Engineering Mechanics - Statics Episode 2 Part 1 ppsx

Engineering Mechanics - Statics Chapter 5
Given:
a 3ft=
b 2ft=
c 1ft=
d 3ft=
e 9ft=
T 300 lb=
Solution:
Initial guesses:
F 10 lb= M
Ax
10 lbft= M
Az
10 lbft=
A
x
10 lb= A
y
10 lb= A
z
10 lb=
Given
A
x
A
y
A
z
⎛
⎜

⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
0
0
TF−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+ 0=
M
Ax
0
M
Az
⎛
⎜
⎜
⎜
⎝

⎞
⎟
⎟
⎟
⎠
a
c−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
0
F−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
×+
e
b− c−

0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
0
T
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
×+ 0=
F
A
x
A
y
A
z
M
Ax

M
Az
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find FA
x
, A
y
, A
z
, M
Ax
, M
Az

,
()
=
M
Ax
M
Az
⎛
⎜
⎝
⎞
⎟
⎠
0
0
⎛
⎜
⎝
⎞
⎟
⎠
lbft=
A
x
A
y
A
z
⎛
⎜

⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
0
0
600
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb= F 900lb=
Problem 5-70
The boom AB is held in equilibrium by a ball-and-socket joint A and a pulley and cord system as
shown. Determine the x, y, z components of reaction at A and the tension in cable DEC.
401
© 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 5
Given:
F

0
0
1500−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
a 5ft=
b 4ft=
cb=
d 5ft=
e 5ft=
f 2ft=
Solution:
α
atan
a
de+
⎛
⎜
⎝
⎞
⎟
⎠
=

La
2
de+()
2
+=
β
atan
b
L
fL
de+
−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
Guesses T
BE
1lb= T
DEC
1lb=
A
x
1lb= A
y

1lb= A
z
1lb=
Given 2 T
DEC
cos
β
()
T
BE
=
0
d
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F×
0
de+
0
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
0
T
BE
− cos
α
()
T
BE
sin
α
()
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
×+ 0=
A
x
A
y

A
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
F+ T
BE
0
cos
α
()
−
sin
α
()
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠

+ 0=
402
© 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 5
A
x
A
y
A
z
T
BE
T
DEC
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟

⎟
⎠
Find A
x
A
y
, A
z
, T
BE
, T
DEC
,
()
= T
DEC
919lb=
A
x
A
y
A
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟

⎟
⎟
⎠
0
1.5 10
3
×
750
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
lb=
Problem 5-71
The cable CED can sustain a maximum tension T
max
before it fails. Determine the greatest
vertical force F that can be applied to the boom. Also, what are the x, y, z components of
reaction at the ball-and-socket joint A?
Given:
T
max
800 lb=
a 5ft=

b 4ft=
cb=
d 5ft=
e 5ft=
f 2ft=
Solution:
α
atan
a
de+
⎛
⎜
⎝
⎞
⎟
⎠
=
La
2
de+()
2
+=
β
atan
b
L
fL
de+
−
⎛

⎜
⎜
⎝
⎞
⎟
⎟
⎠
= T
DEC
T
max
=
Guesses T
BE
1lb= F 1lb=
A
x
1lb= A
y
1lb= A
z
1lb=
Given 2 T
DEC
cos
β
()
T
BE
=

403
© 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 5
0
d
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
0
F−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
×
0
de+

0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
T
BE
− cos
α
()
T
BE
sin
α
()
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠

×+ 0=
A
x
A
y
A
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
0
0
F−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+ T
BE

0
cos
α
()
−
sin
α
()
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+ 0=
A
x
A
y
A
z
T
BE
F
⎛
⎜
⎜
⎜

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find A
x
A
y
, A
z
, T
BE
, F,
()
= F 1306lb=
A
x
A
y
A
z
⎛

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
0
1.306 10
3
×
653.197
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
lb=
Problem 5-72
The uniform table has a weight W and is supported by the framework shown. Determine the
smallest vertical force
P
that can be applied to its surface that will cause it to tip over. Where

should this force be applied?
Given:
W 20 lb=
a 3.5 ft=
b 2.5 ft=
c 3ft=
e 1.5 ft=
f 1ft=
Solution:
θ
atan
f
e
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
33.69 deg=
desin
θ
()
= d 0.832ft=
φ
atan
a
2

e−
b
2
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
φ
11.31 deg=
404
© 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 5
d'
a
2
e−
⎛
⎜
⎝
⎞
⎟

⎠
2
b
2
⎛
⎜
⎝
⎞
⎟
⎠
2
+= d' 1.275 ft=
Tipping will occur about the g - g axis.
Require P to be applied at the corner of the
table for P
min
.
Wd Pd'sin 90 deg
φ
−
θ
+
()
=
PW
d
d' sin 90 deg
φ
−
θ

+
()
= P 14.1 lb=
Problem 5-73
The windlass is subjected to load W. Determine the horizontal force P needed to hold the handle
in the position shown, and the components of reaction at the ball-and-socket joint A and the
smooth journal bearing B. The bearing at B is in proper alignment and exerts only force
reactions perpendicular to the shaft on the windlass.
Given:
W 150 lb=
a 2ft=
b 2ft=
c 1ft=
d 1ft=
e 1ft=
f 0.5 ft=
Solution:
Σ
M
y
= 0;
Wf Pd− 0=
P
Wf
d
= P 75lb=
Σ
F
y
= 0;

A
y
0lb= A
y
0=
Σ
M
x
= 0;
W− aB
z
ab+()+ 0=
405
© 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 5
B
z
Wa
ab+
= B
z
75lb=
Σ
F
z
= 0;
A
z

B
z
+ W− 0=
A
z
WB
z
−= A
z
75lb=
Σ
M
z
= 0;
B
x
ab+()ab+ c+ e+()P− 0=
B
x
Pa b+ c+ e+()
ab+
= B
x
112lb=
Σ
F
x
= 0;
A
x

B
x
− P+ 0=
A
x
B
x
P−= A
x
37.5 lb=
Problem 5-74
A ball of mass M rests between the grooves A and B of the

incline and against a vertical wall at
C. If all three surfaces of contact are smooth, determine the reactions of the surfaces on the
ball. Hint: Use the x, y, z axes, with origin at the center of the ball, and the z axis inclined as
shown.
Given:
M 2kg=
θ
1
10 deg=
θ
2
45 deg=
Solution:
Σ
F
x
= 0;

F
c
cos
θ
1
()
Mgsin
θ
1
()
− 0=
F
c
Mgtan
θ
1
()
⋅=
F
c
0.32 kg m⋅=
406
© 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 5
Σ
F
y
= 0;

N
A
cos
θ
2
()
N
B
cos
θ
2
()
− 0=
N
A
N
B
=
Σ
F
z
= 0;
2 N
A
sin
θ
2
()
Mgcos
θ

1
()
− F
c
sin
θ
1
()
− 0=
N
A
1
2
Mgcos
θ
1
()
⋅ F
c
sin
θ
1
()
⋅+
sin
θ
2
()
⋅=
N

A
1.3 kg m⋅=
NN
A
= N
B
=
Problem 5-75
Member AB is supported by cable BC and at A by a square rod which fits loosely through the
square hole at the end joint of the member as shown. Determine the components of reaction at
A and the tension in the cable needed to hold the cylinder of weight W in equilibrium.
Units Used:
kip 10
3
lb=
Given:
W 800 lb=
a 2ft=
b 6ft=
c 3ft=
Solution:
Σ
F
x
= 0
F
BC
c
c
2

b
2
+ a
2
+
⎛
⎜
⎝
⎞
⎟
⎠
0= F
BC
0lb=
Σ
F
y
= 0
A
y
0= A
y
0lb= A
y
0lb=
407
© 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 5

Σ
F
z
= 0
A
z
W− 0= A
z
W= A
z
800lb=
Σ
M
x
= 0
M
Ax
Wb− 0=
M
Ax
Wb= M
Ax
4.80 kip ft⋅=
Σ
M
y
= 0
M
Ay
0lbft= M

Ay
0lb ft⋅=
Σ
M
z
= 0
M
Az
0lbft= M
Az
0lb ft⋅=
Problem 5-76
The pipe assembly supports the vertical loads shown. Determine the components of reaction at
the ball-and-socket joint A and the tension in the supporting cables BC and BD.
Units Used:
kN 10
3
N=
Given:
F
1
3kN= d 2m=
F
2
4kN= e 1.5 m=
a 1m= g 1m=
b 1.5 m= h 3m=
c 3m= i 2m=
fce−= j 2m=
Solution:

The initial guesses are:
T
BD
1kN= T
BC
1kN=
A
x
1kN= A
y
1kN= A
z
1kN=
The vectors
408
© 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 5
r
1
0
ab+ f+
d
⎛
⎜
⎜
⎝
⎞
⎟

⎟
⎠
= r
2
0
ab+ c+
d
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
r
BC
i
a−
hg−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= r

BD
j−
a−
hg−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= r
AB
0
a
g
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
u
BC
r
BC

r
BC
= u
BD
r
BD
r
BD
= i
1
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= j
0
1
0
⎛
⎜
⎜
⎝
⎞
⎟

⎟
⎠
= k
0
0
1
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
Given A
x
iA
y
j+ A
z
k+ F
1
k− F
2
k− T
BD
u
BD
+ T

BC
u
BC
+ 0=
r
AB
T
BD
u
BD
T
BC
u
BC
+
()
× r
1
F
1
− k
()
×+ r
2
F
2
− k
()
×+ 0=
T

BD
T
BC
A
x
A
y
A
z
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find T
BD
T
BC
, A

x
, A
y
, A
z
,
()
=
T
BD
T
BC
⎛
⎜
⎝
⎞
⎟
⎠
17
17
⎛
⎜
⎝
⎞
⎟
⎠
kN=
A
x
A

y
A
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
0
11.333
15.667−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
Problem 5-77
The hatch door has a weight W and center of gravity at G. If the force
F
applied to the handle
at C has coordinate direction angles of

α
,
β
and
γ
,

determine the magnitude of
F
needed to
hold the door slightly open as shown. The hinges are in proper alignment and exert only force
reactions on the door. Determine the components of these reactions if A exerts only x and z
components of force and B exerts x, y, z force components.
Given:
W 80 lb=
α
60 deg=
409
© 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 5
β
45 deg=
γ
60 deg=
a 3ft=
b 2ft=
c 4ft=
d 3ft=

Solution:
Initial Guesses:
A
x
1lb= A
z
1lb= F 1lb=
B
x
1lb= B
y
1lb= B
z
1lb=
Given
A
x
0
A
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠

B
x
B
y
B
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
+ F
cos
α
()
cos
β
()
cos
γ
()
⎛
⎜
⎜
⎜

⎝
⎞
⎟
⎟
⎟
⎠
+
0
0
W−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+ 0=
ab+
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠

F
cos
α
()
cos
β
()
cos
γ
()
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
⎡
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎦

×
a
c
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
0
W−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
×+
0
cd+
0
⎛
⎜
⎜

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

B
z
F
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find A
x
A
z
, B
x
, B
y
, B

z
, F,
()
=
A
x
A
z
⎛
⎜
⎝
⎞
⎟
⎠
96.5−
13.7−
⎛
⎜
⎝
⎞
⎟
⎠
lb=
B
x
B
y
B
z
⎛

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
48.5
67.9−
45.7
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
F 96lb=
410
© 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 5
Problem 5-78
The hatch door has a weight W and center of gravity at G. If the force
F

applied to the handle at
C has coordinate direction angles
α
,
β
,
γ
determine the magnitude of
F
needed to hold the door
slightly open as shown. If the hinge at A becomes loose from its attachment and is ineffective,
what are the x, y, z components of reaction at hinge B?
Given:
W 80 lb=
α
60 deg=
β
45 deg=
γ
60 deg=
a 3ft=
b 2ft=
c 4ft=
d 3ft=
Solution:
Σ
M
y
= 0;
FW

a
cos
γ
()
ab+()
= F 96lb=
Σ
F
x
= 0;

B
x
F cos
α
()
+ 0=
B
x
F− cos
α
()
= B
x
48− lb=
Σ
F
y
= 0;

B
y
F cos
β
()
+ 0=
B
y
F− cos
β
()
= B
y
67.9− lb=
Σ
F
z
= 0;
B
z
W− F cos
γ
()
+ 0=
B
z
WFcos
γ
()
−= B

z
32lb=
Σ
M
x
= 0;
M
Bx
Wd+ Fcos
γ
()
cd+()− 0=
M
Bx
W− dFcos
γ
()
cd+()+= M
Bx
96lb ft⋅=
Σ
M
z
= 0;
M
Bz
F cos
α
()
cd+()+ F cos

β
()
ab+()+ 0=
M
Bz
F− cos
α
()
cd+()F cos
β
()
ab+()−= M
Bz
675− lb ft⋅=
411
© 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 5
Problem 5-79
The bent rod is supported at A, B, and C by smooth journal bearings. Compute the x, y, z
components of reaction at the bearings if the rod is subjected to forces
F
1

and
F
2
.
F

1
lies in the
y-z plane. The bearings are in proper alignment and exert only force reactions on the rod.
Given:
F
1
300 lb= d 3ft=
F
2
250 lb= e 5ft=
a 1ft=
α
30 deg=
b 4ft=
β
45 deg=
c 2ft=
θ
45 deg=
Solution:
The initial guesses:
A
x
100 lb= A
y
200 lb=
B
x
300 lb= B
z

400 lb=
C
y
500 lb= C
z
600 lb=
Given
A
x
B
x
+ F
2
cos
β
()
sin
α
()
+ 0=
A
y
C
y
+ F
1
cos
θ
()
− F

2
cos
β
()
cos
α
()
+ 0=
B
z
C
z
+ F
1
sin
θ
()
− F
2
sin
β
()
− 0=
F
1
cos
θ
()
ab+()F
1

sin
θ
()
cd+()+ B
z
d− A
y
b− 0=
A
x
bC
z
e+ 0=
A
x
cd+()B
x
d+ C
y
e− 0=
412
© 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 5
A
x
A
y
B

x
B
z
C
y
C
z
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find A
x
A

y
, B
x
, B
z
, C
y
, C
z
,
()
=
A
x
A
y
⎛
⎜
⎝
⎞
⎟
⎠
632.883
141.081−
⎛
⎜
⎝
⎞
⎟
⎠

lb=
B
x
B
z
⎛
⎜
⎝
⎞
⎟
⎠
721.271−
895.215
⎛
⎜
⎝
⎞
⎟
⎠
lb=
C
y
C
z
⎛
⎜
⎝
⎞
⎟
⎠

200.12
506.306−
⎛
⎜
⎝
⎞
⎟
⎠
lb=
Problem 5-80
The bent rod is supported at A, B, and C by smooth journal bearings. Determine the magnitude
of
F
2
which will cause the reaction
C
y
at the bearing C to be equal to zero. The bearings are in
proper alignment and exert only force reactions on the rod.
Given:
F
1
300 lb= d 3ft=
C
y
0lb= e 5ft=
a 1ft=
α
30 deg=
b 4ft=

β
45 deg=
c 2ft=
θ
45 deg=
Solution:
The initial guesses:
A
x
100 lb= A
y
200 lb=
B
x
300 lb= B
z
400 lb=
F
2
500 lb= C
z
600 lb=
Given
A
x
B
x
+ F
2
cos

β
()
sin
α
()
+ 0=
413
© 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 5
A
y
C
y
+ F
1
cos
θ
()
− F
2
cos
β
()
cos
α
(
)
+ 0=

B
z
C
z
+ F
1
sin
θ
()
− F
2
sin
β
()
− 0=
F
1
cos
θ
()
ab+()F
1
sin
θ
()
cd+()+ B
z
d− A
y
b− 0=

A
x
bC
z
e+ 0=
A
x
cd+()B
x
d+ C
y
e− 0=
A
x
A
y
B
x
B
z
C
z
F
2
⎛
⎜
⎜
⎜
⎜
⎜

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find A
x
A
y
, B
x
, B
z
, C
z
, F
2
,
()
= F
2

673.704 lb=
Problem 5-81
Determine the tension in cables BD and CD and the x, y, z components of reaction at the
ball-and-socket joint at A.
Given:
F 300 N=
a 3m=
b 1m=
c 0.5 m=
d 1.5 m=
Solution:
r
BD
b−
d
a
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
414
© 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 5

r
CD
b−
d−
a
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
Initial Guesses:
T
BD
1N= T
CD
1N= A
x
1N= A
y
1N= A
z
1N=
Given
A
x
A

y
A
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
T
BD
r
BD
r
BD
+ T
CD
r
CD
r
CD
+
0
0
F−
⎛

⎜
⎜
⎝
⎞
⎟
⎟
⎠
+ 0=
d
d−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
T
BD
r
BD
r
BD
⎛
⎜
⎝
⎞
⎟

⎠
×
d
d
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
T
CD
r
CD
r
CD
⎛
⎜
⎝
⎞
⎟
⎠
×+
dc−
0
0
⎛

⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
0
F−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
×+ 0=
T
BD
T
CD
A
x
A
y
A
z
⎛

⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find T
BD
T
CD
, A
x
, A
y
, A
z
,
()
=
T
BD

T
CD
⎛
⎜
⎝
⎞
⎟
⎠
116.7
116.7
⎛
⎜
⎝
⎞
⎟
⎠
N=
A
x
A
y
A
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟

⎟
⎟
⎠
66.7
0
100
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
Problem 5-82
Determine the tensions in the cables and the components of reaction acting on the smooth collar at
A necessary to hold the sign of weight W in equilibrium. The center of gravity for the sign is at G.
Given:
W 50 lb= f 2.5 ft=
a 4ft= g 1ft=
b 3ft= h 1ft=
415
© 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 5
c 2ft= i 2ft=
d 2ft= j 2ft=
e 2.5 ft= k 3ft=

Solution:
The initial guesses are:
T
BC
10 lb= A
x
10 lb= M
Ax
10 lb ft⋅=
T
DE
10 lb= A
y
10 lb= M
Ay
10 lb ft⋅=
Given
ak−()
T
DE
ak−()
2
i
2
+ j
2
+
b− d+()
T
BC

db−()
2
i
2
+ c
2
+
+ A
x
+ 0=
i−
T
DE
ak−()
2
i
2
+ j
2
+
i
T
BC
db−()
2
i
2
+ c
2
+

− A
y
+ 0=
j
T
DE
ak−()
2
i
2
+ j
2
+
c
T
BC
db−()
2
i
2
+ c
2
+
+ W− 0=
M
Ax
T
DE
j
i

ak−()
2
i
2
+ j
2
+
+ cT
BC
i
db−()
2
i
2
+ c
2
+
+ Wi− 0=
M
Ay
T
DE
k
j
ak−()
2
i
2
+ j
2

+
− T
BC
c
d
db−()
2
i
2
+ c
2
+
+ Wk f−()+ 0=
416
© 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 5
T
DE
− a
i
ak−()
2
i
2
+ j
2
+
T

BC
b
i
db−()
2
i
2
+ c
2
+
+ 0=
M
Ax
M
Ay
T
BC
T
DE
A
x
A
y
⎛
⎜
⎜
⎜
⎜
⎜
⎜

⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find M
Ax
M
Ay
, T
BC
, T
DE
, A
x
, A
y
,
()
=
T
BC

T
DE
⎛
⎜
⎝
⎞
⎟
⎠
42.857
32.143
⎛
⎜
⎝
⎞
⎟
⎠
lb=
A
x
A
y
⎛
⎜
⎝
⎞
⎟
⎠
3.571
50
⎛

⎜
⎝
⎞
⎟
⎠
lb=
M
Ax
M
Ay
⎛
⎜
⎝
⎞
⎟
⎠
2.698 10
13−
×
17.857−
⎛
⎜
⎝
⎞
⎟
⎠
lb ft⋅=
Problem 5-83
The member is supported by a pin at A and a cable BC. If the load at D is W, determine the x, y,
z components of reaction at these supports.

Units Used:
kip 10
3
lb=
Given:
W 300 lb=
a 1ft=
b 2ft=
c 6ft=
d 2ft=
e 2ft=
f 2ft=
417
© 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 5
Initial Guesses:
T
BC
1lb= A
x
1lb=
A
y
1lb= A
z
1lb=
M
Ay

1lbft= M
Az
1lbft=
Given
A
x
A
y
A
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
T
BC
b
2
c
2
+ ef+ a−()
2
+
ef+ a−

c−
b
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+
0
0
W−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+ 0=
0
M
Ay
M
Az
⎛
⎜

⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
e−
c
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
0
W−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠

×+
a−
0
b
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
T
BC
b
2
c
2
+ ef+ a−()
2
+
ef+ a−
c−
b
⎛
⎜
⎜
⎝
⎞
⎟

⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+ 0=
T
BC
A
x
A
y
A
z
M
Ay
M
Az
⎛
⎜
⎜
⎜
⎜
⎜
⎜

⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find T
BC
A
x
, A
y
, A
z
, M
Ay
, M
Az
,
()
= T
BC
1.05 kip=

A
x
A
y
A
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
450−
900
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
M
Ay

M
Az
⎛
⎜
⎝
⎞
⎟
⎠
600−
900−
⎛
⎜
⎝
⎞
⎟
⎠
lb ft⋅=
Problem 5-84
Determine the x, y, z components of reaction at the pin A and the tension in the cable BC
necessary for equilibrium of the rod.
418
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 5
Given:
F 350 lb= e 12 ft=
a 4ft= f 4ft=
b 5ft= g 10 ft=

c 4ft= h 4ft=
d 2ft= i 10 ft=
Solution:
Initial Guesses:
F
BC
1lb=
A
y
1lb=
M
Ay
1lbft⋅=
A
x
1lb=
A
z
1lb=
M
Az
1lbft⋅=
Given
0
M
Ay
M
Az
⎛
⎜

⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
d
c
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
gd−()
2
ec−()
2
+ f
2
+
gd−
ec−
f−

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+
a−
0
b
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
BC

ah−()
2
e
2
+ b
2
+
ah−
e−
b
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
×+
0=
A
x

A
y
A
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
F
gd−()
2
ec−()
2
+ f
2
+
gd−
ec−
f−
⎛
⎜
⎜
⎝
⎞

⎟
⎟
⎠
+
F
BC
ah−()
2
e
2
+ b
2
+
ah−
e−
b
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+ 0=
419
© 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 5

F
BC
A
x
A
y
A
z
M
Ay
M
Az
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟

⎟
⎠
Find F
BC
A
x
, A
y
, A
z
, M
Ay
, M
Az
,
()
= F
BC
101lb=
A
x
A
y
A
z
⎛
⎜
⎜
⎜
⎝

⎞
⎟
⎟
⎟
⎠
233.3−
140−
77.8
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
M
Ay
M
Az
⎛
⎜
⎝
⎞
⎟
⎠
388.9−
93.3
⎛

⎜
⎝
⎞
⎟
⎠
lb ft⋅=
Problem 5-85
Rod AB is supported by a ball-and-socket joint at A and a cable at B. Determine the x, y, z
components of reaction at these supports if the rod is subjected to a vertical force
F
as shown.
Given:
F 50 lb=
a 2ft= c 2ft=
b 4ft= d 2ft=
Solution:
T
B
10 lb= A
x
10 lb=
A
y
10 lb= A
z
10 lb=
B
y
10 lb=
Given

Σ
F
x
= 0;

T
B
− A
x
+ 0=
Σ
F
y
= 0;

A
y
B
y
+ 0=
420
© 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 5
Σ
F
z
= 0;
F− A

z
+ 0=
Σ
M
Ax
= 0;
Fc() B
y
b()− 0=
Σ
M
Ay
= 0;
Fa() T
B
b()− 0=
Σ
M
Az
= 0;
B
y
a() T
B
c()− 0=
Solving,

T
B
A

x
A
y
A
z
B
y
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find T
B
A
x
, A
y
, A

z
, B
y
,
()
=
T
B
A
x
A
y
A
z
B
y
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟

⎟
⎠
25
25
25−
50
25
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
lb=
Problem 5-86
The member is supported by a square rod which fits loosely through a smooth square hole of
the attached collar at A and by a roller at B. Determine the x, y, z components of reaction at
these supports when the member is subjected to the loading shown.
Given:
M 50 lb ft⋅=
F
20

40−
30−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
a 2ft=
b 1ft=
c 2ft=
Solution:
Initial Guesses
A
x
1lb= A
y
1lb=
421
© 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 5
M
Ax
1lbft= M
Ay

1lbft=
M
Az
1lbft= B
z
1lb=
Given
A
x
A
y
0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
0
0
B
z
⎛
⎜
⎜
⎜

⎝
⎞
⎟
⎟
⎟
⎠
+ F+ 0=
M
Ax
M
Ay
M
Az
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
0
a
0
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
0
0
B
z
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
×+
0
ab+
c−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠

F×
0
0
M−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
+ 0=
A
x
A
y
M
Ax
M
Ay

M
Az
B
z
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find A
x
A
y
, M
Ax

, M
Ay
, M
Az
, B
z
,
()
=
A
x
A
y
⎛
⎜
⎝
⎞
⎟
⎠
20−
40
⎛
⎜
⎝
⎞
⎟
⎠
lb=
B
z

30lb=
M
Ax
M
Ay
M
Az
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
110
40
110
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb ft⋅=
Problem 5-87

The platform has mass M and center of mass located at G. If it is lifted using the three cables,
determine the force in each of these cables.
Units Used:
Mg 10
3
kg= kN 10
3
N= g 9.81
m
s
2
=
422
© 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 5
Given:
M 3Mg=
a 4m=
b 3m=
c 3m=
d 4m=
e 2m=
Solution:
The initial guesses are:
F
AC
10 N= F
BC

10 N= F
DE
10 N=
Given
bF
AC
()
a
2
b
2
+
cF
BC
()
a
2
c
2
+
− 0=
Mge F
AC
()
a
de+
a
2
b
2

+
− F
BC
ad e+()
a
2
c
2
+
− 0=
a
a
2
c
2
+
F
BC
bc+()Mgb− F
DE
b+ 0=
F
AC
F
BC
F
DE
⎛
⎜
⎜

⎜
⎝
⎞
⎟
⎟
⎟
⎠
Find F
AC
F
BC
, F
DE
,
()
=
F
AC
F
BC
F
DE
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟

⎟
⎠
Find F
AC
F
BC
, F
DE
,
()
=
F
AC
F
BC
F
DE
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
kN=
F
BC

423
© 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 5
The platform has a mass of M and center of mass located at G. If it is lifted using the three
cables, determine the force in each of the cables. Solve for each force by using a single
moment equation of equilibrium.
Units Used:
Mg 1000 kg=
kN 10
3
N=
g 9.81
m
s
2
=
Given:
M 2Mg= c 3m=
a 4m= d 4m=
b 3m= e 2m=
Solution:
r
BC
0
c−
a
⎛
⎜

⎜
⎝
⎞
⎟
⎟
⎠
= r
AC
0
b
a
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
r
AD
e− d−
b
0
⎛
⎜
⎜
⎝
⎞

⎟
⎟
⎠
= r
BD
d− e−
c−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
First find F
DE
.
F
DE
de+()Mgd− 0= F
DE
Mgd
de+
= F
DE
4.1 s
2

kN=
Σ
M
y
' = 0;
Next find F
BC
.
Guess F
BC
1kN=
Given
e
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
0
M− g
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
×
ed+
c
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
BC
r
BC
r
BC
⎛
⎜
⎝
⎞
⎟
⎠
×+
⎡

⎢
⎢
⎣
⎤
⎥
⎥
⎦
r
AD
0= F
BC
Find F
BC
()
=F
BC
Find F
BC
()
=
F
BC
kN=F
BC
Now find F
AC
.
Guess F
AC
1kN=

424
Problem 5-88
© 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 5
Given
e
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0
0
M− g
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
×

ed+
b−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
AC
r
AC
r
AC
⎛
⎜
⎝
⎞
⎟
⎠
×+
⎡
⎢
⎢
⎣
⎤
⎥

⎥
⎦
r
BD
0= F
AC
Find F
AC
()
=F
AC
Find F
AC
()
=
F
AC
kN=F
AC
Problem 5-89
The cables exert the forces shown on the pole. Assuming the pole is supported by a
ball-and-socket joint at its base, determine the components of reaction at A. The forces
F
1
and
F
2
lie in a horizontal plane.
Given:
F

1
140 lb=
F
2
75 lb=
θ
30 deg=
a 5ft=
b 10 ft=
c 15 ft=
Solution:
The initial guesses are
T
BC
100 lb= T
BD
100 lb= A
x
100 lb= A
y
100 lb= A
z
100 lb=
Given
F
1
cos
θ
()
F

2
+
()
cT
BC
a
c
a
2
b
2
+ c
2
+
⎛
⎜
⎝
⎞
⎟
⎠
− T
BD
a
c
a
2
c
2
+
⎛

⎜
⎝
⎞
⎟
⎠
− 0=
F
1
sin
θ
()
cbT
BC
c
a
2
b
2
+ c
2
+
⎛
⎜
⎝
⎞
⎟
⎠
− 0=
425
© 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.