Engineering Mechanics - Statics Episode 2 Part 6 pot

Engineering Mechanics - Statics Chapter 6
W
1
γ
L
L
2
a−
⎛
⎜
⎝
⎞
⎟
⎠
a
= W
1
3lb=
+
↑
Σ
F
y
= 0;
R
A
W
1
−
γ
L− 0=

R
A
W
1
γ
L+= R
A
9lb=
Σ
M
B
= 0;
W
2
acos
φ
()
γ
Lcos
φ
()
L
2
a−
⎛
⎜
⎝
⎞
⎟
⎠

− R
A
La−( ) cos
φ
()
− 0=
W
2
γ
L
L
2
a−
⎛
⎜
⎝
⎞
⎟
⎠
R
A
La−()+
a
= W
2
21lb=
+
↑
R
B

W
2
− R
A
−
γ
L− 0=
Σ
F
y
= 0;
R
B
W
2
R
A
+
γ
L+= R
B
36lb=
Σ
M
C
= 0;
R
B
La−( ) cos
φ

()
γ
L
L
2
a−
⎛
⎜
⎝
⎞
⎟
⎠
cos
φ
()
+ W
3
acos φ()− 0=
W
3
R
B
La−()
γ
L
L
2
a−
⎛
⎜

⎝
⎞
⎟
⎠
+
a
= W
3
75lb=
Problem 6-124
The three-member frame is connected at its ends using ball-and-socket joints. Determine the x,
y, z components of reaction at B and the tension in member ED. The force acting at D is
F
.
Given:
F
135
200
180−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
a 6ft= e 3ft=
b 4ft= f 1ft=

601
© 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
d 6ft= g 2ft=
cgf+=
Solution:
AC and DE are two-force members.
Define some vectors
r
DE
e−
b− g−
a
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= u
DE
r
DE
r
DE
=

r
AC
d− e−
b−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= u
AC
r
AC
r
AC
=
r
BD
e
f−
0
⎛
⎜
⎜
⎝
⎞

⎟
⎟
⎠
= r
BA
ed+
c−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
Guesses
B
x
1lb= B
y
1lb= B
z
1lb= F
DE
1lb= F
AC
1lb=
Given

B
x
B
y
B
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
F
DE
u
DE
+ F
AC
u
AC
+ F+ 0= r
BD
F
DE
u
DE

F+
()
× r
BA
F
AC
u
AC
()
×+ 0=
B
x
B
y
B
z
F
DE
F
AC
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞

⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find B
x
B
y
, B
z
, F
DE
, F
AC
,
()
=
B
x
B
y
B
z
⎛
⎜
⎜

⎜
⎝
⎞
⎟
⎟
⎟
⎠
30−
13.333−
3.039 10
12−
×
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
lb=
F
DE
F
AC
⎛
⎜
⎜

⎝
⎞
⎟
⎟
⎠
270
16.415
⎛
⎜
⎝
⎞
⎟
⎠
lb=
602
© 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
The four-member "A" frame is supported at A and E by smooth collars and at G by a pin. All
the other joints are ball-and-sockets. If the pin at G will fail when the resultant force there is
F
max
, determine the largest vertical force P that can be supported by the frame. Also, what
are the x, y, z force components which member BD exerts on members EDC and ABC? The
collars at A and E and the pon at G only exert force components on the frame.
Given:
F
max
800 N=

a 300 mm=
b 600 mm=
c 600 mm=
Solution:
Σ
M
x
= 0;
P− 2 c
b
b
2
c
2
+
F
max
c+ 0=
P
F
max
b
2 b
2
c
2
+
= P 282.843 N=
B
z

D
z
+ F
max
c
b
2
c
2
+
− 0= D
z
B
z
=
B
z
F
max
c
2 b
2
c
2
+
= D
z
B
z
= B

z
283 N=
D
z
283 N=
B
y
D
y
+ F
max
b
b
2
c
2
+
− 0= D
y
B
y
=
B
y
F
max
b
2 b
2
c

2
+
= D
y
B
y
= B
y
283 N= D
y
283 N=
B
x
D
x
= 0=
Problem 6-126
The structure is subjected to the loading shown. Member AD is supported by a cable AB and a
roller at
C
and fits throu
g
h a smooth circular hole at
D
. Member ED is su
pp
orted b
y
a roller at
603

Problem 6-125
© 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
g
pp y
D and a pole that fits in a smooth snug circular hole at E. Determine the x, y, z components of
reaction at E and the tension in cable AB.
Units Used:
kN 10
3
N=
Given:
F
0
0
2.5−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
a 0.5 m= d 0.3 m=
b 0.4 m= e 0.8 m=
c 0.3 m=

Solution:
AB
c− d−
0
e
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
Guesses
F
AB
1kN= D
x
1kN= D
z
1kN=
D
z2
1kN= E
x
1kN= E
y
1kN=
M

Dx
1kNm⋅= M
Dz
1kNm⋅= C
x
1kN=
M
Ex
1kNm⋅= M
Ey
1kN⋅ m=
Given
F F
AB
AB
AB
+
C
x
0
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟

⎠
+
D
x
0
D
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+ 0=
M
Dx
0
M
Dz
⎛
⎜
⎜
⎜
⎝
⎞
⎟

⎟
⎟
⎠
0
b
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
C
x
0
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
×+
d

b
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F×+
cd+
b
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
AB
AB
AB
⎛
⎜
⎝
⎞

⎟
⎠
×+ 0=
D
x
−
0
D
z2
D
z
−
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
E
x
E
y
0
⎛
⎜
⎜

⎜
⎝
⎞
⎟
⎟
⎟
⎠
+ 0=
M
Dx
−
0
M
Dz
−
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
M
Ex
M
Ey
0

⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+
0
a
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
D
x
0
D
z
D
z2
−

⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
×+ 0=
604
© 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
C
x
D
x
D
z
D
z2
E
x
E
y
F
AB

M
Dx
M
Dz
M
Ex
M
Ey
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟

⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find C
x
D
x
, D
z
, D
z2
, E
x
, E
y
, F
AB

, M
Dx
, M
Dz
, M
Ex
, M
Ey
,
()
=
C
x
D
x
D
z
D
z2
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟

⎟
⎟
⎠
0.937
0
1.25
1.25
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
kN=
M
Dx
M
Dz
⎛
⎜
⎜
⎝
⎞
⎟

⎟
⎠
0.5
0
⎛
⎜
⎝
⎞
⎟
⎠
kN m⋅=
E
x
E
y
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
0
⎛
⎜
⎝
⎞
⎟

⎠
kN=
M
Ex
M
Ey
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0.5
0
⎛
⎜
⎝
⎞
⎟
⎠
kN m⋅= F
AB
1.562 kN=
Problem 6-127
The structure is subjected to the loadings shown.Member AB is supported by a ball-and-socket
at A and smooth collar at B. Member CD is supported by a pin at C. Determine the x, y, z
components of reaction at A and C.
605

© 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 2m= M 800 N m⋅=
b 1.5 m= F 250 N=
c 3m=
θ
1
60 deg=
d 4m=
θ
2
45 deg=
θ
3
60 deg=
Solution:

Guesses
B
x
1N= B
y
1N=
A
x
1N= A
y

1N= A
z
1N=
C
x
1N= C
y
1N= C
z
1N=
M
Bx
1Nm⋅= M
By
1Nm⋅=
M
Cy
1Nm⋅= M
Cz
1Nm⋅=
Given
A
x
A
y
A
z
⎛
⎜
⎜

⎜
⎝
⎞
⎟
⎟
⎟
⎠
B
x
B
y
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+ 0=
c
a
0
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
B
x
B
y
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
×
M
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟

⎠
+
M
Bx
−
M
By
−
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+ 0=
F
cos
θ
1
()
cos
θ
2
()
cos

θ
3
()
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
B
x
−
B
y
−
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠

+
C
x
C
y
C
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+ 0=
0
0
bd+
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F

cos
θ
1
()
cos
θ
2
()
cos
θ
3
()
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
⎡
⎢
⎢
⎢
⎣
⎤
⎥
⎥

⎥
⎦
×
0
0
b
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
B
x
−
B
y
−
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟

⎠
×+
M
Bx
M
By
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+
0
M
Cy
M
Cz
⎛
⎜
⎜
⎜
⎝
⎞
⎟

⎟
⎟
⎠
+ 0=
606
© 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
A
y
A
z
C
x
C
y
C
z
B
x
B
y
M
Bx
M
By
M

Cy
M
Cz
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟

⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find A
x
A
y
, A
z
, C
x
, C
y
, C
z
, B
x
, B

y
, M
Bx
, M
By
, M
Cy
, M
Cz
,
()
=
A
x
A
y
A
z
C
x
C
y
C
z
⎛
⎜
⎜
⎜
⎜
⎜

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
172.3−
114.8−
0
47.3
61.9−
125−
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞

⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
N=
M
Cy
M
Cz
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
429−
0
⎛
⎜
⎝
⎞
⎟
⎠
Nm⋅=

Problem 6-128
Determine the resultant forces at pins B and C on member ABC of the four-member frame.
Given:
w 150
lb
ft
=
a 5ft=
b 2ft=
607
© 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
c 2ft=
e 4ft=
dab+ c−=
Solution:
The initial guesses are
F
CD
20 lb= F
BE
40 lb=
Given
F
CD
cd+()F
BE
ec

db−()
2
e
2
+
− 0=
w− ab+()
ab+
2
⎛
⎜
⎝
⎞
⎟
⎠
F
BE
e
db−()
2
e
2
+
a+ F
CD
ab+()− 0=
F
CD
F
BE

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
Find F
CD
F
BE
,
()
=
F
CD
F
BE
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
350
1531
⎛

⎜
⎝
⎞
⎟
⎠
lb=
Problem 6-129
The mechanism consists of identical meshed gears A and B and arms which are fixed to the
gears. The spring attached to the ends of the arms has an unstretched length
δ
and a stiffness
k. If a torque M is applied to gear A, determine the angle
θ
through which each arm rotates.
The gears are each pinned to fixed supports at their centers.
608
© 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:
δ
100 mm=
k 250
N
m
=
M 6Nm⋅=
r
δ

2
=
a 150 mm=
Solution:
Σ
M
A
= 0;
F− rPacos
θ
()
− M+ 0=
Σ
M
B
= 0;
Pacos
θ
()
Fr− 0=
2Pacos
θ
()
M=
2k 2a( ) sin
θ
()
acos
θ
()

M=
2ka
2
sin 2
θ
()
M=
θ
1
2
asin
M
2ka
2
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
16.1 deg=
Problem 6-130
Determine the force in each member of the truss and state if the members are in tension or
compression.
Units Used:
kN 1000 N=
609
© 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
1
20 kN=
F
2
10 kN=
a 1.5 m=
b 2m=
Solution:
θ
atan
b
a
⎛
⎜
⎝
⎞
⎟
⎠
=
Guesses F
AG
1kN= F
BG
1kN=
F

GC
1kN= F
GF
1kN= F
AB
1kN=
F
BC
1kN= F
CD
1kN= F
CF
1kN=
F
DF
1kN= F
DE
1kN= F
EF
1kN=
Given
F
AB
− cos
θ
()
F
BC
+ 0=
F

AB
− sin
θ
()
F
BG
− 0=
F
GC
cos
θ
()
F
GF
+ F
AG
− 0=
F
GC
sin
θ
()
F
BG
+ F
1
− 0=
F
BC
− F

CD
+ F
GC
cos
θ
()
− F
CF
cos
θ
()
+ 0=
F
GC
− sin
θ
()
F
CF
sin
θ
()
− 0=
F
CD
− F
DE
cos
θ
()

+ 0=
F
DF
− F
DE
sin
θ
()
− 0=
F
GF
− F
CF
cos
θ
()
− F
EF
+ 0=
610
© 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
DF
F
CF
sin
θ

()
+ F
2
− 0=
F
DE
− cos
θ
()
F
EF
− 0=
611
© 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
AG
F
BG
F
GC
F
GF
F
AB
F
BC
F

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

⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find F
AG
F
BG
, F
GC
, F
GF
, F
AB
, F
BC
, F

CD
, F
CF
, F
DF
, F
DE
, F
EF
,
()
=
F
AG
F
BG
F
GC
F
GF
F
AB
F
BC
F
CD
F
CF
F
DF

F
DE
F
EF
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟

⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
13.13
17.50
3.13
11.25
21.88−
13.13−
9.37−
3.13−
12.50
15.62−
9.37
⎛
⎜
⎜
⎜
⎜
⎜
⎜

⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
kN=
Positive (T)
Negative (C)
Problem 6-131
The spring has an unstretched length

δ
. Determine the angle
θ
for equilibrium if the uniform
links each have a mass m
link
.
612
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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Engineering Mechanics - Statics Chapter 6
Given:
m
link
5kg=
δ
0.3 m=
k 400
N
m
=
a 0.1 m=
b 0.6 m=
g 9.81
m
s
2
=
Solution:

Guesses
θ
10 deg=
F
BD
1N=
E
x
1N=
Given
m
link
g
ab+
2
cos
θ
()
F
BD
bcos
θ
()
− E
x
bsin
θ
()
+ 0=
2− m

link
g
ab+
2
cos
θ
()
E
x
2 bsin
θ
()
+ 0=
F
BD
k 2 bsin
θ
()
δ
−
()
=
F
BD
E
x
θ
⎛
⎜
⎜

⎜
⎝
⎞
⎟
⎟
⎟
⎠
Find F
BD
E
x
,
θ
,
()
=
θ
21.7 deg=
Problem 6-132
The spring has an unstretched length
δ
. Determine the mass m
link
of each uniform link if the
angle for equilibrium is
θ
.
Given:
δ
0.3 m=

613
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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Engineering Mechanics - Statics Chapter 6
θ
20 deg=
k 400
N
m
=
a 0.1 m=
b 0.6 m=
g 9.81
m
s
2
=
Solution:
Guesses E
y
1N= m
link
1kg=
F
s
1N=
Given
F
s

2 bsin
θ
()
δ
−
()
k=
m
link
g
ab+
2
cos
θ
()
F
s
bcos
θ
()
− E
y
bsin
θ
()
+ 0=
2− m
link
g
ab+

2
cos
θ
()
E
y
2bsin
θ
()
+ 0=
m
link
F
s
E
y
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
Find m
link
F
s

, E
y
,
()
=
m
link
3.859 kg=
y
2 b()
sin
θ
()
=
y 2 bsin
θ
()
=
F
s
y
δ
−
()
k()=
F
s
44.17 N=
614
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
Σ
M
A
= 0;
E
y
2()ab+( ) sin
θ
()
2 M()g()−
⎡
⎣
⎤
⎦
ab
+
2
cos
θ
()()
0=
E
y
2( ) sin
θ
()
2 m()

g
2
−
⎡
⎢
⎣
⎤
⎥
⎦
cos
θ
()
0=
(
⇒
E
y
2 sin
θ
()()
mg( ) cos
θ
()()
=
E
y
mg( ) cos
θ
()()
2 sin

θ
()
=
Σ
M
C
= 0;
mg( ) cos
θ
()()
2 sin
θ
()
ab+( ) sin
θ
()
mg()
ab+
2
⎛
⎜
⎝
⎞
⎟
⎠
cos
θ
()
+ F
s

bcos
θ
()()
− 0=
mF
s
b
ga b+()
=
m 3.859 kg=
Problem 6-133
Determine the horizontal and vertical components of force that the pins A and B exert on the
two-member frame.
Given:
w 400
N
m
=
a 1.5 m=
b 1m=
c 1m=
F 0N=
θ
60 deg=
Solution:
Guesses
A
x
1N= A
y

1N= B
x
1N=
615
© 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
y
1N= C
x
1N= C
y
1N=
Given
w− a
a
2
C
x
asin
θ
()
+ C
y
acos
θ
()
− 0=

Fc C
x
c− C
y
b− 0= A
y
C
y
− wacos
θ
()
− 0=
A
x
− C
x
− wasin
θ
()
+ 0= C
x
B
x
− F− 0=
C
y
B
y
+ 0=
A

x
A
y
B
x
B
y
C
x
C
y
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟

⎠
Find A
x
A
y
, B
x
, B
y
, C
x
, C
y
,
()
=
A
x
A
y
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
300.0
80.4

⎛
⎜
⎝
⎞
⎟
⎠
N=
B
x
B
y
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
220
220
⎛
⎜
⎝
⎞
⎟
⎠
N=
Problem 6-134
Determine the horizontal and vertical components of force that the pins A and B exert on the

two-member frame.
Given:
w 400
N
m
=
a 1.5 m=
b 1m=
c 1m=
F 500 N=
θ
60 deg=
616
© 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:
Guesses
A
x
1N= A
y
1N= B
x
1N=
B
y
1N= C
x

1N= C
y
1N=
Given
w− a
a
2
C
x
asin
θ
()
+ C
y
acos
θ
()
− 0= A
y
C
y
− wacos
θ
()
− 0=
C
x
B
x
+ F− 0=

Fc C
x
c− C
y
b− 0=
A
x
− C
x
− wasin
θ
()
+ 0= C
y
B
y
− 0=
A
x
A
y
B
x
B
y
C
x
C
y
⎛

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

y
,
()
=
A
x
A
y
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
117.0
397.4
⎛
⎜
⎝
⎞
⎟
⎠
N=
B
x
B
y
⎛

⎜
⎜
⎝
⎞
⎟
⎟
⎠
97.4
97.4
⎛
⎜
⎝
⎞
⎟
⎠
N=
Problem 6-135
Determine the force in each member of the truss and
indicate whether the members are in tension or
compression.
Units Used:

kip 1000 lb=
Given:

F
1
1000 lb= b 8ft=
F
2

500 lb= c 4ft=
a 4ft=
Solution:

Joint B:

617
© 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
Initial Guesses:
Given
F
BC
100 lb= F
BA
150 lb=
Σ
F
x
= 0;

F
1
F
BC
cos atan
b
a

⎛
⎜
⎝
⎞
⎟
⎠
⎛
⎜
⎝
⎞
⎟
⎠
− F
BA
cos atan
c
a
⎛
⎜
⎝
⎞
⎟
⎠
⎛
⎜
⎝
⎞
⎟
⎠
− 0=

Σ
F
y
= 0;

F
BC
− sin atan
b
a
⎛
⎜
⎝
⎞
⎟
⎠
⎛
⎜
⎝
⎞
⎟
⎠
F
BA
sin atan
c
a
⎛
⎜
⎝

⎞
⎟
⎠
⎛
⎜
⎝
⎞
⎟
⎠
+ F
2
− 0=
F
BC
F
BA
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
Find F
BC
F
BA
,
()

=
F
BC
373lb=
(C)

F
BA
1178.51 lb=
F
BA
1.179 kip=
(C)

Joint A:
F
AC
F
BA
c
a
a
2
c
2
+
a
2
=
Σ

F
y
= 0;

F
AC
F
BA
()
sin atan
c
a
⎛
⎜
⎝
⎞
⎟
⎠
⎛
⎜
⎝
⎞
⎟
⎠
− 0=
F
AC
833lb=
(T)

Problem 6-136
Determine the force in each
member of the truss and state if
the members are in tension or
compression.
Units Used:
kip 10
3
lb=
Given:
F 1000 lb=
a 10 ft=
b 10 ft=
618
© 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:
θ
atan
b
a
⎛
⎜
⎝
⎞
⎟
⎠
=

Guesses
F
AB
1lb= F
AG
1lb= F
BC
1lb= F
BG
1lb=
F
CD
1lb= F
CE
1lb= F
CG
1lb= F
DE
1lb=
F
EG
1lb=
Given
F
AB
F
AG
cos
θ
()

+ 0=
F
AB
− F
BC
+ 0=
F
BG
0=
F
CD
F
BC
− F
CG
cos
θ
()
− 0=
F
CE
F− F
CG
sin
θ
()
+ 0=
F
CD
− F

DE
cos
θ
()
− 0=
F
DE
cos
θ
()
F
EG
− 0=
F
CE
− F
DE
sin
θ
()
− 0=
F
EG
F
AG
cos
θ
()
− F
CG

cos
θ
()
+ 0=
F
AG
− sin
θ
()
F
BG
− F
CG
sin
θ
()
− 0=
F
AB
F
AG
F
BC
F
BG
F
CD
F
CE
F

CG
F
DE
F
EG
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟

⎟
⎟
⎟
⎟
⎠
Find F
AB
F
AG
, F
BC
, F
BG
, F
CD
, F
CE
, F
CG
, F
DE
, F
EG
,
()
=
619
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics Chapter 6
F
AB
F
AG
F
BC
F
BG
F
CD
F
CE
F
CG
F
DE
F
EG
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
333
471−
333
0
667
667
471
943−
667−
⎛
⎜

⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
lb=
Positive (T), Negative (C)
Problem 6-137
Determine the force in members AB, AD, and AC of the space truss and state if the members are
in tension or compression. The force
F

is vertical.
Units Used:
kip 10
3
lb=
Given:
F 600 lb=
a 1.5 ft=
b 2ft=
c 8ft=
Solution:
AB
a
c−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= AC
a−
c−
0
⎛
⎜
⎜

⎝
⎞
⎟
⎟
⎠
=
AD
0
c−
b
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
Guesses F
AB
1lb= F
AC
1lb= F
AD
1lb=
620
© 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
AB
AB
AB
F
AC
AC
AC
+ F
AD
AD
AD
+
0
0
F−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+ 0=
F
AB
F
AC

F
AD
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
Find F
AB
F
AC
, F
AD
,
()
=
F
AB
F
AC
F
AD
⎛
⎜
⎜

⎜
⎝
⎞
⎟
⎟
⎟
⎠
1.221−
1.221−
2.474
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kip=
Positive (T)
Negative (C)
621
© 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 7-1
The column is fixed to the floor and is subjected to the loads shown. Determine the internal normal
force, shear force, and moment at points A and B.
Units Used:

kN 10
3
N=
Given:
F
1
6
kN=
F
2
6
kN=
F
3
8
kN=
a 150 mm=
b 150 mm=
c 150 mm=
Solution:
Free body Diagram: The support reaction need not be computed in this case.
Internal Forces: Applying equations of equillibrium to the top segment sectioned through
point A, we have
+
→
Σ
F
x
= 0;
V

A
0
= V
A
0
=
+
↑
Σ
F
y
= 0;
N
A
F
1
− F
2
− 0= N
A
F
1
F
2
+= N
A
12.0
kN=
F
1

aF
2
b− M
A
− 0= M
A
F
1
aF
2
b−= M
A
0
kN m⋅=
Σ
M
A
= 0;
Applying equations of equillibrium to the top segment sectioned through point B, we have
+
→
Σ
F
x
= 0;
V
B
0
= V
B

0
=
+
↑
Σ
F
y
= 0;
N
B
F
1
− F
2
− F
3
− 0= N
B
F
1
F
2
+ F
3
+= N
B
20.0
kN=
F
1

aF
2
b− F
3
c− M
B
+ 0= M
B
F
1
− aF
2
b+ F
3
c+= M
B
1.20
kN m⋅=
+Σ
M
B
= 0;
Problem 7-2
The axial forces act on the shaft as shown. Determine the internal normal forces at points A
and B.
622
© 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:
F
1
20
lb=
F
2
50
lb=
F
3
10
lb=
Solution:
Section A:
Σ
F
z
= 0;
F
2
2
F
1
− N
A
− 0=
N
A
F

2
2
F
1
−=
N
A
10.00
lb=
Section B:
Σ
F
z
= 0;
F
2
2
F
1
− N
A
− N
B
+ 0=
N
B
F
2
− 2F
1

+ N
A
+=
N
B
0.00
lb=
Problem 7-3
The shaft is supported by smooth bearings at A and B and subjected to the torques shown.
Determine the internal torque at points C, D, and E.
Given:
M
1
400
Nm⋅=
M
2
150
Nm⋅=
M
3
550
Nm⋅=
623
© 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:
Section C:

Σ
M
x
= 0;
T
C
0
=
Section D:
Σ
M
x
= 0;
T
D
M
1
− 0=
T
D
M
1
=
T
D
400.00
Nm⋅=
Section E:
Σ
M

x
= 0;
M
1
M
2
+ T
E
− 0=
T
E
M
1
M
2
+=
T
E
550.00
Nm⋅=
Problem 7-4
Three torques act on the shaft. Determine the internal torque at points A, B, C, and D.
Given:
M
1
300
Nm⋅=
M
2
400

Nm⋅=
M
3
200
Nm⋅=
Solution:
Section A:
Σ
Μ
x
= 0;
T
A
− M
1
+ M
2
− M
3
+ 0=
T
A
M
1
M
2
− M
3
+=
T

A
100.00
Nm⋅=
Section B:
Σ
M
x
= 0;
T
B
M
3
+ M
2
− 0=
624
© 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
B
M
3
− M
2
+=
T
B
200.00

Nm⋅=
Section C:
Σ
Μ
x
= 0;
T
C
− M
3
+ 0=
T
C
M
3
=
T
C
200.00
Nm⋅=
Section D:
Σ
Μ
x
= 0;
T
D
0
=
Problem 7-5

The shaft is supported by a journal bearing at A and a thrust bearing at B. Determine the normal
force, shear force, and moment at a section passing through (a) point C, which is just to the
right of the bearing at A, and (b) point D, which is just to the left of the force F
2
.
Units Used:
kip 10
3
lb=
Given:
F
1
2.5
kip=
F
2
3
kip=
w 75
lb
ft
=
b 12 ft=
c 2 ft=
a 6 ft=
Solution:
Σ
M
B
= 0;

625
© 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.