Engineering Mechanics - Statics Episode 2 Part 9 ppsx

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

ab+ x−()M
0
−
1
2
w
ab+ x−
b
⎛
⎜
⎝
⎞
⎟
⎠
ab+ x−()
ab+ x−
3
⎛
⎜
⎝
⎞
⎟
⎠
−
⎡
⎢
⎣
⎤
⎥
⎦

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

x
2
()
V
3
x
3
()
x
1
ft
x
2
ft
,
x
3
ft
,
0 5 10 15
10
5
0
Distance (ft)
Moment (kip-ft)
M
1
x
1
()

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

© 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
Problem 7-86
Draw the shear and moment diagrams
for the beam.
Units Used:
kN 10
3
N=
Given:
w 2
kN
m
= a 3 m= b 3 m=
Solution:
x
1
0 0.01
a, a =
V
1
x()
1
2
wb
1
2
w

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

⎞
⎟
⎠
1
2
w
ax−
a
⎛
⎜
⎝
⎞
⎟
⎠
ax−()
ax−
3
⎛
⎜
⎝
⎞
⎟
⎠
−
⎡
⎢
⎣
⎤
⎥
⎦

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

1
kN
=
M
2
x()
1−
2
wb a
2b
3
+ x−
⎛
⎜
⎝
⎞
⎟
⎠
1
2
w
xa−
b
⎛
⎜
⎝
⎞
⎟
⎠
xa−()

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

V
2
x
2
()
x
1
x
2
,
723
© 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
0123456
20
10
0
Distance (m)
Moment (kN-m)
M
1
x
1
()
M
2
x
2

()
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

2
⎛
⎜
⎝
⎞
⎟
⎠
b
a
3
+
⎛
⎜
⎝
⎞
⎟
⎠
+ wb
b
2
⎛
⎜
⎝
⎞
⎟
⎠
+ w
c
2
⎛

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

2
⎛
⎜
⎝
⎞
⎟
⎠
b
a
3
+
⎛
⎜
⎝
⎞
⎟
⎠
+ w
b
2
2
⎛
⎜
⎝
⎞
⎟
⎠
+ w
c
2

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

40.00
⎛
⎜
⎝
⎞
⎟
⎠
kip=
x
1
0 0.01
a, a =
V
1
x() w−
x
a
⎛
⎜
⎝
⎞
⎟
⎠
x
2
1
kip
=
M
1p

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

x
2
a 1.01a, ab+ =
V
2
x() Aw
a
2
− wx a−()−
⎡
⎢
⎣
⎤
⎥
⎦
1
kip
=
M
2p
x() M
1
− w
a
2
x
2a
3
−
⎛

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

ab+ c+ x−
c
⎛
⎜
⎝
⎞
⎟
⎠
ab+ c+ x−
2
⎛
⎜
⎝
⎞
⎟
⎠
1
kip
=
M
3p
x() w−
ab+ c+ x−
c
⎛
⎜
⎝
⎞
⎟
⎠

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

20
0
20
40
Distance (ft)
Force (lb)
V
1
x
1
()
V
2
x
2
()
V
3
x
3
()
x
1
ft
x
2
ft
,
x
3

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
()

x
1
ft
x
2
ft
,
x
3
ft
,
Problem 7-88
Draw the shear and moment diagrams for
the beam.
Units Used:
kip 10
3
lb=
Given:
w
1
2
kip
ft
= w
2
1
kip
ft
= a 15 ft=

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

w
1
x
a
⎛
⎜
⎝
⎞
⎟
⎠
x
2
x
3
−
⎡
⎢
⎣
⎤
⎥
⎦
1
kip ft⋅
=
726
© 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 2 4 6 8 101214

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.

Engineering Mechanics - Statics Chapter 7
Given:
a 4 m= F
1
4
kN=
b 6 m= F
2
6
kN=
c 3 m=
d 2 m=
e 3 m=
Solution:
Initial guesses:
y
B
1
m= P 1 kN= T
AB
1
kN= T
BC
1
kN= T
CD
1
kN= T
DE
1

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

⎜
⎝
⎞
⎟
⎟
⎠
T
AB
F
1
− 0=
T
BC
−
c
c
2
y
B
e−
()
2
+
⎡
⎢
⎣
⎤
⎥
⎦
T

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

()
2
+
⎡
⎢
⎣
⎤
⎥
⎦
T
CD
d
d
2
e
2
+
⎛
⎜
⎝
⎞
⎟
⎠
T
DE
+ 0=
y
B
e−
()

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

⎠
T
DE
+ F
2
− 0=
y
B
P
T
AB
T
BC
T
CD
T
DE
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟

⎟
⎟
⎟
⎟
⎠
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

CD
, T
DE
,
()
= y
B
3.53
m
=
P 800.00 N=
T
max
8.17
kN=
Problem 7-90
Cable ABCD supports the lamp of mass M
1
and the lamp of mass M
2
. Determine the maximum
tension in the cable and the sag of point B.
Given:
M
1
10
kg=
M
2
15

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
b
b
2
y
B
d−
()
2
+
⎡
⎢
⎣
⎤
⎥
⎦
T
BC
+ 0=
y
B
a
2
y
B
2

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

T
BC
+ M
1
g− 0=
b−
b
2
y
B
d−
()
2
+
⎡
⎢
⎣
⎤
⎥
⎦
T
BC
c
c
2
d
2
+
⎛
⎜

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

d
c
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
T
CD
+ M
2
g− 0=
729
© 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
y
B
T
AB
T
BC
T
CD

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

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
100.163
38.524
157.243
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
T
max
max 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

D
1
ft=
T
AB
1
lb= T
BC
1
lb=
T
CD
1
lb= T
DE
1
lb=
Given
b−
b
2
y
B
2
+
⎛
⎜
⎝
⎞
⎟

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

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

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

⎢
⎣
⎤
⎥
⎦
T
CD
+ 0=
730
© 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
fy
B
−
c
2
fy
B
−
()
2
+
⎡
⎢
⎢
⎣
⎤
⎥

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

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

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

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

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

, y
B
, y
D
,
()
=
T
AB
T
BC
T
CD
T
DE
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
675.89
566.90
603.86

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

⎟
⎠
ft=
Problem 7-92
The cable supports the three loads shown. Determine the magnitude of
P
1
and find the sag y
D
for
the given data.
Given:
P
2
300
lb= c 20 ft=
y
B
8
ft= d 15 ft=
a 4 ft= e 12 ft=
b 12 ft= f 14 ft=
Solution:
Guesses P
1
1
lb= T
AB
1
lb=

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

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

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

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

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

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

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

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

⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find T
AB
T
BC
, T
CD
, T
DE
, P
1
, y

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

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
lb=
P
1
658
lb=
y
D
6.44
ft=
Problem 7-93
The cable supports the loading shown. Determine the distance x
B
the force at point B acts from A.
732
© 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
P 40 lb= c 2 ft=
F 30 lb= d 3 ft=

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
+
⎛
⎜

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

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

T
BC
− 0=
x
B
d−
x
B
d−
()
2
b
2
+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
T
BC
d
c
2
d
2
+

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

+
⎡
⎢
⎣
⎤
⎥
⎦
T
BC
c
c
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
T
CD
−
e
e
2
f
2
+

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

CD
, T
BC
, x
B
,
()
=
T
AB
T
CD
T
BC
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
50.90
36.70
38.91
⎛
⎜
⎜

⎝
⎞
⎟
⎟
⎠
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

10
lb= T
CD
30
lb=
T
BC
20
lb= P 10 lb=
Given
x
B
−
x
B
2
a
2
+
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
T
AB
x

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

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

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

f
e
2
f
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F+ 0=
c−
c
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
T
CD
b
b
2

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

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

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

Problem 7-95
Determine the forces P
1
and P
2
needed to hold the cable in the
position shown, i.e., so segment CD
remains horizontal. Also, compute the
maximum tension in the cable.
Given:
kN 10
3
N=
F 5 kN= d 4 m=
a 1.5 m= e 5 m=
b 1 m= f 4 m=
c 2 m=
Solution:
Guesses
F
AB
1
kN= F
BC
1
kN=
F
CD
1
kN= F

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

+
⎛
⎜
⎝
⎞
⎟
⎠
F
BC
+ 0=
735
© 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
a
a
2
c
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
AB
b

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

+ 0=
b
b
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
BC
P
1
− 0=
F
CD
−
f
f
2
ab+()
2
+
⎡
⎢
⎣

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

DE
P
1
P
2
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find F
AB
F
BC
, F

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

12.50
10.31
10.00
11.79
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
kN=
P
1
P
2
⎛
⎜
⎝
⎞
⎟
⎠
2.50
6.25
⎛

⎜
⎝
⎞
⎟
⎠
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

, F
DE
,
()
= F
max
12.50
kN=
736
© 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
The cable supports the loading shown. Determine the distance x
B
from the wall to point B.
Given:
W
1
8
lb=
W
2
15
lb=
a 5 ft=
b 8 ft=
c 2 ft=
d 3 ft=
Solution:

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
+
⎛
⎜
⎜
⎝
⎞

⎟
⎟
⎠
T
AB
x
B
d−
b
2
x
B
d−
()
2
+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
T
BC
− W
2
+ 0=
a

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

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

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

+
⎛
⎜
⎝
⎞
⎟
⎠
T
CD
− W
1
− 0=
T
AB
T
BC
T
CD
x
B
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟

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

10.82
4.09
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb= x
B
5.65
ft=
Problem 7-97
Determine the maximum uniform loadin
g
w, measured in lb/ft, that the cable can su
pp
ort if it is
737
Problem 7-96
© 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
g
pp
capable of sustaining a maximum tension T
max

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

H
⎛
⎜
⎝
⎞
⎟
⎠
x
2
=
x
a
2
= yb= F
H
wa
2
8b
=
dy
dx
⎛
⎜
⎝
⎞
⎟
⎠
tan
θ
max

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

max
25.64
deg=
T
max
F
H
cos
θ
max
()
=
wa
2
8bcos
θ
max
()
= w
T
max
8
bcos
θ
max
()
a
2
= w 51.93
lb

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

H
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
= F
H
wa
2
8b
= F
H
13021
lb=
tan
θ
max
()
x
y
d
d
x
a
2

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

⎟
⎠
=
θ
max
25.64
deg=
T
max
F
H
cos
θ
max
()
= T
max
14.44
kip=
The minimum tension occurs at
θ
0 deg=
T
min
F
H
= T
min
13.0
kip=

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
⌠
⎮

⎮
⌡
d=
y
1
F
H
wx
3
6b
c
1
x+ c
2
+
⎛
⎜
⎝
⎞
⎟
⎠
=
739
© 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
Apply boundary conditions
yx= 0=
and

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

6a
= F
H
1.875
kip=
θ
max
atan
wb
2
2F
H
b
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
max
63.43
deg=
T
max
F
H
cos
θ

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

γ
⌠
⎮
⌡
d
⌠
⎮
⌡
d=
y
γ
x
2
2F
H
=
x
y
d
d
γ
x
F
H
=
Guesses x
1
1
ft= F
H

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
⎛
⎜

⎝
⎞
⎟
⎠
Find x
1
F
H
,
()
= F
H
36.46
kip=
tan
θ
A
()
γ
F
H
x
1
b−
()
=
θ
A
atan
γ

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

γ
F
H
x
1
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
C
44.00
deg=
T
A
F
H
cos
θ
A
()
= T
B
F
H
= T
C

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

⎞
⎟
⎟
⎠
kip=
Problem 7-101
The cable is subjected to the triangular
loading. If the slope of the cable at
point O is zero, determine the equation
of the curve y = f(x) which defines the
cable shape OB, and the maximum
tension developed in the cable.
Units used:
kip 10
3
lb=
Given:
w 500
lb
ft
= b 8 ft=
a 15 ft=
741
© 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
y
1
F

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

3
6
⎛
⎜
⎝
⎞
⎟
⎠
C
1
x+ C
2
+
⎡
⎢
⎣
⎤
⎥
⎦
=
x
y
d
d
1
F
H
wx
2
2a

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

At x = 0, y = 0,

C
2
0
=
y
wx
3
6aF
H
=
x
y
d
d
wx
2
2aF
H
=
At x = a, y = b
b
wa
3
6aF
H
= F
H
1
6
w

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

2aF
H
=
θ
max
()
atan
wa
2
2aF
H
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
max
()
57.99 deg=
T
max
F
H
cos
θ
max
()

= 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

−
⎡
⎢
⎣
⎤
⎥
⎦
=
a 100 ft= w
0
150
lb
ft
=
b 20 ft=
Solution:
y
1
F
H
xxwx()
⌠
⎮
⌡
d
⌠
⎮
⌡
d=
y

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

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

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

x 0=
dy
dx
0= C
1
0
=
At
x 0= y 0= C
2
0
=
Thus
y
1
F
H
w
0
x
2
2
x
4
w
0
3
a
2
−

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

⎟
⎠
=
At
x
a
2
=
we have
b
1
F
H
w
0
2
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
w
0
3
a
2

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

⎝
⎞
⎟
⎟
⎠
= F
H
5
w
0
a
2
48b
= F
H
7812.50
lb=
tan
θ
max
()
x
y
d
d
a
2
⎛
⎜
⎝

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

3
−
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
=
w
0
a
3F
H
=
θ
max
atan
w
0
a
3F
H
⎛
⎜
⎝
⎞
⎟

⎠
=
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

max
10
kN=
w 600
N
m
=
a 25 m=
Solution:
The equation of the cable:
y
1
F
H
xxw
⌠
⎮
⌡
d
⌠
⎮
⌡
d= y
1
F
H
wx
2
2
C

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

F
H
C
2
()
= C
2
0
=
x
y
d
d
= 0 at x = 0, then from Eq.[2]
0
1
F
H
C
1
()
= C
1
0
=
Thus,
y
w
2F
H

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

2h
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
=
tan
θ
max
()
wa
2F
H
= cos
θ
max
()
2F
H
4
F
H
2
wa()
2

+
=
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

2
16h
2
1+= h Find h()= h 7.09 m=
Problem 7-104
A fiber optic cable is suspended over the poles so that the angle at the supports is
θ
.
Determine the minimum tension in the cable and the sag. The cable has a mass density
ρ
and
the supports are at the same elevation.
Given:
θ
22 deg=
ρ
0.9
kg
m
=
a 30 m=
g 9.81
m
s
2
=
Solution:
θ
max
θ

=
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.