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Solution Manual to Accompany
Engineering Mechanics: Dynamics, 3
rd Edition Andrew Pytel and Jaan Kiusalaas NOTE: For the Complete File, Download link at the of this File) 1 / 3
Chapter 11 11.1 (a)m= 30lb
5:32ft/s
2
= 5:639slugsJ
(b)W=mg= 5:639(32:2) = 181:6lbJ 11.2 WSI=gV= 7850(9:81)(0:06 2
)(0:120) = 104:51N
WU S=WSI
0:2248lb
1:0N
= 104:51 (0:2248) = 23:5lbJ 11.3 (a)100kN/m 2 = 10010 3 N m 2
0:2248lb
1:0N
1:0m
2 1550in.2
= 14:50lb/in.
2 J (b)30m/s= 30m s
3:281ft
1:0m
1:0mi
5280ft
3600s
1:0h
= 67:1mi/hJ
(c)800slugs= 800slugs
14:593kg
1:0slug
= 11:6710
3
kg= 11:67MgJ
(d)20lb/ft 2 = 20lb ft 2
4:448N
1:0lb
1:0ft
2
0:092 903 04m
2 = 958N/m 2 J 11.4 I= 20kgm 2 = 20kgm 2
0:06853slugs
1:0kg
10:764ft
2
1:0m
2
= 14:75slugsft
2 But1:0slug= 1:0lbs 2 =ft
I= 14:75
lbs 2 ft ft 2
= 14:75lbfts
2 J 11.5 KE= 1 2 mv 2 + 1 2 mk 2 !2 Since the dimensions each term must be the same, we have
[KE] = [M]
L 2 T 2
= [M]
k 2
1 T 2
Therefore, [k] = [L] 1© 2010. Cengage Learning, Engineering. All Rights Reserved. 2 / 3
(a)In the SI system
[KE] = [M]
L 2 T 2
= kgm 2 s 2 J [k] =mJ (b)In the US system
[KE] = [M]
L 2 T 2
=
F T 2 L
L 2 T 2
= [F L] =lbftJ [k] =ftJ 11.6 [g] [k] [x]
1 W
=
L T 2
F L
[L]
1 F
=
L T 2
= [a]Q.E.D.
11.7 [F] = [k] [x]
ML T 2
= [k][L] [k] =
M T 2
J 11.8 (a)
mv 2
=
F T 2 L
L 2 T 2
= [F L]J
(b)[mv] =
F T 2 L
L T
= [F T]J
(c)[ma] =
F T 2 L
L T 2
= [F]J
11.9
Rewrite the equation asy= 1:0x
2
[y] = [1:0]
x 2
[L] = [1:0]
L 2
[1:0] =
1 L
y=x 2 can be dimensionally correct only if the units of the implied constant
1:0are in.
1 .J 11.10 (a)[I] =
mR 2
=
F T 2 L
L 2
=
F LT 2
J 2© 2010. Cengage Learning, Engineering. All Rights Reserved.
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