Engineering Mechanics - Statics Chapter 1
Problem 1-1
Represent each of the following combinations of units in the correct SI form using an
appropriate prefix:
(a)
m/ms
(b)
μ
km
(c)
ks/mg
(d)
km μN⋅
Units Used:
μN10
6−
N=
μkm 10
6−
km=
Gs 10
9
s=
ks 10
3
s=
mN 10
3−
N=
ms 10
3−

ks
mg
1
Gs
kg
=
d() km μN⋅ 110
3−
× mN=
km μN⋅ 1mmN⋅=
1
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Engineering Mechanics - Statics Chapter 1
Problem 1-2
Wood has a density d. What is its density expressed in SI units?
Units Used:
Mg 1000 kg=
Given:
d 4.70
slug
ft
3
=
Solution:
1slug 14.594 kg=
d 2.42
Mg
m

m
=
Mg
mm
Gg
m
=
b()
mN
μs
10
3−
N
10
6−
s
=
10
3
N
s
=
kN
s
=
mN
μs
kN
s
=

Mg
ms
10
3
kg
10
3−
s
=
10
6
kg
s
=
Gg
s
=
Mg
ms
Gg
s
=
b()
N
mm
1N
10
3−
m
= 10

Problem 1-5
Represent each of the following with SI units having an appropriate prefix:
(a)
S
1
,
(b)
S
2
,
(c)
S
3
.
Units Used:
kg 1000 g= ms 10
3−
s= kN 10
3
N=
Given:
S
1
8653 ms=
S
2
8368 N=
S
3
0.893 kg=

× gm=
kN 10
3
N=
Given:
x 45320 kN=
y 568 10
5
×
()
mm=
z 0.00563 mg=
Solution:
a() x 45.3MN=
b() y 56.8km=
c() z 5.63μg=
Problem 1-7
Evaluate (
ab⋅
)/c to three significant figures and express the answer in SI units using an
appropriate prefix.
Units Used:
μm10
6−
m=
Given:
a 204 mm()=
4
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1
to
Nm⋅
, (b) S
2
to kN/m
3
, (c) S
3
to mm/s. Express the result to three
significant figures. Use an appropriate prefix.
Units Used:
kN 10
3
N=
Given:
S
1
200g lb ft⋅=
S
2
350g
lb
ft
3
=
S
3
8
ft

3
? Express
the result to three significant figures. Use an appropriate prefix.
Units Used:
Mg 10
3
kg=
mN 10
3−
N=
kN 10
3
N=
Given:
m
1
10 kg=
m
2
0.5 gm=
m
3
4.50 Mg=
Solution:
a() Wm
1
g=
W 98.1 N=
b() Wm
2

ft
3
=
Solution:
ρ
8.33
Mg
m
3
=
Problem 1-13
A concrete column has diameter d and length L. If the density (mass/volume) of concrete is
ρ
,
determine the weight of the column in pounds.
Units Used:
Mg 10
3
kg=
kip 10
3
lb=
Given:
d 350 mm=
L 2m=
7
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 1

5.26
slug
ft
3
=
Solution:
ρ
2.17
Mg
m
3
=
Problem 1-15
Determine your own mass in kilograms, your weight in newtons, and your height in meters.
Solution:
Example
W 150 lb=
mW= m 68.039 kg=
Wg 667.233 N=
h 72 in= h 1.829 m=
8
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Engineering Mechanics - Statics Chapter 1
Problem 1-16
Two particles have masses m
1
and m
2

2
=
F 10.0 nN=
W
1
m
1
g= W
1
78.5 N=
W
1
F
7.85 10
9
×=
W
2
m
2
g= W
2
118 N=
W
2
F
1.18 10
10
×=
Problem 1-17

1
150 kg=
r 275 mm=
Solution:
F
Gm
1
2
2r()
2
=
F 4.96 μN=
Since the force F

is measured in Newtons, then the equation is dimensionally homogeneous.
Problem 1-18
Evaluate each of the following to three significant figures and express each answer in SI units
using an appropriate prefix: (a) x, (b) y, (c) z.
Units Used:
MN 10
6
N=
kN 10
3
N=
μm10
6−
m=
Given:
x 200 kN()

2
b
2
/c
2
, (c) a
3
b
3
.
Units Used:
μm10
6−
m= Mm 10
6
m=
Mg 10
6
gm= kg 10
3
gm=
ms 10
3−
s=
Given:
a
1
684 μm=
b
1

2
c
2
3.69 Mm
s
kg
=
c() a
3
b
3
1.14 km kg⋅=
Problem 1-20
Evaluate each of the following to three significant figures and express each answer in SI units
using an appropriate prefix:
(a)
a
1
/b
1
2
(b)
a
2
2
b
2
3
.
Units Used:

kg
2
=
b() a
2
2
b
2
3
135.48 kg
3
m
2
⋅=
12
© 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 2-1
Determine the magnitude of the resultant force
F
R

= F
1

+
F
2

2
F
2
2
+ 2 F
1
F
2
cos
ψ
()
−=
F
R
867 N=
F
R
sin
ψ
()
F
2
sin
θ
()
=
θ
asin F
2
sin

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Engineering Mechanics - Statics Chapter 2
θ
120 deg=
Solution:
F
R
F
1
2
F
2
2
+ 2 F
1
F
2
cos 180 deg
θ
−
()
−=
F
R
72.1 lb=
β
asin F
1
sin 180 deg
θ

2
375 lb=
θ
30 deg=
φ
45 deg=
Solution:
F
R
F
1
2
F
2
2
+ 2 F
1
F
2
cos 90 deg
θ
+
φ
−
()
−=
F
R
178 kg=
F

⎠
=
β
37.89 deg=
14
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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
Angle measured ccw from x axis
360 deg
φ
−
β
+ 353deg=
Problem 2-4
Determine the magnitude of the resultant force
F
R

= F
1
+ F
2

and its direction, measured
counterclockwise from the positive u axis.
Given:
F
1

+
()
−=
F
R
605 N=
F
R
sin 180 deg
β
−
γ
−
α
+
()
F
2
sin
θ
()
=
θ
asin F
2
sin 180 deg
β
−
γ
−

30 deg=
15
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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
500 N=
β
45 deg=
γ
70 deg=
Solution:
F
1u
sin
γα
−
()
F
1
sin 180 deg
γ
−
()
=
F
1u
F

α
()
sin 180 deg
γ
−
()
=
F
1v
160 N=
Problem 2-6
Resolve the force
F
2
into components acting along the u and v axes and determine the
magnitudes of the components.
Given:
F
1
300 N=
F
2
500 N=
α
30 deg=
β
45 deg=
γ
70 deg=
Solution:

βγ
+
()
−
⎡
⎣
⎤
⎦
sin
γ
()
⎡
⎢
⎣
⎤
⎥
⎦
=
F
2v
482.2 N=
Problem 2-7
Determine the magnitude of the resultant force
F
R
= F
1
+ F
2
and its direction measured

2
2
F
1
2
+ 2 F
1
F
2
cos
α
()
−=
F
R
61.4 lb=
sin
θ
'
()
F
2
sin
α
()
F
R
=
θ
' asin sin

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Engineering Mechanics - Statics Chapter 2
Given:
F
1
25 lb=
F
2
50 lb=
θ
1
30 deg=
θ
2
30 deg=
θ
3
45 deg=
Solution:
F
u
−
sin
θ
3
θ
2
−
()
F

−
()
F
1
sin
θ
2
()
=
F
v
F
1
sin 180 deg
θ
3
−
()
sin
θ
2
()
=
F
v
35.4 lb=
Problem 2-9
Resolve the force
F
2

2
+
()
−
⎡
⎣
⎤
⎦
F
2
sin
θ
2
()
=
F
u
F
2
sin 180 deg
θ
1
θ
2
+
()
−
⎡
⎣
⎤

()
sin
θ
2
()
=
F
v
50− lb=
Problem 2-10
Determine the components of the
F
force acting along the u and v axes.
Given:
θ
1
70 deg=
θ
2
45 deg=
θ
3
60 deg=
F 250 N=
Solution:
F
u
sin 180 deg
θ
1

θ
2
()
= F
u
320 N=
19
© 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
v
sin
θ
1
()
F
sin
θ
2
()
=
F
v
F sin
θ
1
()
sin

−
⎡
⎣
⎤
⎦
F
a
sin
θ
1
()
=
F
a
F sin
θ
1
()
sin 180 deg
θ
1
θ
2
+
()
−
⎡
⎣
⎤
⎦

θ
1
θ
2
+
()
−
⎡
⎣
⎤
⎦
= F
b
26.9 lb=
Problem 2-12
The component of force
F
acting along line aa is required to be F
a
. Determine the magnitude
of
F
and its component along line bb.
20
© 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

a
sin 180 deg
θ
1
−
θ
2
−
()
sin
θ
1
()
⎛
⎜
⎝
⎞
⎟
⎠
= F 19.6 lb=
F
a
sin
θ
1
()
F
b
sin
θ

2
40 deg=
Solution:
F
AB
sin
θ
1
()
F
sin 180 deg
θ
1
θ
2
+
()
−
⎡
⎣
⎤
⎦
=
21
© 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
F
AB

θ
2
()
F
sin 180 deg
θ
1
θ
2
+
()
−
⎡
⎣
⎤
⎦
=
F
AC
F
sin
θ
2
()
sin 180 deg
θ
1
θ
2
+

.
Given:
F
R
1200 lb=
F
1
600 lb=
θ
1
60 deg=
Solution:
TF
1
2
F
R
2
+ 2 F
1
F
R
cos 90 deg
θ
1
−
()
−=
T 744lb=
sin

© 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 2-15
Resolve the force
F
1

into components acting along the u and v axes and determine the magnitudes
of the components.
Given:
F
1
250 N=
F
2
150 N=
θ
1
30 deg=
θ
2
30 deg=
θ
3
105 deg=
Solution:
F
1v

F
1v
129N=
F
1u
sin 180 deg
θ
1
−
θ
3
−
()
F
1
sin
θ
3
()
=
F
1u
F
1
sin 180 deg
θ
1
−
θ
3

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Engineering Mechanics - Statics Chapter 2
F
2
150 N=
θ
1
30 deg=
θ
2
30 deg=
θ
3
105 deg=
Solution:
F
1v
sin
θ
1
()
F
2
sin 180 deg
θ
3
−
()
=
F

2
sin 180 deg
θ
3
−
()
=
F
2u
F
2
sin 180 deg
θ
3
−
()
sin 180 deg
θ
3
−
()
⎛
⎜
⎝
⎞
⎟
⎠
=
F
2u

cos 180 deg
φ
−
()
−=
24
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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
Since

cos 180deg
φ
−
()
cos−
φ
()
=
,
F
R
F
1
2
F
2
2
+ 2 F

θ
of line AB on the tailboard block.
Given:
F
1
400 N=
θ
1
30 deg=
Solution:
F
R
F
1
2
F
1
2
+ 2F
1
F
1
cos 90 deg
θ
1
−
()
−=
F
R

−=
θ
60deg=
25
© 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.