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

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

⎞
⎟
⎟
⎠
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
⎛
⎜
⎜
⎝

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

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

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=

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
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Engineering Mechanics - Statics Chapter 2
Solution:
r
cdcot
θ
()
+
da−
⎛
⎜
⎝
⎞
⎟
⎠
= r
2.09

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

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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Engineering Mechanics - Statics Chapter 2
Given:
a 3ft=
b 6ft=
c 4ft=
Solution:
r
AB
c
b
a−
⎛

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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Engineering Mechanics - Statics Chapter 2
Given:
a 400=
b 125=
θ
25 deg=
Solution:
r
AB
absin
θ
()
+

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

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

φ
()
+
()
2
+ ccos
φ
()()
2
+=
r 2m=
87
Solution:
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Engineering Mechanics - Statics Chapter 2
fF
ccos
φ
(
)
(
)
−
r
⎡
⎢
⎣
⎤

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

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

⎝
⎞
⎟
⎟
⎠
deg=
88
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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
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

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

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

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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−

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

52.1−
156.2−
364.5
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
lb=
90
© 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
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=

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

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

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−
⎛
⎜
⎜
⎝

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

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

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
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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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
⎛
⎜

⎠
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=

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

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

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

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
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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
y 1ft=
Given
F
x

⎝
⎞
⎟
⎠
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

⎠
kN=
98
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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

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

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

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

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

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−
⎛
⎜
⎜

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

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

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
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

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+

z
+
()
k+
⎡
⎣
⎤
⎦
=

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

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=

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
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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Engineering Mechanics - Statics Chapter 2
r
1
9m=
r
2
6m=
α
60 deg=

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

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

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

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.
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.