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
3
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
th an Th
sa eir d i is w
le co s p or
w of a urs rov k is
ill
de ny es a ided pro
st pa nd s te
ro rt
o c
y of as lel ted
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)
+ ©Fy¿ = may¿ ;
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
T1 + a U1 - 2 = T2
3
1
1
0 + WA ¢ ¢sA ≤ - Ff ¢sA - WB ¢sB = mAv2A + mB v2B
5
2
2
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
Eqs. (1) and (2) yields
vA = 3.52 ft>s
(2)
Ans.
vB = 7.033 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,
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B
*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
0 + 2[F1 - 3(9.81)] + 4[8(9.81) - F2] =
A
1
1
(3)v2A + (8)v2B
2
2
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
Ans.
th an Th
sa eir d i is w
le co s p or
w of a urs rov k is
ill
de ny es a ided pro
st pa nd s te
ro rt
o c
y of as lel ted
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)
vA = 3.82 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.
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.
A
5 km/h
B
25 km/h
C
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
A vB B 1 = c 5(103)
1h
m
da
b = 5.556 m>s
h 3600 s
m
1h
da
b = 1.389 m>s,
h 3600 s
and A vC B 1 = c 25(103)
m
1h
da
b = 6.944 m>s
h 3600 s
For the first case,
+ B
A:
th an Th
sa eir d i is w
le co s p or
w of a urs rov k is
ill
de ny es a ided pro
st pa nd s te
ro rt
o c
y of as lel ted
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)
A vA B 1 = c 20(103)
mA(vA)1 + mB(vB)1 = (mA + mB)v2
10000(5.556) + 5000(1.389) = (10000 + 5000)vAB
vAB = 4.167 m>s :
Using the result of vAB and considering the second case,
+ B
A:
(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 ;
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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15–55.
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
(vT)1
B
T
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
T1 + V1 = T2 + V2
th an Th
sa eir d i is w
le co s p or
w of a urs rov k is
ill
de ny es a ided pro
st pa nd s te
ro rt
o c
y of as lel ted
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)
v2 = 3.059 m>s
T1 =
1
1
(19 000)(4.167)2 + (75 000)(2.778)2 = 454.282 kJ
2
2
T2 =
1
(19 000 + 75 000)(3.059)2 = 439.661 kJ
2
Hence,
454.282(103) + 0 = 439.661(103) +
x = 0.221 m
1
(600)(103)x 2
2
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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