Engineering Mechanics - Statics Chapter 8
μ
pyg
pp
p
can be held.
Given:
θ
20 deg=
Solution:
Paper:
+
↑
Σ
F
y
= 0;
FF+ W− 0= F
W
2
=
F
μ
N= N
W
2
μ
=
Cylinder:
F' r
W

− 0= N'
W
2
sin
θ
()
1
μ
cos
θ
()
+
⎛
⎜
⎝
⎞
⎟
⎠
=
F'
μ
N'=
W
2
μ
W
2
⎛
⎜
⎝

θ
()
−
sin
θ
()
=
μ
0.176=
Problem 8-43
The crate has a weight W
1
and a center of gravity at G. If the coefficient of static friction
between the crate and the floor is
μ
s
, determine if the man of weight W
2
can push the crate to
the left. The coefficient of static friction between his shoes and the floor is
μ
'
s
. Assume the
man exerts only a horizontal force on the crate.
801
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 8

F
x
= 0;
μ
s
N
C
P− 0= P
μ
s
N
C
=
Σ
Μ
O
= 0;
W
1
− xPd+ 0= x
Pd
W
1
=
Since
x 0.90 ft=
<
a 4.50 ft=
there will not be any tipping.
Σ

N
m
= F
mmax
70.00 lb=
Since
F
m
60.00 lb=
<
F
mmax
70.00 lb=
then the
man can push the crate.
Problem 8-44
The crate has a weight W
1
and a center of gravity at G. If the coefficient of static friction
between the crate and the floor is
μ
s
, determine the smallest weight of the man so that he

can
push the crate to the left. The coefficient of static friction between his shoes and the floor is
μ
'
s
. Assume the man exerts only a horizontal force on the crate.

− 0= N
C
W
1
=
Σ
F
x
= 0;
μ
s
N
C
P− 0= P
μ
s
N
C
=
Σ
Μ
O
= 0;
W
1
− xPd+ 0= x
Pd
W
1
=

= N
m
171.4 lb=
Σ
F
y
= 0;
N
m
W
2
− 0= W
2
N
m
= W
2
171.4 lb=
Problem 8-45
The wheel has weight W
A
and rests on a surface for which the coefficient of friction is
μ
B
. A
cord wrapped around the wheel is attached to the top of the homogeneous block of weight W
C
.
803
Given:

D
1lb= F
D
1lb=
x 1ft=
Given N
B
W
A
− P− 0= TF
B
− 0= PT− F
B
−
()
h
2
0=
F
B
μ
B
N
B
= T− F
D
+ 0= N
D
W
C

⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find PN
B
, F
B
, T, N
D
, F
D
, x,
()
=
P
N
B
F
B
T
N
D
F
D
⎛
⎜

⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
lb= x 0.67 ft=
Now checke the assumptions
F
Dmax
μ
D
N
D
=
Since
F
D
6.67 lb=
<
F
Dmax
9.00 lb=
then the block does not slip
804
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

W
2
30 lb= b 3ft=
μ
B
0.2= c 1.5 ft=
μ
D
0.3=
Solution:
For the wheel : Assume slipping occurs,
Σ
F
y
= 0;
N
B
W
1
− 0= N
B
W
1
= N
B
20.00 lb=
Σ
F
x
= 0;

D
30.00 lb=
Σ
F
x
= 0;
F
D
T− 0= F
D
T= F
D
4.00 lb=
Σ
M
O
= 0;
Tb N
D
x− 0= xT
b
N
D
= x 0.40 ft=
F
Dmax
μ
D
N
D

and negligible
thickness. Determine the minimum force
P
needed to move the post. The coefficients of static
friction at B and C are
μ
B
and
μ
C
respectively.
Given:
m
p
50 kg= a 2m=
μ
B
0.4= b 400 mm=
μ
C
0.2= c 300 mm=
w 800
N
m
= d 3=
g 9.81
m
s
2
= e 4=

wa= N
B
533.33 N=
806
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 8
Post:
Assume slipping occurs at C:
F
C
μ
C
N
C
=
The initial guesses are
P 1N= N
C
1N= F
B
1N=
Given
e−
d
2
e
2
+

P
N
C
F
B
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
Find PN
C
, F
B
,
()
= P 354.79 N=
Now check to see if the post slips at B.
F
Bmax
μ
B
N
B
=

N
m
=
a 2m=
b 400 mm=
c 300 mm=
d 3=
e 4=
Solution: Member AB:
1
2
− wa
2a
3
N
B
a+ 0=
N
B
1
3
wa= N
B
533.33 N=
Post:
Guesses N
C
1N=
μ
B

+
⎛
⎜
⎝
⎞
⎟
⎠
P
μ
C
N
C
−
μ
B
N
B
− 0=
e−
d
2
e
2
+
⎛
⎜
⎝
⎞
⎟
⎠

()
=
μ
B
μ
C
⎛
⎜
⎝
⎞
⎟
⎠
0.0964
0.0734
⎛
⎜
⎝
⎞
⎟
⎠
=
808
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 8
Problem 8-49
The block of weight W is being pulled up the inclined plane of slope
α
using a force

()
=
Σ
F
x
= 0;
P cos
φ
()
W sin
α
()
−
μ
N− 0=
Σ
F
y
= 0;
NWcos
α
()
− P sin
φ
()
+ 0=
P cos
φ
()
W sin

⎛
⎜
⎝
⎞
⎟
⎠
= W
sin
α
()
tan
θ
()
cos
α
()
+
cos
φ
()
tan
θ
()
sin
φ
()
+
⎛
⎜
⎝

+
⎛
⎜
⎝
⎞
⎟
⎠
=
PW
sin
αθ
+
()
cos
φθ
−
()
⎛
⎜
⎝
⎞
⎟
⎠
= QED()
Problem 8-50
Determine the angle
φ
at which
P
should act on the block so that the magnitude of

Σ
F
y
= 0;
NWcos
α
()
− P sin
φ
()
+ 0=
P cos
φ
()
W sin
α
()
−
μ
W cos
α
()
P sin
φ
()
− 0=
(
−
PW
sin

()
+
cos
φ
()
tan
θ
()
sin
φ
()
+
⎛
⎜
⎝
⎞
⎟
⎠
=
PW
sin
α
()
cos
θ
()
sin
θ
()
cos

()
=
dP
d
φ
W
sin
αθ
+
()
sin
φθ
−
()
cos
2
φθ
−
()
⎡
⎢
⎣
⎤
⎥
⎦
= 0=
sin
αθ
+
()

M 6kg=
μ
s
0.5=
θ
1
30 deg=
θ
2
30 deg=
Solution:
Guesses N
A
1N= F
A
1N= F
AC
1N=
P 1N= N
B
1N= F
B
1N= F
BC
1N=
Assume that A slips first
Given
F
AC
cos

− 0=
F
BC
cos
θ
1
()
Mgsin
θ
1
()
− F
B
− 0=
F
BC
− sin
θ
1
()
Mgcos
θ
1
()
− N
B
+ 0=
F
A
μ

⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find PN
A
, F
A
, N
B
, F
B
, F
AC
, F
BC
,
()
= P
1
23.9 N=
Assume that B slips first
811
Given:

F
A
F
AC
cos
θ
2
()
− 0=
F
BC
cos
θ
1
()
Mgsin
θ
1
()
− F
B
− 0=
F
BC
− sin
θ
1
()
Mgcos
θ

⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find PN
A
, F
A
, N
B
, F
B
, F
AC
, F
BC

m
s
2
=
m
c
50 kg=
m
s
40 kg=
μ
A
0.3=
812
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 8
μ
B
0.6= r
1
0.2 m= r
2
0.4 m=
Solution: Assume that the spool slips at A but not at B.
The initial guesses are
F
B
2N= P 3N= N

P
N
B
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
Find F
B
P, N
B
,
()
=
F
B
P
N
B
⎛
⎜
⎜
⎜
⎝

441 N=
<
F
Bmax
530 N=
then our assumptions are correct.
P 589 N=
Problem 8-53
A board of weight W
1
is placed across the channel and a boy of weight W
2
attempts to walk
across. If the coefficient of static friction at A and B
μ
s
, determine if he can make the crossing;
and if not, how far will he get from A before the board slips?
813
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 8
Given:
W
1
50 lb=
W
2
100 lb=

2
+
⎛
⎜
⎝
⎞
⎟
⎠
+ N
B
c
b
2
c
2
+
⎛
⎜
⎝
⎞
⎟
⎠
− 0=
Σ
F
y
= 0;
N
A
W

⎝
⎞
⎟
⎠
+ 0=
Σ
M
B
= 0;
W
1
a
2
⎛
⎜
⎝
⎞
⎟
⎠
W
2
ad−()+ N
A
a− 0=
N
A
N
B
d
⎛

⎝
⎞
⎟
⎠
lb= d 6.47 ft=
Since
d 6.47 ft=
<
a 10.00 ft=
then the board will slip
Problem 8-54
Determine the minimum force P needed to push the tube E up the incline. The tube has a
mass of M
1
and the roller D has a mass of M
2
. The force acts parallel to the plane, and the
coefficients of static friction at the contacting surfaces are
μ
A
,
μ
B
and
μ
C
. Each cylinder has a
radius of r.
814
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

A
100 N= F
A
10 N= P 100 N=
N
B
200 N= F
B
20 N=
N
C
300 N= F
C
30 N=
Given
For roller D
Σ
F
x'
= 0;PN
A
− F
C
− M
2
gsin
θ
N
A
F
B
− M
1
gsin
θ
()
− 0=
Σ
F
y'
= 0;
N
B
F
A
− M
1
gcos
θ
()
− 0=
Σ
M
0'
= 0;

P
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find N
A
N
B
, N
C
, F

⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
526
795
692
158
158
158
1174
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝

⎝
⎞
⎟
⎠
199
277
⎛
⎜
⎝
⎞
⎟
⎠
N=
Since
F
B
158 N=
<
F
Bmax
199 N=
and
F
C
158 N=
<
F
Cmax
277 N=
then our

Σ
F
x
= 0;
F− cos
θ
()
Nsin
θ
()
+ F'− 0=
Σ
F
y
= 0;
N' W− Ncos
θ
()
− F sin
θ
()
− 0=
Σ
M
O
= 0;
F r F' r− 0=
Solving,
FF'=
1

1+
⎛
⎜
⎝
⎞
⎟
⎠
≥
μ
s
sin
θ
()
cos
θ
()
1+
≥
For Pipe C:
μ
'
s
N'
1
2
W sin
θ
()
cos
θ


θ
at which
the pole can be placed without slipping.
Solution:
Σ
F
x
= 0;
N
A
Tsin
θ
2
⎛
⎜
⎝
⎞
⎟
⎠
− 0=
Σ
F
y
= 0;
μ
s
N
A
W− Tcos

⎝
⎞
⎟
⎠
sin
θ
()
+ 0=
817
Solution:
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 8
Solving we find
N
A
Tsin
θ
2
⎛
⎜
⎝
⎞
⎟
⎠
=
μ
s
Tsin

2
⎛
⎜
⎝
⎞
⎟
⎠
+
⎛
⎜
⎝
⎞
⎟
⎠
=
Tsin
θ
2
⎛
⎜
⎝
⎞
⎟
⎠
Lcos
θ
()
μ
s
Lsin

⎠
L
2
sin
θ
()
=
sin
θ
2
⎛
⎜
⎝
⎞
⎟
⎠
cos
θ
()
μ
s
sin
θ
()
+
()
1
2
sin
θ

s
2
sin
θ
2
⎛
⎜
⎝
⎞
⎟
⎠
sin
θ
()
1
2
sin
θ
()
cos
θ
2
⎛
⎜
⎝
⎞
⎟
⎠
sin
θ

⎝
⎞
⎟
⎠
cos
θ
()
−
sin
θ
2
⎛
⎜
⎝
⎞
⎟
⎠
sin
θ
()
=
μ
s
cos
θ
2
⎛
⎜
⎝
⎞

2
−
⎛
⎜
⎝
⎞
⎟
⎠
2sin
θ
2
⎛
⎜
⎝
⎞
⎟
⎠
cos
θ
2
⎛
⎜
⎝
⎞
⎟
⎠
−= cot
θ
2
⎛

⎜
⎝
⎞
⎟
⎠
=
θ
2atan
μ
()
=
Problem 8-57
The carpenter slowly pushes the uniform board horizontally over the top of the saw horse. The
board has a uniform weight density
γ
and the saw horse has a weight W and a center of gravity
at G. Determine if the saw horse will stay in position, slip, or tip if the board is pushed forward
at the given distance d. The coefficients of static friction are shown in the figure.
Given:
γ
3
lb
ft
=
818
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 8
L 18 ft=

⎞
⎟
⎠
= N 48.60 lb=
To cause slipping of the board on the saw horse:
P
xb
μ
N= P
xb
24.30 lb=
To cause slipping at the ground:
P
xg
μ
'N W+()= P
xg
19.08 lb=
To cause tipping
NW+()bP
xt
a− 0=
P
xt
NW+()b
a
= P
xt
21.20 lb=
Choose the critical case

Engineering Mechanics - Statics Chapter 8
L 18 ft=
W 15 lb=
a 3ft=
b 1ft=
μ
0.5=
μ
' 0.3=
d 14 ft=
Solution:
Board:
L−
γ
L
2
⎛
⎜
⎝
⎞
⎟
⎠
Nd+ 0= N
1
2
L
2
γ
d
⎛

16.57 lb=
Choose the critical case
P
x
min P
xb
P
xg
, P
xt
,
()
= P
x
14.91 lb=
Problem 8-59
The disk of mass m
o
rests on the surface for which the coefficient of static friction is
μ
A
Determine the largest couple moment M that can be applied to the bar without causing motion.
820
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 8
Given:
m
o

N
A
m
o
g− B
y
− 0=
B
y
r
μ
A
N
A
r− 0=
M
N
A
B
x
B
y
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟

110.36
110.36
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N= M 77.3 N m⋅=
Problem 8-60
The disk of mass m
0
rests on the surface for which the coefficient of static friction is
μ
A
Determine the friction force at A.
821
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 8
Given:
M 50 N m⋅=
m
o
45 kg=
μ
A

A
− 0=
B
y
rF
A
r− 0=
B
x
B
y
N
A
F
A
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
Find B
x
B

μ
A
N
A
=
Since
F
A
71.4 N=
<
F
Amax
102.6 N=
then our assumption is good.
F
A
71.4 N=
Problem 8-61
A block of weight W is attached to a light rod AD that pivots at pin A. If the coefficient of static
friction between the plane and the block is
μ
s
, determine the minimum angle
θ
at which the
b
lock ma
y
be
p

b−
c
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= r
BA
b−
0
a
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
n
1
r
BC
r

⎞
⎟
⎟
⎠
= r
CA
0
c−
a
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= n
2
r
CA
r
CA
= n
3
n
1
n
2
×=

+
()
− Wk− 0=
N
T
θ
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
Find NT,
θ
,
()
=
N
T
⎛
⎜
⎝
⎞
⎟
⎠
2.30
5.46
⎛

10 deg=
W 100 lb=
μ
s
0.3=
Solution:
Initial guesses:
N' 10 lb= N
A
15 lb=
N
B
20 lb= P 5lb=
Given
Wedge B:
Σ
F
x
= 0;
N' sin
θ
()
N
B
− 0=
Σ
F
y
= 0;
N' cos

()
− 0=
N'
N
B
N
A
P
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
Find N' N
B
, N
A
, P,
()
=
N'
N
B

friction between the wedge and the two surfaces of contact is
μ
s
. Neglect the size and weight
824
© 2007 R. C. Hibbeler. Published by 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.
Engineering Mechanics - Statics Chapter 8
of the wedge and the thickness of the beam.
Units Used:
kN 10
3
N=
Given:
F
1
2kN= a 3m=
F
2
4kN= b 2m=
F
3
4kN= c 3m=
F
4
2kN=
θ
15 deg=
μ

sin
θ
()
− N
1
− 0=
μ
s
N
1
μ
s
N
2
cos
θ
()
+ N
2
sin
θ
()
+ P− 0=
N
1
N
2
P
⎛
⎜

⎞
⎟
⎠
kN=
P 4.83 kN=
Problem 8-64
The three stone blocks have weights W
A
, W
B,
and W
C
. Determine the smallest horizontal force
P
that must be applied to block C in order to move this block.The coefficient of static friction
between the blocks is
μ
s,
and between the floor and each block
μ
'
s.
825
© 2007 R. C. Hibbeler. Published by 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.