Engineering Mechanics - Statics Chapter 7
V
2
x() B
y
−
1
2
w
ab+ x−
b
⎛
⎜
⎝
⎞
⎟
⎠
ab+ x−()+
⎡
⎢
⎣
⎤
⎥
⎦
1
lb
=
M
2
x() B
y

1
kip ft⋅
=
x
3
ab+ 1.01 ab+(), ab+ c+ =
V
3
x() 0= M
3
x() M
0
−
1
kip ft⋅
=
0 2 4 6 8 10121416
1500
1000
500
0
Distance (ft)
Force (lb)
V
1
x
1
()
V
2

M
2
x
2
()
M
3
x
3
()
x
1
ft
x
2
ft
,
x
3
ft
,
721
© 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 7
Draw the shear and moment diagrams for the beam.
Solution:
722
Problem 7-85

ax−
a
⎛
⎜
⎝
⎞
⎟
⎠
ax−()+
⎡
⎢
⎣
⎤
⎥
⎦
1
kN
=
M
1
x()
1−
2
wb
2b
3
a+ x−
⎛
⎜
⎝

1
kN m⋅
=
x
2
a 1.01a, ab+ =
V
2
x()
1
2
wb
1
2
w
xa−
a
⎛
⎜
⎝
⎞
⎟
⎠
xa−()−
⎡
⎢
⎣
⎤
⎥
⎦

xa−
3
⎛
⎜
⎝
⎞
⎟
⎠
−
⎡
⎢
⎣
⎤
⎥
⎦
1
kN m⋅
=
0123456
5
0
5
10
Distance (m)
Force (kN)
V
1
x
1
()

()
x
1
x
2
,
Problem 7-87
Draw the shear and moment diagrams for the beam.
Units Used:
kip 10
3
lb=
Given:
w 5
kip
ft
= M
1
15
kip ft⋅= M
2
15
kip ft⋅= a 6 ft= b 10 ft= c 6 ft=
Solution:
M
1
Ab− M
2
− w
a

⎜
⎝
⎞
⎟
⎠
c
3
⎛
⎜
⎝
⎞
⎟
⎠
− 0= AB+ wb− w
ac+
2
⎛
⎜
⎝
⎞
⎟
⎠
− 0=
A
M
1
M
2
− w
a

6
⎛
⎜
⎝
⎞
⎟
⎠
−
b
=
Bwb
ac+
2
+
⎛
⎜
⎝
⎞
⎟
⎠
A−=
A
B
⎛
⎜
⎝
⎞
⎟
⎠
40.00

x() w−
x
a
⎛
⎜
⎝
⎞
⎟
⎠
x
2
x
3
M
1
−
⎡
⎢
⎣
⎤
⎥
⎦
1
kip ft⋅
=
724
© 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 7

⎜
⎝
⎞
⎟
⎠
− Ax a−()+ wx a−()
xa−
2
⎛
⎜
⎝
⎞
⎟
⎠
−
⎡
⎢
⎣
⎤
⎥
⎦
1
kip ft⋅
=
x
3
ab+ 1.01 ab+(), ab+ c+ =
V
3
x() w

ab+ c+ x−
2
⎛
⎜
⎝
⎞
⎟
⎠
ab+ c+ x−
3
⎛
⎜
⎝
⎞
⎟
⎠
M
2
−
⎡
⎢
⎣
⎤
⎥
⎦
1
kip ft⋅
=
0 5 10 15 20
40

ft
,
725
© 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 7
0 5 10 15 20
60
40
20
0
20
Distance (ft)
Moment (kip-ft)
M
1p
x
1
()
M
2p
x
2
()
M
3p
x
3
()

Solution:
x 0 0.01a, a =
Vx() w
2
xw
1
x
a
⎛
⎜
⎝
⎞
⎟
⎠
x
2
−
⎡
⎢
⎣
⎤
⎥
⎦
1
kip
= Mx() w
2
x
x
2

0
2
4
Distance (ft)
Force (kip)
Vx()
x
ft
0 2 4 6 8 101214
0
20
40
Distance (ft)
Moment (kip-ft)
Mx()
x
ft
Problem 7-89
Determine the force P needed to hold the cable in the position shown, i.e., so segment BC
remains horizontal. Also, compute the sag y
B
and the maximum tension in the cable.
Units Used:
kN 10
3
N=
727
© 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.

kN=
Given
a−
a
2
y
B
2
+
⎛
⎜
⎝
⎞
⎟
⎠
T
AB
T
BC
+ 0=
y
B
a
2
y
B
2
+
⎛
⎜

CD
+ 0=
y
B
e−
c
2
y
B
e−
()
2
+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
T
CD
P− 0=
c−
c
2
y
B
e−

−
c
2
y
B
e−
()
2
+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
T
CD
e
e
2
d
2
+
⎛
⎜
⎝
⎞
⎟

⎟
⎟
⎟
⎟
⎠
Find y
B
P, T
AB
, T
BC
, T
CD
, T
DE
,
()
=
728
© 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 7
T
max
max T
AB
T
BC
, T

kg=
a 1 m=
b 3 m=
c 0.5 m=
d 2 m=
Solution:
Guesses y
B
1
m= T
AB
1
N= T
BC
1
N= T
CD
1
N=
Given
a−
a
2
y
B
2
+
⎛
⎜
⎝

+
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
T
AB
y
B
d−
b
2
y
B
d−
()
2
+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦

⎝
⎞
⎟
⎠
T
CD
+ 0=
y
B
d−
()
−
b
2
y
B
d−
()
2
+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
T
BC

⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
Find y
B
T
AB
, T
BC
, T
CD
,
()
=
T
AB
T
BC
T
CD
⎛

()
= T
max
157.2
N= y
B
2.43
m=
Problem 7-91
The cable supports the three loads shown. Determine the sags y
B
and y
D
of points B and D.
Given:
a 4 ft= e 12 ft=
b 12 ft= f 14 ft=
c 20 ft= P
1
400
lb=
d 15 ft= P
2
250
lb=
Solution:
Guesses y
B
1
ft= y

⎠
T
AB
c
c
2
fy
B
−
()
2
+
⎡
⎢
⎣
⎤
⎥
⎦
T
BC
+ 0=
y
B
b
2
y
B
2
+
⎛

− P
2
− 0=
c−
c
2
fy
B
−
()
2
+
⎡
⎢
⎣
⎤
⎥
⎦
T
BC
d
d
2
fy
D
−
()
2
+
⎡

⎥
⎦
T
BC
fy
D
−
d
2
fy
D
−
()
2
+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
T
CD
+ P
1
− 0=
d−
d

T
DE
+ 0=
fy
D
−
()
−
d
2
fy
D
−
()
2
+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
T
CD
ay
D
+
e

y
D
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find T
AB
T
BC
, T
CD
, T
DE

744.44
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
lb=
y
B
y
D
⎛
⎜
⎝
⎞
⎟
⎠
8.67
7.04
⎛
⎜
⎝
⎞

T
BC
1
lb= T
CD
1
lb=
731
© 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 7
T
DE
1
lb= y
D
1
ft=
Given
b−
b
2
y
B
2
+
⎛
⎜
⎝

+
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
T
AB
fy
B
−
c
2
fy
B
−
()
2
+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦

+
⎡
⎢
⎣
⎤
⎥
⎦
T
CD
+ 0=
fy
B
−
c
2
fy
B
−
()
2
+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
T

−
()
2
+
⎡
⎢
⎣
⎤
⎥
⎦
T
CD
e
e
2
ay
D
+
()
2
+
⎡
⎢
⎣
⎤
⎥
⎦
T
DE
+ 0=

+
()
2
+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
T
DE
+ P
2
− 0=
T
AB
T
BC
T
CD
T
DE
P
1
y
D
⎛

D
,
()
=
T
AB
T
BC
T
CD
T
DE
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
983.33
854.21
916.11
1084.68
⎛
⎜

a 5 ft= e 3=
b 8 ft= f 4=
Solution:
The initial guesses:
T
AB
10
lb= T
CD
30
lb=
T
BC
20
lb= x
B
5
ft=
Given
x
B
−
x
B
2
a
2
+
⎛
⎜

a
x
B
2
a
2
+
⎛
⎜
⎝
⎞
⎟
⎠
T
AB
b
x
B
d−
()
2
b
2
+
⎡
⎢
⎣
⎤
⎥
⎦

⎛
⎜
⎝
⎞
⎟
⎠
T
CD
−
f
e
2
f
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F+ 0=
b
x
B
d−
()
2
b
2

⎛
⎜
⎝
⎞
⎟
⎠
F− 0=
T
AB
T
CD
T
BC
x
B
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
Find T
AB
T

⎝
⎞
⎟
⎟
⎠
lb= x
B
4.36
ft=
Problem 7-94
The cable supports the loading shown. Determine the magnitude of the horizontal force
P
.
733
Given:
© 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 7
F 30 lb= c 2 ft=
x
B
6
ft= d 3 ft=
a 5 ft= e 3=
b 8 ft= f 4=
Solution:
The initial guesses:
T
AB

B
d−
x
B
d−
()
2
b
2
+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
T
BC
− P+ 0=
a
a
2
x
B
2
+
⎛
⎜

⎛
⎜
⎝
⎞
⎟
⎠
T
CD
x
B
d−
b
2
x
B
d−
()
2
+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
T
BC
+

x
B
d−
()
2
+
⎡
⎢
⎣
⎤
⎥
⎦
T
BC
+
e
e
2
f
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F− 0=
T
AB

T
CD
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
70.81
48.42
49.28
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb= P 71.40 lb=
734
Given:
© 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 7

DE
1
kN=
P
1
1
kN= P
2
1
kN=
Given
c−
a
2
c
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
AB
d
b
2
d
2

b
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
BC
− F− 0=
d−
b
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
BC
F
CD

⎤
⎥
⎦
F
DE
+ 0=
ab+
f
2
ab+()
2
+
⎡
⎢
⎣
⎤
⎥
⎦
F
DE
P
2
− 0=
F
AB
F
BC
F
CD
F

CD
, F
DE
, P
1
, P
2
,
()
=
F
AB
F
BC
F
CD
F
DE
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠

⎜
⎝
⎞
⎟
⎠
kN=
T
max
max F
AB
F
BC
, F
CD
, F
DE
,
()
= T
max
12.50
kN=
F
max
max F
AB
F
BC
, F
CD

Guesses
T
AB
1
lb= T
BC
1
lb=
T
CD
1
lb= x
B
1
ft=
Given
x
B
−
a
2
x
B
2
+
⎛
⎜
⎜
⎝
⎞

a
2
x
B
2
+
⎛
⎜
⎝
⎞
⎟
⎠
T
AB
b
b
2
x
B
d−
()
2
+
⎡
⎢
⎣
⎤
⎥
⎦
T

⎜
⎝
⎞
⎟
⎠
T
CD
− 0=
b
b
2
x
B
d−
()
2
+
⎡
⎢
⎣
⎤
⎥
⎦
T
BC
c
c
2
d
2

⎟
⎠
Find T
AB
T
BC
, T
CD
, x
B
,
()
=
T
AB
T
BC
T
CD
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
15.49

before it will break.
Given:
T
max
3000
lb=
a 50 ft=
b 6 ft=
Solution:
y
1
F
H
xxw
⌠
⎮
⌡
d
⌠
⎮
⌡
d=
wx
2
2F
H
=
y
w
2F

()
=
w
F
H
a
2
⎛
⎜
⎝
⎞
⎟
⎠
=
4b
a
=
θ
max
atan
4b
a
⎛
⎜
⎝
⎞
⎟
⎠
=
θ

ft
=
Problem 7-98
The cable is subjected to a uniform loading w. Determine the maximum and minimum tension in
the cable.
Units Used:
kip 10
3
lb=
Given:
w 250
lb
ft
=
a 50 ft=
b 6 ft=
738
© 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 7
Solution:
y
wx
2
2F
H
= b
w
2F

=
⎛
⎜
⎝
⎞
⎟
⎠
=
w
F
H
a
2
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
max
atan
wa
2F
H
⎛
⎜
⎝
⎞

Problem 7-99
The cable is subjected to the triangular loading. If the slope of the cable at A is zero, determine
the equation of the curve y = f(x) which defines the cable shape AB, and the maximum
tension developed in the cable.
Units Used:
kip 10
3
lb=
Given:
w 250
lb
ft
=
a 20 ft=
b 30 ft=
Solution:
y
1
F
H
xx
wx
b
⌠
⎮
⎮
⌡
d
⌠
⎮

x
y
d
d
0= x 0=,
Thus
C
1
C
2
= 0=
y
wx
3
6F
H
b
=
set
ya= xb= a
wb
3
6F
H
b
=
F
H
wb
2

max
()
= T
max
4.19
kip=
Problem 7-100
The cable supports a girder which has weight density
γ
. Determine the tension in the cable at points
A, B, and C.
Units used:
kip 10
3
lb=
Given:
γ
850
lb
ft
=
a 40 ft=
b 100 ft=
c 20 ft=
Solution:
y
1
F
H
xx

1
lb=
740
© 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 7
Given c
γ
x
1
2
2F
H
= a
γ
2F
H
bx
1
−
()
2
=
x
1
F
H
⎛
⎜

F
H
x
1
b−
()
⎡
⎢
⎣
⎤
⎥
⎦
=
θ
A
53.79
− deg=
tan
θ
C
()
γ
F
H
x
1
=
θ
C
atan

F
H
cos
θ
C
()
=
T
A
T
B
T
C
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
61.71
36.46
50.68
⎛
⎜
⎜
⎝

H
xx
wx
a
⌠
⎮
⎮
⌡
d
⌠
⎮
⎮
⌡
d
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
y
1
F
H
w
a
x

⎛
⎜
⎝
⎞
⎟
⎠
C
1
F
H
⎛
⎜
⎝
⎞
⎟
⎠
+=
At x = 0,
x
y
d
d
0=
,
C
1
0
=

At x = 0, y = 0,

a
2
b
⎛
⎜
⎝
⎞
⎟
⎠
= F
H
2343.75
lb=
742
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 7
x
y
d
d
tan
θ
max
()
=
wa
2

= T
max
4.42
kip=
Problem 7-102
The cable is subjected to the parabolic loading w = w
0
(1
−
(2x/a)
2
). Determine the equation y = f(x)
which defines the cable shape AB and the maximum tension in the cable.
Units Used:
kip 10
3
lb=
Given:
ww
0
1
2x
a
⎛
⎜
⎝
⎞
⎟
⎠
2

1
F
H
xw
0
x
4x
3
3a
2
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
C
1
+
⌠
⎮
⎮
⎮
⌡
d=
y
1

1
F
H
w
0
x
4w
0
x
3
3a
2
− C
1
+
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
743
© 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 7
At

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
dy
dx
1
F
H
w
0
x
4w
0
x
3
3a
2
−
⎛
⎜
⎜
⎝
⎞
⎟

a
2
⎛
⎜
⎝
⎞
⎟
⎠
4
−
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
=
5
48
w
0
a
2
F
H
⎛
⎜
⎜

⎞
⎟
⎠
=
1
F
H
w
0
a
2
⎛
⎜
⎝
⎞
⎟
⎠
4w
0
3
a
2
a
2
⎛
⎜
⎝
⎞
⎟
⎠

⎠
=
θ
max
32.62
deg=
T
max
F
H
cos
θ
max
()
= T
max
9.28
kip=
Problem 7-103
The cable will break when the maximum tension reaches T
max
.

Determine the minimum sag h if
it supports the uniform distributed load w.
Given:
kN 10
3
N=
T

1
x+ C
2
+
⎛
⎜
⎝
⎞
⎟
⎠
=
dy
dx
1
F
H
wx C
1
+
()
=
744
© 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 7
Boundary Conditions:
y = 0 at x = 0, then from Eq.[1]
0
1

⎛
⎜
⎝
⎞
⎟
⎠
x
2
=
dy
dx
w
F
H
x= h
w
2F
H
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
= F
H
w

+
=
T
max
F
H
cos
θ
max
()
= F
H
2
wa()
2
4
+=
wa
2
a
2
16h
2
1+=
Guess h 1 m=
Given T
max
wa
2
a

=
745
© 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.