Engineering Mechanics - Statics Episode 1 Part 4 ppsx

Engineering Mechanics - Statics Chapter 2
Given:
F 100 lb=
a 2ft=
b 8ft=
c 6ft=
d 4ft=
e 2ft=
Solution:

r
CD
c−
b
e
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= u
CD
r
CD
r
CD
= r
CB

c−
d−
e
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= u
CB
r
CB
r
CB
=
F
BC
Fu
CD
()
u
CB
⋅= F
BC
10.5 lb=
Problem 2-129
Determine the angle

θ
between pipe segments BA and BC.
Given:

F 100 lb=
a 3ft=
b 8ft=
c 6ft=
d 4ft=
e 2ft=
Solution:

r
BC
c
d
e−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= r
BA
a−
0
0

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
θ
acos
r
BC
r
BA
⋅
r
BC
r
BA
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
143.3 deg=
121

© 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
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Engineering Mechanics - Statics Chapter 2
Problem 2-130
Determine the angles
θ


and
φ
made between the axes OA of the flag pole and AB and AC,
respectively, of each cable.
Given:
F
B
55 N= c 2m=
F
c
40 N= d 4m=
a 6m= e 4m=
b 1.5 m= f 3m=
Solution:
r
AO
0
e−
f−
⎛
⎜

⎜
⎝
⎞
⎟
⎟
⎠
= r
AB
b
e−
af−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= r
AC
c−
e−
df−
⎛
⎜
⎜
⎝
⎞
⎟

⎟
⎠
=
θ
acos
r
AB
r
AO
⋅
r
AB
r
AO
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
74.4 deg=
φ
acos
r
AC
r
AO
⋅

r
AC
r
AO
⎛
⎜
⎝
⎞
⎟
⎠
=
φ
55.4 deg=
Problem 2-131
Determine the magnitude and coordinate direction angles of
F
3
so that resultant of the three
forces acts along the positive y axis and has magnitude F
R
.
122
© 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 2
Given:
F
R
600 lb=

F
1
180 lb=
F
2
300 lb=
φ
40 deg=
θ
30 deg=
Solution:
The initial guesses
:
F
3
100 lb=
β
30 deg=
α
10 deg=
γ
60 deg=
Given
F
Rx
=
Σ
F
x
;

F
1
− F
2
cos
θ
()
sin
φ
()
+ F
3
cos
α
()
+ 0=
F
Ry

=
Σ
F
y
;
F
2
cos
θ
()
cos

φ
()
F
3
cos
β
()
+ F
R
=
F
Rz
=
Σ
F
z
;
F
2
− sin
θ
()
F
3
cos
γ
()
+ 0=
cos
α

()
2
cos
β
()
2
+ cos
γ
()
2
+ 1=
Solving:
F
3
α
β
γ
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
Find F

3
α
,
β
,
γ
,
()
=
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
88.3
20.6
69.5
⎛
⎜
⎜
⎝
⎞

⎟
⎟
⎠
deg= F
3
428.3 lb=
123
© 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 2
Determine the magnitude and coordinate direction angles of
F
3
so that resultant of the three
forces is zero.
Given:
F
1
180 lb=
F
2
300 lb=
φ
40 deg=
θ
30 deg=
Solution:
The initial guesses
:

α
10 deg=
β
30 deg=
γ
60 deg=
F
3
100 lb=
Given
F
Rx
=
Σ
F
x
;
F
1
− F
2
cos
θ
()
sin
φ
()
+ F
3
cos

α
()
+ 0=
F
Ry
=
Σ
F
y
;
F
2
cos
θ
()
cos
φ
()
F
3
cos
β
()
+ 0=
F
Rz
=
Σ
F
z

;
F
2
− sin
θ
()
F
3
cos
γ
()
+ 0=
cos
α
()
2
cos
β
()
2
+ cos
γ
()
2
+ 1=
Solving:
F
3
α
β

γ
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
Find F
3
α
,
β
,
γ
,
()
=
α
β
γ
⎛
⎜
⎜
⎜

⎝
⎞
⎟
⎟
⎟
⎠
87
142.9
53.1
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
deg= F
3
249.6 lb=
124
Problem 2-132
© 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 2
Problem 2-133
Resolve the force
F
into two components, one acting parallel and the other acting perpendicular

to the u axis.
Given:

F 600 lb=
θ
1
60 deg=
θ
2
20 deg=
Solution:

F
perpendicular
F cos
θ
1
θ
2
−
()
=
F
perpendicular
460lb=
F
parallel
F sin
θ
1

θ
2
−
()
=
F
parallel
386lb=
Problem 2-134
The force
F
has a magnitude F and acts at the midpoint C of the thin rod. Express the force as
a Cartesian vector.
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 2
Given:
F 80 lb=
a 2ft=
b 3ft=
c 6ft=
Solution:
r
CO
b−
2
a
2

c−
2
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
=
F
v
F
r
CO
r
CO
= F
v
34.3−
22.9

68.6−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
Problem 2-135
Determine the magnitude and direction of the resultant
F
R

=
F
1
+
F
2
+
F
3
of the three forces by
first finding the resultant
F
' =
F
1

+
F
3
and then forming
F
R
=
F
' +
F
2
. Specify its direction
measured counterclockwise from the positive x axis.
126
© 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 2
Given:
F
1
80 N=
F
2
75 N=
F
3
50 N=
θ
1

30 deg=
θ
2
30 deg=
θ
3
45 deg=
Solution:

F
1v
F
1
sin
θ
1
()
−
cos
θ
1
()
⎛
⎜
⎝
⎞
⎟
⎠
= F
2v

F
2
cos
θ
2
θ
3
+
()
sin
θ
2
θ
3
+
()
⎛
⎜
⎝
⎞
⎟
⎠
= F
3v
F
3
cos
θ
3
()

sin
θ
3
()
⎛
⎜
⎝
⎞
⎟
⎠
=
F' F
1v
F
3v
+= F'
4.6−
104.6
⎛
⎜
⎝
⎞
⎟
⎠
N= i
1
0
⎛
⎜
⎝

⎞
⎟
⎠
= j
0
1
⎛
⎜
⎝
⎞
⎟
⎠
=
F
R
F' F
2v
+= F
R
14.8
177.1
⎛
⎜
⎝
⎞
⎟
⎠
N= F
R
177.7 N=

θ
atan
F
R
j
F
R
i
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
85.2 deg=
Problem 2-136
The leg is held in position by the quadriceps AB, which is attached to the pelvis at A. If the force
exerted on this muscle by the pelvis is
F
, in the direction shown, determine the stabilizing force
component acting along the positive y axis and the supporting force component acting along the
negative x axis.
Given:
F 85 N=
127
© 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 2
θ
1
55 deg=
θ
2
45 deg=
Solution:
F
x
F cos
θ
1
θ
2
−
()
=
F
x
83.7 N=
F
y
F sin
θ
1
θ
2
−
()

=
F
y
14.8 N=
Problem 2-137
Determine the magnitudes of the projected components of the force
F
in the direction of the
cables AB and AC .
Given:
F
60
12
40−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
a 3m=
b 1.5 m=
c 1m=
d 0.75 m=
e 1m=
128
© 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 2
Find the unit vectors, then use the dot product
r
AB
a−
d−
e
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= r
AB
3−
0.8−
1
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠

m= u
AB
r
AB
r
AB
= u
AB
0.9−
0.2−
0.3
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
r
AC
a−
c
b
⎛
⎜
⎜
⎝
⎞

⎟
⎟
⎠
= r
AC
3−
1
1.5
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
m= u
AC
r
AC
r
AC
= u
AC
0.9−
0.3
0.4
⎛
⎜
⎜

⎝
⎞
⎟
⎟
⎠
=
F
AB
Fu
AB
= F
AB
78.5−
19.6−
26.2
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N= F
AC
Fu
AC
= F
AC
72.9−

24.3
36.4
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
Problem 2-138
Determine the magnitude and
coordinate direction angles of
F
3
so
that resultant of the three forces is
zero.
Given:
F
1
180 lb=
φ
40 deg=
F
2
300 lb=
θ
30 deg=

Solution:
The initial guesses
:
α
10 deg=
β
30 deg=
γ
60 deg=
F
3
100 lb=
Given
F
Rx

=
Σ
F
x
;
F
1
− F
2
cos
θ
()
sin
φ

()
+ F
3
cos
α
()
+ 0=
F
Ry
=
Σ
F
y
;
F
2
cos
θ
()
cos
φ
()
F
3
cos
β
()
+ 0=
129
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 2
F
Rz
=
Σ
F
z
;
F
2
− sin
θ
()
F
3
cos
γ
()
+ 0=
cos
α
()
2
cos
β
()
2

+ cos
γ
()
2
+ 1=
Solving:
F
3
α
β
γ
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
Find F
3
α
,
β
,
γ

,
()
=
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
87
142.9
53.1
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
deg= F
3
249.6 lb=

Problem 2-139
Determine the angles
θ
and
φ
so that the resultant force is directed along the positive x axis
and has magnitude F
R.
.
Given:
F
1
30 lb=
F
2
30 lb=
F
R
20 lb=
Solution:

Initial Guesses:
θ
20 deg=
φ
20 deg=
Given
F
1
sin

φ
()
F
2
sin
θ
()
=
F
R
2
F
1
2
F
2
2
+ 2 F
1
F
2
cos 180 deg
θ
−
φ
−
()
−=
130
© 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 2
θ
φ
⎛
⎜
⎝
⎞
⎟
⎠
Find
θφ
,
()
=
θ
70.5 deg=
φ
70.5 deg=
Problem 2-140
Determine the magnitude of the resultant force and its direction measured counterclockwise
from the x axis.
Given:
F
1
300 lb=
F
2
200 lb=

θ
1
40 deg=
θ
2
100 deg=
Solution:
F
Rx
F
1
cos 180 deg
θ
2
−
()
F
2
cos
θ
1
()
+= F
Rx
205.3 lb=
F
Ry
F
1
sin 180 deg

θ
2
−
()
F
2
sin
θ
1
()
−= F
Ry
166.9 lb=
F
R
F
Rx
2
F
Ry
2
+= F
R
265lb=
θ
atan
F
Ry
F
Rx

⎛
⎜
⎝
⎞
⎟
⎠
=
θ
39.1 deg=
131
© 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 3
Problem 3-1
Determine the magnitudes of
F
1

and
F
2

so that the particle is in equilibrium.
Given:
F 500 N=
θ
1
45 deg=
θ

2
30deg=
Solution:
Initial Guesses
F
1
1N= F
2
1N=
Given
+
→
Σ
F
x
= 0;
F
1
cos
θ
1
()
F
2
cos
θ
2
()
+ F− 0=
+

↑
Σ
F
y
= 0;
F
1
sin
θ
1
()
F
2
sin
θ
2
()
− 0=
F
1
F
2
⎛
⎜
⎝
⎞
⎟
⎠
Find F
1

F
2
,
()
=
F
1
F
2
⎛
⎜
⎝
⎞
⎟
⎠
259
366
⎛
⎜
⎝
⎞
⎟
⎠
N=
Problem 3-2
Determine the magnitude and direction
θ

of
F

so that
the particle is in equilibrium.
Units Used:
kN 10
3
N=
Given:
F
1
7kN=
F
2
3kN=
c 4=
d 3=
132
© 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 3
Solution:
The initial guesses:
F 1kN=
θ
30deg=
Given
Equations of equilibrium:
+
→
Σ

F
x
= 0;
d−
c
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
1
F cos
θ
()
+ 0=
+
↑
Σ
F
y
= 0
;
c
c

2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
1
F
2
− F sin
θ
()
− 0=
F
θ
⎛
⎜
⎝
⎞
⎟
⎠
Find F
θ
,
()

= F 4.94 kN=
θ
31.8 deg=
Problem 3-3
Determine the magnitude of
F
and the orientation
θ
of the force
F
3
so that the particle is in equilibrium.
Given:
F
1
700 N=
F
2
450 N=
F
3
750 N=
θ
1
15 deg=
θ
2
30 deg=
Solution:
Initial Guesses:

F 1N=
θ
10deg=
Given
+
→
Σ
F
x
= 0;
F
1
cos
θ
1
()
F
2
sin
θ
2
()
− F
3
cos
θ
()
− 0=
133
© 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 3
+
↑
Σ
F
y
= 0;
FF
2
cos
θ
2
()
+ F
1
sin
θ
1
()
+ F
3
sin
θ
()
− 0=
F
θ
⎛

⎜
⎝
⎞
⎟
⎠
Find F
θ
,
()
= F 28.25 N=
θ
53.02 deg=
Problem 3-4
Determine the magnitude and angle
θ
of
F
so that the particle is in equilibrium.
Units Used:
kN 10
3
N=
Given:
F
1
4.5 kN=
F
2
7.5 kN=
F

3
2.25 kN=
α
60 deg=
φ
30 deg=
Solution:
Guesses:
F 1kN=
θ
1=
Given
Equations

of

Equilibrium:
+
→
Σ
F
x
= 0;
F cos
θ
()
F
2
sin
φ

()
− F
1
− F
3
cos
α
()
+ 0=
+
↑
Σ
F
y
= 0;
F sin
θ
()
F
2
cos
φ
()
− F
3
sin
α
()
− 0=
F

θ
⎛
⎜
⎝
⎞
⎟
⎠
Find F
θ
,
()
= F 11.05 kN=
θ
49.84 deg=
134
© 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 3
Problem 3-5
The members of a truss are connected to the gusset plate. If the forces are concurrent at point O,
determine the magnitudes of
F
and
T
for equilibrium.
Units Used:
kN 10
3
N=

Given:
F
1
8kN=
F
2
5kN=
θ
1
45 deg=
θ
30 deg=
Solution:
+
→
Σ
F
x
= 0;
T− cos
θ
()
F
1
+ F
2
sin
θ
1
()

+ 0=
T
F
1
F
2
sin
θ
1
()
+
cos
θ
()
=
T 13.3 kN=
+
↑
Σ
F
y
= 0;
FTsin
θ
()
− F
2
cos
θ
1

()
− 0=
FTsin
θ
()
F
2
cos
θ
1
()
+=
F 10.2 kN=
Problem 3-6
The gusset plate is subjected to the forces of four members. Determine the force in member B and its
proper orientation
θ
for equilibrium. The forces are concurrent at point O.
Units Used:
kN 10
3
N=
135
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 3
Given:
F 12 kN=
F

1
8kN=
F
2
5kN=
θ
1
45 deg=
Solution:
Initial Guesses
T 1kN=
θ
10deg=
Given
+
→
Σ
F
x
= 0;
F
1
Tcos
θ
()
− F
2
sin
θ
1

()
+ 0=
+
↑
Σ
F
y
= 0;
T− sin
θ
()
F
2
cos
θ
1
()
− F+ 0=
T
θ
⎛
⎜
⎝
⎞
⎟
⎠
Find T
θ
,
()

= T 14.31 kN=
θ
36.27 deg=
Problem 3-7
Determine the maximum weight of the engine that
can be supported without exceeding a tension of
T
1
in chain AB and
T
2
in chain AC.
Given:
θ
30 deg=
T
1
450 lb=
T
2
480 lb=
Solution:
Initial Guesses
F
AB
T
1
=
F
AC

T
2
=
W 1lb=
136
© 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 3
Given
Assuming cable AB reaches the maximum
tension F
AB
= T
1
.
+
→
F
AC
cos
θ
()
F
AB
− 0=
Σ
F
x
= 0;

+
↑
Σ
F
y
= 0;
F
AC
sin
θ
()
W− 0=
F
AC1
W
1
⎛
⎜
⎝
⎞
⎟
⎠
Find F
AC
W,
()
= W
1
259.81 lb=
Given

Assuming cable AC reaches the maximum tension F
AC
= T
2
.
+
→
F
AC
cos
θ
()
F
AB
− 0=
Σ
F
x
= 0;
+
↑
Σ
F
y
= 0;
F
AC
sin
θ
()

W− 0=
F
AB2
W
2
⎛
⎜
⎝
⎞
⎟
⎠
Find F
AB
W,
()
= W
2
240.00 lb=
W min W
1
W
2
,
()
= W 240.00 lb=
Problem 3-8
The engine of mass M is suspended from a vertical chain at A. A second
chain is wrapped around the engine and held in position by the spreader
bar BC. Determine the compressive force acting along the axis of the bar
and the tension forces in segments BA and CA of the chain. Hint:

Analyze equilibrium first at A, then at B.
Units Used:
kN 10
3
N=
Given:
M 200 kg=
θ
1
55 deg=
g 9.81
m
s
2
=
137
© 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 3
Solution:
Initial guesses:
F
BA
1kN= F
CA
2kN=
Given
Point A
+

→
Σ
F
x
= 0;
F
CA
cos
θ
1
()
F
BA
cos
θ
1
()
− 0=
+
↑
Σ
F
y
= 0;
Mg() F
CA
sin
θ
1
()

− F
BA
sin
θ
1
()
− 0=
F
BA
F
CA
⎛
⎜
⎝
⎞
⎟
⎠
Find F
BA
F
CA
,
()
=
F
BA
F
CA
⎛
⎜

⎝
⎞
⎟
⎠
1.20
1.20
⎛
⎜
⎝
⎞
⎟
⎠
kN=
At point B:
+
→
Σ
F
x
= 0;
F
BA
cos
θ
1
()
F
BC
− 0=
F

BC
F
BA
cos
θ
1
()
= F
BC
687 N=
Problem 3-9
Cords AB and AC can each sustain a maximum tension T. If the drum has weight W,
determine the smallest angle
θ

at

which they can be attached to the drum.
Given:
T 800 lb=
W 900 lb=
Solution:
+
↑
Σ
F
y
= 0;
W 2Tsin
θ

()
− 0=
θ
asin
W
2T
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
34.2 deg=
138
© 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 3
Problem 3-10
The crate of weight W is hoisted using the ropes AB and AC. Each rope can withstand a
maximum tension T before it breaks. If AB always remains horizontal, determine the smallest
angle
θ
to which the crate can be hoisted.
Given:
W 500 lb=
T 2500 lb=
Solution:

Case 1:

Assume
T
AB
T=
The initial guess
θ
30 deg= T
AC
2000 lb=
Given
+
→
Σ
F
x
= 0;
T
AB
T
AC
cos
θ
()
− 0=
+
↑
Σ
F

y
= 0;
T
AC
sin
θ
()
W− 0=
T
AC1
θ
1
⎛
⎜
⎝
⎞
⎟
⎠
Find T
AC
θ
,
()
=
θ
1
11.31 deg= T
AC1
2550lb=
Case 1:

Assume
T
AC
T=
The initial guess
θ
30 deg= T
AB
2000 lb=
Given
+
→
Σ
F
x
= 0;
T
AB
T
AC
cos
θ
()
− 0=
+
↑
Σ
F
y

= 0;
T
AC
sin
θ
()
W− 0=
T
AB2
θ
2
⎛
⎜
⎝
⎞
⎟
⎠
Find T
AB
θ
,
()
=
θ
2
11.54 deg= T
AB2
2449lb=
θ
max

θ
1
θ
2
,
()
=
θ
11.54 deg=
139
© 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 3
Problem 3-11
Two electrically charged pith balls, each having mass M, are suspended from light threads of equal
length. Determine the resultant horizontal force of repulsion, F, acting on each ball if the measured
distance between them is r.
Given:
M 0.2 gm=
r 200 mm=
l 150 mm=
d 50 mm=
Solution:
The initial guesses
:
T 200 N= F 200 N=
Given
+
↑

Σ
F
x
= 0;
FT
rd−
2l
⎛
⎜
⎝
⎞
⎟
⎠
− 0=
+
→
Σ
F
y
= 0;
T
l
2
rd−
2
⎛
⎜
⎝
⎞
⎟

⎠
2
−
l
⎡
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎦
Mg− 0=
T
F
⎛
⎜
⎝
⎞
⎟
⎠
Find TF,()= F 1.13 10
3−
× N= T 0.00 N=
Problem 3-12
The towing pendant AB is subjected to the force
F
which is developed from a tugboat. Determine

the force that is in each of the bridles, BC and BD, if the ship is moving forward with constant
velocity.
Units Used:
kN 10
3
N=
140
© 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 3
Given:
F 50 kN=
θ
1
20 deg=
θ
2
30 deg=
Solution:
Initial guesses:
T
BC
1kN= T
BD
2kN=
Given
+
→
Σ

F
x
= 0;
T
BC
sin
θ
2
()
T
BD
sin
θ
1
()
− 0=
+
↑
Σ
F
y
= 0;
T
BC
cos
θ
2
()
T
BD

cos
θ
1
()
+ F− 0=
T
BC
T
BD
⎛
⎜
⎝
⎞
⎟
⎠
Find T
BC
T
BD
,
()
=
T
BC
T
BD
⎛
⎜
⎝
⎞

⎟
⎠
22.32
32.64
⎛
⎜
⎝
⎞
⎟
⎠
kN=
Problem 3-13
Determine the stretch in each s
p
rin
g
for
141
equilibrium of the block of mass M. The
springs are shown in the equilibrium position.
© 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 3
Given:
M 2kg=
a 3m=
b 3m=
c 4m=
k

AB
30
N
m
=
k
AC
20
N
m
=
k
AD
40
N
m
=
g 9.81
m
s
2
=
Solution:
The initial guesses:
F
AB
1N= F
AC
1N=
Given

+
→
F
AB
c
a
2
c
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
AC
b
a
2
b
2
+
⎛
⎜
⎝
⎞
⎟
⎠

− 0=
Σ
F
x
= 0;
+
↑
F
AC
a
a
2
b
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
AB
a
a
2
c
2
+
⎛

⎜
⎝
⎞
⎟
⎠
+ Mg− 0=
Σ
F
y
= 0;
F
AC
F
AB
⎛
⎜
⎝
⎞
⎟
⎠
Find F
AC
F
AB
,
()
=
F
AC
F

AB
⎛
⎜
⎝
⎞
⎟
⎠
15.86
14.01
⎛
⎜
⎝
⎞
⎟
⎠
N=
x
AC
F
AC
k
AC
= x
AC
0.79 m=
x
AB
F
AB
k

AB
= x
AB
0.47 m=
142
© 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 3
Problem 3-14
The unstretched length of spring AB is
δ
. If the block is held in the equilibrium position shown,
determine the mass of the block at D.
Given:
δ
2m=
a 3m=
b 3m=
c 4m=
k
AB
30
N
m
=
k
AC
20
N

m
=
k
AD
40
N
m
=
g 9.81
m
s
2
=
Solution:
F
AB
k
AB
a
2
c
2
+
δ
−
()
=
The initial guesses:
m
D

1kg= F
AC
1N=
Given
+
→
F
AB
c
a
2
c
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
AC
b
a
2
b
2
+
⎛
⎜

⎝
⎞
⎟
⎠
− 0=
Σ
F
x
= 0;
+
↑
F
AC
a
a
2
b
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
AB
a
a
2

c
2
+
⎛
⎜
⎝
⎞
⎟
⎠
+ m
D
g− 0=
Σ
F
y
= 0;
F
AC
m
D
⎛
⎜
⎝
⎞
⎟
⎠
Find F
AC
m
D

,
()
= F
AC
101.8 N= m
D
12.8 kg=
143
© 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 3
Problem 3-15
The springs AB and BC have stiffness k and unstretched
lengths l/2. Determine the horizontal force
F
applied to
the cord which is attached to the small pulley B so that
the displacement of the pulley from the wall is d.
Given:
l 6m=
k 500
N
m
=
d 1.5 m=
Solution:
Tk
l
2

⎛
⎜
⎝
⎞
⎟
⎠
2
d
2
+
l
2
−
⎡
⎢
⎣
⎤
⎥
⎦
= T 177.05 N=
+
→
Σ
F
x
= 0;
d
d
2
l

2
⎛
⎜
⎝
⎞
⎟
⎠
2
+
2T()F− 0=
F
d
l
2
⎛
⎜
⎝
⎞
⎟
⎠
2
d
2
+
2T()= F 158.36 N=
Problem 3-16
The springs AB and BC have stiffness k and an unstretched length of l. Determine the
displacement d of the cord from the wall when a force F is applied to the cord.
144
© 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 3
Given:
l 6m=
k 500
N
m
=
F 175 N=
Solution:
The initial guesses:
d 1m= T 1N=
Given
+
→
Σ
F
x
= 0;
F− 2T()
d
d
2
l
2
⎛
⎜
⎝
⎞

⎟
⎠
2
+
+ 0=
Spring Tkd
2
l
2
⎛
⎜
⎝
⎞
⎟
⎠
2
+
l
2
−
⎡
⎢
⎣
⎤
⎥
⎦
=
T
d
⎛

⎜
⎝
⎞
⎟
⎠
Find Td,()= T 189.96 N= d 1.56 m=
Problem 3-17
Determine the force in each cable and the force
F
needed to hold the lamp of mass M in the position
shown. Hint: First analyze the equilibrium at B; then, using the result for the force in BC, analyze the
equilibrium at C.
Given:
M 4kg=
θ
1
30 deg=
θ
2
60 deg=
θ
3
30 deg=
Solution:
Initial guesses:
145
© 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.