An Instructor’s Solution Manual to Accompany Engineering Mechanics: Dynamics, 3rd Edition Andrew Pytel and Jaan Kiusalaas

An Instructor’s Solution Manual to Accompany
Engineering Mechanics: Dynamics, 3rd Edition
Andrew Pytel and Jaan Kiusalaas

, An Instructor’s Solution Manual
for
Pytel and Kiusalaas’s

Engineering Mechanics: Dynamics
Third Edition




Andrew Pytel
Pennsylvania State University




Jaan Kiusalaas
Pennsylvania State University

,Chapter 11
11.1
30 lb
(a) m = 2 = 5:639 slugs J
5:32 ft/s

(b) W = mg = 5:639(32:2) = 181:6 lb J

11.2
WSI = gV = 7850(9:81) (0:062 )(0:120) = 104: 51 N
0:2248 lb
WUS = WSI = 104:51 (0:2248) = 23:5 lb J
1:0 N

11.3
100 103 N 0:2248 lb 1:0 m2
(a) 100 kN/m2 = = 14:50 lb/in.2 J
m2 1:0 N 1550 in.2

30 m 3:281 ft 1:0 mi 3600 s
(b) 30 m/s = = 67:1 mi/h J
s 1:0 m 5280 ft 1:0 h

14:593 kg
(c) 800 slugs = 800 slugs = 11:67 103 kg = 11:67 Mg J
1:0 slug

20 lb 4:448 N 1:0 ft2
(d) 20 lb/ft2 = = 958 N/m2 J
ft2 1:0 lb 0:092 903 04 m2

11.4
0:06853 slugs 10:764 ft2
I = 20 kg m2 = 20 kg m2 = 14:75 slugs ft2
1:0 kg 1:0 m2
But 1:0 slug = 1:0 lb s2 =ft

lb s2 2
I = 14:75 ft = 14:75 lb ft s2 J
ft

11.5
1 1
mv 2 + mk 2 ! 2
KE =
2 2
Since the dimensions each term must be the same, we have
L2 1
[KE] = [M ] = [M ] k 2
T2 T2
Therefore,
[k] = [L]


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© 2010. Cengage Learning, Engineering. All Rights Reserved.

, (a) In the SI system

L2 kg m2
[KE] = [M ] 2
= J [k] = m J
T s2

(b) In the US system

L2 FT2 L2
[KE] = [M ] 2
= = [F L] = lb ft J
T L T2
[k] = ft J

11.6
1 L F 1 L
[g] [k] [x] = [L] = = [a] Q.E.D.
W T2 L F T2

11.7
ML M
[F ] = [k] [x] = [k][L] [k] = J
T2 T2

11.8
FT2 L2
(a) mv 2 = = [F L] J
L T2

FT2 L
(b) [mv] = = [F T ] J
L T

FT2 L
(c) [ma] = = [F ] J
L T2

11.9
Rewrite the equation as y = 1:0 x2

1
[y] = [1:0] x2 [L] = [1:0] L2 [1:0] =
L

y = x2 can be dimensionally correct only if the units of the implied constant
1:0 are in. 1 . J

11.10
FT2
(a) [I] = mR2 = L2 = F LT 2 J
L



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© 2010. Cengage Learning, Engineering. All Rights Reserved.