Solution Manual for Engineering Mechanics Statics in SI Units, 14th Edition

Name: Solution Manual for Engineering Mechanics Statics in SI Units, 14th Edition
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Author: Lucy Gray

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1
1–1.
Evaluate each of the following and express with an
appropriate prefix: (a) (430 kg) 2, (b) (0.002 mg)2 , and
(c) (230 m) 3.
Solution
a) (430 kg) 2 = 0.185(106
) kg2 = 0.185 Mg2 Ans.
b) (0.002 mg) 2 = [2(10- 6
) g] 2 = 4 mg2 Ans.
c) (230 m) 3 = [0.23(10 ) m]3 3 = 0.0122 km3 Ans.
Ans:
0.185 Mg2
4 mg2
0.0122 km3

2
1–2.
Represent each of the following combinations of units in the
correct SI form: (a) Mg>ms, (b) N>mm, (c) mN>(kg # ms).
Solution
(a) Mg
ms = 103 kg
10-3 s = 106 kg>s = Gg>s Ans.
(b) N
mm = 1 N
10-3 m = 103 N>m = kN>m Ans.
(c) mN
(kg # ms) = 10-3 N
10-6 kg # s = kN>(kg # s) Ans.
Ans:
Gg>s
kN>m
kN>(kg # s)

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11
1–11.
Using the SI system of units, show that Eq. 1–2 is a
dimensionally homogeneous equation which gives F in
newtons. Determine to three significant figures the
gravitational force acting between two spheres that are
touching each other. The mass of each sphere is 200 kg and
the radius is 300 mm.
Ans:
7.41 mN
SOLUTION
Using Eq. 1–2,
Ans.= 7.41A10- 6 B N = 7.41 mN
= 66.73A10- 12 B c 200(200)
0.6 2 d
F = G m1 m2
r2
N = a m 3
kg # s2 b a kg # kg
m 2 b = kg # m
s2 (Q.E.D.)
F = G m1 m2
r2

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13
1–13.
A rocket has a mass of 3.529(10 ) kg on earth. Specify
(a) its mass in SI units and (b) its weight in SI units. If the
rocket is on the moon, where the acceleration due to gravity
is gm = 1.61 m/s 2,determine to three significant figures
(c) its weight in SI units and (d) its mass in SI units.
Solution
a) 3.529(106
) kg = 3.53 Gg Ans.
b) W e = mg = 33.5291106 2kg4 19.81 m>s 2 2
= 34.6191106 2 kg # m>s 2
= 34.6 MN Ans.
c) W m = mgm = 33.5291106 2 kg4 11.61 m>s 2 2
= 5.6821106 2N = 5.68 MN Ans.
Or
W m = W e a gm
g b = (34.619 MN) a 1.61 m>s 2
9.81 m>s 2 b = 5.68 MN
d) Since the mass is independent of tis location, then
mm = me = 3.53 Gg Ans.
Ans:
3.53 Gg
34.6 MN
5.68 MN
mm = me = 3.53 Gg
6

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14
1–14.
and (c) 435 MN 23.2 mm.>
(a) 354 mg(45 km) (0.0356 kN), (b) (0.004 53 Mg) (201 ms),>
SOLUTION
a)
Ans.
b)
Ans.
c) Ans.435 MN>23.2 mm = 435A10 6 B N
23.2A 10-3 B m = 18.75A109 B N
m = 18.8 GN>m
= 0.911 kg # s
(0.00453 Mg)(201 ms) = C 4.53A10-3 B A103 B kgD C 201A 10-3 B sD
= 0.447 kg # m>N
= 0.447 A103 B g # m
N
(354 mg)(45 km)>(0.0356 kN) = C354A10-3 B gD C45A103 B mD
0.0356 A103 B N
Evaluate each of the following to three significant figures and
express each answer in Sl units using an appropriate prefix:
Ans:
0.447 kg # m>N
0.911 kg # s
18.8 GN>m

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15
1–15.
Evaluate each of the following to three significant figures and
express each answer in SI units using an appropriate prefix:
(a) (212 mN) , (b)2 (52 800
Solution
(a) (212 mN) 2 = 3212(10)-3 N42 = 0.0449 N 2 = 44.9(10)-3 N 2 Ans.
(b) (52 800 ms) 2 = 352 800(10)-3 42 s2 = 2788 s2 = 2.79(103 ) s2 Ans.
(c) 3548(10) 6 41
2 ms = (23 409)(10)-3 s = 23.4(10) 3
(10)-3 s = 23.4 s Ans.
Ans:
44.9(10)-3 N 2
2.79(103 ) s2
23.4 s
ms) , and (c)2 [548(10 )]6 1>2 ms.

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Solution Manual for Engineering Mechanics Statics in SI Units, 14th Edition

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