Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 05

5–1.
Draw the free-body diagram of the dumpster D of the
truck, which has a weight of 5000 lb and a center of gravity
at G. It is supported by a pin at A and a pin-connected
hydraulic cylinder BC (short link). Explain the significance
of each force on the diagram. (See Fig. 5–7b.)

1.5 m

G

D

1m

3m
A

B
20

30

C

SOLUTION
The Significance of Each Force:
W is the effect of gravity (weight) on the dumpster.
Ay and Ax are the pin A reactions on the dumpster.

T

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w le co ro is
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y ar d le d
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e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g

itt id tio
ed e n
. We or
b)

FBC is the hydraulic cylinder BC reaction on the dumpster.

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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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5–2.
Draw the free-body diagram of member ABC which is
supported by a smooth collar at A, rocker at B, and short link
CD. Explain the significance of each force acting on the
diagram. (See Fig. 5–7b.)

SOLUTION

3m

60
A

4 kN m
B

45
4m

D

C

2.5 kN

6m

The Significance of Each Force:
NA is the smooth collar reaction on member ABC.
NB is the rocker support B reaction on member ABC.
FCD is the short link reaction on member ABC.
2.5 kN is the effect of external applied force on member ABC.

T
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th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s

in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

4 kN # m is the effect of external applied couple moment on member ABC.

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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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5–3.
Draw the free-body diagram of the beam which supports the
80-kg load and is supported by the pin at A and a cable which
wraps around the pulley at D. Explain the significance of
each force on the diagram. (See Fig. 5–7b.)

D

5

4
3

A

B

E

SOLUTION

C

T force of cable on beam.

2m

2m

1.5 m

Ax, Ay force of pin on beam.

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sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la

is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

80(9.81)N force of load on beam.

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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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*5–4.
Draw the free-body diagram of the hand punch, which is
pinned at A and bears down on the smooth surface at B.

F ϭ 8 lb

1.5 ft

SOLUTION

A
B

2 ft

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w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la

is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

0.2 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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5–5.
Draw the free-body diagram of the uniform bar, which has a
mass of 100 kg and a center of mass at G. The supports A, B,
and C are smooth.

0.5 m

0.2 m

1.25 m

C
G

1.75 m

SOLUTION

A

0.1 m
30Њ

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th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
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ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat

th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

B

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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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5–6.
Draw the free-body diagram of the beam,which is pin
supported at A and rests on the smooth incline at B.

800 lb

800 lb
600 lb

600 lb

400 lb
3 ft

3 ft

3 ft

3 ft

0.6 ft
1.2 ft
A
0.6 ft

SOLUTION

B
5

3

T

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ro p an o te
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th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g

itt id tio
ed e n
. We or
b)

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,
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5–7.
Draw the free-body diagram of the beam, which is pin
connected at A and rocker-supported at B.

500 N

SOLUTION

800 Nиm

B

5m

A
4m

T
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th d wo
sa eir is p rk
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e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
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or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in

er ld
m W ina g
itt id tio
ed e n
. We or
b)

8m

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained 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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*5–8.
Draw the free-body diagram of the bar, which has a
negligible thickness and smooth points of contact at A, B,
and C. Explain the significance of each force on the
diagram. (See Fig. 5–7b.)

3 in.
30
C

5 in.
B
A
8 in.

SOLUTION

10 lb

NA, NB, NC force of wood on bar.

30

T
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th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p

or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

10 lb force of hand on bar.

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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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5–9.
Draw the free-body diagram of the jib crane AB, which is pin
connected at A and supported by member (link) BC.

C

5

3
4

SOLUTION

B

0.4 m
A

4m

3m

T
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e
s
in f th se for Un

te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

8 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:
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5–10.
4 kN

Determine the horizontal and vertical components of
reaction at the pin A and the reaction of the rocker B on
the beam.

B

A

30Њ

SOLUTION
6m

2m

Equations of Equilibrium: From the free-body diagram of the beam, Fig. a, NB can
be obtained by writing the moment equation of equilibrium about point A.
a
+ ©MA = 0;

NB cos 30°(8) - 4(6) = 0
NB = 3.464 kN = 3.46 kN

Ans.

Using this result and writing the force equations of equilibrium along the x and

y axes, we have
+ ©F = 0;
:
x

A x - 3.464 sin 30° = 0
A x = 1.73 kN
A y + 3.464 cos 30° - 4 = 0

T
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th d wo
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ill o u vi pr
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st ny s d s ec
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th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p

or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

+ c ©Fy = 0;

Ans.

A y = 1.00 kN

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:
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5–11.
Determine the magnitude of the reactions on the beam at A
and B. Neglect the thickness of the beam.

600 N

3

15Њ
400 N

5
4

B

A
4m

8m

SOLUTION
a + ©MA = 0;

By (12) - (400 cos 15°)(12) - 600(4) = 0
By = 586.37 = 586 N

+ ©F = 0;
:
x

Ans.

Ax - 400 sin 15° = 0
Ax = 103.528 N

+ c ©Fy = 0;

Ay - 600 - 400 cos 15° + 586.37 = 0

T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co

w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

Ay = 400 N
FA = 2(103.528)2 + (400)2 = 413 N

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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*5–12.
Determine the components of the support reactions at the

fixed support A on the cantilevered beam.

6 kN

30Њ

SOLUTION

30Њ

Equations of Equilibrium: From the free-body diagram of the cantilever beam, Fig. a,
Ax, Ay, and MA can be obtained by writing the moment equation of equilibrium about
point A.
+ ©F = 0;
:
x

1.5 m

4 kN

1.5 m

4 cos 30° - A x = 0
A x = 3.46 kN

+ c ©Fy = 0;

1.5 m

A

Ans.

A y - 6 - 4 sin 30° = 0
A y = 8 kN

Ans.

a+ ©MA = 0;MA - 6(1.5) - 4 cos 30° (1.5 sin 30°) - 4 sin 30°(3 + 1.5 cos 30°) = 0
Ans.

T
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th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat

th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

MA = 20.2 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:
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5–13.
The 75-kg gate has a center of mass located at G. If A
supports only a horizontal force and B can be assumed as a

pin, determine the components of reaction at these supports.

A

1.25 m
G

1m
B

SOLUTION
Equations of Equilibrium: From the free-body diagram of the gate, Fig. a, By and
Ax can be obtained by writing the force equation of equilibrium along the y axis and
the moment equation of equilibrium about point B.
+ c ©Fy = 0;

By - 75(9.81) = 0
By = 735.75 N = 736 N

a+©MB = 0;

Ans.

A x(1) - 75(9.81)(1.25) = 0
Ans.

T
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th d wo
sa eir is p rk

w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or

b)

A x = 919.69 N = 920 N

Using the result Ax = 919.69 N and writing the force equation of equilibrium along
the x axis, we have
+ ©F = 0;
:
x

Bx - 919.69 = 0

Bx = 919.69 N = 920 N

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:
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5–14.
The overhanging beam is supported by a pin at A and
the two-force strut BC. Determine the horizontal and
vertical components of reaction at A and the reaction at B
on the beam.

1m

600 N
1m

A

SOLUTION

B

900 Nиm

1.5 m

Equations of Equilibrium: Since line BC is a two-force member, it will exert a force FBC
directed along its axis on the beam as shown on the free-body diagram, Fig. a. From the
free-body diagram, FBC can be obtained by writing the moment equation of equilibrium
about point A.
a+ ©MA = 0;

800 N
2m

C

3
FBC a b (2) - 600(1) - 800(4) - 900 = 0
5
FBC = 3916.67 N = 3.92 kN

Ans.

Using this result and writing the force equations of equilibrium along the x and
y axes, we have
4
3916.67a b - A x = 0
5

T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y

k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

+
: ©Fx = 0;

A x = 3133.33 N = 3.13 kN
+ c ©Fy = 0;

Ans.

3
- A y - 600 - 800 + 3916.67a b = 0
5
A y = 950 N

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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5–15.
Determine the horizontal and vertical components of
reaction at the pin at A and the reaction of the roller at B on
the lever.

14 in.
30Њ
F ϭ 50 lb

A

SOLUTION
Equations of Equilibrium: From the free-body diagram, FB and A x can be obtained
by writing the moment equation of equilibrium about point A and the force
equation of equilibrium along the x axis, respectively.
a+ ©MA = 0;

18 in.

50 cos 30°(20) + 50 sin 30°(14) - FB(18) = 0
FB = 67.56 lb = 67.6 lb

+ ©F = 0;
:

x

20 in.

B

Ans.

A x - 50 sin 30° = 0
A x = 25 lb

Ans.

+c ©Fy = 0;

T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite

gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

Using the result FB = 67.56 lb and writing the force equation of equilibrium along
the y axis, we have
A y - 50 cos 30° - 67.56 = 0
A y = 110.86 lb = 111 lb

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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*5–16.
Determine the components of reaction at the supports A and
B on the rod.

P
L
––
2

SOLUTION
Equations of Equilibrium: Since the roller at A offers no resistance to vertical
movement, the vertical component of reaction at support A is equal to zero. From
the free-body diagram, Ax, By, and MA can be obtained by writing the force
equations of equilibrium along the x and y axes and the moment equation of
equilibrium about point B, respectively.
+ ©F = 0;
:
x

Ax = 0

+ c ©Fy = 0;

By - P = 0
By = P

a + ©MB = 0;

Pa

A

L
––
2
B

Ans.

Ans.

L
b - MA = 0
2
PL
2

Ans.

T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr

de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

MA =

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained 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.


5–17.
If the wheelbarrow and its contents have a mass of 60 kg and
center of mass at G, determine the magnitude of the resultant
force which the man must exert on each of the two handles in
order to hold the wheelbarrow in equilibrium.
B

G
0.6 m

0.5 m
A

SOLUTION
a +©MB = 0;

0.5 m

0.9 m

- Ay (1.4) + 60(9.81)(0.9) = 0

Ay = 378.39 N

+ c ©Fy = 0;

378.39 - 60(9.81) + 2By = 0
By = 105.11 N

+ ©F = 0;
:
x

Bx = 0
Ans.

T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e

ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

FB = 105 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.


5–18.

Determine the tension in the cable and the horizontal and
vertical components of reaction of the pin A. The pulley at
D is frictionless and the cylinder weighs 80 lb.

D
2
1

A

B
C

SOLUTION

5 ft

5 ft

3 ft

Equations of Equilibrium: The tension force developed in the cable is the same
throughout the whole cable. The force in the cable can be obtained directly by
summing moments about point A.

+ ©F = 0;
:
x

2

≤ 1102 - 801132 = 0
25
T = 74.583 lb = 74.6 lb

T152 + T ¢

Ax - 74.583 ¢

1
25

≤ = 0

Ax = 33.4 lb

+ c ©Fy = 0;

Ans.

T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by

e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

a + ©MA = 0;

74.583 + 74.583

2

25

Ans.

- 80 - By = 0

Ay = 61.3 lb

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.


5–19.
The shelf supports the electric motor which has a mass of
15 kg and mass center at Gm. The platform upon which it
rests has a mass of 4 kg and mass center at Gp. Assuming that
a single bolt B holds the shelf up and the bracket bears
against the smooth wall at A, determine this normal force
at A and the horizontal and vertical components of reaction
of the bolt on the bracket.

150 mm

Gm

50 mm
40 mm
B

Gp

60 mm

SOLUTION
a + ©MA = 0;

200 mm

Bx (60) - 4(9.81)(200) - 15(9.81)(350) = 0

A

Bx = 989.18 = 989 N

Ans.

+ ©F = 0;
:
x

Ax = 989.18 = 989 N

Ans.

+ c ©Fy = 0;

By = 4(9.81) + 15(9.81)
Ans.

T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la

is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

By = 186.39 = 186 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.


*5–20.
The pad footing is used to support the load of 12 000 lb.
Determine the intensities w1 and w2 of the distributed
loading acting on the base of the footing for the
equilibrium.

12 000 lb

5 in.

9 in.

9 in.

w2

w1

SOLUTION

35 in.

Equations of Equilibrium: The load intensity w2 can be determined directly by
summing moments about point A.
w2 a

35
b 117.5 - 11.672 - 12114 - 11.672 = 0
12

T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d

th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

a + ©MA = 0;

w2 = 1.646 kip>ft = 1.65 kip>ft
+ c ©Fy = 0;

Ans.

35
35
1
1w - 1.6462a b + 1.646 a b - 12 = 0
2 1
12
12
w1 = 6.58 kip>ft

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.


5–21.
When holding the 5-lb stone in equilibrium, the humerus H,
assumed to be smooth, exerts normal forces FC and FA on
the radius C and ulna A as shown. Determine these forces
and the force FB that the biceps B exerts on the radius for
equilibrium. The stone has a center of mass at G. Neglect the
weight of the arm.

B

H
H
FB

SOLUTION
a + ©MB = 0;

75Њ
FC

- 5(12) + FA (2) = 0
FA = 30 lb

+ c ©Fy = 0;

Ans.

A
FA
2 in.

FB sin 75° - 5 - 30 = 0
FB = 36.2 lb

+ ©F = 0;
:
x

0.8 in.

G
C

14 in.

Ans.

FC - 36.2 cos 75° = 0
Ans.

T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co

w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

FC = 9.38 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.


5–22.
The smooth disks D and E have a weight of 200 lb and 100 lb,
respectively. If a horizontal force of P = 200 lb is applied to
the center of disk E, determine the normal reactions at the
points of contact with the ground at A, B, and C.

1.5 ft

5

4

1 ft

3

P

D
E

A
B

C

SOLUTION
For disk E:
224
= 0
5

+ âF = 0;
:
x

- P + N Â

+ c ©Fy = 0;

1
NC - 100 - N¿ a b = 0
5

For disk D:
4
224
NA a b - N¿ ¢
≤ = 0
5
5

+ c ©Fy = 0;

3
1
NA a b + NB - 200 + N¿ a b = 0
5
5

Set P = 200 lb and solve:

T
an his
th d wo

sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n

. We or
b)

+ ©F = 0;
:
x

N¿ = 204.12 lb

NA = 250 lb

Ans.

NB = 9.18 lb

Ans.

NC = 141 lb

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.


5–23.
The smooth disks D and E have a weight of 200 lb and 100 lb,

respectively. Determine the largest horizontal force P that
can be applied to the center of disk E without causing the
disk D to move up the incline.

1.5 ft

5

4

1 ft

3

P

D
E

A
B

C

SOLUTION
For disk E:
224
≤ = 0
5

+ ©F = 0;
:
x

- P + N Â

+ c âFy = 0;

1
NC - 100 - N a b = 0
5

For disk D:
4
224
NA a b - N¿ ¢
≤ = 0
5
5

+ c ©Fy = 0;

3
1
NA a b + NB - 200 + N¿ a b = 0
5
5

T
an his

th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio

ed e n
. We or
b)

+ ©F = 0;
:
x

Require NB = 0 for Pmax. Solving,

N¿ = 214 lb

Pmax = 210 lb

Ans.

NA = 262 lb
NC = 143 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.


*5–24.
The man is pulling a load of 8 lb with one arm held as
shown. Determine the force FH this exerts on the humerus
bone H, and the tension developed in the biceps muscle B.

Neglect the weight of the man’s arm.

8 lb

13 in.

SOLUTION
a + ©MB = 0;

H

- 81132 + FH11.752 = 0
FH = 59.43 = 59.4 lb

+ ©F = 0;
:
x

B TB 1.75 in.

Ans.
FH

8 - TB + 59.43 = 0
Ans.

T
an his
th d wo
sa eir is p rk

w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e
ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or

b)

TB = 67.4 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.


5–25.
Determine the magnitude of force at the pin A and in the
cable BC needed to support the 500-lb load. Neglect the
weight of the boom AB.

B

8 ft
C

22Њ

A

35Њ

SOLUTION
Equations of Equilibrium: The force in cable BC can be obtained directly by
summing moments about point A.

a + ©MA = 0;

FBC sin 13°(8) - 500 cos 35°(8) = 0
FBC = 1820.7 lb = 1.82 kip

+
Q ©Fx = 0;

Ans.

A x - 1820.7 cos 13° - 500 sin 35° = 0

a + ©Fy = 0;

T
an his
th d wo
sa eir is p rk
w le co ro is
ill o u vi pr
de f a rse de ot
st ny s d s ec
ro p an o te
y ar d le d
th t o a ly by
e
s
in f th se for Un
te is ss th ite
gr w in e

ity o g us d S
of rk ( stu e o tat
th inc de f i es
e lu nt ns co
w d le tr p
or in a uc y
k g rn to rig
an on in rs h
d th g. in t la
is e D t w
no W iss ea s
t p or em ch
in
er ld
m W ina g
itt id tio
ed e n
. We or
b)

A x = 2060.9 lb
A y + 1820.7 sin 13° - 500 cos 35° = 0
Ay = 0

Thus,

FA = A x = 2060.9 lb = 2.06 kip

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.