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
(a) m =
30 lb
5:32 ft/s
= 5:639 slugs J
2
(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
= 104:51 (0:2248) = 23:5 lb J
WUS = WSI
1:0 N
11.3
(a) 100 kN/m2 =
(b) 30 m/s =
100
30 m
s
103 N
m2
0:2248 lb
1:0 N
3:281 ft
1:0 m
1:0 mi
5280 ft
20 lb
ft2
3600 s
= 67:1 mi/h J
1:0 h
14:593 kg
= 11:67
1:0 slug
(c) 800 slugs = 800 slugs
(d) 20 lb/ft2 =
1:0 m2
= 14:50 lb/in.2 J
1550 in.2
4:448 N
1:0 lb
103 kg = 11:67 Mg J
1:0 ft2
= 958 N/m2 J
0:092 903 04 m2
11.4
I = 20 kg m2 = 20 kg m2
0:06853 slugs
1:0 kg
10:764 ft2
= 14:75 slugs ft2
1:0 m2
But 1:0 slug = 1:0 lb s2 =ft
I = 14:75
lb s2 2
ft = 14:75 lb ft s2 J
ft
11.5
1
1
mv 2 + mk 2 ! 2
2
2
Since the dimensions each term must be the same, we have
KE =
[KE] = [M ]
L2
= [M ] k 2
T2
1
T2
Therefore,
[k] = [L]
1
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(a) In the SI system
[KE] = [M ]
kg m2
L2
=
J
2
T
s2
[k] = m J
(b) In the US system
[KE]
= [M ]
FT2
L2
=
2
T
L
L2
= [F L] = lb ft J
T2
= ft J
[k]
11.6
[g] [k] [x]
1
W
=
L
T2
F
1
[L]
L
F
=
L
= [a] Q.E.D.
T2
11.7
[F ] = [k] [x]
ML
= [k][L]
T2
[k] =
M
T2
J
11.8
(a)
mv 2 =
FT2
L
L2
= [F L] J
T2
(b) [mv] =
FT2
L
L
= [F T ] J
T
(c) [ma] =
FT2
L
L
= [F ] J
T2
11.9
Rewrite the equation as y = 1:0 x2
[y] = [1:0] x2
[L] = [1:0] L2
[1:0] =
1
L
y = x2 can be dimensionally correct only if the units of the implied constant
1:0 are in. 1 . J
11.10
(a) [I] = mR2 =
FT2
L
L2 = F LT 2 J
2
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(b) [I] = mR2 = M L2 J
11.11
(a)
v 3 = [A] x2 + [B] [v] t2
L
J
T3
[A] =
(b)
h
x2 = [A] t2
2
e[B][t ]
[A] =
i
L
L3
= [A] L2 + [B]
T3
T
L2
[B] =
J
T4
L2 = [A] T 2 [1]
L2
T2
J
[B] =
T2
[B] T 2 = [1]
1
T2
J
11.12
m
d2 x
dx
+c
+ kx = P0 sin !t
dt2
dt
[m]
FT2
d2 x
=
2
dt
L
L
= [F ]
T2
Therefore, the dimension of each term in the expression is [F ].
+ [c]
dx
L
= [c]
= [F ]
dt
T
[c] =
FT
L
[k] [x] = [k] [L] = [F ]
[k] =
F
J
L
[P0 ] [sin !t] = [P0 ] [1] = [F ]
[!] [t] = [!] [T ] = [1]
J
[P0 ] = [F ] J
[!] =
1
T
J
11.13
F =G
(a) [G] =
(b) [G] =
mA mB
R2
[F ] L2
2
[F T 2 =L]
=
L4
FT4
G=
F R2
mA mb
[G] =
[F ] L2
[M 2 ]
J
M L=T 2 L2
L3
=
2
[M ]
MT2
J
3
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11.14
(a) [E] = [m] c2 =
FT2
L
L2
= [F L] J
T2
M L2
L2
=
2
T
T2
(b) [E] = [m] c2 = [M ]
J
11.15
2
m2
11 10
=
6:67
10
= 2:668 10
R2
0:52
= mg = 10(9:81) = 98:1 N
2:668 10 8
=
100% = 2:72 10 8 % J
98:1
F
= G
W
F
W
100%
8
N
11.16
F =G
m2
= 3:44
R2
2
8 (3=32:2)
2
10
(18=12)
= 1:327
10
10
lb J
11.17
m=
W R2
2000(6378 + 1800)2
=
GMe
(6:67 10 11 ) (5:9742
106
= 336 kg J
1024 )
11.18
gm =
gm
Mm
=
ge
Me
Re
Rm
2
=
GMm
2
Rm
ge =
0:07348
5:9742
6378
1737
GMe
Re2
2
= 0:1658 t
1
Q.E.D
6
11.19
Me = 5:9742
Re = 6378
W =G
Me m
103 m
= 3:44
10
Ms m
= 6:67
R2
10
2
(2Re )
0:06853 slugs
= 0:4094
1:0 kg
1024 kg
3:281 ft
= 20:93
1:0 m
(2
106 ft
1024 )(150=32:2)
8 (0:4094
2
106 )
20:93
1024 slugs
= 37:4 lb J
11.20
F =G
11
1:9891
(149:6
1030 (1:0)
2
109 )
= 0:00593 N J
4
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11.21
G
5:9742
1:9891
Me m
r2
r
R
Me
r2
Me
Ms
1024
1030
0
Ms m
(R r)2
= distance from the earth
= distance between the earth and the sun
= G
=
=
=
=
Ms
r)2
r2
2
R
2Rr + r2
(R
(149:6
2:238
r = 259
109 )2
1022
r2
2(149:6
2:992
106 m = 259
109 )r + r2
1011 r
3:329 4
105 r2
103 km J
5
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6
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Chapter 12
12.1
=
0:13t4 + 4:1t3 + 0:12t2 ft
= y_ = 0:52t3 + 12:3t2 + 0:24t ft/s
y
v
a = v_ =
1:56t2 + 24:6t + 0:24 ft/s
2
At maximum velocity:
a=0
vmax
y=
=
=
1:56t2 + 24:6t + 0:24 = 0
t = 15:779 s
0:52(15:779)3 + 12:3(15:779)2 + 0:24(15:779)
1023 ft/s J
0:13(15:779)4 + 4:1(15:779)3 + 0:12(15:779)2 = 8080 ft J
12.2
1
7
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12.3
12.4
x = t3 3t2
v = x_ = 3t2
45t in.
6t 45 in./s
2
a = v_ = 6t
6 in./s
At t = 8 s:
x = 83 3(8)2 45(8) = 40 in. J
v = 3(8)2 6(8) 45 = 99 in./s J
a =
6(8)
6 = 42 in./s
2
J
Note that v(0) < 0 and v(8 s) > 0. The reversal of velocity occurs when
v = 3t2 6t 45 = 0
t = 5:0 s
3
2
x = 5
3(5)
45(5) = 175 in.
2
8
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At t = 8 s:
s = 175 + (175
40) = 310 in. J
175
40
5s
8s
0 x (in.)
12.5
(a) x = t2
t3
ft
90
(b) a = v_ = 2
v = x_ = 2t
t2
ft/s
30
xmax = 602
603
= 1200 ft J
90
t
ft/s2
15
v = 0 when t = 60 s
a = 0 when t = 30 s
vmax = 2(30)
302
= 30 ft/s J
30
12.6
12.7
(a) x = 2t2 10t in.
When x = 30 in.:
v = x_ = 4t
2t2
(b) v = 0 when 4t
t = 7:110 s J
10t = 30
10 = 0
2.5 s
-12.5
10 in./s
t = 2:5 s
x = 2(2:5)2
10(2:5) =
12:5 in
7.11 s x (in.)
30
0
s = 2(12:5) + 30 = 55:0 in. J
3
9
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12.8
x =
y
=
4t2 2 mm
16t4 16t2 + 4
4t4
x2
=
=
12
12
4t2 + 1
mm
3
When t = 2 s:
vx
vy
v
ax
= x_ = 8t = 8(2) = 16 mm/s
16t3 8t
16(2)3 8(2)
= y_ =
=
= 37:33 mm/s
3
3
q
p
=
vx2 + vy2 = 162 + 37:332 = 40:6 mm/s J
2
= v_ x = 8 mm/s
48(2)2 8
48t2 8
2
=
= 61:33 mm/s
ay = v_ y =
3
3
q
p
2
a2x + a2y = 82 + 61:332 = 61:9 mm/s J
a =
12.9
12.10
4
10
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12.11
12.12
x =
15
2t2 m
y
15
10t + t2 m
=
(a) At t = 0:
vx = x_ =
vy = y_ =
10j m/s J
v=
(b) At t = 5 s:
4t m/s
10 + 2t m/s
a=
20i m/s J
v=
ax = v_ x =
4 m/s
2
ay = v_ y = 2 m/s
2
4i + 2j m/s2 J
4i + 2j m/s2 J
a=
12.13
x =
y =
ay
(a) a =
66t m
vx = x_ = 66 m/s
ax = v_ x = 0
2
86t 4:91t m
vy = y_ = 86 9:82t m/s
= v_ y =
9:82 m/s
2
9:82j m/s2 J
_
(b) When t = 0:
v = 66i + 86j m/s J
(c) When y = h:
vy = 0
(d) When y =
86
9:82t = 0
t = 8:758 s
h = 86(8:758)
4:91(8:758)2 = 377 m J
120 m:
4:91t2 =
86t
120
t = 18:814 s
L = 66(18:814) = 1242 m J
5
11
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12.14
12.15
6
12
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12.16
12.17
7
13
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12.18
12.19
8
14
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12.20
12.21
12.22
9
15
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12.23
12.24
12.25
_ = 1200 rev
1:0 min
r
2 rad
1:0 rev
1:0 min
= 125:66 rad/s
60 s
=
55 + 10 cos + 5 cos 2
dr _
= ( 10 sin
v = r_ =
d
dv _
a = v_ =
= ( 10 cos
d
jajmax = 30(125:66)2 = 474 000
mm
10 sin 2 ) (125:66) mm/s
20 cos 2 )(125:66)2 mm/s
2
mm/s = 474 m/s (at t = 0) J
2
2
10
16
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*12.26
12.27
T
ma
=
mg
FBD
v = 4t m/s
MAD
a = v_ = 4 m/s
F = ma + "
T
2
mg = ma
T = m(g + a) = 50(9:81 + 4) = 691 N J
11
17
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12.28
y
mg
x
F = µk N
FBD
N
MAD
1000 m/km
3600 s/h
v0 = 100 km/h = 100 km/h
Fy
=
Fx
= ma
0 +"
v
N
x =
kN
Z
v dt + C2 =
k gt
1
2
kN
)a=
= ma
a dt + C1 =
Z
= 27:78 m/s
) N = mg
mg = 0
+
!
=
ma
=
m
=
kg
+ C1
2
k gt
+ C1 t + C2
When t = 0 (initial conditions):
x=0
) C2 = 0
)x=
1
2
2
k gt
+ v0 t
When v = 0:
k gt
1
2
x =
+ v0 = 0
v0
kg
kg
) C1 = v0
v = v0
v=
k gt
)t=
v0
kg
2
+ v0
v0
kg
+ v0
=
v02
2
kg
2
=
27:78
= 60:5 m J
2(0:65)(9:81)
12.29
mg 5o
y
x
=
F = µk N
FBD
ma
MAD
N
v0 = 100 km/h = 27:78 m/s
12
18
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= 0 +"
N mg cos 5 = 0
) N = mg cos 5
N
+
mg
sin
5
=
ma
= ma +!
k
kN
)a =
+ g sin 5 = (sin 5
k cos 5 )g
m
2
= (sin 5
0:65 cos 5 )9:81 = 5:497 m/s
Fy
Fx
v
=
x =
Z
Z
a dt + C1 =
5:497t + C1
v dt + C2 =
2:749t2 + C1 t + C2
When t = 0 (initial conditions):
) C2 = 0
x=0
v = v0
) C1 = v0 = 27:78 m/s
2:749t2 + 27:78t m
5:497t + 27:78 m/s
x =
v =
When v = 0:
) t = 5:054 s
5:497t + 27:78 = 0
x=
2:749(5:054) + 27:78(5:054) = 70:2 m J
2
12.30
1:2t
F
2
=
= 12t m/s
m
0:1
Z
v =
ax dt + C1 = 6t2 + C1 m/s
Z
x =
vx dt + C2 = 2t3 + C1 t + C2 m
a =
When t = 0 (initial conditions):
x = 0 ) C2 = 0
)x=
v = 64 m/s ) C1 = 64 m/s
2t3 + 64t m
v=
6t2 + 64 m/s
When t = 4 s:
x=
2(4)3 + 64(4) = 128 m
When v = 0 :
6t2 + 64 = 0
x=
t = 3:266 s
2(3:266)3 + 64(3:266) = 139:35 m
13
19
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Distance traveled:
128 = 150:7 m J
d = 2(139:35)
139.35
0
x (m)
128
12.31
a =
dt
=
p
p
0:04 v
F
2
=
= 4 v m/s
m
0:01
p
Z
Z
v
dv
dv
dv
p + C1 =
t=
+ C1 =
+ C1
a
a
2
4 v
When t = 0:6 s (initial condition):
) 0:6 =
v = 0:16 m/s
)t=
p
0:16
+ C1
2
C1 = 0:4 s
p
v
2
+ 0:4 s
) v = (2t 0:8) = 4t2 3:2t + 0:64 m/s
2
Z
4
x = v dt + C2 = t3 1:6t2 + 0:64t + C2
3
When t = 0 (initial condition):
x=0
) C2 = 0
)x=
4 3
t
3
1:6t2 + 0:64t m
When t = 0:8 s:
x=
4
(0:8)3
3
1:6(0:8)2 + 0:64(0:8) = 0:1707 m J
14
20
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12.32
12.33
4t 4
F
=
=t 1
4
Zm
1
=
a dt + C1 = t2
2
Z
1 3
=
v dt + C2 = t
6
a =
v
y
m/s
2
t + C1
1 2
t + C1 t + C2
2
When t = 0 (initial condition):
y = 0 ) C2 = 0
)y=
1 3
t
6
When t = 8 s:
y=
When v = 0:
1 2
t
2
1
3
(8)
6
1 2
t
2
y=
v=
1
3
(5:123)
6
t
8 m/s ) C1 =
1 2
t
2
8t m
v=
1
2
(8)
2
8 (8) =
t
t = 5:123 s
1
2
(5:123)
2
8 (5:123) =
d = 2(31:70)
8 m/s
10:67 m
8=0
Distance traveled:
8 m/s
31:70 m
10:67 = 52:7 m J
15
21
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-31.70
0
y (m)
-10.67
12.34
12.35
12.36
y
mg
20o
NA
ma
x =
FA = 0.4 NA
FBD
MAD
Assume impending sliding (FA = 0:4NA )
Fy = 0 + "
NA cos 20
0:4NA sin 20
mg = 0
NA = 1:2455 mg
Fx = max
+
!
NA sin 20 + 0:4NA cos 20 = ma
1:2455 mg(sin 20 + 0:4 cos 20 ) = ma
a = 0:894g J
16
22
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12.37
Let y be measured up from the base of the cli¤.
y
ma
x
=
FBD
MAD
mg
Fy = ma + "
a=
dv
v
dy
) v dv =
At impact y = 0
)
1 2
v
2
1 2
v
2
1 2
v =
2
g dy
Initial condition: v = v0 when y = h: ) C =
)
)a=
mg = ma
gy + C
1 2
v + gh
2 0
v02 = g(h
)v=
v02 = gh
g
y)
p
v02 + 2gh J
12.38
12.39
20o
20 lb
8t
y
FBD
20 a
g
=
x
MAD
N
Fx = ma + %
8t
20 sin 20 =
20
20
a=
a
g
32:2
17
23
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a=
v
=
x =
32:2
(8t)
20
Z
Z
32:2 sin 20 = 12:88t
a dt + C1 = 6:44t2
11:013 ft/s
2
11:013t + C1 ft/s
v dt + C2 = 2:147t3
5:507t2 + C1 t + C2 ft
) C1 = C2 = 0
Initial conditions: x = v = 0 at t = 0:
(a) When v = 0:
6:44t2
) t = 1:7101 s
11:013t = 0
) x = 2:147(1:7101)3
5:507(1:7101)2 =
5:37 ft J
(b) When x = 0:
2:147t3
) t = 2:565 s
5:507t2 = 0
) v = 6:44(2:565)2
11:013(2:565) = 14:12 ft/s J
12.40
x
4
ma
3
P = 8 - 2t
FBD mg N
=
4P
5m
v
=
x =
3
4 8 2t
g=
5
5 5=32:2
Z
Z
Initial conditions: v =
When x = 0 :
10:95t2
3
mg = ma
5
4
P
5
Fx = ma + %
a=
MAD
3
(32:2) = 21:90
5
a dt + C1 = 21:90t
v dt + C2 = 10:95t2
10:304t ft/s
5:152t2 + C1
1:7173t3 + C1 t + C2
10 ft/s, x = 0 at t = 0. ) C1 =
1:7173t3
v = 21:90(1:1049 )
2
10t = 0
5:152(1:1049 )2
10 ft/s
C2 = 0
t = 1:1049 s J
10 = 7:91 ft/s J
18
24
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12.41
(mA + mB)g
A
y
θ
x
B
=
(mA + mB) a
NB FBD
Fx = ma + &
MAD
(mA + mB )g sin = (mA + mB )a
a = g sin
Assume impending slipping between A and B:
mA g
y
Fx = ma + &
=
A
x
F =µsNA
NA FBD
(mA g
s
θ ma
A
MAD
NA ) sin +
s NA
cos = mA g sin
= tan J
12.42
19
25
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12.43
12.44
W
F
=
µkΝ
N
FBD
ma
MAD
20
26
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Fy
Fx
= 0
= 0
F
a =
v
+" N W =0
N = W = 3000 lb
+! F
k N = ma
1000e 0:2t 0:05(3000)
kN
=
m
3000=32:2
=
10:733e
=
Z
=
0:2t
a dt + C =
53:67e
0:2t
1:61 ft/s
Z
2
0:2t
(10:733e
1:61)dt + C
1:61t + C ft/s
Initial condition: v = 0 when t = 0. ) C = 53:67 ft/s
v = 53:67(1
e
0:2t
)
1:61t ft/s
Maximum velocity occurs at t = 4 s (end of powered travel)
h
i
vmax = 53:67 1 e 0:2(4)
1:61(4) = 23:1 ft/s J
12.45
0.2(9.81) N
P
=
y
P N
FBD 20 o
0.2(1.5) N
x
MAD
(a)
Fx = max
+%
P
[P + 0:2(9:81)] sin 20 = 0:2(1:5)
P = 1:476 N J
(b) Reverse the direction of max , i.e., use max =
Fx = max
+%
P
0:2(1:5) N.
[P + 0:2(9:81)] sin 20 =
0:2(1:5)
P = 0:564 N J
12.46
ma
40o P
N
mg µN
FBD
y
=
x
MAD
21
27
© 2010. Cengage Learning, Engineering. All Rights Reserved.
) N = P sin 40
(P sin 40 ) = ma
= 0 +!
P sin 40
N =0
= may + "
P cos 40
mg
Fx
Fy
P
(cos 40
m
When motion impends: a = 0 and =
sin 40 )
a=
0=
P
(cos 40
5
s
0:5 sin 40 )
= 0:5
9:81
When collar begins to slide: P = 110:31 N and
a=
110:31
(cos 40
5
g
0:4 sin 40 )
P = 110:31 N
=
k
= 0:4
9:81 = 1:418 m/s
2
J
12.47
12.48
22
28
© 2010. Cengage Learning, Engineering. All Rights Reserved.
12.49
12.50
v0 = 8 km/h = 8
1000
3600
mg
kx
= 2:222 m/s
x
= ma
MAD
FBD N
Fx = max
kx = ma
v dv = a dx =
k
x dx
m
Initial condition: v = v0 when x = 0:
)C=
a=
dv
v
dx
+
) v 2 = v02
)a=
1 2
v =
2
k
x
m
k 2
x +C
2m
1 2
v
2 0
k 2
x
m
23
29
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Stopping condition: v = 0 when x = 0:4 m.
0 = 2:2222
k
20
103
(0:4)2
k = 6:17
105 N/m J
12.51
24
30
© 2010. Cengage Learning, Engineering. All Rights Reserved.
12.52
*12.53
F = ma
a=v
x=
dv
=
dx
Z
T
cD v 2 = ma
1
(T + cD v 2 )
m
mv
dv = dx
T + cD v 2
mv
dv + C =
T + cD v 2
m
T + cD v 2
ln
+C
2cD
cD
Initial condition: v = v0 when x = 0: ) C =
)x=
m=
T + cD v02
m
ln
2cD
cD
m
T + cD v02
ln
2cD T + cD v 2
2500
= 77:64 slugs
32:2
v0 = (90)
5280
= 132 ft/s
3600
When v = 0:
x=
T + cD v02
77:64
450 + 0:006(132)2
m
ln
=
ln
= 1352 ft J
2cD
T
2(0:006)
450
25
31
© 2010. Cengage Learning, Engineering. All Rights Reserved.
*12.54
.
x
5 lb
30o
4 ft
θ
8 lb
FBD
Ν
5 a
32.2
MAD
Fx = ma + .
5 sin 30
32:2
(5 sin 30
5
8 cos ) = 16:1
)a=
cos = p
a=v
=
51:52 cos ft/s
2
x
x
=p
2
2
+4
x + 16
x2
51:52x
p
x2 + 16
dv
= 16:1
dx
5
a
32:2
8 cos =
v dv =
51:52x
p
x2 + 16
16:1
dx
p
1 2
v = 16:1x 51:52 x2 + 16 + C
2
Initial condition: v = 0 when x = 0: ) C = 51:51(4) = 206:0 (ft/s)2
When a = 0 (v = vmax ):
16:1
1 2
v
2 max
=
=
*12.55
51:52x
p
=0
x2 + 16
16:1(1:3159)
2
) x = 1:3159 ft J
p
51:52 (1:3159)2 + 16 + 206:0
10:241 (ft/s)
p
vmax = 2(10:241) = 4:53 ft/s J
F = ma + #
dv
=g
a=
dt
t=
1
ln
cd =m
g
mg
cd
v
m
dt =
(cd =m)v
+C =
cd =m
Initial condition: v = 0 when t = 0:
)t=
cD v = ma
)C=
dv
(cd =m) v
g
m
ln
cD
m
ln
cD
mg
+v +C
cD
mg
cD
m
mg=cD
ln
cD mg=cd v
26
32
© 2010. Cengage Learning, Engineering. All Rights Reserved.
When v = v1 (terminal velocity), a = 0:
mg
When v = 0:9v1 = 0:9
:
cd
t=
) v1 =
mg
cd
m
1
m
m
=
ln
ln 10 = 2:30
J
cd 1 0:9
cD
cd
12.56
7
v
2
+
m/s
4 16
Letting x1 = x, x2 = v, the equivalent …rst-order equations and initial conditions
are
7 x2
+
x_ 1 = x2
x_ 2 =
4 16
x1 (0) = 0
x2 (0) = 20 m/s
a=
The MATLAB program that integrates the equations is
function problem12_56
[t,x] =ode45(@f,[0:0.5:25],[0 20]);
printSol(t,x)
function dxdt = f(t,x)
dxdt = [x(2)
-7/4 + x(2)/16];
end
end
The 2 lines of output that span the instant where v = 0 are
t
2.0000e+001
2.0500e+001
x1
x2
2.4124e+002 7.7255e-002
2.4105e+002 -8.0911e-001
By inspection of output, the stopping distance is x = 241 m J
12.57
v2
10
Letting x1 = x, x2 = v, the equivalent …rst-order equations and initial conditions
are
x22
x_ 1 = x2
x_ 2 =
10
x1 (0) = 0
x2 (0) = 20 in/.s
a=
The MATLAB program that integrates the equations is
27
33
© 2010. Cengage Learning, Engineering. All Rights Reserved.
function problem12_57
[t,x] =ode45(@f,(0:0.01:0.51),[0 20]);
printSol(t,x)
function dxdt = f(t,x)
dxdt = [x(2)
-x(2)^2/10];
end
end
The 3 lines of output that span the instant where v = 10 in./s are
t
4.9000e-001
5.0000e-001
5.1000e-001
x1
6.8310e+000
6.9315e+000
7.0310e+000
x2
1.0101e+001
1.0000e+001
9.9010e+000
By inspection, when v = 10 in./s, t = 0:500 s J
12.58
Letting x1 = x, x2 = v, the equivalent …rst-order equations and initial conditions
are
x_ 1 = x2
x_ 2 = 32:2(1 32:3 10 6 x22 )
x1 (0) = 0
x2 (0) = 0
The MATLAB program that integrates the equations is
function problem12_58
[t,x] =ode45(@f,(0:0.1:7),[0 0]);
printSol(t,x)
function dxdt = f(t,x)
dxdt = [x(2)
32.2*(1-32.3e-6*x(2)^2)];
end
end
The 2 lines of output that span the instant where v = 100 mi/h = 146:67 ft/s
are
t
6.5000e+000
6.6000e+000
x1
5.6244e+002
5.7710e+002
x2
1.4612e+002
1.4710e+002
Linear interpolation:
6:6
147:10
6:5
t 6:5
=
146:12
146:67 146:12
t = 6:56 s J
28
34
© 2010. Cengage Learning, Engineering. All Rights Reserved.
12.59
a=
5796 1
p
5
x2
+9
x in./s
2
Letting x1 = x, x2 = v, the equivalent …rst-order equations and initial conditions
are
!
5
x1
x_ 1 = x2
x_ 2 = 5796 1 p 2
x1 + 9
x1 (0) = 8 in.
x2 (0) = 0
The MATLAB program that integrates the equations is
function problem12_59
[t,x] =ode45(@f,(0:0.001:0.042),[8 0]);
printSol(t,x)
function dxdt = f(t,x)
dxdt = [x(2)
-5796*(1 - 5/sqrt(x(1)^2 + 9))*x(1)];
end
end
The 2 lines of output that span the instant where x = 0 are
t
x1
x2
3.9000e-002 3.7400e-002 -2.2275e+002
4.0000e-002 -1.8542e-001 -2.2304e+002
By inspection, when x = 0, v = 223 in./s J
12.60
a=
32:2 1
6:24
10
4 2
v exp( 3:211
10
5
x) ft/s
2
(a) Letting x1 = x, x2 = v, the equivalent …rst-order equations and initial
conditions are
x_ 1 = x2
x_ 2 =
32:2 1
6:24
x1 (0) = 30 000 ft
10
4 2
x2 e 3:211 10
5
x1
x2 (0) = 0
The MATLAB program that integrates the equations is
function problem12_59
function problem12_60
[t,x] = ode45(@f,(0:0.2:10),[30000 0]);
printSol(t,x)
axes(’fontsize’,14)
plot(x(:,1),x(:,2),’linewidth’,1.5)
29
35
© 2010. Cengage Learning, Engineering. All Rights Reserved.
grid on
xlabel(’x (ft)’); ylabel(’v (ft/s)’)
function dxdt = f(t,x)
dxdt = [x(2)
-32.2*(1-6.24e-4*x(2)^2*exp(-3.211e-5*x(1)))];
end
end
The 3 lines of output that span the instant where v = vmax are
t
7.4000e+000
7.6000e+000
7.8000e+000
x1
x2
2.9612e+004 -6.4384e+001
2.9599e+004 -6.4385e+001
2.9586e+004 -6.4384e+001
By inspection, vmax = 64:4 ft/s J at x = 29 600 ft J
(b)
0
-10
v (ft/s)
-20
-30
-40
-50
-60
-70
2.94
2.95
2.96
2.97
x (ft)
2.98
2.99
3
x 10
4
12.61
(a)
mg y
P = kx
v
F=µN
=
N
FBD
x
ma
MAD
The FBD shown is valid only if v > 0 (block is moving to the right.) If v < 0
(block is moving to the left), the direction of the friction force F must be
30
36
© 2010. Cengage Learning, Engineering. All Rights Reserved.
reversed.
Fy
Fx
)
=
a=
= 0 +"
= ma +!
k
x
m
30
x
1:6
N
N
sign(v) =
m
0:2(9:81) sign(v) =
mg = 0
) N = mg
kx
N sign(v) = ma
k
m
g sign(v)
18:75x
1:962 sign(v) m/s
2
J
(b) With the notation x1 = x and x2 = v, the equivalent …rst-order equations
are
x_ 1 = x2
x_ 2 = 18:75x1 1:962 sign(x2 )
subject to the initial conditions x1 = 0, x2 = 6 m/s at t = 0.
The corresponding MATLAB program is:
function problem12_61
[t,x] =ode45(@f,[0:0.02:1.2],[0 6]);
printSol(t,x)
axes(’fontsize’,14)
plot(x(:,1),x(:,2),’linewidth’,1.5)
grid on
xlabel(’x (m)’); ylabel(’v (m/s)’)
function dxdt = f(t,x)
dxdt = [x(2)
-18.75*x(1)-1.962*sign(x(2))];
end
end
The block stops twice during the period 0 < t < 1:2 s. Only the lines of output
that span the instant where v = 0 are shown below.
t
3.4000e-001
3.6000e-001
x1
x2
1.2844e+000 1.3160e-001
1.2822e+000 -3.5272e-001
Linear interpolation:
3:6 3:4
t1 3:4
=
0:35272 0:13160
0 0:13160
t1 = 3:45 s J
1.0600e+000 -1.0745e+000 -2.1421e-001
1.0800e+000 -1.0745e+000 2.0038e-001
Linear interpolation:
t2 1:06
1:08 1:06
=
0:20038 ( 0:21421)
0 ( 0:21421)
t2 = 1:070 s J
31
37
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(c)
6
4
v (m/s)
2
0
-2
-4
-6
-1.5
-1
-0.5
0
x (m)
0.5
1
1.5
12.62
(a)
y
mg
P(t)
kx
=
x
N
FBD
Fx = ma
P (t)
=
=
)
=
+
!
ma
MAD
P (t)
kx = ma
25t N when t
25 N when t
a=
P (t)
m
k
x
m
1s
1s
(12:5t + 12:5) + (12:5t
12:5) sgn (1
t)
(12:5t + 12:5) + (12:5t 12:5) sgn (1 t) 25
x
2
2
2
6:25 [(t + 1) + (t 1) sgn (1 t)] 12:5x m/s J
a=
(b) With the notation x1 = x and x2 = v, the equivalent …rst-order equations
are
x_ 1 = x2
x_ 2 = 6:25 [(t + 1) + (t 1) sgn (1 t)] 12:5x1
subject to the initial conditions x1 = x2 = 0 at t = 0.
The corresponding MATLAB program is:
32
38
© 2010. Cengage Learning, Engineering. All Rights Reserved.
function problem12_62
[t,x] = ode45(@f,(0:0.05:3),[0 0]);
printSol(t,x)
axes(’fontsize’,14)
plot(x(:,1),x(:,2),’linewidth’,1.5)
grid on
xlabel(’x (m)’); ylabel(’v (m/s)’)
function dxdt = f(t,x)
dxdt = [x(2)
6.25*(t+1 + (t-1)*sign(1-t)) - 12.5*x(1)];
end
end
Below are partial printouts that span vmax and xmax .
t
8.0000e-001
9.0000e-001
1.0000e+000
x1
7.1290e-001
9.1150e-001
1.1087e+000
x2
1.9518e+000
2.0003e+000
1.9205e+000
By inspection, vmax = 2:00 m/s J
1.3000e+000
1.4000e+000
1.5000e+000
1.5271e+000 6.0188e-001
1.5535e+000 -8.0559e-002
1.5113e+000 -7.5304e-001
By inspection, xmax = 1:554 m J
(c)
2
1.5
1
v (m/s)
0.5
0
-0.5
-1
-1.5
-2
0
0.5
1
x (m)
1.5
2
33
39
© 2010. Cengage Learning, Engineering. All Rights Reserved.
12.63
a = 80
16v 1:5 ft/s
2
With the notation x1 = x and x2 = v, the equivalent …rst-order equations are
x_ 1 = x2
x_ 2 = 80
16x1
subject to the initial conditions x1 = x2 = 0 at t = 0.
The corresponding MATLAB program is:
function problem12_63
[t,x] =ode45(@f,[0:0.005:0.15],[0 0]);
printSol(t,x)
function dxdt = f(t,x)
dxdt = [x(2)
80 - 16*x(2)^1.5];
end
end
Only the 4 lines of output that span x = 0:25 ft are shown below.
t
1.0500e-001
1.1000e-001
1.1500e-001
1.2000e-001
x1
2.2384e-001
2.3822e-001
2.5263e-001
2.6709e-001
x2
2.8693e+000
2.8794e+000
2.8876e+000
2.8944e+000
Use linear interpolation to …nd v at x = 0:25 ft:
2:8876
0:25263
2:8794
v 2:8794
=
0:23822
0:25 0:23822
v = 2:89 ft/s J
12.64
x =
x_ =
x
• =
Fx
Fy
=
=
1
(y 12)2 m
y_ = 10 m/s (const.)
40
1
(y 12)y_ m/s
20
1
1
2
(y 12)•
y + y_ 2 =
0 + 102 = 5 m/s
20
20
m•
x = 0:5(5) = 2:5 N J
my = 0 J
34
40
© 2010. Cengage Learning, Engineering. All Rights Reserved.
12.65
y = 64:4(4
x_ =
y_ =
128:8t ft/s
y• =
128:8 ft/s
x
•=
Fx
Fy
t ft
2
10 sin t ft/s
2
2
5 2 cos t ft/s
2
t2 ) ft
x = 20 cos
2
4
5 2 cos t = 6:130 cos t lb
32:2
2
2
4
= m•
y=
( 128:8) = 16:00 lb
32:2
= m•
x=
t (s)
0
1
2
Fx (lb)
6:13
0
6:13
Fy (lb)
16:0
16:0
16:0
J
12.66
12.67
x = b sin
2 t
t0
y=
2 b
2 t
cos
t0
t0
2 t
4 2b
sin
ax =
t20
t0
vx =
b
4
vy =
ay =
4 t
t0
b
4 t
sin
t0
t0
2
4 t
4 b
cos
t20
t0
1 + cos
At point B: x = 0 ) t = 0
) ax = 0
) ay =
4
2
t20
b
=
4
2
(1:2)
=
0:82
74:02 m/s
2
1
41
© 2010. Cengage Learning, Engineering. All Rights Reserved.
N
0.2N
y
x
F =
0.5(74.02) N
MAD
0.5(9.81) N
FBD
Fy = may
+"
N
0:5(9:81) =
0:5(74:02)
N = 32:11 N
Fx = 0
+
!
F
0:2N = 0
F = 0:2N = 0:2(32:11) = 6:42 N J
12.68
2
42
© 2010. Cengage Learning, Engineering. All Rights Reserved.
12.69
x
y
20o
g
Acceleration
diagram
From the acceleration diagram of a water droplet:
ax =
g sin 20 =
11:013 ft/s
2
ay =
g cos 20 =
30:26 ft/s
2
At t = 0 (initial conditions):
x=y=0
vx = 20 cos ft/s
vy = 20 sin ft/s
Integrating and using initial conditions, we get
vx = 11:013t + 20 cos ft/s
x = 5:507t2 + 20t cos ft
vy = 30:26t + 20 sin ft/s
y = 15:13t2 + 20t sin ft
When y = 0:
15:13t2 + 20t sin = 0
t = 1:3219 sin s
3
43
© 2010. Cengage Learning, Engineering. All Rights Reserved.
= jxj = 5:507(1:3219 sin )2 + 20(1:3219 sin ) cos
= j4:81 cos 2 + 13:22 sin 2
4:81j ft J
R
12.70
From Eqs. (e) of Sample Problem 12.11:
x =
y
(v0 cos ) t = (70 cos 65 )t = 29:58t ft
1
1 2
gt + (v0 sin ) t =
(32:2)t2 + (70 sin 65 )t
=
2
2
=
16:1t2 + 63:44t ft
At point B:
x = 60 ft
60 = 29:58t
t = 2:028 s
h = yjt=2:028 = 16:1(2:028)2 + 63:44(2:028) = 62:4 ft J
*12.71
12.72
y
x
g
Acceleration
diagram
At t = 0 (initial conditions):
x = 0
vx = 200 sin 30 = 100 m/s
y = 1200 m
vy = 200 cos 30 = 173:21 m/s
4
44
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Integrating acceleration and applying initial conditions:
ax = 0
vx = 100 m/s
x = 100t m
2
ay = 9:81 m/s
vy = 9:81t 173:21 m/s
y = 4:905t2 173:21t + 1200 m
When y = 0:
4:905t2
173:21t + 1200 = 0
t = 5:932 s
x = 100(5:932) = 593:2 m
d = 1200 tan 30
593:2 = 99:6 m J
12.73
From Eqs. (e) of Sample Problem 12.11:
x = (v0 cos ) t
y=
1 2
gt + (v0 sin ) t
2
(a) At point B:
y
=
0
x = R
1 2
gt + (v0 sin ) t = 0
2
2v0
sin
R = (v0 cos )
g
sin cos =
Rg
2v02
t=
=
sin 2 =
2v0
sin
g
2v02
sin cos
g
Rg
J
v02
(b)
sin 2 =
2
2
1
2
3000(9:81)
= 0:4709
2502
= sin 1 0:4709 = 28:09
= 180
28:09 = 151:91
1
= 14:05 J
J
2 = 76:0
12.74
From Eqs. (e) of Sample Problem 12.11:
x = (v0 cos ) t
y=
1 2
gt + (v0 sin ) t
2
5
45
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(a)
x =
y
(42 cos 28 ) t = 37:08t ft
)t=
1
(32:2)t2 + (42 sin 28 ) t =
2
=
x
s
37:08
16:1t2 + 19:718t
Substituting for t:
y
=
=
2
x
x
+ 19:718
37:08
37:08
0:01171x2 + 0:5318x ft J
16:1
(b) Check if ball hits the ceiling.
dy
dx
=
0:02342x + 0:5318 = 0
ymax
=
0:01171(22:71)2 + 0:5318(22:71) = 6:04 ft
x = 22:71 ft
Since ymax < 25 ft, the ball will not hit the ceiling.
Check if ball clears the net. When x = 22 ft:
y=
0:01171(22)2 + 0:5318(22) = 6:03 ft
Since y > 5 ft, the ball clears the net. J
When x = 42 ft:
y=
0:01171(42)2 + 0:5318(42) = 1:679 ft
Since y > 0, the ball lands behind the baseline. J
12.75
From Eqs. (e) of Sample Problem 12.11:
x = (v0 cos ) t
y
vy
y=
1 2
gt + (v0 sin ) t
2
1
(32:2)t2 + (v0 sin 70 )t =
2
= y_ = 32:2t + 0:9397v0 ft/s
=
16:1t2 + 0:9397v0 t ft
When y = ymax
vy = 0
ymax = 27 ft
32:2t + 0:9397v0 = 0
t = 0:02918v0 s
16:1(0:02918v0 )2 + 0:9397v0 (0:02918v0 ) = 27
v0 = 44:4 ft/s J
0:013712v02 = 27
6
46
© 2010. Cengage Learning, Engineering. All Rights Reserved.
12.76
12.77
7
47
© 2010. Cengage Learning, Engineering. All Rights Reserved.
*12.78
12.79
(a)
FD(vx /v)
FD
y
mg vx
v
vy
may
x =
max
FD(vy /v)
MAD
FBD
8
48
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Fx = max
ax
=
=
=
=
FD
vx
= max
v
0:0005v 2 vx
FD vx
=
= 0:005vvx
m v q
0:1
v
2
0:005vx vx2 + vy2 m/s J
Fy = may
ay
+
!
+"
FD
vy
v
mg = may
FD vy
0:0005v 2 vy
g=
9:81 =
m v q
0:1
v
2
0:005vy vx2 + vy2 9:81 m/s J
0:005vvy
9:81
(b) Letting x1 = x, x2 = y, x3 = vx and x4 = vy , the equivalent …rst-order
equations are
x_ 1 = x3
x_ 2 = x4
q
q
x_ 3 = 0:005x3 x23 + x24
x_ 4 = 0:005x4 x23 + x24 9:81
The initial conditions are
x1 (0) = 0
x2 (0) = 2 m/s
x3 (0) = 30 cos 50 = 19:284 m/s
x4 (0) = 30 sin 50 = 22:981 m/s
The following MATLAB program was used to integrate the equations:
function problem12_79
[t,x] = ode45(@f,(0:0.05:2),[0 2 19.284 22.981]);
printSol(t,x)
function dxdt = f(t,x)
v = sqrt(x(3)^2 + x(4)^2);
dxdt = [x(3)
x(4)
-0.005*x(3)*v
-0.005*x(4)*v-9.81];
end
end
The two lines of output that span x = 30 m are
t
1.7000e+000
1.7500e+000
x1
2.9607e+001
3.0405e+001
x2
2.3944e+001
2.4117e+001
x3
1.5990e+001
1.5925e+001
x4
3.6997e+000
3.1952e+000
9
49
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Linear interpolation for h:
30:405
24:117
29:607
30
=
23:944
h
29:607
23:944
h = 24:0 m J
Linear interpolation for vx and vy :
30:405
15:925
29:607
30
=
15:990
vx
29:607
15:990
vx = 15:958 m/s
30 29:607
29:607
=
vy = 3:451 m/s
3:6997
vy 3:6997
p
) v = 15:9582 + 3:4512 = 16:33 m/s J
30:405
3:1952
12.80
(a)
F y
F(y/d)
may
x
max
F(x/d) =
FBD
Fx = max
ax =
+
!
x
= max
d
0:5x
2
m/s J
=
3=2
(x2 + y 2 )
y
F = may
d
0:5y
2
=
m/s J
3=2
(x2 + y 2 )
F
1 x
1 0:005 x
x
F =
= 0:5 3
2
m d
0:01 d d
d
Fy = may
ay =
MAD
+"
1 0:005 y
y
1 y
F =
= 0:5 3
m d
0:01 d2 d
d
The initial conditions are:
x = 0:3 m y = 0:4 m vx = 0
vy =
2 m/s at t = 0 J
(b) Letting x1 = x, x2 = y, x3 = vx and x4 = vy , the equivalent …rst-order
equations are
x_ 1 = x3
x_ 2 = x4
x_ 3 =
0:5x1
(x21 +
3=2
x22 )
x_ 4 =
0:5x2
3=2
(x21 + x22 )
The MATLAB program for solving the equations is
10
50
© 2010. Cengage Learning, Engineering. All Rights Reserved.
function problem12_80
[t,x] = ode45(@f,(0:0.005:0.25),[0.3 0.4 0 -2]);
printSol(t,x)
function dxdt = f(t,x)
d3 = (sqrt(x(1)^2 + x(2)^2))^3;
dxdt = [x(3)
x(4)
0.5*x(1)/d3
0.5*x(2)/d3];
end
end
The two output lines spanning y = 0 are shown below.
t
2.2000e-001
2.2500e-001
x1
x2
3.5879e-001 3.0450e-003
3.6213e-001 -5.2884e-003
x3
x4
6.5959e-001 -1.6667e+000
6.7883e-001 -1.6668e+000
Linear interpolation for x at y = 0:
0:36213 0:35879
x 0:35879
=
0:0052884 0:0030450
0 0:0030450
x = 0:360 m J
Linear interpolation for vx :
vx :
0:67883 0:65959
vx 0:65959
=
0:0052884 0:0030450
0 0:0030450
vx = 0:6666 m/s
By inspection vy =
1:6667 m/s.
p
) v = 0:66662 + 1:66672 = 1:795 m/s J
12.81
(a) The signs of ax and ay in the solution of Prob. 12.80 must be reversed.
ax =
0:5x
(x2
+
3=2
y2 )
m/s
2
J
ay =
0:5y
(x2
+
3=2
y2 )
m/s
2
J
The initial conditions are the same as in Problem 12.80:
x = 0:3 m y = 0:4 m vx = 0 vy =
2 m/s
at t = 0: J
(b) MATLAB program:
function problem12_81
[t,x] = ode45(@f,(0:0.005:0.2),[0.3 0.4 0 -2]);
printSol(t,x)
function dxdt = f(t,x)
11
51
© 2010. Cengage Learning, Engineering. All Rights Reserved.
d3 = (sqrt(x(1)^2 + x(2)^2))^3;
dxdt = [x(3)
x(4)
-0.5*x(1)/d3
-0.5*x(2)/d3];
end
end
The two lines of output spanning y = 0 are:
t
1.8000e-001
1.8500e-001
x1
x2
x3
x4
2.5952e-001 8.5117e-003 -6.3935e-001 -2.3329e+000
2.5623e-001 -3.1544e-003 -6.7692e-001 -2.3333e+000
Linear interpolation for x at y = 0:
0:25623 0:25952
x 0:25952
=
0:0031544 0:0085117
0 0:0085177
Linear interpolation for vx :
vx
0:67692 ( 0:63935)
=
0:0031544 0:0085117
0
By inspection, vy =
12.82
x = 0:257 m J
( 0:63935)
0:0085177
vx =
0:6668 m/s
2:3332 m/s.
p
v = 0:66682 + 2:33322 = 2:43 m/s J
FD (vx/v)
FD
mg vx v y
vy
may
x
max
=
FD (vy/v)
FBD
MAD
vx
vx
= max
cD v 1:5
= max
v
v
cD p
) ax =
vx v
m
cD
0:0012
0:5
=
= 0:06869 (ft s)
m
(9=16)(32:2)
Fx = max
+
!
) ax =
Fy = may
+"
FD
0:06869vx vx2 + vy2
vy
FD
mg = may
y
cD p
vy v g =
m
The initial conditions are:
) ay =
x=0
y = 6 ft
0:25
0:06869vy vx2 + vy2
vx = 120 ft/s
J
vy
cD v 1:5
y
ft/s
2
0:25
vy = 0
mg = may
32:2 ft/s
2
J
at t = 0 J
12
52
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(b) Letting x1 = x, x2 = y, x3 = vx and x4 = vy , the equivalent …rst-order
equations are
x_ 1
= x3
x_ 2 = x4
x_ 3
=
0:06869x3 x23 + x24
0:25
x4
=
0:06869x4 x23 + x24
0:25
32:2
The MATLAB program is
function problem12_82
[t,x] = ode45(@f,(0:0.02:0.7),[0 6 120 0]);
printSol(t,x)
function dxdt = f(t,x)
v25 = sqrt((sqrt(x(3)^2 + x(4)^2)));
dxdt = [x(3)
x(4)
-0.06869*x(3)*v25
-0.06869*x(4)*v25-32.2];
end
end
The two lines of output that span y = 0 are:
t
6.4000e-001
6.6000e-001
x1
x2
6.1878e+001 2.6073e-001
6.3426e+001 -8.0650e-002
x3
x4
7.7840e+001 -1.6851e+001
7.6894e+001 -1.7286e+001
Linear interpolation for x at y = 0:
R 61:878
63:426 61:878
=
0:080650 0:26073
0 0:26072
R = 63:1 ft J
Linear interpolation for t at y = 0:
t 0:64
0:66 0:64
=
0:080650 0:26073
0 0:26072
t = 0:655 s J
12.83
ax =
10
0:5vx m/s
2
ay =
9:81
0:5vy m/s
2
(a) Letting x1 = x, x2 = y, x3 = vx and x4 = vy , the equivalent …rst-order
equations are
x_ 1 = x3
x_ 2 = x4
x_ 3 =
10
0:5x3
x_ 4 =
9:81
0:5x4
At t = 0 (initial conditions):
x1 = x2 = 0
x3 = 30 cos 40 = 22:98 m/s
x4 = 30 sin 40 = 19:284 m/s
MATLAB program:
13
53
© 2010. Cengage Learning, Engineering. All Rights Reserved.
function problem12_83
[t,x] = ode45(@f,(0:0.05:3.5),[0 0 22.98 19.284]);
printSol(t,x)
axes(’fontsize’,14)
plot(x(:,1),x(:,2),’linewidth’,1.5)
grid on
xlabel(’x (ft)’); ylabel(’y (ft)’)
function dxdt = f(t,x)
dxdt = [x(3)
x(4)
-10-0.5*x(3)
-9.81-0.5*x(4)];
end
end
Two lines of output that span y = 0:
t
3.1000e+000
3.1500e+000
x1
x2
x3
x4
5.7152e+000 4.7141e-001 -1.0878e+001 -1.1363e+001
5.1656e+000 -1.0184e-001 -1.1103e+001 -1.1567e+001
Linear interpolation for x at y = 0:
5:1656 5:7152
b
=
0:10184 0:47141
0
5:7152
0:47141
b = 5:26 m J
3:15 3:10
t 3:10
=
0:10184 0:47141
0 0:47141
t = 3:14 s J
Linear interpolation for t at y = 0:
(b)
15
y (ft)
10
5
0
0
5
10
x (ft)
15
20
14
54
© 2010. Cengage Learning, Engineering. All Rights Reserved.
12.84
(a)
y
mg
x
R
y
θ
Spring force is F = k(R
x
F
MAD
FBD
+
!
p
x2 + y 2 :
F cos = max
k(R L0 ) x
F
cos =
=
m
m
R
10
0:5
1
x = 40 1
0:25
R
=
=
Fy = may
ay
max
=
L0 ), where R =
Fx = max
ax
θ
may
=
F
sin
m
=
40 1
+"
k(R
g=
0:5
R
y
F sin
L0 ) y
m
R
9:81 m/s
2
k
L0
1
x
m
R
0:5
2
x m/s J
R
mg = may
g=
k
m
1
L0
R
y
g
J
The initial conditions are:
x = 0:5 m y =
0:5 m vx = vy = 0 at t = 0 J
(b) Letting x1 = x, x2 = y, x3 = vx and x4 = vy , the equivalent …rst-order
equations are
x_ 1
= x3
x_ 2 = x4
0:5
40 1
x1
R
x_ 3
=
x_ 4 =
40 1
0:5
R
x2
9:81
p
where R = x21 + x22 .
The MATLAB program is:
function problem12_84
[t,x] = ode45(@f,(0:0.02:2),[0.5 -0.5 0 0]);
axes(’fontsize’,14)
plot(x(:,1),x(:,2),’linewidth’,1.5)
grid on
xlabel(’x (m)’); ylabel(’y (m)’)
function dxdt = f(t,x)
15
55
© 2010. Cengage Learning, Engineering. All Rights Reserved.
rr = 1-0.5/sqrt(x(1)^2 + x(2)^2);
dxdt = [x(3)
x(4)
-40*rr*x(1)
-40*rr*x(2)-9.81];
end
end
-0.4
-0.5
y (m)
-0.6
-0.7
-0.8
-0.9
-1
-1.1
-0.5 -0.4 -0.3 -0.2 -0.1
0 0.1 0.2 0.3 0.4 0.5
x (m)
12.85
(a) The expressions for the accelerations in Prob. 12.84 are now valid only
when the spring is in tension. If the spring is not in tension, the spring force is
zero. Therefore, we have
8
0:5
<
2
x m/s if R > 0:5 m
40 1
ax =
J
R
:
0
if R 0:5 m
8
< 40 1 0:5 y 9:81 m/s2 if R > 0:5 m
J
ay =
R
:
2
9:81 m/s
if R 0:5 m
The initial conditions are:
x = y = 0:5m vy = vy = 0 at t = 0 J
(b) MATLAB program:
function problem12_85
[t,x] = ode45(@f,(0:0.02:2),[0.5 0.5 0 0]);
16
56
© 2010. Cengage Learning, Engineering. All Rights Reserved.
axes(’fontsize’,14)
plot(x(:,1),x(:,2),’linewidth’,1.5)
grid on
xlabel(’x (m)’); ylabel(’y (m)’)
function dxdt = f(t,x)
rr = 1-0.5/sqrt(x(1)^2 + x(2)^2);
if rr < 0; rr = 0; end
dxdt = [x(3)
x(4)
-40*rr*x(1)
-40*rr*x(2)-9.81];
end
end
1
0.5
y (m)
0
-0.5
-1
-1.5
-0.4
-0.2
0
0.2
0.4
0.6
x (m)
12.86
(a)
ax =
aD cos + aL sin
ay =
aD sin
aL cos
g
Substitute
sin
aL
where v =
vx2
+
=
=
vx
vy
cos =
aD = 0:05v 2
v
v
0:16!v = 0:16(10)v = 1:6v
vy2 .
vx
vy
+ 1:6v
=
v
v
vy
vx
ay = 0:05v 2
1:6v
32:2 =
v
v
The initial conditions at t = 0 are:
ax =
x=y=0
0:05v 2
0:05vvx + 1:6vy ft/s
0:05vvy
vx = 60 cos 60 = 30 ft/s
1:6vx
2
J
32:2 ft/s
2
J
vy = 60 sin 60 = 51:96 ft/s
17
57
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(b) Letting x1 = x, x2 = y, x3 = vx and x4 = vy , the equivalent …rst-order
equations are
x_ 1
x_ 3
= x3
x_ 2 = x4
=
0:05vx3 + 16x4
x_ 4 =
0:05vx4
16x3
32:2
MATLAB program:
function problem12_86
[t,x] = ode45(@f,(0:0.02:1.2),[0 0 30 51.96]);
printSol(t,x)
axes(’fontsize’,14)
plot(x(:,1),x(:,2),’linewidth’,1.5)
grid on
xlabel(’x (ft)’); ylabel(’y (ft)’)
function dxdt = f(t,x)
v = sqrt(x(3)^2 + x(4)^2);
dxdt = [x(3)
x(4)
-0.05*v*x(3)+1.6*x(4)
-0.05*v*x(4)-1.6*x(3)-32.2];
end
end
The two lines of output that span y = 0 are
t
1.0400e+000
1.0600e+000
x1
x2
1.9377e+001 2.5868e-001
1.9388e+001 -1.9335e-001
x3
x4
9.6888e-001 -2.2523e+001
2.3210e-001 -2.2675e+001
Linear interpolation for t at y = 0:
1:06 1:04
t 1:04
=
0:19335 0:25868
0 0:25868
t = 1:051 s J
Linear interpolation for x at y = 0:
19:388 19:377
x 19:377
=
0:19335 0:25868
0 0:25868
x = 19:38 ft J
18
58
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(c)
10
8
y (ft)
6
4
2
0
0
5
10
x (ft)
15
20
12.87
a (ft/s2)
3
60
150 170 190 t (s)
-20
-40
0 20
0
-1
-2
v (ft/s)
60
40
7800
1000
600
400
0
t (s)
x (ft)
9800
9400
8400
600
t (s)
0
d = 9800 ft J
19
59
© 2010. Cengage Learning, Engineering. All Rights Reserved.
12.88
a=
F
m
g=
60 103
2000
9:81 = 20:19 m/s
2
a (m/s2)
20.19
403.8
0
20 t (s)
0
v (m/s)
403.8
4038
0
20
0
y (m)
4038
0
0
t (s)
20 t (s)
yjt=20 = 4038 m J
12.89
It is su¢ cient to consider only vertical motion: a =
Initial conditions:
vjt=0 = v0 sin = 0:9397v0 ft/s
a (ft/s2)
0 0
g=
32:2 ft/s2 :
yjt=0 = 3 ft
t1
t (s)
-32.2t1
-32.2
v (ft/s)
0.9397v0
0
0.4698v0 t1
t1
0
t (s)
y (ft)
27
0
0
t1 t (s)
20
60
© 2010. Cengage Learning, Engineering. All Rights Reserved.
End conditions (t1 is the time when the ball is at its maximum height):
vjt=t1 = 0
) 0:9397v0
yjt=t1 = 27 ft
32:2t1 = 0
t1 = 0:02918v0
) 0:4698v0 t1 = 27
0:4698(0:02918)v02 = 27
v0 = 44:4 ft/s J
12.90
Horizontal motion: ax = 0
vx jt=0 = 70 cos 65 = 29:58 ft/s
ax (ft/s2)
0
t1 t (s)
0
vx (ft/s)
29.58
0
0
x (ft)
60
0
0
29.58t1
t1
Vertical motion: ay = 32:2 ft/s
vy jt=0 = 70 sin 65 = 63:44 ft/s
ay (ft/s2)
0
0
-32.2
vy (ft/s)
63.44
0
0
y (ft)
62.4
t (s)
2.028
-65.30
62.49
00
t1 t (s)
) 29:58t1 = 60
End condition: xjt=t1 = 60 ft
2
t (s)
2.028
t (s)
1.970
-0.05
2.028
t (s)
t1 = 2:028 s
h = yjt=2:028 = 62:4 ft J
21
61
© 2010. Cengage Learning, Engineering. All Rights Reserved.
12.91
Horizontal motion: ax = 0 Vertical motion: ay =
ax (m/s2)
0
t (s)
0
vx (m/s)
200
0
ay (m/s2)
0
0
-9.81t1
-9.81
9:81 m/s
t1
2
t (s)
vy (m/s)
200 t1
t1
0
7
00
t (s)
t (s)
-68.67
y (m)
h
x (m)
1400
0
0
0
0
t1 t (s)
End condition: xjt=t1 = 1400 m
7
200t1 = 1400
h = area under vy diagram =
t (s)
t1 = 7:0 s
240:3 m J
12.92
Horizontal motion: ax = 0 Vertical motion: ay =
ay (ft/s2)
0
0
ax (ft/s2)
0
0
10
t (s)
-32.2
vx (ft/s)
v0
0
0
10
10
00
t (s)
t (s)
-1610
t (s)
-322
y (ft)
1610
x (ft)
10v0
0
0
0
0
10 t (s)
= tan 20
t=10
10
-322
2
vy (ft/s)
10v0
vy
vx
32:2 ft/s
10
322
= tan 20
v0
R = xjt=10 = 10v0 = 8850 ft J
t (s)
v0 = 885 ft/s J
h = yjt=0 = 1610 ft J
22
62
© 2010. Cengage Learning, Engineering. All Rights Reserved.
12.93
y
22.6o
x
g
Acceleration
diagram
ax
= g sin 22:6 = 32:2 sin 22:6 = 12:374 ft/s
ay
=
g cos 22:6 =
32:2 cos 22:6 =
2
29:73 ft/s
2
At t = 0 (initial conditions):
x = 0
y = 0
ax (ft/s)
12.374
0
0
vx = 260 cos 22:6 = 240:0 ft/s
vy = 260 sin 22:6 = 99:92 ft/s
t1 = 3.361 s
6.722 t (s)
ay (ft/s)
83.18
6.722
0
t (s)
-29.73 t1
= -99.92
-29.73
vx (ft/s)
323.2
240.0
0
vy (ft/s)
1892.8
x (ft)
1892.8
167.92
99.92
t (s)
0
-99.92
t (s)
y (ft)
167.92
0
vy = 0 at t = t1 .
h = 167:9 ft J
t (s)
) 99:92
t (s)
0
29:73t1 = 0
R = 1893 ft J
) t1 = 3:361 s
time of ‡ight = 6:72 s J
23
63
© 2010. Cengage Learning, Engineering. All Rights Reserved.
12.94
ax (m/s2)
0
0
vx (m/s)
190.52
0
0
16.091
t (s)
t1
-9.81t1
t (s)
vy (m/s)
110
3066
16.091
x (m)
3066
0
0
ay (m/s2)
0
0
-9.81
16.091
110 - 9.81t1
t (s)
0
(110 - 4.905t1)t1
0
t1 t (s)
y (m)
(110 - 4.905t1)t1
0
0
t (s)
End condition: yjt t1 = 500 m. ) (110
The larger root is t1 = 16:091 s J
R = xjt=16:091 = 3066 m J
t1
t (s)
4:905t1 )t1 = 500
12.95
24
64
© 2010. Cengage Learning, Engineering. All Rights Reserved.
12.96
2 mi = 2(5280) = 10 560 ft
45 mi/h = 45
5280
= 66 ft/s
3600
Accelerate at the maximum rate (6.6 ft/s2 ) until maximum allowable speed (66
ft/s) is reached at time t1 . Then maintain this speed until time t2 . Finally,
decelerate at the maximum rate (5.5 ft/s2 ) until the train stops at time t3 . The
distance traveled during this time must be 10 560 ft.
a (ft/s2)
6.6 0
6.6t1
0
t2
t1
t3
t (s)
-5.5(t3 - t2)
-5.5
v (ft/s)
66
0
0
330 66(t2 - 10) 396
t (s)
10
t2 + 12
t2
x (ft)
10 560
0
0
10
159
171
t (s)
From a and v diagrams:
6:6t1 = 66 ft/s
5:5(t3 t2 ) = 66 ft/s
) t1 = 10 s
) t3 t2 = 12 s
From v and x diagrams:
330 + 66(t2
10) + 396 = 10 560
) t2 = 159:0 s
t3 = t2 + 12 = 159:0 + 12 = 171:0 s J
25
65
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12.97
26
66
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12.98
a (m/s 2)
8 + t1 = 11
8
t (s)
00 3
-1.8t1
-1.8 -9.6
-25
t1
-3.2
-5.0
v (m/s)
40
30.4
105.6
89.5
5.4
0
8.1
t (s)
x (m)
203.2
195.1
105.6
t (s)
v = 0 when 40
9:6
25
1:8t1 = 0
) t1 = 3:0s
After touchdown, the plane travels 203 m J
12.99
27
67
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12.100
28
68
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12.101
29
69
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12.102
a (m/s2)
0.40
0.09756
00
t1 = 0.4878
1.0
t (s)
-0.10756
-0.42
v (m/s)
0.09756
0.03173
0
-0.01
0.4878
0.03173
0.9756
t (s)
y (m)
0.06346
0.03173
t (s)
0
From similar triangles on the a-diagram:
1:0
t1
=
0:40
0:82
) t1 = 0:4878 m
vmax = 0:0976 m/s J
ymax = 0:0635 m J
12.103
a (ft/s2)
60
t1
t2
14
t (s)
0 0273.3
-32.2(
t
14)
2
-32.2
-78.7
= -194.6
v (ft/s)
273.3
194.6
1660 1208
588
0
t (s)
20.04
0
9.111 14
y (ft)
3456
2868
1660
t (s)
0
30
70
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From similar triangles on a-diagram:
t1
14
=
60
60 + 32:2
) t1 = 9:111 s
Let t2 be the time when v = 0: Therefore,
194:6
32:2(t2
vmax = 273 ft/s J
) t2 = 20:04 s
14) = 0
ymax = 3460 ft J occurring at t = 20:0 s J
12.104
v = 2x3
a=v
ajx=2 = (2(8)
dv
= 2x3
dx
8x2 + 12x mm/s
8x2 + 12x
8(4) + 12(2)) (6(4)
6x2
16x + 12
16(2) + 12) = 32:0 mm/s
2
J
12.105
a = At + B
When t = 0: a = 0 ) B = 0
When t = 6 ft/s: a = 8 ft/s ) 8 = A(6)
)a=
4
2
t ft/s
3
v=
When t = 0: v = v0 ) C = v0
When t = 6 s: v = 16 ft/s
)
2
(36) + v0 = 16
3
Z
)A=
4
ft/s3
3
a dt + C =
) v0 =
2 2
t +C
3
8:0 ft/s J
12.106
Car B
a = 2 ft/s2
v = 2t + C3 ft/s
vjt=0 = 60 ft/s ) C3 = 60 ft/s
v = 2t + 60 ft/s
x = t2 + 60t + C4 ft
xjt=0 = 0 ) C4 = 0
x = t2 + 60t ft
Car A
a = 4 ft/s2
v = 4t + C1 ft/s
vjt=0 = 30 ft/s ) C1 = 30 ft/s
v = 4t + 30 ft/s
x = 2t2 + 30t + C2 ft
xjt=0 = 400 ft ) C2 = 400 ft
x = 2t2 + 30t 400 ft
Car A overtakes car B when xA = xB :
2t2 + 30t
3t2 30t
400 =
400 =
t2 + 60t
0
t = 17:58 s J
31
71
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12.107
(a)
x = 3t3 9t + 4 in.
x = xjt=2 xjt=0 = [3(8)
(b)
v = x_ = 9t2
) v = 0 when t = 1:0 s
9 in./s
xjt=0 = 4 in.
-2
xjt=1 =
0
4 = 6:0 in. J
9(2) + 4]
2 in.
xjt=2 = 10 in.
4
x (in.)
10
d = 6 + 12 = 18 in. J
12.108
Fall of the stone:
a = 32:2 ft/s
) v = 32:2t + C1 ft/s
) y = 16:1t2 + C1 t + C2 ft
2
When t = 0: v = x = 0 ) C1 = C2 = 0
Let t1 be the time of fall and h the depth of the well. ) h = 16:1t21 ft
Travel of the sound:
Let t2 be the time for sound to travel the distance h: ) h = 1120t2 ft
) t2 = 4
t 1 + t2 = 4
16:1t21 = 1120(4
16:1t21 = 1120t2
) t1 = 3:793 s
t1
t1 )
) h = 16:1(3:793) = 232 ft J
2
12.109
2
12t 6t2 ft/s
Z
v =
a dt + C1 = 6t2
Z
x =
v dt + C2 = 2t3
a =
When t = 0: x = v = 0:
2t3 + C1 ft/s
0:5t4 + C1 t + C2 ft
) C1 = C2 = 0
) x = 2t3
) v = 6t2
0:5t4 ft
2t3 ft/s
32
72
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(a)
xjt=0 = 2(5)3
x = xjt=5
0:5(5)4
62:5 ft J
0=
(b) When v = 0:
6t2
2t3 = 0
xjt=0 = 0 xjt=3 = 2(3)
3
t = 3:0 s
4
0:5(3) = 13:5 ft xjt=5 =
13.5
0
-62.5
62:5 ft
x (ft)
d = 2(13:5) + 62:5 = 89:5 ft J
12.110
v = 16
x
m/s
4
a=v
dv
= (16
dx
x
)
4
1
4
=
4+
x
2
m/s J
16
12.111
x
m/s
4
v = 16
16
dx
= dt
x=4
dx
x
= 16
dt
4
x
4 ln 16
=t+C
4
Initial condition: x = 0 when t = 0. ) C =
) ln
16
16
t
=
x=4
4
16
16
= et=4
x=4
4 ln 16
x = 64(1
e
t=4
)m J
12.112
33
73
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12.113
12.114
F = (50
+"
a =
v
=
y
=
F
m
Z
Z
g=
(50
t)
103 N
t) 103
1400
a dt + C1 = 25:90t
v dt + C2 = 12:95t2
9:81 = 25:90
0:7143t m/s
2
0:3571t2 + C1
0:11903t3 + C1 t + C2
Initial conditions: y = v = 0 when t = 0. ) C1 = C2 = 0:
yjt=20 = 12:95(20)2
0:11903(20)3 = 4230 m J
34
74
© 2010. Cengage Learning, Engineering. All Rights Reserved.
12.115
12.116
12.117
From Eqs. (d) and (e) of Sample Problem 12.11:
x = v0 t cos
0
y=
1 2
gt + v0 t sin
2
(a)
0
Let t = t1 when the ball hits the inclined surface. At t = t1 :
x = 220 cos 20 = 206:7 ft
y=
220 sin 20 =
75:24 ft
35
75
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Using Eqs. (a):
206:7
= v0 t1 cos 25
75:24
=
v0 =
228:1
t1
1
(32:2)t21 + v0 t1 sin 25
2
) 228:1 = 38:09t21 178:02
228:1
= 69:9 ft/s J
) v0 =
3:265
v0 =
38:09t21
178:02
t1
t1 = 3:265 s
12.118
0.4 lb y
x
F 8=
FBD N
73
0.4 a
32.2
3
MAD
When x = 8 in. the elongation of the spring and the spring force are
p
F = k = 6(4:544) = 27:26 lb
= 32 + 82 4 = 4:544 in.
Fx = ma
0:4
8
p F =
a
32:2
73
+
!
0:4
8
p (27:26) =
a
32:2
73
a = 2055 ft/s
2
J
12.119
36
76
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12.120
12.121
From Eqs. (d) and (e) of Sample Problem 12.11:
x = v0 t cos
0
y=
1 2
gt + v0 t sin
2
(a) Let t = t1 when the ball hits the fairway at y =
1
(9:81)t21 + 45t1 sin 40
2
) R = 45(6:162) cos 40 = 212 m J
)
8=
0
8 m, x = R.
t1 = 6:162 s
(b) At t = 6:162 s:
vx
vy
= x_ = v0 cos 0 = 45 cos 40 = 34:47 m/s
= y_ = gt + v0 sin 0 = 9:81(6:162) + 45 sin 40 =
p
v = 34:472 + 31:522 = 46:7 m/s J
31:52 m/s
37
77
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12.122
38
78
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Chapter 13
13.1
2
1
5
3
4
13.2
2
2
at = 0:8 m/s
a = 1:5 m/s
q
p
2
an = a2 a2t = 1:52 0:82 = 1:2689 m/s
an
=
v
=
v2
)v=
15:930
m
s
p
an =
1 km
1000 m
p
1:2689(200) = 15:930 m/s
3600 s
= 57:3 km/h J
1h
13.3
athrust
an
30o
g
a
an = g cos 30 = 32:2 cos 30 = 27:89 ft/s
an =
v2
)
=
2
v2
8002
= 22 900 ft J
=
an
27:89
13.4
At point A:
at = 0
v=
p
an =
v = 31:32
m
s
p
(0:2
an =
v2
9:81) (500) = 31:32 m/s
1 km
1000 m
3600 s
= 112:8 km/h J
1h
1
79
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13.5
At point B:
v = vx = v0 cos
an =
v
2
2
=
an = g
v
=
an
2
vA
cos2
g
J
13.6
at =
dv
dv ds
=
v
ds dt
ds
v dv = at ds
)
1 2
v = at s + C
2
At point A: v = 3 m/s at s = 0. ) C = 4:5 (m/s)2
R
(4)
At point B: v = 6 m/s at s =
=
=2 m
2
2
)
1 2
(6) = at (2 ) + 4:5
2
) at = 2:149 m/s
2
v2
62
2
=
= 9 m/s
R
4
q
p
2
a = a2t + a2n = 2:1492 + 92 = 9:25 m/s J
an =
13.7
(a)
at =
dv
dv ds
=
v
ds dt
ds
v dv = at ds
)
1 2
v = at s + C
2
At point A: v = 60 in./s at s = 0. ) C = 1800 (in./s)2
At point B: v = 20 in./s at s = 40 in.
)
1
(20)2 = at (40) + 1800
2
) at =
2
40 in./s
J
(b) At point B:
an =
an =
q
a2
v2
a2t =
)
=
p
502
( 40)2 = 30 in./s
2
v2
202
2
=
= 13:33 in./s J
an
30
2
80
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.8
at
)
1 2
v
2
=
=
dv ds
dv
=
v
ds
Zds dt
at ds + C =
v dv = at ds
Z
0:05s ds + C = 0:025s2 + C
At point O: v = 20 in./s at s = 0. ) C = 200 (in./s)2
At point B: s = 80 in.
1 2
v = 0:025(80)2 + 200
2
an =
13.9
v2
=
) v 2 = 720 (in./s)
2
720
2
2
= 6:0 in./s
at = 0:05s = 0:05(80) = 4:0 in./s
120
q
p
2
a = a2n + a2t = 62 + 42 = 7:21 in./s J
13.10
3
81
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13.11
13.12
4
82
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13.13
13.14
13.15
5
83
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.16
13.17
13.18
y=
x2
m
80
y0 =
x
40
y 00 =
1
m
40
1
At point A (x = 10 m):
y0 =
1
an =
=h
v2
10
= 0:25 m
40
y 00
1+
2
(y 0 )
i3=2 =
y 00 = 0:025 m
0:025
3=2
(1 + 0:252 )
2
1
= 0:02283 m
1
= 122 (0:02283) = 3:288 m/s
at = v_ = 4 m/s
q
p
2
a = a2n + a2t = 3:2882 + 42 = 5:18 m/s J
2
6
84
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.19
at
o
20
an
g
2
at
= g sin 20 = 9:78 sin 20 = 3:345 m/s
an
= g cos 20 = 9:78 cos 20 = 9:190 m/s
an =
v2
v=
p
an =
2
p
9:190(5000) = 214 m/s J
v_ = at = 3:35 m/s
2
J
*13.20
Curve Car travels at constant speed v1 , determined by an = amax .
amax =
v12
v1 =
t1 =
p
amax =
p
5(200) = 31:62 m/s
s1
200
= 19:871 s
=
v1
31:62
Straightaway Car accelerates for 500 m at the rate v_ = amax and then brakes
for the next 500 m at the rate v_ = amax : .
For the …rst 500 m:
v
= v1 + amax t
1
s = v1 t + amax t2
2
1
500 = 31:62t2 + (5)t22
2
t2 = 9:168 s
Total time to complete the circuit is
t = 2t1 + 4t2 = 2(19:871) + 4(9:168) = 76:4 s J
7
85
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13.21
v
θ
v0
B
R
y
θ
O
v0
)_=
sec
R
v0
v2
• = v0 sec tan _ = v0 sec tan
sec
= 02 sec2 tan
R
R
R
R
v = v0 sec = 6 sec 60 = 12:0 in./s J
) y_ = v0 = R cos
y = R sin
2
an = R _ = R
v0
sec
R
= R• = R
at
2
=
_
v02
62
2
sec2 =
sec2 60 = 8:0 in./s
R
18
v02
sec2 tan
R2
=
v02
sec2 tan
R
62
2
sec2 60 tan 60 = 13:856 in./s
18
q
p
a = a2t + a2n = 13:8562 + 8:02 = 16:0 in./s J
=
*13.22
x3
ft
12 800
At point A (x = 40 ft):
y0 =
1
=
402
=
4267
0:3750
y 00
[1 +
3=2
(y 0 )2 ]
an
=
x2
4267
y0 =
y=
=
y 00 =
y 00 =
40
=
2133
0:018753
3=2
0:3750)2 ]
[1 + (
=
x
ft
2133
1
0:018 753 ft
1
0:015394 ft
1
v2
2
= 202 (0:015394) = 6:158 ft/s
j j
2
at = v_ = 4 ft/s
q
p
2
a = a2n + a2t = 6:1582 + 42 = 7:34 ft/s J
8
86
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.23
13.24
13.25
9
87
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.26
v
vθ
θ
vR cot
)_=
R
v = R _ = vR cot
•
= R
•
R
)a =
aθ
θ
aR
v = vR csc = 350 csc 40 = 545 m/s J
vR = R_ = 350 m/s
aR
a
vR
=
350 cot 40
5000
= 0:08342 rad/s
2
R _ = a sin
2
R_
100
=
sin
5000(0:08342)2
2
= 101:4 m/s J
sin 40
13.27
R = 0:75 + 0:5t2 m
R_ = 1:0t m/s
• = 1:0 m/s2
R
t2 rad
2
_ = t rad/s
• = rad/s2
=
At t = 2 s:
R = 0:75 + 0:5(2)2 = 2:75 m
R_ = 1:0(2) = 2 m/s
• = 1:0 m/s2
R
(2)2 = 2 rad
2
_ = (2) = 2 rad/s
• = rad/s2
=
vR = R_ = 2 m/s
v = R _ = 2:75(2 ) = 17:279 m/s
q
p
2 + v2 =
22 + 17:2792 = 17:39 m/s J
v = vR
aR
a
2
2
R _ = 1:0 2:75(2 )2 = 107:57 m/s
2
= R• + 2R_ _ = 2:75( ) + 2(2)(2 ) = 33:77 m/s
•
= R
10
88
© 2010. Cengage Learning, Engineering. All Rights Reserved.
a=
q
a2R + a2 =
13.28
p
2
( 107:57)2 + 33:772 = 112:7 m/s J
R = 4 + 2 sin ft
R_ = 2 cos _ = 3 cos ft/s
• = 3 sin _ = 4:5 sin ft/s2
R
_ = 1:5 rad/s
•=0
(a) At point A ( = 0):
R = 4 ft
R_ = 3 ft/s
vR = R_ = 3 ft/s J
aR
a
•=0
R
v = R _ = 4(1:5) = 6 ft/s J
2
R _ = 0 4(1:5)2 = 9 ft/s J
2
= R• + 2R_ _ = 4(0) + 2(3)(1:5) = 9 ft/s J
•
= R
2
(b) At point B ( = 90 )·
:
R_ = 0
R = 6 ft
vR = R_ = 0 J
aR
a
•=
R
4:5 ft/s
2
v = R _ = 6(1:5) = 9 ft/s J
_
• R _ 2 = 4:5 6(1:5)2 = 18 ft/s2 J
= R
= R• + 2R_ _ = 6(0) + 2(0)(1:5) = 0 J
*13.29
At point A ( = 0):
R
R_
= 4 + 2 sin = 4 ft
= 2 cos _ = 2 _ ft/s
•
R
=
2 cos •
2 sin
_ 2 = 2• ft/s2
Computation of _ :
vR = R_ = 2 _ ft/s
v = R _ = 4 _ ft/s
42 = (2 _ )2 + (4 _ )2
2
+ v2
v 2 = vR
2
2
) _ = 0:8 (rad/s)
Computation of •:
aR
a
_
2
2
2
R _ = 2• 4 _ = 2• 4(0:8) = 2• 3:2 ft/s
2
= R• + 2R_ _ = 4• + 2(2 _ ) _ = 4• + 4(0:8) = 4• + 3:2 ft/s
•
= R
11
89
© 2010. Cengage Learning, Engineering. All Rights Reserved.
vθ
v
13.30
A
(2• 3:2)
4_
=
=2
4• + 3:2
2_
v
aR
=
a
vR
a=
α aθ
a
aR
A α vR
3:2 =
2
•=
0:320 rad/s
aR
=
2( 0:320)
3:84 ft/s
a
=
4( 0:320) + 3:2 = 1:920 ft/s
2
2
q
p
2
a2R + a2 = ( 3:84)2 + 1:9202 = 4:29 ft/s J
v = R_
vR = R_ = 600 ft/s
Find _ from acceleration:
aθ θ
-g
aR
θ
_=
s
• + g sin
R
R
•
aR = R
r
=
2_
R_ =
g sin
50 + 32:2 sin 30
8000
=
0:09090 rad/s
By inspection of the ‡ight path we see that _ is negative; hence _ =
rad/s.
) v = 8000( 0:09090) = 727:2 ft/s
q
p
2 + v2 =
6002 + ( 727:2)2 = 943 ft/s J
v = vR
0:09090
13.31
_ a = R.
• To …nd R_ and R,
•
Velocity and acceleration of the follower are v = R,
di¤erentiate cam pro…le twice:
R2 + 4R cos
12 = 0
_
_
2RR + 4R cos
4R sin _ = 0
• + 4R
• cos
2R_ 2 + 2RR
4R_ sin _ 4R_ sin _
2
4R cos _
4R sin • = 0
(a)
(b)
(c)
12
90
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Substituting
= 0, _ = 3 rad/s, R = 2 in. into Eq. (b) yields
2(2)R_ + 4R_
) v = R_ = 0 J
0=0
and Eq. (c) becomes
• + 4R
•
0 + 2(2)R
0
0
4(2)(3)2
0=0
R = 0:3
0:4
• = 9:0 in./s2 J
)a=R
13.32
_
R_ =
0:4
=
0:4
•=
R
(a) When
2
•
(b) When
•=0
=2
= 0:1 m
v = R _ = 0:1(2) = 0:2 m/s J
2_
R _ = 0 0:1(2)2 = 0:4 m/s J
2
= R• + 2R_ _ = 0 + 2( 0:2546)(2) = 1:018 m/s J
•
= R
2
= =3:
0:255 m/s J
vR = R_ =
a
0:4
0:255 m/s J
R = 0:3
aR
=0
_ = 2 rad/s
= =2:
vR = R_ =
a
0:2546 m/s
0:4
R = 0:3
aR
=
m
0:4
=3
= 0:16667 m
v = R _ = 0:16667(2) = 0:333 m/s J
2_
R _ = 0 0:16667(2)2 = 0:667 m/s J
2
= R• + 2R_ _ = 0 + 2( 0:2546)(2) = 1:018 m/s J
•
= R
2
13.33
• = 0, _ = 30 2 = rad/s, • = 0.
Given: R = 0:6 ft, R_ = 3 ft/s, R
60
Note that in the position = 0 we have eR = i and e = j.
(a)
v
_ R + R _ e = 3eR + 0:6 e
= Re
= 3eR + 1:885e ft/s = 3i + 1:885j ft/s J
13
91
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(b)
•
a = (R
= 0
=
2
R _ )eR + (R• + 2R_ _ )e
0:6 2 eR + [0:6(0) + 2(3) ] e
2
5:92eR + 18:85e ft/s =
5:92i + 18:85j ft/s J
2
13.34
13.35
14
92
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.36
13.37
13.38
15
93
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.39
16
94
© 2010. Cengage Learning, Engineering. All Rights Reserved.
*13.40
13.41
13.42
h = R sin = 8500 sin 40 = 5464 ft J
h_ = R_ sin + R _ cos = 0
0 = R_ sin 40 + 8500(0:04) cos 40
R_ =
405:2 ft/s
17
95
© 2010. Cengage Learning, Engineering. All Rights Reserved.
vR
v
13.43
= R_ = 405:2 ft/s
v = R _ = 8500(0:04) = 340:0 ft/s
q
p
2 + v2 =
=
vR
( 405:2)2 + 340:02 = 529 ft/s J
13.44
13.45
•=0
R = 0:4 m
R_ = R
z =
0:2
z_ = 0:2 _ = 0:2(6) =
z• =
0:2• = 0:2( 10) = 2 m/s
v
v
1:2 m/s
= R_ eR + R _ e + z_ ez = 0 + 0:4(6)e + ( 1:2)ez = 2:4e
p
2:42 + ( 1:2)2 = 2:68 m/s J
=
1:2ez m/s
18
96
© 2010. Cengage Learning, Engineering. All Rights Reserved.
• R _ 2 )eR + (R• + 2R_ _ )e + z•ez
a = (R
= 0 0:4(6)2 eR + [(0:4)( 10) + 0] e + 2ez
2
=
14:4eR 4e + 2ez m/s
p
2
a =
( 14:4)2 + ( 4)2 + 22 = 15:08 m/s J
13.46
R
_
z
z_
•=0
= 4 in. (const.)
) R_ = R
= 0:8 rad/s (const.)
)•=0
= 0:5 sin 4 in.
= 2 _ cos 4 = 2(0:8) cos 4 = 1:6 cos 4 in./s
z• =
=
2
8 _ sin 4 + 2• cos 4 =
8(0:8)2 sin 4 + 0
2
5:12 sin 4 in./s
v = R _ = 4(0:8) = 3:2 in./s
vz = z_ = 1:6 cos 4 in./s
p
n
= 3:22 + 1:62 = 3:58 in./s at =
, n = 0; 1; 2; : : : J
4
vR = R_ = 0
) vmax
aR
) amax
2
2
R _ = 4(0:8)2 = 2:56 in./s
2
= R• + 2R_ _ = 0
az = z• = 5:12 sin 4 in./s
•
= R
a
p
2
= 2:562 + 5:122 = 5:72 in./s at
=
8
+
n
, n = 0; 1; 2; : : : J
4
13.47
19
97
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13.48
*13.49
20
98
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.50
13.51
z
ρ
mg
= man
µsN
FBD
Fz
=
Fn
= man
v
13.52
=
0
p
+"
s
N
N
mg = 0
MAD
) N = mg
+
s N = man
s mg = m
p
g = 0:4(600)(32:2) = 87:9 ft/s J
v = 60 km/h =
v2
60 000
m/s = 16:667 m/s
3600
1
99
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Assume impending sliding of crate.
Mg
R
y
v
MR
2
x
15o
µsΝ
N
FBD
Fy
= may
N
= M
= M
Fx = max
+-
N
M g cos 15 = M
v2
sin 15
R
16:6672
9:81 cos 15 +
sin 15
55
v2
sin 15
R
g cos 15 +
+.
s
MAD
sN
+ M g sin 15
= 10:783M
v2
cos 15
R
16:6672
= M
cos 15
55
= 0:217 J
= M
(10:783M ) + M (9:81) sin 15
s
13.53
W
T
maR
o
35
N
FR
T cos 35
N
= maR
=
60
0
32:2
Fz
60 + 90:99 sin 35
FBD
MAD
+ !
T cos 35 =
10(2)2
= 0
= 0
W •
(R
g
2
R_ )
T = 90:99 lb J
+ " N W + T sin 35 = 0
N = 7:81 N J
13.54
ρ =4m
N
µsN
z
z
mρv
mg
FBD
2
MAD
2
100
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Fz
Fn
v
=
0
+"
sN
)N =
mg = 0
v2
)
+! N =m
r
r
g
4(9:81)
=
= 8:09 m/s J
=
0:6
s
= man
mg
mg
s
=m
v2
s
13.55
13.56
mg
mρv
n
2
.
mv
t
µN
N
FBD
MAD
(a)
Fn
= man
+"
N
= m g+
v2
N
mg = m
= 7:5 9:81 +
v2
2:52
2
= 97:01 N J
(b)
Ft
v_
= mat
+!
N = mv_
N
0:3(97:01)
2
=
=
= 3:88 m/s J
m
7:5
13.57
T
FBD
mg
maθ
MAD
maR
=
θ
3
101
© 2010. Cengage Learning, Engineering. All Rights Reserved.
• = 0, we have aR =
Since R_ = R
F
= ma
g sin
• =
R
FR
_2
_
2
R _ and a = R•
+%
mg sin = mR•
9:81 sin 25
2
=
= 2:07 rad/s J
2:0
2
= maR
+ & mg cos
T = mR _
1
8:5
1 T
g cos
=
9:81 cos 25
=
R m
2:0 0:5
p
=
4:055 = 2:01 rad/s J
2
= 4:055 (rad/s)
13.58
13.59
4
102
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13.60
13.61
Critical position is
= 90 .
N
FBD
MAD
man
W
Fn
_
W _2
R
= man
+# W =
g
r
r
g
32:2
=
= 4:01 rad/s J
=
R
2
Ft = mat
+
N=
W •
R = 0 (since • = 0)
g
Because N = 0 in the critical position, friction would not change the result.
13.62
y
N
F
45o
t
n
W = 0.5 lb
FBD
v = vt = vx cos 45
v=
x
man
mat
ma
MAD
vx
6
=
= 8:485 ft/s
cos 45
cos 45
5
103
© 2010. Cengage Learning, Engineering. All Rights Reserved.
an =
v2
=
8:4852
2
= 48:0 ft/s
1:5
Because vx = constant, ax = 0, so that a is vertical. Therefore (see MAD),
an = at .
Ft = mat
+&
F cos 45 + W sin 45
F cos 45 + 0:5 sin 45
F
= mat
0:5
(48:0)
=
32:2
= 0:554 lb J
13.63
x
N
F
FBD
MAD
β R
man β
O
Fx
= max
+
kx0
x =
k
m_
2
=
F = man sin
2
k(x
2
x0 ) = m(R _ )
x
R
2(0:5)
= 2:24 ft J
(2=32:2) (5)2
13.64
.
θ
R = 3 ft
W = 1.5 lb
FBD
MAD
maR
F = 2 lb
(a) Rotation in the horizontal plane: delete W from the FBD (it acts perpendicular to the plane of motion).
FR
= maR
1:5
2 =
(0
32:2
+#
2
3_ )
•
F = m(R
2
R_ )
_ = 3:78 rad/s J
(b) Rotation in the vertical plane: use FBD as shown.
FR
= maR
1:5
(0
2 + 1:5 =
32:2
+#
2
3_ )
•
F + W = m(R
2
R_ )
_ = 1:892 rad/s J
6
104
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.65
F 45o
N
F
aR
maθ
p
= k = 18(0:8 2
•
= R
2
R_ = 0
maR
MAD
FBD
mg
0:4) = 13:165 N (spring force)
R
2
v0
R
=
0:8
2:4
0:8
2
=
7:20 m/s
2
F
= ma
+ " F cos 45
mg = ma
F
13:165
2
a =
cos 45
g=
cos 45
9:81 = 13:463 m/s
m
0:4
FR = maR
+ ! N F sin 45 = maR
N = F sin 45 + maR = 13:165 sin 45 + 0:4( 7:20) = 6:43 N J
q
q
2
2
a =
a2R + a2 = ( 7:20) + 13:4632 = 15:27 m/s J
13.66
mg
R
T
maR
.
θ
MAD
N FBD
FR = maR
+ !
• = 0. ) T =
(a) If string is intact, R
(b) After string breaks, T = 0. ) 0 =
• = 560 m/s2 J
)R
T =
2:5 0
h
•
2:5 R
•
m(R
2
R_ )
1:4(20)2 = 1400 N J
1:4(20)2
i
7
105
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.67
*13.68
8
106
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.69
13.70
9
107
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.71
13.72
13.73
10
108
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.74
11
109
© 2010. Cengage Learning, Engineering. All Rights Reserved.
*13.75
13.76
aR
a
θ
aθ
R_
R_ cos
R _ sin
R θ 2m
O
=
4 m/s (given)
=
0
R cos = 2 m
1
cos
=
R
2
m
1
R_ = R _ tan = 4 tan
12
110
© 2010. Cengage Learning, Engineering. All Rights Reserved.
a
=
=
a =
1 d(4R)
1
1 d(R2 _ )
=
= 4 R_
R dt
R dt
R
cos
2
4
(4 tan ) = 8 sin m/s
2
8 sin
a
2
=
= 8 tan m/s
cos
cos
F = ma = 0:6(8 tan ) = 4:8 tan N J
13.77
We must show that a = 0.
= constant ) R2 • + 2RR_ _ = 0
) a = R• + 2R_ _ = 0 Q.E.D
R2 _
R• + 2R_ _ = 0
*13.78
.
θ
N
FBD
maθ
R
maR
MAD
(a)
FR
dR_ _
R
dR
= maR
= R_
2
R_ )
•
0 = m(R
2
R_ dR_ = _ R dR
2
• = R_2
R
1 _2
1 2
R = _ R2 + C
2
2
Initial condition: R_ = 2 m/s when R = 0:5 m.
)
)
1
1
2
2
(2) = (8)2 (0:5)2 + C
C = 6:0 (m/s)
2
2
1
1 _2
1 _2
2
R = (8)2 R2 6
R = 32R2 6 (m/s)
2
2
2
When R = 1:0 m:
1 _2
R
2
v
=
32(1:0)2
6
R_ = 7:211 m/s
_ R + R_e
= vR eR + v e = Re
= 7:21eR + 8:0e m/s J
(b)
F
N
= ma
+ " N = m(R• + 2R_ _ )
= 0:2 [0 + 2(7:211)(8)] = 23:1 N J
13
111
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.79
*13.80
14
112
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.81
13.82
15
113
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.83
16
114
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.84
13.85
Initial conditions:
Letting x1 =
are
9:81
sin = sin
9:81
rad _ = 0 at t = 0
g
sin =
L
•=
Equation of motion:
= 60 =
3
and x2 = _ , the …rst-order equations and the initial conditions
x_ 1
x1 (0)
= x2
x_ 2 =
=
x2 (0) = 0
3
sin x1
The MATLAB program for solving the equations at intervals of 0.1 s is:
function problem13_85
17
115
© 2010. Cengage Learning, Engineering. All Rights Reserved.
[t,x] = ode45(@f,(0:0.1:2),[pi/3,0]);
printSol(t,x)
function dxdt = f(t,x)
dxdt = [x(2)
-sin(x(1))];
end
end
The two lines of printout that span x1 = 0 are:
t
x1
x2
1.6000e+000 8.5645e-002 -9.9633e-001
1.7000e+000 -1.4249e-002 -9.9990e-001
Linear interpolation:
t 1:6
=
1:7 1:6
0 0:085 645
0:014 249 0:085 645
t = 1:686 s J
which agrees with the analytical solution
13.86
2
R _ and a = R•
• = 0, we have aR =
(a) Since R_ = R
θ mg
FBD
=
µN
maR
N
FR
F
= maR
= ma
+.
+&
mg sin
mg cos
maθ
MAD
2
N = mR _
N = mR•
Elimination of N yields
mg cos
mg sin
= mR• + mR_ 2
• = g (cos
sin )
R
_2
Q.E.D.
(b) Substituting numerical values, the equation of motion becomes
Letting x1 =
conditions are
2
• = 9:81 (cos
0:3 sin ) 0:3 _
2
_
and x2 = , the equivalent …rst-order equations and the initial
x_ 1 = x2
x1 (0) = 0
x_ 2 = 4:905(cos
x2 (0) = 0
0:3 sin )
0:3x22
The corresponding MATLAB program is
18
116
© 2010. Cengage Learning, Engineering. All Rights Reserved.
function problem13_86
[t,x] = ode45(@f,(0:0.02:1.6),[0,0]);
printSol(t,x)
function dxdt = f(t,x)
dxdt = [x(2)
4.905*(cos(x(1)) - 0.3*sin(x(1))) - 0.3*x(2)^2];
end
end
The two lines of output spanning x2 = 0 are
t
1.5200e+000
1.5400e+000
x1
x2
2.0982e+000 1.1455e-002
2.0977e+000 -6.3349e-002
Linear interpolation:
x1 2:0982
2:0977 2:0982
x1
=
=
0 ( 0:063349)
0:011455 ( 0:063349)
= 2:098 rad = 120:2 J
13.87
(a)
θ mg
maθ
maR
=
FBD
=
!
0
)_=
cos !t
=
rad/s
FR
= maR
•
)R
•
R
R(0)
= R_
!
0
sin !t
g sin = R( !
= 15 =
0
sin !t)2
12
rad
2
R_ )
g sin(
0
cos !t)
2
2
12
0
•
mg sin = m(R
+%
2
= R
MAD
N
sin t
32:2 sin
12
cos t
J
_
= 2 ft, R(0)
=0 J
_ the equivalent …rst-order equations and the
(b) Letting x1 = R and x2 = R,
initial conditions are
2
2
x_ 1
x1 (0)
= x2
= 2 ft
x_ 2 = x1
12
x2 (0) = 0
sin t
32:2 sin
12
cos t
The corresponding MATLAB program is
19
117
© 2010. Cengage Learning, Engineering. All Rights Reserved.
function problem13_87
[t,x] = ode45(@f,(0:0.025:3.5),[2,0]);
plot(t,x(:,1),’linewidth’,1.5)
xlabel(’t (s)’); ylabel(’R (ft)’)
grid on
printSol(t,x)
function dxdt = f(t,x)
dxdt = [x(2)
x(1)*(pi^2/12*sin(pi*t))^2-32.2*sin(pi/12*cos(pi*t))];
end
which produces the following plot:
4.5
4
3.5
R (ft)
3
2.5
2
1.5
1
0.5
0
0.5
1
1.5
2
2.5
3
3.5
t (s)
(c) The two lines lines of output spanning R = 4 ft are:
t
3.4000e+000
3.4250e+000
x1
3.9520e+000
4.0718e+000
x2
4.7323e+000
4.8523e+000
Linear interpolation:
4 3:9529
4:0718 3:9529
=
3:4250 3:4
t 3:4
4:8523 4:7323
R_ 4:7323
=
3:410 3:4
3:4250 3:4
t = 3:410 s J
R_ = 4:78 ft/s J
20
118
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.88
(a) Kinematics:
R
aR
a
• = z• tan
= z tan
R_ = z_ tan
R
2
2
• R _ = (•
= R
z z _ ) tan
= R• + 2R_ _ = (z • + 2z_ _ ) tan
az = z•
6 tanβ
R
6 ft
β
s
z
FBD
β
mg N
=
maz
maθ
s
maR
β MAD
Kinetics:
0 = m(z • + 2z_ _ ) tan
F
= ma
+
Fs
= mas
+%
g
= az + aR tan
mg cos
= m(az cos
2z_ _
Q.E.D.
z
+ aR sin )
)•=
2
2
z _ ) tan2
= z• + (•
z
z• =
z _ tan2
1 + tan2
g
Q.E.D.
Initial conditions:
z(0) = 6 ft
(b) With
z(0)
_
=0
(0) = 0
_ (0) =
v1
6 tan
=
10
6 tan
J
= 20 and g = 32:2 ft/s2 , we have
z• = 0:116978z _
2
28:433 ft/s
2
•=
2z_ _
z
_ (0) = 4:5791 rad/s
Letting x1 = , x2 = _ , x3 = z and x4 = z,
_ the equivalent …rst-order equation
and the initial conditions are
x_ 1
x_ 3
= x2
= x4
x_ 2 = 2x4 x2 =x3
x_ 4 = 0:116978x3 x22 28:433
10
x1 (0) = 0
x2 (0) =
= 4:5791 rad/s
6 tan 20
x3 (0) = 6 ft
x4 (0) = 0
The MATLAB program for solving the equation of motion is
function problem13_88
[t,x] = ode45(@f,(0:0.02:2),[0,4.5791,6,0]);
plot(x(:,1),x(:,3),’linewidth’,1.5)
21
119
© 2010. Cengage Learning, Engineering. All Rights Reserved.
xlabel(’theta (rad)’); ylabel(’z (ft)’)
grid on
printSol(t,x)
function dxdt = f(t,x)
dxdt = [x(2)
-2*x(4)*x(2)/x(3)
x(4)
0.116978*x(3)*x(2)^2-28.433];
end
end
6.5
6
z (ft)
5.5
5
4.5
4
3.5
0
2
4
6
8
theta (rad)
10
12
14
(c) The two lines of output spanning x4 = 0 are:
t
7.6000e-001
7.8000e-001
x1
5.2777e+000
5.4915e+000
x2
1.0695e+001
1.0671e+001
x3
x4
3.9257e+000 -1.3783e-002
3.9302e+000 4.6720e-001
By inspection, minimum value of x3 is 3:93 ft, so that
hmin = 6
3:93 = 2:07 ft J
22
120
© 2010. Cengage Learning, Engineering. All Rights Reserved.
13.89
(a)
k(R − L0)
FBD θ
=
maθ
mg
FR
F
MAD
maR
•
+ & mg sin
k(R L0 ) = m(R
k
• = R_2
) R
(R L0 ) + g sin
m
= ma
+ . mg cos = m(R• + 2R_ _ )
1
) • = (g cos
2R_ _ )
R
= maR
2
R_ )
Substituting given data, the equations of motion become
• = R_2
R
40(R
0:5) + 9:81 sin
J
• = 1 (9:81 cos
R
2R_ _ ) J
The initial conditions are
R(0) = 0:5 m
_
R(0)
= (0) = _ (0) = 0 J
_ x3 = and x4 = _ , the equivalent …rst-order
(b) Letting x1 = R, x2 = R,
equation and the initial conditions are
x_ 1 = x2
x_ 2 = x1 x24 40(x1 0:5) + 9:81 sin x3
x_ 3 = x4
x_ 4 = (9:81 cos x3 2x2 x4 )=x1
x1 (0) = 0:5 m
x2 (0) = x3 (0) = x4 (0) = 0
The MATLAB program for numerical integration becomes
function problem13_89
[t,x] = ode45(@f,(0:0.01:0.75),[0.5,0,0,0]);
plot(x(:,3)*180/pi,x(:,1),’linewidth’,1.5)
xlabel(’theta (deg)’); ylabel(’R (ft)’)
grid on
printSol(t,x)
function dxdt = f(t,x)
dxdt = [x(2)
x(1)*x(4)^2-40*(x(1)-0.5)+9.81*sin(x(3))
x(4)
(9.81*cos(x(3))-2*x(2)*x(4))/x(1)];
end
end
23
121
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1.3
1.2
1.1
R (ft)
1
0.9
0.8
0.7
0.6
0.5
0
20
40
60
theta (deg)
80
100
120
(c) From the partial printout shown below, we see that Rmax = 1:186 m J
t
6.5000e-001
6.6000e-001
6.7000e-001
x1
x2
1.1859e+000 1.4175e-002
1.1854e+000 -1.2450e-001
1.1834e+000 -2.6261e-001
x3
1.5646e+000
1.5823e+000
1.6002e+000
x4
1.7783e+000
1.7798e+000
1.7839e+000
13.90
(a) The equations of motion are identical to those derived in Prob. 13.89:
• = R_2
R
40(R
0:5) + 9:81 sin
J
• = 1 (9:81 cos
R
2R_ _ ) J
The initial conditions are also the same, except for R(0):
R(0) = 0:75 m
_
R(0)
= (0) = _ (0) = 0 J
(b) The only changes in the MATLAB program are in the arguments of
ode45— changing R(0) to 0:75 and extending the integration period to 1:1 s,
as shown below
24
122
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[t,x] = ode45(@f,(0:0.01:1.1),[0.75,0,0,0]);
1.6
1.4
1.2
R (ft)
1
0.8
0.6
0.4
0.2
0
25
50
75
theta (deg)
100
125
150
(c) Partial printout of the results:
t
6.7000e-001
6.8000e-001
6.9000e-001
x1
x2
1.2815e+000 1.4977e-001
1.2820e+000 -5.1364e-002
1.2805e+000 -2.5326e-001
x3
1.8433e+000
1.8549e+000
1.8663e+000
x4
1.1749e+000
1.1529e+000
1.1337e+000
1.0000e+000
1.0100e+000
5.2318e-001 -3.0221e+000
4.9342e-001 -2.9271e+000
2.2686e+000
2.2902e+000
2.1048e+000
2.2322e+000
By inspection, Rmax = 1:282 m J
The cord becomes slack when R = 0:5 m. Using linear interpolation:
0:49342
2:2902
0:52318
0:5
=
2:2686
0:52318
2:2686
= 2:285 rad = 130:9
J
25
123
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13.91
F1
F2 = mat
Path
man
MAD
W
FBD
Since the speed is constant, we have at = 0
Ft
= mat
+
Fn
= man
+#
F1
= W
1+
F2 = 0
v2
g
F1 + W =
= 175
W v2
g
1+
2102
= 24:7 lb
32:2(1200)
) F = 24:7 lb # J
13.92
26
124
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13.93
13.94
Consider a water particle about to exit the pipe at C. Using polar coordinates,
we have
R
_
0:12 m
R_ = 0:6 m/s
2
= 160
= 16:755 rad/s
60
=
2
•=0
R
•=0
2
R _ = 0 0:12(16:755)2 = 33:69 m/s
2
a = R• + 2R_ _ = 0 + 2(0:6)(16:755) = 20:11 m/s
q
p
2
a2R + a2 = ( 33:69)2 + 20:112 = 39:2 m/s J
a =
aR
•
= R
27
125
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13.95
13.96
28
126
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13.97
13.98
29
127
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13.99
13.100
mg
β man
z
µsN
Fz
=
N
=
Fn
v
0
cos
N n
FBD
+"
mg
N cos
s
sin
MAD
sN
sin
mg = 0
m(32:2)
=
= 39:66m lb
cos 12
0:8 sin 12
mv 2
= man
+
=
s N cos + N sin
r
N
( cos + sin )
=
m s
r
250(39:66m)
=
(0:8 cos 12 + sin 12 ) = 99:1 ft/s J
m
30
128
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13.101
13.102
θ mg
maR
FBD
N
F
= ma
_ d_
=
g
sin d
R
Initial condition: _ = 0 when
1 2
) _
2
v
FR
When N
cos
mg sin = m(R• + 2R_ _ )
d_ _
g
• = g sin
= sin
R
d
R
1 _2
g
=
cos + C
2
R
+.
= R• + 0
g sin
MAD
maθ
= 0. ) C = g=R.
g
=
(1 cos )
R
p
= R _ = 2gR(1
= maR
=
0:
=
2(1
+-
2g
(1
R
cos ) J
_=
N
g cos = 0
cos )
r
cos )
•
mg cos = m(R
2
R_ =
= cos
1
2
R_ )
2g(1 cos )
2
= 48:2 J
3
31
129
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13.103
T ββT
T
FBD
FBD
β mg
mg
Before cut
=
n
MAD
mat
After cut
(a) Due to symmetry, TAB = TCB = T
Fy
T
=
0
=
mg
2 cos
+#
mg 2T cos = 0
mg
=
= 0:610mg J
2 cos 35
(b) Immediately after release v = 0; hence an = 0
Fn
T
= man
= mg cos
+ % T mg cos = 0
= mg cos 35 = 0:819mg J
(c) The results would be equal if
mg
2 cos
= mg cos
cos2
= 0:5
= 45
J
13.104
32
130
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13.105
13.106
13.107
33
131
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13.108
13.109
34
132
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Chapter 14
14.1
14.2
1
133
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14.3
14.4
14.5
(a) When L0 = b:
A
UA
B
=
0
B
1
k
2
=
=
p
2
B
2b
2
A
b = 0:4142b
1 h
2
k (0:4142b)
=
2
i
0 =
0:0858kb2 J
(b) When L0 = 0:8b:
UA
A
= b
B
=
0:8b = 0:2b
1
k
2
2
B
2
A
p
= 2b 0:8b = 0:6142b
i
1 h
2
k (0:6142b)
(0:2b)2 =
2
B
=
0:1686kb2 J
2
134
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14.6
14.7
200 lb
FBD
µkN
40 lb
20o
N
Fy = 0
U1
2
+"
N
200 + 40 sin 20 = 0
N = 186:32 lb
1
mv 2 0
2 2
1 200
[ 0:18(186:32) + 40 cos 20 ] x =
(4:5)2
2 32:2
x = 15:53 ft J
= T2
T1
(
B
= TB
TA
d
=
kN
+ 40 cos 20 ) x =
14.8
UA
mghA
k mgd
=0
1
mv 2
2 A
2
2ghA + vA
2(32:2)(4) + 102
=
= 13:88 ft J
2g k
2(32:2)(0:4)
3
135
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14.9
U1
2
= T2
T1
(
k mg) x
1
k
2
2
=
1
mv 2
2 1
0
1
1
(150)(x 8)2 =
(25)(8)2
2
2
75x2 1151x + 4000 = 0
The larger root is x = 10:03 m
0:2(25)(9:81)x
14.10
v1 = 45 mi/h = 66 ft/s
U1
2
v2
14.11
1
= T2 T1
mgh2 = m(v22 v12 )
2
q
p
2
2
v1 2gh2 = 66
2(32:2)(30) = 49:2 ft/s = 33:6 mi/h J
=
1
2
1.2 m
0.5 m
1.3 m
h = 0.8 m
U1
2
F
= T2
=
T1
Fh =
1
mv 2
2 2
0
mv22
0:8(6)2
=
= 18:0 N J
2h
2(0:8)
14.12
4
136
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*14.13
14.14
5
137
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14.15
14.16
In the limiting case, package will arrive at D with zero velocity.
) TA = TD = 0.
UA D = area under P -x diagram
10(9:81)(2) = 0
P = 98:1 N J
2P
mghD = 0
14.17
We must add the work of the friction force to UA
UA
D
= 2P
10(9:81)(2) + (UA
D
C )fric
in Prob.14.16:
+ (UC
D )fric
y' β mg
x'
µkN
Fy 0 = 0
+-
On AC we have
) (UA
) UA
UA
C )fric
D
mg cos
=0
N = mg cos
= 98:1 cos
=
kN
AC =
0:15(98:1)(6) =
88:29 N m
= 25 , so that N = 98:1 cos 25 = 88:91 N
D )fric
D
N
= 0, so that N = 98:1 N
On CD we have
) (UC
N
ma
MAD
=
FBD
=
kN
CD =
0:15(88:91)(2 csc 25 ) =
= 2P 10(9:81)(2) 88:29
= 0 yields P = 173:8 N J
63:11 = 2P
63:11 N m
347:6 N m
6
138
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14.18
U1
2
= T2
1
k
2
T1
1
(2500)(0:15)2
2
30o
2
1
mv 2 0
2 2
1
cos 30 ) =
(0:25)v22
2
2
v22 = 219:7 (m/s)
mg (1
cos 30 )
0:25(9:81)(2)(1
n
=
mg
ma n
t
ma t
MAD
FBD
N
Fn
= man
+-
N
= m g cos 30 +
N
mg cos 30 = m
v22
v22
= 0:25 9:81 cos 30 +
219:7
2
= 29:6 N J
14.19
mg 35o y
µk N
FBD
x
N
Fy
N
= 0
+ % N mg cos 35 = 0
= mg cos 35 = 5(9:81) cos 35 = 40:18 N
The distance traveled by the block is 3 +
spring.
U1
2
= T2
T2
mg(3 + ) sin 35
5(9:81)(3 + ) sin 35
m, where
k N (3
+ )
= deformation of the
1
k
2
2
1
(4000) 2
2
2000 2 + 18:089
144:27
The positive root is
0:25(40:18)(3 + )
=
=
=
=
0
1
mv 2
2 1
1
(5)(6)2
2
0
0:2641 m
Fmax = k = 4000(0:2641) = 1056 N J
7
139
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14.20
14.21
8
140
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14.22
14.23
9
141
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.24
Position 1 = position at entry (x = 0); Position 2 = position at exit (x = 54
in.).
U1
2
=
=
=
T2
T1
(area under the F -x diagram)
18 + 24
24 + 16
10 + 18
(24) +
(6) +
(24)
2
2
2
942 lb in. = 78:5 lb ft
=
=
U1
2
v2
1 3:2=16 2
1
m(v22 v12 ) =
(v
2102 )
2
2 32:2 2
3:106 10 3 v22 136:96 lb ft
= T2 T1
78:5 = 3:106
= 137:2 ft/s J
10
3 2
v2
136:96
14.25
14.26
Position 1 = release position ( = 0); choose as datum for gravitational potential
energy.
10
142
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Position 2 = position where spring has maximum deformation ( =
V1 = T 1 = 0
T1 + V1
V2 =
= T 2 + V2
Fmax
= k
max
W
0=
max
W
1
+ k
2
max
= 2W J
2
max
1
+ k
2
max ).
T2 = 0
2
max
max
=
2W
k
14.27
V1
=
0 (datum)
T1 = T2 = 0
1
1 2
=
mgy + k = 0:1(9:81)y + (20)(0:2
2
2
= 10y 2 4:981y + 0:4
V2
T1 + V1 = T2 + V2
The smaller root is y
y)2
10y 2 4:981y + 0:4 = 0
0:1006 m = 100:6 mm J
=
14.28
Position 1 = position just before car hits the spring.
Position 2 = position where car comes to a stop.
T1 + V1
1
(20
2
103 )
8(1000)
3600
1
1
mv12 + 0 = 0 + k
2
2
= T2 + V2
2
=
1
k(0:4)2
2
k = 617
2
103 N/m J
14.29
2 δ2 1
δ1
0.2 m
Spring 1
Spring 2
Position 1 = position just before the car hits the spring
Position 2= position where the car comes to a stop
2
= 0:4 m
T 1 + V1
1
(20 000)
2
8000
3600
=
1
= T2 + V2
2
=
2
0:2 = 0:2 m
1
1
mv 2 + 0 = 0 + k(
2
2
1
k(0:22 + 0:42 )
2
k = 494
2
1
+
2
2)
103 N/m J
11
143
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14.30
14.31
C
Datum
A
B
VB
VC
h
δ = 6 in.
1:0
1
1
(0:5) + k(0:5)2 =
W + k 2=
2
16
2
1:0
(50) = 3:125 lb ft
= Wh =
16
=
T B + VB
k
= TC + VC
0
= 25:3 lb/ft J
0:03125 + 0:125k lb ft
0:03125 + 0:125k = 0 + 3:125
14.32
12
144
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14.33
A
0.4 m
0.2 m
0.1 m
Datum
B
Position A:
p
=
0:42 + 0:12 0:15 = 0:2623 m
A
1
1
VA = mg(2R) + k 2A = 0:6(9:81)(0:4) + (200)(0:2623)2 = 9:235 N m
2
2
1
1
2
2
TA =
mv = (0:6)(3) = 2:70 N m
2 A
2
Position B:
B
VB
TB
T A + VA = T B + VB
=
0:3 0:15 = 0:15 m
1 2
1
=
k = (200)(0:15)2 = 2:25 N m
2 B
2
1
1
2
2
=
mv 2 = (0:6)vB
= 0:3vB
2 B
2
2
2:70 + 9:235 = 0:3vB
+ 2:25
vB = 5:68 m/s J
13
145
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14.34
14.35
14.36
Position 1 = position where = max ; choose as datum for gravitational potential energy.
Position 2 = position where = 0.
T 1 + V1
v22
= T2 + V2
=
2Lg(1
1
mv 2 mgL(1 cos 50 )
2 2
2
cos 50 ) = 2(2)(9:81)(1 cos 50 ) = 14:017 (m/s)
0+0=
14
146
© 2010. Cengage Learning, Engineering. All Rights Reserved.
T
man
FBD
MAD
mg
Fn
= man
T
= m g+
+"
v22
L
v22
L
14:017
= 0:5 9:81 +
2
T
mg = m
= 8:41 N J
14.37
The potential energy of a spring in terms of the spring force F is
Ve =
1
k
2
2
=
1
k
2
F
k
2
=
1 F2
2 k
Position 1 = position just before drum stops (choose as datum for gravitational
potential).
Position 2 = position of maximum tension in cable.
In Position 1 the cable force is the weight W of the elevator.
) V1
=
T1
=
1 W2
1 12002
=
= 1:0909 N m
2 k
2 660 103
1W 2
1 1200 2
v =
(8) = 1192:5 N m
2 g 1
2 32:2
(Ve )1 =
Let F be the cable force in Position 2
V2 = (Ve )2 + (Vg )2 =
=
F2
1
2 660 103
T 1 + V1 = T 2 + V 2
7:576
F
1 F2
W
2 k
k
F
F (F 2400)
1200
=
660 103
1:32 106
)
10
7
F2
1192:5 + 1:0909
1:8182
10
3
F
=
0+
F (F 2400)
1:32 106
1193:6 = 0
F = 40 900 lb J
14.38
The potential energy of a spring in terms of the spring force F is
Ve =
1
k
2
2
=
1
k
2
F
k
2
=
1 F2
2 k
15
147
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Position 1 = position just before drum stops (choose as datum for gravitational
potential).
Position 2 = position of maximum tension in cable.
In Position 1 the cable force is the weight W of the elevator.
) V1
=
=
T1
=
1 W2
1 W2
+
2 kcable
2 kspring
1
1
1
+
(1200)2
= 151:09 N m
3
2
660 10
4800
1 1200 2
1W 2
v1 =
(8) = 1192:5 N m
2 g
2 32:2
(Ve )1 =
Let F be the cable force in Position 2.
) V2
1 F2
F
F
1 F2
+
W
+
2 kcable
2 kspring
kcable
kspring
1
1
1
1
+
+
F 2 1200
F
660 103
4800
660 103
4800
=
(Ve )2 + (Vg )2 =
=
1
2
=
1:0492
T1 + V 1 = T 2
1:0492 10
V2
4 2
F
104 F 2
0:2518F
1192:5 + 151:09 = 0 + 1:0492 104 F 2 0:2518F
0:2518F 1343:6 = 0
F = 4970 lb J
14.39
1
3 in.
2
1
2
V1
V2
T2
12 in.
.
93 in
12.36
= 0:12 in. = 0:01 ft
= (12:3693 12) + 0:12 = 0:4893 in. = 0:040 78 ft
1 2
= 2
k
= k 21 = 30(0:01)2 = 0:003 lb ft
2 1
1 2
k
= 0:75(0:25) + 30(0:040 78)2 =
=
Wy + 2
2 2
1 0:75 2
1
mv 2 =
v = 0:011646v22
=
2 2
2 32:2 2
V1 + T1
v2
= V2 + T2
0:003 + 0 =
= 3:47 ft/s J
0:13761 lb ft
0:13761 + 0:011646v22
16
148
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14.40
Datum
5 ft
1.0 ft
6 ft
Position 1
V1
=
1:2(1:0)(0:5) =
V2
=
1:2(6)(3) =
T 1 + V1
v2
Position 2
0:6 lb ft
21:6 lb ft
= T2 + V2
0
= 13:71 ft/s J
v2
T1 = 0
1 1:2(6) 2
T2 =
v = 0:11180v22
2 32:2 2
0:6 = 0:11180v22
21:6
14.41
Choose line AC as the datum for gravitational potential.
i2
1 h
V1 = (Vg )1 + (Ve )1 = mgR sin 30 + k R(45 30)
2
180
2
1
15
= 0:21(9:81)(0:5) sin 30 + (80) 0:5
= 1:2004 N m
2
180
V2 = mgR = 0:21(9:81)(0:5) = 1:0301 N m
1
1
mv 2 = (0:21)v22 = 0:105v22
T2 =
2 2
2
T 1 + V1
v2
= T2 + V2
0 + 1:2004 = 0:105v22 + 1:0301
= 1:274 m/s J
14.42
17
149
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14.43
18
150
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.44
14.45
19
151
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.46
14.47
20
152
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.48
On surface of earth (R = Re ):
V1
=
=
6:673
GMe m
=
Re
7:50 1010 J
10
11
(5:974 1024 )(1200)
6378 103
In outer space (R ! 1), V2 = 0.
) Energy required = V2
V1 = 7:50
1010 J J
14.49
14.50
1
153
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14.51
The propelling force is F = ma: Hence the power of F is P = F v, or
P
19 250
= mav
=
35(550) =
1366:5a
170 + 580
32:2
a = 14:09 m/s
2
a 40
5280
3600
J
14.52
Let F be the driving force.
P
= F v = (ma) v =
P
dx = v 2 dv
m
)
dv dx
dv
v = mv 2
dx dt
dx
P
1
x = v3 + C
m
3
m
Initial condition: v = 30 mi/h = 44 ft/s when x = 0.
1
2
(44)3 (ft/s)
3
)C=
When v = 60 mi/h = 88 ft/s: x =
x=
m 3
(v
3P
3600=32:2
(883
3(40 550)
443 )
443 ) = 1010 ft J
14.53
Let F be the driving force.
P
v dv
dv
v
dt
1 2
P
v = t+C
2
m
= F v = (ma) v = m
=
P
dt
m
Initial condition: v = v0 when t = 0. ) C = v02 =2
r
1 2
P
P
1 2
) v = t + v0
v = 2 t + v02
2
m
2
m
When t = 60 s:
v=
s
2
450
150
103
(60) + 102 = 21:4 m/s J
103
2
154
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.54
P =
dW h
15
= 600
= 12 857 lb ft/s = 23:4 hp J
dt
0:7
14.55
14.56
Pin
Pout
240 hp = 132 103 lb ft/s
62:4 1600
dW
h=
(58) = 96:51 103 lb ft/s
=
dt
60
Pout
96:51 103
100% = 73:1% J
=
100% =
Pin
132 103
=
14.57
3
155
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.58
14.59
14.60
14.61
(a)
Pout
=
(mg)v = 500(9:81)(0:68) = 3335 W
Pout
3335
=
100% =
100% = 79:4% J
Pin
4200
4
156
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(b)
v=
Pout
3335
=
= 0:425 m/s J
mg
800(9:81)
14.62
14.63
14.64
The propelling force during acceleration is
F = ma + FD = ma + 0:12v 2
The power of F is
P = F v = mav + 0:12v 3
)a=
P
0:12v 3
mv
(a) When v = 60 km/h = 16:667 m/s:
a=
150
103 0:12(16:667)3
2
= 4:98 m/s J
1800(16:667)
5
157
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(b) When v = 120 km/h = 33:33 m/s:
a=
150
103 0:12(33:33)3
2
= 2:43 m/s J
1800(33:33)
14.65
) P = F v = F0 v + cv 3
F = F0 + cv 2
19:3 = F0 (50) + c(50)3
32:3 = F0 (60) + c(60)3
The solution is: c = 1:384 8
When v = 70 mi/h, we have
10
4
19:3 = 50F0 + 1:25
32:3 = 60F0 + 2:16
105
105
, F0 = 0:03979
P = 0:03979(70) + (1:3848
10
4
)(703 ) = 50:3 hp J
14.66
*14.67
6
158
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.68
T0
v1
L0
1
L0
1
14.69
L1
2
=
Z
0
1
T1
v2
1
mv02
T1 = mv12
)
= 12 = 0:85
2
2
T0
v0
p
p
=
0:85v0 = 0:85(20) = 18:439 m/s
=
= p1
+ " L0 1 = mv1 ( mv0 )
0:25
(20 + 18:439) = 0:298 lb s J
= m(v0 + v1 ) =
32:2
5s
F dt =
p0
Z
5s
(2ti
0:6t2 j)dt = t2 i
0:2t3 j
0
mv1 + L1
2
v2
= mv2
= 22:5i
2(10i) + 25(i
12:5j m/s J
5s
0
= 25(i
j) N s
j) = 2v2
14.70
14.71
14.72
W
FBD
F =µsW
N=W
7
159
© 2010. Cengage Learning, Engineering. All Rights Reserved.
The limiting force between the tires and the road is
F =
L0
1
t
sW
= 0:65(2800) = 1820 lb
= p1 p0
+ !
F t = mv1 mv0
m
2800=32:2
5280
(v0 v1 ) =
60
0 = 4:20 s J
=
F
1820
3600
14.73
14.74
8
160
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.75
mv2
W
62o
y
x
70o
mv1
Momentum just
before impact
N
FBD during
contact
Momentum just
after impact
(a)
(L1
2 )x
v2
= m [(v2 )x (v1 )x ]
0 = m (v2 cos 62
cos 70
cos 70
= 30
= v1
= 21:86 ft/s J
cos 62
cos 62
v1 cos 70 )
(b)
(L1
2 )y
= m [(v2 )y (v1 )y ] = m (v2 sin 62 + v1 sin 70 )
2:8=16
(21:86 sin 62 + 30 sin 70 ) = 0:258 lb s J
=
32:2
14.76
W
x
N
FBD during
contact
L1 2
1:8j
θ2
y
45o
mv1
Momentum just
before impact
= m(v2 v1)
= 0:05 [v2 (cos
2i
+ sin
2 j)
mv2
Momentum just
after impact
20(cos 45 i
sin 45 j)]
Equating like components:
0 = v2 cos
1:8
= v2 sin
0:05
The solution is v2 cos
) v2
2
2
2
20 cos 45
+ 20 sin 45
= 14:142, v2 sin 2 = 21:86.
p
=
14:1422 + 21:862 = 26:0 m/s J
21:86
= tan 1
= 57:1 J
14:142
2
9
161
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.77
14.78
14.79
10
162
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.80
y
m
L0 1
(mg sin 15 ) t j
) v1
mg
v0
60o x
15
= m(v1
= m [v1
o
n15
mg si
o
15
o
v0 )
v0 (cos 60 i + sin 60 j)]
= v0 (cos 60 i + sin 60 j) g t sin 15 j
= 2(cos 60 i + sin 60 j) 9:81(0:5) sin 15 j
= 1:0i + 0:463j m/s J
14.81
L1
2
= m(v2 v1 ) = 12 [4:8i 6 (i cos 35
=
1:379i + 41:3j N s J
j sin 35 )]
14.82
F
) L1
L1
2
2
(4)2
= 50:27(1 e 0:5t ) lb
= pA = 4(1 e 0:5t )
4
Z 10
Z 10
=
F dt = 50:27
(1 e 0:5t )dt = 402:8 lb s
= m(v2
0
v1 )
0
402:8 =
500
(v2
32:2
0)
v2 = 25:9 ft/s J
14.83
11
163
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.84
14.85
14.86
L1
2
=
Z
10
F dt =
0
L1
2
v2
10
(289t
17:4t2 )dt = 8650 lb s
0
3200
(v2 0)
32:2
3600 s/h
= 59:3 mi/h J
87:04 ft/s = 87:04 ft/s
5280 ft/mi
= m(v2
=
Z
v1 )
8650 =
12
164
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.87
Impulse of F
=
(L1
2 )F
=
Z
1:0 + 1:2
1:0 + 1:2
(0:75) + 1:2(1:25) +
(1:0) W
2
2
3:425W "
=
=
Impulse of W = (L1
(L1
2 )F
F dt = area under F -t diagram
(L1
3:425W
2 )W
3W
2 )W
=W
t = 3W #
= m(v2 v1 )
W
(v2 0)
=
9:81
v2 = 4:17 m/s J
14.88
14.89
14.90
p = mv = 0:5(2i + 4j + 6k) N s
rB=A = 0:2i + 0:8j + 0:6k m/s
hA = rB=A
p=
i
0:2
1:0
j
0:8
2:0
k
0:6
3:0
= 1:2i
0:4k N m s J
14.91
y
A
O
mv sin40o
mv
mv cos40o
x
R
40o
13
165
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(a)
hO = mvR =
0:25
(15)(2) = 0:233 lb ft s
32:2
J
(b)
hA
= mv cos 40 (R cos 40 ) mv sin 40 (R R sin 40 )
= mvR(cos2 40 + sin2 40
sin 40 ) = mvR(1 sin 40 )
0:25
=
(15)(2)(1 sin 40 ) = 0:0832 lb ft s
J
32:2
14.92
2θ
v
y
m
r
O
θ
R
2θ
r = 2R cos (i cos + j sin )
h = r
=
x
v = v( i sin 2 + j cos 2 )
i
cos
sin 2
(mv) = 2mvR cos
j
sin
cos 2
k
0
0
2mvR cos (cos cos 2 + sin sin 2 )k = 2mvR cos2
kJ
14.93
14
166
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.94
14.95
14.96
.z
θ
R
M
m 12 N.m
M
=
h1
=
Z
t2
M dt
=
0
h2 = h1 +
Z
0
0
t2 t
1.0 s
mv1 Rk = m(R _ 1 )Rk = mR2 _ 1 k
12
(3)2 (5)k = 16:770k lb ft s
32:2
1
(12)(1:0) + 12(t2 1:0) k = (12t2 6) k lb ft s
2
t2
M dt
0 = [ 16:770 + (12t2
0
6)] k
t2 = 1:898 s J
1
167
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.97
z
θ
L
R
ω
h
h1
h2
h1
=
=
=
=
m
L
mRvk = m(L sin )(!L sin )k = mL2 ! sin2 k
0:6(0:8)2 (15) sin2 70 k = 5:086k N m s
0:6(0:8)2 ! 2 sin2 30 k = 0:0960! 2 k N m s
h2
5:086k =0:0960! 2 k
! 2 = 53:0 rad/s J
14.98
(hO )A
vB
=
(hO )B
mvA RA = mvB RB
1 + 0:161 cos 180
RA
= 4340 m/s J
= 6000
= vA
RB
1 + 0:161 cos 0
14.99
2
168
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.100
14.101
(a) Energy is conserved. Choose Position 2 as the datum for gravitational
potential energy
z
R
N
T 1 + V1
v2
mg
FBD
1
1
= T 2 + V2
mv12 + mgh = mv22 + 0
2
2
q
p
2
2
v1 + 2gh = 6 + 2(9:81)(0:5) = 6:768 m/s J
=
(b) Angular momentum about the z-axis is conserved
(hz )1
cos
=
(hz )2
(mv1 ) R = (mv2 cos ) R
v1
6
=
= 0:8865
= 27:6 J
=
v2
6:768
14.102
3
169
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.103
14.104
4
170
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.105
14.106
5
171
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.107
h20
GMe (1 + e cos )
R=
12 106
1 + e cos 70
1 + e cos 70
RA
=
=
6
RB
1 + e cos 0
18 10
1+e
Since e > 1, the trajectory is a hyperbola J
)
12
RA
=
106
=
h0
=
h0 = RA v0
e = 1:0268
h20
GMe (1 + e cos 0)
h20
10 11 )(5:9742
(6:673
9:847
v0 =
10
10
1024 )(1 + 1:0268)
2
m =s
9:847 1010
h0
= 8210 m/s J
=
RA
12 106
14.108
a =
413:7
109
h0
=
=
=
=
=
h20
GMsun (1
(6:673
7:385
e2 )
h20
10 11 ) (1:9884
10
15
1:4513
0:082 )
m =s
2 h30
(GMsun )2 (1 e2 )3=2
2 (7:385
((6:673
1030 ) (1
2
2
1030 )) (1 0:082 )3=2
1:453 108 s
108 s =
= 1682 days J
(24 h/day) (3600 s/h)
10
11 ) (1:9884
1015 )3
6
172
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.109
14.110
7
173
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.111
14.112
GMm oon
=
Rm oon
=
6:673
1737
10
11
(0:07348
1024 ) = 4:903
1012 m3 =s2
103 m
Rmin
Rmax
= Rm oon + hmin = (1737 + 140)
= Rm oon + hmax = (1737 + 340)
2:077 1:877
Rmax Rmin
=
e =
Rmax + Rmin
2:077 + 1:877
s
s
4:903
GMm oon (1 + e)
vmax =
=
Rmin
103 = 1:877
103 = 2:077
106 m
106 m
= 0:0506
1012 (1 + 0:0506)
= 1656:6 m/s
1:877 106
The total energy of the spacecraft in the elliptical orbit is
E
= mE0 = m
=
14
103
1 2
v
2 max
1
1656:62
2
GMm o on
Rmin
4:903 1012
1:877 106
=
1:736 0
1010 N m
The escape trajectory is parabolic, for which E = 0. Hence the energy required
to traverse from the elliptical to the parabolic orbit is
1010 N m J
Eesc = 1:736
8
174
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.113
Find speed at A just before retrorockets are …red. The orbit is circular (e = 0).
R
) h0
h0
h20
GM
p e
p
=
RGMe = (6688 103 )(6:673 10 11 )(5:974
= 5:163 1010 m2 =s
h0
5:163 1010
= 7720 m/s
=
= Rv1
v1 =
R
6688 103
=
1024 )
Find speed at A just after the retrorockets are …red. The new orbit is elliptic
with Rmax = R and Rmin = Re
e=
6378
Rmax Rmin
6688 6378
= 0:02373
=
Rmax + Rmin
6688 + 6378
Rmin
=
103
=
h0
=
h0 = Rv2
L1
2
= m(v2
h20
GMe (1 + e)
(6:673
5:102
v2 =
h20
10 11 )(5:974 1024 )(1 + 0:02373)
1010 m2 =s
5:102
h0
=
R
6688
v1 ) = 6000(7629
1010
= 7629 m/s
103
7720) =
5:5
105 N s J
The answer is accurate to only 2 signi…cant …gures because 2 …gures were lost
in the subtraction.
14.114
9
175
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.115
10
176
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.116
14.117
(a) At point A:
R
E0
= Re + H = (6:378 + 0:240) 106 = 6:618 106 m
1 2 GMe
1
3:987 1014
=
v
= (11:4 103 )2
= 4:735
2
R
2
6:618 106
2
106 (m/s)
Since E0 > 0 , the trjectory is a hyperbola. Q.E.D.
11
177
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(b) When R ! 1, then E0 !
) v1 =
14.118
p
2E0 =
1 2
v
2 1
p
2 (4:735
106 ) = 3080 m/s J
12
178
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.119
13
179
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.120
14.121
(a)
maθ
R
O
F
θ FBD
maR
=
O
MAD
14
180
© 2010. Cengage Learning, Engineering. All Rights Reserved.
FR
= maR
•
R
= R_
2
•
R
= R_
2
•
F = m(R
+%
3:439
2
GMe
= R_
R2
1:4078 1016
J
R2
F = ma
GmMe
•
= m(R
R2
10 8 (409:37 1021 )
R2
2
R_ )
0 = m(R• + 2R_ _ )
+-
2
R_ )
2R_ _
J
R
•=
= Re + H0 = (3963 + 480) (5280) = 2:346 107 ft J
5280
_
R(0)
= v0 sin 5 = 17 500
sin 5 = 2237 ft/s J
3600
(0) = 0 J
cos 5
_ (0) = v0 cos 5 = 17 500 5280
R(0)
3600 2:346 107
R(0)
=
1:0899
10
3
rad/s J
_ x3 = and x4 = _ , the equivalent …rst-order
(b) Letting x1 = R, x2 = R,
equations and the initial conditions are:
x_ =
x2
x(0) =
x1 x24
2:346
1:4078 1016
x21
107
2237
0
x4
1:0899
2x2 x4
x1
10
3
T
T
The corresponding MATLAB program is:
function problem14_121
[t,x] = ode45(@f,(0:50:7500),[2.346e7,2237,0,1.0899e-3]);
printSol(t,x)
plot(t,x(:,1),’linewidth’,1.5)
xlabel(’t (s)’); ylabel(’R (ft)’)
grid on
printSol(t,x)
function dxdt = f(t,x)
dxdt = [x(2)
x(1)*x(4)^2-1.4078e16/x(1)^2
x(4)
-2*x(2)*x(4)/x(1)];
end
end
15
181
© 2010. Cengage Learning, Engineering. All Rights Reserved.
7
3
x 10
2.9
2.8
R (ft)
2.7
2.6
2.5
2.4
2.3
2.2
0
1000
2000
3000
4000
t (s)
5000
6000
(c) The following lines of printout span the time when
t
7.0000e+003
7.0500e+003
x1
2.3407e+007
2.3518e+007
x2
2.1743e+003
2.2885e+003
Use linear interpolation to …nd t at
6:3118
7050
6:2573
2
=
7000
t
7000
8000
=2 :
x3
6.2573e+000
6.3118e+000
x4
1.0949e-003
1.0845e-003
=2 :
6:2573
7000
t = 7020 s J
(d) The partial printouts shown below span Rmax and Rmin :
t
2.7500e+003
2.8000e+003
2.8500e+003
x1
x2
2.9403e+007 1.0405e+002
2.9406e+007 -2.2806e+000
2.9403e+007 -1.0861e+002
x3
2.2919e+000
2.3266e+000
2.3613e+000
x4
6.9388e-004
6.9376e-004
6.9390e-004
6.2500e+003
6.3000e+003
6.3500e+003
2.2614e+007 -2.2121e+002
2.2608e+007 -4.1752e+001
2.2610e+007 1.3794e+002
5.3960e+000
5.4547e+000
5.5134e+000
1.1729e-003
1.1736e-003
1.1733e-003
16
182
© 2010. Cengage Learning, Engineering. All Rights Reserved.
By inspection, we …nd that
Rmax
= 29:41 106 ft = 5570 mi
Rmin = 22:61 106 ft = 4282 mi
) Hmax = Rmax Re = 5570 3963 = 1607 mi J
) Hmin = Rmin Re = 4282 3963 = 319 mi J
14.122
(a) The equations of motion were derived in the solution of Prob. 14.121:
1:4078 1016
J
R2
• = R_2
R
•=
2R_ _
J
R
The initial conditions are:
= Re + H0 = (3963 + 480) (5280) = 2:346 107 ft J
5280
_
sin 5 = 1917:4 ft/s J
R(0)
= v0 sin 5 = 15 000
3600
(0) = 0 J
cos 5
_ (0) = v0 cos 5 = 15 000 5280
R(0)
3600 2:346 107
R(0)
=
0:9342
10
3
rad/s J
_ x3 = and x4 = _ , the equivalent …rst-order
(b) Letting x1 = R, x2 = R,
equations and the initial conditions are:
x_ =
x(0) =
x2
x1 x24
2:346
1:4078 1016
x21
107
1917:4
0
x4
0:9342
T
2x2 x4
x1
10
3
T
The corresponding MATLAB program is:
function problem14_122
[t,x] = ode45(@f,(0:10:1500),[2.346e7,1917.4,0,0.9342e-3]);
printSol(t,x);
plot(x(:,3)*180/pi,x(:,1)/5280,’linewidth’,1.5)
xlabel(’theta (deg)’); ylabel(’R (mi)’)
grid on
printSol(t,x)
function dxdt = f(t,x)
dxdt = [x(2)
x(1)*x(4)^2-1.4078e16/x(1)^2
x(4)
-2*x(2)*x(4)/x(1)];
end
end
17
183
© 2010. Cengage Learning, Engineering. All Rights Reserved.
4600
4500
4400
R (mi)
4300
4200
4100
4000
3900
0
10
20
30
40
50
theta (deg)
60
(c) Partial printout spanning R = Re = 2:0926
t
1.4400e+003
1.4500e+003
x1
x2
2.0939e+007 -5.0550e+003
2.0888e+007 -5.0878e+003
70
80
90
107 ft:
x3
1.3986e+000
1.4103e+000
x4
1.1727e-003
1.1784e-003
Linear interpolation:
2:0888
1:4103
2:0939
2:0926 2:0939
=
1:3986
1:3986
= 1:4016 rad = 80:3
J
14.123
y 20 lb
8 lb
FBD
o
0.3N
Fy
L1 2
[8 cos 30
0:3(16)] t
30
x
N
= 0
+ " N 20 + 8 sin 30
N = 16 lb
= m(v2 v1 )
+ !
Fx t = m(v2 v1 )
20
(40 10)
t = 8:76 s J
=
32:2
18
184
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.124
14.125
(a)
T A + VA
= TB + VB = TC + VC
1
1
2
2
mvB
+ 0 = mvC
+ 8mg
=
2
2
0 + 24mg
p
2(24)(9:81) = 21:7 m/s J
p
=
2(16)(9:81) = 17:72 m/s J
vB
=
vC
(b)
mg
MAD
FBD
man
N
Fn = man
+#
mg
N =m
2
vC
=
2
mvC
mg N
Contact is lost when N = 0.
)
=
2
17:722
vC
=
= 32:0 m J
g
9:81
14.126
U1
2
= T2
T1
k W (d
+ )
0:3(40)(6 + ) +
1
k
2
1
(300)
2
2
2
1W 2
v
2 g 1
1 40
=
(15)2
2 32:2
= 0:6333 ft
=
0
F = k = 300(0:6333) = 190:0 lb J
19
185
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14.127
GMe
Re
= (6:673 10 11 )(5:974
= 6378 103 m
v1
1024 ) = 3:987
1014 m3 =s2
h2
h1
O
v2
R1
R1 v1
) R2
R2
= R2 v2 (angular momentum about O is conserved)
v1
6
=
R1 = 0:8R1
R1 =
v2
7:5
1 2 GMe
1 2 GMe
v
v
=
(energy is conserved)
2 1
R1
2 2
R2
1
1
1 2
v22 ) = GMe
(v
2 1
R1
R2
1
1
1
(60002 75002 ) = 3:987 1014
2
R1
0:8R1
R1
h1
h2
= 9:844 106 m
R2 = 0:8(9:844 106 ) = 7:875 106 m
= R1 Re = (9:844 6:378) 106 = 3:466 106 m J
= R2 Re = (7:875 6:378) 106 = 1:497 106 m J
20
186
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14.128
14.129
y mg
P(N)
80
FBD Ps
µsmg
x
N = mg
0
0
5
10
15
t(s)
First determine if the crate will move. The smallest force that will move the
crate is
Ps = s mg = 0:25(18)(9:81) = 44:15 N
Since Ps < 80 N, the crate will move
L1
2
L1
2
= (area under P -t diagram)
t
k mg
= 5(80 + 60) 0:2(18)(9:81)(10) = 346:8 N s
= m(v2 v1 )
346:8 = 18(v2 0)
v2 = 19:27 m/s J
14.130
Energy is conserved. Choose horizontal plane at O as the datum for Vg .
p
= AB L0 = 52 + 42 + 32 2:5 = 4:571 ft
A
1
1
VA = W OA + k 2A = 8(5) + (4:2)(4:571)2 = 83:88 lb ft
2
2
p
= OB L0 = 42 + 32 2:5 = 2:50 ft
O
1 2
1
k = (4:2)(2:50)2 = 13:125 lb ft
VO =
2 O
2
1 8 2
1
2
lb ft
TO =
mv 2 =
v = 0:12422vO
2 O
2 32:2 O
TA + VA
vO
2
+ 13:125
= TO + VO
0 + 83:88 = 0:12422vO
= 23:9 ft/s J
21
187
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14.131
14.132
14.133
Angular momentum about AB is conserved:
m(R1 ! 1 )R1 = m(R2 ! 2 )R2
T
= T2
T1 =
T
T1
=
R2 ! 2
R1 ! 1
% energy lost
=
T
T1
2
1
m(R2 ! 2 )2
2
R2
1=
R1
"
100% = 1
!2
=
!1
R1
R2
2
1
m(R1 ! 1 )2
2
2
4
R1
R1
1=
R2
R2
#
2
R1
100% J
R2
2
1
22
188
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14.134
Energy is conserved:
TA + VA
vB
1W 2
1
1
1W 2
= TB + VB
v + k 2 =
v + k 2
2 g A 2 A
2 g B 2 B
r
r
gk 2
32:2(2:5)
2
2
( A
(0:52 1:02 )
vA +
252 +
=
B) =
W
1:2
= 23:97 ft/s J
Angular momentum about O is conserved:
RA vA = RB (v )B
Rate of elongation
14.135
(v )B =
=
RA
2
vA =
(25) = 20 ft/s
RB
2:5
(vR )B =
p
23:972
=
q
v22
2
(v )B
202 = 13:21 ft/s J
14.136
23
189
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14.137
14.138
24
190
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14.139
LA
A
m
0.4
30o 0.2 m
Datum
L2A = 0:42 + 0:22
VA
=
=
=
VB
LA = 0:2479 m
1
k(LA L0 )2 + mgR cos 30
2
1
(110)(0:2479 0:08)2 + 0:6(9:81)(0:4) cos 30
2
3:589 N m
1
k(LB L0 )2 + mgR
2
1
(110)(0:2 0:08)2 + 0:6(9:81)(0:4) = 3:146 N m
2
1
1
2
2
= 0:3vB
mv 2 = (0:6)vB
2 B
2
=
=
TB
2(0:4)(0:2) cos 30
=
2
+ 3:146
0 + 3:589 = 0:3vB
TA + VA = TB + VB
vB = 1:215 m/s J
14.140
Choose the horizontal line through O as the datum for Vg .
V1 =
V2
T2
1
k(L1
2
L0 )2 =
1
(200)
2
52
42
1
k(L2
2
=
1
(200)
2
=
1 40 2
1W 2
v2 =
v = 0:6211v22 lb ft
2 g
2 32:2 2
T 1 + V1
v2
p
= 69:44 lb ft
12
=
L0 )2
2
Wh
402 + 122
12
42
!2
40
48
12
=
= T2 + V 2
0 + 69:44 = 0:6211v22
= 19:22 ft/s J
159:96 lb ft
159:96
25
191
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26
192
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Chapter 15
15.1
vB=A
= vB vA
= 260(i cos 30
j sin 30 )
= 445i 601j mi/h J
520( i cos 65 + j sin 65 )
15.2
aB=A = aB aA = 0 0:5 = 0:5 ft/s J
Z
vB=A = aB=A dt + C1 = 0:5t + C1 ft/s
2
At t = 0, vB=A = 45
30 = 15 mi/h = 22 ft/s. ) C1 = 22 ft/s
0:5t + 22 ft/s J
vB=A =
xB=A =
Z
vB=A dt + C2 =
0:25t2 + 22t + C2 ft
At t = 0, xB=A = 2 mi = 10 560 ft. ) C2 = 10 560 ft
xB=A =
xB=A jt=120 s =
0:25t2 + 22t + 10 560 ft J
0:25(120)2 + 22(120) + 10 560 = 9600 ft J
15.3
For aB=A to be zero, aB must be parallel to aA (vertical).
(at )B
(an)B
o
60
(an )B
=
aB
=
aB=A = aB
aB
162
= 1:280 m/s
200
(an )B
1:280
2
=
= 1:478 m/s
sin 60
sin 60
2
vB
aA = 0
=
aA = aB = 1:478 m/s # J
2
1
193
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15.4
15.5
Let W refer to the water and B to the boat
vW
vB
θ
vB/W
vB = vW + vB=W
(a)
sin =
vW
vB=W
=
10
= 0:4167
24
= 24:6
J
(b)
vB
=
t
=
q
2
vB=W
2 =
vW
p
242
102 = 21:82 km/h
4:8
d
=
= 0:220 h = 13:2 min J
vB
21:82
2
194
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15.6
15.7
15.8
3
195
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15.9
15.10
4
196
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.11
5
197
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.12
15.13
vA
aA
aA
aB
aB=A
=
70 km/h = 19:444 m/s
2
19:4442
vA
2
=
= 0:9452 m/s
=
R
400
= 0:9452(i sin 40
j cos 40 ) = 0:6076i
0:7241j m/s
2
2
=
1:8j m/s
= aB aA =
=
0:608i
1:8j
(0:6076i
0:7241j)
1:076j m/s J
2
15.14
Position (a): vwind
Position (b): vwind
= (vb oat )a + vwind/b oat a
= 5j + va ( i cos 6 + j sin 6 )
= (vb oat )b + vwind/b oat b
= 5i + vb ( i cos 30 + j sin 30 )
Equating like components of vwind :
va cos 6 = 5 vb cos 30
5 + va sin 6 = vb sin 30
6
198
© 2010. Cengage Learning, Engineering. All Rights Reserved.
The solution is
va = 4:50 mi/h, vb = 10:941 mi/h.
4:48i + 5:47j mi/h J
vwind = 5j + 4:50( i cos 6 + j sin 6 ) =
15.15
A
B
xA
xB
Length of cable:
dL
dt
vB
L = 3xA + 2xB + constant
=
0:
=
3vA + 2vB = 0
1:5vA =
1:5(4) =
6 in./s = 6 in./s ! J
15.16
A
xA
yB
B
Length of cable
dL
dt
vA
= L = xA + 3yB + constant
= vA + 3vB = 0
=
3vB =
3( 0:4) = 1:2 m/s
! J
15.17
yB
yA
B A
Length of cable:
dL
dt
vA=B
vB
= vA
=
vB
1
(160) =
2
L = 2yB + yA + constant
=
0: 2vB + vA = 0
1
vA
2
240 = vA
vB =
1
vA
2
vA = 160 mm/s # J
80 mm/s = 80 mm/s " J
7
199
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.18
15.19
15.20
yA
yB
A
B
yC
C
8
200
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Length of cable
= L = (yC yA ) + 2yC + (yC
= 4yC yA yB + constant
dL
dt
=
vC
=
vC
=
4vC
vA
yB ) + constant
vB = 0
3 + 1:2
vA + vB
=
=
4
4
0:45 ft/s " J
0:45 ft/s
15.21
2 ft
yB
yA
B
A
Length of cable
=
dL
dt
=
vB
=
vB
=
2yB +
q
2 + 22 + constant
yA
yA vA
2vB + p 2
=0
yA + 22
1
1
y
3
p A
p
v =
(1:8) =
2 + 22 A
2 yA
2 32 + 22
0:749 ft/s
0:749 ft/s " J
15.22
2.4 m
2.4 m
yB
yA
A
Length of cable
dL
dt
vB
B
q
2 + 2:42 + constant
= L = y A + 2 yB
yB
= vA + 2 p 2
vB = 0
yB + 2:42
p
2 + 2:42
yB
=
vA
2yB
p
22 + 2:42
=
( 3:6) = 2:81 m/s " J
2(2)
9
201
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.23
20
0
300
B
β
θ
xB
300 cos + 200 cos
300 sin
200 sin
= xB
= 0
(a)
(b)
200
200
sin =
sin 60 = 0:5774
= 35:26
300
300
From (b) : 300 _ cos
200 _ cos = 0
_
_ = 200 cos = 200(2) cos 60 = 0:8165 rad/s
300 cos
300 cos 35:26
_
From (a) :
300 sin
200 _ sin = vB
vB =
300(0:8165) sin 35:26
200(2) sin 60 = 488 m/s
vB = 488 m/s ! J
From (b)
:
sin
=
15.24
10
202
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.25
15.26
11
203
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.27
15.28
12
204
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.29
15.30
13
205
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.31
15.32
y
vA/B
35o
x
x
_A
xB
G
x
De…nition of mass center:
mA xA + mB xB = (mA + mB )x
Since there are no external forces acting on the system in the x-direction,
dx=dt = 0.
) mA x_ A + mB x_ B = 0
vA=B = vA
vB
x_ B =
mA
x_ A =
mB
140
x_ A =
50
18(i cos 35 + j sin 35 ) = (x_ A i + y_ A j)
2:8x_ A
( 2:8x_ A i)
Equating like components:
18 cos 35
18 sin 35
vA
vB
= 3:8x_ A
x_ A = 3:880 ft/s
= y_ A
y_ A = 10:324 ft/s
= x_ A i + y_ A j = 3:88i + 10:32j ft/s J
= x_ B i = 2:8x_ A i = 2:8(3:880)i = 10:86i ft/s J
14
206
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.33
15.34
The motion of the mass center is not changed by the explosion.
ax
x
ay
y
= 0
vx = 240 cos 60 = 120 m/s
= 120t m
=
g
vy = gt + 240 sin 60 = 9:81t + 207:9 m/s
=
4:905t2 + 207:9t m
When t = 35 s:
x =
y =
120(35) = 4200 m
4:905(35)2 + 207:9(35) = 1267:9 m
15
207
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(mA + mB )x
(20 + 40)(4200)
(mA + mB )y
(20 + 40)(1267:9)
=
=
=
=
mA xA + mB xB
20(3820) + 40xB
mA yA + mB yB
20(1960) + 40yB
xB = 4390 m J
yB = 922 m J
15.35
16
208
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.36
15.37
System (crate + cart):
Fx = max
+ !
Crate:
P =
100 + 300
a
32:2
100 lb
FBD P
0.2N
a = 0:0805P
100 a MAD
32.2
=
N = 100 lb
Fx
P
0:2(100)
= max
=
+ !
100
(0:0805P )
32:2
P
0:2N =
100
a
32:2
P = 26:7 lb J
17
209
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.38
15.39
18
210
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.40
19
211
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.41
20
212
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.42
21
213
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.43
15.44
Kinematics: aB =
1
aA = 2 m/s.
2
WA
T
FBD's
P
A
N
y
T T
A
mA aA
MAD's
mBaB
x
B
B
WB
Kinetics:
Block B:
Fy = ma
2T 8(9:81) = 8(2)
+ " 2T WB = mB aB
T = 47:24 N
Block A:
Fx = ma
P 47:24 = 3(4)
+ ! P T = mA aA
P = 59:2 N J
22
214
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.45
15.46
o
15o mg 45o
mg 15
o
T
A 15
B t
t
T
NB
NA
FBD's
Ft = mat
+&
A
man
B
mat
mat
man
MAD's
mg sin 15 + T cos 15
mg sin 45
T cos 15
= mat
= mat
(Block A)
(Block B)
Subtracting 2nd equation from 1st:
mg(sin 15
12(9:81)(sin 15
sin 45 ) + 2T cos 15
sin 45 ) + 2T cos 15
T
= 0
= 0
= 27:3 N J
15.47
40ο
WA
A
T
40ο
µANA
NA
T
FBD's
y
x
µBNB
Α
WB
B
NB
mAa
Β
mBa
MAD's
23
215
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Block A:
Fy = 0
Fx = 0
+ - NA WA cos 40
NA 40 cos 40 = 0
NA
+ % T + A NA WA sin 40
T + 0:15(30:64)
= 0
= 30:64 lb
=
mA a
40
40 sin 40 =
a
32:2
T 21:12 =
1:2422a
(a)
Block B:
Fy = 0
Fx = 0
+ - NB WB cos 40
NB 60 cos 40 = 0
NB
+%
T + B NB WB sin 40
T + 0:3(45:96)
The solution of (a) and (b) is
= 0
= 45:96 lb
=
mB a
60
a
60 sin 40 =
32:2
T 24:78 =
1:8634a
(b)
a = 14:78 ft/s2 and T = 2:76 N J
15.48
B
x
m
0m
0
2
A
2 kg
θ
8 kg
The x-coordinate of the mass center remains constant:
mA xA + mB xB = constant
mA xA + mB (xA + 0:2 cos ) = constant
) mA vA + mB (vA 0:2 _ sin ) = 0
8vA + 2 [vA 0:2(2:4) sin ] = 0
vA
aA
0:0960 sin m/s ! J
2
= 0:0960 _ cos = 0:0960(2:4) cos = 0:230 cos m/s ! J
=
24
216
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.49
yC
yA
C
1
2
yB
W B
W A
Kinematics:
Cable 1 :
Cable 2 :
Kinetics:
yA + (yA yC ) = constant ) 2aA aC = 0
yB + 2yC = constant
) aB + 2aC = 0
aC = 2aA
aB = 4aA
F = ma + #
T1
T1
T2
T2
C
A
Pulley C
:
T1
Block A :
W
Block B
W
:
B
T1
FBD's
W
T2
W
2T2 = 0
T1 = 2T2
W
W
aA
W 4T2 =
aA
2T1 =
g
g
W
W
T2 =
aB
W T2 = 4 aA
g
g
(a)
(b)
Solution of (a) and (b) is
aA =
3
g
17
T2 =
5
W J
17
) T1 = 2
5
W
17
=
10
W J
17
25
217
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*15.50
26
218
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.51
15.52
(a)
mg
2
N
k(x1 − x2)
FBD's
mg
1
F
=
N
2
ma 2
1
ma 1
MAD's
27
219
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Block 1:
Fx = max
+ ! F k(x1 x2 ) = ma1
32:2
1
[F k(x1 x2 )] =
[1:2 (2 12) (x1 x2 )]
=
m
5
2
= 7:728 154:56(x1 x2 ) ft/s J
a1
a1
Block 2:
a2
=
Fx = max
+ ! k(x1 x2 ) = ma2
k
2 12
(x1 x2 ) =
(x1 x2 ) = 154:56(x1
m
5=32:2
x2 ) ft/s
2
J
Initial conditions:
x1 (0) = x_ 1 (0) = x2 (0) = x_ 2 (0) = 0 J
(b) Letting x = x1 x2 x_ 1
and the initial conditions are
x_
=
x3
x(0)
=
0
x4
0
7:728
0
0
T
x_ 2
, the equivalent …rst-order equations
154:56(x1
x2 )
154:56(x1
x2 )
T
T
The MATLAB program for integrating these equations is
function problem15_52
[t,x] = ode45(@f,[0,0.1],[0,0,0,0]);
printSol(t,x);
function dxdt = f(t,x)
dxdt = [x(3)
x(4)
7.728-154.56*(x(1)-x(2))
154.56*(x(1)-x(2))];
end
end
From the last line of output
t
1.0000e-001
x1
3.4149e-002
x2
4.4914e-003
x3
6.0233e-001
x4
1.7047e-001
we deduce that
v1
P
= 0:602 33 ft/s = 7:23 in./s J
= k(x1 x2 ) = (2 12)(34:149
v2 = 0:170 47 ft/s = 2:05 in./s J
4:491) 10 3 = 0:712 lb J
28
220
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15.53
(a)
xB
xA
d
F F
A
FBD's
Particle A
:
aA
=
mAaA
=
B
mBaB
MAD's
Fx = max
+ !
F = mA aA
2
F
c=d
0:005=(xB xA )2
=
=
mA
mA
0:015
1
2
m/s J
3(xB xA )2
=
Particle B
:
aB
=
=
Fx = max
+ !
0:005=(xB xA )2
F
=
mB
0:01
1
2
m/s J
2(xB xA )2
F = mB aB
Initial conditions:
xA (0) = 0
xB (0) = 0:5 m
(b) Letting x = xA xB x_ A
and the initial conditions are
x_
=
x3
x(0)
=
0
x4
0:5 m
x_ A (0) = 0
x_ B
T
x_ B (0) =
, the equivalent …rst-order equations
1
1
3(x2
0
2 m/s J
x1 )2
2(x2
x1 )2
2 m/s
The MATLAB program that solves the equations numerically is
function problem15_53
[t,x] = ode45(@f,[0:0.005:0.25],[0,0.5,0,-2]);
printSol(t,x);
function dxdt = f(t,x)
xx =1/(x(2)-x(1))^2;
dxdt = [x(3)
x(4)
-xx/3
xx/2];
end
end
29
221
© 2010. Cengage Learning, Engineering. All Rights Reserved.
The condition for minimizing d is d_ = x_ B x_ A = 0. Hence x_ B = x_ A when d
is minimized. The following two lines of the output span the instance where
x_ B = x_ A :
t
x1
2.1500e-001 -6.2911e-002
2.2000e-001 -6.6964e-002
When t
When t
x2
x3
x4
1.6437e-001 -7.9442e-001 -8.0837e-001
1.6045e-001 -8.2667e-001 -7.5999e-001
= 0:215 s, d = 0:16437
= 0:220 s, d = 0:16045
) dmin = 0:227 m J
( 0:06291) = 0:2273 m
( 0:06696) = 0:2274 m
We …nd x_ A = x_ B = v by linear interpolation.
0:75999
0:82667
( 0:80837)
v
=
( 0:79442)
v
( 0:80837)
( 0:79442)
Therefore,
x_ A = x_ B = 0:800 m/s
v=
0:800 m/s
J
15.54
(a)
F F
A
B
mBaB
=
mAaA
FBD's
MAD's
(Only horizontal forces are shown)
Block:
aA
Bullet:
aB
=
=
Fx = max
+ ! F = mA aA
10
50(vB vA )
F
2
=
(vB vA ) m/s J
=
mA
15
3
Fx = max
+ !
F = mB aB
50(vB vA )
F
2
=
= 2000(vB vA ) m/s J
mB
0:025
Initial conditions:
xA (0) = xB (0) = vA (0) = 0
vB (0) = 600 m/s J
30
222
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(b) Letting x = xA xB vA
and the initial conditions are
x_
=
x3
x(0)
=
0
vB
T
, the equivalent …rst-order equations
10
(x4 x3 )
3
0 600 m/s
x4
0
2000(x4
x3 )
The corresponding MATLAB program is
function problem15_54
[t,x] = ode45(@f,[0:0.005e-3:1e-3],[0,0,0,600]);
printSol(t,x);
function dxdt = f(t,x)
dxdt = [x(3)
x(4)
(x(4)-x(3))*10/3
-(x(4)-x(3))*2000];
end
end
The last line of output is shown below
t
1.0000e-003
x1
5.6722e-004
x2
2.5967e-001
x3
8.6368e-001
x4
8.1795e+001
We see that at t = 0:001s we have
xA = 5:67
4
10
m = 0:567 mm J
vB = 81:8 m/s J
15.55
(a)
Let xA and xB be the displacements of the cars, measured from the position
where the bumpers just touch.
xA
FF
A
xB
B
mAaA
=
mBaB
MAD's
FBD's
(Only horizontal forces are shown)
Car A
:
aA
=
Fx = max
+ !
F = mA aA
F
20 000(xA xB )
=
= 53:67(xA
mA
12 000=32:2
Car B
:
aB
=
xB ) ft/s
2
J
Fx = max
+ ! F = mB aB
F
20 000(xA xB )
2
=
= 35:78(xA xB ) ft/s J
mB
18 000=32:2
31
223
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Initial conditions:
xA (0) = xB (0) = 0
(b) Letting x = xA xB vA
and the initial conditions are
x_
=
x3
x(0)
=
0
x4
0
vB (0) = 5 ft/s J
vA (0) = 8 ft/s
T
vB
, the equivalent …rst-order equations
53:67(x1
8 ft/s
x2 )
35:78(x1
x2 )
5 ft/s
The MATLAB program shown below was used for integration. Note that only
t and the contact force F were printed out (x1 on the printout represents F )
function problem15_55
[t,x] = ode45(@f,[0:0.005:0.4],[0,0,8,5]);
F = 20000*(x(:,1)-x(:,2));
printSol(t,F)
function dxdt = f(t,x)
dxdt = [x(3)
x(4)
-53.67*(x(1)-x(2))
35.78*(x(1)-x(2))];
end
end
(c) The following partial printout spans Fmax :
t
1.6000e-001
1.6500e-001
1.7000e-001
x1
6.3335e+003
6.3436e+003
6.3396e+003
By inspection we determine that Fmax = 6340 lb J
The two lines of output spanning F = 0 are:
3.3000e-001 1.3013e+002
3.3500e-001 -1.6984e+002
We use linear interpolation to …nd t when contact is lost:
169:84 130:13
0
=
0:335 0:330
t
130:13
0:330
t = 0:332 s J
32
224
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.56
(a) Let xA and xB be the displacements of the blocks from the equilibrium
position (where the spring is unstretched)
mAg
FBD's
xB
B
mAaA
P
A
F
N
N F
mBg
=
MAD's
mBaB
NB
F =
kN
Block A
:
aA
=
k mA g
sgn (vA
vB )
P = mA aA
k
vB )
xA
mA
2000
xA
0:3(9:81) sgn (vA vB )
2
2
2:943 sgn (vA vB ) 1000xA m/s J
=
aB
vB ) =
Fx = max
+ !
F P
=
k g sgn (vA
mA
=
Block B
sgn (vA
F
:
Fx = max
+ ! F = mB aB
mA
F
=
= k
g sgn (vA vB )
mB
mB
2
(9:81) sgn (vA vB )
= 0:3
4
=
1:4715 sgn (vA
vB ) m/s
2
J
Initial conditions:
xA (0) = xB (0) =
(b) Letting x = xA xB vA
and the initial conditions are
x_
=
x(0)
=
x3
x4
0:02 m
T
vB
2:943 sgn (x3
0:02 m
vA (0) = vB (0) = 0 J
0:02 m
0
, the equivalent …rst-order equations
x4 )
1000x1
1:4715 sgn (x3
x4 )
0
The ‡ollowing MATLAB program was used for the plot:
33
225
© 2010. Cengage Learning, Engineering. All Rights Reserved.
function problem15_56
[t,x] = ode45(@f,[0:0.0025:0.2],[-0.02,-0.02,0,0]);
axes(’fontsize’,13)
plot(t,x(:,3),t,x(:,4),’linewidth’,1.5)
xlabel(’t (s)’); ylabel(’v (m/s’)
gtext(’A’);gtext(’B’) % Creates mouse-movable text
grid on
function dxdt = f(t,x)
dxdt = [x(3)
x(4)
-2.943*sign(x(3)-x(4))-1000*x(1)
1.4715*sign(x(3)-x(4))];
end
end
0.6
A
0.4
v (m/s
0.2
B
0
-0.2
-0.4
0
0.05
0.1
t (s)
0.15
0.2
34
226
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.57
(a) Let v be the velocity of A relative to the disk
q
) vR = R_
v = R( _ !)
v = R_ 2 + R2 ( _
!
=
2
60
45
The friction force F =
) FR = F
FR
•
R
F
= 4:7124 rad/s
O
FR
FBD
F
maθ
R
= θ
MAD
maR
Fθ
k mg
vR
=
v
opposes the relative velocity vector v
k mg
R_
v
= maR
= R_
2
= R_
2
where
v=
Initial conditions:
x2
x(0)
=
1:0 ft 0
9:66
0
where
v=
q
9:66
!)
v
2
R_ )
R_
J
v
4:7124)2
_
R(0)
= (0) = _ (0) = 0 J
_
x1 x24
=
•
FR = m(R
R_
kg
v
R( _
F = m(R• + 2R_ _ )
_ !
F
2R_ _
=
kg
mR
R
v
_ 4:7124
9:66
J
v
R(0) = 1:0 ft
x_
k mg
+%
q
R_ 2 + R2 ( _
(b) Letting x = R R_
the initial conditions are
v
=
v
2
FR
= R_
m
2
R_
0:3(32:2) = R _
v
2R_ _
R
2R_ _
R
=
F =F
+&
= ma
• =
!)2
T
x2
v
, the equivalent …rst-order equations and
x4
2x2 x4
x1
9:66
x4
4:7124
v
0
x22 + x21 (x4
4:7124)2
The corresponding MATLAB program is
1
227
© 2010. Cengage Learning, Engineering. All Rights Reserved.
function problem15_57
[t,x] = ode45(@f,[0:0.01:1],[1,0,0,0]);
printSol(t,x)
function dxdt = f(t,x)
v = sqrt(x(2)^2 + x(1)^2*(x(4) - 4.7124 )^2);
dxdt = [x(2)
x(1)*x(4)^2 - 9.66*x(2)/v
x(4)
-2*x(2)*x(4)/x(1) - 9.66*(x(4) - 4.7124)/v];
end
end
The two lines of output spanning R = 2:5 ft are:
t
9.4000e-001
9.5000e-001
x1
2.4831e+000
2.5203e+000
x2
3.6938e+000
3.7325e+000
x3
2.0583e+000
2.0765e+000
x4
1.8289e+000
1.8092e+000
We …nd R_ and _ at R = 2:5 ft by linear interpolation:
3:7325
2:5203
3:6938
R_ 3:6938
=
2:4831
2:5 2:4831
_ 1:8289
1:8092 1:8289
=
2:5203 2:4831
2:5 2:4831
q
q
) vA = R_ 2 + (R _ )2 = 3:7112 + (2:5
R_ = 3:711 ft/s
_ = 1:8200 rad/s
1:8200) = 5:87 ft/s J
2
15.58
2
228
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.59
vA = 18 ft/s
U1
2
vB = 9 ft/s "
)
xA = 2
yB
= T2 T1
1
1
1
2
2
2
+ mB vB
k( 22
xA WB yB = 0
mA vA
k WA
1)
2
2
2
1
k(42 0) 0:3(6)(4) 8(2)
2
1
1
6
8
=
(18)2
(9)2
k = 2:13 lb/ft J
2 32:2
2 32:2
15.60
3
229
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.61
15.62
Horizontal momentum is conserved:
p1 = p2 + !
0 = mb oat vb oat + mb oy vb oy cos 35
0 = 60vb oat + 40(10 cos 35 )
vb oat = 5:46 m/s
vb oat = 5:46 m/s
J
15.63
4
230
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.64
Horizontal momentum is conserved:
p1
= p2
+ !
mA
vA =
mB
vB
=
0 = mA vA + mB vB
12
vA = 1:5vA
8
Energy is conserved:
T 1 + V1
1
(2
2
12)
1:8
2 12
1
1
1
2
2
+ mB vB
+0
0 + k 2 = mA vA
2
2
2
1 0:5
1 0:75 2
v +
( 1:5vA )2
2 32:2 A 2 32:2
= T2 + V2
2
=
0:0675 =
vB =
vA = 1:523 ft/s
2
vA = 1:5228 ft/s
0:029 11vA
1:5( 1:5228) = 2:28 ft/s
J
vB = 2:28 ft/s ! J
15.65
5
231
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.66
) 2vA + vB = 0
Constraint: 2yA + xB = constant
T
vB =
2vA
T
T
B
A
6 lb
Block A: L1
(6
2
2T ) t
Block B: L1
2
Tt
= p2
=
+#
6
(5)
32:2
= p2
=
p1
p2
(6
(6
2T ) t =
6
vA
32:2
2T ) t = 0:9317
+ !
Tt =
(a)
10
vB
32:2
10
( 10) = 3:1056
32:2
(b)
Substituting (b) into (a):
6t
2(3:1056) = 0:9317
t = 1:190 s J
15.67
Let vB = …nal velocity of block and vC = …nal velocity of cart, both positive to
the right.
Mometum is conserved:
p1
vC
= p2 + !
mB
vB =
=
mC
0 = mC vC + mB vB
8
vB = 0:4vB
20
6
232
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Energy is conserved:
T 2 + V2 = T 1 + V1
vC
vB=C
vB=C
1
1
2
2
+ mB vB
+0 =
mC vC
2
2
1 2
1
=
(20)( 0:4vB )2 + 8vB
2
2
vB =
=
0:4( 3:273) = 1:309 m/s
= vB vC = 3:273 1:309 =
= 4:58 m/s
J
1
0+ k 2
2
1
(12 000) (0:1)2
2
3:273 m/s
4:582 m/s
15.68
) 2vA + vB = 0
Constraint: 2yA + yB = constant
T
T
2
= p2
p1
B
8 lb
+"
Block A :
(2T
WA )t = mA vA
0
(2T
Block B
(T
WB ) t = mB vB
0
(T
:
2vA
T
A
6 lb
L1
vB =
6
vA
32:2
8
8)(5) =
( 2vA )
32:2
6)(5) =
The solution is T = 3:790 lb and
vA = 42:4 ft/s " J
15.69
7
233
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.70
p1
vA
= p2 + !
mB
=
vB =
mA
0 = mA vA + mB vB
8
vB = 2vB
4
WA
3P
U1
2
= T2
T1
(3P
[3(1:5)
A
µkWA
N = WA
1
1
2
2
+ mB vB
mA vA
2
2
1 4
1 8 2
0:36(4)] (1:8) =
( 2vB )2 +
v
2 32:2
2 32:2 B
vB = 3:84 ft/s J
k WA )
xA=B
=
15.71
8
234
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.72
15.73
9
235
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.74
10
236
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.75
11
237
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.76
L
mωL
L
O
Momentum
diagram
L/3
mωL
G
2L/3
mωL
hG = 2(m!L)L + (m!L)
8
2L
= m!L2
3
3
J
15.77
z
8 m/s
2m
O
m
2 12 m 3
/s
y
x 1 4 3m
m
4m
A
i
1
2
3
ri
4i
4i + 4j + 3k
3i + 2j
(a) hO
(b) hA
ri=A
4j
3k
7i 2j
=
X
=
340i
= m
vi
8k
12j
6i
ri
6 m/s
3
ri
vi
32j
36i + 48k
12k
36i 32j + 60k
vi = 5( 36i
ri=A vi
32i
36i
12k
68i 12k
32j + 60k)
180i 160j + 300k N m s J
X
= m
ri=A vi = 5( 68i 12k)
60k N m s J
15.78
(a)
(AO )1
6
2
=
(hO )2
t
=
130 3(0:4)2 + 6(0:2)2
(b)
+
!=
(hO )1
Z
+
C0 t = 0
_ ( mA r2
A
2
mB rB
)
t = 15:60 s J
dt + C1 = t + C1
(a is constant)
When t = 0, ! = 130 rad/s. ) C1 = 130 rad/s
When t = 15:60 s, ! = 0. ) 0 = (15:60) 130 )
= 8:333 m/s2
12
238
© 2010. Cengage Learning, Engineering. All Rights Reserved.
+
d
When
= 0, ! =
When ! = 0,
=
=
= ! d!
d! d
d!
d!
=
=
!
dt
d dt
d
1
8:333 = ! 2 + C2
2
1
130 rad/s. ) C2 =
( 130)2 = 8450 (rad/s)2
2
8450
1014:0
= 1014:0 rad. ) =
= 161:4 rev
8:333
2
J
15.79
13
239
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.80
14
240
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.81
A
mv0
3L/4 m(vA)y
3L/4 L/4 3m(vB)y
B
A
mv
L/4
3mvx
G
x
3mv0
G
B
Initial momenta
(p1 )x
=
(p2 )x
+ !
(p1 )y
=
(p2 )y
+"
(hG )1
=
(hG )2
+
3
v0
2
=
3
[(vB )y
4
)
(vA )y =
3
) vA =
1
i
2
Final momenta
3mv0
mv0 = 3mvx + mvx
0 = m(vA )y + 3m(vB )y
mv0
(vA )y ] =
1
v0
2
L
3L
+ 3mv0 =
4
4
vB =
0.25 m
0.25 m
(vA )y =
3L
L
+ 3m(vB )y
4
4
1
(vB )y = v0
2
m(vA )y
3
[(vB )y + 3(vB )y ]
4
3
=
v0
2
3
j v0 J
2
1
v0
2
3(vB )y
vx =
1
1
i + j v0 J
2
2
15.82
yA
yB
A
B
= L
) vA
=
When yB
=
yA = L
2y v
p B B
2 + 0:252
yB
0:25 m: vA =
B 2m
∆yA
A
m
A
m
q
2 + 0:252
2 yB
2(0:25)vB
p =
0:25 2
p
2vB
0.25 m
q
2 + 0:252
yA + 2 y B
B 2m
Position 2
Position 1 (Datum)
15
241
© 2010. Cengage Learning, Engineering. All Rights Reserved.
yA
= yA jyB =0 yA jyB =0:25 m = [L
p
= 0:5( 2 1) m
2(0:25)]
h
L
i
p
2 2(0:25)
1
1
2
+ mg yA 2mg(0:25)
mv 2 + (2m) vB
2 A 2
h
i
p
1 p
2
+ 9:81 0:5( 2 1)
2vB )2 + vB
2(9:81)(0:25)
0 =
(
2
vB = 1:199 m/s # J
T 1 + V1
= T2 + V2
0+0=
15.83
16
242
© 2010. Cengage Learning, Engineering. All Rights Reserved.
*15.84
15.85
17
243
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.86
#15.87
18
244
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.88
19
245
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.89
6g
Before
impact
p 1 = p2
45o
420 m/s
4 kg
After
v' impact
o
45
v 0 = 0:445 m/s J
0:006(420 cos 45 ) = 4:006v 0
+ !
15.90
p1
v…nal
= p2 + !
mA vA + mB vB = (mA + mB )v…nal
40(5) + 55(3)
mA vA + mB vB
=
= 3:842 m/s ! J
=
mA + mB
40 + 55
% energy lost
=
=
T2
T1
1
1
100% = 1
95(3:842)2
40(5)2 + 55(3)2
2
(mA + mB ) v…nal
2 + m v2
mA vA
B B
100%
100% = 6:20% J
15.91
Position 1 = just before impact; Position 2 = just after impact; Position 3 =
max. compression of spring
p1
v…nal
= p2
=
T 2 + V 2 = T 3 + V3
+ !
mA vA = (mA + mB ) v…nal
mA
0:0075
vA =
vA = 0:003736vA
mA + mB
2:0075
1
2
+0
(mA + mB )v…nal
2
1
2
(2:0075) (0:003736vA )
2
vA
=
=
=
1
0+ k 2
2
1
(7500)(0:049)2
2
802 m/s J
20
246
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.92
Position 1 = release position; Position 2 = just before impact; Position 3 = just
after impact; Position 4 = …nal (rest) position
U1
1 WA 2
= T2 T1
WA h =
v
2 g 2
p
p
2gh = 2(32:2)(8) = 22:70 ft/s
=
2
v2
U3
p3
= p3
v3
=
4
= T4
d
=
+
mA v2 = (mA + mB )v3
mA
8
(22:70) = 6:985 ft/s
v2 =
mA + mB
26
T3
k (WA
+ WB )d = 0
1
(mA + mB )v32
2
(mA + mB )v32
v2
6:9852
= 3 =
= 3:79 ft J
2 k (WA + WB )
2 kg
2(0:2)(32:2)
15.93
(a) Position 1 = just before impact; Position 2 = just after impact; Position
3 = …nal (rest) position
p1 = p2 + !
0:02(600) = (10 + 0:02)v2
U2
k (WA
3
= T3
+ WB )d
=
0:25(10 + 0:02)(9:81)d
=
0
mB v1 = (mA + mB )v2
v2 = 1:1976 m/s
T2
1
(mA + mB )v22
2
1
(10 + 0:02)(1:1976)2
2
d = 0:292 m J
(b)
T1
=
T2
=
% energy loss
=
1
1
mB v12 = (0:02)(600)2 = 3600 J
2
2
1
1
(mA + mB )v22 = (10 + 0:02)(1:1976)2 = 7:186 J
2
2
T1 T2
3600 7:186
100% =
100% = 99:8% J
T1
3600
15.94
(a) Horizontal momentum is conserved:
p1
v…nal
= p2
=
+ !
mB vB cos 30 = (mA + mB )v…nal
mB
60
vB cos 30 =
(8) cos 30 = 1:4846 ft/s J
mA + mB
280
21
247
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(b)
% energy lost =
1
=
1
2
(mA + mB ) v…nal
2
mB vB
T2
T1
100% = 1
!
2
280 (1:4846)
60 (8)
2
100%
100% = 83:9% J
15.95
Position 1 = just before collision; Position2 = just after collision; Position3 =
…nal (rest) position
p1
= p2
v…nal
U2
+ !
mA vA = (mA + mB )v…nal
5000
mA
vA = 0:5814vA
vA =
mA + mB
8600
=
3
= T3
0:8(3600)(22) =
T2
k WB d
1 8600
2
(0:5814vA )
2 32:2
=0
1
2
(mA + mB )v…nal
2
vA = 37:5 ft/s J
15.96
Choose Position 2 as the datum for Vg . Energy beween positions 1 and 2 is
conserved:
6(9:81)(2)(1
T1 + V1
= T 2 + V2
cos 40 )
=
0 + mgh1 =
1
(6)v22
2
1
mv 2 + 0
2 2
v2 = 3:030 m/s
Horizontal momentum is conserved during impact:
mv2 = mtot v20
p2 = p02 + !
v20 = 2:273 m/s
6(3:030) = 8v20
Energy beween positions 2 and 3 is conserved:
T 2 + V2
= T3 + V 3
1
2
mtot (v20 ) + 0 = 0 + mtot gh3
2
1
(6 + 2)(2:273)2 + 0 = 0 + (6 + 2)(9:81)2(1 cos )
2
cos
= 0:8683
= 29:7 J
22
248
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.97
(a) Momentum is conserved during impact between carts (note that the parcel
keeps going at v = 15 ft/s):
+ !
(mA + mC )v = (mA + mB )v 0 + mC v
mA
50
15 = 7:5 ft/s J
v=
mA + mB
100
p1
= p01
v0
=
(b) Impact between the front of cart A and the parcel conserves the momentum:
p01
v…nal
= p…nal + !
(mA + mB ) v 0 + mC v = (mA + mB + mC )v…nal
100(7:5) + 35(15)
(mA + mB ) v 0 + mC v
= 9:44 ft/s J
=
=
mA + mB + mC
135
15.98
23
249
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.99
p1
p2
= mA vA (i cos 25 + j sin 25 ) mB vB j
3:5=16
10
=
(120)(i cos 25 + j sin 25 )
(12)j
32:2
32:2
= 0:7388i 3:382j slug ft/s
3:5=16 + 10
= (mA + mB )v2 =
v2 = 0:3174v2
32:2
p1
v2
= p2
0:7388i 3:382j =0:3174v2
= 2:33i 10:66j ft/s J
15.100
24
250
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.101
15.102
25
251
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.103
26
252
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.104
15.105
27
253
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.106
28
254
© 2010. Cengage Learning, Engineering. All Rights Reserved.
*15.107
15.108
1
255
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.109
p1
= p2 + !
mA (vA )1 + mB (vB )1 = mA (vA )2 + mB (vB )2 (a)
(vB )2 (vA )2
e(vA )1 e(vB )1 = (vB )2 (vA )2
(b)
e =
(vA )1 (vB )1
Divide (a) by mB and subtract (b):
mA
e (vA )1 + (1 + e)(vB )1
mB
(vA )1 (mA =mB e) + (vB )1 (1 + e)
1 + mA =mB
=
mA
+ 1 (vA )2
mB
=
(vA )2 Q.E.D.
=
(vA )2 +
=
1+
Divide (a) by mA and add (b):
(vA )1 +
mB
(vB )1
mA
mB
e (vB )1
mA
(vA )1 (1 + e) + (vB )1 (mB =mA e)
1 + mB =mA
(1 + e)(vA )1 +
=
mB
(vB )2
mA
mB
mA
(vB )2
(vB )2 Q.E.D.
15.110
With mA = mB = m and e = 1=2 the formulas in Prob. 15.109 become
(vA )2 =
3
1
(vA )1 + (vB )1
4
4
(vB )2 =
3
1
(vA )1 + (vB )1
4
4
(vB )2 =
3
v0
4
After impact of A and B:
(vA )2 =
1
v0
4
2
256
© 2010. Cengage Learning, Engineering. All Rights Reserved.
After impact of B and C:
(vB )3
=
(vC )3
=
1
1
(vB )2 =
4
4
3
3
(vB )2 =
4
4
3
v0
16
9
=
v0
16
3
v0
4
3
v0
4
=
After second impact of A and B:
(vA )4
=
(vB )4
=
1
(vA )2 +
4
3
(vA )2 +
4
Final velocities are:
13
v0 J
vA =
64
3
1
(vB )3 =
4
4
1
3
(vB )3 =
4
4
vB =
1
v0
4
1
v0
4
15
v0 J
64
3
4
1
+
4
+
3
v0
16
3
v0
16
vC =
13
v0
64
15
=
v0
64
=
9
v0 J
16
15.111
With e = 1=2, mA = mc = m, mB = 0:6m the formulas in Prob. 15.109 become
Between A and B: (vA )2
(vB )2
Between B and C: (vB )2
(vC )2
(vA )1 (1=0:6 0:5) + (vB )1 (1:5)
1 + 1=0:6
= 0:4375(vA )1 + 0:5625(vB )1
(vA )1 (1:5) + (vB )1 (0:6 0:5)
=
1 + 0:6
= 0:9375(vA )1 + 0:0625(vB )1
(vB )1 (0:6 0:5) + (vC )1 (1:5)
=
1 + 0:6
= 0:0625(vB )1 + 0:9375(vC )1
(vB )1 (1:5) + (vC )1 (1=0:6 0:5)
=
1 + 1=0:6
= 0:5625(vB )1 + 0:4375(vC )1
=
After impact of A and B:
(vA )2 = 0:4375v0
(vB )2 = 0:9375v0
After impact of B and C:
(vB )3
(vC )3
= 0:0625(0:9375v0 ) = 0:05859v0
= 0:5625(0:9375v0 ) = 0:5273v0
After second impact between A and B:
(vA )4
(vB )4
= 0:4375(0:4375v0 ) + 0:5625(0:05859v0 ) = 0:2244v0
= 0:9375(0:4375v0 ) + 0:0625(0:05859v0 ) = 0:4138v0
3
257
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Final velocities are:
vA = 0:224v0 J
vB = 0:414v0 J
vC = 0:527v0 J
15.112
15.113
p1 = p2 + !
mA (vA )1 = mA (vA )2 + mB (vB )2
0:25(5) = 0:25( 2) + 1:25(vB )2
(vB )2 = 1:40 m/s !
e=
vsep
(vB )2 (vA )2
1:40 ( 2)
=
=
= 0:68 J
vapp
(vA )1
5
4
258
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.114
5
259
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.115
15.116
Since the impact force is in the y-direction, the x-components of the velocities
are unchanged.
) (vAx )2 = 0
(vBx )2 = 20 ft/s
(py )1 = (py )2
e=
+"
(vBy )2
2800(30) = 2800(vAy )2 + 3600(vBy )2
(vAy )2
0:2 =
vA
(vBy )2
(vAy )2
30
(a)
(b)
Solution of (a) and (b) is (vAy )2 = 9:75 ft/s, (vBy )2 = 15:75 ft/s. Thus the
velocity vectors immediately after the impact are
(vA )2 = 9:75j ft/s J
(vB )2 = 20i + 15:75j ft/s J
15.117
The x-component of the momentum is conserved for each disk:
(vAx )2 = 12 cos 70 = 4:104 m/s
(vBx )2 = 0
Momentum of the system is conserved in the y-direction:
+ " mA (vAy )1 = mA (vAy )2 + mB (vBy )2
m(12 sin 70 ) = m(vAy )2 + (2m)(vBy )2
(vAy )2 + 2(vBy )2 = 11:276 m/s
(a)
6
260
© 2010. Cengage Learning, Engineering. All Rights Reserved.
e=
(vBy )2 (vAy )2
vsep
=
vapp
(vAy )1
(vBy )2 (vAy )2
(vBy )2 (vAy )2
12 sin 70
9:585 m/s
0:85 =
=
(b)
Solution of (a) and (b) is
(vAy )2 =
2:631 m/s
) (vA )2 = 4:10i
(vBy )2 = 6:954 m/s
2:63j m/s J
(vB )2 = 6:95j m/s J
15.118
Let v2 = velocity of pellet and ! 2 = angular velocity of the bar after impact.
(hO )1 = (hO )2 +
mp ellet v0 L
/ = mp ellet v2 L
/ + 2m(L! 2 )L
/
0:16
0:16
(300) =
v2 + 2 (0:5) (0:75)! 2
3 = 0:01v2 + 0:75! 2
(a)
16
16
e=
L! 2 v2
v0
Solution of (a) and (b) is v2 =
0:75 =
0:75! 2 v2
300
(b)
220 ft/s and ! 2 = 6:93 rad/s J
15.119
Since the release positions are the same, the velocities of the pendulums are
equal just before the impact:
(vA )1 = (vB )1 = v
Because A returns to its release position, its velocity just after the impact equals
the velocity just before the impact:
(vA )2 = (vA )1 = v
mv 3mv
A
B
mv
3mv'
mv + 3mv = mv + 3mv 0
+
e=
B
Momenta after
impact
Momenta before
impact
p 1 = p2
A
v0 =
1
v
3
v v0
v v=3
1
=
=
J
v+v
2v
3
7
261
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.120
Position
Position
Position
Position
Position
1
2
3
4
5
=
=
=
=
=
release position
just before A hits the ground (choose as the datum position)
just after A hits the ground but before impacting with B
just after A and B impact
B is at its maximum elevation
Position 1 to 2:
T1 + V1
(vA )2
=
Position 2 to 3:
(vA )3
(vB )3
1
0 + mgh = mv22 (valid for each ball)
2
p
p
(vB )2 = v2 = 2gh = 2(32:2)(5) = 17:944 ft/s
= T 2 + V2
= e(vA )2
(vA )3 = 0:85(17:944) = 15:252 ft/s
= (vB )2 = 17:944 ft/s
Position 3 to 4:
B
Before
impact
B
m(vB)3
10m(vA)3
After
10m(vA)4 impact
A
A
p3 = p4
e=
+"
(vB )4 (vA )4
(vA )3 + (vB )3
m(vB)4
10m(vA )3 m(vB )3 = 10m(vA )4 + m(vB )4
10(15:252) 17:944 = 10(vA )4 + (vB )4
134:58 = 10(vA )4 + (vB )4
0:85 =
(vB )4 (vA )4
15:252 + 17:944
28:22 = (vB )4
(a)
(vA )4 (b)
Solution of (a) and (b) is (vA )4 = 9:669 ft/s, (vB )4 = 37:89 ft/s.
Position 4 to 5:
Ball B :
1
mB (vB )24 + 0 = 0 + mB g (h
2
37:892
(vB )24
=5+
= 27:3 ft J
= dA +
2g
2(32:2)
T 4 + V 4 = T 5 + V5
h
dA )
8
262
© 2010. Cengage Learning, Engineering. All Rights Reserved.
*15.121
15.122
y
4 N.s
0
15
θ
80
B
A
32.23o
3.2 N.s
0.4(vA)y
x
Just before impact
0.8(vB)y
0.4(vA)x
0.8(vB)x
Just after impact
Geometry:
80
= 32:23
150
The x-component of the impulse of each disk is unchanged
= sin
Disk A :
Disk B :
1
4 sin 32:23 = 0:4(vA )x
3:2 sin 32:23 = 0:8(vB )x
(vA )x = 5:333 m/s
(vB )x = 2:133 m/s
9
263
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Momentum of the system is unchanged in the y-direction
3:2 cos 32:23
4 cos 32:23
0:4(vA )y + 0:8(vB )y
vsep
vapp
(vA )y
e=
0:7
=
(vB )y
=
=
=
0:8(vB )y + 0:4(vA )y
0:6767
(vA )y (vB )y
10 cos 32:23 + 4 cos 32:23
8:290
(a)
(b)
Solution of (a) and (b) is
(vA )y = 4:963 m/s
(vB )y =
3:327 m/s
The …nal speeds are
vA
vB
15.123
q
2
=
( 5:333) + 4:9632 = 7:29 m/s J
p
=
2:1332 + ( 3:327)2 = 3:95 m/s J
15.124
10
264
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.125
15.126
15.127
Choose the water jet outside the hose as the control volume
vin = 50 m/s
A=
1 2
1
d =
(0:04)2 = 1:2566
4
4
10
3
m2
11
265
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(a) Stationary plate: vout = 0
m
_ = Av = 1000(1:2566
10
3
)(50) = 62:83 kg/s
The force acting on control volume:
F = m(v
_ out
vin ) = 62:83(0
Force acting on plate:
50) =
3142 N
F = 3140 N J
P =
(b) Moving plate: vout = 4 m/s
m
_ = A(vin
F
P
vout ) = 1000(1:2566
10
3
= m(v
_ out vin ) = 57:80(4
=
F = 2660 N J
)(50
50) =
4) = 57:80 kg/s
2659 N
15.128
y
x
vin
vout
30o
Control volume
m
_ =
vin
=
210(8:34)
= 0:9065 slugs/s
32:2(60)
50i ft/s
vout = 50(i cos 30 + j sin 30 ) = 43:30i + 25:0j ft/s
Force acting on control volume:
F = m(v
_ out
vin ) = 0:9065(43:30i + 25:0j 50i) =
6:074i + 22:66j lb
Force acting on vane:
P =
F = 6:07i
22:7j lb J
12
266
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.129
15.130
13
267
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.131
15.132
14
268
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.133
T
A
ρgL
L
FBD
B
Let be the mass of the chain per unit length.
Choose the portion AB of the chain as the control volume (the bottom link lies
on the ground).
We have steady ‡ow, where the force acting on the control volume is
F = m(v
_ out
vin )
where
vin =
m
_ =
0 (bottom link has no velocity)
vout = 3:2(1:5) = 4:8 kg/s
) F = 4:8((1:5
vout = 1:5 m/s
0) = 7:2 N
From FBD:
F =T
) T = F + gL = 7:2 + 3:2(9:81)(4) = 132:8 N J
gL
15.134
vout
40
y
vin
Control volume
Mg
o
P
x
N
Let the xy axes be attached to the control volume.
vin = 3i ft/s
vout = 30( i cos 40 + j sin 40 ) =
20
( 22:98i + 19:284j 3i)
32:2
16:137i + 11:978j lb
( F)x = 16:14 lb J
F = m(v
_ out
P
=
=
22:98i + 19:284j ft/s
vin ) =
15
269
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.135
d
a
5 in.
P
25.4 ft/s
a
d2 62:4
d2
vout =
(25:4) = 38:66d2 slugs/s
4
4 32:2
D2
(5=12)2
= Aa a p =
p=
(624) = 85:09 lb
4
4
= P = m(v
_ out vin )
85:09 = 38:66d2 (25:4 0)
= 0:2944 ft = 3:53 in. J
m
_ =
P
F
d
15.136
W
y
FBD
x
θ
2T
Thrust of one engine = T = m
_ out u = (80 + 1:6)(660) = 53 860 N
Fy
=
0 +"
2T sin
W =0
8000(9:81)
W
= sin 1
= 46:77
= sin 1
2T
2(53 860)
J
Fx
= max + !
2T cos = ma
2T cos
2(53 860) cos 46:77
2
a =
=
= 9:22 m/s J
m
8000
15.137
y
x
FBD
T
T =m
_ out u =
Fy
Mg
θ
250
(1500) = 11 646 lb
32:2
M=
5000
= 155:28 slugs
32:2
=
0 +"
T sin
Mg = 0
Mg
155:28(28:8)
= sin 1
= sin 1
= 22:6
T
11 646
J
16
270
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.138
Mg
FBD
5 km
A
T
Control volume = rocket + wire above A.
M = 80 + 0:005(5000) = 105 kg
+
d(M v)
= M_ v + M a = ( v) v + M a
dt
0:005(250)2 + 105a = 312:5 + 105a N
"
p_V =
=
+" F =
p_V
=
a =
F
+"
Mg
T =
105(9:81)
312:5 + 105a =
20 =
1050:1
1050:1 N
a=
12:98 m/s
2
12:98 m/s # J
2
15.139
Choose the entire chain as the control volume.
+
p_V
! pV = ( x)x_
=
F
+ !
) p_V = (x_ 2 + x•
x) = x_ 2 + 0
2
x_ 2 = P
P =
(3)2 = 0:559 lb J
32:2
15.140
Mg
30 in.
FBD
P
17
271
© 2010. Cengage Learning, Engineering. All Rights Reserved.
m
_ =
"
0:075
(80)
vin Ain =
32:2
4
10
12
2
#
= 0:101 63 slugs/s
2
20
30
= 4:909p
4 12
vin ) + #
M g P = m(0
_
P
= pAout = p
F
= m(v
_ out
4:909p
=
0:101 63(80)
p = 5:73 lb/ft
2
vin )
J
15.141
15.142
18
272
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.143
15.144
19
273
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.145
p1 = p 2 +
4000(25) = 36 000v2
mcar v1 = mtot v2
v2 = 2:78 mi/h J
15.146
p1 = p 2
+ !
0
0
+ mB vB
mA vA + mB vB = mA vA
0
+ 5000(5:2)
8000(6) + 5000(2) = 8000vA
0
vA = 4:0 mi/h J
15.147
yA
yB
A
B
L_ = 4vA + vB = 0
vB = 8 m/s " J
L = 4yA + yB + constant
4(2) + vB = 0
vB = 8 m/s
15.148
Position 1 = just before impact
Position 2 = just after impact (datum position for Vg )
Position 3 = maximum displacement of block
Momentum parallel to the incline is conserved:
p1
0:75
v0 cos 20
16
= p2 + %
mbullet v0 cos 20 = mtot v1
0:75
=
+ 6 v1
v1 = 7:284 10 3 v0
16
Energy is conserved after the impact :
T2 + V2 = T3 + V3
1
mtot v12 + 0
2
1
(7:284
2
10
3
=
0 + mtot gd sin 20
v0 )2
=
32:2(1:4) sin 20
v0
=
762 ft/s J
20
274
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.149
15.150
y
µsW
A
250
ω T
FBD
T
B
400
µsW
x
Block B tends to slide to the right and block A tends to slide downward. The
friction forces shown oppose impending sliding.
Block B:
Fx
0:25(2 9:81) T
Block A:
Fy
0:25(2 9:81) T
= max + !
T = mrB ! 2
sW
=
2(0:4)! 2
4:905 + T = +0:8! 2
= may + "
=
2(0:25)! 2
T = mrA ! 2
4:905 T = 0:5! 2
(a)
sW
(b)
Solution of (a) and (b) is T = 21:3 N and ! = 5:72 rad/s J
15.151
21
275
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15.152
22
276
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.153
23
277
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.154
24
278
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.155
15.156
25
279
© 2010. Cengage Learning, Engineering. All Rights Reserved.
15.157
15.158
Position 1 = just before shell is …red
Position 2 = just after the shell is …red
26
280
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Position 3 = maximum compression of spring
System: p1 = p2 + !
0 = mA (vA )2 mB (vB )2
0 = 26(2000) 700 (vB )2
(vB )2 = 74:29 ft/s
Barrel only: T2 + V2
1
2
700
32:2
(74:29)2
= T3
1
(28
2
=
1
1
2
mB (vB )2 + 0 = 0 + k
2
2
V3
103 )
2
3
3
2
3
= 2:070 ft J
15.159
Block A:
3 lb
y
25o
=
=
Fx
= max
a =
0
+"
NA cos 25
+ !
3 a
32.2
MAD
0.2NA
FBD NA
Fy
x
0:2NA sin 25
3=0
NA = 3:651 lb
3
a
NA sin 25 + 0:2NA cos 25 =
32:2
32:2
2
(3:651) (sin 25 + 0:2 cos 25 ) = 23:66 ft/s
3
System (block A and wedge B):
9 lb
P
9 a
32.2
=
FBD 0.4NB NB
Fy
=
Fx
= max
+ !
=
9
(23:66) = 10:21 lb J
32:2
P
0
MAD
+"
0:4(9) +
NB
9=0
P
NB = 9 lb
9
a
0:4NB =
32:2
27
281
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15.160
28
282
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15.161
29
283
© 2010. Cengage Learning, Engineering. All Rights Reserved.
30
284
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Chapter 16
16.1
16.2
= 10e
0:5t
!=
20e
0:5t
+ C1
= 40e
0:5t
+ C1 t + C2
= 0 when t = 0. ) C1 = 20 rad/s, C2 =
Initial conditions: ! =
)!=
20e
0:5t
+ 20 rad/s
= 40e
0:5t
+ 20t
40 rad
40 rad
When t ! 1, ! ! 20 rad/s. ) ! 1 = 20 rad/s J
When ! = 0:5! 1 :
10
=
=
20e 0:5t + 20
e
40(0:5) + 20(1:3863)
0:5t
= 0:5
t = 1:3863 s
40 = 7:726 rad = 1:230 rev J
16.3
=
) ! d!
=
d! d
d!
d!
=
=
!
dt
d dt
d
1 2
d
! =
+ C1
2
Initial condition: ! = 8000 rev/min when
)
When
1 2
(!
2
( is constant)
= 0. ) C1 = 80002 =2 (rev/min)
2
80002 ) =
= 3200 rev, ! = 4000 rev/min.
)
1
(40002
2
80002 ) = (3200)
=
7500 rev/min
2
d!
dt = d!
t = ! + C2
dt
Initial condition: ! = 8000 rev/min when t = 0. ) C2 = 8000 rev/min
When ! = 0:
=
t = C2
7500t =
t = 1:0667 min = 64:0 s J
8000
1
285
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16.4
=
4t2 + 24t
10 rad
!=
8t + 24 rad/s
=
8 rad/s
2
(a) When t = 4 s:
!=
8(4) + 24 =
8 rad/s J
=
8 rad/s
2
J
(b) Note that the rotation reverses direction when t = 3 s (obtained by setting
! = 0).
When t = 0, = 10 rad
When t = 3 s, = 4(3)2 + 24(3) 10 = 26 rad
When t = 4 s, = 4(4)2 + 24(4) 10 = 22 rad
The total angle turned through is tot = (10 + 26) + (26 22) = 40 rad J
16.5
Z
4 + 6t
!=
dt = 4t + 3t2 + C1
Z
=
! dt = 2t2 + t3 + C1 t + C2
=
Initial conditions: ! = 0 and
= 0 when t = 0: ) C1 = C2 = 0
) ! = 4t + 3t2
= 2t2 + t3
When ! = 24 rad/s:
24 = 4t + 3t2
t = 2:239 s
)
= 2(2:239)2 + 2:2393 = 21:3 rad J
16.6
2
286
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.7
16.8
2
= 12 rad/s
Initial conditions:
C2 = 0.
= 6t2 + C1 t + C2 rad
! = 12t + C1 rad/s
= 0, ! =
) ! = 12t
24 rad/s when t = 0. ) C1 =
24 rad/s
= 6t2
24 rad/s and
24t rad
Note that the rotation reverses direction when t = 2 s (obtained by setting
! = 0).
When t = 0, = 0:
When t = 2 s, = 6(2)2 24(2) = 24:0 rad
When t = 4 s, = 6(4)2 24(4) = 0
The total angle turned through is tot = 48 rad J
3
287
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.9
!
=
=
4t1=2
=
jt=6 s
Z
! dt =
jt=0 s =
8 3=2
t +C
3
8 3=2
(6) + C
3
J
C = 39:2 rad
16.10
16.11
Pulley B:
v
=
a =
(RB )o ! B
(RB )o
16 = 0:25! B
9 = 0:25
B
! B = 64 rad/s J
B
B
=
36 rad/s
2
J
Belt between B and C:
v0
=
(RB )i ! B = 0:1(64) = 6:4 m/s
0
=
(RB )i
a
B
= 0:1( 36) =
3:6 m/s
2
Pulley C:
v0
0
a
= RC ! C
= RC
C
! C = 21:3 rad/s J
6:4 = 0:3! C
3:6 = 0:3
C
C
=
2
12 rad/s
J
4
288
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.12
Left pulley:
vA
=
(aA )n
=
a2A
=
(RA )i ! A
12 = 0:75! A
! A = 16 rad/s
2
2
2
(aA )t = (RA )o A =
(RA )o ! A = 2(16) = 512 ft/s
2
2
2
2
(aA )n + (aA )t
600 = 512 + 4 2A
A = 156:41
2
A
2
rad/s
Right pulley:
vB
= RB ! B
(aB )n
=
RB ! 2B
(aB )t
= ab elt = (RA )i A = 0:75(156:41) = 117:31 ft/s
q
p
2
(aB )2n + (aB )2t = 115:22 + 117:312 = 164:4 ft/s J
=
aB
16.13
12 = 1:25! B
! B = 9:6 rad/s
2
= 1:25(9:6) = 115:2 ft/s
2
2
For point A:
(aA )t
=
60 sin 40 = 38:57 m/s
(aA )t
= OA
(aA )n
= OA! 2
2
(aA )n = 60 cos 40 = 45:96 m/s
38:57 = 0:12
= 321:4 rad/s
2
2
2
45:96 = 0:12! 2
! 2 = 383 (rad/s)
For point C:
OC
=
p
(120=2)2 + (90=2)2 = 75 mm
(aC )t
= OC = 0:75(321:4) = 241:1 m/s
(aC )n
= OC! 2 = 0:75(383) = 287:3 m/s
y
an
5
7
O
(aC )x
=
(aC )y
=
aC
=
2
2
C x
at
45
60
60
45
(aC )n + (aC )t =
75
75
45
60
(aC )n
(aC )t =
75
75
2
85:2i 365j m/s J
60(287:3) + 45(241:1)
=
75
45(287:3) 60(241:1)
=
75
85:2 m/s
365 m/s
2
2
5
289
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16.14
16.15
6
290
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.16
AC
vB
= !
vB
=
B
0:24i + 0:36j 0:2k
= 0:5035i + 0:7553j
=p
0:242 + 0:362 + 0:22
rB=C = 2
0:3021i
AC
( 0:36j) = 2
i
0:5035
0
0:4196k
j
0:7553
0:36
k
0:4196
0
0:3625k m/s J
=
= 7
=
=
rB=C + ! (! rB=C ) =
rB=C + ! vB
( 0:36j) + 2 AC vB
AC
i
j
k
i
j
0:4196 + 2 0:5035 0:7553
7 0:5035 0:7553
0
0:36
0
0:3021
0
1:605i + 0:619j
0:812k m/s
2
k
0:4196
0:3625
J
7
291
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16.17
16.18
!
rB=A
vB
= !
=
12i
= !
15j + 9k
=
= 25 p
152 + 92
9k in.
21:44j + 12:862k rad/s
CA
i
0
12
rB=A =
j
21:44
0
k
12:862
9
= 192:96i + 154:34j + 257:3k in./s
= 16:08i + 12:86j + 21:4k ft/s J
aB
= !
vB =
i
0
192:96
j
21:44
154:34
=
7502i + 2482j + 4137k in./s
=
625i + 207j + 345k ft/s
2
k
12:862
257:3
2
J
16.19
v)
( B y
B
v)
( B x
=
A
8 ft/s
60o + ω
A
2ω
2 ft
B
vB = vA + vB=A
8
292
© 2010. Cengage Learning, Engineering. All Rights Reserved.
+
!
)
+
!
"
=
(vB )x = 8 cos 60 = 4 ft/s
q
p
2
(vB )2x = 62
(vB )y = vB
42 = 4:472 ft/s
(vB )y = 8 sin 60 + 2!
4:472 = 8 sin 60 + 2!
1:228 rad/s
! = 1:228 rad/s
J
16.20
Wheel rolls without slipping: vC = R! = 1:75!
8 ft/s
B
+"
0.75ω
= 1.75ω +
B
0.75 ft
(vB)x C
ω
C
vB = vC + vB=C
8 = 0:75!
! = 10:667 rad/s
J
vC = 1:75! = 1:75(10:667) = 18:67 ft/s
! J
16.21
Wheel rolls without slipping: vC = R!
vA
A =
Rω
Rω ω
AB
ω
R
+ C
C
B +
B
2
2R
45o
A
2 2RωAB
vA = vC + vB=C + vA=B
p
! AB = 0:5!
J
+ " 0 = R! 2 2R! AB sin 45
p
p
+
vA = R! + 2 2R! AB cos 45 = R! + 2 2R(0:5!) cos 45
vA = 2R!
J
16.22
0.6 m/s
A
90 mm
C
150 mm
B
0.8 m/s
9
293
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0.24 m
A
0.6 m/s
0.8 m/s
=
B +
A
0.24ω
ω
B
vA = vB + vA=B
0:6 =
0:8 + 0:24!
! = 5:833 rad/s
C
vC
C
0.8 m/s
+
=
B
J
0.875 m/s
0.15 m
+ !
5.833 rad/s
B
vC = vB + vC=B
+ !
vC =
0:8 + 0:875 = 0:075 m/s ! J
16.23
10
294
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16.24
11
295
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16.25
16.26
8 rad/s
A
30 D D' 24 24 E
B ωB
vD vD'= vD
vD
= vA + vD=A = 0 + ! A AD = 8(30) = 240mm/s
vD0
= vE + vD0 =E = 0 + ! B D0 E
240 = ! B 48
! B = 5 rad/s
J
12
296
© 2010. Cengage Learning, Engineering. All Rights Reserved.
vB
vB
! AB
= vE + vB=E = 0 + ! B BE = 5(24) = 120 # mm/s
= vA + vA=B = 0 + ! AB AB
= 2:22 rad/s
J
120 = ! AB (54)
16.27
13
297
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16.28
16.29
vC
! CD rC=D
! CD k 60i
60! CD j
= vB + vC=B
= ! AB rB=A + ! BC rC=B
=
2:8k ( 30i) + ! BC k (30i + 60j)
= 84j + ! BC (30j 60i)
Equating like components:
60! BC
60! CD
= 0
= 84
) ! BC = 0 J
) ! CD = 1:40 rad/s
J
14
298
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.30
y
B
=
rB=A =
( 6i) =
36j =
vB
! AB
6k
D 6
15
8
E
A
x
vD + vB=D
! DE rD=E + ! BD rB=D
! DE k ( 6i + 8j) + ! BD k ( 15i
! DE ( 6j 8i) + ! BD ( 15j + 8i)
8j)
Equating like components:
0
! BD
=
8! DE + 8! BD
= ! DE = 1:714 rad/s
36 =
J
6! DE
15! BD
16.31
15
299
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.32
B
A
Geometry:
rB=A
rC=B
vC
vC j
12
30o x
18
y
20
φ
C
12 cos 30 + 18 cos = 20
= 57:74
= 12(i cos 30 + j sin 30 ) = 10:392i + 6j in.
= 18(i cos 57:74
j sin 57:74 ) = 9:608i 15:221j in.
= vB + vC=B
= ! AB rB=A + ! BC rC=B
= 16k (10:392i + 6j) + ! BC k (9:608i
= 166:27j 96i + ! BC (9:608j + 15:221i)
15:221j)
Equating like components:
0
vC
=
96 + 15:221! BC
! BC = 6:307 rad/s
= 166:27 + 6:307(9:608) = 227 in./s " J
16.33
30o
Β
vB
Α
=
2 m/s
ω
+
20o
0.5ω
Α 0.5 m Β
vB = vA + vB=A
16
300
© 2010. Cengage Learning, Engineering. All Rights Reserved.
+ ! vB sin 30
+ " vB cos 30
(0:3420!) cos 30
= 0:5! sin 20
vB = 0:3420!
=
2 + 0:5! cos 20
=
2 + 0:5! cos 20
! = 11:516 rad/s
(vC)y
Α
(vC)x =
C
11.516 rad/s
+
2 m/s
20o
Α 1.0 m
J
11.516 m/s
C
vC = vA + vC=A
16.34
+
+
!
"
vC
=
(vC )x = 11:516 sin 20 = 3:939 m/s
(vC )y = 2 + 11:516 cos 20 = 8:822 m/s
p
3:9392 + 8:8222 = 9:66 m/s J
17
301
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.35
16.36
18
302
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.37
vB
A
vB
30j
30j
80
B
160 o
y
30
vA
D
x
= vA + vB=A
30j = vA i + ! BD rB=A
= vA i + ! BD k 160(i cos 30 + j sin 30 )
= vA i + ! BD (138:56j 80i)
Equating y-components:
30 = 138:56! BD
vD
vD
16.38
! BD = 0:2165 rad/s
= vB + vD=B = 30j + ! BD rD=B
= 30j + 0:2165k 80(i cos 30 + j sin 30 )
= 30j + 15:0j 8:660i = 45:0j 8:660i mm/s
p
45:02 + ( 8:660)2 = 45:8 mm/s J
=
19
303
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16.39
20
304
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.40
16.41
y
vB
0.25 m
B
x y
β
ω
x
C
Point C is the I.C. of the wheel. ) vB = BC !
(vB )x
=
BC ! cos
2:4
=
8(0:25 + y)
(vB )y
= BC ! sin
0:7
=
8x
R+y
= !(R + y)
BC
y = 0:05 m = 50 mm J
=
BC !
x
= !x
BC
x = 0:0875 m = 87:5 mm J
= BC !
16.42
21
305
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.43
16.44
22
306
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.45
16.46
vC
24 rad/s
80
A
vB
vB
vB
120
C
B
= vA + vA=B = 0 + ! AB AB
= 4:8(200) = 960 mm/s
vC
vC
= vA + vC=A = 0 + ! A AC
= 24(80) = 1920 mm/s
ωB
I.C. of B
(A and B on the arm AB)
(A and C on gear A)
The velocities vC and vB establish the instant center for velocities of gear B.
) !B =
vC
1920
=
= 8 rad/s
240
240
J
1
307
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.47
0.6 m/s
0.6 m/s
A
a
I.C.
90 mm
C
240 _ a
150 mm
B
0.8 m/s
0.8 m/s
Velocities of A and B establish the instant center for velocities of the pinion
gear:
240 a
a
=
a = 102:86 mm
0:6
0:8
!
=
vC
=
vA
600
=
= 5:833 rad/s
J
a
102:86
(a AC)! = (102:86 90)(5:833) = 75:0 mm/s ! J
16.48
vB
B
30
6 in
.
o
6 rad/s
A
Geometry: BE =
vB
! BC
vC
! CD
vC
8 in.
60o ωCD
ωBC
13.856 in.
16
in.
E
C
4 in.
D
8
= 16:0 in.
cos 60
CE = 8 tan 60 = 13:856 in.
= ! AB AB = 6(6) = 36 in./s
vB
36
=
=
= 2:25 rad/s
16
BE
= ! BC CE = 2:25(13:856) = 31:18 in./s
vC
31:18
=
=
= 7:80 rad/s
J
4
CD
2
308
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.49
16.50
16.51
3
309
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.52
16.53
C
50o
A
50o
ωAB
70o
vA L 60o
30o
B
vB
Point C is the I.C. of bar AB
L
sin 50
L
sin 50
=
=
BC
sin 70
AC ! A
sin 60
sin 70
= 1:2267L
sin 50
sin 60
= 1:1305L
AC = L
sin 50
BC = L
4
310
© 2010. Cengage Learning, Engineering. All Rights Reserved.
vA = AC ! AB = AC
0:8
vB
= 0:7373 m/s J
= (1:1305L)
1:2267L
BC
16.54
D
C
12 i
n.
ωBC
B
vB
Geometry:
AC
BD
! BC
vB
! AB
γ
30 in./s
β
6 i 45o
n.
ωAB A
12
6
=
= 20:70
sin
sin 45
= 180
45
20:70 = 114:3
p
2
= CD = 12 + 62 2(12)(6) cos 114:3 = 15:468 in.
AC
sin 45
=
6=
15:468
sin 45
6 = 15:875 in.
vC
30
= 1:940 rad/s
=
15:468
CD
= ! BC BD = 1:940(15:875) = 30:80 in./s
vB
30:80
=
=
= 5:13 rad/s
J
6
AB
=
16.55
A
12 rad/s
27 in.
60o
in.
15
vB
vD
B
E
D
ωBC F
Geometry: BF =
AE
cos 60
AB =
27
cos 60
15 = 39 in.
5
311
© 2010. Cengage Learning, Engineering. All Rights Reserved.
vB
! BC
= ! AB AB = 12(15) = 180 in./s
vB
180
= 4:62 rad/s
J
=
=
39
BF
16.56
13 in.
12 rad/s
B
A
15 in.
vB
D
BF
= tan
DF
= BF + DF cot = (27
1
= tan
BE
vB
! BC
E
β
β
27 in.
Geometry:
ωBC
F
vD
1
27
15
= 42:71
13
15) + 13 cot 42:71 = 26:08 in.
= ! AB AB = 12(15) = 180 in./s
vB
180
=
= 6:90 rad/s
J
=
26:08
BE
vC
C
60
0
16.57
E
450
F
ωBCD
B
A
0
60
0
50 o
35 vB
72 rad/s
450
D
Geometry: BF
BE
DE
vB
! BCD
vD
=
q
BC
2
2
CF =
p
6002
vD
4502 = 396:9 mm
396:9
BF
=
= 484:5 mm
cos 35
cos 35
= DF + BF tan 35 = 450 + 396:9 tan 35 = 727:9 mm
=
= ! AB AB = 72(500) = 36 000 mm/s
vB
36 000
=
=
= 74:30 rad/s
J
484:5
BE
= ! BCD DE = 74:30(727:9) = 54 100 mm/s = 54:1 m/s ! J
6
312
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.58
ωBC
vB
E
ωAB = ωBD = 8 rad/s
600
B
o
A
45
600
vD
o
60
vC C
o
30
D
Note that A is the I.C. of BD as well as AB:
Geometry: AC
AD
AB
BE
vB
! BC
vC
vD
= EC = 600 + 600 cot 60 = 946:4 mm
= 600 + 600 cot 30 = 1639:2 mm
600
= 848:5 mm
=
sin 45
AC
946:4
=
848:5 = 489:9 mm
AB =
cos 45
cos 45
= ! AB AB = 8(848:5) = 6788 mm/s
6788
vB
=
= 13:856 rad/s
=
489:9
BE
= ! BC EC = 13:856(946:4) = 13 110 mm/s = 13:11 m/s
= ! AB AD = 8(1639:2) = 13 110 mm/s = 13:11 m/s " J
J
16.59
7
313
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.60
vB
B
ωAB
8
E
vA
O 16 in./s
A 3
5
4
C ωdisk
Geometry: OB
BE
AE
q
p
2
2
=
AB
AO = 82 32 = 7:416 in.
3
3
= AO + OB = 3 + (7:416) = 8:562 in.
4
4
5
5
=
OB = (7:416) = 9:270 in.
4
4
vO
16
= 4 rad/s
vA = ! disk AC = 4(5) = 20 in./s
=
4
OC
20
vA
=
=
= 2:1575 rad/s
9:270
AE
= ! AB BE = 2:1575(8:562) = 18:47 in./s " J
! disk
=
! AB
vB
16.61
y
B
o
30
8 m/s
=
6 m/s2
B
2
x
+
30o
A
A
0.2ωAB2
m
0.2 o
30
0.2αAB
αAB
ωAB (sense indeterminate)
aB = aA + aB=A
x+ %
y+ -
6 cos 30 = 8 sin 30
0:2! 2AB
6 sin 30 = 8 cos 30
0:2
AB
! AB = 6:78 rad/s J
AB
= 49:6 rad/s
2
J
8
314
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.62
16.63
9
315
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.64
(a) and (b)
10
316
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.65
A
= B
1.2(2)2 m/s2
30o
α
B ω = 2 rad/s
8 m/s2 +
m
1.2
aA
A
1.2α
aA = aB + aA=B
+
+
The solution is
"
!
aA = 1:2 sin 30
1:2(2)2 cos 30
0 = 8 1:2 cos 30 + 1:2(2)2 sin 30
= 10:007 rad/s2 and aA =
10:161 m/s2
) aA = 10:16 m/s # J
2
16.66
ω
A
!
vB
aB y
B 50o
4 ft α
50o
vA = 5 ft/s
aA =14 ft/s2
= !k
= k
) ! rB=A = 4!j
vB
0
vB (i cos 50 + j sin 50 )
rB=A = 4i ft
rB=A = 4 j
= vA + !
=
x
5(i cos 50
rB=A
j sin 50 ) + 4!j
11
317
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Equating like components:
vB cos 50
vB sin 50
!
= 5 cos 50
vB = 5 ft/s %
=
5 sin 50 + 4!
5 sin 50 =
= 1:9151 rad/s
= aA +
aB
0
aB (i cos 50 + j sin 50 )
=
rB=A + !
14(i cos 50
5 sin 50 + 4!
(!
rB=A )
j sin 50 ) + 4 j + !k
4!j
Equating like components:
4! 2 = 14 cos 50
4(1:9151)2
aB cos 50
=
14 cos 50
aB
aB sin 50
=
=
=
8:823 ft/s = 8:823 ft/s . J
14 sin 50 + 4
8:823 sin 50 =
2
0:991 rad/s
J
2
2
14 sin 50 + 4
16.67
β
20 rad/s
6 in. A
! BC = 0 (vB and vC are parallel)
2
aC =
aC
+ !
+"
0.5(20 )ft/s2
0.5 ft 20 rad/s +
B
A
= sin
1
6
= 30
12
1.0αBC C
30o
αBC
1.0
ft
vB
B
12
in.
vC
C
B
= aB + aC=B
0
=
0:5(202 )
aC
=
1:0
1:0
BC
cos 30
2
BC
= 230:9 rad/s
sin 30 = 1:0(230:9) sin 30 = 115:5 ft/s " J
2
BC
12
318
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.68
3 rad/s
A
γ ft
4
vC
! BC
sin
B
vB
C
= ! AB = 3 rad/s (A is the I.C. for BC)
2
1
=
= 2 sin 1 = 38:94
6
3
38:94
= 90
= 70:53
=
+2
= 90
2
2
2
180
3 rad/s
A
12 rad/s2
t
6f
aC
t
6f
β
6 ft
β
C =
B
+
6(12) ft/s2 +
2
γ
4(32) ft/s2
B
4 ft
C
αBC
3 rad/s
2
6(3 ) ft/s
= aB + aC=B
aC
+"
4αBC
0 =
0 =
+
=
BC
aC =
=
+
aC
6(32 ) cos + 6(12) sin + 4(32 ) cos + 4 BC sin
54 cos 38:94 + 72 sin 38:94 + 36 cos 70:53
4 BC sin 70:53
2
26:32 rad/s
6(32 ) sin
6(12) cos
4(32 ) sin + 4 BC cos
54 sin 38:94
72 cos 38:94
36 sin 70:53
4( 26:32) cos 70:53
2
=
91:1 ft/s = 91:1 ft/s
2
! J
16.69
By inspection: b! = 2b! BC
aC
ω, α
C =
A
bα
b
B
bω +
) ! BC =
1
2!
ω/2, αBC
2
B
2b(ω/2)2
2b
C
2bαBC
13
319
© 2010. Cengage Learning, Engineering. All Rights Reserved.
aC
+
aC
+"
0
= aB + aC=B
!
= b! 2 + 2b
2
= b
2b
2
=
BC
3 2
b!
2
BC
J
=
1
2
J
16.70
B
4 m/s
o
D
! BC =
6m
0.1
30
C
vC
ωBC
vB
4
= 28:87 rad/s (D is the I.C. for BC)
=
0:16 cos 30
BD
C
28.87 rad/s
B
6m
0.1
αBC
aC =
30o
C
0.16(28.872) m/s
0.16αBC
aC
+
0
= aB + aC=B (note that aB = 0)
=
0:16
BC
cos 30 + 0:16(28:872 ) sin 30
2
aC
=
481:2 rad/s
= 0:16 BC sin 30
0:16(28:872 ) cos 30
= 0:16( 481:2) sin 30
0:16(28:872 ) cos 30
aC
=
BC
+#
154:0 m/s = 154:0 m/s " J
2
2
16.71
vA = 2 m/s
A
0.5
0.4
E 0.3
ωBC
vB
ωAB B
0.6
C
14
320
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Point E is the I.C. of bar AB
! AB
vB
! BC
0.6αBC
A
ωBC = 2.5 rad/s 1.2 m/s2
=
C
αBC
A + 0.4
ωAB = 5 rad/s
αAB
0.5
0.6(2.5)2 rad/s
B
0.6
vA
2
= 5:0 rad/s
=
0:4
EA
= EB ! AB = 0:3(5) = 1:5 m/s
1:5
vB
=
= 2:5 rad/s
=
0:6
BC
=
0.3
0.5αAB
0.5(5)2 rad/s2
B
aB = aA + aB=A
+
!
+
"
The solution is
)
AB
0:6
BC
= 31:125 rad/s2 and
2
AB
4
3
1:2 + (0:5 AB )
(0:5)(5)2
5
5
4
3
= (0:5)(5)2 + (0:5 AB )
5
5
0:6(2:5)2 =
= 31:1 rad/s
J
BC
=
32:23 rad/s2
2
BC
= 32:2 rad/s
J
16.72
15
321
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.73
16
322
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.74
Acceleration Analysis
Assume
AB
to be counterclockwise.
16.75
A ωAB
12
B
D
vB
.21
21
30
45o
C
ωCD
vC
24 rad/s
17
323
© 2010. Cengage Learning, Engineering. All Rights Reserved.
vB = 21:21(24) = 509:0 in./s
A
42.42 rad/s
αAB
12
B
12αCD
αCD
12
+
! AB = ! CD =
12αAB +
509:0
vB
=
= 42:42 rad/s
AB
12
B
30(242) in./s2
30 C
24 rad/s
12(42.42 ) in./s2
2
12(42.422) in./s2
D
=0
42.42 rad/s C
aD
=
+ %
=
+ -
AB
CD
aB + aC=B + aD=C = 0
12
30(242 ) sin 45
AB
12(42:422 ) = 0
2820 rad/s
J
12(42:422 ) + 30(242 ) cos 45 + 12
2
2
=
2
2820 rad/s = 2820 rad/s
CD
=0
J
16.76
in.
5
2
15 in.
B
C
vC
! AB
A
ωAB
D
vB
6
rad/s
in.
15
vC
= vB = ! CD CD = 6(15) = 90 in./s
vB
90
=
=
= 3:6 rad/s
! BC = 0 (vB and vC are parallel)
25
AB
αAB
A
3.6 rad/s
45o
25
B
25αAB
25(3.62) in./s2
20 rad/s2
D
αBC
45o 6 rad/s
+ 15
=
C 15
15(20) in./s2
C
2
15αBC 15(6 ) in./s2
B
18
324
© 2010. Cengage Learning, Engineering. All Rights Reserved.
aB + aC=B
2
+%
25(3:6 ) + 15
BC
= aC
sin 45
=
= 20:36 rad/s
= 15(20)
= 15(20)
BC
+&
25
25
+ 15 BC cos 45
+ 15(20:36) cos 45
AB
AB
15(62 )
=
AB
J
2
J
2
3:36 rad/s
16.77
D
B
vB
15 in.
ωBD
8 in.
in.
17 vD
6 in.
E
Point E is the I.C. of bar BD
vB
! BD
= AB ! AB = 6(3) = 18 in./s #
vB
18
=
= 0:8571 rad/s
=
21
BE
D
2
10αDE D 10ωDE
ωAB
=
A
10
6ωAB2
B 6
! DE = ! BD = 0:8571 rad/s
8
6
ωDE
αDE
E
+
17
2
17ωBD
B
15
17αBD
ωBD
αBD
8
aB = aD + aB=D
+ !
+"
6
8
15
8
(10! 2DE )
(10 DE ) + (17! 2BD ) + (17
10
10
17
17
6(3)2 = 6(0:8571)2 8 DE + 15(0:8571)2 + 8 BD
38:57 = 8 BD 8 DE
6! 2AB
0
=
0 =
0 =
=
8
6
8
(10! 2DE )
(10 DE ) + (17! 2BD )
10
10
17
8(0:8571)2 6 DE + 8(0:8571)2 15
15 BD 6 DE
Solution of (a) and (b) is
)
BD
= 1:3775 rad/s2 and
2
BD
= 1:378 rad/s
J
DE
DE
=
15
(17
17
BD )
(a)
BD )
(b)
3:444 rad/s2
= 3:44 rad/s
J
19
325
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.78
16.79
20
326
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.80
21
327
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.81
22
328
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.82
16.83
y
O
ω
! = 30
rA/O
0.6 ft A, A'
2
60
=
x
rad/s
23
329
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Let A0 be a point on OB that is coincident with A at the instant under consideration
aA
= aA0 + aA=OB + aC
= ! (! rA=O ) + 0 + 2! vA=OB
=
k ( k 0:6i) + 2( k 3i)
=
k (0:6 j) + 6 j = 0:6 2 i + 6 j
aA
=
5:92i + 18:85j ft/s
2
J
16.84
16.85
24
330
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.86
6 rad/s2
A
y
!
rP=A
rP=A
vP
aP
B
3 rad/s
x
!
r P/Ao
30
36 in./s
P
15 in.
2
= 3k rad/s
= 6k rad/s
vP=B = 36j in./s
= 15(i + j tan 30 ) = 15i + 8:660j in.
= 3k (15i + 8:660j) = 45j 25:98i in./s
= vA + vP 0 =A + vP=B = 0 + ! rP=A + vP=B
= 45j 25:98i + 36j = 51:96i + 126:0j
=
26:0i + 81:0j in./s J
= aA + aP 0 =A + aP=B + aC
= 0+
rP=A + ! (! rP=A ) + 0 + 2! vP=B
=
6k (15i + 8:660j) + 3k (45j 25:98i) + 2(3k)
= ( 90j + 51:96i) + ( 135i 77:94j) 216i
=
299i
167:9j in./s
2
36j
J
25
331
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.87
y
vB
ω
2 m/s
70o
B
A
x
0.75 m
= vA + vB 0 =A + vB=B
= 0 + ! rB=A + vB=B
vB (i cos 70 + j sin 70 ) = !k 0:75i + 2i = 0:75!j + 2i
vB
Equating like components:
vB cos 70
vB sin 70
= 2
vB = 5:848 m/s
= 0:75!
5:848 sin 70 = 0:75!
! = 7:327 rad/s J
= aA + aB 0 =A + aB=B + aC
= 0 + ! (! rB=A ) + 0 + 2! vB=B
= 7:327k 0:75(7:327)j + 2(7:327k) 2i
=
40:3i + 29:3j m/s J
aB
16.88
vP
P
vP
+
+"
ft/s
45o 4(1.6971)
P'
45o
=
+
1
7
vP/AB
9
4 rad/s
.6
1
A
= vA + vP 0 =A + vP=AB = 0 + vP 0 =A + vP=AB
0
=
vP
vP
=
=
4(1:6971)
1.6971(12) rad/s2
45o
(aP)x
76.8 ft/s2
=
vP=AB = 6:788 ft/s
4(1:6971) + vP=AB cos 45
[4(1:6971) + 6:788] cos 45 = 9:60 ft/s J
(aP )y =
P
vP=AB sin 45
vP2
=
R
9:602
2
= 76:8 ft/s
1:2
1.6971(42) rad/s2
ω
45o
45o
+ vP/AB
aP/AB
2(4)(6.788) ft/s2
P'
1
4 rad/s 697
+
1.
A
12 rad/s2
26
332
© 2010. Cengage Learning, Engineering. All Rights Reserved.
= aA + aP 0 =A + aP=AB + aC = 0 + aP 0 =A + aP=AB + aC
aP
+"
76:8
+ !
42 )
=
1:6971(12
aP=AB
=
2
(aP )x
=
1:6971(12 + 42 )
(aP )x
=
28:8 ft/s
2
) aP
=
28:8i
47:52 ft/s
76:8j ft/s
aP=AB
2(4)(6:788) sin 45
47:52 + 2(4)(6:788) cos 45
2
J
16.89
ωAB
0.2
309
m
0.2309ωAB
P'
vP/AB x
30o
1.2 m/s
+
=
60o
P
y
A
= vA + vP 0 =A + vP=AB = 0 + vP 0 =A + vP=AB
= 0:2309! AB
! AB = 4:50 rad/s
J
= vP=AB
vP=AB = 0:6 m/s
vP
1:2 cos 30
1:2 sin 30
+x +y .
0.2309 (4.52) m/s2
P'
0.2309αΑΒ
0 =
4.50 rad/s
A
0.2
309
m
30o
+ 60 o
ωAB
+ v
P/AB
aP/AB
30o
2(4.50)(0.6) m/s2
αΑΒ
aP = aA + aP 0 =A + aP=AB + aC
0 = 0 + aP 0 =A + aP=AB + aC
+x -
0
=
0:2309
AB
2(4:50)(0:6)
AB
= 23:4 rad/s
2
J
16.90
aO
vP=disk
20
2
(2) = 3:333 ft/s
12
2
6 ft/s #
aP=disk = 28 ft/s #
= R =
=
1
333
© 2010. Cengage Learning, Engineering. All Rights Reserved.
aP = 3.333 ft/s
O
2
1.0(52) ft/s2
P'
1.0 ft
1.0(2) ft/s
2
+
O
2(5)(6) ft/s2
ω
+
28 ft/s2
vP/disk
+
5 rad/s
2 rad/s2
aP = aO + aP 0 =O + aP=disk + aC
+
+
!
"
)
(aP )x =
3:333
(aP )y =
1:0(52 )
aP =
65:3i
1:0(2)
2(5)(6) =
28 =
53:0j ft/s
2
53:0 ft/s
65:3 ft/s
2
2
J
*16.91
2
334
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.92
! = 4k rad/s
=0
rP 0 =A = 0:24(i cot 60 + j) = 0:13856i + 0:24j m
(a)
= ! rP 0 =A = 4k (0:13856i + 0:24j) = 0:5542j 0:96i m/s
= vP=AB (i cos 60 + j sin 60 ) = vP=AB (0:5i + 0:8660j) m/s
vP 0 =A
vP=AB
vP
vP i
= vA + vP 0 =A + vP=AB
= 0 + 0:5542j 0:96i+vP=AB (0:5i + 0:8660j)
Equating y-components:
+
"
vP=AB
=
0 = 0:5542 + 0:8660vP=AB
0:640 m/s
60o
vP=AB =
0:6400 m/s
J
(b)
aP 0 =A
aP=AB
aC
= !
(!
rP 0 =A ) = 4k
(0:5542j
= aP=AB (0:5i + 0:8660j) m/s
=
aP
aP i
2!
vP=AB = 2(4k)
0:96i) =
2:217i
3:84j m/s
2
2
( 0:32i
0:5542j) =
= aA + aP 0 =A + aP=AB + aC
= 0 2:217i 3:84j+aP=AB (0:5i + 0:8660j)
2:56j + 4:434i m/s
2:56j + 4:434i
Equating y-components:
+"
0=
3:84 + 0:8660aP=AB
2:56
aP=AB = 7:390 m/s
2
60o J
16.93
3
335
© 2010. Cengage Learning, Engineering. All Rights Reserved.
2
16.94
4
336
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.95
16.96
5
337
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.97
B
P
390
A
150
2 rad/s
22.62o
360
D
6
338
© 2010. Cengage Learning, Engineering. All Rights Reserved.
P
0.3 m
0.3(2) m/s
m
0.39 o
22.62
= ωAB
A
D
2 rad/s
0.39ωAB
22.62o
P'
+ v
P/AB
x
y
vP = vA + vP 0 =A + vP=AB = 0 + vP 0 =A + vP=AB
+x 0:3(2) sin 22:62 = 0:39! AB
! AB = 0:5917 rad/s
+y .
0:3(2) cos 22:62 = vP=AB
vP=AB = 0:5538 m/s
0.39αAB
P
0.15 m
0.15(22) m/s2
2 rad/s
O
aP
A
+
22.62
ωAB
22.62o
22.62 +
vP/AB
vP/AB
2(0.5917)(0.5538)
o
0.5917 rad/s
=
+x AB
αAB 0.39 m P'
o
=
0.39(0.59172) rad/s2
J
=
aA + aP 0 =A + aP=AB + aC = 0 + aP 0 =A + aP=AB + aC
0:15(22 ) cos 22:62 = 0:39
0:260 rad/s
2
AB
2(0:5917)(0:5538)
J
16.98
7
339
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.99
A
9i
n.
θ
xD
xD
vD
aD
=
=
9 cos in.
9 _ sin =
B
9 (8) sin 40 =
2
=
9• sin
9 _ cos in./s
=
9(140) sin 40
46:3 in./s
2
9(8)2 cos 40 =
) vD = 46:3 in./s ! J
1251 in./s
aD = 1251 in./s
2
2
! J
16.100
A
yA
θ L
B
xB
2
Constraint: x2B + yA
= L2
(a) Di¤erentiating with respect to time:
2xB vB + 2yA vA
xB
vB
) vA =
yA
) vA
=
=
=
0
(a)
vB tan =
1:4 tan 20 =
0:5096 m/s
0:510 m/s # J
(b) Di¤erentiating (a) with respect to time:
2
2
+ yA aA
+ xB aB + vA
vB
2
2
vB + vA + xB aB
) aA =
yA
aA
=
=
=
0
1:42 + ( 0:5096)2 + 0
=
1:8 cos 20
1:312 m/s
2
1:312 m/s # J
2
8
340
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.101
xA
A
in.
12
yB θ
B
xB
xB
x
•B
Substituting
0
=
0
=
= xA + 12 sin
x_ B = x_ A + 12 cos
2
= x
•A + 12(cos • sin _ )
yB
=
y•B
=
12 cos
y_ B =
12 sin
_2
12(sin • + cos
_
(a)
_
(b)
= 25 , x
•B = y•B = 0 and x
•A = 36 in./s2 into (a) and (b), we get
36 + 12(cos 25 •
2
sin 25 _ )
0 = 36 + 10:876•
2
12(sin 25 • + cos 25 _ )
0=
5:071•
5:071 _
10:876 _
2
2
2
2:719 rad/s2 , _ = 1:2677 (rad/s)2 :
The solution is: • =
) • = 2:72 rad/s
2
J
_ = 1:126 rad/s (sense unknown) J
16.102
B
0.25 m
5m
0.7
0.5
m
C
φ
x θ
D
Geometry:
x = 0:75 cos + 0:5 cos
0:25 = 0:75 sin
0:5 sin
(a)
Di¤erentiation with respect to time:
x_ =
0 =
0:75 _ sin
0:75 _ cos
0:5 _ sin
0:5 _ cos
(b)
(c)
9
341
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Substitute
= 60 and x_ = 1:2 m/s:
From (a):
From (b):
From (c):
0:25 = 0:75 sin 60
0:5 sin
= 53:04
1:2 = 0:75 _ sin 60
0:5 _ sin 53:04
0 = 0:75 _ cos 60
0:5 _ cos 53:04
Solving (d) and (e) simultaneously, we get _ =
rad/s. Therefore,
! CD = 1:045 rad/s
1:302 rad/s and _ =
(d)
(e)
1:045
J
16.103
10
342
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.104
16.105
11
343
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.106
16.107
12
344
© 2010. Cengage Learning, Engineering. All Rights Reserved.
*16.108
A
θ
9 ft
6 ft
φ
C
L
B
Geometry:
6 sin
L sin
6 cos + L cos
= 0
= 9 ft
(a)
(b)
Di¤erentiating with respect to time:
6 _ cos
6 _ sin
Substituting
= 20 and L_ =
L _ cos
L_ sin
L _ sin + L_ cos
6 sin 20
L sin = 0
6 cos 20 + L cos = 9 ft
The solution is L = 3:939 ft and
The solution is _ =
(c)
(d)
2 ft/s, we get
From (a):
From (b):
From (c): 6 _ cos 20
From (d):
6 _ sin 20
= 0
= 0
= 31:40 .
3:939 _ cos 31:40
( 2) sin 31:40
_
3:939 sin 31:40 + ( 2) cos 31:40
0:4053 rad/s and _ =
! AB = 0:427 rad/s
= 0
= 0
0:4265 rad/s. Therefore,
J
13
345
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.109
D
y
A
y
y_
θ
θ
C
m
0.5
m
B
5
0.
θ
= 3(0:5) sin = 1:5 sin m
= 1:5 _ cos m/s
y• =
Substitute
0.5
m
(a)
2
2
1:5 _ sin + 1:5• cos m/s
(b)
= 30 , y_ = 3 m/s and y• = 0:
_ = 2:309 rad/s
1:5(2:309)2 sin 30 + 1:5• cos 30
From (a):
3 = 1:5 _ cos 30
From (b):
0=
! ABC = 2:31 rad/s
J
• = 3:078 rad/s2
2
ABC
= 3:08 rad/s
J
16.110
14
346
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.111
15
347
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.112
x
G
2R θ
A
x = 2R tan
x_ = 2R _ sec2
2
x• = 2R _ (2 sec )(sec tan ) + 2R• sec2
= 4R! 2 sec2 tan
(note that _ = ! and • = 0)
= 4R! 2 sec2 50 tan 50 = 11:537R! 2
)
gear
=
x•
= 11:54! 2
R
J
16.113
16
348
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.114
17
349
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.115
16.116
16.117
y
at y
o
6.28 m/s2 25 A x
B
an
25o
0.72 m/s2
an
at x
18
350
© 2010. Cengage Learning, Engineering. All Rights Reserved.
At point A:
an
=
6:28 cos 25 + 0:72 sin 25 = 5:996 m/s
2
at
=
6:28 sin 25
2
0:72 cos 25 = 2:002 m/s
Any point on the rim of the disk has the same an and at . Therefore, at point
B we have
2
aB = 2:00i + 6:00j m/s J
16.118
Because the two arms form a parallelogram linkage, member BC translates (BC
remains horizontal). Hence the velocity and acceleration vectors of B and C are
identical.
rB=A
=
2(i cos 30 + j sin 30 ) = 1:7321i + 1:0j m
! AB
=
2k rad/s
AB
=
1:5k rad/s
2
= vB = ! AB rB=A = 2k (1:7321i + 1:0j)
=
2:0i + 3:464j m/s J
= aB = ! AB vB + AB rB=A
= 2k ( 2:0i + 3:464j) + ( 1:5k) (1:7321i + 1:0j)
=
4:0j 6:928i 2:598j + 1:5i
2
=
5:43i 6:60j m/s J
vC
aC
16.119
60 in./s
= ωAB
A
in.
10
C
17ωAB
Bω
BC
B
.
n
i
17
8 in. + 8 in.
10ωBC
15 in.
6 in.C
= vB + vC=B
8
8
+ 10! BC
= 8! AB + 8! BC
+ ! 0 =
17! AB
17
10
15
6
+ " 60 = 17! AB
+ 10! BC
= 15! AB + 6! BC
17
10
vC
The solution is
! AB = ! BC = 2:86 rad/s
J
19
351
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.120
16.121
ωAB E
2
vB
B 1.2
vD D
2
vA
2
60o
60o
ωBC
ωBD
A
C
F
Point E is the I.C. of bar AB, and point F is the I.C. of bar BD
! BD
vB
! AB
vA
5
vD
=
= 2:406 rad/s
1:2
tan
60
FD
= F B ! BD = (1:2 sec 60 ) (2:406) = 5:774 ft/s
vB
5:774
=
=
= 2:887 rad/s
2
BE
= ! AB EA = 2:887(4 sin 60 ) = 10:0 ft/s
J
=
20
352
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.122
16.123
21
353
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.124
15
0
A
805
.0
E
4
3
A
201
.2
2
60
Geometry: AD
AE
! AB
vB
! BC
1
360
vA
D 7 rad/s
D 90
vA
720
2
120
O
1
ωAB
C
B
vB
180
ωBC
p
902 + 1802 = 201:2 mm
p
=
3602 + 7202 = 805:0 mm
=
2
= AD! OA = 201:2(7) = 1408:4 mm/s
vA
1408:4
=
=
= 1:7496 rad/s
805:0
AE
= BE! AB = 720(1:7496) = 1259:7 mm/s
vB
1259:7
=
= 7:00 rad/s
J
=
180
BC
16.125
22
354
© 2010. Cengage Learning, Engineering. All Rights Reserved.
vC
C
60
0
16.126
F
B
E
vB
0
60
00
72 rad/s 5 o
35
450
A
D
Geometry: BE
BF
DF
vB
! BCD
vD
ωBCD
vD
p
6002 4502 = 396:9 mm
BE
396:9
=
=
= 484:5 mm
cos 35
cos 35
= DE + BE tan 35 = 450 + 396:9 tan 35 = 727:9 mm
=
= ! AB AB = 72(500) = 36 000 mm/s
vB
36 000
=
=
= 74:30 rad/s
J
484:5
BF
= ! BCD DF = 74:30(727:9) = 54 080 mm/s = 54:1 m/s ! J
23
355
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.127
Let D0 be a point on rod AB that is coincident with D at the instant under
consideration
aD
= aD0 + aD=AB + aC
=
rD0 =A + ! (! rD0 =A ) + aD=AB + 2! vD=AB
= 18k (12i 5j) + 6k [6k (12i 5j)] + 48j + 2(6k)
= (216j + 90i) + 6k (72j + 30i) + 48j + 432i
=
216j + 90i
432i + 180j + 48j + 432i = 90i + 444j in./s
( 36j)
2
J
16.128
24
356
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16.129
16.130
(a)
O
ωAB = 6 rad/s
o
30
B
o
vB 40
o
60 800
Geometry: BC
OC
OA
C
y
x
A
vA
= 800 cos 40 = 612:8 mm
= BC tan 30 = 612:8 tan 30 = 353:8 mm
= OC + AB sin 40 = 353:8 + 800 sin 40 = 868:0 mm
vA = ! AB OA = 6(868:0) = 5208 mm/s = 5:21 m/s ! J
25
357
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(b)
! AB
aA
aA i
aA i
! AB
aA
rA=B
rA=B
=
=
=
=
2
6k rad/s
AB = 8k rad/s
aA i
aB = aB (i cos 60
j sin 60 ) = aB (0:5i 0:8660j)
BC i AC j = 612:8i 800j sin 40 = 612:8i 514:2j mm
6k (612:8i 514:2j) = 3677j + 3085i
= aB + AB rA=B + ! AB (! AB rA=B )
= aB (0:5i 0:8660j) + 8k (612:8i 514:2j) + 6k (3677j + 3085i)
= aB (0:5i 0:8660j) + (4902j + 4114i) + ( 22 060i + 18 510j)
Equating like components:
aA = 0:5aB + 4114 22 060
0 =
0:8660aB + 4902 + 18 510
The solution is
aB = 27 050 mm/s2 and aA =
aA = 4:42 m/s
2
4423 mm/s2 . Therefore,
J
16.131
26
358
© 2010. Cengage Learning, Engineering. All Rights Reserved.
27
359
© 2010. Cengage Learning, Engineering. All Rights Reserved.
28
360
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Chapter 17
17.1
17.2
1
6 ft
1.5 ft
z
=
R12 h1 =
z
(0:9317)(1:5)2 (6) = 13:172 slugs
=
m2
=
Iz
=
(Iz )1 + (Iz )2 =
=
3
13:172(1:5)2
10
3
2
4 ft
30
3
= 0:9317 slug/ft
32:2
m1
3
1.0 ft
3
R22 h2 =
3
(0:9317)(1:0)2 (4) =
3:903 slugs
3
(m1 R12 + m2 R22 )
10
3:903(1:0)2 = 7:72 slug ft2 J
17.3
1
361
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.4
17.5
2
362
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.6
"
Iz
=
1
mro d b2 + mro d
3
12
) Iz
=
1
1
mrod b2 = mb2 J
2
6
2
b
p
2 3
#
where mro d =
m
3
17.7
"Thin" implies that t
b, so that t can be neglected in comparison to b.
One side (mass = m=4): (Iz )1 =
Bottom (mass = m=2): (Iz )2 =
Iz = 2(Iz )1 + (Iz )2 = 2
1 m 2 m
b +
12 4
4
m
1 m 2
(b + b2 ) +
12 2
2
1
mb2
12
2
b
2
=
p b
2
2
1
mb2
12
2
=
1 2
mb
3
1
1
+ mb2 = mb2 J
3
2
17.8
2
75
y
1
180
150
x
m1
=
m2
=
m3
=
60
y
3
x
Thickness = 80
0:15(0:18)(0:08)(2650) = 5:724 kg
2
(0:075)2 (0:08)(2650) = 1:8732 kg
(0:06)2 (0:08)(2650) =
2:398 kg
3
363
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(Ix )1
=
(Ix )2
=
(Ix )3
=
Ix
=
1
(5:724)(0:152 + 0:082 ) = 0:013 785 kg m2
12
1
(1:8732) 3(0:075)2 + 0:082 = 0:003 63 kg m2
12
1
(2:398) 3(0:06)2 + 0:082 = 0:003 44 kg m2
12
0:003 44 = 0:013 98 kg m2 J
i (Ix )i = 0:013 785 + 0:003 63
(Iz )1
=
(Iz )2
=
(Iz )3
=
Iz
=
1
(5:724)(0:152 + 0:182 ) + 5:724(0:09)2 = 0:072 55 kg m2
12
1
(1:8732)(0:075)2 = 0:002 63 kg m2
4
1
(2:398)(0:06)2 = 0:004 32 kg m2
2
0:004 32 = 0:0709 kg m2 J
i (Iz )i = 0:072 55 + 0:002 63
17.9
y
y
50
100
50
z
x
200
50
100
150
100
Block:
Cylinder:
Hole:
(Iz )1
=
(Iz )2
=
(Iz )3
=
Iz =
200
m1 = 0:2(0:4)(0:1)(7850) = 62:80 kg
m2 = (0:05)2 (0:15)(7850) = 9:248 kg
m3 =
(0:05)2 (0:1)(7850) = 6:165 kg
62:80
(0:22 + 0:42 ) + 62:80(0:1)2 = 1:6747 kg m2
12
9:248
(0:05)2 + 9:248(0:2)2 = 0:3815 kg m2
2
6:165
(0:05)2 = 0:0077 kg m2
2
i (Iz )i
= 1:6747 + 0:3815
0:0077 = 2:05 kg m2 J
4
364
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.10
y
1
12 in.
x
_
r
G
2
9 in.
0:6 12
= 0:010 648 slugs
32:2 21
m1 =
(Iz )1
=
(Iz )2
=
=
Iz
=
x =
y
=
r2
Iz
=
=
=
m2 =
0:6 9
= 0:007 986 slugs
32:2 21
1
1
m1 L21 = (0:010 648)(1:0)2 = 0:003 549 slug ft2
3
3 "
#
2
L
1
1
2
m2 L22 + m2 L21 +
= m2 (3L21 + L22 )
12
2
3
1
(0:007 986) 3(1:0)2 + 0:752 = 0:009 483 slug ft2
3
(Iz )1 + (Iz )2 = 0:003 549 + 0:009 483 = 0:013 032 slug ft2 J
mi xi
Li xi
12(0) + 9(4:5)
= 1:9286 in.
=
=
mi
Li
21
mi yi
Li yi
12( 6) + 9( 12)
= 8:571 in.
=
=
mi
Li
21
x2 + y 2 = 1:92862 + ( 8:571)2 = 77:18 in.2 = 0:5360 ft2
Iz mr2 = 0:013 032 (0:010 648 + 0:007 986)(0:5360)
0:003 04 slug ft2 J
17.11
5
365
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.12
17.13
(a) Sphere:
Ix
=
=
2
mR2 + m(L + R)2
5
2
130
5
2 130
+
5 32:2
12
32:2
60 + 5
12
2
= 118:73 slug ft2
Rod:
Ix
Pendulum:
=
1
mL2 + m
12
=
1
3
25
32:2
2
L
2
60
12
=
1
mL2
3
2
= 6:47 slug ft2
Ix = 118:73 + 6:47 = 125:20 slug ft2 J
(b) Using the speci…ed approximation:
2
Ix
130 60 + 5
= 118:45 slug ft2
32:2
12
118:45 125:20
100% = 5:39% J
125:20
= m(L + R)2 =
% error =
6
366
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.14
I = IO mx2
408:5 120x2
)I
= IC m(2 x)2
= 145:5 120(2 x)2
x = 1:5479 m J
2
= 408:5 120(1:5479) = 121:0 kg m2 J
17.15
*17.16
7
367
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.17
17.18
8
368
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.19
B
20 lb
O
ma
O
=
2'
P
A
N
( MO )FBD
P
B
2'
A
60o
MAD
FBD
= ( MO )M AD
= 23:1 lb J
+
P (2 sin 60 )
20(4 cos 60 ) = 0
17.20
50 lb
190 lb
B
t
2f
B
50o
Ax
A
Ay
t
2f
FBD
80 a
32.2
=
t
2f
80 lb
y
190 a
32.2
=
A
FBD N1
MAD
Bar:
P
x
N2
MAD
( MA )FBD = ( MA )M AD
+
50(4 sin 50 )
a =
13:231 ft/s
Cart + bar:
+
!
80(2 cos 50 ) =
80
a(2 sin 50 )
32:2
2
Fx = max
190
190
a=
(13:231) = 78:1 lb J
P =
32:2
32:2
9
369
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.21
17.22
10
370
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.23
17.24
11
371
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.25
y
80 lb
FBD
80 a
32.2 n
o
= 30
MAD
x
F
N
The box translates along a circular path of 5-ft radius
) an = R! 2 = 5(2)2 = 20 ft/s
Fy
= may
+"
N
Fx
= max
+ !
2
80
(20) sin 30
32:2
80
F =
(20) cos 30
32:2
N = 104:8 lb J
80 =
F = 43:0 lb J
17.26
L/2
A
N
mg
( MA )FBD
L
+
mg
2
=
aB
B = A
FBD
L/2
m Lα
2
( MA )M AD
L
L mL2
=
m
+
2
2
12
3
= L = g #J
2
mL2α
12
B
MAD
=
3g
2L
12
372
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.27
Ay
N2 =
5a
y
x
N1
37 o
T
MAD
FBD
A
Ax
1.5 m
37
A
o
0.75 m
5(9.81) N
= 37 o
2(9.81) m 2a
B
B FBD
MAD
System:
Fx = max
+.
5(9:81) sin 37 = 5a
a = 5:904 m/s
2
Bar:
( MA )FBD
T
= ( MA )M AD
+
(T sin 37 ) 1:5 = (2a cos 37 ) 0:75
= a cot 37 = 5:904 cot 37 = 7:83 N J
17.28
mg
B
A
L/2 R L/2
FBD's
mg
F
D
L/2
L/2
1 2
B 12 mL αAC
C
C = A
L/2
L/2
T
MAD's
1 mL2
T
12 αDE
E = D
E
L/2
L/2
F Lα
m 2 ( AC + αDE)
Kinematics:
# aE = aC =
L
2
AC
# aF = aE +
L
2
DE
=
L
(
2
AC
+
DE )
Kinetics— bar AC:
( MB )FBD = ( MB )M AD
L
1
T =
mL2
2
12
+
AC
AC
6T
mL
=
Kinetics— barDE:
( MF )FBD
Fy
T
=
( MF )M AD
= may
=
+#
+
mg
L
1
T =
mL2
2
12
L
T = m ( AC +
2
DE
DE )
DE
=m
L
2
6T
mL
6T
2
mL
=
1
mg J
7
13
373
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.29
y
TA
1.0'
A
G
T
1.0' B B
=
35o
mg
x
FBD
A
G
ma
MAD
B
35o
Bar AB translates, so that all points on the bar have the same acceleration
( MG )FBD
Fy
2TA
= ( MG )M AD
+
(TB TA )(1:0 cos 35 ) = 0
) T A = TB
= may
+ % TA + TB mg cos 35 = 0
= 6 cos 35
) TA = TB = 2:46 lb J
17.30
14
374
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.31
17.32
20(0.32)α
20(9.81) N
G 0.4 m
0.4 m 0.4 m
=
G
T
600 N
R
T
FBD's
A
40(9.81) N
MAD's
40(0.4α)
A
System:
+
600(0:4)
( MG )FBD = ( MG )M AD
40(9:81)(0:4) = 20(0:32 ) + 40(0:4 )(0:4)
=
2
10:127 rad/s
J
Block A:
Fy
T
= may
+ " T 40(9:81) = 40(0:4 )
= 40(9:81) + 40(0:4)(10:127) = 554 N J
15
375
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.33
FBD
mg
2 ft
A
B
=
2 ft
µs NA
B
NB 4.5 ft 3 ft NA
Fx
( MB )FBD
ma
4:5mg 7:5
= max
+ !
=
( MB )M AD
=
2ma
MAD
G
s NA
ma A
4.5 ft 3 ft
= ma
NA =
ma
s
+
4:5mg 7:5NA = 2ma
a
2
7:5
= 2a
a = 12:74 ft/s J
0:8
4:5(32:2)
s
17.34
T
T
A mg R FC B
mg
FD
C NC
D
N
D
FBD's
Disk A
:
1mR2α
2
= A
mRα
B
D
MAD's
( MC )FBD = ( MC )M AD
1
T R + mg(R sin 45 ) = mR2 + mR (R)
2
+
Disk B: ( MD )FBD = ( MD )M AD
)
T
1mR2α
2
45o mRα
C
=
+
TR =
1
mR2 + mR (R)
2
T R + mg(R sin 45 ) = T R
mg sin 45
30 sin 45
=
= 10:61 lb J
2
2
17.35
FA
FA
3'
C
2'
FBD
FB
2(aC + 3α)
A
A
1180 lb.ft =
5aC
20α
FB
B
4(-aC +2α)
B MAD
Only horizontal forces are shown
Kinematics:
! aA = aC + 3
aB =
aC + 2
16
376
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Kinetics— racks:
FA = 2(aC + 3 )
FB = 4( aC + 2 )
Kinetics— gear:
2
= 5(2)2 = 20 slug ft2
IC = mkC
( MC )FBD = ( MC )M AD
1180
+
1180 3FA 2FB = 20
6(aC + 3 ) 8( aC + 2 ) = 20
2aC 54
=
1180 (a)
Fx = max
+ ! FB FA = 5aC
4( aC + 2 ) 2(aC + 3 ) = 5aC
2
11aC = 0
and aC = 4:0 ft/s2 ! J
= 22:0 rad/s2
Solution of (a) and (b) is
(b)
17.36
α 4'
x
4'
G
aA
3' T
B
aB
y
B
G
2mα
mg
A
3'
A
B
1
2
12 m(5 )α
1.5mα
=
A
N FBD
MAD
Kinematics:
aG
aG
=
=
)
=
)
k
rG=A = rG=B = 1:5i + 2j ft
rG=A = ( 1:5j + 2i)
aA + aG=A = aA i +
rG=A = aA i + ( 1:5j + 2i)
(aG )y = 1:5
aB + aG=B = aB j +
rG=B = aB j + (1:5j 2i)
(aG )x = 2
Kinetics:
Fx
( MA )F BD
+
1:5mg
4T
1:5mg
4T
= max
+
T = 2m
= ( MA )M AD
1
m(52 ) + 1:5m (1:5) 2m (2)
=
12
= 0:3333m
(a)
(b)
Solution of (a) and (b) is
T
= 0:180g = 0:180(32:2) = 5:80 rad/s
= 0:360mg = 0:360(20) = 7:20 lb J
2
J
17
377
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.37
18
378
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.38
19
379
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.39
17.40
R2 h = (0:12)2 (0:36)(7850) = 42:62 kg
3
3
= mg = 42:62(9:81) = 418:1 N
3
3
m(4R2 + h2 ) =
(42:62) 4(0:12)2 + 0:362 = 0:2992 kg m2
=
80
80
m =
W
I
22o
0.2
7m
G
=
418.1 N
F
A
N FBD
0.2
7m
0.2992α
A
=
Fx
11.507α
y
x
MAD
( MA )FBD = ( MA )M AD
418:1(0:27 sin 22 ) = 0:2992 + 11:507 (0:27)
+
Fy
G
=
)
=
)
12:416 rad/s
2
J
may
+ " N 418:1 = 11:507 sin 22
N = 418:1 11:507(12:416) sin 22 = 365 N " J
max
+ ! F = 11:507 cos 22
F = 11:507(12:416) cos 22 = 132:5 N ! J
20
380
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.41
17.42
21
381
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.43
22
382
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.44
17.45
23
383
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.46
24
384
© 2010. Cengage Learning, Engineering. All Rights Reserved.
14.47
25
385
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.48
26
386
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.49
20o
FBD
y
Mg
Iα
x
G
R
=
Ma-
0.075N C
N
Fy
N
Fx = max
G
MAD
C
= 0
+ - N M g cos 20 = 0
= M (9:81) cos 20 = 9:218M
+ % 0:075N M g sin 20
0:075(9:218M ) M (9:81) sin 20
=
=
a =
Ma
Ma
2:66 m/s
2
J
17.50
27
387
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.51
28
388
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.52
1
389
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.53
2
390
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.54
17.55
Kinematics:
y
1.2 m
ω, α
O G
x
0.4 m
3
391
© 2010. Cengage Learning, Engineering. All Rights Reserved.
a = aO + aG=O = R i +
rG=O + !
! rG=O
= 1:2(3)i + ( 3k) 0:4i + ( !k) ( !k 0:4i)
= 3:6i 1:2j + ( !k) ( 0:4!j)
=
0:4! 2 i
3:6
1:2j m/s
Kinetics:
2
Iα
O G
0.4 m Mg
FBD N
Ma-y
O G
0.4 m
=
C F
MAD
Ma-x
1.2 m
C
I = M k 2 = M (0:4)2 = 0:16M
( MC )FBD = ( MC )M AD
+
0:4M g = 1:2M ax 0:4M ay + I
0:4M (9:81) = 1:2M 3:6 0:4! 2
0:4M ( 1:2) + 0:16M (3)
3:924
=
5:280
0:48! 2
! = 1:681 rad/s J
17.56
4
392
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.57
5
393
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.58
17.59
6
394
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.60
7
395
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.61
8
396
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.62
9
397
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.63
17.64
10
398
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.65
11
399
© 2010. Cengage Learning, Engineering. All Rights Reserved.
*17.66
12
400
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.67
13
401
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.68
14
402
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.69
(a)
0.8 m mg
FBD
Fx
20v
220
FD
N d
P
0.3 m
=
C y
= max
= 40a
ma
0.4 m
C
x
MAD
+ !
a = 5:5
P FD = ma
2
0:5v m/s Q.E.D.
(a)
(b)
dv
Zdt
a =
t
=
)
dv
= dt
a
5:5
dv
+C =
5:5 0:5v
dv
= dt
0:5v
2:0 ln(5:5
0:5v) + C
Initial condition: v = 0 when t = 0. ) C = 2:0 ln 5:5
5:5 0:5v
= 2:0 ln 1
) t = 2:0 ln
5:5
When tipping impends, we have d = 0 on the FBD
v
11:0
(b)
( MC )FBD = ( MC )M AD
+
0:8P 0:3mg = 0:4ma
2
0:8(220) 0:3(40)(9:81) = 0:4(40)a
a = 3:643 m/s
Eq. (a) :
3:643 = 5:5
Eq. (b)
t=
:
0:5v
v = 3:714 m/s
3:714
t = 0:824 s J
2:0 ln 1
11:0
17.70
(a)
W
I
= mg = 2:4(9:81) = 23:54 N
1
1
=
mR2 = (2:4)(0:18)2 = 0:03888 kg m2
2
2
23.54 N
O 45o
0.18 m
0.03888α
=
y
NO
C 0.15NC
FBD
NC
C
O
x
MAD
15
403
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Fy
NC
= may
+%
= 20:47 N
NC cos 45 + 0:15NC sin 45
23:54 cos 45 = 0
( MO )FBD
=
( MO )M AD
+
0:15NC (0:18) = 0:03888
0:15(20:47)(0:18)
=
0:03888
= 14:215 rad/s
2
J
(b)
=
d!
!
d
)
d = ! d!
Initial condition: ! = 0 when
14:215 d = ! d!
14:215 =
1 2
! +C
2
= 0: ) C = 0
) 14:215 =
1 2
!
2
)
= 0:03517! 2
Final angular speed of the disk is
! 1 = v=R = 6=0:18 = 33:33 rad/s
= 0:03517(33:33)2 = 39:07 rad = 6:22 rev J
When ! = ! 1 :
17.71
(a)
Wheel: I = mR2 =
Ay
A
1.0'
6 lb
1.0' 20o
18
(2)2 = 2:236 slug ft2
32:2
N
Ax
0.75N
2' 18 lb
C Cx
Cy
o
40
0.75N B FBD
N
Bar AB :
+
N =
Wheel
+
:
=
2.236α
20o
=
C
MAD
FBD
MA = 0
6(1:0 sin 20 ) + 0:75N (2 cos 40 )
15:033 lb
N (2 sin 40 ) = 0
( MC )FBD = ( MC )M AD
0:75N (2) = 2:236
0:75(15:033)(2) =
2
10:085 rad/s = 10:085 rad/s
2
2:236
J
16
404
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(b)
=
d!
dt
) d! =
dt =
)!=
10:085 dt
10:085t + C
Initial condition: ! = 400 rev/min = 41:89 rad/s when t = 0
) C = 41:89 rad/s
)!=
When ! = 0: t =
10:085t + 41:89
41:89
= 4:15 s J
10:085
17.72
(a)
Wheel: I = mR2 =
Ay
A
1.0'
6 lb
1.0' 20o
N
Ax
0.75N
Wheel
+
:
=
2.236α
20o
2' 18 lb
C Cx
Cy
o
40
0.75N B FBD
N
Bar AB :
+
N =
18
(2)2 = 2:236
32:2
=
C
MAD
FBD
MA = 0
6(1:0 sin 20 )
0:8429 lb
0:75N (2 cos 40 )
N (2 sin 40 ) = 0
( MC )FBD = ( MC )M AD
0:75N (2) = 2:236
0:75(0:8429)(2) =
2
0:5654 rad/s = 0:5654 rad/s
2
2:236
J
(b)
=
d!
dt
) d! =
dt =
0:5654 dt
)!=
0:5654t + C
Initial condition: ! = 400 rev/min = 41:89 rad/s when t = 0
) C = 41:89 rad/s
)!=
When ! = 0: t =
0:5654t + 41:89
41:89
= 74:1 s J
0:5654
17
405
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.73
17.74
18
406
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.75
19
407
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.76
20
408
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.77
17.78
(a)
F A
N
A
L/2
θ
FBD
L/2
mg
=
B
mL ω2
θ
2
L
m α
2
mg
1 mL2α
12
MAD
B
21
409
© 2010. Cengage Learning, Engineering. All Rights Reserved.
+
( MA )FBD
=
L
sin
2
=
mg
( MA )M AD
1
L
mL2 + m
12
2
3g
(cos
sin )
2L
=
2
L
cos
2
mg
J
(b)
=
d!
!
d
) ! d!
=
1 2
!
2
=
3g
(cos
2L
d =
sin ) d
3g
(sin + cos ) + C
2L
Initial condition: ! = 0 when = 0: ) C = 3g=2L
r
3
g (sin + cos
1) J
)!=
L
(c)
Fx = max
Maximum
L
F = mg + m ! 2 sin
2
+
=
L
cos
2
occurs when ! = 0:
sin + cos
When
m
max
1=0
max
= 90
J
= 90 :
L
F = mg + m ! 2 = mg + 0 = mg J
2
17.79
A
θ
L/2
L/2
θ
FBD
mg
2
+
m L
g( + L) cos
2 2
=
1m 2
12 2 L α
MAD
mg
2
( MA )FBD
1m 2
12 2 L α
=
=
=
A
m Lω2
22
mLα
22
mLω2
2
mLα
2
( MA )M AD
m L
2 2
18 g
cos
17 L
2
+
m 2
L +2
2
1 m 2
L
12 2
J
22
410
© 2010. Cengage Learning, Engineering. All Rights Reserved.
=
d!
!
d
) ! d! =
Initial condition: ! = 0 when
= 0:
r
)!=
d
=
1 2
!
2
=
18 g
cos d
17 L
18 g
sin + C
17 L
)C=0
36 g
sin
17 L
J
17.80
An
A
3 ft
At
t
Bt
n
240 lb.ft B
FBD
Bn
mat
man
60 lb
At 1.0 ft 1.0 ft
θ A
C
Cn
An
ma t
= A
θ
C
man
MAD
FBD
60
(3) = 5:590 lb
32:2
= mL! 2 = 5:590! 2 lb
= mL =
(a)
Bar AB:
Bar AC
80 60 cos
MB = 0
:
=
+
3At
240 = 0
Ft = mat
+5:590
= 14:311
At = 80 lb
At 60 cos = 5:590
2
10:733 cos rad/s J
(b)
=
)
d!
!
! d! = d = (14:311 10:733 cos )d
d
1 2
! = 14:311
10:733 sin + C
2
Initial condition: ! = 0 when = 0. ) C = 0
p
)!=
28:62
21:47 sin rad/s J
23
411
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.81
24
412
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.82
17.83
25
413
© 2010. Cengage Learning, Engineering. All Rights Reserved.
*17.84
26
414
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.85
(a) Kinematics of bar AB:
aA
L/2
A
L
+ 2α
ω, α
Lω2
2
G
B
a = aA + aG=A
Kinetics:
m2g
Ax
FBD's
Collar:
Bar:
m2aA
=
MAD's
Ay
A
G θ
m1g
Fx = max
Ax
m Lα
= 12
+ !
L/2
L/2
B
P0
N Ay
B
P0
A
m1 Lω2
2
m1aA
I-α
Ax = m2 aA
Ax = P0
L
Ax = m1 aA + m1 ! 2 sin
2
L 2
L
) P0 m2 aA = m1 aA + m1 ! sin
m1
cos
2
2
1
m1 L(! 2 sin + cos )
P0
2
) aA =
m1 + m2
Fx = max
+ !
m2 aA
m1
L
cos
2
27
415
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Bar: ( MA )FBD
L
m1 g sin
2
L
m1 g sin
2
g sin
=
=
=
1
m1 L2
3
=
( MA )M AD
L
= I + m1
2
+
L
2
(m1 aA )
1
m1 L(! 2 sin +
2
m1 + m2
P0
m1
L
cos
2
cos ) L
cos
2
1
P0
m1 L(! 2 sin + cos )
2
2
L
cos
3
m1 + m2
2P0 cos
2g(m1 + m2 ) sin
L! 2 m1 cos sin
(4L=3) (m1 + m2 ) + m1 L cos2
Q.E.D.
(b) Substituting the given data, the equation of motion becomes
=
=
2(12) cos
24 cos
2(9:81)(3:6 + 2:0) sin
0:8! 2 (3:6) cos sin
(4 0:8=3)(3:6 + 2:0) + 3:6(0:8) cos2
109:87 sin
2:88! 2 sin cos
5:973 + 2:88 cos2
The initial conditions are = ! = 0 at t = 0. Letting x1 = and x2 = !, the
equivalent …rst-order equations and the initial conditions are
x_ 1
x1 (0)
= x2
x_ 2 =
24 cos x1
109:87 sin x1 2:88x22 sin x1 cos x1
5:973 + 2:88 cos2 (x1 )
= x2 (0) = 0
The corresponding MATLAB program is:
function problem17_85
[t,x] = ode45(@f,[0:0.02:2],[0,0]);
printSol(t,x*180/pi)
axes(’fontsize’,14)
plot(t,x(:,1)*180/pi,’linewidth’,1.5)
xlabel(’t (s)’); ylabel(’theta (deg)’)
grid on
function dxdt = f(t,x)
s = sin(x(1)); c = cos(x(1));
num = 24*c - 109.87*s - 2.88*x(2)^2*s*c;
den = 5.973 + 2.88*c^2;
dxdt = [x(2); num/den];
end
end
Below is a partial printout spanning the location where
that is in degrees and ! is in deg/s
is maximized. Note
28
416
© 2010. Cengage Learning, Engineering. All Rights Reserved.
t
8.6000e-001
8.8000e-001
x1
x2
2.4173e+001 1.8418e+000
2.4179e+001 -1.3238e+000
By inspection we see that
max
= 24:2 J
(c)
25
theta (deg)
20
15
10
5
0
0
0.5
1
t (s)
1.5
2
17.86
(a) In Prob. 17.85 the di¤erential equation of motion was
=
Setting sin =
=
:
24
0:325 313 ! 2
24 cos
109:87 sin
2:88! 2 sin cos
5:973 + 2:88 cos2
and cos = 1, we obtain the "linearized" form
109:87
2:88! 2
= 2:711
5:973 + 2:88
(0:3253! 2 + 12:411)
12: 410 5 + 2: 710 95
(b) Using x1 = and x2 = !, the equivalent …rst-order equations and the
initial conditions are
x_ 1 = x2
x_ 2 = 2:711 (0:3253x22 + 12:411)x1
x1 (0) = x2 (0) = 0
The corresponding MATLAB program is:
29
417
© 2010. Cengage Learning, Engineering. All Rights Reserved.
function problem17_86
[t,x] = ode45(@f,[0:0.02:2],[0,0]);
printSol(t,x*180/pi)
axes(’fontsize’,14)
plot(t,x(:,1)*180/pi,’linewidth’,1.5)
xlabel(’t (s)’); ylabel(’theta (deg)’)
grid on
function dxdt = f(t,x)
dxdt = [x(2)
2.711 - (0.3253*x(2)^2 + 12.411)*x(1)];
end
end
Below is a partial printout spanning the location where
degrees and ! is in deg/s)
t
8.8000e-001
9.0000e-001
is maximized ( is in
x1
x2
2.4771e+001 1.5291e+000
2.4771e+001 -1.5149e+000
By inspection, we have max = 24:8 J
The error caused by linearization is
% error =
24:2 24:8
24:6
100% =
2:4%
(c)
25
theta (deg)
20
15
10
5
0
0
0.5
1
t (s)
1.5
2
30
418
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.87
(a) Geometry:
B
φ
A
L sin
L
R
θ C
) sin
= R sin
tan
p
1
=
Kinetics:
sin
sin2
=q
P tanφ
θ
R
sin
L
sin
(L=R)2
sin2
Iα
P B
φ
=
R
Cx
=
C
C
Cy
MAD
FBD
( MC )FBD = ( MC )M AD
+
P (R sin ) + (P tan )(R cos ) = I
0
1
sin
A (R cos ) = W k 2
(P0 sin ) (R sin ) + @P0 q
g
2
2
(L=R)
sin
=
0
gP0 R sin @
cos
sin + q
W k2
(L=R)2
sin2
1
A Q.E.D.
(b) Substituting the given data, the equation of motion becomes
0
1
32:2(24)(0:75) sin @
cos
A
=
sin + q
180(0:6)2
(1:5=0:75)2 sin2
!
cos
= 8:944 sin
sin + p
4 sin2
The initial conditions are = =2 rad and ! = 0 at t = 0. Letting x1 = and
x2 = !, the equivalent …rst-order equations and the initial conditions are
!
cos x1
x_ 1 = x2
x_ 2 = 8:944 sin x1 sin x1 + p
4 sin2 x1
x1 (0)
=
2
rad
x2 (0) = 0
31
419
© 2010. Cengage Learning, Engineering. All Rights Reserved.
The corresponding MATLAB program is
function problem17_87
[t,x] = ode45(@f,[0:0.02:2.4],[pi/2,0]);
printSol(t,x)
axes(’fontsize’,14)
plot(t,x(:,2),’linewidth’,1.5)
xlabel(’t (s)’); ylabel(’omega (rad/s)’)
grid on
function dxdt = f(t,x)
s = sin(x(1)); c = cos(x(1));
dxdt = [x(2)
8.944*s*(s + c/sqrt(4-s^2))];
end
end
The two lines of output that span ! = 10 rad/s are:
t
2.2400e+000
2.2600e+000
x1
1.3274e+001
1.3473e+001
x2
9.8875e+000
1.0031e+001
Using linear interpolation to …nd t when ! = 10 rad/s:
2:26
10:031
2:24
t
=
9:8875
10
2:24
9:8975
t = 2:25 s J
(c)
12
omega (rad/s)
10
8
6
4
2
0
0
0.5
1
1.5
2
2.5
t (s)
32
420
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.88
(a) Kinematics (note that AB has two DOF: angle
L/2
A
β +
aA
θ
G
Lα
2
and displacement of A):
A
ω, α
L ω2
2
a = aA + aG=A
Kinetics (note that
FBD
=
):
A
y
B
=
B
Fy
= may
+.
aA
= g sin
L
( sin
2
+
( MA )FBD
L
mg cos
2
L
mg cos
2
1
g cos
2
1
g cos
2
)
=
=
=
=
=
=
MAD I-α
N
L/2
θ β x
mg
G
mg sin
φ
θ
y
A
m L ω2
φ 2
maA m Lα x
2
L
= maA + m ( sin
2
! 2 cos )
! 2 cos )
( MA )M AD
L
L
L
I + m
+ (maA ) sin
2
2
2
L
1
L
L
mL2 + m
+ (maA ) sin
12
2
2
2
1
1
L + aA sin
3
2
1
1
L
L + sin g sin
( sin
! 2 cos )
3
2
2
(2g=L)(cos
sin sin )
4=3 sin2
! 2 sin cos
Q.E.D.
(b) Using the given data, we have
2g
L
=
)
2(32:2)
= 8:050
= 60 = rad
8
3
8:050(cos
sin sin ) ! 2 sin cos
a=
4=3 sin2
=
=
3
33
421
© 2010. Cengage Learning, Engineering. All Rights Reserved.
The initial conditions are = ! = 0 when t = 0. Letting x1 = and x2 = ! 2 ,
the equivalent …rst-order equations and the initial conditions are:
x_ 1
x_ 2
x1 (0)
= x2
0:8050 [cos x1
=
sin sin(
4=3
x22 sin(
x2 )
x1 )]
sin2 (
x1 ) cos(
x1 )
= x2 (0) = 0
The MATLAB program is:
function problem17_88
[t,x] = ode45(@f,[0:0.01:1.0],[0,0]);
printSol(t,x)
function dxdt = f(t,x)
s = sin(pi/3-x(1)); c = cos(pi/3-x(1));
num = 8.050*(cos(x(1))-sin(pi/3)*s)- x(2)^2*c*s;
den =4/3-s^2;
dxdt = [x(2); num/den];
end
end
The two lines of output spanning
t
8.4000e-001
8.5000e-001
x1
1.0249e+000
1.0477e+000
We compute ! at
1:0477
2:2872
= =3 = 1:0472 rad are:
x2
2.2575e+000
2.2872e+000
= =3 by linear interpolation:
1:0249
=3 1:0249
=
2:2575
! 2:2575
! = 2:29 rad/s J
17.89
(a)
..
m1 Lθ
B
2
r
- ..
L/2 G N
L/2 L .I21θ
mg
m1 θ
A θ Ax 1
2
.. . .
= A
m2(rθ + 2rθ)
θ m2g
Ay
.. .
m2(r − rθ 2)
MAD's
FBD's
N
B
+
Bar: ( MA )FBD
L
m1 g cos + N r
2
L
m1 g cos + N r
2
=
=
=
( MA )M AD
L
L
m1 •
2
2
1
1
m1 L2 •
m1 L2 • =
12
4
I1 •
1
m1 L2 •
3
(a)
34
422
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Collar:
F = ma
+- N
•
= m2 (g cos + r + 2r_ _ )
N
m2 g cos = m2 (r• + 2r_ _ )
(b)
Substituting (b) into (a):
L
cos + m2 (g cos + r• + 2r_ _ )r
2
3 m1
gL cos + 2r(g cos + 2r_ _ )
2 m2
3
m1
g cos
L + 2r + 4rr_ _
2
m2
m1 g
3
2
)•=
Collar:
g cos
=
=
m1
L + 2r + 4rr_ _
m2
Q.E.D.
m1 2
L + 3r2
m2
Fr = mar
r• =
1
m1 L2 •
3
m1 2
L + 3r2 •
m2
m1 2
L + 3r2 •
m2
=
2
+%
m2 g sin = m2 (•
r
2
r_ )
g sin + r _ Q.E.D.
(b) Substituting the given data, we get
• =
3 32:2 cos (2(1:5) + 2r) + 4rr_ _
=
2
2(1:5)2 + 3r2
r• =
32:2 sin + r _
The initial conditions are:
_ r r_
Letting x =
are
32:2 cos (3 + 2r) + 4rr_ _
3 + 2r2
2
T
= =3, _ = 0, r = 1:5 ft and r_ = 0 when t = 0
, the …rst-order equations and the initial conditions
x_ 1
= x2
x_ 2 =
32:2 cos x1 (3 + 2x3 ) + 4x3 x4 x2
3 + 2x23
x_ 3
= x4
x_ 4 =
32:2 sin x1 + x3 x22
x(0)
=
=3 rad
0
1:5 ft
0
T
The MATLAB program that produced the plot is
function problem17_89
[t,x] = ode45(@f,[0:0.01:0.5],[pi/3,0,1.5,0]);
axes(’fontsize’,14)
plot(x(:,1),x(:,3),’linewidth’,1.5)
xlabel(’theta (rad)’); ylabel(’r (ft)’)
grid on
function dxdt = f(t,x)
35
423
© 2010. Cengage Learning, Engineering. All Rights Reserved.
num = 32.2*cos(x(1))*(3 + 2*x(3)) + 4*x(3)*x(4)*x(2);
den =3 + 2*x(3)^2;
dxdt = [x(2)
-num/den
x(4)
-32.2*sin(x(1)) + x(3)*x(2)^2];
end
end
1.5
r (ft)
1.25
1
0.75
0.5
-2
-1
0
theta (rad)
1
2
17.90
(a) Kinematics:
aθ
Top view
..
y
r
ω, α
ar
D
C + C
aD = aC + aD=C
36
424
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Kinetics:
B
D
L/2 C Cx N
r
mAB y..
..
Iθ
C
Cy
A
θ
A
maθ
D
mar
D
N
FBD's
A
θ MAD's
y = a sin pt
) y• =
ap2 sin pt
:
( MC )FBD = ( MC )M AD
1 •
I
) N=
r
Collar D
N
..
my
=
k(r − r0)
Rod AB
=
B
+
N r = I•
(a)
:
F = ma
+ - N = ma + m•
y cos
2
•
_
= m(r + 2r_ ) + m( ap sin pt) cos
(b)
Equating (a) and (b), we obtain
1 •
I
= m(r• + 2r_ _ ) + m( ap2 sin pt) cos
r
1
I + mr2 • = m 2r_ _ ap2 sin pt cos
r
r
• =
ap2 sin pt cos
2r_ _ Q.E.D.
I=m + r2
Rod AB
k(r
:
Fr = mar
2
+%
k(r
r0 ) = mar + m•
y sin
= m(•
r r _ ) + m( ap2 sin pt) sin
2
k
r• =
(r r0 ) + r _ + ap2 sin pt sin Q.E.D
m
r0 )
(b) Substituting the given data, we get
• =
r
(312:5
10
r sin 10t cos
=
r• =
=
6 ) =0:125
+ r2
2r_ _
h
0:01(10)2 sin 10t cos
2r_ _
i
0:0025 + r2
2
3:125
(r 0:05) + r _ + 0:01(10)2 sin 10t sin
0:125
2
25(r 0:05) + r _ + sin 10t sin
37
425
© 2010. Cengage Learning, Engineering. All Rights Reserved.
The initial conditions are: = _ = 0, r = 0:05 m and r_ = 0 when t = 0 Letting
_ r r_ T , the …rst-order equations and the initial conditions are
x=
x_ 1
= x2
x_ 2 =
x_ 3
= x4
x_ 4 =
=
x(0)
0
0
x3 (sin 10t cos x1 2x4 x2 )
0:0025 + x23
0:05) + x3 x22 + sin 10t sin x1
25(x3
0:05 m
0
T
The following MATLAB program was used to produce the plot:
function problem17_90
[t,x] = ode45(@f,[0:0.01:3],[0,0,0.05,0]);
axes(’fontsize’,14)
plot(t,x(:,1),’linewidth’,1.5)
xlabel(’time (s)’); ylabel(’theta (rad)’)
grid on
function dxdt = f(t,x)
num = x(3)*(sin(10*t)*cos(x(1)) - 2*x(4)*x(2));
den = 0.0025 + x(3)^2;
dxdt = [x(2)
num/den
x(4)
-25*(x(3) - 0.05) + x(3)*x(2)^2 + sin(10*t)*sin(x(1))];
end
end
2
theta (rad)
1.5
1
0.5
0
0
Since
0.5
1
1.5
time (s)
2
2.5
3
is positive, the rotation is counter-clockwise (as viewed from above)
38
426
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.91
17.92
39
427
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.93
y
30o
x
G
A =
0.4
0.2N N
0.4
0.4
P
mg
FBD
ma
G
MAD
Assume impending tipping
Fy
N
= 0
+ - N mg cos 30 = 0
= mg cos 30 = 50(9:81) cos 30 = 424:8 N
( MG )FBD = ( MG )M AD
+
0:4N 0:4(0:2N ) 0:4P = 0
P = 0:8N = 0:8(424:8) = 340 N J
40
428
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.94
17.95
Assume impending loss of contact at B
W + 24 lb
0.3N
FBD's
N
+
:
A
Ax
=
1.5
'
Bar AB
Ay
1.5
'
1.5
'
A
B
W + 24 a
g
30 lb =
W
MAD's
Wa
g
60o
o
60
( MA )FBD = ( MA )M AD
W
W (1:5 cos 60 ) =
a(1:5 sin 60 )
g
a = 0:5774g
41
429
© 2010. Cengage Learning, Engineering. All Rights Reserved.
System:
Fy = 0
+"
= 0
N = W + 24 lb
W + 24
+ ! 30 0:3N =
a
g
W + 24
(0:5774g)
30 0:3(W + 24) =
g
W = 10:19 lb J
Fx = max
N
(W + 24)
17.96
D
T
G
A
B FBD
y
L/2
L/2 o
45 mg
1 2
aG = 45o B + G L/2 B
D
12mL α
α
L
aB
α
L/2
L/2
2
A
aG = aB + aG/B
maB o G
45 1mLα
2
( MD )FBD = ( MD )M AD
+
mg
L
4
1
1
mL2 + mL
12
2
6g
5L
=
=
Fx
T + mg sin 45
= max
=
1
mL
2
+&
6g
5L
x
B MAD
T + mg sin 45 =
L
4
1
mL sin 45
2
T = 0:283mg J
sin 45
17.97
0.24 m
45o
B
=
( MA )FBD = ( MA )FBD
45o
_
ma
45o
G
m
B
MAD
A
Ax
ma = mCG! 2 = 6
C
0.2
2 4
A
0.2
4
FBD
Ay
2
ω
T
m
0:24
p
2
+
(10)2 = 101:82 N
0:24T
=
T
=
0:24
p
2
72:0 N J
101:82
42
430
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.98
y
279.5α
1800 lb
o
62.11aA MAD
419.3α
40
A
B = x
A 4.5 ft
4.5 ft B
4.5 ft
4.5 ft
0.8 N
N
2000 lb
FBD
I
=
max
=
1
1 2000 2
mL2 =
9 = 419:3 slug ft2
12
12 32:2
L
2000
2000
aA = 62:11aA
may = m
=
32:2
2
32:2
9
2
= 279:5
( MA )F BD = ( MA )M AD
(1800 sin 40 )(9) 2000(4:5) = 419:3 + 279:5 (4:5)
=
Fy = may
Fx = max
+"
N
N
2000 + 1800 sin 40
2000 + 1800 sin 40
N
+ ! 0:8N
0:8(1078:5)
J
2
0:8427 rad/s
1800 cos 40
1800 cos 40
= 279:5
= 279:5(0:8427)
= 1078:5 lb
= 62:11aA
= 62:11aA
aA
=
aA
=
8:309 ft/s
8:31 ft/s
2
2
J
17.99
43
431
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.100
44
432
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.101
17.102
Ax
I
mat
=
40o
A
5 ft
5 ft
G
40o 100 lb
A
B
x
5 ft
FBD
y
5 ft
B
25.88α MAD
15.528 α
15.528ω2
50 lb
1
1 100
mL2 =
(10)2 = 25:88 slug ft
12
12 32:2
L
100 10
L
=
= 15:528
man = m ! 2 = 15:528! 2
= m
2
32:2 2
2
=
( MA )FBD = ( MA )M AD
+
100(5 sin 40 )
=
25:88 + 15:528 (5)
=
3:105 rad/s
2
Fy = may
+ " 50 100 = 15:528 sin 40
15:528! 2 cos 40
50 = 15:528(3:105 sin 40 + ! 2 cos 40 )
! = 1:264 rad/s J
45
433
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.103
17.104
G
100 lb y
300 lb
2 ft
1.0 ft
D
T
FBD
N
0.4 N
The spool "rolls" on point D:
I = mk 2 =
x
=
MAD
) a = DG
300
(2)2 = 37:27 slug ft2
32:2
Fy = 0
+"
G
37.27α
α
18.634
D
N
= 2 ft/s2
ma =
300 = 0
300
(2 ) = 18:634 lb
32:2
N = 300 lb
46
434
© 2010. Cengage Learning, Engineering. All Rights Reserved.
( MD )FBD = ( MD )M AD
100(5) 0:4N (1:0) = 37:27 + 18:634 (2)
+
500
120
=
)a =
= 5:098 rad/s
2(5:098) = 10:20 ft/s
Fx = ma
+ !
T + 0:4(300) = 18:634(5:098)
100
2
74:54
! J
2
100 T + 0:4N = 18:634
T = 125:0 lb J
17.105
n
A An
R
FBD
At
=
θ mg
A
mRω2
MAD
t mRα
mR2α
(a)
( MA )FBD
=
( MA )M AD
g
sin J
=
2R
+
mg(R sin ) =
mR2
mR (R)
(b)
=
)
d!
!
) ! d! = d =
d
1 2
g
! =
cos + C
2
2R
g
sin d
2R
= =2. ) C = 0
r
g
cos J
)!=
R
Initial condition: ! = 0 when
(c)
Fn
An
mg cos
Ft
At + mg sin
A=
s
&
=
2
(2mg cos ) +
= man %
An
g
= mR
cos
R
At + mg sin =
g
sin
mR
2R
1
mg sin
2
2
= mg
mg cos = mR! 2
An = 2mg cos
mR
At =
r
1
mg sin
2
4 cos2 +
1
sin2
4
J
47
435
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.106
B y
3m
20(9.81) N
P
20a
2.25 m
=
1.5 m
A
3m
FBD
NA
A
MAD
= max
+ ! P = 20a
) a = 0:05P
= ( MA )M AD
= 20a(2:25)
= 20(0:05P )(2:25)
P = 785 N J
Fx
( MA )FBD
20(9:81)(3) + 1:5P
20(9:81)(3) + 1:5P
+
B
x
17.107
Kinematics:
_
a
G
3
4
A
=
aA
3
4
G
+ 5α
3
4
5'
A
α
a = aA + aG=A
Kinetics:
y
y
x
G 5'
mg
FBD A
x
Iα
4 5
3
µkΝ
=
m(aΑ + 5α)
G 5'
MAD A
N
400
W
=
= 12:422 slugs
g
32:2
1
1
I =
m(b2 + h2 ) =
(12:422)(62 + 82 ) = 103:52 slug ft2
12
12
4
4
Fy = may
+- N
mg = 0
N = (400) = 320 lb
5
5
m =
( MG )FBD
=
( MG )M AD
0:125(320)(5) =
103:52
Fx = max
+%
0:125(320)
3
(400)
5
aA
=
=
+
kN
(5) = I
2
= 1:9320 rad/s
3
mg =
5
kN
m(aA + 5 )
12:422 [aA + 5(1:9320)]
6:44 ft/s . J
2
48
436
© 2010. Cengage Learning, Engineering. All Rights Reserved.
17.108
(a)
P
A
y
L/2
mg θ
FBD
+
P (L cos )
=
)
L/2
By
x
=
Bx
B
( MB )FBD
L
mg
cos
2
α
mL
2
ω2
mL
1 mL2α
2
12
θ
B
MAD
A
=
( MB )M AD
1
L
L
=
mL2 + m
12
2
2
3 2P
=
g cos J
2L m
3
d!
!
) ! d! = d =
d
2L
3 2P
1 2
! =
g sin + C
2
2L m
2P
m
Initial condition: ! = 0 when = 0: ) C = 0
s
3 2P
)!=
g sin
L m
g cos d
J
(b)
Fx
Bx
L
L
sin
Bx = m ! 2 cos + m
2
2
L 3 2P
L 3 2P
= m
g sin cos + m
g cos
2 L m
2 2L m
9
9
=
(2P mg) sin cos = (2P mg) sin 2 ! J
4
8
= max
+ !
sin
17.109
R
FBD O
x
O
mg
=
1 mR2α
2
O
MAD
µkmg
N = mg
Final angular speed of disk: ! 0 =
v
5
=
= 20 rad/s
R
0:25
49
437
© 2010. Cengage Learning, Engineering. All Rights Reserved.
( MO )FBD
=
)
( MO )M AD
=2
+
k mg(R)
=
1
mR2
2
0:25(9:81)
2
kg
=2
= 19:62 rad/s
R
0:25
d!
) d! = dt = 19:62dt
dt
Initial condition: ! = 0 when t = 0. ) C = 0
=
! = ! 0 when 19:62t = 20
) ! = 19:62t + C
) t = 1:019 s J
50
438
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Chapter 18
18.1
18.2
18.3
18.4
B
3
12 ft
lb
t
3 f lb
12
50 lb.ft
A
k = 15 lb/ft
L0 = 3 ft
θ
C
1
439
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Work of gravity:
(U1
2 )g
=
=
2W h = 2 [12(1:5 sin 45 )]
25:46 lb ft
Work of couple: (U1
2 )c
= C
Work of spring: (U1
2 )e
=
=
2
=
4
= 39:27 lb ft
1
k (L2 L0 )2
2
h
1
(15) (6 cos 45
2
55:92 lb ft
=
Total work: U1
= 50
L0 )2
(L1
2
3)
3)2
(6
i
25:46 + 39:27 + 55:92 = 69:7 lb ft J
18.5
1
300 mm
G
30 N.m
200 mm
2
A
300 mm
Work of gravity: (U1
2 )g
=
Work of couple: (U1
2 )c
= C
Work of spring: (U1
2 )e
=
=
=
Total work: U1
2
mg
h=
= 30
8(9:81)( 0:25) = 19:620 J
2
= 47:12 J
1
2
k( 22
1)
2
p
1
0:52 + 0:22
(360)
2
10:240 J
= 19:620 + 47:12
2
0:3
0
10:240 = 56:5 J J
2
440
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.6
6 rad/s
250 mm
6 rad/s
200 N
B
250 mm
150 mm
150 mm
A
A
200 N
(a)
Both cases: P
Case (a): P
Case (b): P
B
(b)
= F vB = F (AB !) = 200(6)AB = 1200 AB
= 1200(0:4) = 480 W J
= 1200(0:1) = 120 W J
18.7
De…nition of power:
De…nition of e¢ ciency:
(a) C
=
(b) C
=
Pout = C!
Pout = Pin
)C=
Pin
!
0:78(12 103 )
= 49:7 N m J
1800(2 =60)
0:78(12 103 )
= 24:8 N m J
3600(2 =60)
18.8
3
441
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.9
18.10
18.11
4
442
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.12
18.13
5
443
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.14
18.15
I
IC
T
= mk 2 = 60(0:24)2 = 3:456 kg m2
2
= I + m CG = 3:456 + 60(0:252 + 0:122 ) = 8:070 kg m2
1
1
=
IC ! 2 = (8:070)(2)2 = 16:14 J J
2
2
6
444
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.16
18.17
0.3 rad/s C
d
6 ft
G
4 ft
30o
1.8 ft/s
A
Point is the I.C. of the bar. From geometry:
I
IC
T
d2 = (4 sin 30 )2 + (6
4 cos 30 )2 = 10:431 ft2
1
1
mL2 =
12
12
(8)2 = 3:313 slug ft2
=
20
32:2
= I + md2 = 3:313 +
=
20
(10:431) = 9:792 slug ft2
32:2
1
1
IC ! 2 = (9:792)(0:3)2 = 0:441 lb ft J
2
2
18.18
4 rad/s
A
2.4 f
t
10 lb
3.2 ft/s
0.8 ft
3.2 ft/s
C
B 2 lb
15 lb
7
445
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Bar AB is translating with the velocity v = R! = 0:8(4) = 3:2 ft/s
Disk: IC =
T
=
=
1
1
mR2 =
2
2
15
32:2
(0:8)2 = 0:14907 slug ft2
1
1
1
IC ! 2 + mAB v 2 + mB v 2
2
2
2
10
2
1
2
0:14907(4) +
(3:2)2 +
(3:2)2
2
32:2
32:2
= 3:10 lb ft J
18.19
2 m/s
1.2 kg
80 mm
B
160
mm
2.4
kg
2 m/s
A
25 rad/s
C
Bar BC is translating with the velocity v = 2 m/s
! AB
TAB
TBC
T
v
2
= 25 rad/s
=
0:08
AB
1 1
1
1
2
=
IA ! 2AB =
mAB AB ! 2AB = (1:2)(0:08)2 (25)2 = 0:80 J
2
2 3
6
1
1
=
mBC v 2 = (2:4)(2)2 = 4:8 J
2
2
= 0:8 + 4:8 = 5:6 J J
=
18.20
5
= 0:155 28 slugs
32:2
= mR2 = 0:155 28(1:5)2 = 0:3494 slug
1
1
1
mv 2 + I! 2 = (0:155 28)(30)2 +
=
2
2
2
m =
I
T
ft2
1
(0:3494)(102 ) = 87:3 lb ft J
2
8
446
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.21
B
vB
3m
2 m 30o vC
C
7.5 rad/s A
D
30o
ωBC
y
x
E
Point E is the I.C. of bar BC
vB
! BC
vC
vBC
=
=
TBC
=
=
= ! AB AB = 7:5(2) = 15 m/s
vB
15
=
= 2:50 rad/s
=
3 csc 30
EB
= ! BC EC = 2:50(3 cot 30 ) = 12:990 m/s
1
1
(vB + vC ) = [ 15i + 12:990( i cos 30 + j sin 30 )]
2
2
13:125i + 3:248j m/s
1
1
2
IBC ! 2BC + mBC vBC
2
2
1 (12)(3)2
1
(2:5)2 + (12)(13:1252 + 3:2482 ) = 1125 J J
2 12
2
18.22
m
1.0
3 m/s
1.0
m
ωC
60o
30o
A 1.0 m G 1.0 m B
vB
Point C is the I.C. of the bar.
!
=
IC
=
T
=
vA
3
= 3 rad/s
=
1:0
AC
1
1
mL2 + md2 =
80(2)2 + 80(1:0)2 = 106:67 kg m2
12
12
1
1
IC ! 2 = (106:67)(3)2 = 480 J J
2
2
9
447
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.23
18.24
10
448
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.25
Choose the horizontal plane through point A as the datum for gravitational
potential energy
) V ( ) = mg
L
cos
2
T =
1
1
IA ! 2 =
2
2
1
mL2 ! 2
3
1
1
mgL cos + mL2 ! 2 = C
2
6
1
Initial condition: ! = 0 when = 0: ) C = mgL
2
r
1
g
1
1
2 2
) mgL cos + mL ! = mgL
! = 3 (1 cos ) J
2
6
2
L
V + T = C (constant)
18.26
ω
m = 4 kg
k- = 0.18 m
6 kg A
0.25 m
0.15 m
C
vA
vB
B
12 kg
11
449
© 2010. Cengage Learning, Engineering. All Rights Reserved.
The displacements of the blocks during a 90 clockwise turn of the pulley are
U1
T2
2
= RA
= 0:25
yB
= RB
= 0:15
2
= 0:3927 m "
2
= 0:2356 m #
=
mA g yA + mB g yB
= [ 6(0:3927) + 12(0:2356)] 9:81 = 4:621 J
1
1
1
2
2
+ mB vB
+ mC k 2 ! 2
mA vA
=
2
2
2
1
2
=
6(0:25!) + 12(0:15!)2 + 4(0:18)2 ! 2 = 0:3873! 2
2
2
U1
yA
= T2
4:621 = 0:3873! 2
T1
! = 3:45 rad/s J
0
18.27
A
D
5 ft
Datum
C
5 ft
A
B
B
5 ft
5 ft
D
C
2 ft
E
5 ft
Position 2
7 ft
E
Position 1
IB
13 ft
5 ft
5 ft
=
"
1
1
2
2
2
mAC AC + mCD CD + mCD CB +
12
12
=
1
12
=
77:64 slug ft2
100
32:2
(10)2 +
1
12
50
32:2
(5)2 +
= WCD
V2
=
T2
=
T 1 + V1
!
2
#
50 2
(5 + 2:52 )
32:2
CD
= 50(2:5) = 125 lb ft
2
1
1
WCD BC + k 2 = 50(5) + (12)(13
2
2
1
1
2
2
2
IB ! = (77:64)! = 38:82!
2
2
V1
CD
2
7)2 =
= T 2 + V2
0 + 125 = 38:82! 2
= 2:02 rad/s J
34:0 lb ft
34:0
12
450
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.28
18.29
M (lb.ft)
0.88
0.50
0
0
π
θ (rad)
Open
θ
M
Closed
Mass moment of inertia of the jaw about axis AD:
I
= IAB + IBC + IDC = 2IAB + IBC = 2
=
2 2:22 10
3
32:2
=
1:7236
U1
U1
10
3
5
(3)
3
12
2
+
(2:22
=
area under M - diagram =
T2
=
1 2
1
I! = (1:7236
2
2
= T2
!
=
10
32:2
3
)(2)
3
12
+ mBC L2AB
2
slug ft2
2
2
1
mAB L2AB
3
T1
501:6 rad/s
2:168 =
10
5
0:88 + 0:50
2
= 2:168 lb ft
)! 2
1
(1:7236
2
10
) vBC = LAB ! =
5
)! 2
0
3
(501:6) = 125:4 ft/s J
12
13
451
© 2010. Cengage Learning, Engineering. All Rights Reserved.
*18.30
18.31
14
452
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.32
15
453
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.33
18.34
16
454
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.35
M = 0.15P
G
P
C
0.4 m
Replace P by the equivalent force-couple system shown.
Displacement of G:
U1
2
=
T2
=
s=R
= 0:4(2 ) = 0:8 m
M
+ P s = 0:15P (2 ) + P (0:8 ) = 0:5 P N m
1
1
1
IC ! 2 =
I + mR2 ! 2 =
1:6 + 40(0:4)2 (6)2 = 144:0 N m
2
2
2
U1
2
= T2
T1
0:5 P = 144:0
0
P = 91:7 N J
18.36
17
455
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.37
18.38
(a)
A
0.6
m
30o
Datum
0.15 m
B
vA
A
0.6 m
B
ωAB
Position 2
Position 1
In Position 2 point B is stationary. ) Point B is the I.C. of barAB
V1
T2
= mAB g(0:3 sin 30 ) = 4:5(9:81)(0:3 sin 30 ) = 6:622 J
1 1
1
1
=
(IAB )B ! 2AB =
mAB L2AB ! 2AB = (4:5)(0:6)2 ! 2AB = 0:270! 2AB
2
2 3
6
T1 + V1
vA
= T 2 + V2
0 + 6:622 = 0:270! 2AB + 0
= LAB ! AB = 0:6(4:952) = 2:97 m/s # J
! AB = 4:952 rad/s
18
456
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(b)
Datum
δ
B
0.6 m
A
Position 3
V3 =
mAB g
1
+ k
2
2
T 1 + V1
900
2
22:07
6:622
2
=
1
4:5(9:81) + (1800)
2 2
= T3 + V3
=
0
2
0 + 6:622 = 0 + 900
The positive root is
2
= 900
2
22:07
22:07
= 0:0989 m J
18.39
3 rad/s
G 2.7 ft/s
0.25'
0.65'
L0
C
Position 1
I
v1
V1
T1
V2
T2
Datum
ω2
0.25'
0.4ω2
0.4' G
C
L0 + 0.65π
Position 2
16
(0:25)2 = 0:198 94 slug ft2
32:2
v2 = CG ! 2 = 0:4! 2
= CG ! 1 = 0:9(3) = 2:7 ft/s
= IO
md2 = 0:23
= mge = 16(0:25) = 4:0 lb ft
1
1 2 1
1 16
=
I! 1 + mv12 = (0:198 94)(3)2 +
(2:7)2 = 2:706 lb ft
2
2
2
2 32:2
1
1
=
mge + k 2 = 4:0 + (5)(0:65 )2 = 6:425 lb ft
2
2
1 16
1 2 1
1
=
I! 2 + mv2 = (0:198 94)! 22 +
(0:4! 2 )2 = 0:139 22! 22
2
2
2
2 32:2
T1 + V1
!2
= T 2 + V2
2:706 + 4:0 = 0:139 22! 22 + 6:425
= 1:421 rad/s J
19
457
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.40
B
L/2
B
L/2
Position 1
L/2
30o
ω2
A
A
C
P
L/2
C
Position 2
P
In Position2, point A is the I.C. of the bar
U1
U1
2
2
= P (L sin 30 )
= T2
T1
T2 =
1
1
IA ! 22 =
2
2
1
P L sin 30 = mL2 ! 22
6
1
mL2 ! 22
3
0
!2 =
r
3P
J
mL
18.41
20
458
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.42
18.43
1
459
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.44
18.45
2
460
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.46
18.47
3
461
© 2010. Cengage Learning, Engineering. All Rights Reserved.
4
462
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.48
18.49
5
463
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.50
60 lb
A v2
60 lb
C
240 lb.ft
60o 240 lb.ft
Position 1
D
B
C
3 ft
3 ft
3 ft
3 ft
A
Position 2
D
B
Bar AC undergoes curvilinear translation
U1
= C
2
T2
h = 240
30
180
60(3
3 sin 60 ) = 101:55 lb ft
1
1 60 2
mv22 =
v
2
2 32:2 2
=
U1
mg
2
= T2
T1
101:55 =
1 60 2
v +0
2 32:2 2
v2 = 10:44 ft/s J
18.51
30o
FBD
80(9.81) N
µkN =
T
Fy
kN
U1
x
y
MAD
80a
N
= 0
+ - N 80(9:81) cos 30 = 0
= 0:4(679:7) = 271:9 N
2
=
2mgy
T2
=
2
=
kN x
= 2(80)(9:81)(x sin 30 )
N = 679:7 N
271:9x = 512:9x
1 1
1
1
2
mv 2 + IA ! 2A = mv22 +
mRA
2 2
2
2 2
1
1
1+
mv22 = 1 +
80(5)2 = 2500 J
4
4
v2
RA
2
6
464
© 2010. Cengage Learning, Engineering. All Rights Reserved.
U1
2
= T2
T1
512:9x = 2500
0
x = 4:87 m J
18.52
C
A
0.9 m
Position 2
Position 1
1.5
m
C
0.24 m
P ωC
1.2 m B
Datum
A
1.5 m ωAB B
3 m/s
P
In Position 2, point B is the I.C. of bar AB
IC
=
(IAB )B
=
! AB
=
V1
1
1
2
mC RC
= (1:8)(0:24)2 = 0:05184 kg m2
2
2
1
1
2
mAB LAB = (2:7)(1:5)2 = 2:025 kg m2
3
3
vA
3
3
vA
= 2 rad/s
!C =
= 12:5 rad/s
=
=
LAB
1:5
RC
0:24
1
= mC g(0:9) + mAB g(0:45) + k
2
2
1
1
1:8(9:81)(0:9) + 2:7(9:81)(0:45) + (58)(1:2 0:3)2 = 51:30 J
2
1
1 2
k
P x = (58)(2:1 0:3)2 P (1:5 1:2)
=
2 2
2
= 93:96 0:3P J
1
1
1
2
=
+ (IAB )B ! 2AB
IC ! 2C + mC vA
2
2
2
1
1
1
=
(0:05184)(12:5)2 + (1:8)(3)2 + (2:025)(2)2 = 16:20 J
2
2
2
=
V2
T2
T1 + V1
P
= T 2 + V2
0 + 51:30 = 16:20 + (93:96
= 196:2 N J
0:3P )
7
465
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.53
*18.54
8
466
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.55
vC = 2vA
15"
vA
ωA
vA
!A
= vB = 0:5vC
) sA = sB = 0:5
vC
vC
=
= 0:4vC
= !B =
2R
30=12
250
32:2
IA = IB = mR2 =
15
12
sC
2
= 12:131 slug ft2
Cylinder A reaches corner of the slab when
sC
T2
sA = 20 in.
=
=
U1
2
U1
1
2
+2
mC vC
2
1 1500
2
vC
2 32:2
= P
2
sC
= T2
sC = 180
T1
0:5 sC = 20 in.
sC = 40 in.
1
1
2
IA ! 2A + mA vA
2
2
+ 12:131(0:4vC )2 +
40
12
250
2
(0:5vC )2 = 27:174vC
32:2
= 600 lb ft
2
600 = 27:174vC
0
vC = 4:70 ft/s J
18.56
9
467
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.57
18.58
10
468
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.59
mABv-AB
IABω
A 1.0'
m v- BC BC
IBCω
B
2.75'
(a)
mv-
d
C
=
B
A
C
(b)
Momentum diagram (a):
IAB !
=
IBC !
=
mAB vAB
=
mBC vBC
=
) hO
=
=
1
mAB L2AB ! =
12
1
mBC L2BC ! =
12
1 2:5
(2)2 ! = 0:02588!
12 32:2
1 5
(1:5)2 ! = 0:02912!
12 32:2
2:5
mAB (rAB !) =
(1:0!) = 0:07764!
32:2
5
mBC (rBC !) =
(2:75!) = 0:4270!
32:2
0:02588! + 0:02912! + 0:07764!(1:0) + 0:4270!(2:75)
1:3069!
Momentum diagram (b):
mv =
7:5
(!d) = 0:2329!d
32:2
) hO = 0:2329!d2
Diagrams (a) and (b) are equivalent:
(hO )a = (hO )b :
1:3069! = 0:2329!d2
d = 2:37 ft J
11
469
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.60
18.61
18.62
(AO )1
2
=
Z
60 s
C(t)dt = 12
0
=
(hO )1
(AO )1
2
=
60 s
1
2t
e
0
714:0 lb ft s
(hO )2 = mk 2 ! 2 =
0
= (hO )2
Z
(hO )1 :
40
32:2
9
12
714:0 = 0:6988! 2
1
dt = 12 t + e
2
60 s
2t
0
2
! 2 = 0:6988! 2
0
! 2 = 1022 rad/s J
18.63
T
C 0.2 m A
mAg
FBD
75 N
C 0.2 m A
mAv1
IAω1
Initial
momentum
diagram
B
B
mBg
mBv1
12
470
© 2010. Cengage Learning, Engineering. All Rights Reserved.
From FBD:
(mA + mB )g = (8 + 6)(9:81) = 137:34 N
+
MC = 137:34(0:2)
75(0:4) =
2:532 N m
Initial momentum diagram:
+
+
#
mv1 = (mA + mB )R! 1 = (8 + 6)(0:2)(3) = 8:40 N s
2
! 1 = 8(0:15)2 (3) = 0:54 N m s
IA ! 1 = mA kA
Impulse-momentum principle:
+
t =
MC t = (hC )2
0:877 s J
(hC )1 :
2:532t = 0
[8:40(0:2) + 0:54]
18.64
h1
A1
2
=
=
0
Z
0
A1
2
!1
= h2
=
h1
t
h2 = I!
Z t
M (t)dt =
M0 e
t=t0
dt = M0 t0 1
e
t=t0
0
M 0 t0 1
e
t=t0
= I!
0
!=
M 0 t0
1
I
e
t=t0
M 0 t0
4:6(3:8)
= 24:3 rad/s J
=
0:72
I
18.65
13
471
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.66
400
P
W
3P 3P 200 Ox
O
Oy
3Pµk
200
FBD's
2P
+
(AO )1
2
=
Z
12 s
(3P
k) R
dt = 3
k R(area
under P -t diagram)
0
=
3(0:3)(0:2)(6P0 ) = 1:08P0
+
(AO )1
(hO )1 =
2
= (hO )2
mk 2 ! 1 =
(hO )1
20(0:16)2 400
1:08P0 = 0
2
60
=
( 21:45)
21:45 N m s
P0 = 19:86 N J
18.67
14
472
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.68
I-ω
mg
0.3 m
FBD
O
O
µkmg
Initial
mv- momentum
diagram
mg
Let t be the time when cylinder stops
+
!
t
=
+
L1 2 = p2 p1
k mgt = 0
2
v
=
= 0:3398 s
0:6(9:81)
kg
(AO )1
k mgRt
=
!
=
2
= (hO )2
(hO )1
mv
k mgRt
1
mR2 !
2
2 k gt
2(0:6)(9:81)(0:3398)
=
= 13:33 rad/s
R
0:3
=0
I!
J
18.69
15
473
© 2010. Cengage Learning, Engineering. All Rights Reserved.
*18.70
16
474
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.71
B
A
mB Lω2
2
I-
IAω1
10" 10"
Final momenta
IA
=
1 6
1
mA R2 =
2
2 32:2
IB
=
1
1 4
mB L2 =
12
12 32:2
(hO )1
=
0:052 41(16)
=
0:8386
=
(hO )2
"
9
12
2
= 0:052 41 slug ft2
20
12
2
= 0:028 76 slug ft2
IA ! 1 = IA ! 2 + IB ! 2 + mA L2 ! 2 + mB
0:052 41 + 0:028 76 +
0:6850! 2
IAω2
ω
B 2
Initial momenta
+
mALω2
6
32:2
20
12
2
+
4
32:2
10
12
2
L
2
#
2
!2
!2
! 2 = 1:224 rad/s J
18.72
17
475
© 2010. Cengage Learning, Engineering. All Rights Reserved.
*18.73
18.74
18
476
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.75
18.76
18.77
O
0.1mω1
O
Iω1
0.1 m
Initial momenta
I=
0.3 m
0.3mω2
Iω2
Final momenta
1
1
mL2 =
(15)(0:6)2 = 0:45kg m2
12
12
+
(hO )1 = (hO )2
I! 1 + (0:1)2 m! 1 = I! 2 + (0:3)2 m! 2
0:45 + (0:1)2 (15) (12) = 0:45 + (0:3)2 (15) ! 2
! 2 = 4:0 rad/s J
19
477
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.78
Let IO = moment of inertia of the ring , M = mass of the ring, and m = mass
of the slider
IO = I + M R2 = M R2 + M R2 = 2M R2
+
(hO )1
2
IO + m OA ! 1
i
h
p
2M R2 + m( 2R)2 ! 1
!2
=
(hO )2
2
=
IO + m OB
=
2M R2 + m(2R)2 ! 2
=
M +m
2:2 + 0:35
!1 =
40 = 35:2 rad/s J
M + 2m
2:2 + 0:70
!2
*18.79
20
478
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.80
18.81
1
479
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.82
Ι-AB ω1 I-1ω1 m(2.5ω1)
I1ω1
m(2.5ω1)
2.5' O 2.5'
Initial momenta
m(2.5ω2)
I2ω2
m(2.5ω2)
ΙAB ω2
I2ω2
A
O B
Final momenta
One plate:
1
1 12
mb2 =
(2)2 = 0:124 22 slug ft2
12
12 32:2
1
I2 =
m(2b2 ) = 2I1 = 0:2484 slug ft2
12
In position 1:
I1 =
In position 2:
IAB =
1 8
1
mAB L2 =
(3)2 = 0:186 34 slug ft2
12
12 32:2
+
(hO )1
2 I1 ! 1 + m(2:5) ! 1 + IAB ! 1
2
= (hO )2
= 2 I2 ! 2 + m(2:5)2 ! 2 + IAB ! 2
12
(2:5)2 + 0:186 34 (12)
32:2
12
2 0:2484 +
(2:5)2 + 0:186 34 ! 2
32:2
5:342! 2
! 2 = 11:44 rad/s J
2 0:124 22 +
=
61:12
=
18.83
2
480
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.84
o
O
Iω1 R sin50
R mv-1
^y
O
Iω2
O O
^x
R
y
O
mv2 R
x
50o
A N^
FBD during
impact
Momenta before
impact
Momenta after
impact
(a)
m=
Z
+
(AO )1
2
^ dt R sin 50
N
30
= 0:9317 slugs
32:2
=
(hO )2
(hO )1
=
(I! 2 + mv2 R)
( I! 1
mv1 R)
1
1
mR2 ! 2 + m! 2 R2 + mR2 ! 1 + m! 1 R2
2
2
3
2
=
mR (! 2 + ! 1 )
2
3(0:9317)(0:5)
3mR
(! 2 + ! 1 ) =
(2 + 4)
=
2 sin 50
2 sin 50
= 5:473 lb s J
=
Z
^ dt
N
(b)
Z
Z
Z
Z
+
!
^ x dt
O
^ x dt
O
=
(Lx )1 2 = (px )2 (px )1
Z
^ dt sin 50 = mv2 mv1 =
N
Z
^ dt sin 50
N
mR(! 2 + ! 1 )
=
5:473 sin 50
+
"
^ y dt
O
+
^ y dt
O
=
(Ly )1 2 = (py )2 (py )1
Z
^ dt cos 50 = 0 0
N
Z
^ dt cos 50 = 5:473 cos 50 =
N
Z
^ dt =
O
mR(! 2 + ! 1 )
0:9317(0:5)(2 + 4) = 1:397 lb s
3:518 lb s
p
1:3972 + ( 3:518)2 = 3:79 lb s J
3
481
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.85
18.86
(a)
A
^
F
B
Momenta before
impact
+
L/2
^
R
C
D mv
D1
^
F
d
Iω2
(mAB + mD)ω2d
L/2
FBD during
impact
Momenta after
impact
0:14
(1800)(0:5) = 3:913 lb ft s
32:2
(hC )2 = (mAB + mD )! 2 d2 + I! 2
1 26
26 + 0:14
! 2 (0:5)2 +
(5)2 ! 2 = 1:8851! 2
32:2
12 32:2
(hC )1 = mD v1 d =
+
=
(hC )1 = (hC )2
3:913 = 1:8851! 2
! 2 = 2:0758 rad/s J
4
482
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(b)
Velocities
after impact d
T2
V2
V3
θ
ω2 Datum
ω2d
Position of
maximum
displacement
C
G
1 2 1
2
I! + (mD + mAB ) (! 2 d)
2 2 2
1 1 26
1 0:14 + 26
=
(5)2 (2:0758)2 +
(2:0758 0:5)2
2 12 32:2
2
32:2
= 4:061 lb ft
=
(WD + WAB )d = (0:14 + 26)(0:5) = 13:070 lb ft
=
(WD + WAB )d cos = 13:070 cos
=
T 2 + V2 = T 3 + V3
4:061
13:070 = 0
13:070 cos
= 46:4
J
18.87
P^
A
L/2 Iω
mv-
A
A
ω
L/2
FBD during
impact
+
(hA )1 = (h2 )2 :
vC = v
!
Momenta after
impact
0 = I!
L
2
y
=v
mv
L
2
6
v
L
C
y
vvC
Velocities after
impact
1
L
mL2 ! = mv
12
2
L
2
y
=0
y=
!=6
v
L
L
J
3
18.88
mv1
0.25mω2
G
5m
0.2
G
0.2 m
Iω2
A^x
A
A
A ^
Ay
1 Momentum
FBD during 2 Momenta
before impact impact
after impact
0.25 m
A Datum
3 Imminent
tipping
5
483
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(a)
I=
1
m(0:32 + 0:42 ) = 0:02083m
12
+
(hA )1 = (hA )2 :
mv1 (0:2) = I! 2 + (0:25)2 m! 2
0:2mv1 = 0:02083m! 2 + (0:25)2 m! 2
! 2 = 2:40v1 J
(b)
1 2 1
I! + m(0:25! 2 )2 + 0:2mg = 0 + 0:25mg
2 2 2
1
1
(0:02083m)(2:40v1 )2 + m(0:25 2:40v1 )2 + 0:2m(9:81)
2
2
= 0:25m(9:81)
= 0:4905
v1 = 1:430 m/s J
T2 + V2
= T 3 + V3 :
0:240v12
18.89
(a)
+
(hA )1 = (hA )2 :
!2
=
mv1
L
= IA ! 2
2
3
3
v1 =
(5) = 9:375 rad/s
2L
2(0:8)
mv1
1
L
= mL2 ! 2
2
3
J
(b) Let horizontal plane through A be the datum for gravitational potential
energy
L 1
L 1
+ IA ! 22 = mg + IA ! 23
2
2
2
2
L
1 1
1
1
+
mL2 ! 22 = mg +
mL2 ! 23
2 3
2
2 3
s
r
g
9:81
=
! 22 6 = 9:3752 6
= 3:78 rad/s
L
0:8
V2 + T 2
mg
= V3 + T3 :
L
2
!3
mg
J
18.90
1 mR2ω
1
2
1 mR2ω
2
2
mRω1
20o R
A
Momenta before
impact
mRω2
R
A
^
^
N
F
FBD during impact
A
Momenta after
impact
6
484
© 2010. Cengage Learning, Engineering. All Rights Reserved.
+
(hA )1 = (hA )2 :
1
1
mR2 ! 1 + (m! 1 R) R cos 20 =
mR2 ! 2 + (m! 2 R) R
2
2
! 1 (1 + 2 cos 20 ) = 3! 2
!2 =
1 + 2 cos 20
1 + 2 cos 20
!1 =
(4) = 3:84 rad/s
3
3
J
18.91
18.92
7
485
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.93
8
486
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.94
18.95
(a)
R/2
R
G
Momenta before
impact
FBD during
impact
F^
mv-2
G
Iω2
Momenta after
impact
9
487
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Z
F^ dt = mv2 = 2:8m
Z
R
(AG )1 2 = (hG )2 (hG )1
+
F^ dt = I! 2 0
2
2
3:5
3:5
R
(2:8m) =
mR2 ! 2
!2 =
=
= 33:60 rad/s J
2
5
R
1:25=12
L1
= p2
2
p1
+
(b)
mg
mv-2
Iω2
G
R
C
Momenta after
impact
(hC )1
mv2 R
2
mR2 ! 2
5
!3
=
G
R
F
N
FBD during
slipping
m(Rω3)
G
R
C
C
Iω3
Momenta during
rolling
(hC )1
+
mv2 R I! 2 = (mR! 3 ) R + I! 3
2
= mR2 ! 3 + mR2 ! 3
5
5v2 2R! 2
5(2:8) 2(1:25=12)(33:60)
=
=
7R
7(1:25=12)
= 9:60 rad/s J
18.96
mbullv1
d O
Momentum before
impact
I
=
=
R
O^y O^x
O
FBD during
impact
mbullv2
d
Iω2
O
Momenta after
impact
1
mR2 + md2
4
1
2
(2:5) (0:2)2 + 2:5(0:375)2 = 0:7531 kg m2
4
2
(hO )1 = (hO )2 :
mbull v1 d = mbull v2 d + I! 2
0:0097(850)(0:375) = 0:0097 (v2 ) (0:375) + 0:7531(2:61)
v2 = 310 m/s J
10
488
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.97
IA
=
=
=
1
2
2
2
2
+ mc AB + RC
mAB AB + mC RC
3
5
2
2
1 2
2 5
5
20
3
23
+
+
3 32:2 12
5 32:2 12
32:2 12
2
0:6318 slug ft2
Let v2 be the velocity of ball D after the impact
+
0:6318! 1
=
(hA )1 = (hA )2 :
IA ! 1 = mD (AB + RC )v2
W
23
v2
! 1 = 0:094 21W v2
32:2 12
(a)
Since impact is elastic, energy is conserved
T1
1
(0:6318) ! 21
2
= T2 :
=
1
2
1
1
IA ! 21 = mD v22
2
2
W
32:2
v22
p
! 1 = 0:2217 W v2
(b)
Equating (a) and (b):
p
0:094 21W = 0:2217 W
W = 5:54 lb J
18.98
11
489
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.99
12
490
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.100
18.101
13
491
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.102
Point O is the I.C. of bar BC
O
2m
ωBC
d
60o
vC
A
vB
OC
d2
=
=
=
=
G
m
1.0
ωAB
B
m
1.0
2m
vB
30o
C
AB ! AB = 2(1:2) = 2:4 m/s
! BC = ! AB = 1:2 rad/s
OA cos 60 = 4 sin 60 = 3:464 m
2
2
CG + OC
2(CG)(OC) cos 30
2
2
1:0 + 3:464
2(1:0)(3:464) cos 30 = 6:999 m2
14
492
© 2010. Cengage Learning, Engineering. All Rights Reserved.
T
=
=
=
=
1
1
(IAB )A ! 2AB + (IBC )O ! 2BC
2
2
1 1
1 1
2
2
mAB AB ! 2AB +
mBC BC + mBC d2 ! 2BC
2 3
2 12
1 1
1 1
(30)(2)2 (1:2)2 +
(30)(2)2 + 30(6:999) (1:2)2
2 3
2 12
187:2 J J
18.103
18.104
15
493
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.105
18.106
16
494
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.107
18.108
2 ft
2 ft
2 ft
L - 4.472'
2 ft
C
2'
4.47
A
B
2 ft
B
L - 2'
ωAB
A
Position 1
Position 2
C
vC
vC = AB ! AB = 2! AB
Let L be the length of the rope.
U1
2
=
=
T2
=
=
U1
2
AB
+ WC [(L 2) (L 4:472)]
2
W (1:0) + W (2:472) = 1:472W
1
1
2
(IAB )A ! 2AB + mC vC
2
2
1 W
1 1 W
(2)2 ! 2AB +
(2! AB )2 = 0:08282W ! 2AB
2 3 32:2
2 32:2
WAB
= T2 T1
1:472W = 0:08282W ! 2AB
) vC = 2(4:216) = 8:43 ft/s # J
0
! AB = 4:216 rad/s
17
495
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.109
^
N
T^A
C
A
G
^
N
45o 45o
FBD during
impact
45o 45o
Momentum
before
impact
2.4 m
2.68
3m
G
O
T^B
O 2.4m ω 1.2 m
C
ΑΒ
C
.
30 N s
Iω
2.683mCω
B A
B
G
1.2 m
A
B
O
Momenta
after
impact
Point O is the I.C. of bar AB.
I=
1
1
mAB L2 =
(15)(4:8)2 = 28:80 kg m2
12
12
+
(30)(1:2)
(hO )1 = (hO )2
2
= I! + (2:4)2 mAB ! + (2:863) mC !
2
(30)(1:2) = 28:80! + (2:4)2 (15)! + (2:683) (5)!
! = 0:238 rad/s
J
18.110
18
496
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.111
18.112
19
497
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.113
18.114
^y
O
L
mv1
A
Momentum
before impact
I=
O
^
O
x
L
F^
Iω2
F^
FBD during
impact
1
1
mOA L2 =
12
12
L/2
O
A
O
Datum
Lm ω
2 OA 2
θ
L/2
L/2
A
mv2
Momenta after Position of max.
displacement
impact
40
32:2
(4)2 = 1:6563 slug ft2
20
498
© 2010. Cengage Learning, Engineering. All Rights Reserved.
System:
"
I + mOA
40
32:2
1:6563 +
+
mv1 L
#
(hO )1 = (hO )2
=
I! 2 + mOA
!2
=
mL(v2
(2)2 ! 2
=
!2
=
2
L
2
! 2 + mv2 L
2
L
2
v1)
6
(4)(2000
16 32:2
3:516
rad/s
Bar only: T2 + V2
= T3 + V3
2
1 2 1
L
L
I! 2 + mOA
! 22 WOA
2
2
2
2
"
#
2
L
1
! 22
I + mOA
2
2
1
(1:6563) +
2
40
32:2
(2)2 (3:516)2
1500)
=
0
WOA
L
cos
2
L
= WOA (1
2
=
40(2)(1
=
60:8
cos )
cos )
J
18.115
IAω1
Α^y Α^x
Α 1.5'
IAωA2
IA
=
IB
=
Momenta before
impact
F^
F^
1
mA L2A =
12
1
mB LB =
12
Β^y Β^
x
1.2' Β
IBωB2
FBD's during
impact
Momenta after
impact
1 8
(3)2 = 0:18634 slug ft2
12 32:2
1 6
(2:4)2 = 0:08944 slug ft2
12 32:2
Kinematics (plastic impact):
1:5! A2 = 1:2! B2
! B2 = 1:25! A2
21
499
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Bar A:
+
1:5
Z
Z
A1
=
2
(hA )2
(hA )1
F^ dt = IA (! A2
F^ dt =
! A1 ) = 0:18634 (! A2
0:12423 (! A2
5)
5)
Bar B:
+
1:2
Z
A1
=
2
(hA )2
(hA )1
F^ dt = IB (! B2
Z
F^ dt =
! B1 ) = 0:08944 (1:25! A2 )
0:09317! A2
0:12423 (! A2
J
! A2 = 2:857 rad/s
0
5) = 0:09317! A2
! B2 = 1:25(2:857) = 3:57 rad/s
J
18.116
0.08 m
mg
FBD
C
0.4N2
N2
0.4N1
N1
Fx
Fy
= 0:
= 0:
+
+"
N2 0:4N1 = 0
N1 + 0:4N2 = mg = 12(9:81)
The solution is
N1 = 101:48 N
+
=
A1 2 = MA t = [C 0:4(N1 + N2 )R] t
[5 0:4(101:48 + 40:59)(0:08)] 3 = 1:3613 N m s
(hA )2 = I! 2 =
A1
2
= (hA )2
N2 = 40:59 N
(hA )1 :
2
2
mR2 ! 2 = (12)(0:08)2 ! 2 = 0:030 73! 2
5
5
1:3613 = 0:030 72! 2 0
! 2 = 44:3
rad/s J
22
500
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18.117
Position 2
Position 1
1.8 m
1.2
m
θ
0.6 m
Datum
2.5 rad/s
1.5 m/s
!1 =
V2
T1
v1
1:5
=
= 2:5 rad/s
R
0:6
= W h = (9:81m) [1:2(1 cos )] = 11:772(1
1
1 1 2
1
mv12 + I! 21 = mv12 +
mr ! 21
=
2
2
2 2
1
1
2
2
=
m (1:5) + (0:6) (2:5)2 = 1:6875m
2
2
V1 + T1
cos
cos )m
= V2 + T 2 :
0 + 1:6875m = 11:772(1 cos )m + 0
1:6875
= 1
= 0:8567
= 31:1 J
11:772
18.118
Ox
Oy
ωA
A
C
O
2 ft
O
B
(mΒ + mC)vICωC
IΒωΒ
Final momenta
FBD
Kinematics:
! C = ! A (C and A are rigidly connected)
! B = ! A + ! B=A = ! A 25 rad/s
Final momentum diagram:
+
"
20 + 12
(2! A ) = 1:987 6! A lb s
32:2
2
10
20
2
(! A 25) = 0:4313! A
= mB kB
!B =
32:2 12
(mB + mC )v =
+
IB ! B
+
2
IC ! C = mC kC
!A =
12
32:2
3
12
10:783
2
! A = 0:023 29! A
23
501
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Conservation of angular momentum:
1:987 6! A (2) + (0:4313! A
+
(hO )2
10:783) + 0:023 29! A
!A
= (hO )1
= 0
= 2:43 rad/s J
24
502
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Chapter 19
19.1
19.2
From Sample Problem 19.4:
aC=Q
= ( 623:6 + 141:42 y + 200:0 z )i
+ (720:2 141:42 x + 173:21 z )j
+ ( 509:0 200:0 z 173:21 y )k
1
503
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Equating like components of aC = aC k = aQ + aC=Q , we get
0 = 1621:4 + 141:42 y + 200:0 z
0 = 720:2 141:42 x + 173:21 z
aC = 787:0 200:0 x 173:21 y
x
z
2
=
1:020 rad/s
=
2
4:99 rad/s
J
2
J
=
(a)
(b)
(c)
4:41 rad/s
aC = 1755 mm/s
2
2
J
J
19.3
19.4
!
20i + 30j + 50k
202 + 302 + 502
= ! ( 0:3244i + 0:4867j + 0:8111k)
= !
OC
rB =
= !p
i
0:3244
0
j
0:4867
30
k
0:8111
50
vB
= !
vB
= !(0:002i + 16:220j 9:732k)
p
= ! 0:0022 + 16:2202 + 9:7322 = 18:916!
160
vB
) !=
=
= 8:46 rad/s J
18:916
18:916
2
504
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.5
19.6
3
505
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.7
Let the y-axis be embedded in cone A, so that the coordinate system rotates
about the z-axis with the angular velocity
(a) From the velocity diagram of cone A:
z
!A
ω1 = 4 rad/s y
40o Ω
ωA
= ! 1 tan 40 = 2 tan 40 = 1:6782 rad/s
= !1 j
k = 2j 1:6782k rad/s J
(b)
=
=
d! A
dt
+
!A = !
_1+
!A
=xyz
7j + ( 1:6782k)
(2j
1:6782k) = 3:36i
2
7j rad/s
J
19.8
4
506
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.9
19.10
z
ω1 = 40 rad/s
x
ω2 = 16 rad/s
23.6o
(a) Let the xyz-axes be attached to the frame B.
!1
!2
!
= 40 (i cos 23:6 + k sin 23:6 ) = 36:65i + 16:014k rad/s
= 16k rad/s
= ! 1 + ! 2 = (36:65i + 16:014k) +16k
= 36:65i + 32:01k rad/s (valid only at the instant) J
5
507
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(b) Since the xyz-axes are attached to the frame, we have
)
=
=
d!
dt
400k + 16k
+ !2
!=!
_ 2 + !2
= !2
!
=B
2
(36:65i + 32:01k) = 400k + 586j rad/s
J
19.11
19.12
6
508
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.13
z
A
1.3'
1.2' y
'
2.1
3
4 5
1.5'
B
x
3'
rB=A = 2:1i
vB=A
vA
= !
rB=A =
i
!x
2:1
3j
j
!y
3
1:3k ft
k
!z
1:3
= ( 1:3! y + 3! z )i + (1:3! x + 2:1! z )j + ( 3! x
= 4k ft/s
vB = vB ( 0:6i + 0:8k) ft/s
2:1! y )k
vB = vA + vB=A
Equating like components, we get
0:6vB =
1:3! y + 3! z
0 = 1:3! x + 2:1! z
0:8vB = 4 3! x 2:1! y
(a)
(b)
(c)
Assuming the spin velocity of AB to be zero, we have ! rB=A = 0
2:1! x
3! y
1:3! z = 0
(d)
7
509
© 2010. Cengage Learning, Engineering. All Rights Reserved.
0:6vB = 1:3! y + 3! z
0 = 1:3! x + 2:1! z
0:8vB = 4 3! x 2:1! y
2:1! x 3! y 1:3! z = 0
The solution of (a)-(d) is
!x
vB
= 0:435 rad/s
= 2: 26 ft/s J
! y = 0:422 rad/s
!z =
0:270 rad/s
19.14
8
510
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.15
19.16
9
511
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.17
rB=A
vA
vB=A
= !
=
rB=A =
(0:18! y
=
=
0:5i + 0:6j + 0:18k m
1:2i m/s
vB = vB j
i
!x
0:5
j
!y
0:6
k
!z
0:18
0:6! z )i + ( 0:18! x
0:5! z )j + (0:6! x + 0:5! y ) k
vB = vA + vB=A and ! rB=A = 0 (no spin velocity) yield
0 = 1:2 + 0:18! y 0:6! z
vB = 0:18! x 0:5! z
0 = 0:6! x + 0:5! y
0 = 0:5! x + 0:6! y + 0:18! z
The solution is
!x
vB
=
=
0:280 rad/s
1:0 m/s J
! y = 0:336 rad/s
!z =
1:899 rad/s
19.18
z
A
0.8
m
θ
y
cos
=
)
When
=
y
0:8
B
y
_ sin = y_ = 0:2 = 0:25
0:8
0:8
0:25
cos
_=
) • = 0:25
sin
sin2
• = 0:8660 rad/s2
30 : _ = 0:5 rad/s
)
10
512
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Let the xyz-coordinates be attached to the supporting frame. )
!1
= !1
= 1:5k rad/s
! 2 = _ i = 0:5i rad/s
= !
_ =!
_1+!
_ 2 = (!
_ 1 )=xyz + ! 1 ! 1 + (!
_ 2 )=xyz + ! 1
= 0+
•i + 1:5k
0:5i =
0:866i + 0:75j rad/s
2
!2
J
19.19
19.20
11
513
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.21
12
514
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.22
(a) Let the xyz-axes be attached to arm AB, so that
! 1 = 3k rad/s
! 2 = 10i rad/s
= !
_ =!
_1+!
_ 2 = (!
_ 1 )=xyz + ! 1
=
(0 + 0) + (0 + ! 1
= !1
! 2 ) = 3k
! 1 + (!
_ 2 )=xyz + ! 1
10i = 30j rad/s
!2
J
2
(b)
!
rQ=B
aQ=B
= (10i + 3k) ( 10k) = 100j in./s
= ! (! rQ=B ) +
rQ=B
= (10i + 3k) 100j + 30j ( 10k)
2
aQ
= (1000k 300i) 300i = 1000k 600i in./s
= aB + aQ=B = AB! 21 j + aQ=B = 30(3)2 j + 1000k
=
600i
270j + 1000k in./s
2
600i
J
19.23
(a) Let the xyz-axes be attached to arm OA, so that
!1
!
= !1
= 4k rad/s
! 2 = 6 (j cos 35 + k sin 35 ) = 4:915j + 3:441k rad/s
= ! 1 + ! 2 = 4:915j + 7:441k rad/s
= !
_ = (!)
_ =xyz +
!=
2
=
15k + 19:66i rad/s
15k + 4k
(4:915j + 7:441k)
J
(b)
rA=O
vA
rP=A
vP=A
vP
=
=
=
=
= 0:6 (j cos 35 + k sin 35 ) = 0:4915j + 0:3441k
= ! 1 rA=O = 4k (0:4915j + 0:3441k)
=
1:9660i m/s
0:25 ( j sin 35 + k cos 35 ) = 0:14339j + 0:2048k m
! rP=A = (4:915j + 7:441k) ( 0:14339j + 0:2048k)
2:074i m/s
vA + vP=A = 1:966i + 2:074i = 0:108i m/s J
13
515
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.24
(a) Let the xyz axes be embedded in the arm AB, so that
! 1 = 12k rad/s
= !1
! 2 = 9i rad/s
For arm AB:
!
= ! 1 + ! 2 = 9i + 12k rad/s
= !
_1+!
_ 2 = (!
_ 1 )=xyz + ! 1
= 0 + (0 + 12k
! 1 + (!
_ 2 )=xyz + ! 1
9i) = 108j rad/s
2
!2
J
(b)
= 1:2 (j cos 38 + k sin 38 ) = 0:9456j + 0:7388k m
!
= (9i + 12k) (0:9456j + 0:7388k)
=
11:347i 6:649j + 8:510k m/s
(! rB=A ) = (9i + 12k) ( 11:347i 6:649j + 8:510k)
rB=A
rB=A
!
rB=A
aB
=
79:79i
=
108j
212:75j
59:84k m/s
2
(0:9456j + 0:7388k) =79:79i m/s
2
= aB=A = ! (! rB=A ) +
rB=A
= (79:79i 212:75j 59:84k) + 79:79i
=
159:6i
212:8j
59:8k m/s
2
J
19.25
z z'
O
ω
β y
y'
(a) Let the x0 y 0 z 0 axes be the principal axes of the rod
) Ix0
= Iz0 =
Iy0 = 0
! x0
=
hO
= Ix0 ! x0 i0 + Iy0 ! y0 j0 + Iz0 ! z0 k0 =
=
0
mL2
12
! y0 =
mL2
! (k cos
12
! sin
! z0 = ! cos
+ j sin ) cos
mL2 0
!k cos
12
J
14
516
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(b)
T
=
=
1
1
mL2
! h0 = (!k)
! (k cos
2
2
12
1
mL2 ! 2 cos2 J
24
+ j sin ) cos
19.26
z
y
5m
0.1
0.1
5m
O
G
O
x
G
^
P
Let ! 2 = ! x i + ! y j + ! z k
From solution of Sample Problem 19.7:
(hO )2 = 0:12! x i + (0:06! y
(AO )1
2
= rG=O
=
0:48i
(AO )1
2
0:045! z )j + ( 0:045! y + 0:06! z )k
Z
i
j
0
0:15
2:2
1:4
0:33j + 0:33k N m s
k
0:15
1:8
^ dt =
P
= (hO )2
(hO )1 = (hO )2
0
Equating like components:
0:48 = 0:12! x
0:33 = 0:06! y 0:045! z
0:33 = 0:045! y + 0:06! z
The solution is
!x
T2
= 4:0 rad/s
) ! 2 = 4:0i
=
=
! y = 3:143 rad/s
! z = 3:143 rad/s
3:143j + 3:143k rad/s J
1
1
! 2 (hO )2 = ! 2 (AO )1 2
2
2
1
[4(0:48) + ( 3:143)( 0:33) + 3:143(0:33)] = 1:997 J J
2
15
517
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.27
!x = 0
!y =
20 rad/s
!z = 0
19.28
(a) The xyz-axes are the principal axes of the disk
! = ! 1 + ! 2 = 60j + 20k rad/s
hO
1
1
mR2 = (12)(0:15)2 = 0:0675 kg m2
4
2
2Ix = 0:1350 kg m2
Ix
= Iz =
Iy
=
= Ix ! x i + Iy ! y j + Iz ! z k = 0i + 0:1350(60)j + 0:0675(20)k
= 8:10j + 1:35k N m s J
16
518
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(b)
T
=
=
1
1
! hO + mv2
2
2
1
1
[60(8:10) + 20(1:35)] + (12)(20
2
2
0:4)2 = 641 J J
19.29
(a)
!x
) hO
Ixy
=
Iy
=
= !z = 0
!y = !0
=
Ixy ! 0 i + Iy ! 0 j Iyz ! 0 k
mxy = mR( R) + m( R)R = 2mR2
1
Iy + mx2 = 2
mR2 + mR2 = 2:5mR2
4
Iyz = 0
) hO = mR2 ! 0 (2i + 2:5j) J
(b) Since O is the mass center and vO = 0, we have
hA = hO + rO=A
vO = hO + 0 = mR2 ! 0 (2i + 2:5j) J
19.30
0.3 m
Position 2
A
x
0.4 m
ω y
m
0.5
Position 1
B
δ
ωrod
In Position 2, the spring is undeformed and the rod is rotating about B with
the angular velocity
vB
0:15!
= 0:3!
! ro d =
=
0:5
AB
.
1
1
T2 =
(Idisk )O ! 2 + (Iro d )B ! 2ro d
2
2
1 1
1 1
2 2
=
mdisk R ! +
mro d L2 ! 2ro d
2 2
2 3
1 1
1 1
=
(6)(0:15)2 ! 2 +
(2)(0:5)2 (0:3!)2
2 2
2 3
0:04125! 2
1 2
1
=
k = (2000)(0:1)2 = 10 N m
2
2
=
V1
17
519
© 2010. Cengage Learning, Engineering. All Rights Reserved.
T1 + V1
!
= T 2 + V2
0 + 10 = 0:04125! 2 + 0
= 15:57 rad/s J
19.31
19.32
18
520
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.33
19
521
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.34
19.35
(a)
OB
rB=A
vA
vB
= vA + !
p
=
1:82 1:22 0:92 = 0:9950 m
= 0:9950i 1:2j 0:9k m
=
1:4j m/s
vB = vB i
rB=A
vB i =
1:4j +
i
!x
0:9950
j
!y
1:2
) vB =
0:9! y + 1:2! z
0 = 0:9! x + 0:9950! z 1:4
0 =
1:2! x 0:9950! y
k
!z
0:9
(a)
(b)
(c)
Since the moment of inertia of AB about its axis is negligible, the spin velocity
of AB is irrelevant. We assume the spin velocity to be zero, i.e.,
! rB=A = 0
0:9950! x
1:2! y
0:9! z = 0
(d)
20
522
© 2010. Cengage Learning, Engineering. All Rights Reserved.
The solution of (a)-(d) is
vB
!y
= 1:6884 m/s
! x = 0:3889 rad/s
=
0:4690 rad/s
! z = 1: 0553 rad/s
12(1:8)2
(0:3889i 0:4690j + 1: 0553k)
12
1:260i 1:520j + 3:419k N m s J
h = I! =
=
(b)
v
=
T
=
=
=
vA + vB
1:4j + 1:6884i
=
= 0:8442i 0:7j m/s
2
2
1
(I! 2 + mv 2 )
2
1 12(1:8)2
(0:38892 + 0:46902 + 1: 05532 ) + 12(0:84422 + 0:72 )
2
12
9:62 N m J
19.36
21
523
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.37
z
ω1
A
R
O
2R
ω
ω2
B
ω2 = 2ω1
ω
y
2
1
ω1
! = ( 2j + k)! 1
For the disk:
v
Iy
Ix
U1
2R! 1 = 2(0:5)! 1 = ! 1
1
1
14
=
mR2 =
(0:5)2 = 0:05435 slug ft2
2
2 32:2
1
= Iz = Iy = 0:02717 slug ft2
2
1
1
Iy ! 2y + Iz ! 2z + mv 2
2
2
1
1 14 2
0:05435( 2! 1 )2 + 0:02717! 21 +
! = 0:3397! 21
=
2
2 32:2 1
= C0 (4 ) = 0:35(4 ) = 4:398 lb ft
T2
U1
=
=
2
2
= T2
T1
4:398 = 0:3397! 21
0
! 1 = 3:60 rad/s J
19.38
22
524
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.39
23
525
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.40
Ix
=
Iy
=
Iz
=
Ixy
=
Iyz
(hO )2
(AO )1
2
m 2
1 m
2
b +
(2b)2 = mb2
2
12 2
3
1 m
1
(2b)2 = mb2
12 2
6
1 m
m
b
2
(2b) +
b2 +
12 2
2
2
mb
m
( b) +
4 2
4
= Izx = 0
b
2
b=
2
!
=
19 2
mb
24
1 2
mb
4
=
(Ix ! x Ixy ! y )i + ( Ixy ! x + Iy ! y )j + Iz ! z k
1
1
2
1
19
!x + !y i +
! x + ! y j + ! z k mb2
=
3
4
4
6
24
Z
Z
Z
^
^
= rAO k P dt = b(i j) k P dt = b P^ dt (i + j)
(AO )1
2
= (hO )2
(hO )1
Equating like components, we get
Z
2
1
!x +
P^ dt =
mb
3
Z
1
1
!x +
P^ dt =
mb
4
0 = !z
1
!y
4
1
!y
6
24
526
© 2010. Cengage Learning, Engineering. All Rights Reserved.
The solution is
!x
=
)
Z
Z
12 1
60 1
P^ dt
!y =
P^ dt
7 mb
7 mb
Z
12
!=
P^ dt (i 5j) J
7mb
!z = 0
19.41
(a)
L1
2
v2
= mv2 mv1 :
=
0:125i m/s J
0:15i = 1:2v2
0
(b)
1
1
mR2 = (1:2)(0:1)2 = 0:006 kg m2
2
2
1
= Iz = mR2 = 0:003 kg m2
4
= Ix ! x i + Iy ! y j + Iz ! z k = 0:003(2! x i + ! y j + ! z k)
Ix
=
Iy
h2
A1
2
= rA=G L1 2 = 0:1(j cos 30
k sin 30 )
= 0:0075j + 0:012 99k N m s
( 0:15i)
A1 2 = h2 h1 = h2 0
0:0075j + 0:012 99k = 0:003(2! x i + ! y j + ! z k)
) !x = 0
T
! y = 2:5 rad/s
! z = 4:330 rad/s
1
1
1
1
h2 ! 2 + mv22 = A1 2 ! 2 + mv22
2
2
2
2
1
1
=
[0:0075(2:5) + 0:012 99(4:330)] + (1:2)(0:125)2
2
2
= 0:0469 N m J
=
19.42
25
527
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.43
26
528
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.44
27
529
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.45
19.46
28
530
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.47
19.48
1
531
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.49
19.50
B
FBD
L mg
R
Ay z ω
A
C
Az = mg
x
O
y
Rotation about …xed axis:
Mx
C
= Iyz ! 2z
=
C = myz! 2 = m( R)
L
2
!2
1
mRL! 2 J
2
2
532
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.51
y
x
0.015 m
RA A
G
0.6
m
0.4
m
B
RB
z
FBD (only horizontal forces are shown)
We have rotation about a …xed axis
) Mx = Iyz ! 2z
My =
Ixz ! 2z
With origin of xyz-axes at A:
Iyz
Ixz
My
= myz = 24(0)(0:6) = 0
= mxz = 24(0:015)(0:6) = 0:2160 kg m2
=
Ixz ! 2z
1:0RB = 0:2160(20)2 =
) RB = 86:4 N J
86:4 N
With origin of the xyz-axes at B:
Iyz
Ixz
My
= myz = 24(0)( 0:4) = 0
= mxz = 24(0:015)( 0:4) = 0:1440 kg m2
1:0RA = ( 0:1440) (20)2 = 57:6 N
=
Ixz ! 2z
) RA = 57:6 N J
19.52
3
533
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.53
5m
0.0
W1
y
z
x 0
.1 m R
B
W2
0
RA
B .1 m
m
0.1
A
0.8 N. m
W1
FBD
m1
m2
W1
1:8
m
=
= 0:3 kg (0.1 m 0.05 m plate)
6
6
4(1:8)
4m
=
= 1:2 kg (0.2 m 0.1 m plate)
=
6
6
= 0:3(9:81) = 2:943 N
W2 = 1:2(9:81) = 11:772 N
=
Ixz
=
Iyz
=
0
X
mi yi zi = 0:3(0:1)( 0:025) + 0:3( 0:1)( 0:075)
0:0015 kg m2
X
=
(Iz )i + mi d2i
=
Iz
5m
0.0
=
1
1
(1:2)(0:2)2 + 2
(0:3)(0:1)2 + 0:3(0:12 + 0:052 )
12
12
=
0:012 kg m2
4
534
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Rotation about …xed axis:
Mz = Iz !_ z
My =
0:8 = 0:012!_ z
Iyz !_ z
(2W1 + W2 )(0:05)
(2 2:943 + 11:772) (0:05)
Fx = 0
(2
2
!_ z =
0:1RB
0:1RB
RB
66:67 rad/s
=
=
=
(2W1 + W2 ) + RA + RB = 0
2:943 + 11:772) + RA + 7:83 = 0
) RA = 9:83i N J
J
Iyz !_ z
0:0015( 66:67)
7:83 N
RA = 9:83 N
RB = 7:83i N J
19.54
x
RB
B
m
0.1
y
RA
.
z
A
FBD (only horizontal forces are shown)
From solution of Problem 19.53:
Iyz = 0:0015 kg m2
Izx = 0
Rotation about a …xed axis:
Mx
Fy
= Iyz ! 2z
0:1RB = 0:0015(10)2
RB = 1:5 N
= 0
RA + R B = 0
RA = RB = 1:5 N
) RA =
1:5j N J
RB = 1:5j N J
19.55
x
A
y
RA
1.2
5'
B
FBD
0.7
5'
RB
C
'
2 lb
5 .
0.7 ω z
D .75' 3 lb
0
5
535
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Use Eqs. (19.37) with ! z = 0 , !_ z = !_
Ixz
=
0
Iyz = myz =
Iz
=
1
1
mL2 =
3
3
Mz = Iz !_
1:25RB
My =
2 =
RA + 2:80
) RA =
3(0:75)
3
32:2
3
(0:75)(2) = 0:13975 slug ft2
32:2
(1:5)2 = 0:06988 slug ft2
2(1:5) = 0:06988!_
!_ =
2
10:733 rad/s
Iyz !_
1:25RB 3(2) + 2(2) = 0:139 75!_
0:139 75( 10:733)
RB = 2:80 lb
Fx
= max
1:0
=
1:050i lb J
RA + R B
3 L
!_
32:2 2
1:0 =
3
(0:75)( 10:733)
32:2
RB = 2:80i lb J
RA =
!
_ =
1:050 lb
10:733k rad/s
2
J
19.56
z
C
R
O
FBD
x
y
mg
Use modi…ed Euler equations with the xyz-axes be attached to the fork
!1
!
Ix
Iy
=
180i rad/s
! 2 = 40k rad/s
= ! 2 = 40k rad/s
= ! 1 + ! 2 = 180i + 40k rad/s
= mR2 = 0:15(0:08)2 = 9:6 10 4 kg m2
1
Ix = 4:8 10 4 kg m2
=
2
Mx = Iz
My = Ix
Mz = Iy
y !z
z !y
z !x
Iy
Iz
x !z
Cx
Cy
x !y
Ix
y !x
Cz
)C=
= 0 0=0
= (9:6 10 4 )(40)( 180)
=
6:91 N m
= 0 0=0
0
6:91k N m J
6
536
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.57
19.58
z
O
ω0
Oy y
O
m
0.2
m
0.1
ωz
z
Oz
y
ωy
9.81m
25o
A
G FBD
25o
A
Let the xyz-axes be attached to the bar. These axes are the principal axes of
the bar at O:
!
Ix
Iy
= ! 0 (j sin 25 + k cos 25 ) = ! 0 (0:4226j + 0:9063k)
1
1
mL2 = m(0:2)2 = 0:013 333m
=
3
3
= Ix = 0:013 333m
Iz = 0
Euler equation:
Mx = ! y ! z (Iz Iy )
(9:81m) (0:1 sin 25 ) = (0:4226! 0 ) (0:9063! 0 )(0
! 0 = 9:01 rad/s J
0:013 333m)
7
537
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.59
19.60
y
A
x
50o
T
B
FBD
Ay
Ax A
0.9
m
ωy
ωx
0.9
m
ω
20(9.81) N
50o
Let the xyz-axes be attached to the clevis so that the z-axis coincides with the
pin
!x
!y
Ix
= ! sin 50 = 6 sin 50 = 4:596 rad/s
= ! cos 50 = 6 cos 50 = 3:857 rad/s
!z = 0
1
1
= 0
Iy = Iz = mL2 = (20)(1:8)2 = 21:60 kg m2
3
3
8
538
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Euler equation:
+
Mz = ! x ! y (Iy Ix )
20(9:81)(0:9 cos 50 ) = (4:596)(3:857)(21:60
T = 360 N J
T (1:8 sin 50 )
0)
19.61
A
L/4
H
ω
A
=
G
mr ω2
L/
2
V
L/
2
45o
G
mg y'
x'
FBD
B
From FBD-MAD:
r=
Fx0
Fy 0
B
r
MAD
L
sin 45
2
= max0
= 0
V
y
ω x
o
45
G
A
B
L
= 0:103 55L
4
H = mr! 2 = 0:103 55mL! 2
mg = 0
V = mg
Euler equation (xyz-axes embedded in bar AB):
Mz
= ! x ! y (Iy Ix )
L
sin 45
+
V
2
=
(! sin 45 ) (! cos 45 )
mg
=
H
L
sin 45
2
L
cos 45
2
1
mL2 0
12
0:103 55mL! 2
L
cos 45
2
1
mL2 ! 2 sin 45 cos 45
12
The solution is
r
r
g
32:2
! = 2:126
= 2:126
= 8:53 rad/s J
L
2
9
539
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.62
CA
RA
A x
b/2
b
FBD
b
G
z
mg
y
Mass center G of the disk travels a circular path of radius rG=A about A. Hence
the (normal) acceleration of G is
a = !2
(! 2
= !0 i
rG=A ) = ! 0 i
(! 0 bk
! 0 bj) =
F = ma:
RA =
[! 0 i
! 20 bj
(bj + bk)]
! 20 bk
m! 20 b(j + k) J
Note that RA is directed along the line AG; thus its moment about G is zero.
Use modi…ed Euler equations with xyz-axes attached to arm AG
)
= !2 i = !0 i
= ! 2 i + ! 1 k = ! 0 (i + 2k) (angular velocity of disk)
!
Modi…ed Euler equations (note that !
_ = 0):
Mx
= Iz
My
=
Mz
= Iy
)
y !z
Iz
Iy
x !z
x !y
CA =
z !y :
+ Ix
(CA )x = 0
z !x :
Ix
y !x :
2 2
m! 0 b
8
(CA )y =
m(b=2)2
(! 0 )(2! 0 ) =
2
mb2 ! 20
4
(CA )z = 0
j J
19.63
4 rad/s Q
ωx
θ ωz
x
12 rad/s
P
z
Q
R
z
CA θ
x
C
D
P
FBD
CD
Let the xyz-axes be attached to P Q (the y-axis coinciding with the shaft of the
motor at B)
! x = 4 sin rad/s
!y =
) !_ x = (4 cos ) _
12 rad/s
!_ y = 0
! z = 4 cos rad/s
!_ z = ( 4 sin ) _
10
540
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Substituting _ = 12 rad/s, we get
!_ x = 48 cos rad/s
2
!_ y = 0
!_ z =
2
48 sin rad/s
1
1
mL2 =
(16)(2:4)2 = 7:680 kg m2
12
12
Euler equations with D as the moment center:
Ix = Iy =
Mx
Cx
My
Cy
= Ix !_ x + ! y ! z (Iz Iy )
= 7:680(48 cos ) + ( 12)( 4 cos )(0
= Iy !_ y + ! z ! x (Ix Iz )
= 7:680(0) + (4 cos )(4 sin )(7:680
Mz
Cz
) CD
CA
= Iz ! z + ! x ! y (Iy Ix )
= 0 + (4 cos )( 12)(7:680
Iz = 0
7:680) = 0
0) = 122:88 sin cos
7:680) = 0
= Cy = 122:88 sin cos = 61:4 sin 2 N m J
= Cx sin + Cz cos = 0 J
19.64
11
541
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.65
The no precession condition is
= Iz =I = 1
p
h
= 3 J
R
1
1
mR2 =
m(3R2 + h2 )
2
12
) Iz = I
19.66
Z,y
z
.
φ
R
8 ft
O
x
ψ.
mg
A
FBD
0.75 ft
F
_ = 20 rad/s
_ = _ R = 20 0:75 = 1:875 rad/s
L
8
Since the precession axis is perpendicular to the spin axis, Eq. (19.48) applies:
= Iz _ _
Mx
(F
5)(8)
=
1
2
(F
5
32:2
mg)L =
1
mR2 _ _
2
F = 5:20 lb J
(0:75)2 ( 20)( 1:875)
19.67
z
x
Z,y .
φ
mg
B
ψ.
O
0.5
m
F
A
mAg
0.5
m
FBD's
O
0.5
m
R
F
A
Since the precession axis is perpendicular to the spin axis, Eq. (19.48) applies
to the disk:
1
Mx = Iz _ _
(F mA g) 0:5 = mA R2 _ _
2
where F is the force applied by the shaft on the disk. Substituting _ = 2 rad/s,
and _ = 40 rad/s, we get
[F
25(9:81)] 0:5 =
1
(25)(0:3)2 (2)( 40)
2
F = 65:25 N
From the FBD of the shaft:
Mx = 0
0:5mg
0:5F = 0
m=
F
65:25
=
= 6:65 kg J
g
9:81
12
542
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.68
z Z
5o
β ω
y
1
1
Iz
=2
tan = tan = tan
2
I
= tan 1 (2 tan ) = tan 1 (2 tan 5 ) = 9:925
(19.52):
=
)
_ =1
(10.53c):
)
_ =
_ cos = 1
2
2
(10) cos 5 =
4:98 rad/s
4:98k rad/s J
_
( 4:98)
=
= 5:056 rad/s
(1
) cos
(1 2) cos 9:925
= !(j sin + k cos ) = 5:056(j sin 9:925 + k cos 9:925 )
= 0:872j + 4:980k rad/s J
(19.53a):
!=
!
)!
19.69
x
_
=
_
=
Z,y
C φ.
z
ψ.
15 000(2 )
= 1570:8 rad/s
60
600(5280)
1
v
=
= 0:083 33 rad/s
R
3600 2(5280)
Since the precession axis is perpendicular to the spin axis, Eq. (19.48) applies
to the rotor:
Mx
C
= Iz _ _
500
= mk 2 _ _ =
(1:2)2 (0:083 33)(1570:8) = 2930 lb ft J
32:2
13
543
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.70
z
RO = 2mg
y
O mg
L/2
L/2
FBD
mg Z
θ
Iz
Iz
=
2
"
1
mR2
2
R
= mR2
I
=
1
2 mR2 + m
4
I
=
1
mR2 1
2
L
2
2
#
=
L2
1
mR2 1 + 2
2
R
L2
R2
We have steady precession with _ = 2 rad/s, _ = 3 rad/s and
(19.46):
+
Mx = (Iz
0
=
1
mR2 1
2
0
=
0:8988 1
= 32
2
I) _ sin cos + Iz _ _ sin
L2
(2)2 sin 32 cos 32 + mR2 (2)(3) sin 32
R2
L
L2
+ 3:180
= 2:13 J
R2
R
19.71
14
544
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.72
Z
y
L
θ
B1
B B2
β
FBD
R
A
z mg C
N
Let the xyz-axes be attached to disk. The motion is steady precession with
_ = ! = 4 rad/s
= tan
1
R
= tan
L
1
4
= 26:57
8
Iz
=
1
1
mR2 =
2
2
I
=
1
1
mR2 + mL2 =
4
4
=
0:117 32 slug ft2
8
32:2
N
= 153:4
2
= 0:013 803 slug ft2
8
32:2
4
12
2
+
8
32:2
8
12
2
(Iz
2
I) _ sin cos + Iz _ _ sin
L
cos
=
(Iz
2
I) _ sin cos + Iz _ _ sin
8
sin 26:57
12
N
mg(L sin )
8
4
12
= 180
=
Mx
+
_ = _ =0
N
= (0:013 803
= 2:31 lb J
8=12
cos 26:57
0:117 32) (4)2 sin 153:4 cos 153:4 + 0
15
545
© 2010. Cengage Learning, Engineering. All Rights Reserved.
#19.73
16
546
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#19.74
17
547
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.75
Z, y
1.5 ft
1.5 ft
A
RA
'_
=
Iz
=
1:0 rad/s
1
1
mR2 =
2
2
Fy = 0
Mx = Iz '_ _
+
Bz
G
420 lb
RB
_ = 200 rad/s
420
(0:5)2 = 1:6304 slug ft2
32:2
+"
RA + RB
1:5RA 1:5RB
R A RB
420 = 0
= 1:6304(1:0)(200)
= 217:5
(a)
(b)
Solution of (a) and (b) is
RA = 319 lb J
RB = 101:3 lb J
18
548
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.76
#19.77
19
549
© 2010. Cengage Learning, Engineering. All Rights Reserved.
#19.78
20
550
© 2010. Cengage Learning, Engineering. All Rights Reserved.
21
551
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.79
19.80
#19.81
22
552
© 2010. Cengage Learning, Engineering. All Rights Reserved.
23
553
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.82
19.83
24
554
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.84
19.85
(b)
Z
z
60 rad/s
Space cone
Body cone
30
o
49.1o
ω
25
555
© 2010. Cengage Learning, Engineering. All Rights Reserved.
19.86
19.87
26
556
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Chapter 20
20.1
(a)
x(t)
E
(b)
r
v(t)
r
250
= E sin(pt + )
p=
= 5 rad/s
10
q
p
x20 + (v0 =p)2 = 0:042 + ( 0:08=5)2 = 0:0431 m J
=
=
tan
1
x0 p
v0
= tan
1
= x(t)
_
= pE cos(pt + )
1
t=
p
)
2
1
=
5
k
=
m
0:04(5)
=
0:08
1:1903 rad
v(t) = 0 when pt +
2
=
2
+ 1:1903 = 0:552 s J
20.2
(a)
x = E sin(pt + ) where
E=
s
x20 +
v0
p
2
1:5
v0
=p
p= p
= 153:90 rad/s
0:0122 0:0072
E 2 x20
153:90
p
=
= 24:5 Hz J
=
2
2
)
f
(b)
tan
=
x =
x0 p
0:007(153:90)
= 0:7182
=
v0
1:5
0:012 sin (153:90t + 0:6228) m J
= 0:6228 rad
20.3
1
557
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.4
20.5
2
558
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.6
20.7
3
559
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.8
20.9
(a)
F
k
(b)
∆
F
k'
∆'
First …nd the equivalent spring sti¤ness k 0 . In case (a) the spring force is
F=2, resulting in the elongation = F=(2k) of the spring. The corresponding
displacement of the pulley is = =2 = F=(4k). In case (b) we have 0 = F=k 0 .
Therefore, (a) and (b) have the same sti¤ness if 0 = , or k 0 = 4k.
r
r
r
k0
4k
k
=
)p=
=2
J
m
m
m
4
560
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.10
Lθ T
x
Lθ T
FBD
Fx = max
For small , we have sin
mx..
MAD
+ !
2T sin = m•
x
tan = x=L, so that the equation of motion becomes
2T
x
= m•
x
L
x
•+
2T
x=0 J
mL
Since this equation has the formpx
• + p2 x = 0, the motion is simple harmonic
with the circular frequency p = 2T =mL.
20.11
..
mLθ
FBD L
mg
kbθ
=
θ
b
.
mLθ 2
MAD
O
O
Use small angle approximations: sin
and cos
( MO )FBD = ( MO )M AD
mg(L )
+
• + kb
If
1
kb (b)
= mL•(L)
2
mgL
mL2
=
0
= 1:0 s, then p = 2 = = 2 rad/s
kb2 mgL
= 4
mL2
4 2 mL2 + mgL kb2 = 0
4 2 (0:2)L2 + 0:2(9:81)L 250(0:06)2 = 0
7:896L2 + 1:9620L 0:9 = 0
) p2 =
2
The positive root is L = 0:236 m J
5
561
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.12
a
mg
θ
O
b
R
k(bθ +∆) =
FBD
Use small angle approximations: sin
( MO )FBD = ( MO )M AD
O
a
.2
maθ
..
maθ MAD
and cos
+
1
mga + k(b +
)b =
ma2 •
where is the static elongation of the spring. Substituting the static equilibrium
equation mga + k b = 0, we get
kb2
)p
2
• + kb
=0
ma2 •
ma2
s
r
k b
16
10
=
= 7:178 rad/s
=
ma
2:5=32:2 20
=
=
2
2
=
= 0:875 s J
p
7:178
20.13
6
562
© 2010. Cengage Learning, Engineering. All Rights Reserved.
#20.14
#20.15
W
Water level after displacement x
Water level at equilibrium
∆
ρ
ρw
Area = A
h =
Ww
FBD
..
(W/g)x
MAD
7
563
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.16
8
564
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.17
20.18
(a)
s
p
=
r
ccr
=
2mp = 2
=
c
8
=
= 0:4379 < 1
ccr
18:270
k
=
m
16(12)
= 21:01 rad/s
14=32:2
14
32:2
(21:01) = 18:270 lb s/ft
) oscillator is underdamped J
9
565
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(b)
ln
xn+1
xn
xn+1
xn
=
= e
2
p
1
3:060
2
=
2 (0:4379)
p
=
1 0:43792
3:060
= 0:0469 J
20.19
20.20
c
= p
=1
2 km
p
p
) c = 2 km = 2 80(3) = 31:0 N s/m J
20.21
=
)
q
2
2
2
p
=
0:5
1
= 0:5
)
d
2
p
p 1
p
p
= 0:8660
c = 2 km = 2(0:8660) 80(3) = 26:8 N s/m J
0:5
10
566
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.22
p
=
x(t)
=
r
r
k
80
=
= 5:164 rad/s
m
3
(A1 + A2 t)e pt
x(t)
_
= p(A1 + A2 t)e
pt
+ A2 e
pt
Initial conditions:
=
0:01 m ) A1 = 0:01 m
=
0:2 m/s ) pA1 + A2 = 0:2
5:164(0:01) + A2 =
0:2
A2 = 0:14836 m/s
x(0)
x(0)
_
) x(0:1) = [0:01
5:164(0:1)
0:14836(0:1)] e
=
2:89
10
3
m J
20.23
(a)
x
.
2.5x
FBD
.
2.5x
mg
75x
N
Equivalent spring constant is
k=
1
1
= 75 lb/ft
=
1=k1 + 1=k2
1=200 + 1=120
From the FBD:
Fx = max
+ !
75x
2(2:5)x_ =
0:02485•
x + 5x_ + 75x =
p
=
=
0:8
x
•
32:2
0
r
r
k
75
=
= 54:94 rad/s
m
0:02485
c
5
=
= 1:8312 overdamped J
2mp
2(0:02485)(54:94)
(b)
+
q
q
2
1 p
=
1:8312 +
2
1 p
=
1:8312
p
1:83122
p
1:83122
1 54:94 =
16:326 s
1
1 54:94 =
184:89 s
1
11
567
© 2010. Cengage Learning, Engineering. All Rights Reserved.
) x(t) = A1 e
16:326t
+ A2 e
184:89
Initial conditions:
x(0) =
x(0)
_
=0
1
1
ft
A1 + A2 =
12
12
16:326A1 + 184:89A2 = 0
The solution is A1 = 0:091 40 ft, A2 =
) x(t) = 91:40e
16:326t
0:008 071 ft/s
8:071e
184:89
10
3
ft J
20.24
2x
.
2cx
bx
kx
b Oy
FBD
O
Ox
b
x
MAD
=
O
b
mg
(MO )FBD =
(MO )M AD
+
mg
m•
x + 4cx_ + k +
x
=
b
p
=
=
d
=
..
mx
(kx)b + 2cx(2b)
_
+ mgx =
mxb
•
0
r
k + mg=b
2000 + 4(9:81)=(0:3)
=
= 23:08 rad/s
m
4
4(36)
4c
=
= 0:7799
2mp
2(4)(23:08)
2
2
p
p
=
= 0:435 s J
2
23:08
1
0:77992
p 1
r
12
568
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.25
20.26
13
569
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.27
14
570
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.28
#20.29
15
571
© 2010. Cengage Learning, Engineering. All Rights Reserved.
16
572
© 2010. Cengage Learning, Engineering. All Rights Reserved.
#20.30
#20.31
17
573
© 2010. Cengage Learning, Engineering. All Rights Reserved.
18
574
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.32
19
575
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.33
20.34
r
p
=
Z
= Y
k
=
m
s
!
p
24
= 39:31 rad/s
0:5=32:2
(!=p)2
= 0:5
1 (!=p)2
0:7927
1 0:7927
2
35
39:31
=
2
= 0:7927
= 1:912 in. J
20.35
With
= 0, Eq. (20.26) is
X=
1
P0 =k
=
(!=p)2
1
P0 =k
! 2 =(k=m)
)k=
P0
+ m! 2
X
If !=p < 1 (X = +7:5 mm):
0:25
+ 4(62 ) = 177:3 N/m J
0:0075
7:5 mm):
k=
If !=p > 1 (X =
k=
0:25
+ 4(62 ) = 110:7 N/m J
0:0075
20.36
(a) Using Eq. (20.26) with
X1
X2
=
5
=
= 0:
1 (! 2 =p)2
20
1 (10=p)2
=
2
1 (! 1 =p)
4
1 (5=p)2
p2 100
p = 2:5 rad/s J
p2 25
20
576
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(b)
X1
=
P0
k
P0 =k
j1 (! 1 =p)2 j
=
0:02 1
0:02 =
j1
P0 =k
(5=2:5)2 j
(5=2:5)2 = 0:06 m = 60 mm J
#20.37
21
577
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.38
20.39
k
m
m
4k Equivalent
systems
It was shown in the solution of Prob. 20.9 that the e¤ective spring constant
of the system is k 0 = 4k. Hence we can replace the mass-spring-pulley system
with the equivalent mass-spring system shown. With = 0 and Z = 5Y; Eq.
(20.31) yields
r
(!=p)2
!
5
5Y = Y
) =
1 (!=p)2
p
6
r ! r
r !
r
r
r
5
5
k0
5
4k
k
) !=
p=
=
= 1:826
J
6
6
m
6
m
m
22
578
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.40
20.41
23
579
© 2010. Cengage Learning, Engineering. All Rights Reserved.
#20.42
20.43
m
k Equivalent
system
The system is equivalent to the mass-spring system shown, where
k
=
mg
=
m(32:2)
= 483:0m
0:8=12
2
k
182
!
2
= 483:0 (rad/s)
=
= 0:6708
m
p
483:0
(!=p)2
0:6708
= Y
= 0:2
= 0:4075 ft = 4:89 in. J
1 (!=p)2
1 0:6708
)
Z
p2 =
24
580
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.44
20.45
20.46
25
581
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.47
X
=
!
= 0:5
p
8
c
c
c
p
= p
=
= p
= 0:5350
2mp
2 km
2m k=m)
2 60(30=32:2)
X
=
=
0:25
P0
3 in. = 0:25 ft
=
=
q
[1
q
(1
P0 =k
2
(!=p)2 ] + (2 !=p)2
P0 =60
2
0:52 )
0:25 =
2
+ (2(0:5350)(0:5))
P0 =60
0:9213
13:82 lb J
#20.48
26
582
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27
583
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.49
20.50
28
584
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.51
20.52
29
585
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.53
20.54
30
586
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.54
20.55
1
587
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.56
20.57
1.5θ
1.5 ft
1.5θ
F2
P(t)
θ
FBD 2θ
Let
O
=
2 ft
2 ft
1.5 ft
F1
Oy
O O
x
ma y
ma x
MAD
mg
be the (small) angular displacement of the bar.
F2 = cv = 2:5(1:5 _ ) = 3:75 _
F1 = k = 10(1:5 ) = 15
2
588
© 2010. Cengage Learning, Engineering. All Rights Reserved.
+
MO = 2max
1:5F1 1:5F2 mg(2 ) + 1:5P (t) = 2m(2•)
6 •
6(2 ) + 1:5(0:5 sin 5t) = 4
1:5 3:75 _
32:2
0:7453• + 5:625 _ + 34:5 = 0:75 sin 5t
1:5(15 )
Comparing with M • + C _ + K = P0 sin !t, we have
r
r
5
K
34:5
=
= 6:804 rad/s
!=p =
= 0:7349
p =
M
0:7453
6:804
C
5: 625
=
=
= 0:5546
2M p
2(0:7453)(6:804)
From Eq. (20.26), the angular amplitude is
=
P0 =K
q
2
=
0:5=34:5
q
=
24
= 0:015 485 rad
2
0:73492 )2 + [2(0:5546)(0:7349)]
(1
XC
2
(!=p)2 ] + (2 !=p)
[1
= 24(0:015 485) = 0:372 in. J
20.58
(a)
5x.
mg
100(0.005sinωt - x)
FBD
50x
N
Fx = max
+
!
100(0:005 sin !t
x)
5x_
50x = 10•
x
10•
x + 5x_ + 150x = 0:5 sin !t J
(b) Comparing with M x + Cx + Kx = P0 sin !t, we obtain
r
r
K
150
p =
=
= 3:873 rad/s
M
10
5
C
=
= 0:06455
=
2M p
2(10)(3:873)
With ! = p Eqs. (20.26) and (20.27) yield
X
=
P0 =K
0:5=150
=
= 0:02582 m = 25:82 mm
2
2(0:06455)
=
tan
1
1 = =2 rad
3
589
© 2010. Cengage Learning, Engineering. All Rights Reserved.
x(t) = 25:82 sin(3:873t + =2) mm J
20.59
(a)
Ox O
P0 sinωt
Oy
b
FBD 2b θ
.
2cbθ
s
mg − kb(θs − θ)
2bθ
mg
Let
bθ
be the static (angular) displacement.
MO = IO •
+
[mg
s
+ )] b
2cb _ 2b + (P0 sin !t) b
(mg)2b
mb2 + m(2b)2 •
=
Substituting
kb(
s
= mg=(kb) and dividing by b2 , we get
2mg
5m• + 4c _ + k +
b
=
P0
sin !t J
b
(b) Comparing with M • + C _ + K = P0 sin !t, we obtain
K
p
!
p
2mg
2(0:6)(9:81)
= k+
= 150 +
= 173:54 N
b s
0:5
r
K
173:54
=
= 7:606 rad/s
=
M
5(0:6)
=
=
8
= 1:0518
7:606
4c
4(4)
C
=
=
= 0:3506
2M p
2(5m)p
10(0:6)(7:606)
4
590
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From Eq. (20.26):
=
=
=
rh
q
(P0 =b) =k
i2
2
(!=p)
+ (2 !=p)2
1
(2=0:5) =150
2
1:05182 )2 + [2(0:3506)(1:0518)]
(1
0:035 79 rad = 2:051 J
From Eq. (20.27):
tan
=
=
2 !=p
1
tan
2
(!=p)
1
=
2(0:3506)(1:0518)
=
1 1:05182
( 6:926) =
81:8
6:926
J
#20.60
5
591
© 2010. Cengage Learning, Engineering. All Rights Reserved.
#20.61
6
592
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.62
mg
Let
s
Ox O L/2
L/2
Oy
kL(θs + θ)
θ
FBD
be the static angular displacement
= IO •
1
[kL( s + )] L =
mL2 •
3
MO
+
Upon substituting
becomes
(mg)
s
L
2
= mg=(2kL) and cancelling L2 , the equation of motion
1 •
m +k =0
3
7
593
© 2010. Cengage Learning, Engineering. All Rights Reserved.
)p=
s
k
m=3
p
1
f=
=
2
2
r
r
3k
k
= 0:276
Hz J
m
m
20.63
20.64
8
594
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20.65
Use formula given in Prob. 20.64:
p
2
r
gy
+ y2
gy
k2 = 2 y2
p
=
)
Pin at A :
y = 0:2 m
k2 =
From (b):
Pin at B
From (a):
= 1:6 s
9:81(0:2)
3:9272
(a)
k2
p=
(b)
2
=
2
= 3:927 rad/s
1:6
0:22 = 0:08723 m2
:
y = 0:15 m
9:81(0:15)
2
p2 =
= 48:87 (rad/s)
0:087232 + 0:152
2
2
=p
= 0:899 s J
)
=
p
48:87
20.66
(a) Let
s
be the static angular displacement
mg
Ox O L/2
.
c2Lθ/2
Oy
+
c2 L _
2
mg
Upon substituting
becomes
s
L/2
L
2
h
θ
FBD
.
kL(θs + θ) + c1Lθ
MO = IO •
i
1
kL( s + ) + c1 L _ L =
mL2 •
3
= mg=(2kL) and cancelling L2 , the equation of motion
1 •
c2
m + c1 +
3
4
_
k =0 J
(b) Comparing with M • + C _ + K = 0, we obtain
s
s
r
K
k
80
p =
=
=
= 3:464 rad/s
M
m=3
20=3
Ccr
=
=
m
20
3:464 = 46:19 N s/m
p=2
3
3
C
c1 + c2 =4
25 + 16=4
=
=
= 0:628 J
Ccr
Ccr
46:19
2M p = 2
9
595
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.67
Oy
O O
R
x FBD
3mg
θ
mg
MO = IO •
+
(mg) R
=
2:833R• + g
=
1
1
(3m) R2 + m(2R)2 •
2
3
0
Comparing with M • + K = 0, we obtain
r
r
r
K
g
g
=
= 0:5941
p =
M
2:833R
R
r
r
p
0:5941 g
g
f =
=
= 0:0946
Hz J
2
2
R
R
20.68
2
= I + m OG =
=
Letting
s
θ
8(9.81) N
G
m
25
1
.
0
O 0.15 m
Oy
Ox
IO
F2
0.2 m
F1
1
m(a2 + b2 ) + m
12
FBD
b2
a2
+
4
4
=
1
m(a2 + b2 )
3
1
(8)(0:152 + 0:22 ) = 0:166 67 kg m2
3
be the static angular displacement, the spring forces are
F1
F2
= k1
= k2
1
2
= 300 [0:2( s + )] = 60( s + )
= 600 [0:15( s + )] = 90( s + )
MO = IO •
+
8(9:81)(0:075) F1 (0:2) F2 (0:15) = 0:166 67•
8(9:81)(0:075) 60( s + )(0:2) 90( s + )(0:15) = 0:166 67•
10
596
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Substituting the static equilibrium equation
8(9:81)(0:075)
60 s (0:2)
90 s (0:15) = 0
we get for the equation of motion
0:166 67• + 25:5 = 0
Comparing with M • + K = 0, we obtain
r
r
p
1
1
K
25:5
)f =
=
=
= 1:969 Hz J
2
2
M
2
0:166 67
20.69
θ
Fs
A
10/12 ft FBD
G
30 lb
F C
2 ft
N
(a) The spring force is
Fs = k = 500
MC = IC •
+
500
Fs
34
12
y
34
12
34
12
34
12
4014 + 1416:7y
0:5
4014 + 1416:7
sin 25t
12
5:182• + 4014
y
= m k 2 + R2 •
=
30
(1:252 + 22 )•
32:2
5:182•
=
5:182•
=
59:03 sin 25t J
=
(b) Comparing with M • + K = F0 sin !t, we obtain
r
r
K
4014
p =
=
= 27:83 rad/s
M
5:182
59:03=4014
F0 =K
=
=
= 0:07618 rad (amplitude of )
2
1 (!=p)
1 (25=27:83)2
XG = 24 = 24(0:07618) = 1:828 in. J
11
597
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20.70
20.71
12
598
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.72
20.73
(a)
Oy
O
Ox
b
θ FBD
.
2cbθ
b/2
kb(θs + θ)
IO =
2b
mg
1
m(3b)2 + m
12
b
2
2
= mb2
MO
+
mg
b
2
2cb _ (2b)
The static angular displacement
s
(mg)
kb(
s
+ )b
= IO •
= mb2 •
is given by
b
2
kb2
s
=0
(a)
13
599
© 2010. Cengage Learning, Engineering. All Rights Reserved.
The equation of motion becomes, after utilizing Eq. (a) and cancelling b2 ,
m• + 4c _ + k = 0 J
(b)
Comparing with M • + C _ + K = 0, we obtain
r
r
k
3000
p =
=
= 22:36 rad/s
m
6
4c
mp
6(22:36)
=
ccr =
=
= 67:1 N s/m J
2mp
2
2(1)
20.74
.
chθ
A
mg
x-
h
Bx
MB = IB
B
By
+
FBD
θ
C
kb(θs + θ)
b
mgx
ch2 _
kb2 (
s
+ ) = IB •
Substituting the static angular displacement s = mgx=(kb2 ), the equation of
motion becomes
IB • + ch2 _ + kb2 = 0
Comparison with M • + C _ + K = 0 yields
p2
=
=
K
kb2
=
M
IB
(20 12)(2:5)2
kb2
2
=
2 = 7:709 slug ft
2
p
(2
2:22)
ch2
c(1:0)2
C
=
=
= 0:004 650c
2M p
2IB p
2(7:709)(2
2:22)
Damping is critical when
) IB =
=1
) ccr =
1
= 215 lb s/ft J
0:004 650
14
600
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.75
#20.76
15
601
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.77
16
602
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.78
17
603
© 2010. Cengage Learning, Engineering. All Rights Reserved.
*20.79
*20.80
18
604
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.81
k1Rθ
Oy
O
θ
R
k2Rθ
Vg
=
Ve
=
V
)
=
W
Ox
R/2
G
FBD
W
R
cos
2
W
R
2
1
1
2
2
1
1
k1 (R )2 + k2 (R )2
2
2
WR 1 WR
+
= Vg + Ve =
+ (k1 + k2 )R2
2
2
2
2
WR
40(2:2)
+ (k1 + k2 )R2 =
+ (30 + 40)(2:2)2
2
2
382:8 lb ft
K=
T
)M
1 _2
I
2
=
= I = mk 2 =
)p=
r
K
=
M
r
40
(1:5)2 = 2:795 lb ft s2
32:2
382:8
= 11:70 rad/s J
2:795
20.82
Datum
A
L
V
=
T
=
=
θ
B
Lθ
C
L
1
1
L
k(L )2 2mg
= kL2 2 mgL
) K = kL2
2
2
2
2
2
2
1
1
1
) M = mL2
(IA + IC ) • =
2
mL2 •
2
2
3
3
r
r
r
2
2
M
2mL =3
m
=2
=2
= 5:13
J
2
p
K
kL
k
19
605
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.83
(a)
m
θa
Datum
V
T
b
θ
1 2
1
+
= mga cos + k(b )2 mga 1
2
2
1
kb2 mga 2
) K = kb2
= mga +
2
1
m(a _ )2
) M = ma2
=
2
r
r
K
kb2 mga
p=
=
J
M
ma2
(b)
kb2
mga > 0
k>
1 2
kb
2
2
mga
mga
J
b2
20.84
20
606
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.85
20.86
21
607
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.87
22
608
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.88
Datum
L
θ
R
m
V
=
T
=
1 2
) K = mgL
2
2
_ 2 = 1 m L2 + R
_2
2
2
mg(L cos )
mgL 1
1
1
m(L _ )2 +
2
2
1
mR2
2
R2
) M = m L2 +
2
s
s
r
m L2 + R2 =2
M
2
L2 + R2 =2
=
=2
=2
=2
p
K
mgL
gL
The condition
= 1:4 s yields
s
L2 + R2 =2
=
2
gL
R2
1:42 g
L
+
4 2
2
=
0
1:5987L + 0:055 56
=
0
L2
L2
1:4
4
2
2L
+ R2 =2
= 1:42
gL
1:42 (32:2)
1
L+
2
4
2
L2
1
3
2
=0
The two solutions are
L1 = 0:03554 ft = 0:426 in. J
L2 = 1:5632 ft = 18:76 in. J
*20.89
Datum
α
h
G G'
B
L/2 cosθ
G
θ
L/2
Side view
B
Top view
G'
23
609
© 2010. Cengage Learning, Engineering. All Rights Reserved.
From side view:
h
=
L
cos
2
sin
V
=
mgh =
1
mgL cos sin
2
1
mgL 1
2
1
2
1
mgL sin
2
2
1 _2
1 1
1
T =
IB =
mL2 _
) M = mL2
2
2 3
3
r
r
r
p
K
3g sin
g
1
1
f=
=
=
= 0:1949
sin
2
2
M
2
2L
L
)
2
sin
K=
J
20.90
20.91
Datum position (spring undeformed)
β
R
x+
∆
24
610
© 2010. Cengage Learning, Engineering. All Rights Reserved.
Let
be the static elongation of the spring
V
=
=
)
T
=
)
f=
1
k(x + )2 mg(x + ) sin
2
1
1
k 2 + mg sin + (k
mg sin )x + kx2
2
2
K=k
2
1
x_
1 3
1 1
+ mx_ 2 =
mR2
m x_ 2
2 2
R
2
2 2
3
M= m
2
r
r
r
1
1
K
2k
k
p
=
=
= 0:1299
J
2
2
M
2
3m
m
20.92
Area = A
x
Water level
h
F
The force F of buoyancy is equal to the weight of water displaced:
F =A
water x
Buoyancy acts like a spring of sti¤ness k = A
V
=
T
=
p
water
1
A
x2 mwo o d gx
) K = A water
2 water
1
1
mwood x_ 2 = Ah wo o d x_ 2
) M = Ah wo o d
2
2
g
g
s
=
r
=
2
2
=
= 0:612 s J
p
10:266
K
=
M
r
g
h
water
wo o d
=
32:2(0:036)
= 10:266 rad/s
0:5(0:022)
25
611
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.93
20.94
26
612
© 2010. Cengage Learning, Engineering. All Rights Reserved.
#20.95
27
613
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.96
R
R−r
Datum
ω v
θ
r
v = (R
V
=
T
=
mg(R
r) cos
r) _
!=
mg(R
v
R r_
=
r
r
r) 1
1
2
2
) K = mg(R
r)
2
=
1 2 1
12 2 R r
_ 2 + 1 m(R r)2 _ 2
I! + mv 2 =
mr
2
2
25
r
2
2
1 7
7
) M = m(R r)2
m(R r)2 _
2 5
5
s
r
r
1
K
1
5g
g
p
=
=
= 0:1345
J
f=
2
2
M
2
7(R r)
R r
28
614
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.97
Use Rayleigh’s method
.
θmax
Datum θmax
Position
of Vmax
IC
=
mgh =
=
Tmax
=
R
mg [r + (R
r) 1
r) cos
1 2
2 max
max ]
=
mgR
2
1
1 _2
IC max =
2mR2 _ max = mR2 p2
2
2
Tmax
p
Position
of Tmax
= r + (R r) cos max
= I + mR2 = mR2 + mR2 = 2mR2
mg r + (R
V0
C
R
h
Vmax
R−r
r
h
= Vmax V0
mR2 p2
r
g(R r)
=
J
2R2
2
max
=
1
mgR + mg(R
2
r)
2
max
2
max
1
mg(R
2
r)
2
max
20.98
(a)
x
Fx = max
(b)
p=
p
kx
mg
FBD
N
+ !
40 = 6:325 rad/s
kx
2kx = m•
x
)f =
2(200)x = 10•
x
x
• + 40x = 0 J
p
6:325
=
= 1:007 Hz J
2
2
(0)
29
615
© 2010. Cengage Learning, Engineering. All Rights Reserved.
(c)
x = E sin(pt + )
s
s
2
2
v
( 110)
E =
= 26:50 mm
x20 + 02 = 202 +
p
40
p !
20
x
p
40
0
= tan 1
= 0:8550 rad
= tan 1
v0
110
)
x = 26:50 sin(6:325t
0:8550) mm J
20.99
20.100
30
616
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.101
20.102
20.103
31
617
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.104
20.105
32
618
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.106
20.107
33
619
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.108
20.109
x
mg
k1x
cx.
k2(y − x)
N
Fx = max
FBD
+ ! k2 (y x) k1 x cx_ = m•
x
k2 Y sin !t (k1 + k2 ) x cx_ = m•
x
m•
x + cx_ + (k1 + k2 ) x = k2 Y sin !t
This has the form m•
x + cx_ + kx = P0 sin !t, where
k
P0
p
= k1 + k2 = 10 + 15 = 25 kN/m
= k2 Y = 15(0:028) = 0:420 m
=
=
!
p
=
r
r
k
25 000
=
= 64:55 rad/s
m
6
c
60
=
= 0:07746
2mp
2(6)(64:55)
20
= 0:3098
64:55
34
620
© 2010. Cengage Learning, Engineering. All Rights Reserved.
The steady-state displacement is x(t) = X sin(!t
X
=
=
=
)
q
q
), where
P0 =k
2
2
(!=p)2 ] + (2 !=p)
[1
0:420=25
2
= 0:01856 m
2
0:30982 ) + [2(0:07746)(0:3098)]
(1
2(0:07746)(0:3098)
2 !=p
= tan 1
= 0:0530 rad
2
1 (!=p)
1 0:30982
x(t) = 0:01856(sin 20t 0:0530) m J
x(t) lags y(t) by 0:0530 rad (3:04 ) J
tan
1
20.110
35
621
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.111
36
622
© 2010. Cengage Learning, Engineering. All Rights Reserved.
20.112
20.113
b
az
F
b
FBD
θ
z
a
Oy
x
y,z
O
Ox
=
may
mg
F =k =k
b
+ z
a
ax = a _
2
MAD
a
ma x
O
ay = a• + y• = z• + y•
37
623
© 2010. Cengage Learning, Engineering. All Rights Reserved.
( MO )FBD
F b (mg) a
b
k
+ z b mga
a
b2
mga + k z + ma•
z
a
= ( MO )M AD
=
(may ) a
+
kb
But kb
is
=
m(•
z + y•)a
= maY ! 2 sin !t
mga = 0 (static equilibrium equation), so that the equation of motion
z• +
k b2
z = Y ! 2 sin !t J
m a2
20.114
From Eq. (20.31):
rh
1
2
(!=p)
i2
+ (2 !=p)2
Y
= Z
!
p
18
= 0:9549
2 (3)
q
2
2
(1 0:95492 ) + [2(0:25)(0:9549)]
= 10
0:95492
= 5:32 mm J
)Y
(!=p)2
=
38
624
© 2010. Cengage Learning, Engineering. All Rights Reserved.