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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
vB
30
O
SOLUTION
45
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 (
6.714
) = 18.0° b
20.61
Ans.
Ans.
(800)2 = (20 t)2 + (15 t)2 - 2(20 t)(15 t) cos 75°
t = 36.9 s
Also
t =
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Ans.
800
800
= 36.9 s
=
vA>B
21.68
Ans.
An
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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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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
250 ft
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
- 15i -
19022
300
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sa eir d i is w
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w of a urs rov k is
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st pa nd s te
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th t se y
e his s fo by
s r
in
te wo ing the Uni
gr rk s u te
ity ( tu s d
of inc de e o Sta
th lud nt f in te
e in lea s s c
w g r tru o
or o ni c p
k n ng to yri
an th . rs gh
d e W Dis in t l
is
t a
no orl sem eac ws
t p d W in hi
er id ati ng
m e on
itt W o
ed e r
. b)
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
A
aA>B = {10.69i + 16.70j} ft>s2
aA>B = 2(10.69)2 + (16.70)2 = 19.8 ft>s2
Ans.
A
ns.
16.70
b = 57.4° a
10.69
Ans.
An
u = tan - 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,
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Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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),
N - 80(9.81) + 360 sin 20° = 80(0)
th an Th
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s r
in
te wo ing the Uni
gr rk s u te
ity ( tu s d
of inc de e o Sta
th lud nt f in te
e in lea s s c
w g r t ru o
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k n ng to yri
an th . rs gh
d e W Dis in t l
is
t a
no orl sem eac ws
t p d W in hi
er id ati ng
m e on
itt W o
ed e r
. b)
+ c ©Fy = may ;
N = 661.67 N
+ ©F = ma ;
:
x
x
360 cos 20° - 0.3(661.67) = 80a
a = 1.75 m>s2
Ans.
A 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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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
Kinematic: Applying equation s = s0 + v0 t +
(a + )
1
0.75 = 0 + 0 + aB A 22 B
2
1
a t2, we have
2 c
A
60
aB = 0.375 m>s2
Establishing the position - coordinate equation, we have
2sA + (sA - sB) = l
3sA - sB = l
Taking time derivative twice yields
From Eq.(1),
3aA - 0.375 = 0
(1)
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s r
in
te wo ing the Uni
gr rk s u te
ity ( tu s d
of inc de e o Sta
th lud nt f in te
e in lea s s c
w g r t ru o
or o ni c p
k n ng to yri
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d e W Dis in t l
is
t a
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er id ati ng
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itt W o
ed e r
. b)
3aA - aB = 0
aA = 0.125 m>s2
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),
a + ©Fy¿ = may¿ ;
T - 5(9.81) sin 60° = 5(0.375)
T = 44.35 N
From FBD(a),
+ c ©Fy = may ;
3(44.35) - 9.81m
9.81mA = mA ( -0.125)
mA = 13.7 kg
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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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
F
A
B
C
kCB
3 kN/m
kAB
5 kN
2 kN/m
SOLUTION
FCB = kCBx = 3000x
+ ©F = ma ;
;
x
x
FAB = kABx = 2000x
5000 - 3000x - 2000x = 2a
2500 - 2500x = a
a dx - v dy
(2500 - 2500x) dx =
2500(0.2) - ¢
v = 30 m>s
L0
v
2500(0.2)2
v2
≤ =
2
2
v dv
th an Th
sa eir d i is w
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w of a urs rov k is
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de ny es a ided pro
st pa nd s te
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e his s fo by
s r
in
te wo ing the Uni
gr rk s u te
ity ( tu s d
of inc de e o Sta
th lud nt f in te
e in lea s s c
w g r t ru o
or o ni c p
k n ng to yri
an th . rs gh
d e W Dis in t l
is
t a
no orl sem eac ws
t p d W in hi
er id ati ng
m e on
itt W o
ed e r
. b)
L0
0.2
A
Ans.
ns.
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained 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.