Engineering Mechanics - Statics Chapter 8
Check: If
F
A
604 N=
<
F
Amax
664 N=
then our no-slip assumption is good.
Problem 8-10
The block brake is used to stop the wheel from rotating when the wheel is subjected to a
couple moment M
0
If the coefficient of static friction between the wheel and the block is
μ
s
,
determine the smallest force
P
that should be applied.
Solution:
Σ
M
C
= 0;
Pa Nb−
μ
s
Nc+ 0=
N

Pa
b
μ
s
c−
=
μ
s
Nr M
O
− 0=
Σ
M
O
= 0;
μ
s
Par
b
μ
s
c−
M
O
=
P
M
O
b
μ

s
c−
()
μ
s
ra
=
Problem 8-11
The block brake is used to stop the wheel from rotating when the wheel is subjected to a couple
771
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Engineering Mechanics - Statics Chapter 8
moment M
0
If the coefficient of static friction between the wheel and the block is
μ
s
, show that the
brake is self locking, i. e.,
P 0≤
, provided
b
c
μ
s
≤

Solution:

Σ
M
C
= 0;
Pa Nb−
μ
s
Nc+ 0=
N
Pa
b
μ
s
c−
=

Σ
M
O
= 0;
μ
s
Nr M
O
− 0=
μ
s
Par
b
μ

s
c−
M
O
=
P
M
O
b
μ
s
c−
()
μ
s
ra
=
P < 0 if
b
μ
s
c−
()
0<
i.e. if
b
c
μ
s
<

Problem 8-12
The block brake is used to stop the wheel from rotating when the wheel is subjected to a couple
moment M
0
If the coefficient of static friction between the wheel and the block is
μ
s
, determine the
smallest force
P
that should be applied if the couple moment
M
O
is applied
counterclockwise
.
772
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Engineering Mechanics - Statics Chapter 8
Solution:
Σ
M
C
= 0;
Pa Nb−
μ
s
Nc− 0=

N
Pa
b
μ
s
c+
=
Σ
M
O
= 0;
μ
s
− Nr M
O
+ 0=
μ
s
Par
b
μ
s
c+
M
O
=
P
M
O
b

μ
s
c+
()
μ
s
ra
=
Problem 8-13
The block brake consists of a pin-connected lever and friction block at B. The coefficient of static
friction between the wheel and the lever is
μ
s
and a torque
M
is applied to the wheel. Determine if
the brake can hold the wheel stationary when the force applied to the lever is (a) P
1
(b) P
2
.
773
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Engineering Mechanics - Statics Chapter 8
μ
s
0.3=
M 5Nm⋅=

a 50 mm=
b 200 mm=
c 400 mm=
r 150 mm=
P
1
30 N=
P
2
70 N=
Solution: To hold lever:
Σ
M
O
= 0;
F
B
rM− 0=
F
B
M
r
= F
B
33.333 N=
Require
N
B
F
B

μ
s
= N
B
111.1 N=
Lever,
Σ
M
A
= 0;
P
Reqd
bc+()N
B
b− F
B
a− 0=
P
Reqd
N
B
bF
B
a+
bc+
= P
Reqd
39.8 N=
(a) If
P

1
30.00 N=
>
P
Reqd
39.81 N=
then the break will hold the wheel
(b) If
P
2
70.00 N=
>
P
Reqd
39.81 N=
then the break will hold the wheel
Problem 8-14
The block brake consists of a pin-connected lever and friction block at B. The coefficient of static
friction between the wheel and the lever is
μ
s
and a torque
M
is applied to the wheel. Determine if
the brake can hold the wheel stationary when the force applied to the lever is (a) P
1
(b) P
2.
Assume that the torque
M

is applied
counter-clockwise.
774
Given:
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Engineering Mechanics - Statics Chapter 8
μ
s
0.3=
M 5Nm⋅=
a 50 mm=
b 200 mm=
c 400 mm=
r 150 mm=
P
1
30 N=
P
2
70 N=
Solution: To hold lever:
Σ
M
O
= 0;
F
B
rM− 0=

F
B
M
r
= F
B
33.333 N=
Require
N
B
F
B
μ
s
= N
B
111.1 N=
Lever,
Σ
M
A
= 0;
P
Reqd
bc+()N
B
b− F
B
a+ 0=
P

Reqd
N
B
bF
B
a−
bc+
= P
Reqd
34.3 N=
(a) If
P
1
30.00 N=
>
P
Reqd
34.26 N=
then the break will hold the wheel
(b) If
P
2
70.00 N=
>
P
Reqd
34.26 N=
then the break will hold the wheel
775
Given:

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Engineering Mechanics - Statics Chapter 8
Problem 8-15
The doorstop of negligible weight is pin connected
at A and the coefficient of static friction at B is
μ
s
.
Determine the required distance s from A to the
floor so that the stop will resist opening of the
door for any force
P
applied to the handle.
Given:
μ
s
0.3=
a 1.5 in=
Solution:
Σ
F
y
= 0;
N
B
s
s
2

a
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
A
− 0=
Σ
F
x
= 0;
μ
s
N
B
a
s
2
a
2
+
⎛
⎜
⎝
⎞

⎟
⎠
F
A
− 0=
μ
s
s
s
2
a
2
+
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
A
a
s
2
a
2
+
⎛

⎜
⎝
⎞
⎟
⎠
F
A
− 0=
μ
s
sa= s
a
μ
s
= s 5.00 in=
Problem 8-16
The chair has a weight W and center of gravity at G. It is propped against the door as shown.
If the coefficient of static friction at A is
μ
A
,

determine the smallest force
P
that must be
applied to the handle to open the door.
776
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Engineering Mechanics - Statics Chapter 8
Given:
μ
A
0.3=
a 1.20 ft=
b 0.75 ft=
c 3ft=
θ
30 deg=
W 10 lb=
Solution:
Guesses B
y
1lb= N
A
1lb= P 1lb=
Given
Σ
F
x
= 0;
P−
μ
A
N
A
+ 0=
Σ
F

y
= 0;
N
A
W− B
y
− 0=
Σ
Μ
Β
= 0;
μ
A
N
A
c cos
θ
()
N
A
c sin
θ
()
− Wca−( )sin
θ
()
bcos
θ
()
+

⎡
⎣
⎤
⎦
+ 0=
B
y
N
A
P
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
Find B
y
N
A
, P,
()
=
B
y
11.5 lb=

N
A
21.5 lb=
P 6.45 lb=
Problem 8-17
The uniform hoop of weight W is suspended from the peg at A and a horizontal force
P
is
slowly applied at B. If the hoop begins to slip at A when the angle is
θ ,
determine the
coefficient of static friction between the hoop and the peg.
777
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Engineering Mechanics - Statics Chapter 8
Given:
θ
30 deg=
Solution:
Σ
F
x
= 0;
μ
N
A
cos
θ

()
P+ N
A
sin
θ
()
− 0=
P
μ
cos
θ
()
sin
θ
()
−
()
N
A
=
Σ
F
y
= 0;
μ
N
A
sin
θ
()

W− N
A
cos
θ
()
+ 0=
W
μ
sin
θ
()
cos
θ
()
+
()
N
A
=
Σ
Μ
Α
= 0;
W− r sin
θ
()
Pr rcos
θ
()
+

()
+ 0=
W sin
θ
()
P 1 cos
θ
()
+
()
=
μ
sin
θ
()
cos
θ
()
+
()
sin
θ
()
sin
θ
()
μ
cos
θ
()

−
()
1 cos
θ
()
+
()
=
μ
sin
θ
()
1 cos
θ
()
+
=
μ
0.27=
Problem 8-18
The uniform hoop of weight W is suspended from the peg at A and a horizontal force
P
is
slowly applied at B. If the coefficient of static friction between the hoop and peg is
μ
s
,
determine if it is possible for the hoop to reach an angle
θ
before the hoop begins to slip.

778
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Engineering Mechanics - Statics Chapter 8
Given:
μ
s
0.2=
θ
30 deg=
Solution:
Σ
F
x
= 0;
μ
N
A
cos
θ
()
P+ N
A
sin
θ
()
− 0=
P
μ

cos
θ
()
sin
θ
()
−
()
N
A
=
Σ
F
y
= 0;
μ
N
A
sin
θ
()
W− N
A
cos
θ
()
+ 0=
W
μ
sin

θ
()
cos
θ
()
+
()
N
A
=
Σ
Μ
Α
= 0;
W− r sin
θ
()
Pr rcos
θ
()
+
()
+ 0=
W sin
θ
()
P 1 cos
θ
()
+

()
=
μ
sin
θ
()
cos
θ
()
+
()
sin
θ
()
sin
θ
()
μ
cos
θ
()
−
()
1 cos
θ
()
+
()
=
μ

sin
θ
()
1 cos
θ
()
+
=
μ
0.27=
If
μ
s
0.20=
<
μ
0.27=
then it is not possible to reach
θ
30.00 deg=
.
Problem 8-19
The coefficient of static friction between the shoes at A and B of the tongs and the pallet is
μ
s1
and
between the pallet and the floor
μ
s2
. If a horizontal towing force

P
is applied to the tongs,
determine the largest mass that can be towed.
779
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Engineering Mechanics - Statics Chapter 8
μ
s1
0.5= a 75 mm=
μ
s2
0.4= b 20 mm=
P 300 N= c 30 mm=
g 9.81
m
s
2
=
θ
60 deg=
Solution:
Assume that we are on the verge of
slipping at every surface.
Guesses
T 1N= N
A
1N=
F 1N= N

ground
1N=
F
A
1N= mass 1kg=
Given
2 T sin
θ
()
P− 0=
T− sin
θ
()
bc+()T cos
θ
()
a− F
A
b− N
A
a+ 0=
F
A
μ
s1
N
A
=
2 F
A

F− 0=
N
ground
massg− 0=
F
μ
s2
N
ground
=
T
N
A
F
A
F
N
ground
mass
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟

⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find TN
A
, F
A
, F, N
ground
, mass,
()
=
780
Given:
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Engineering Mechanics - Statics Chapter 8
T
N
A
F
A
F
N
ground

⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
173.21
215.31
107.66
215.31
538.28
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟

⎟
⎟
⎟
⎟
⎠
N= mass 54.9 kg=
Problem *8-20
The pipe is hoisted using the tongs. If the coefficient of static friction at A and B is
μ
s
,

determine the
smallest dimension b so that any pipe of inner diameter d can be lifted.
Solution:
W 2 F
B
− 0=
W
2
⎛
⎜
⎝
⎞
⎟
⎠
bN
B
h− F
B

d
2
⎛
⎜
⎝
⎞
⎟
⎠
− 0=
Thus
F
B
W
2
=
N
B
W 2 bd−()
4h
=
Require
F
B
μ
s
N
B
≤
W
2

μ
s
W 2bd−()
4 h
≤ 2 h
μ
s
2bd−()≤ b
h
μ
s
d
2
+>
Problem 8-21
A very thin bookmark having a width a. is in the middle of a dictionary of weight W. If the
pages are b by c, determine the force
P
needed to start to pull the bookmark out.The coefficient
of static friction between the bookmark and the paper is
μ
s
. Assume the pressure on each page
and the bookmark is uniform.
781
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Engineering Mechanics - Statics Chapter 8
Given:

a 1in=
W 10 lb=
b 8in=
c 10 in=
μ
s
0.7=
Solution:
Pressure on book mark :
P
1
2
W
bc
= P 0.06
lb
in
2
=
Normal force on bookmark:
NPca=
F
μ
s
N= F 0.44 lb=
Σ
F
x
= 0;
P 2F− 0= P 2F= P 0.88 lb=

Problem 8-22
The uniform dresser has weight W and rests on a tile floor for which the coefficient of friction
is
μ
s
. If the man pushes on it in the direction
θ
, determine the smallest magnitude of force
F
needed to move the dresser. Also, if the man has a weight W
man,
, determine the smallest
coefficient of static friction between his shoes and the floor so that he does not slip.
Given:
W 90 lb=
μ
s
0.25=
782
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Engineering Mechanics - Statics Chapter 8
W
man
150 lb=
θ
0 deg=
Solution:
Dresser:

Guesses N
D
1lb= F 1lb=
Given
+
↑
Σ
F
y
= 0;
N
D
W− F sin
θ
()
− 0=
+
→
Σ
F
x
= 0;
F cos
θ
()
μ
s
N
D
− 0=

N
D
F
⎛
⎜
⎝
⎞
⎟
⎠
Find N
D
F,
()
= F 22.50 lb=
Man:
Guesses N
m
1lb=
μ
m
0.2=
Given
+
↑
N
m
W
man
− F sin
θ

()
+ 0=
Σ
F
y
= 0;
+
→
Σ
F
x
= 0;
F− cos
θ
()
μ
m
N
m
+ 0=
N
m
μ
m
⎛
⎜
⎝
⎞
⎟
⎠

Find N
m
μ
m
,
()
=
μ
m
0.15=
Problem 8-23
The uniform dresser has weight W and rests on a tile floor for which the coefficient of friction
is
μ
s
. If the man pushes on it in the direction
θ
, determine the smallest magnitude of force
F
needed to move the dresser. Also, if the man has a weight W
man
, determine the smallest
coefficient of static friction between his shoes and the floor so that he does not slip.
Given:
W 90 lb=
783
© 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
s
0.25=
W
man
150 lb=
θ
30 deg=
Solution:
Dresser:
Guesses N
D
1lb= F 1lb=
Given
+
↑
Σ
F
y
= 0;
N
D
W− F sin
θ
()
− 0=
+
→
Σ
F

x
= 0;
F cos
θ
()
μ
s
N
D
− 0=
N
D
F
⎛
⎜
⎝
⎞
⎟
⎠
Find N
D
F,
()
= F 30.36 lb=
Guesses N
m
1lb=
μ
m
0.2=

Man:
Given
+
↑
N
m
W
man
− F sin
θ
()
+ 0=
Σ
F
y
= 0;
+
→
Σ
F
x
= 0;
F− cos
θ
()
μ
m
N
m
+ 0=

N
m
μ
m
⎛
⎜
⎝
⎞
⎟
⎠
Find N
m
μ
m
,
()
=
μ
m
0.195=
Problem 8-24
The cam is subjected to a couple moment of
M
. Determine the minimum force
P
that should be
applied to the follower in order to hold the cam in the position shown.The coefficient of static
friction between the cam and the follower is
μ
s

. The guide at A is smooth.
784
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 8
Given:
a 10 mm=
b 60 mm=
M 5Nm⋅=
μ
s
0.4=
Solution:
Σ
M
0
= 0;
M
μ
s
N
B
b− aN
B
− 0=
N
B
M
μ

s
ba+
=
N
B
147.06 N=
Follower:
Σ
F
y
= 0;
N
B
P− 0=
PN
B
=
P 147 N=
Problem 8-25
The board can be adjusted vertically by tilting it up and sliding the smooth pin A along the vertical
guide G. When placed horizontally, the bottom C then bears along the edge of the guide, where the
coefficient of friction is
μ
s
. Determine the largest dimension d which will support any applied
force
F
without causing the board to slip downward.
Given:
μ

s
0.4=
a 0.75 in=
b 6in=
785
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 8
Solution:
+
↑
Σ
F
y
= 0;
μ
s
N
C
F− 0=
Σ
M
A
= 0;
F− bdN
C
+
μ
s

N
C
a− 0=
Solving we find
μ
s
− bd+
μ
s
a− 0= d
μ
s
ab+()= d 2.70in=
Problem 8-26
The homogeneous semicylinder has a mass m and mass center at G. Determine the largest angle
θ
of the inclined plane upon which it rests so that it does not slip down the plane. The coefficient of
static friction between the plane and the cylinder is
μ
s
. Also, what is the angle
φ
for this case?
Given:
μ
s
0.3=
Solution:
The semicylinder is a two-force member:
Since

F
μ
N=
tan
θ
()
μ
s
N
N
=
μ
S
=
θ
atan
μ
s
()
=
786
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Engineering Mechanics - Statics Chapter 8
θ
16.7 deg=
Law of sines
r
sin 180 deg

φ
−
()
4r
3π
sin
θ
()
=
φ
asin
3π
4
sin
θ
()
⎛
⎜
⎝
⎞
⎟
⎠
=
φ
42.6 deg=
Problem 8-27
A chain having a length L and weight W rests on a street for which the coefficient of static
friction is
μ
s

. If a crane is used to hoist the chain, determine the force
P
it applies to the chain
if the length of chain remaining on the ground begins to slip when the horizontal component is
P
x
. What length of chain remains on the ground?
Given:
L 20 ft=
W 8
lb
ft
=
μ
s
0.2=
P
x
10 lb=
Solution:
Σ
F
x
= 0;
P
x
−
μ
s
N

c
+ 0=
N
c
P
x
μ
s
=
N
c
50.00 lb=
Σ
F
y
= 0;
P
y
WL− N
c
+ 0=
787
© 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
P
y
WL N
c

−=
P
y
110.00 lb=
PP
x
2
P
y
2
+=
P 110lb=
The length on the ground is supported by
N
c
50.00 lb=
thus
L
N
c
W
=
L 6.25 ft=
Problem 8-28
The fork lift has a weight W
1
and center of gravity at G. If the rear wheels are powered,
whereas the front wheels are free to roll, determine the maximum number of crates, each of
weight W
2

that the fork lift can push forward. The coefficient of static friction between the
wheels and the ground is
μ
s
and between each crate and the ground is
μ
'
s
.
Given:
W
1
2400 lb=
W
2
300 lb=
μ
s
0.4=
μ
'
s
0.35=
a 2.5 ft=
b 1.25 ft=
c 3.50 ft=
Solution:
Fork lift:
Σ
M

B
= 0;
W
1
cN
A
bc+()− 0=
N
A
W
1
c
bc+
⎛
⎜
⎝
⎞
⎟
⎠
= N
A
1768.4 lb=
Σ
F
x
= 0;
μ
s
N
A

P− 0=
788
© 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
P
μ
s
N
A
= P 707.37 lb=
Crate:
N
c
W
2
− 0=
Σ
F
y
= 0;
N
c
W
2
= N
c
300.00 lb=
Σ

F
x
= 0;
P'
μ
'
s
N
c
− 0=
P'
μ
'
s
N
c
= P' 105.00 lb=
Thus
n
P
P'
= n 6.74= n floor n()= n 6.00=
Problem 8-29
The brake is to be designed to be self locking, that is, it will not rotate when no load P

is applied
to it when the disk is subjected to a clockwise couple moment
M
O
. Determine the distance d of

the lever that will allow this to happen. The coefficient of static friction at B is
μ
s
.
Given:
a 1.5 ft=
b 1ft=
μ
s
0.5=
Solution:
Σ
M
0
= 0;
M
0
μ
s
N
B
b− 0=
N
B
M
0
μ
s
b
=

Σ
M
A
= 0;
P 2 aN
B
a−
μ
s
N
B
d+ 0=
789
© 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
P 0=
d
a
μ
s
=
d 3.00ft=
Problem 8-30
The concrete pipe of weight W is being lowered from the truck bed when it is in the position
shown. If the coefficient of static friction at the points of support A and B is
μ
s
determine

where it begins to slip first: at A or B, or both at A and B.
Given:
W 800 lb= a 30 in=
μ
s
0.4= b 18 in=
θ
30 deg= c 5in=
r 15 in=
Solution:
initial guesses are
N
A
10 lb= N
B
10 lb= F
A
10 lb= F
B
10 lb=
Given
Assume slipping at A:
Σ
F
x
= 0;
N
A
F
B

+ W sin
θ
()
− 0=
Σ
F
y
= 0;
F
A
N
B
+ W cos
θ
()
− 0=
Σ
M
0
= 0;
F
B
rF
A
r− 0=
F
A
μ
s
N

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

,
()
=
N
A
N
B
F
A
F
B
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
285.71
578.53
114.29
114.29
⎛
⎜
⎜

⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
lb=
790
© 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
At B,
F
Bmax
μ
s
N
B
=
Since
F
B
114.29 lb=
<
F
Bmax

231.41 lb=
then we conclude that slipping begins at A.
Problem 8-31
A wedge of mass M is placed in the grooved slot of an inclined plane. Determine the maximum
angle
θ
for the incline without causing the wedge to slip. The coefficient of static friction
between the wedge and the surfaces of contact is
μ
s
.
Given:
M 5kg=
μ
s
0.2=
φ
60 deg=
g 9.81
m
s
2
=
Solution:
Initial guesses:
N
W
10 N=
θ
10 deg=

Given
Σ
F
x
= 0;
Mgsin
θ
()
2
μ
s
N
W
− 0=
Σ
F
z
= 0;
2 N
W
sin
φ
2
⎛
⎜
⎝
⎞
⎟
⎠
Mgcos

θ
()
− 0=
Solving,
N
W
θ
⎛
⎜
⎝
⎞
⎟
⎠
Find N
W
θ
,
()
=
N
W
45.5 N=
θ
21.8 deg=
791
© 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 8-32

A roll of paper has a uniform weight W and is suspended from the wire hanger so that it rests
against the wall. If the hanger has a negligible weight and the bearing at O can be considered
frictionless, determine the force
P
needed to start turning the roll. The coefficient of static
friction between the wall and the paper is
μ
s
.
Given:
W 0.75 lb=
θ
30 deg=
φ
30 deg=
μ
s
0.25=
a 3in=
Solution:
Initial guesses:
R 100 lb= N
A
100 lb= P 100 lb=
Given
Σ
F
x
= 0;
N

A
R sin
φ
()
− P sin
θ
()
+ 0=
Σ
F
y
= 0;
R cos
φ
()
W− P cos
θ
()
−
μ
s
N
A
− 0=
Σ
M
0
= 0;
μ
s

N
A
aPa− 0=
Solving for P,
R
N
A
P
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
Find RN
A
, P,
()
=
R 1.14 lb= N
A
0.51 lb= P 0.13 lb=
Problem 8-33
A roll of paper has a uniform weight W and is suspended from the wire hanger so that it rests
against the wall. If the hanger has a negligible weight and the bearing at O can be considered
frictionless, determine the minimum force

P
and the associated angle
θ
needed to start turning
the roll. The coefficient of static friction between the wall and the paper is
μ
s
.
792
© 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 0.75 lb=
φ
30 deg=
μ
s
0.25=
r 3in=
Solution:
Σ
F
x
= 0;
N
A
R sin
φ

()
− P sin
θ
()
+ 0=
Σ
F
y
= 0;
R cos
φ
()
W− P cos
θ
()
−
μ
s
N
A
− 0=
Σ
M
0
= 0;
μ
s
N
A
rPr− 0=

Solving for P,
P
μ
s
W sin
φ
()
cos
φ
()
μ
sin
θφ
−
()
+
μ
sin
φ
()
−
=
For minimum P we must have
dP
d
θ
μ
s
2
− W sin

φ
()
cos
θφ
−
()
cos
φ
()
μ
s
sin
θφ
−
()
+
μ
s
sin
φ
()
−
()
2
= 0=
Implies
cos
θφ
−
()

0=
One answer is
θφ
90 deg+=
θ
120.00 deg=
P
μ
s
W sin
φ
()
cos
φ
()
μ
s
sin
θφ
−
()
+
μ
s
sin
φ
()
−
= P 0.0946 lb=
Problem 8-34

The door brace AB is to be designed to prevent opening the door. If the brace forms a pin
connection under the doorknob and the coefficient of static friction with the floor is
μ
s
determine
the largest length L the brace can have to prevent the door from being opened. Neglect the weight
of the brace.
Given:
μ
s
0.5=
793
© 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
a 3ft=
Solution:
The brace is a two-force member.
μ
s
N
N
L
2
a
2
−
a
=

μ
s
aL
2
a
2
−=
La1
μ
s
2
+=
L 3.35 ft=
Problem 8-35
The man has a weight W, and the coefficient of static friction between his shoes and the floor is
μ
s
.

Determine where he should position his center of gravity G at d in order to exert the maximum
horizontal force on the door. What is this force?
Given:
W 200 lb=
μ
s
0.5=
h 3ft=
Solution:
NW− 0= NW= N 200.00lb=
F

max
μ
s
N= F
max
100lb=
+
→
Σ
F
x
= 0;
PF
max
− 0=
PF
max
= P 100lb=
Σ
M
O
= 0;
Wd Ph− 0= dP
h
W
= d 1.50ft=
794
© 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 8-36
In an effort to move the two crates, each of weight W, which are stacked on top of one
another, the man pushes horizontally on them at the bottom of crate A as shown. Determine
the smallest force
P
that must be applied in order to cause impending motion. Explain what
happens. The coefficient of static friction between the crates is
μ
s
and between the bottom
crate and the floor is
μ
s
'.
Given:
W 100 lb=
μ
s
0.8=
μ
'
s
0.3=
a 2ft=
b 3ft=
Solution:
Assume crate A slips:
Σ
F

y
= 0;
N
A
W− 0= N
A
W= N
A
100.00 lb=
Σ
F
x
= 0;
P
μ
s
N
A
− 0= P
1
μ
s
N
A
= P
1
80.00 lb=
Assume crate B slips:
Σ
F

y
= 0;
N
B
2 W− 0= N
B
2 W= N
B
200.00 lb=
Σ
F
x
= 0;
P
μ
'
s
N
B
− 0= P
2
μ
'
s
N
B
= P
2
60.00 lb=
Assume both crates A and B tip:

Σ
M = 0;
2 W
a
2
⎛
⎜
⎝
⎞
⎟
⎠
Pb− 0= P
3
W
a
b
⎛
⎜
⎝
⎞
⎟
⎠
= P
3
66.7 lb=
P min P
1
P
2
, P

3
,
()
= P 60.00 lb=
Problem 8-37
The man having a weight of W
1
pushes horizontally on the bottom of crate A, which is stacked
on top of crate B. Each crate has a weight W
2
. If the coefficient of static friction between each
crate is
μ
s
and between the bottom crate, his shoes, and the floor is
μ
'
s
, determine if he can
cause impending motion.
795
© 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.