Engineering Mechanics - Statics Episode 2 Part 4 pdf

Engineering Mechanics - Statics Chapter 6
Positive (T)
Negative (C)
F
AB
F
AC
F
AD
F
AE
F
BC
F
BE
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟

⎟
⎟
⎟
⎠
300−
583.095
333.333
666.667−
0
500
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
lb=
F

BF
F
CD
F
CF
F
DE
F
DF
F
EF
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟

⎠
0
300−
300−
0
424.264
300−
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
lb=
Problem 6-59
The space truss is supported by
a ball-and-socket joint at D and
short links at C and E. Determine

the force in each member and
state if the members are in
tension or compression.
Given:
F
1
200
300
500−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
F
2
0
400
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟

⎠
lb=
a 4ft=
b 3ft=
c 3ft=
Solution:
Find the external reactions
Guesses
E
y
1lb= C
y
1lb= C
z
1lb=
D
x
1lb= D
y
1lb= D
z
1lb=
521
© 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 6
Given
D
x

D
y
D
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
0
E
y
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+
0

C
y
C
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+ F
1
+ F
2
+ 0=
0
a
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠

F
1
×
b−
a
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2
×+
0
0
c
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
D
x

D
y
D
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
×+
b−
0
c
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
C
y
C

z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
×+ 0=
E
y
C
y
C
z
D
x
D
y
D
z
⎛
⎜
⎜
⎜
⎜
⎜

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find E
y
C
y
, C
z
, D
x
, D
y
, D
z
,
()
=
E

y
C
y
C
z
D
x
D
y
D
z
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟

⎠
33.333−
666.667−
200
200−
1.253− 10
13−
×
300
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
lb=
Now find the force in each member.
AB

b−
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= AC
b−
a−
c
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= AD
0
a−
c
⎛
⎜
⎜

⎝
⎞
⎟
⎟
⎠
= AE
0
a−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
BC
0
a−
c
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠

= BE
b
a−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= BF
0
a−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= CD
b
0
0
⎛
⎜

⎜
⎝
⎞
⎟
⎟
⎠
=
CF
0
0
c−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= DE
0
0
c−
⎛
⎜
⎜
⎝
⎞
⎟
⎟

⎠
= DF
b−
0
c−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= EF
b−
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
Guesses
F
AB
1lb= F

AC
1lb= F
AD
1lb= F
AE
1lb=
F
BC
1lb= F
BE
1lb= F
BF
1lb= F
CD
1lb=
F
CF
1lb= F
DE
1lb= F
DF
1lb= F
EF
1lb=
Given
522
© 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 6

F
1
F
AB
AB
AB
+ F
AC
AC
AC
+ F
AD
AD
AD
+ F
AE
AE
AE
+ 0=
F
2
F
BC
BC
BC
+ F
BF
BF
BF
+ F

BE
BE
BE
+ F
AB
AB−
AB
+ 0=
0
E
y
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
F
AE
AE−
AE
+ F
BE
BE−
BE

+ F
EF
EF
EF
+ F
DE
DE−
DE
+ 0=
F
BF
BF−
BF
F
CF
CF−
CF
+ F
DF
DF−
DF
+ F
EF
EF−
EF
+ 0=
0
C
y
C

z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
F
BC
BC−
BC
+ F
AC
AC−
AC
+ F
CD
CD
CD
+ F
CF
CF
CF
+ 0=
F
AB

F
AC
F
AD
F
AE
F
BC
F
BE
F
BF
F
CD
F
CF
F
DE
F
DF
F
EF
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜

⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟

⎟
⎟
⎟
⎠
Find F
AB
F
AC
, F
AD
, F
AE
, F
BC
, F
BE
, F
BF
, F
CD
, F
CF
, F
DE
, F
DF
, F
EF
,
()

=
523
© 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 6
Positive (T)
Negative (C)
F
AB
F
AC
F
AD
F
AE
F
BC
F
BE
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝

⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
300−
971.825
1.121 10
11−
×
366.667−
0
500
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟

⎟
⎟
⎟
⎟
⎟
⎠
lb=
F
BF
F
CD
F
CF
F
DE
F
DF
F
EF
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞

⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
0
500−
300−
0
424.264
300−
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟

⎟
⎟
⎠
lb=
Problem 6-60
Determine the force in each member of the space truss and state if the members are in tension or
compression. The truss is supported by a ball-and-socket joints at A, B, and E. Hint: The
support reaction at E acts along member EC. Why?
Given:
a 2m=
F
200−
400
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
b 1.5 m=
c 5m=
d 1m=
e 2m=
Solution:
AC
0

ab+
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= AD
d
a
e
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
BC
c− d−
0
0
⎛
⎜
⎜

⎝
⎞
⎟
⎟
⎠
= BD
c−
b−
e
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= CD
d
b−
e
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=

Guesses F
AC
1N= F
AD
1N= F
BC
1N=
F
CD
1N= F
EC
1N= F
BD
1N=
Given
F F
AD
AD−
AD
+ F
BD
BD−
BD
+ F
CD
CD−
CD
+ 0=
524
© 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 6
F
CD
CD
CD
F
BC
BC−
BC
+ F
AC
AC−
AC
+
0
0
F
EC
−
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟

⎠
+ 0=
Positive (T)
Negative (C)
F
AC
F
AD
F
BC
F
BD
F
CD
F
EC
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟

⎟
⎟
⎟
⎟
⎟
⎠
Find F
AC
F
AD
, F
BC
, F
BD
, F
CD
, F
EC
,
()
=
F
AC
F
AD
F
BC
F
BD
F

CD
F
EC
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
221
343
148
186
397−
295−
⎛

⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
N=
Problem 6-61
Determine the force in each member of the space truss and state if the members are in tension or
compression. The truss is supported by ball-and-socket joints at C, D, E, and G.
Units Used:
kN 10
3
N=
Given:
F 3kN=
a 2m=
b 1.5 m=
c 2m=

d 1m=
e 1m=
Solution:
F
v
F
b
2
c
2
+
0
b
c−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
525
© 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 6
u
AG

1
a
2
e
2
+
e−
a−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
u
AE
1
a
2
d
2
+
d
a−
0
⎛

⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
u
BC
u
AE
=
u
AB
0
0
1−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
u
BD
u

AG
=
u
BE
1
a
2
c
2
+ d
2
+
d
a−
c
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
u
BG
1
a
2
e

2
+ c
2
+
e−
a−
c
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
Guesses
F
AB
1kN= F
AE
1kN= F
AG
1kN=
F
BC
1kN= F
BD
1kN=
F

BE
1kN= F
BG
1kN=
Given
c−
c
2
b
2
+
Fa()
a
a
2
d
2
+
F
BC
c()−
a
a
2
e
2
+
F
BD
c()− 0=

e
a
2
e
2
+
F
BD
a()
d
a
2
d
2
+
F
BC
a()− 0=
F
v
F
AE
u
AE
+ F
AG
u
AG
+ F
AB

u
AB
+ 0=
F
AB
− u
AB
F
BG
u
BG
+ F
BE
u
BE
+ F
BC
u
BC
+ F
BD
u
BD
+ 0=
526
© 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 6
F

AB
F
AE
F
AG
F
BC
F
BD
F
BE
F
BG
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟

⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find F
AB
F
AE
, F
AG
, F
BC
, F
BD
, F
BE
, F
BG
,
()
=
F
AB
F
AE
F
AG

F
BC
F
BD
F
BE
F
BG
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟

⎟
⎠
2.4−
1.006
1.006
1.342−
1.342−
1.8
1.8
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
kN=

Positive (T)
Negative (C)
Problem 6-62
Determine the force in
members BD, AD, and AF of
the space truss and state if the
members are in tension or
compression. The truss is
supported by short links at A,
B, D, and F.
Given:
F
0
250
250−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
a 6ft=
b 6ft=
θ
60 deg=
Solution:

Find the external reactions
hbsin
θ
()
=
Guesses
A
x
1lb= B
y
1lb= B
z
1lb=
D
y
1lb= F
y
1lb= F
z
1lb=
Given
⎛
⎞
⎛
⎞
527
© 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 6

F
A
x
0
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+
0
B
y
B
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟

⎠
+
0
D
y
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+
0
F
y
F
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟

⎟
⎠
+ 0=
0.5b
a
h
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F×
b
a
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
A
x
0
0

⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
×+
b
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
B
y
B
z
⎛
⎜
⎜

⎜
⎝
⎞
⎟
⎟
⎟
⎠
×+
0.5b
0
h
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
D
y
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟

⎟
⎟
⎠
×+
0
a
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
F
y
F
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠

×+ 0=
A
x
B
y
B
z
D
y
F
y
F
z
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟

⎟
⎟
⎠
Find A
x
B
y
, B
z
, D
y
, F
y
, F
z
,
()
=
A
x
B
y
B
z
D
y
F
y
F
z

⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
0
72
125
394−
72
125
⎛
⎜
⎜
⎜

⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
lb=
Now find the forces in the members
AB
0
a−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= AC
b−

a−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= AD
0.5− b
a−
h
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
AE
0.5− b
0
h
⎛
⎜
⎜

⎝
⎞
⎟
⎟
⎠
= AF
b−
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= BC
b−
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=

BD
0.5− b
0
h
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= CD
0.5b
0
h
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= CF
0
a
0
⎛
⎜

⎜
⎝
⎞
⎟
⎟
⎠
=
DE
0
a
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= DF
0.5− b
a
h−
⎛
⎜
⎜
⎝
⎞
⎟
⎟

⎠
= EF
0.5− b
0
h−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
Guesses
F
AB
1lb= F
AC
1lb= F
AD
1lb=
F
AE
1lb= F
AF
1lb= F
BC
1lb=
F

BD
1lb= F
CD
1lb= F
CF
1lb=
528
© 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 6
F
DE
1lb= F
DF
1lb= F
EF
1lb
=
Given
F F
DE
DE−
DE
+ F
AE
AE−
AE
+ F
EF

EF
EF
+ 0=
F
CF
CF
CF
F
CD
CD
CD
+ F
BC
BC−
BC
+ F
AC
AC−
AC
+ 0=
F
DE
DE
DE
F
DF
DF
DF
+ F
AD

AD−
AD
+ F
BD
BD−
BD
+ F
CD
CD−
CD
+
0
D
y
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+ 0=
F
AB
AB−
AB

F
BC
BC
BC
+ F
BD
BD
BD
+
0
B
y
B
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+ 0=
F
AB
AB
AB
F

AC
AC
AC
+ F
AF
AF
AF
+ F
AD
AD
AD
+ F
AE
AE
AE
+
A
x
0
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠

+ 0=
F
AB
F
AC
F
AD
F
AE
F
AF
F
BC
F
BD
F
CD
F
CF
F
DE
F
DF
F
EF
⎛
⎜
⎜
⎜
⎜

⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟

⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find F
AB
F
AC
, F
AD
, F
AE
, F
AF
, F
BC
, F
BD
, F
CD
, F
CF
, F
DE
, F
DF
, F

EF
,
()
=
529
© 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 6
Positive (T)
Negative (C)
F
BD
F
AD
F
AF
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
144.3−
204.1
72.2

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
Problem 6-63
Determine the force in members CF, EF, and DF of the space truss and state if the members
are in tension or compression. The truss is supported by short links at A, B, D, and F.
Given:
F
0
250
250−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
a 6ft=
b 6ft=
θ
60 deg=

Solution:

Find the external reactions
hbsin
θ
()
=
Guesses
A
x
1lb= B
y
1lb= B
z
1lb=
D
y
1lb= F
y
1lb= F
z
1lb=
Given
F
A
x
0
0
⎛
⎜

⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+
0
B
y
B
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+
0
D
y
0
⎛

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+
0
F
y
F
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+ 0=
0.5b
a
h
⎛

⎜
⎜
⎝
⎞
⎟
⎟
⎠
F×
b
a
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
A
x
0
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟

⎟
⎟
⎠
×+
b
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
B
y
B
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠

×+
0.5b
0
h
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
D
y
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
×+
0
a
0

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
F
y
F
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
×+ 0=
530
© 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 6
A

x
B
y
B
z
D
y
F
y
F
z
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟

⎠
Find A
x
B
y
, B
z
, D
y
, F
y
, F
z
,
()
=
A
x
B
y
B
z
D
y
F
y
F
z
⎛
⎜

⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
0
72
125
394−
72
125
⎛
⎜
⎜
⎜
⎜
⎜

⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
lb=
Now find the forces in the members
AB
0
a−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= AC
b−
a−
0

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= AD
0.5− b
a−
h
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
AE
0.5− b
0
h
⎛
⎜
⎜
⎝
⎞

⎟
⎟
⎠
= AF
b−
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= BC
b−
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
BD
0.5− b

0
h
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= CD
0.5b
0
h
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= CF
0
a
0
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
=
DE
0
a
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= DF
0.5− b
a
h−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= EF

0.5− b
0
h−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
Guesses
F
AB
1lb= F
AC
1lb= F
AD
1lb=
F
AE
1lb= F
AF
1lb= F
BC
1lb=
F
BD
1lb= F

CD
1lb= F
CF
1lb=
F
DE
1lb= F
DF
1lb= F
EF
1lb=
Given
F F
DE
DE−
DE
+ F
AE
AE−
AE
+ F
EF
EF
EF
+ 0=
F
CF
CF
CF
F

CD
CD
CD
+ F
BC
BC−
BC
+ F
AC
AC−
AC
+ 0=
531
© 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 6
F
DE
DE
DE
F
DF
DF
DF
+ F
AD
AD−
AD
+ F

BD
BD−
BD
+ F
CD
CD−
CD
+
0
D
y
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+ 0=
F
AB
AB−
AB
F
BC
BC

BC
+ F
BD
BD
BD
+
0
B
y
B
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+ 0=
F
AB
AB
AB
F
AC
AC
AC

+ F
AF
AF
AF
+ F
AD
AD
AD
+ F
AE
AE
AE
+
A
x
0
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+ 0=
F
AB

F
AC
F
AD
F
AE
F
AF
F
BC
F
BD
F
CD
F
CF
F
DE
F
DF
F
EF
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜

⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟

⎟
⎟
⎟
⎠
Find F
AB
F
AC
, F
AD
, F
AE
, F
AF
, F
BC
, F
BD
, F
CD
, F
CF
, F
DE
, F
DF
, F
EF
,
()

=
Positive (T)
Negative (C)
F
CF
F
EF
F
DF
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
72.2
144.3−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠

lb=
532
© 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 6
Determine the force developed in each member of the space truss and state if the members are
in tension or compression. The crate has weight W.
Given:
W 150 lb=
a 6ft=
b 6ft=
c 6ft=
Solution: Unit Vectors
hc
2
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
−=
u
AD
1
h

2
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
+
a−
2
0
h
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
=
u
BD

1
h
2
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
+
a
2
0
h
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
=

u
AC
1
h
2
b
2
+
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
+
a
2
−
b
h
⎛
⎜
⎜
⎜
⎜
⎝
⎞

⎟
⎟
⎟
⎟
⎠
=
u
BC
1
h
2
b
2
+
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
+
a
2
b
h
⎛
⎜

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
=
Guesses B
y
1lb= A
x
1lb= A
y
1lb=
F
AB
1lb= F
AC
1lb= F
AD
1lb= F
BC
1lb= F
BD
1lb= F
CD

1lb=
533
Problem 6-64
© 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 6
Given
0
F
CD
−
W−
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
F
AC
u
AC
− F
BC
u

BC
− 0=
F
AB
B
y
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
F
BC
u
BC
+ F
BD
u
BD
+ 0=
A
x
F
AB

−
A
y
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
F
AC
u
AC
+ F
AD
u
AD
+ 0=
A
x
A
y
B
y
F

AB
F
AC
F
AD
F
BC
F
BD
F
CD
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟

⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find A
x
A
y
, B
y
, F
AB
, F
AC
, F
AD
, F
BC
, F
BD
, F
CD
,

()
=
Positive (T)
Negative (C)
F
AB
F
AC
F
AD
F
BC
F
BD
F
CD
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟

⎟
⎟
⎟
⎟
⎟
⎠
0.0
122.5−
86.6
122.5−
86.6
173.2
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠

lb=
Problem 6-65
The space truss is used to support vertical forces at joints B, C, and D. Determine the force in
each member and state if the members are in tension or compression.
534
© 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 6
Units Used:
kN 10
3
N=
Given:
F
1
6kN= a 0.75 m=
F
2
8kN= b 1.00 m=
F
3
9kN= c 1.5 m=
Solution:
Assume that the connections at A, E,
and F are rollers
Guesses
F
BC
1kN= F

CF
1kN=
F
CD
1kN= F
AD
1kN=
F
DF
1kN= F
DE
1kN=
F
BD
1kN= F
BA
1kN=
F
EF
1kN= F
AE
1kN=
F
AF
1kN=
Given
Joint C
F
BC
0= F

CD
0=
F
2
− F
CF
− 0=
Joint D
a
a
2
b
2
+
F
BD
a
a
2
b
2
+ c
2
+
F
AD
+ 0=
535
© 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 6
F
CD
−
b
a
2
b
2
+
F
BD
−
b−
a
2
b
2
+ c
2
+
F
AD
b
b
2
c
2
+

F
DF
−+
0=
F
3
− F
DE
−
c
b
2
c
2
+
F
DF
−
c−
a
2
b
2
+ c
2
+
F
AD
+
0=

Joint B
F
BC
−
a
a
2
b
2
+
F
BD
− 0=
b
a
2
b
2
+
F
BD
0= F
1
− F
BA
− 0=
Joint E
a
a
2

b
2
+
F
AE
0= F
EF
−
b
a
2
b
2
+
F
AE
− 0=
F
BC
F
CF
F
CD
F
AD
F
DF
F
DE
F

BD
F
BA
F
EF
F
AE
F
AF
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟

⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find F
BC
F
CF
, F
CD
, F
AD
, F
DF
, F
DE
, F

BD
, F
BA
, F
EF
, F
AE
, F
AF
,
()
=
536
© 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 6
Positive (T)
Negative (C)
F
BC
F
CF
F
CD
F
AD
F
DF
F

DE
F
BD
F
BA
F
EF
F
AE
F
AF
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝

⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
0.00
8.00−
0.00
0.00
0.00
9.00−
0.00
6.00−
0.00
0.00
0.00

⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟

⎟
⎠
kN=
Problem 6-66
A force
P
is applied to the handles of the pliers. Determine the force developed on the smooth
bolt B and the reaction that pin A exerts on its attached members.
Given:
P 8lb=
a 1.25 in=
b 5in=
c 1.5 in=
Solution:
Σ
M
A
= 0
;
R
B
− cPb+ 0=
R
B
P
b
c
=
R
B

26.7 lb=
Σ
F
x
= 0;
A
x
0=
Σ
F
y
= 0;
A
y
P− R
B
− 0=
537
© 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 6
A
y
PR
B
+=
A
y
34.7 lb=

Problem 6-67
The eye hook has a positive locking latch when it supports the load because its two parts are
pin-connected at A and they bear against one another along the smooth surface at B. Determine
the resultant force at the pin and the normal force at B when the eye hook supports load
F
.
Given:
F 800 lb=
a 0.25 in=
b 3in=
c 2in=
θ
30 deg=
Solution:
Σ
M
A
= 0;
F
B
− cos 90 deg
θ
−
()
b() F
B
sin 90 deg
θ
−
()

c()− Fa+ 0=
F
B
F
a
cos 90 deg
θ
−
()
b sin 90 deg
θ
−
()
c+
= F
B
61.9 lb=
+
↑
Σ
F
y
= 0;
F− F
B
sin 90 deg
θ
−
()
− A

y
+ 0=
A
y
FF
B
sin 90 deg
θ
−
()
+= A
y
854lb=
+
→
Σ
F
x
= 0;
A
x
F
B
cos 90 deg
θ
−
()
− 0=
A
x

F
B
cos 90 deg
θ
−
()
= A
x
30.9 lb=
F
A
A
x
2
A
y
2
+= F
A
854lb=
Problem 6-68
Determine the force
P
needed to hold the block of mass F in equilibrium.
538
© 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 6
Given:

F 20 lb=
Solution:
Pulley B:
Σ
F
y
= 0;
2PT− 0=
Pulley A:
Σ
F
y
= 0;
2TF− 0=
T
1
2
F= T 10lb=
2PT= P
1
2
T=
P 5lb=
Problem 6-69
The link is used to hold the rod in place. Determine the required axial force on the screw at E if
the largest force to be exerted on the rod at B, C or D is to be
F
max
. Also, find the magnitude of
the force reaction at pin A. Assume all surfaces of contact are smooth.

Given:
F
max
100 lb=
a 100 mm=
b 80 mm=
c 50 mm=
θ
45 deg=
539
© 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 6
Solution:
Assign an initial value for R
E
. This will be scaled
at the end of the problem.
Guesses A
x
1lb= A
y
1lb= R
B
1lb=
R
C
1lb= R
D

1lb= R
E
1lb=
Given A
x
− R
E
+ R
B
cos
θ
()
− 0= A
y
R
B
sin
θ
()
− 0=
R
E
aR
B
cos
θ
()
ab+ ccos
θ
()

+
()
− R
B
sin
θ
()
csin
θ
()
− 0=
R
B
cos
θ
()
R
D
− 0= R
B
sin
θ
()
R
C
− 0=
A
x
A
y

R
B
R
C
R
D
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find A
x
A
y
, R
B

, R
C
, R
D
,
()
=
A
x
A
y
R
B
R
C
R
D
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟

⎟
⎟
⎟
⎟
⎠
0.601
0.399
0.564
0.399
0.399
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
lb=
Now find the critical load and scale the problem
ans
R
B
R

C
R
D
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
= F
scale
F
max
max ans()
= R
E
F
scale
R
E
= R
E
177.3 lb=
F
A
F

scale
A
x
2
A
y
2
+= F
A
127.9 lb=
Problem 6-70
The man of weight W
1
attempts to lift himself and the seat of weight W
2
using the rope and
pulley system shown. Determine the force at A needed to do so, and also find his reaction on
the seat.
540
© 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 6
Given:
W
1
150 lb=
W
2
10 lb=

Solution:
Pulley C:
Σ
F
y
= 0;
3TR− 0=
Pulley B:
Σ
F
y
= 0;
3RP− 0=
Thus,
P 9T=
Man and seat:
Σ
F
y
= 0;
TP+ W
1
− W
2
− 0=
10TW
1
W
2
+=

T
W
1
W
2
+
10
= T 16lb=
P 9T= P 144lb=
Seat:
Σ
F
y
= 0;
PN− W
2
− 0=
NPW
2
−= N 134lb=
Problem 6-71
Determine the horizontal and vertical components of force that pins A and C exert on the frame.
Given:
F 500 N=
a 0.8 m= d 0.4 m=
541
© 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 6

b 0.9 m= e 1.2 m=
c 0.5 m=
θ
45deg=
Solution:
BC is a two-force member
Member AB :
Σ
M
A
= 0;
F− cF
BC
e
a
2
e
2
+
b+ F
BC
a
a
2
e
2
+
cd+()+ 0=
F
BC

Fc
a
2
e
2
+
eb ac+ ad+
= F
BC
200.3 N=
Thus,
C
x
F
BC
e
a
2
e
2
+
= C
x
167 N=
C
y
F
BC
a
a

2
e
2
+
= C
y
111 N=
Σ
F
x
= 0;
A
x
F
BC
e
a
2
e
2
+
− 0= A
x
F
BC
e
a
2
e
2

+
= A
x
167 N=
Σ
F
y
= 0;
A
y
F− F
BC
a
a
2
e
2
+
+ 0=
A
y
FF
BC
a
a
2
e
2
+
−= A

y
389 N=
Problem 6-72
Determine the horizontal and vertical components of force that pins A and C exert on the
frame.
Units Used:
kN 10
3
N=
Given:
F
1
1kN=
542
© 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 6
F
2
500 N=
θ
45 deg=
a 0.2 m=
b 0.2 m=
c 0.4 m=
d 0.4 m=
Solution:
Guesses
A

x
1N= A
y
1N=
C
x
1N= C
y
1N=
Given
A
x
C
x
− 0= A
y
C
y
+ F
1
− F
2
− 0=
F
1
aA
y
2 a− A
x
d+ 0= F

2
− bC
y
bc+()+ C
x
d− 0=
A
x
A
y
C
x
C
y
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
Find A
x

A
y
, C
x
, C
y
,
()
=
A
x
A
y
C
x
C
y
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟

⎠
500
1000
500
500
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
N=
Problem 6-73
The truck exerts the three forces shown on the girders of the bridge. Determine the reactions at
the supports when the truck is in the position shown. The girders are connected together by a
short vertical link DC.
Units Used:
kip 10
3
lb=
543
© 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 6
Given:
a 55 ft= f 12 ft=
b 10 ft= F
1
5 kip=
c 48 ft= F
2
4 kip=
d 5ft= F
3
2 kip=
e 2ft=
Solution:
Member CE:
Σ
M
C
= 0;
F
3
− eE
y
ec+()+ 0= E
y
F
3
e
ec+
= E

y
80lb=
Σ
F
y
= 0;
C
y
F
3
− E
y
+ 0= C
y
F
3
E
y
−= C
y
1920lb=
Member ABD:
Σ
M
A
= 0;
F
1
− aF
2

da+()− C
y
ad+ b+()− B
y
ad+()+ 0=
B
y
F
1
aF
2
da+()+ C
y
ad+ b+()+
da+
= B
y
10.8 kip=
Σ
F
y
= 0;
A
y
F
1
− B
y
+ F
2

− C
y
− 0=
A
y
C
y
F
1
B
y
−+ F
2
+= A
y
96.7 lb=
Problem 6-74
Determine the greatest force
P
that can be applied to the frame if the largest force resultant
acting at A can have a magnitude F
max
.
Units Used:
kN 10
3
N=
544
© 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 6
Given:
F
max
2kN=
a 0.75 m=
b 0.75 m=
c 0.5 m=
d 0.1 m=
Solution:
Σ
M
A
= 0;
Tc d+()Pa b+()− 0=
+
→
Σ
F
x
= 0;
A
x
T− 0=
+
↑
Σ
F
y

= 0;
A
y
P− 0=
Thus,
T
ab+
cd+
P= A
y
P= A
x
ab+
cd+
P=
Require,
F
max
A
x
2
A
y
2
+=
P
F
max
ab+
cd+

⎛
⎜
⎝
⎞
⎟
⎠
2
1+
=
P 743 N=
Problem 6-75
The compound beam is pin supported at
B and supported by rockers at A and C.
There is a hinge (pin) at D. Determine
the reactions at the supports.
Units Used:
kN 10
3
N=
545
© 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.