0.3NB are required to act up the plane to oppose the motion of
the blocks which are down the plane. Since the blocks are connected, they have a
common acceleration a.
Equations of Motion: By referring to Figs. (a) and (b),
+Q©Fy¿ = may¿ ;
NA - 10(9.81) cos 30° = 10(0)
NA = 84.96 N
R + ©Fx¿ = max¿ ;
10(9.81) sin 30° - 0.1(84.96) - F = 10a
(1)
40.55 - F = 10a
and
+Q©Fy¿ = may¿ ;
NB - 6(9.81) cos 30° = 6(0)
NB = 50.97 N
R + ©Fx¿ = max¿ ;
F + 6(9.81) sin 30° - 0.3(50.97) = 6a
(2)
F + 14.14 = 6a
Solving Eqs. (1) and (2) yields
a = 3.42 m>s2
Ans.
F = 6.37 N
Ans:
F = 6.37 N
249
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13–6.
The 10-lb block has a speed of 4 ft>s when the force of
F = (8t2) lb is applied. Determine the velocity of the block
when t = 2 s. The coefficient of kinetic friction at the surface
is mk = 0.2.
v 4 ft/s
F (8t2) lb
Solution
Equations of Motion. Here the friction is Ff = mk N = 0.2N. Referring to the FBD
of the block shown in Fig. a,
10
(0) N = 10 lb
32.2
10
+ ΣFx = max; 8t 2 - 0.2(10) =
a
S
32.2
+ c ΣFy = may; N - 10 =
a = 3.22(8t 2 - 2) ft>s2
Kinematics. The velocity of the block as a function of t can be determined by
integrating dv = a dt using the initial condition v = 4 ft>s at t = 0.
v
L4 ft>s
dv =
L0
t
3.22 (8t 2 - 2)dt
8
v - 4 = 3.22 a t 3 - 2tb
3
When t = 2 s,
v = 5 8.5867t3 - 6.44t + 46ft>s
v = 8.5867(23) - 6.44(2) + 4
= 59.81 ft>s
Ans.
= 59.8 ft>s
Ans:
v = 59.8 ft>s
250
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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
SOLUTION
Fs = kx;
k
Fs = 4 A 21 + s2 - 1 B
-4 A 21 + s2 - 1 B ¢
+ ©F = ma ;
:
x
x
1
-
L0
4 lb/ft
¢ 4s ds -
4s ds
21 + s
1
- C 2s2 - 4 31 + s2 D 0 =
2
≤ =
v
L15
a
s
21 + s2
≤ = a
2
dv
b av b
32.2
ds
2
b v dv
32.2
1
A v2 - 152 B
32.2
Ans.
v = 14.6 ft>s
Ans:
v = 14.6 ft>s
280
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*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) = 15a
v2
b
5
Ans.
v = 3.13 m>s
Ans:
v = 3.13 m>s
296
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
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).
Equations of Motion: Realizing that an =
©Fn = man;
147.15 = 2 a
A
v2
102
and referring to Fig. (a),
=
r
r
102
b
r
Ans.
r = 1.36 m
Ans:
r = 1.36 m
297
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*13–132.
The rocket is traveling around the earth in free flight along
the elliptical orbit AC. Determine its change in speed when
it reaches A so that it travels along the elliptical orbit AB.
C
A
B
Solution
8 Mm
Applying Eq. 13–27,
ra =
(2GMe >rpv2p) - 1
rpv2p
2GMe
vp =
10 Mm
rp
2GMe
rav2p
8 Mm
- 1 =
=
rp
ra
rp + ra
ra
2GMe ra
B rp(rp + ra)
For orbit AC, rp = 10 ( 106 ) m, ra = 16 ( 106 ) m and vp = (vA)AC. Then
(vA)AC =
D
2 366.73 ( 10-12 )4 35.976 ( 1024 )4 316 ( 106 )4
10 ( 106 ) 310 ( 106 ) + 16 ( 106 )4
= 7005.74 m>s
For orbit AB, rp = 8 ( 106 ) m, ra = 10 ( 106 ) m and vp = vB. Then
vB =
D
2 366.73 ( 10-12 )4 35.976 ( 1024 )4 310 ( 106 )4
8 ( 106 ) 38 ( 106 ) + 10 ( 106 )4
= 7442.17 m>s
Since h is constant at any position of the orbit,
h = rpvp = rava
8 ( 106 ) (7442.17) = 10 ( 106 ) (vA)AB
(vA)AB = 5953.74 m>s
Thus, the required change in speed is
∆v = (vA)AB - (vA)AC = 5953.74 - 7005.74
= -1052.01 m>s = - 1.05 km>s
Ans.
The negative sign indicates that the speed must be decreased.
Ans:
∆v = - 1.05 km>s
376
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14–9.
The “air spring” A is used to protect the support B and
prevent damage to the conveyor-belt tensioning weight C
in the event of a belt failure D. The force developed by
the air spring as a function of its deflection is shown by the
graph. If the block has a mass of 20 kg and is suspended
a height d = 0.4 m above the top of the spring, determine
the maximum deformation of the spring in the event the
conveyor belt fails. Neglect the mass of the pulley and belt.
F (N)
D
1500
C
d
0.2
s (m)
A
B
Solution
Work. Referring to the FBD of the tensioning weight, Fig. a, W does positive
work whereas force F does negative work. Here the weight displaces downward
SW = 0.4 + xmax where xmax is the maximum compression of the air spring. Thus
UW = 20(9.81) ( 0.4 + xmax ) = 196.2 ( 0.4 + xmax )
The work of F is equal to the area under the F-S graph shown shaded in Fig. b, Here
F
1500
; F = 7500xmax. Thus
=
xmax
0.2
1
UF = - (7500 xmax)(xmax) = - 3750x2max
2
Principle of Work And Energy. Since the block is at rest initially and is required
to stop momentarily when the spring is compressed to the maximum, T1 = T2 = 0.
Applying Eq. 14–7,
T1 + ΣU1 - 2 = T2
0 + 196.2(0.4 + xmax) + ( - 3750x2max ) = 0
3750x2max - 196.2xmax - 78.48 = 0
xmax = 0.1732 m = 0.173 m 6 0.2 m
(O.K!)
Ans.
Ans:
xmax = 0.173 m
385
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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
Energy: Here, the speed of the car is y = a
5280 ft
1h
35 mi
b * a
b * a
b =
h
1 mi
3600 s
51.33 ft>s. Thus, the kinetic energy of the car is
U =
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,
t =
Pbulb = 100 W * a
550 ft # lb>s
1 hp
b * a
b =
746 W
1 hp
204.59(103)
U
= 2774.98 s = 46.2 min
=
Pbulb
73.73
Ans.
Ans:
t = 46.2 min
422
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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.
15 ft
SOLUTION
Step height: 0.125 m
The number of steps:
4
= 32
0.125
Total load: 32(150)(9.81) = 47 088 N
If load is placed at the center height, h =
4
= 2 m, then
2
4
U = 47 088 a b = 94.18 kJ
2
vy = v sin u = 0.6 ¢
4
2 ( 32(0.25) ) 2 + 42
t =
h
2
= 7.454 s
=
vy
0.2683
P =
U
94.18
=
= 12.6 kW
t
7.454
≤ = 0.2683 m>s
Ans.
Also,
P = F # v = 47 088(0.2683) = 12.6 kW
Ans.
Ans:
P = 12.6 kW
423
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15–14.
A tankcar has a mass of 20 Mg and is freely rolling to the
right with a speed of 0.75 m>s. If it strikes the barrier,
determine the horizontal impulse needed to stop the car if
the spring in the bumper B has a stiffness (a) k S ∞
(bumper is rigid), and (b) k = 15 kN>m.
v 0.75 m/s
k
B
Solution
+
a) b) ( S )
mv1 + Σ
L
F dt = mv2
20 ( 103 ) (0.75) L
L
F dt = 0
F dt = 15 kN # s
Ans.
The impulse is the same for both cases. For the spring having a stiffness k = 15 kN>m,
the impulse is applied over a longer period of time than for k S ∞ .
Ans:
I = 15 kN # s in both cases.
487
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15–11.
During operation the jack hammer strikes the concrete surface
with a force which is indicated in the graph. To achieve this the
2-kg spike S is fired into the surface at 90 m>s. Determine the
speed of the spike just after rebounding.
F (kN)
1500
S
Solution
0
Principle of Impulse and Momentum. The impulse of the force F is equal to the
area under the F–t graph. Referring to the FBD of the spike, Fig. a
( + c ) m(vy)1 + Σ
2( - 90) +
Lt1
0
0.1
0.4
t (ms)
t2
Fy dt = m(vy)2
1
3 0.4 ( 10-3 ) 4 3 1500 ( 103 ) 4 = 2v
2
v = 60.0 m>s c Ans.
Ans:
v = 60.0 m>s
484
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15–34.
The 0.15-kg baseball has a speed of v = 30 m>s just before
it is struck by the bat. It then travels along the trajectory
shown before the outfielder catches it. Determine the
magnitude of the average impulsive force imparted to the
ball if it is in contact with the bat for 0.75 ms.
v2 15
v1 30 m/s
15
0.75 m
2.5 m
100 m
SOLUTION
+ )
(:
mA (vA)1 + mB(vB)1 = (mA + mB)v2
3000
7500
4500
(3) (6) =
v
32.2
32.2
32.2 2
v2 = -0.600 ft>s = 0.600 ft>s ;
Ans.
Ans:
v = 0.6 ft>s d
509