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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
no orl sem eac ws
er id ati ng
ed e r
itt W o
t a
t p d W in hi
. b)
m e on
d e W Dis in t l
an th . rs gh
k n ng to yri
Chapter 17 Structural Analysis or o ni c p
w g r tru o 833
e in lea s s c
th lud nt f in te
of inc de e o Sta
Chapter 18 Internal Forces 953
ity ( tu s d
gr rk s u te
is
te wo ing the Uni
e his s fo by
Chapter 19 Friction 1023
y of as lel ted
s r
o c
st pa nd s te
th t se y
de ny es a ided pro
Review 2 Planar Kinematics and Kinetics of a Rigid Body 1080
w of a urs rov k is
le co s p or
sa eir d i is w
in
ro rt
th an Th
Chapter 20 Center of Gravity and Centroid 1131
ill
Chapter 21 Moments of Inertia 1190
Chapter 22 Virtual Work 1270
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained 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:
Full file at http://testbankcart.eu/Solution-Manual-for-Engineering-Mechanics-Dynamics-13th-Edition-by-Hibbeler
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no orl sem eac ws
er id ati ng
ed e r
itt W o
t a
t p d W in hi
. b)
m e on
d e W Dis in t l
an th . rs gh
k n ng to yri
or o ni c p
w g r tru o
e in lea s s c
th lud nt f in te
of inc de e o Sta
ity ( tu s d
gr rk s u te
is
te wo ing the Uni
e his s fo by
y of as lel ted
s r
o c
st pa nd s te
th t se y
de ny es a ided pro
w of a urs rov k is
le co s p or
sa eir d i is w
in
ro rt
th an Th
ill
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained 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:
Full file at http://testbankcart.eu/Solution-Manual-for-Engineering-Mechanics-Dynamics-13th-Edition-by-Hibbeler
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Education, Inc., Upper Saddle River, NJ 07458. https://www.stuvia.com/user/gabrielucky
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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 Ans.
no orl sem eac ws
er id ati ng
ed e r
itt W o
t a
t p d W in hi
. b)
m e on
d e W Dis in t l
an th . rs gh
k n ng to yri
or o ni c p
w g r tru o
e in lea s s c
th lud nt f in te
of inc de e o Sta
ity ( tu s d
gr rk s u te
is
te wo ing the Uni
e his s fo by
y of as lel ted
s r
o c
st pa nd s te
th t se y
de ny es a ided pro
w of a urs rov k is
le co s p or
sa eir d i is w
in
ro rt
th an Th
ill
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained 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:
Full file at http://testbankcart.eu/Solution-Manual-for-Engineering-Mechanics-Dynamics-13th-Edition-by-Hibbeler
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Rights and Permissions Department, Pearson https://www.stuvia.com/user/gabrielucky
Education, Inc., Upper Saddle River, NJ 07458. https://www.stuvia.com/user/gabrielucky
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12–2.
When a train is traveling along a straight track at 2 m/s, it v
begins to accelerate at a = 160 v-42 m>s2, where v is in m/s.
s
Determine its velocity v and the position 3 s after the
acceleration.
SOLUTION
dv
a =
dt
dv
dt =
a
3 v
L0 L2 60v
dv
dt = -4
1
3 = (v5 - 32)
300
v = 3.925 m>s = 3.93 m>s
no orl sem eac ws
Ans.
er id ati ng
ed e r
itt W o
t a
t p d W in hi
. b)
m e on
d e W Dis in t l
ads = vdv
an th . rs gh
k n ng to yri
or o ni c p
w g r tru o
vdv 1 5
e in lea s s c
ds = = v dv
th lud nt f in te
a 60
of inc de e o Sta
ity ( tu s d
s 3.925
L0 60 L2
1
gr rk s u te
is
ds = v5 dv
te wo ing the Uni
e his s fo by
y of as lel ted
a b`
1 v6 3.925
s r
s =
o c
st pa nd s te
th t se y
60 6 2
de ny es a ided pro
w of a urs rov k is
= 9.98 m Ans.
le co s p or
sa eir d i is w
in
ro rt
th an Th
ill
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained 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:
Full file at http://testbankcart.eu/Solution-Manual-for-Engineering-Mechanics-Dynamics-13th-Edition-by-Hibbeler
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Education, Inc., Upper Saddle River, NJ 07458. https://www.stuvia.com/user/gabrielucky
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
no orl sem eac ws
er id ati ng
ed e r
itt W o
t a
t p d W in hi
. b)
m e on
d e W Dis in t l
an th . rs gh
k n ng to yri
Chapter 17 Structural Analysis or o ni c p
w g r tru o 833
e in lea s s c
th lud nt f in te
of inc de e o Sta
Chapter 18 Internal Forces 953
ity ( tu s d
gr rk s u te
is
te wo ing the Uni
e his s fo by
Chapter 19 Friction 1023
y of as lel ted
s r
o c
st pa nd s te
th t se y
de ny es a ided pro
Review 2 Planar Kinematics and Kinetics of a Rigid Body 1080
w of a urs rov k is
le co s p or
sa eir d i is w
in
ro rt
th an Th
Chapter 20 Center of Gravity and Centroid 1131
ill
Chapter 21 Moments of Inertia 1190
Chapter 22 Virtual Work 1270
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained 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:
Full file at http://testbankcart.eu/Solution-Manual-for-Engineering-Mechanics-Dynamics-13th-Edition-by-Hibbeler
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Rights and Permissions Department, Pearson https://www.stuvia.com/user/gabrielucky
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no orl sem eac ws
er id ati ng
ed e r
itt W o
t a
t p d W in hi
. b)
m e on
d e W Dis in t l
an th . rs gh
k n ng to yri
or o ni c p
w g r tru o
e in lea s s c
th lud nt f in te
of inc de e o Sta
ity ( tu s d
gr rk s u te
is
te wo ing the Uni
e his s fo by
y of as lel ted
s r
o c
st pa nd s te
th t se y
de ny es a ided pro
w of a urs rov k is
le co s p or
sa eir d i is w
in
ro rt
th an Th
ill
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained 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:
Full file at http://testbankcart.eu/Solution-Manual-for-Engineering-Mechanics-Dynamics-13th-Edition-by-Hibbeler
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Rights and Permissions Department, Pearson https://www.stuvia.com/user/gabrielucky
Education, Inc., Upper Saddle River, NJ 07458. https://www.stuvia.com/user/gabrielucky
, https://www.stuvia.com/user/gabrielucky https://www.stuvia.com/user/gabrielucky https://www.stuvia.com/user/gabrielucky
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 Ans.
no orl sem eac ws
er id ati ng
ed e r
itt W o
t a
t p d W in hi
. b)
m e on
d e W Dis in t l
an th . rs gh
k n ng to yri
or o ni c p
w g r tru o
e in lea s s c
th lud nt f in te
of inc de e o Sta
ity ( tu s d
gr rk s u te
is
te wo ing the Uni
e his s fo by
y of as lel ted
s r
o c
st pa nd s te
th t se y
de ny es a ided pro
w of a urs rov k is
le co s p or
sa eir d i is w
in
ro rt
th an Th
ill
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained 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:
Full file at http://testbankcart.eu/Solution-Manual-for-Engineering-Mechanics-Dynamics-13th-Edition-by-Hibbeler
https://www.stuvia.com/user/gabrielucky
Rights and Permissions Department, Pearson https://www.stuvia.com/user/gabrielucky
Education, Inc., Upper Saddle River, NJ 07458. https://www.stuvia.com/user/gabrielucky
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12–2.
When a train is traveling along a straight track at 2 m/s, it v
begins to accelerate at a = 160 v-42 m>s2, where v is in m/s.
s
Determine its velocity v and the position 3 s after the
acceleration.
SOLUTION
dv
a =
dt
dv
dt =
a
3 v
L0 L2 60v
dv
dt = -4
1
3 = (v5 - 32)
300
v = 3.925 m>s = 3.93 m>s
no orl sem eac ws
Ans.
er id ati ng
ed e r
itt W o
t a
t p d W in hi
. b)
m e on
d e W Dis in t l
ads = vdv
an th . rs gh
k n ng to yri
or o ni c p
w g r tru o
vdv 1 5
e in lea s s c
ds = = v dv
th lud nt f in te
a 60
of inc de e o Sta
ity ( tu s d
s 3.925
L0 60 L2
1
gr rk s u te
is
ds = v5 dv
te wo ing the Uni
e his s fo by
y of as lel ted
a b`
1 v6 3.925
s r
s =
o c
st pa nd s te
th t se y
60 6 2
de ny es a ided pro
w of a urs rov k is
= 9.98 m Ans.
le co s p or
sa eir d i is w
in
ro rt
th an Th
ill
© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained 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:
Full file at http://testbankcart.eu/Solution-Manual-for-Engineering-Mechanics-Dynamics-13th-Edition-by-Hibbeler
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Education, Inc., Upper Saddle River, NJ 07458. https://www.stuvia.com/user/gabrielucky
