Engineering Mechanics - Statics Episode 1 Part 3 pptx

Engineering Mechanics - Statics Chapter 2
α
acos
da−
r
⎛
⎜
⎝
⎞
⎟
⎠
=
α
76.9 deg=
β
acos
eb−
r
⎛
⎜
⎝
⎞
⎟
⎠
=
β
142deg=
γ
acos
fc−
r

⎛
⎜
⎝
⎞
⎟
⎠
=
γ
124deg=
Problem 2-81
A position vector extends from the origin to point A(a, b, c). Determine the angles
α
,
β
,
γ
which
the tail of the vector makes with the x, y, z axes, respectively.
Given:

a 2m= b 3m= c 6m=
Solution:
r
a
b
c
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
= r
2
3
6
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
m=
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠

acos
r
r
⎛
⎜
⎝
⎞
⎟
⎠
=
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
73.4
64.6
31.0
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
deg=
Problem 2-82
Express the position vector
r
in Cartesian vector form; then determine its magnitude and
coordinate direction angles.
Given:
a 4m=
81
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Engineering Mechanics - Statics Chapter 2
b 8m=
c 3m=
d 4m=
Solution:

r
c−
d− b−
a
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
= r
3−
12−
4
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
m= r 13 m=
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠

acos
r
r
⎛
⎜
⎝
⎞
⎟
⎠
=
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
103.3
157.4
72.1
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
deg=
Problem 2-83
Express the position vector
r
in Cartesian vector form; then determine its magnitude and coordinate
direction angles.
Given:
a 8ft=
b 2ft=
c 5ft=
θ
30 deg=
φ
20 deg=
Solution:
r
c− cos
φ
()
sin
θ
()
accos
φ
()
cos

θ
()
−
bcsin
φ
()
+
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
= r
2.35−
3.93
3.71
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
ft= r 5.89 ft=

82
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Engineering Mechanics - Statics Chapter 2
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
acos
r
r
⎛
⎜
⎝
⎞
⎟
⎠
=
α
β

γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
113.5
48.2
51
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
deg=
Problem 2-84
Determine the length of the connecting rod AB by first formulating a Cartesian position vector
from A to B and then determining its magnitude.
Given:

b 16 in=
a 5in=

α
30 deg=
Solution:
r
asin
α
()
b+
a− cos
α
()
⎛
⎜
⎝
⎞
⎟
⎠
= r
18.5
4.3−
⎛
⎜
⎝
⎞
⎟
⎠
in= r 19in=
Problem 2-85
Determine the length of member
AB of the truss by first

establishing a Cartesian position
vector from A to B and then
determining its magnitude.
Given:
a 1.2 m=
b 0.8 m=
c 0.3 m=
d 1.5 m=
θ
40 deg=
83
© 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 2
Solution:
r
cdcot
θ
()
+
da−
⎛
⎜
⎝
⎞
⎟
⎠
= r
2.09

0.3
⎛
⎜
⎝
⎞
⎟
⎠
m= r 2.11 m=
Problem 2- 86
The positions of point A on the building and point B on the antenna have been measured relative
to the electronic distance meter (EDM) at O. Determine the distance between A and B. Hint:
Formulate a position vector directed from A to B; then determine its magnitude.
Given:
a 460 m=
b 653 m=
α
60 deg=
β
55 deg=
θ
30 deg=
φ
40 deg=
Solution:
r
OA
a− cos
φ
()
sin

θ
()
acos
φ
()
cos
θ
()
asin
φ
()
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
r
OB
b− cos
β
()
sin
α
()

b− cos
β
()
cos
α
()
bsin
β
()
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
=
r
AB
r
OB
r
OA
−= r
AB
148.2−
492.4−

239.2
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
m= r
AB
567.2 m=
Problem 2-87
Determine the lengths of cords ACB and CO. The knot at C is located midway between A and B.
84
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 2
Given:
a 3ft=
b 6ft=
c 4ft=
Solution:
r
AB
c
b
a−
⎛

⎜
⎜
⎝
⎞
⎟
⎟
⎠
= r
OA
0
0
a
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
r
AC
r
AB
2
=
r
AB
7.8 ft=

r
OC
r
OA
r
AC
+=
r
OC
3.91 ft=
Problem 2-88
Determine the length of the crankshaft AB by first formulating a Cartesian position vector from
A to B and then determining its magnitude.
85
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 2
Given:
a 400=
b 125=
θ
25 deg=
Solution:
r
AB
absin
θ
()
+

bcos
θ
()()
−
0
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
mm= r
AB
467mm=
Problem 2-89
Determine the length of wires AD, BD, and CD. The ring at D is midway between A and B.
Given:
a 0.5 m=
b 1.5 m=
c 2m=
d 2m=
e 0.5 m=
86
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 2
r

AD
c−
2
d
2
e
2
b
2
−
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
= r
AD
1−
1

0.5−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
m= r
AD
1.5 m=
r
BD
r
AD
−=
r
BD
1.5 m=
r
CD
c
2
d
2
a
b
2
+

e
2
−
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
= r
CD
1
1
1
⎛
⎜
⎜
⎝
⎞
⎟

⎟
⎠
m= r
CD
1.7 m=
Problem 2-90
Express force
F
as a Cartesian vector; then determine its coordinate direction angles.
Given:
F 600 lb= c 3ft=
a 1.5 ft=
φ
60 deg=
b 5ft=
Solution:
rbi acsin
φ
()
+
()
j+ 0 ccos
φ
()
−
()
k+=
rb
2
acsin

φ
()
+
()
2
+ ccos
φ
()()
2
+=
r 2m=
87
Solution:
© 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 2
fF
ccos
φ
(
)
(
)
−
r
⎡
⎢
⎣
⎤

⎥
⎦
=
dF
b
r
= eF
acsin
φ
()
+
r
⎛
⎜
⎝
⎞
⎟
⎠
=
d 452lb= e 370lb= f 136− lb=
Fdi ej+ fk+
()
lb=
α
acos
d
F
⎛
⎜
⎝

⎞
⎟
⎠
=
α
41.1 deg=
β
acos
e
F
⎛
⎜
⎝
⎞
⎟
⎠
=
β
51.9 deg=
γ
acos
f
F
⎛
⎜
⎝
⎞
⎟
⎠
=

γ
103deg=
Problem 2-91
Express force
F
as a Cartesian vector; then
determine its coordinate direction angles.
Given:
a 1.5 ft=
b 5ft=
c 3ft=
θ
60 deg=
F 600 lb=
Solution:
r
b
acsin
θ
()
+
c− cos
θ
()
⎛
⎜
⎜
⎝
⎞
⎟

⎟
⎠
= F F
r
r
= F
452
370
136−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟

⎠
acos
F
F
⎛
⎜
⎝
⎞
⎟
⎠
=
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
41.1
51.9
103.1
⎛
⎜
⎜

⎝
⎞
⎟
⎟
⎠
deg=
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 2
Determine the magnitude and coordinate direction angles of the resultant force acting at point A.
Given:
F
1
150 N=
F
2
200 N=
a 1.5 m=
b 4m=
c 3m=
d 2m=
e 3m=
θ
60 deg=
Solution:
Define the position vectors and then the forces
r
AB

ecos
θ
()
aesin
θ
()
+
b−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F
1v
F
1
r
AB
r
AB
= F
1v
38
103.8
101.4−
⎛

⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
r
AC
c
ad−
b−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F
2v
F
2
r
AC
r
AC
= F

2v
119.4
19.9−
159.2−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
Add the forces and find the magnitude of the resultant
F
R
F
1v
F
2v
+= F
R
157.4
83.9
260.6−
⎛
⎜
⎜
⎝
⎞

⎟
⎟
⎠
N= F
R
316 N=
Find the direction cosine angles
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
acos
F
R
F
R
⎛
⎜
⎝
⎞
⎟

⎠
=
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
60.1
74.6
145.6
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
deg=
89
Problem 2-92
© 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 2-93
The plate is suspended using the three
cables which exert the forces shown.
Express each force as a Cartesian vector.
Given:

F
BA
350 lb=
F
CA
500 lb=
F
DA
400 lb=
a 3ft=
b 3ft=
c 6ft=
d 14 ft=
e 3ft=
f 3ft=
g 2ft=
Solution:

r
BA
e− g−

ab+
d
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F
BAv
F
BA
r
BA
r
BA
= F
BAv
109.2−
131
305.7
⎛
⎜
⎜
⎝
⎞
⎟
⎟

⎠
lb=
r
CA
f
b
d
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F
CAv
F
CA
r
CA
r
CA
= F
CAv
102.5
102.5
478.5
⎛
⎜

⎜
⎝
⎞
⎟
⎟
⎠
lb=
r
DA
g−
c−
d
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F
DAv
F
DA
r
DA
r
DA
= F
DAv

52.1−
156.2−
364.5
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
90
© 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 2-94
The engine of the lightweight plane
is supported by struts that are
connected to the space truss that
makes up the structure of the
plane. The anticipated loading in
two of the struts is shown.
Express each of these forces as a
Cartesian vector.
Given:
F
1
400 lb=

F
2
600 lb=
a 0.5 ft=
b 0.5 ft=
c 3.0 ft=
d 2.0 ft=
e 0.5 ft=
f 3.0 ft=
Solution:
r
CD
c
b−
a
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F
1v
F
1
r
CD
r

CD
= F
1v
389.3
64.9−
64.9
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
r
AB
c−
b
e−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F
2v

F
2
r
AB
r
AB
= F
2v
584.0−
97.3
97.3−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
Problem 2-95
The window is held open by cable AB. Determine the length of the cable and express the force
F

acting at A along the cable as a Cartesian vector.
91
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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:

a 300 mm=
b 500 mm=
c 150 mm=
d 250 mm=
θ
30 deg=
F 30 N=
Solution:
r
AB
a− cos
θ
()
cb−
dasin
θ
()
+
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= r
AB

591.6 mm=
F
v
F
r
AB
r
AB
= F
v
13.2−
17.7−
20.3
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
Problem 2-96
The force acting on the man, caused by his pulling on the anchor cord, is
F
. If the length of
the cord is L, determine the coordinates A(x, y, z) of the anchor.
92
© 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
40
20
50−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
L 25 m=
Solution:
r L
F
F
=
r
14.9
7.5
18.6−
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
m=
Problem 2-97
Express each of the forces in Cartesian vector form and determine the magnitude and coordinate
direction angles of the resultant force.
Given:
F
1
80 lb= c 4ft=
93
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 2
F
2
50 lb= d 2.5 ft=
a 6ft= e 12=
b 2ft= f 5=
Solution:
r
AC
d−
c−
e
d
f

⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
= F
1v
F
1
r
AC
r
AC
= F
1v
26.2−
41.9−
62.9
⎛
⎜
⎜
⎝
⎞

⎟
⎟
⎠
lb=
r
AB
b
c−
a−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F
2v
F
2
r
AB
r
AB
= F
2v
6.1
12.1−
18.2−

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kg=
F
R
F
1v
F
2v
+= F
R
12.8−
68.7−
22.8
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb= F
R

73.5 lb=
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
acos
F
R
F
R
⎛
⎜
⎝
⎞
⎟
⎠
=
α
β
γ
⎛

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
1.7
2.8
1.3
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
Problem 2-98
The cable attached to the tractor at B exerts force
F
on the framework. Express this force as a
Cartesian vector
94
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Engineering Mechanics - Statics Chapter 2
Given:
F 350 lb=
a 35 ft=
b 50 ft=
θ
20 deg=
Solution:
Find the position vector and then the force
r
AB
bsin
θ
()
bcos
θ
()
a−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F
v
F
r

AB
r
AB
= F
v
98.1
269.4
200.7−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
Problem 2-99
The cable OA exerts force
F
on point O. If the length of the cable is L, what are the coordinates
(x, y, z) of point A?
Given:

F
40
60
70
⎛
⎜

⎜
⎝
⎞
⎟
⎟
⎠
N=
L 3m=
Solution:

r L
F
F
=
95
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 2
r
1.2
1.8
2.1
⎛
⎜
⎜
⎝
⎞
⎟
⎟

⎠
m=
Problem 2-100
Determine the position (x, y, 0) for fixing cable BA so that the resultant of the forces exerted on the
pole is directed along its axis, from B toward O, and has magnitude F
R
. Also, what is the magnitude
of force
F
3
?
Given:
F
1
500 N=
F
2
400 N=
F
R
1000 N=
a 1m=
b 2m=
c 2m=
d 3m=
Solution:
Initial Guesses
F
3
1N= x 1m= y 1m=

Given
F
R
0
0
1−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
1
c
2
d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
c−
0
d−

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
F
2
a
2
b
2
+ d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
a
b−
d−
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
+
F
3
x
2
y
2
+ d
2
+
⎛
⎜
⎝
⎞
⎟
⎠
x
y
d−
⎛
⎜
⎜
⎝
⎞
⎟
⎟

⎠
+=
F
3
x
y
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
Find F
3
x, y,
()
=
x
y
⎛
⎜
⎝
⎞
⎟
⎠
1.9

2.4
⎛
⎜
⎝
⎞
⎟
⎠
m= F
3
380 N=
96
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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
The cord exerts a force
F
on the hook. If the cord is length L, determine the location x, y of the
point of attachment B, and the height z of the hook.
Given:
F
12
9
8−
⎛
⎜
⎜
⎝
⎞
⎟

⎟
⎠
lb=
L 8ft=
a 2ft=
Solution:
Initial guesses
x 1ft= y 1ft= z 1ft=
Given
xa−
y
z−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
L
F
F
=
x
y
z
⎛
⎜
⎜

⎝
⎞
⎟
⎟
⎠
Find xy, z,()=
x
y
z
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
7.65
4.24
3.76
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
ft=
Problem 2-102

The cord exerts a force of magnitude F on the hook. If the cord length L, the distance z, and the
x component of the force, F
x
,

are given,

determine the location x, y of the point of attachment B
of the cord to the ground.
Given:
F 30 lb=
L 8ft=
z 4ft=
F
x
25 lb=
a 2ft=
Solution:
Guesses
x 1ft=
97
Problem 2-101
© 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
y 1ft=
Given
F
x

xa−
L
⎛
⎜
⎝
⎞
⎟
⎠
F= L
2
xa−()
2
y
2
+ z
2
+=
x
y
⎛
⎜
⎝
⎞
⎟
⎠
Find xy,()=
x
y
⎛
⎜

⎝
⎞
⎟
⎠
8.67
1.89
⎛
⎜
⎝
⎞
⎟
⎠
ft=
Problem 2-103
Each of the four forces acting at E has magnitude F. Express each force as a Cartesian vector
and determine the resultant force.
Units used:
kN 10
3
N=
Given:
F 28 kN=
a 4m=
b 6m=
c 12 m=
Solution:
Find the position vectors and
then the forces
r
EA

b
a−
c−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F
EA
F
r
EA
r
EA
= F
EA
12
8−
24−
⎛
⎜
⎜
⎝
⎞
⎟
⎟

⎠
kN=
98
© 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 2
r
EB
b
a
c−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F
EB
F
r
EB
r
EB
= F
EB
12

8
24−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
r
EC
b−
a
c−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F
EC
F
r
EC
r

EC
= F
EC
12−
8
24−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
r
ED
b−
a−
c−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F
ED

F
r
ED
r
ED
= F
ED
12−
8−
24−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
Find the resultant sum
F
R
F
EA
F
EB
+ F
EC
+ F
ED

+= F
R
0
0
96−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
Problem 2-104
The tower is held in place by three cables. If the force of each cable acting on the tower is
shown, determine the magnitude and coordinate direction angles
α
,
β
,
γ
of the resultant force.

Units Used:
kN 10
3
N=
Given:
x 20 m= a 16 m=

y 15 m= b 18 m=
F
1
600 N= c 6m=
F
2
400 N= d 4m=
F
3
800 N= e 24 m=
99
© 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 2
Find the position vectors, then the force vectors
r
DC
a
b−
e−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F

1v
F
1
r
DC
r
DC
= F
1v
282.4
317.6−
423.5−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
r
DA
x
y
e−
⎛
⎜
⎜
⎝

⎞
⎟
⎟
⎠
= F
2v
F
2
r
DA
r
DA
= F
2v
230.8
173.1
277−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
N=
r
DB
c−
d

e−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F
3v
F
3
r
DB
r
DB
= F
3v
191.5−
127.7
766.2−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠

N=
Find the resultant, magnitude, and direction angles
F
R
F
1v
F
2v
+ F
3v
+= F
R
0.322
0.017−
1.467−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN= F
R
1.502 kN=
α
β
γ
⎛

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
acos
F
R
F
R
⎛
⎜
⎝
⎞
⎟
⎠
=
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞

⎟
⎟
⎟
⎠
77.6
90.6
167.6
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
deg=
Problem 2-105
The chandelier is supported by three chains which are concurrent at point O. If the force in each
chain has magnitude F, express each force as a Cartesian vector and determine the magnitude and
coordinate direction angles of the resultant force.
100
Solution:
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Engineering Mechanics - Statics Chapter 2
Given:
F 60 lb=
a 6ft=
b 4ft=

θ
1
120 deg=
θ
2
120 deg=
Solution:
θ
3
360 deg
θ
1
−
θ
2
−=
r
OA
bsin
θ
1
()
bcos
θ
1
()
a−
⎛
⎜
⎜

⎜
⎝
⎞
⎟
⎟
⎟
⎠
= F
A
F
r
OA
r
OA
= F
A
28.8
16.6−
49.9−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
r
OB

bsin
θ
1
θ
2
+
()
bcos
θ
1
θ
2
+
()
a−
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
= F
B
F
r
OB

r
OB
= F
B
28.8−
16.6−
49.9−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
r
OC
0
b
a−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= F

C
F
r
OC
r
OC
= F
C
0
33.3
49.9−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
F
R
F
A
F
B
+ F
C
+= F
R

149.8 lb=
α
β
γ
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
acos
F
R
F
R
⎛
⎜
⎝
⎞
⎟
⎠
=
α
β
γ
⎛

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
90
90
180
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
deg=
101
© 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 2-106
The chandelier is supported by three chains which are concurrent at point O. If the resultant
force at O has magnitude F
R

and is directed along the negative z axis, determine the force in
each chain assuming F
A
= F
B
= F
C
= F.
Given:

a 6ft=
b 4ft=
F
R
130 lb=
Solution:

F
a
2
b
2
+
3a
F
R
=
F 52.1 lb=
Problem 2-107
Given the three vectors

A
,
B
,

and
D
,

show that

AB D+
()
⋅ AB⋅
()
AD⋅
()
+=
.
Solution:
Since the component of (
B
+
D)
is equal to the sum of the components of
B
and
D
, then
AB D+

()
⋅ AB⋅ AD⋅+=
(QED)
Also,

AB D+
()
⋅ A
x
i A
y
j+ A
z
k+
()
B
x
D
x
+
()
i B
y
D
y
+
()
j+ B
z
D

z
+
()
k+
⎡
⎣
⎤
⎦
=

=
A
x
B
x
D
x
+
()
A
y
B
y
D
y
+
()
+ A
z
B

z
D
z
+
()
+
102
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 2

=
A
x
B
x
A
y
B
y
+ A
z
B
z
+
()
A
x
D

x
A
y
D
y
+ A
z
D
z
+
()
+
=
AB⋅
()
AD⋅
()
+
(QED)
Problem 2-108
Cable BC exerts force
F

on the top of the flagpole. Determine the projection of this force along
the z axis of the pole.
Given:
F 28 N=
a 12 m=
b 6m=
c 4m=

Solution:

r
BC
b
c−
a−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= k
0
0
1
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
F
v

F
r
BC
r
BC
=
F
z
F
v
− k= F
z
24 N=
Problem 2-109
Determine the angle
θ
between the tails of the two vectors.
103
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Engineering Mechanics - Statics Chapter 2
r
1
9m=
r
2
6m=
α
60 deg=

β
45 deg=
γ
120 deg=
φ
30 deg=
ε
40 deg=
Solution:
Determine the two position vectors and use the dot
product to find the angle
r
1v
r
1
sin
ε
()
cos
φ
()
sin
ε
()
− sin
φ
()
cos
ε
()

⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
= r
2v
r
2
cos
α
()
cos
β
()
cos
γ
()
⎛
⎜
⎜
⎜
⎝
⎞
⎟

⎟
⎟
⎠
=
θ
acos
r
1v
r
2v
⋅
r
1v
r
2v
⎛
⎜
⎝
⎞
⎟
⎠
=
θ
109.4 deg=
Problem 2-110
Determine the magnitude of the projected component of
r
1

along

r
2
, and the projection of
r
2
along
r
1
.
Given:
r
1
9m=
r
2
6m=
α
60 deg=
β
45 deg=
γ
120 deg=
φ
30 deg=
ε
40 deg=
104
Given:
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may

be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 2
Solution:
Write the vectors and unit vectors
r
1v
r
1
sin
ε
()
cos
φ
()
sin
ε
()
− sin
φ
()
cos
ε
()
⎛
⎜
⎜
⎜
⎝
⎞
⎟

⎟
⎟
⎠
= r
1v
5.01
2.89−
6.89
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
m=
r
2v
r
2
cos
α
()
cos
β
()
cos
γ
()

⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
= r
2v
3
4.24
3−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
m=
u
1
r
1v
r
1v

= u
2
r
2v
r
2v
= u
1
0.557
0.321−
0.766
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
= u
2
0.5
0.707
0.5−
⎛
⎜
⎜
⎝
⎞
⎟

⎟
⎠
=
The magnitude of the projection of
r
1
along
r
2
.
r
1v
u
2
⋅ 2.99 m=
The magnitude of the projection of
r
2
along
r
1
.
r
2v
u
1
⋅ 1.99 m=
Problem 2-111
Determine the angles
θ

and
φ
between the wire segments.
Given:

a 0.6=
b 0.8=
c 0.5=
d 0.2=
Solution:
ead−=
105
© 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.