Chapter 11
11.1
24 lb
2 = 4:511 slugs J
5:32 ft/s
= mg = 4:511(32:2) = 145:3 lb J
(a) m =
(b) W
11.2
W
1 2
R h g = (0:0752 )(0:125)(2700)(9:81) = 19:503 N
3
3
0:2284 lb
(19:503 N)
= 4:45 lb J
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 of each term must be the same, we have
KE =
[KE] = [M ]
L2
= [M ] k 2
T2
1
T2
1
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Therefore,
[k] = [L]
(a) In the SI system
kg m2
L2
=
J
T2
s2
[KE] = [M ]
[k] = m J
(b) In the US system
[KE]
= [M ]
[k]
= ft J
FT2
L2
=
T2
L
L2
= [F L] = lb ft J
T2
11.6
[g] [k] [x]
1
W
[ ]=
PL
EA
=
L
T2
1
F
[L]
L
F
=
L
= [a] Q.E.D.
T2
11.7
(a)
[L] =
1
E
FL
L2
[E] =
F
L2
J
(b) Substituting [F ] = M L=T 2 into the result of part (a):
[E] =
ML
T2
1
M
=
2
L
T 2L
J
11.8
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
(a)
2
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
11.9
Rewrite the equation as y = 1:0 x2
[y] = [1:0] x2
[L] = [1:0] L2
1
L
[1:0] =
y = x2 can be dimensionally correct only if the units of the implied constant
1:0 are in. 1 . J
11.10
FT2
L
(a) [I] = mR2 =
L2 = F LT 2 J
(b) [I] = mR2 = M L2 J
11.11
(a)
v 3 = [A] x2 + [B] [v] t2
L
J
T3
[A] =
(b)
x2 = [A] t2
h
2
e[B][t ]
[A] =
i
L2
T2
L3
L
= [A] L2 + [B]
T3
T
L2
[B] =
J
T4
L2 = [A] T 2 [1]
J
[B] =
T2
[B] T 2 = [1]
1
T2
J
11.12
dx
d2 x
+c
+ kx = P0 sin !t
2
dt
dt
d2 x
FT2
L
[m]
=
= [F ]
2
dt
L
T2
m
Therefore, the dimension of each term in the expression is [F ].
[c]
dx
L
= [c]
= [F ]
dt
T
[c] =
[k] [x] = [k] [L] = [F ]
[k] =
[P0 ] [sin !t] = [P0 ] [1] = [F ]
[!] [t] = [!] [T ] = [1]
FT
L
J
F
J
L
[P0 ] = [F ] J
[!] =
1
T
J
3
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11.13
F =G
mA mB
R2
G=
[F ] L2
(a) [G] =
L4
=
2
FT4
[F T 2 =L]
(b) [G] =
M L=T 2 L2
L3
=
2
[M ]
MT2
F R2
mA mb
[G] =
[F ] L2
[M 2 ]
J
J
11.14
Using the base dimensions of an absolute [MLT] system:
ML
M
= [C] 3
2
T
L
[F ] = [C][ ][v 2 ][A]
L2
[L2 ]
T2
[C] = [1] Q.E.D J
11.15
F
W
F
W
100%
82
m2
= (6:67 10 11 ) 2 = 2:668 10 8 N
2
R
0:4
= mg = 8(9:81) = 78:48 N
2:668 10 8
=
100% = 3:40 10 8 % J
78:48
= G
11.16
F =G
2
m2
= 3:44
R2
10 8
(2=32:2)
= 7:46
(16=12)2
10 11 lb J
11.17
m=
W R2
(3000) (6378 + 1600)2 106
=
= 479 kg J
GMe
(6:67 10 11 ) (5:9742 1024 )
11.18
GMm
2
Rm
ge =
0:073483
5:9742
6378
1738
gm =
gm
Mm
=
ge
Me
Re
Rm
2
=
GMe
Re2
2
= 0:1656 t
1
Q.E.D
6
4
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
11.19
Me = 5:9742
3:281 ft
= 20:93
1:0 m
103 m
Re = 6378
W =G
0:06853 slugs
= 0:4094
1:0 kg
1024 kg
Me m
2 = 3:44
(2Re )
(2
106 ft
1024 )(150=32:2)
(0:4094
10 8
2
106 )
20:93
1024 slugs
= 37:4 lb J
11.20
F =G
Ms m
= 6:67
R2
10 11
1:9891
1030 (1:0)
(149:6
2
109 )
= 0:00593 N J
11.21
Me m
r2
Me
Ms
5:9742 1024
1:9891 1030
=
0
=
G
Ms m
Me
Ms
=
(R r)2
r2
(R r)2
r2
2
R
2Rr + r2
r2
9
2
(149:6 10 )
2(149:6 109 )r + r2
= G
=
2:238
r = 259
1022
2:992
106 m = 259
1011 r
3:329 4
105 r2
103 km J
5
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Chapter 12
12.1
y
v
=
0:16t4 + 4:9t3 + 0:14t2 ft
= y_ = 0:64t3 + 14:7t2 + 0:28t ft/s
a = v_ =
1:92t2 + 29:4t + 0:28 ft/s
2
At maximum velocity (a = 0):
1:92t2 + 29:4t + 0:28 = 0
vmax
y
t = 15:322 s
=
0:64(15:3223 ) + 14:7(15:3222 ) + 0:28(15:322)
= 1153 ft/s J
=
0:16(15:3224 ) + 4:9(15:3223 ) + 0:14(15:3222 )
= 8840 ft J
12.2
6
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12.65
y = 64:4(4
x_ =
y_ =
128:8t ft/s
y• =
128:8 ft/s
x
•=
Fx
Fy
t2 ) ft
x = 20 cos
t ft
2
10 sin t ft/s
2
2
5 2 cos t ft/s
2
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 b
4 t
cos
t20
t0
1 + cos
At point B: x = 0 ) t = 0
) ax = 0
) ay =
4 2b
=
t20
4 2 (1:2)
=
0:82
74:02 m/s
2
7
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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
8
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12.69
x
y
g
20o
From the acceleration diagram of a water droplet:
ax
=
g sin 20 =
32:2 sin 20 =
11:013 ft/s
ay
=
g cos 20 =
32:2 cos 20 =
30:26 ft/s
2
2
Initial conditions at t = 0 :
x = y=0
vx = 22 cos 30 = 19:053 ft/s
vy = 22 sin 30 = 11:0 ft/s
Integrating and using initial conditions:
vx =
x =
11:013t + 19:053 ft/s
5:507t2 + 19:053t ft
vy = 30:26t + 11:0 ft/s
y = 15:13t2 + 11:0t ft
Droplet lands when y = 0:
y
R
=
15:13t2 + 11:0t = 0
t = 0:7270 s
= xjt=0:7270s = 5:507(0:72702 ) + 19:053(0:7270) = 10:94 ft J
9
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12.70
From Eqs. (e) of Sample Problem 2.11:
x =
y
=
=
(v0 cos )t = (65 cos 55 ) t = 37:28t ft
1 2
32:2 2
gt + (v0 sin )t =
t + (65 sin 55 )t
2
2
16:1t2 + 53:24t ft
At point B:
x = 60 ft
60 = 37:28t
t = 1:6094 s
h = yjt=1:6094s = 16:1(1:60942 ) + 53:24(1:6094) = 44:0 ft J
*12.71
From Sample Problem 12.12:
x = C1 e ct=m + C2
c
m
C1
mgt
+ C4
c
mg
c
C3 e ct=m
m
c
y = C3 e ct=m
c ct=m
e
m
vy =
vx
=
=
0:0025
= 0:06708 s 1
1:2=32:2
mg
1:2
=
= 480:0 ft/s
c
0:0025
x = C1 e 0:06708t + C2
y = C3 e 0:06708t
vx =
C1 0:06708e 0:06708t
vy
=
C3 0:06708e 0:06708t
v0 sin
v0 cos
=
=
480t + C4
480:0
70 sin 65 = 63:44 ft/s
70 cos 65 = 29:58 ft/s
Initial conditions at t = 0:
x
y
vx
vy
=
=
=
=
0 ) C2 = C1
0 ) C4 = C3
v0 cos
) C1 (0:06708) = 29:58
C1 =
v0 sin
) C3 (0:06708) 480:0 = 63:44
441:0 ft
C3 = 8101 ft
When x = 60 ft:
60 =
441:0e 0:06708t + 441:0
t = 2:180 s
(0:06708)(2:180)
h = yjt=2:180 = 8101e
480(2:180) + (8101) = 55:7 ft J
10
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.72
y
x
g
Acceleration
diagram
At t = 0 (initial conditions):
x =
y =
0
vx = 200 sin 30 = 100 m/s
1200 m
vy = 200 cos 30 = 173:21 m/s
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
593:2 = 99:6 m J
d = 1200 tan 30
12.73
Eqs. (d) and (e) of Sample Problem 12.11:
x = v0 t cos = 2500t cos ft
1 2
y = v0 t sin
gt = 2500t sin
2
Setting x = R = 5280 ft and solving for t:
5280 = 2500t cos
t=
16:1t2 ft
5280
2500 cos
=
2:112
s
cos
2:112
cos
Setting y = 0, we get after dividing by t:
2500 sin
16:1t =
0
2500 sin
16:1
2:112
2500
= 0:013 601
sin cos
=
16:1
1
sin 2
2
=
0:013 601
=
1
sin 1 (0:02720) = 0:779
2
=0
sin 2 = 2(0:013 601) = 0:02720
J
11
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12.74
From Eqs. (e) of Sample Problem 12.11:
x = (v0 cos ) t
1 2
gt + (v0 sin ) t
2
y=
(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
=
ymax
=
0:02342x + 0:5318 = 0
x = 22:71 ft
2
0:01171(22:71) + 0:5318(22:71) = 6:04 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
1 2
gt + (v0 sin ) t
2
y=
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
2
16:1(0:02918v0 ) + 0:9397v0 (0:02918v0 ) = 27
v0 = 44:4 ft/s J
0:013712v02 = 27
12
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12.76
Equations (e) of Sample Problem 12.10:
x =
y
=
=
(v0 cos )t = (30 cos 60 )t = 15t ft
1 2
1
gt + (v0 sin )t =
(32:2)t2 + (30 sin 60 )t
2
2
16:1t2 + 25:98t ft
vx = x_ = 15 ft/s
y
vy = y_ =
32:2t + 25:98 ft/s
B
v
30o
30o
h
y
30o
x tan 30o
30o
x
x
At point B:
v is parallel to the inclined surface
32:2t + 25:98
= tan 30
15
)
vy
= tan 30
vx
t = 0:5379 s
x = 15(0:5379) = 8:069 ft
y =
16:1(0:53792 ) + 25:98(0:5379) = 9:316 ft
h = (y x tan 30 ) cos 30 = (9:316 8:069 tan 30 ) cos 30
= 4:03 ft J
13
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12.77
*12.78
14
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12.79
(a)
y
mg vx
x =
max
FD(vy /v)
MAD
FBD
FD(vx /v)
FD
+
!
Fx = max
ax
=
=
ay
=
=
may
v
vy
FD
0:0005v 2 vx
FD vx
=
=
m v q
0:1
v
0:005vx
vx2 + vy2 m/s
Fy = may
+"
FD vy
g=
m v q
0:0005v 2 vy
0:1
v
0:005vy
vx
= max
v
vx2 + vy2
FD
2
vy
v
9:81 m/s
J
mg = may
9:81 =
2
0:005vvx
0:005vvy
9:81
J
15
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(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
2
2
x_ 4 = 0:005x4 x23 + x24 9:81
x_ 3 = 0:005x3 x3 + x4
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
Linear interpolation for h:
30:405
24:117
29:607
30
=
23:944
h
29:607
23:944
h = 24:0 m J
29:607
15:990
vx = 15:958 m/s
Linear interpolation for vx and vy :
30:405
15:925
29:607
30
=
15:990
vx
30:405
3:1952
29:607
30 29:607
=
vy = 3:451 m/s
3:6997
vy 3:6997
p
) v = 15:9582 + 3:4512 = 16:33 m/s J
16
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.80
(a)
F y
F(y/d)
may
x
max
F(x/d) =
FBD
MAD
x
= max
d
1 x
0:5x
1 0:005 x
x
2
ax = F =
m/s J
= 0:5 3 =
3=2
2
2
m d
0:01 d2 d
d
(x + y )
y
Fy = may + "
F = may
d
1 y
1 0:005 y
y
0:5y
2
m/s J
ay = F =
= 0:5 3 =
3=2
2
2
m d
0:01 d2 d
d
(x + y )
Fx = max
+
!
F
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
3=2
2
(x1 + x22 )
x_ 4 =
0:5x2
3=2
2
(x1 + x22 )
The MATLAB program for solving the equations is
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
17
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
3=2
(x2 + y 2 )
m/s
2
J
ay =
0:5y
3=2
(x2 + y 2 )
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)
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
x = 0:257 m J
18
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Linear interpolation for vx :
0:67692 ( 0:63935)
vx
=
0:0031544 0:0085117
0
By inspection, vy =
12.82
( 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
0:0012
cD
0:5
=
= 0:06869 (ft s)
m
(9=16)(32:2)
Fx = max
Fy = may
+
!
FD
) ax =
0:06869vx vx2 + vy2
+"
FD
vy
y
cD p
vy v g =
m
The initial conditions are:
) ay =
x=0
y = 6 ft
0:25
ft/s
J
cD v 1:5
mg = may
0:06869vy vx2 + vy2
vx = 120 ft/s
2
0:25
vy = 0
vy
y
mg = may
32:2 ft/s
2
J
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
=
0:06869x3 x23 + x24
0:25
x4
=
0:06869x4 x23 + x24
0:25
32:2
The MATLAB program is
19
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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:
0:66 0:64
t 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:
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)
20
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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:
b
5:1656 5:7152
=
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
21
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.84
(a)
y
mg
x
R
y
θ
Spring force is F = k(R
x
F
MAD
L0 ), where R =
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
=
FBD
Fx = max
ax
θ
may
=
F
sin
m
g=
=
40 1
0:5
R
+"
k(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
x1
40 1
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)
22
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
40 1
x m/s if R > 0:5 m
ax =
J
R
:
0
if R 0:5 m
8
< 40 1 0:5 y 9:81 m/s2 if R > 0:5 m
ay =
J
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]);
23
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
ay =
aD sin
0.6
x (m)
12.86
(a)
ax =
aD cos + aL sin
aL cos
g
Substitute
sin
=
aL
=
vy
vx
cos =
aD = 0:05v 2
v
v
0:16!v = 0:16(10)v = 1:6v
where v = vx2 + 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
24
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(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:
t 1:04
1:06 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
25
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(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 0 20
-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
26
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.88
x (m)
x = 10t 2
4000
0
0
v (m/s)
400
t (s)
20
0
0
t (s)
20
a (m/s2 )
20
0
0
v
t (s)
20
= slope of x diagram
a = slope of v diagram
vj20s =
a=
dx
= 400 m/s
dy 20s
400
2
= 20 m/s
20
12.89
It is su¢ cient to consider vertical motion only:
a=
g=
32:2 ft/s
2
Initial conditions:
vjt=0 = v0 sin 60 = 0:8660v0 ft/s
yjt=0 = 0
27
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
a (ft/s2)
0 0
t1
t (s)
-32.2t1
-32.2
v (ft/s)
0.8660v0
0
0.4330v0t1
0
t1
t (s)
y (ft)
27
0
0
t1 t (s)
End conditions (t1 is the time when the ball is at its maximum height):
vjt=t1
yjt=t1
= 0
0:8660v0 32:2t = 0
t1 = 0:026 89v0
2
= 27 ft
0:4330(0:026 89)v0 = 27
v0 = 48:2 ft/s J
28
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.90
a (ft/s2 )
00
t1
6
−24
t (s)
−8 (t1 −6)
-8
v (ft/s)
88
64
96
384
42
0
6
x (ft)
480
t1 = 14 s t (s)
522
0
0
6
14
t (s)
From a and v diagrams
8(t1 6) =
64
t1 = 14:0 s J
From x diagram xjt1 = 522 ft J
29
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.91
Horizontal motion (ax = 0)
Vertical motion (ay =
ax (m/s2)
0
t (s)
0
vx (m/s)
240
0
ay (m/s2)
0
0
-9.81
9:81 m/s2 )
t1
-9.81t1
t (s)
vy (m/s)
240t1
t1
0
5
00
t (s)
t (s)
1
9.81
−49.05
y (m)
h
x (m)
1200
0
0
0
0
t1 t (s)
t (s)
5
End condition:
xjt=t1
=
h
=
1200 m
240t1 = 1200
t1 = 5:0 s
1
[area under vy diagram] = (49:05)(5) = 122:6 m J
2
12.92
Horizontal motion: ax = 0 Vertical motion: ay =
32:2 ft/s
2
30
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
ax (ft/s2)
0
0
10
ay (ft/s2)
00
t (s)
-32.2
vx (ft/s)
v0
0
10
-322
t (s)
vy (ft/s)
10v0
0
10
10
00
t (s)
-1610
t (s)
-322
y (ft)
1610
x (ft)
10v0
0
0
0
0
10 t (s)
vy
= tan 20
vx t=10
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
12.93
y
22.6o
g
x
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
vx = 260 cos 22:6 = 240:0 ft/s
vy = 260 sin 22:6 = 99:92 ft/s
31
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
ax (ft/s)
12.374
0
0
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)
167.92
99.92
t (s)
0
-99.92
1892.8
x (ft)
1892.8
t (s)
y (ft)
167.92
0
t (s)
vy = 0 at t = t1 .
h = 167:9 ft J
) 99:92
t (s)
0
29:73t1 = 0
R = 1893 ft J
) t1 = 3:361 s
time of ‡ight = 6:72 s J
12.94
ax (m/s2)
0
0
16.091
ay (m/s2)
00
t (s)
-9.81
vx (m/s)
190.52
0
0
t (s)
vy (m/s)
110
3066
16.091
x (m)
3066
0
0
t1
-9.81t1
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
32
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.95
33
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
34
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.97
35
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.98
a (m/s2)
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
a (m/s2)
12
0.6
0.6
0
0
0.1
0.2
0.3
0.4
t (s)
0.1
0.2
0.3
0.4
t (s)
v (m/s)
1.2
0.6
00
From the v diagram:
vj0:4s = 1:2 m/s J
36
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
xj0:4s
= area under v diagram
1
= 2 (0:6)(0:1) + 4(0:6)(0:1) = 0:28 m J
3
12.100
37
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.101
38
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.102
a (m/s2)
0.38
0.10182
00
t1 = 0.5359
1.1
t (s)
−0.11282
−0.40
v (m/s)
0.10182
0.03638
0
−0.011
0.5359
0.03638
1.0718
t (s)
y (m)
0.07276
0.03638
t (s)
0
From similar triangles on the a-diagram:
t1
0:38
vmax
=
=
1:1
t1 = 0:5359 s
0:78
0:1018 m/s J
ymax = 0:0728 m J
39
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.103
a (ft/s2 )
60
t1
t2
14
t (s)
0 0273.3
-32.2(
t
2 - 14)
-32.2
-78.7
= -194.6
v (ft/s)
273.3
194.6
1660 1208
588
0
t (s)
20.04
9.111 14
0
y (ft)
3456
2868
1660
t (s)
0
From similar triangles on a-diagram:
14
t1
=
60
60 + 32:2
) t1 = 9:111 s
Let t2 be the time when v = 0: Therefore,
194:6
vmax = 273 ft/s J
32:2(t2
14) = 0
) t2 = 20:04 s
ymax = 3460 ft J occurring at t = 20:0 s J
12.104
v = 2x3
8x2 + 12x mm/s
dv
= 2x3 8x2 + 12x 6x2 16x + 12
dx
2
ajx=2 = (2(8) 8(4) + 12(2)) (6(4) 16(2) + 12) = 32:0 mm/s J
a=v
12.105
a = At + B
When t = 0: a = 0 ) B = 0
4
When t = 6 ft/s: a = 8 ft/s ) 8 = A(6) ) A = ft/s3
3
Z
4
2
2
) a = t ft/s
v = a dt + C = t2 + C
3
3
When t = 0: v = v0 ) C = v0
When t = 6 s: v = 16 ft/s
)
2
(36) + v0 = 16
3
) v0 =
8:0 ft/s J
40
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.106
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 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 overtakes car B when xA = xB :
2t2 + 30t
3t2 30t
400 =
t2 + 60t
400 = 0
t = 17:58 s J
12.107
(a)
x = 3t3 9t + 4 in.
x = xjt=2 xjt=0 = [3(8)
(b)
v = x_ = 9t2
-2
) v = 0 when t = 1:0 s
9 in./s
xjt=0 = 4 in.
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
t 1 + t2 = 3
16:1t21 = 1120t2
) t2 = 3
16:1t21 = 1120(3
t1 )
t1
t1 = 2:881 s
h = yjt=2:881s = 16:1(2:881) = 133:6 ft J
2
41
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12.109
a =
2
When t = 0: x = v = 0:
) C1 = C2 = 0
12t 6t2 ft/s
Z
v =
a dt = 6t2 2t3 + C1 ft/s
Z
x =
v dt = 2t3 0:5t4 + C1 t + C2 ft
) x = 2t3
0:5t4 ft
) v = 6t2
xjt=0 = 2(5)3
0:5(5)4
2t3 ft/s
(a)
x = xjt=5
62:5 ft J
0=
(b) When v = 0:
6t2
xjt=0 = 0
xjt=3 = 2(3)
2t3 = 0
3
t = 3:0 s
4
0:5(3) = 13:5 ft xjt=5 =
0
-62.5
13.5
62:5 ft
x (ft)
d = 2(13:5) + 62:5 = 89:5 ft J
12.110
dv
=
dt
cv
2
1
dv =
v2
c dt
Z
Initial condition: vjt=0 = 800 ft/s
When v = 400 ft/s:
1
v
1
800
1
dv =
v2
)
1
=
400
1
=
v
ct + C1
1
=
v
1
=
800
C1
0:8t
t = 1:5625
)
ct + C1
0:8t
1
800
10 3 s J
1
1
0:8t +
v=
800
0:8t + 1=800
Z
dt
1
1
x =
+ C2 =
ln 0:8t +
0:8t + 1=800
0:8
800
1
1
Initial condition:
xjt=0 = 0 ) 0 =
ln
+ C2
0:8
800
1
1
C2 =
ln
= 8:356 ft 1
0:8
800
1
1
x =
ln 0:8t +
+ 8:356
0:8
800
=
+ C2
42
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When v
=
x =
10 3 s :
400 ft/s t = 1:5625
1
ln 0:8(1:5625
0:8
10 3 ) +
1
+ 8:356 = 0:867 ft J
800
12.111
75e 0:05t ft/s
Z
x =
75e 0:05t dt =
v
Initial condition:
=
1500e 0:05t + C ft
xjt=0 = 0 ) C = 1500 ft
x = 1500(1 e 0:05t ) ft
Setting v = 5 ft/s and solving for t:
5 = 75e 0:05t
t = 54:16 s J
h
i
xjv=5 ft/s = 1500 1 e 0:05(54:16) = 1400 ft J
12.112
may
max
mg
FBD
Fx
Fy
MAD
= max
= may
vx =
Initial condition: vx jt=0 = v0 cos 50
ax = 0
ay = g =
Z
9:81 m/s
2
ax dt = C1
) C1 = v0 cos 50 = 0:6428v0
vx = 0:6428v0
Z
x = vx dt = 0:6428v0 t + C2
Initial condition: xjt=0 = 0
) C2 = 0
x = 0:6428v0 t
Z
vy = ay dt = 9:81t + C3
43
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Initial condition: vy jt=0 = v0 sin 50
) C3 = v0 sin 50 = 0:7660v0
vy = 9:81t + 0:7660v0
Z
y = vy dt = 4:905t2 + 0:7660v0 t + C4
Initial condition: yjt=0 = 0
) C4 = 0
4:905t2 + 0:7660v0 t
y=
At point B:
x = 18 m ) 0:6428v0 t = 18
v0 t = 28:0 m
y = 18 m ) 4:905t2 + 0:7660(28:0) = 18
t = 0:8384 s
28:0
= 33:4 m/s J
v0 =
0:8384
12.113
12.114
F = (50
+"
a =
v
=
y
=
F
m
Z
Z
g=
(50
t)
103 N
t) 103
1400
9:81 = 25:90
a dt = 25:90t
0:3571t2 + C1
v dt = 12:95t2
0:11903t3 + C1 t + C2
0:7143t m/s
2
Initial conditions: y = v = 0 when t = 0. ) C1 = C2 = 0:
yjt=20 = 12:95(20)2
0:11903(20)3 = 4230 m J
44
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12.115
12.116
12.117
From Eqs. (d) and (e) of Sample Problem12.11:
x = v0 t cos = 8t cos 30 = 6:928t
1 2
1
y =
gt + v0 t sin =
(9:81)t2 + 8t sin 30 =
2
2
4:905t2 + 4t
45
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At the landing point
y
=
x =
4:905t2 + 4t = 6:928t tan 20
t = 1:330 s J
9:214
6:928(1:330) = 9:214 m
d=
= 9:81 m J
cos 20
x tan 20
4:905t = 6:522
12.118
0.3 lb y
F6
FBD N
45
x
0.3 a
32.2
3
MAD
When x = 6 in., the elongation of the spring and the spring force are
p
= 32 + 62 3 = 3:708 in.
F = k = 5(3:708) = 18:540 lb
Fx
6
0:3
p F =
a
32:2
45
2
1780 ft/s J
= ma
a =
6
0:3
p (18:540) =
a
32:2
45
12.119
46
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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 0
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=
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
47
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12.122
48
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12.3
x =
v
= x_ = 6
a = v_ =
(a) xmax
=
vmax
=
e t=2
6 1
m
1 t=2
e
= 3e t=2 m/s
2
1 t=2
2
e
= 1:5e t=2 m/s
3
2
6 m at t = 1 J
3 m/s and jajmax = 1:5 m/s both occurring at t = 0 J
2
(b) When x = 3 m: 3 = 6(1 e t=2 )
t =
2 ln(0:5) = 1:3863 s J
v
=
e t=2 = 0:5
3(0:5) = 1:5 m/s J
a=
x = t3 6t2
v = x_ = 3t2
32t in.
12t 32 in./s
a = v_ = 6t
12 in./s
1:5(0:5) =
0:75 m/s
2
J
12.4
2
At t = 10 s:
x = 103 6(102 ) 32(10) = 80 in. J
v = 3(102 ) 12(10) 32 = 148 in./s J
a =
6(10)
2
12 = 48 in./s
J
Reversal of velocity occurs when v = 0 (t 6= 0):
v =
x =
3t2 12t 32 = 0
5:8303 6(5:8302 )
t = 5:830 s
32(5:830) = 192:3 in.
At t = 10 s the distance travelled is
s = 192:3 + (193:3 + 80) = 466 in. J
−192.3 in.
0
5.83 s
80 in.
10 s
49
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
x = 3t2
12t in.
v = x_ = 6t
12 in./s
(a) The bead leaves the wire when x = 40 in.
3t2
t = 6:16 s J
12t = 40
(b) Reversal of velocity occurs when v = 0 (t 6= 0):
v =
x =
6t 10
3(2:02 )
t = 2:0 s
12(2:0) = 12:0 in.
The distance travelled is
s = 2(12) + 40 = 64:0 in. J
−12 in. 0
40 in.
6.16 s
2.0 s
50
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.8
x =
y
=
4t2 2 mm
x2
16t4 16t2 + 4
4t4
=
=
12
12
4t2 + 1
mm
3
When t = 2 s:
vx
vy
v
= 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
ax
= v_ x = 8 mm/s
48t2 8
48(2)2 8
2
ay = v_ y =
=
= 61:33 mm/s
3
3
q
p
2
a =
a2x + a2y = 82 + 61:332 = 61:9 mm/s J
12.9
12.10
y = 50
x=
2t s
6
6
=
in.
y
50 2t
vy = y_ =
vx = x_ =
2 in./s
3
ay = v_ y = 0
in./s
(25
t)2
vy =
2 in./s
ax = vx =
6
(25
3
t)
At t = 20 s:
vx
=
ax
=
3
20)2
(25
6
(25
= 0:12 in./s
3 = 0:048 in./s
20)
2
ay = 0
2j in./s J
v = 0:12i
2
a = 0:048i in./s
J
51
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12.11
12.12
x =
15
2t2 m
y
15
10t + t2 m
=
(a) At t = 0:
v=
(b) At t = 5 s:
v=
vx = x_ =
4t m/s
vy = y_ =
10j m/s J
ax = v_ x =
10 + 2t m/s
a=
20i m/s J
a=
4 m/s
2
ay = v_ y = 2 m/s
2
4i + 2j m/s2 J
4i + 2j m/s2 J
12.13
x =
y
(a)
(b)
(c)
=
58t m
vx = x_ = 58 m/s
78t
2
4:91t m
vy = y_ = 78
9:82t m/s
ay = v_ y =
9:82 m/s
2
a = 9:82j m/s2 J
vjt=0 = 58i + 78j m/s J
y = h when vy = 0:
vy
h
(d)
ax = v_ x = 0
= 78 9:82t = 0
t = 7:943 s
= 78(7:943) 4:91(7:9432 ) = 310 m J
x = L when y =
140 m:
y = 78t 4:91t2 = 140
t = 17:514 s
L = 58(17:514) = 1016 m J
52
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12.14
y=
x2
1000
dy
x
=
dx
500
)2 v0
x
d
/
(dy dy/dx
1+
1
vx
1
= v0 q
2
1 + (dy=dx)
vy
dy=dx
= v0 q
2
1 + (dy=dx)
When x = 100 m:
vx
=
vy
=
v
=
v0
=p
1 + (x=500)2
x=500
=p
1 + (x=500)2
=p
500v0
5002 + x2
=p
xv0
5002 + x2
500(6)
= 5:883 m/s
5002 + 1002
100(6)
p
= 1:1767 m/s
5002 + 1002
5:88i + 1:177j m/s J
p
ax
= v_ x =
dvx dx
dvx
=
vx =
dx dt
dx
ay
= v_ y =
dvy dx
dvy
v
=
vx =
3=2 x
dx dt
dx
(5002 + x2 )
500xv0
v
3=2 x
(5002 + x2 )
5002 v0
When x = 100 m:
ax
=
ay
=
a =
500(100)(6)
3=2
(5002 + 1002 )
5002 (6)
3=2
(5002 + 1002 )
(5:883) =
0:013 31 m/s
(5:883) = 0:0666 m/s
0:013 32i + 0:666j m/s
2
2
2
J
53
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12.15
12.16
54
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.17
12.18
55
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.19
12.20
56
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.21
12.22
12.23
12.24
x = R cos
y = R sin
vy
v0
R cos
v0
vx = ( R sin )
=
R cos
= v0
)
vx = x_ = ( R sin ) _
vy = y_ = (R cos ) _
yields
_=
v0 tan
57
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
ax = v_ x =
_=
v0 sec2
With R = 6 ft, v0 = 2:5 ft/s and
v0 sec2
v0
R cos
= 2:5 ft/s
vx = 2:5 tan 60 =
=
4:33i + 2:5j ft/s J
ay
=
a =
v02
sec3
R
= 60 we get
vy
v
0
=
2:52
sec3 60 =
6
2
8:33i ft/s J
ax =
4:330 ft/s
8:333 ft/s
2
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 mm
dr _
v = r_ =
= ( 10 sin
10 sin 2 ) (125:66) mm/s
d
dv _
2
a = v_ =
= ( 10 cos
20 cos 2 )(125:66)2 mm/s
d
2
2
jajmax = 30(125:66)2 = 474 000 mm/s = 474 m/s (at = 0) J
*12.26
58
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.27
T
ma
=
mg
FBD
MAD
v = 4t m/s
a = v_ = 4 m/s
F = ma + "
T
2
mg = ma
T = m(g + a) = 50(9:81 + 4) = 691 N J
12.28
y
mg
x
F = µk N
ma
=
FBD
N
MAD
1000 m/km
3600 s/h
v0 = 100 km/h = 100 km/h
Fy
=
0 +"
Fx
= ma
+
!
v
=
N
) N = mg
mg = 0
Z
x =
a dt =
kN
)a=
k N = ma
Z
= 27:78 m/s
m
=
kg
k gt + C1
1
2
v dt =
2
k gt + C1 t + C2
When t = 0 (initial conditions):
x=0
)x=
) C2 = 0
v = v0 ) C1 = v0
1
gt2 + v0 t
v=
k gt + v0
2 k
When v = 0:
k gt + v0 = 0
x =
1
2
v0
kg
kg
)t=
v0
kg
v0
kg
=
2
+ v0
v02
2 kg
2
=
27:78
= 60:5 m J
2(0:65)(9:81)
59
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.29
y
mg 5o
x
=
F = µk N
FBD
ma
MAD
N
v0 = 100 km/h = 27:78 m/s
= 0 +"
N mg cos 5 = 0
) N = mg cos 5
+
= ma !
k N + mg sin 5 = ma
N
k
)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 =
5:497t + C1
v dt =
2:749t2 + C1 t + C2
When t = 0 (initial conditions):
x=0
) C2 = 0
v = v0
2:749t2 + 27:78t m
5:497t + 27:78 m/s
x =
v =
When v = 0:
5:497t + 27:78 = 0
x=
) C1 = v0 = 27:78 m/s
) t = 5:054 s
2:749(5:054)2 + 27:78(5:054) = 70:2 m J
12.30
a =
v
=
x =
F
1:2t
=
=
0:1
Zm
Z
12t m/s
2
ax dt =
6t2 + C1 m/s
vx dt =
2t3 + C1 t + C2 m
When t = 0 (initial conditions):
x = 0 ) C2 = 0
v = 64 m/s ) C1 = 64 m/s
60
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
)x=
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
Distance traveled:
d = 2(139:35)
128 = 150:7 m J
139.35
0
x (m)
128
12.31
a =
dt
=
p
p
0:06 v
F
2
=
= 5 v m/s
m
0:012
Z
dv
dv
dv
2p
p =
= p
t=
v + C1
a
5
5 v
5 v
Given v = 0:25 m/s when t = 0:8 s:
0:8 =
2p
0:25 + C1
5
C1 = 0:6 s
2p
v + 0:6 s
5
2
v = (2:5t 1:5) = 6:25t2 7:5t + 2:25 m/s
Z
6:25 3 7:5 2
x =
v dt =
t
t + 2:25t + C2
3
2
t
=
Initial condition: x = 0 when t = 0
xjt=1:2s =
6:25
(1:2)3
3
) C2 = 0
7:5
1:22 + 2:25(1:2) = 0:90 m J
2
12.32
T
cv2
FBD
=
ma
MAD
61
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
ma = T
cv 2
FD = T
a=
cv 2
m
T
Z
Z
m
v
v
dv + C =
x=
dv = m
ln T cv 2 + C
a
T cv 2
2c
Initial condition: v = 0 at x = 0:
m
m
ln (T ) + C
C=
ln (T )
0 =
2c
2c
m
m
m
T
) x=
ln(T cv 2 ) +
ln(T ) =
ln
2c
2c
2c (T cv 2 )
Solve for v:
T
(T cv 2 )
v2
2c
x
m
=
exp
=
T
1
c
cv 2 = T exp
s
T
2c
x
m
exp
v=
Terminal velocity:
v1 = lim v(x) =
x!1
r
2c
x
m
T
1
c
exp
2c
x
m
J
T
J
c
12.33
4t 4
F
2
=
= t 1 m/s
m
4
Z
1
=
a dt = t2 t + C1
2
Z
1 2
1
t + C1 t + C2
=
v dt = t3
6
2
a =
v
y
When t = 0 (initial condition):
y = 0 ) C2 = 0
v=
1 2
t
2
8t m
v=
1
3
(8)
6
1
2
(8)
2
8 (8) =
)y=
1 3
t
6
When t = 8 s:
y=
When v = 0:
1 2
t
2
y=
1
3
(5:123)
6
t
8 m/s ) C1 =
8 m/s
1 2
t
2
8 m/s
t
10:67 m
8=0
t = 5:123 s
1
2
(5:123)
2
8 (5:123) =
Distance traveled:
d = 2(31:70)
31:70 m
10:67 = 52:7 m J
-31.70
0
y (m)
-10.67
62
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
12.37
Let y be measured up from the base of the cli¤.
y
ma
x
=
FBD
MAD
mg
63
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Fy = ma + "
dv
v
dy
)a=
mg = ma
g dy
1 2
v =
2
Initial condition: v = v0 when y = h: ) C =
1 2
v + gh
2 0
a=
) v dv =
)
At impact y = 0
)
1 2
v
2
1 2
v
2
v02 = g(h
)v=
v02 = gh
g
gy + C
y)
p
v02 + 2gh J
12.38
F
=
FBD
F
1 2
v
2
ma
MAD
F
F0 x=b
F0 x=b
=
e
v dv =
e
dx
m
m
m
Z
F0
F0 b x=b
e x=b dx =
e
+C
m
m
= ma
=
a=
Initial condition
:
1 2
v
2
=
v
vjx=1:8 ft =
s
v = 0 at x = 0
)C=
F0 b
m
F0 b
1 e x=b
m
r
2F0 b
1 e x=b
=
m
2(1576)(2)
1
6:62 10 3 =32:2
e 1:8=2:0 = 4270 ft/s J
12.39
20o
18 lb
6t
y
FBD
18 a
g
=
x
MAD
N
64
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Fx
= ma
32:2
(6t
Z18
a =
v
6t
=
Z
x =
18 sin 20 =
18
a
32:2
18 sin 20 ) = 10:733t
11:013 ft/s
a dt = 5:367t2
11:013t + C1 ft/s
v dt = 1:789t3
5:507t2 + C1 t + C2 ft
2
) C1 = C2 = 0
Initial conditions: x = v = 0 at t = 0
(a) When v = 0:
v = 5:367t2 11:013t = 0
t = 2:052 s
x = 1:789 2:0523
5:507(2:0522 ) = 7:73 ft J
(b) When x = 0:
x = 1:789t3 5:507t2 = 0
t = 3:078 s
2
v = 5:367(3:078 ) 11:013(3:078) = 16:95 ft J
12.40
x
4
ma
3
P = 8 - 2t
FBD mg N
=
4
P
5
Fx = ma + %
a=
4P
5m
v
3
4 8 2t
g=
5
5 5=32:2
=
x =
Initial conditions: v =
When x = 0 :
10:95t2
Z
Z
MAD
3
mg = ma
5
3
(32:2) = 21:90
5
10:304t ft/s
a dt = 21:90t
5:152t2 + C1
v dt = 10:95t2
1:7173t3 + C1 t + C2
10 ft/s, x = 0 at t = 0. ) C1 =
1:7173t3
v = 21:90(1:1049 )
10t = 0
5:152(1:1049 )2
10 ft/s
2
C2 = 0
t = 1:1049 s J
10 = 7:91 ft/s J
65
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.41
(mA + mB)g
A
y
θ
x
=
B
(mA + mB) a
NB FBD
MAD
Fx = ma + &
(mA + mB )g sin = (mA + mB )a
a = g sin
Assume impending slipping between A and B:
mA g
y
=
A
x
Fx = ma + &
F =µsNA
NA FBD
(mA g
θ ma
A
MAD
NA ) sin +
s = tan
s NA cos
= mA g sin
J
12.42
(a) Assume impending sliding of crate to the left.
y 20o
60 lb
60 a
g
x
FA = 0.3NA
NA
FBD
Fy
=
Fx
= max
MAD
0
NA cos 20
NA = 67:35 lb
0:3NA sin 20
NA sin 20 + 0:3NA cos 20
67:35(sin 20 + 0:3 cos 20 )
a = 11:54 ft/s
2
60 cos 20 = 0
60 sin 20 =
60 sin 20 =
60
a
32:2
60
a
32:2
J
66
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(b) Assume impending sliding of crate to the right.
y
20 o 60 lb
x
60 a
g
FA = 0.3NA
NA
FBD
Fy
=
Fx
= max
MAD
0
NA cos 20 + 0:3NA sin 20
NA = 54:09 lb
NA sin 20
54:09 (sin 20
0:3NA cos 20
0:3 cos 20 )
a = 9:27 ft/s
2
60 cos 20 = 0
60 sin 20 =
60 sin 20 =
60
a
32:2
60
a
32:2
J
12.43
mg
=
ma
0.2 N
N
FBD
MAD
Fy = 0
N mg = 0
N = mg
Fx = 0
0:2N = ma
0:2mg = ma
2
a =
0:2g = 1:962 m/s
Z
1 2
v dv = a dx
v = 1:962 dx = 1:962x + C
2
1 2
Initial condition: vjx=0 = 6 m/s
(6 ) = C
C = 18 m2 =s2
2
)
v
=
1 2
v = 1:962x + 18
2
0 when
1:962x + 18 = 0
x = 9:17 m J
67
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.44
W
F
ma
=
µkΝ
N
FBD
Fy
Fx
= 0
= 0
F
a =
MAD
+" N W =0
N = W = 3000 lb
+! F
k N = ma
0:2t
N
1000e
0:05(3000)
k
=
m
3000=32:2
=
10:733e 0:2t
v
Z
=
=
1:61 ft/s
a dt =
Z
2
(10:733e 0:2t
53:67e 0:2t
1:61)dt
1:61t + C ft/s
Initial condition: v = 0 when t = 0. ) C = 53:67 ft/s
e 0:2t )
v = 53:67(1
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
mg
kx
30o
µk N
Fy
Fx
= 0
= ma
x
=
ma
N
FBD
MAD
mg cos 30
N =0
mg sin 30
kN
N = mg cos 30
kx = ma
a = g (sin 30
a =
y
9:81(sin 30
k
x
m
25
0:3 cos 30 )
x = 2:356
2:5
k cos 30 )
10x m/s
2
68
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Z
v dv
= a dx
1 2
v =
2
1 2
v
2
=
5x2 + C
2:356x
Initial condition:
(2:356
)C=0
vjx=0 = 0
) x = 0:471 m J
2
v = 0 when 2:356x
10x)dx
5x = 0
12.46
ma
40o P
N
mg µN
FBD
y
=
x
MAD
) 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 =
a=
0=
P
(cos 40
5
sin 40 )
s = 0:5
0:5 sin 40 )
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
69
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12.48
12.49
70
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.50
v0 = 10 km/h = 10
1000
= 2:778 m/s
3600
mg
kx
x
= ma
MAD
FBD N
Fx = max
a =
dv
v
dx
1 2
v =
2
kx = ma
v dv = a dx
a=
v dv =
k
x
m
k
x dx
m
k 2
x +C
2m
) C1 = 12 v02
Initial condition: v = v0 when x = 0:
) v 2 = v02
k 2
x
m
Stopping condition: v = 0 when x = 0:5 m:
2:7782
k
18
103
(0:5)2 = 0
k = 5:56
105 N/m J
12.51
71
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12.52
*12.53
F = ma
dv
=
dx
Z
x=
a=v
T
cD v 2 = ma
1
(T + cD v 2 )
m
mv
dv =
T + cD v 2
mv
dv = dx
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
ln
2cD
cD
m
T + cD v02
ln
2cD T + cD v 2
72
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
2500
= 77:64 slugs
32:2
m=
v0 = (90)
5280
= 132 ft/s
3600
When v = 0:
x=
m
T + cD v02
77:64
450 + 0:006(132)2
ln
=
ln
= 1352 ft J
2cD
T
2(0:006)
450
*12.54
.
x
5 lb
30o
4 ft
θ
= 5
32.2
8 lb
FBD
Ν
MAD
Fx = ma + .
5 sin 30
32:2
(5 sin 30
5
8 cos ) = 16:1
)a=
cos = p
a=v
a
dv
= 16:1
dx
51:52x
p
x2 + 16
v dv =
Initial condition: v = 0 when x = 0:
When a = 0 (v = vmax ):
1 2
v
2 max
5
a
32:2
51:52 cos ft/s
2
x
x
=p
x2 + 42
x2 + 16
1 2
v = 16:1x
2
16:1
8 cos =
16:1
dx
p
51:52 x2 + 16 + C
) C = 51:51(4) = 206:0 (ft/s)2
51:52x
p
=0
x2 + 16
) x = 1:3159 ft J
p
51:52 (1:3159)2 + 16 + 206:0
=
16:1(1:3159)
=
10:241 (ft/s)
p
vmax = 2(10:241) = 4:53 ft/s J
2
51:52x
p
x2 + 16
73
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*12.55
F = ma + #
a=
t=
1
ln
cd =m
dv
=g
dt
g
mg
cd
v
m
dt =
(cd =m)v
+C =
cd =m
Initial condition: v = 0 when t = 0:
)t=
)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
When v = v1 (terminal velocity), a = 0:
mg
When v = 0:9v1 = 0:9
:
cd
t=
cD v = ma
) v1 =
mg
cd
m
1
m
m
=
ln
ln 10 = 2:30
J
cd 1 0:9
cD
cd
12.56
v
7
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
74
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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.
a=
The MATLAB program that integrates the equations is
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
75
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
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 v 2 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 x22 e 3:211 10
5
x1
x2 (0) = 0
The MATLAB program that integrates the equations is
76
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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)
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
77
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12.61
(a)
mg y
P = kx
=
v
F = µN
x
N
FBD
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
reversed.
Fy
Fx
)
=
a=
= 0 +"
= ma +!
k
x
m
30
x
1:6
N
mg = 0
) N = mg
kx
N sign(v) = ma
N
sign(v) =
m
0:2(9:81) sign(v) =
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
78
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
(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)
79
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)
=
(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:
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
80
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(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
a = 80
16v 1:5 ft/s
2
12.63
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 2:8794
v 2:8794
=
0:25263 0:23822
0:25 0:23822
v = 2:89 ft/s J
81
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12.64
y=
x2
400
y_ =
x
x_ =
200
x
v0
200
y• =
1
v0 x_ =
200
v02
200
mg
=
mv02/200
N
FBD
Fy = may
MAD
N
mg =
m
v02
200
v2
Contact is lost when N = 0: g = 0
200
p
p
v0 = 200g = 200(32:2) = 80:3 ft/s J
82
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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
g
an
30o
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
=
v
=
p
an =
ft
109:89
s
an =
v2
p
(0:25 32:2)(1500) = 109:89 ft/s
1 mi
3600 s
= 74:9 mi/h J
5280 ft
1h
83
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13.5
At point B:
v = vx = v0 cos
an =
v
2
2
=
v
=
an
an = g
2
vA
cos2
g
J
13.6
at
dv ds
dv
1 2
=
v
v dv = at ds
v = at s + C
ds dt
ds
2
Initial condition: v = 0 at s = 0 ) C = 0
v2
at =
2s
=
)
At point B:
at
=
an
=
2
52
vB
2
=
= 1:989 ft/s
2( R)
2(2 )
2
vB
52
2
=
= 12:5 ft/s
R
2
q
a2n + a2t =
=
4 cos x
a=
13.7
y
p
2
12:52 + 1:9892 = 12:66 ft/s J
y0 =
0 2
4 sin x
3=2
at
4 cos x
3=2
1 + (y )
1 + 16 sin2 x
jy 00 j
4 cos x
2
2
(2 )(4 cos x)
v
16 cos x
=
=
=
3=2
3=2
2
1 + 16 sin x
1 + 16 sin2 x
= v_ = 0
=
an
y 00 =
=
4
) a = an = 16 m/s J
16 cos( =4)
) a = an =
= 0:419 m/s J
3=2
1 + 16 sin2 ( =4)
2
) a = an = 0 J
At A :
x=0
At B
:
x=
At C
:
x=
2
84
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13.8
at
)
1 2
v
2
=
=
dv ds
dv
=
v
ds
Zds dt
v dv = at ds
at ds + C =
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
(a)
.
.
B
0
75
Dimensions
in mm
A
.
O
θ
vB
300
450 mm/s
85
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vB
(an )A
=
_ = vB = 450 = 0:60 rad/s
RB
750
450 mm/s
2
= RA _ = 300(0:62 ) = 108:0 mm/s
(b)
300
O
450 mm/s
vA = 450 mm/s
(an )A =
2
J
.
A
vA
2
vA
4502
2
=
= 675 mm/s J
RA
300
13.11
13.12
86
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13.13
13.14
13.15
87
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13.16
At point A:
1000
= 9:722 m/s
3600
v2
9:7222
2
=
=
= 0:9452 m/s
100
p
p
a2 a2n =
1:32 0:94522 =
=
v
=
an
at
35
1 2
v =
2
v dv = at ds
0:8925 m/s
2
0:8925s + C
Initial condition:
1
2
(9:7222 ) = 47:26 (m/s)
2
0:8925s + 47:26
s = 53:0 m J
v = 9:722 m/s when s = 0 ) C =
When v = 0:
0=
13.17
13.18
y=
x2
m
80
y0 =
x
40
y 00 =
1
m 1
40
At point A (x = 10 m):
y0 =
1
=h
10
= 0:25 m
40
y 00
0 2
1 + (y )
i3=2 =
y 00 = 0:025 m 1
0:025
3=2
(1 + 0:252 )
= 0:02283 m 1
88
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an =
v2
= 122 (0:02283) = 3:288 m/s
a=
q
a2n + a2t =
13.19
2
at = v_ = 4 m/s
p
2
3:2882 + 42 = 5:18 m/s J
at
o
20
an
g
2
at
= g sin 20 = 9:82 sin 20 = 3:359 m/s
an
= g cos 20 = 9:82 cos 20 = 9:228 m/s
an =
v2
v=
p
2
an =
2
p
9:228(4500) = 204 m/s 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
89
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13.21
v
θ
v0
B
R
y
θ
O
) y_ = v0 = R cos
y = R sin
)_=
_
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
2
v0
sec
R
at
= R• = R
an = R _ = R
2
=
62
v02
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
2
a = a2t + a2n = 13:8562 + 8:02 = 16:0 in./s J
=
13.22
x2
ft
120
y=
x
60
y 00 =
y 00 =
1
ft 1
60
(1=60)
y0 =
1
ft 1
60
At point A (x = 10 ft):
y0
1
an
at
=
=
=
10
1
=
60
6
jy 00 j
=
= 0:015996
3=2
3=2
[1 + (y 0 )2 ]
(1 + ( 1=6)2 )
v2
2
= (102 )(0:015996) = 1:5996 ft/s
2
= v_ = 1:2 ft/s
q
p
2
a = a2n + a2t = 1:59962 + 1:22 = 2:00 ft/s J
90
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13.23
P
2 in./s
θ
v
From velocity diagram: v = 2 sec
R_
v_
an
at
When
= v = 2 sec
=
2 sec
5
= 0:4 sec rad/s
dv _
= (2 sec tan ) (0:4 sec ) = 0:8 tan sec2
d
(2 sec )2
4 sec2
v2
2
=
=
= 0:8 sec2 in./s
=
R
R
5
2
= v_ = 0:8 tan sec2 in./s
=
in./s
= 20 :
v
=
2 sec 20 = 2:13 in./s J
an
=
0:8 sec2 20 = 0:9060 in./s
2
2
0:8 tan 20 sec2 20 = 0:3298 in./s
p
2
a =
0:90602 + 0:32982 = 0:964 in./s J
at
13.24
_ = 2 sec
R
=
13.25
91
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13.26
v
vθ
θ
•
= R
•
R
)a =
aR
aθ
θ
aR
v = vR csc = 350 csc 40 = 545 m/s J
vR = R_ = 350 m/s
)_=
v = R _ = vR cot
a
vR
vR cot
R
=
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 =
v = vR
22 + 17:2792 = 17:39 m/s J
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
92
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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_ = 3 ft/s
R = 4 ft
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
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
R
vR
= 4 + 2 sin ft
= R_ = 2 cos _ ft/s
v = R _ = (4 + 2 sin ) _ ft/s
(a) At point A:
= 0 ) sin = 0 cos = 1
) vR = 2 _ ft/s
v = R _ = 4 _ ft/s
v
r
q
2
2
vR + v =
2_
=
v
=
p
4 ft/s
2
+ 4_
20 _ = 4
2
=
p
20 _
_ = p4 = 0:894 rad/s J
20
(b) At point B:
vR
=
=
=2
0
) sin = 1
v = 6 _ ft/s
cos = 0
93
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
v
=
v
=
q
2 + v 2 = 6 _ ft/s
vR
6_ = 4
4 ft/s
_ = 4 = 0:667 rad/s J
6
13.30
v = R_
vR = R_ = 600 ft/s
Find _ from acceleration:
-g
aθ θ
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 =
v = vR
6002 + ( 727:2)2 = 943 ft/s J
0:09090
13.31
Di¤erentiate cam pro…le twice:
R2 + 6R cos
27 = 0
2RR_ + 6R_ cos
6R sin _ = 0
• + 6R
• cos
2R_ 2 + 2RR
Substituting
_
6R_ sin
6R cos
_2
6R sin • =
0
(a)
(b)
(c)
= 0, _ = 2 rad/s and R = 3 in. into Eq. (b) yields
2(3)R_ + 6R_
0=0
) v = R_ = 0 J
Eq. (c) now becomes
• + 6R
• 0 6(3)(22 ) 0 = 0
0 + 2(3)R
• = 6:0 in./s2 J
)a=R
94
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13.32
R = 0:3
_
R_ =
0:4
=
0:4
•=
R
(a) When
2
•
(b) When
0:4
=2
= 0:1 m
v = R _ = 0:1(2) = 0:2 m/s J
2_
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
= =3:
vR = R_ =
a
•=0
0:255 m/s J
R = 0:3
aR
=0
_ = 2 rad/s
= =2:
vR = R_ =
a
m
0:2546 m/s
0:4
R = 0:3
aR
=
0:4
0:255 m/s J
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 and _ = 15(2 =60) = 0:5 rad/s.
Given: R = 0:7 ft, R_ = 4 ft/s, R
Note that in the position = 0 we have eR = i and e = j.
(a) When = =2:
v
_ R + R _ e = 4eR + 0:7(0:5 )e
= Re
= 4eR + 1:100e = 4i + 1:100j ft/s J
(b)
• R _ 2 )eR + (R• + 2R_ _ )e
a = (R
= 0 0:7(0:5 )2 eR + [0:7(0) + 2(4)(0:5 )] e
=
1:727eR + 12:567e =
1:727i + 12:567j ft/s
2
J
95
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13.34
13.35
_
= 2t rad
= 2 rad/s
• =
When R
vR
v
v
= 2:5 mm: t = cosh 1 2:5 = 1:5668 s
= R_ = sinh(1:5668) = 2:291 mm/s
= R _ = 2:5(2) = 5:0 mm/s
p
2:2912 + 5:02 = 5:50 mm/s J
=
2
2
R _ = 2:5 2:5(22 ) = 7:50 mm/s
2
a = R• + 2R_ _ = 0 + 2 [sinh(1:5668)] (2) = 9:165 mm/s
p
2
a =
7:52 + 9:1652 = 11:84 mm/s J
aR
13.36
0
R = cosh t mm
R_ = sinh t mm/s
• = cosh t mm/s2 = R
R
•
= R
96
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13.37
13.38
13.39
97
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13.40
R2 = b2 sin 2
2RR_ = 2b2 cos 2 _
• + R_ 2 ) = 2b2 ( 2 sin 2 _ 2 + cos 2 •)
2(RR
At point A ( = 45 ):
R
2RR_
•
2bR
= b
= 0 ) R_ = 0
2
•=
= 2b2 ( 2 _ ) ) R
2
R_ =
•
aR = R
v
aR =
aR =
v02
2
2
b_ =
2
3b _
2
v0
v
=
b
b
v0 2
v02
3b
= 3
J
b
b
= R_ = b_
)
an =
2b _
2b _
)_=
v2
v2
3 0 = 0
b
=
b
J
3
13.41
98
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13.42
h = R sin = 8200 sin 35 = 4703 ft J
h_ = R_ sin + R _ cos = 0
0 = R_ sin 35 + 8200(0:03) cos 35
) R_ = 351:3 ft/s
vR
v
13.43
= R_ = 351:3 ft/s
v = R _ = 8200(0:03) = 246:0 ft/s
q
p
2 + v2 =
=
vR
( 351:3)2 + 246:02 = 429 ft/s J
13.44
A
5.25 m
5.25 m
θ
θ
6m
R=
C
A
28.96o
6m
28.96o
B
vR
B
v
C
57.92o
vθ
57.92o
Given: R = 6 m, R_ = 1:2 m/s
cos
vR
5:25
= 28:96
6
= R_ = 1:2 m/s
=
From velocity diagram (note that v is perpendicular to AB):
v=
1:2
vR
=
= 1:416 m/s J
sin 57:92
sin 57:92
99
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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
a =
=
• R _ 2 )eR + (R• + 2R_ _ )e + z•ez
(R
0 0:4(6)2 eR + [(0:4)( 10) + 0] e + 2ez
2
=
13.46
1:2ez m/s
14:4eR 4e + 2ez m/s
p
2
( 14:4)2 + ( 4)2 + 22 = 15:08 m/s J
a =
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
) vmax = 3:22 + 1:62 = 3:58 in./s at =
, n = 0; 1; 2; : : : J
4
vR = R_ = 0
aR
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
) amax = 2:562 + 5:122 = 5:72 in./s at
=
8
+
n
, n = 0; 1; 2; : : : J
4
100
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13.47
13.48
13.49
_ = 3 rad/s
•=0
_ = 2 rad/s
L = 0:8 m
101
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R_ =
R = L cos
z = L sin
When
= 50 and
z_ = L cos
az
z• =
_2
L cos
L sin
_2
= 0:8 cos 40 = 0:6128 m
=
0:8(sin 40 )(2) = 1:0285 m/s
=
0:8(cos 40 )(22 ) = 2:451 m/s
0:8(sin 40 )(22 ) =
z• =
a
_
•=
R
= 40 :
R
R_
•
R
aR
_
L sin
2:057 m/s
2
2
R _ = 2:451 0:6128(32 ) = 7:966 m/s
2
= R• + 2R_ _ = 0 + 2( 1:0285)(3) = 6:171 m/s
•
= R
2
2
= z• = 2:057 m/s
p
2
a = 7:9662 + 6:1712 + 2:0572 = 10:28 m/s J
102
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
13.50
13.51
z
ρ
mg
= man
µsN
FBD
Fz
=
Fn
= man
v
13.52
0
+"
N
N
mg = 0
MAD
) N = mg
+
s N = man
s mg = m
p
p
=
0:4(600)(32:2) = 87:9 ft/s J
s g =
v = 70 km/h =
v2
70 103
m/s = 19:444 m/s
3600
103
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Assume impending sliding of crate.
Mg
R
y
v
MR
2
x
15o
µsΝ
N
FBD
Fy
= may
N
= M
N
M g cos 15 = M
v2
sin 15
R
v2
sin 15
R
19:4442
9:81 cos 15 +
sin 15
50
g cos 15 +
= M
Fx
= max
s
= M
=
MAD
sN
M g sin 15 =
= 11:433M
M
v2
cos 15
R
v 2 =R cos 15
M g sin 15
N
19:4442 =50 cos 15
9:81 sin 15
= 0:417 J
11:433
13.53
W
T
35
N
FR
T cos 35
N
= maR
=
maR
o
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
+ " N W + T sin 35 = 0
N = 7:81 N J
104
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13.54
ρ =4m
N
Fz
µsN
mg
FBD
0
+"
=
z
z
mρv
MAD
sN
mg = 0
)N =
v2
mg
+! N =m
)
r
r
g
4(9:81)
=
=
= 8:09 m/s J
0:6
s
Fn
= man
v
2
mg
s
=m
v2
s
13.55
o
3200 lb 20
n
man
t
F = 0.6N
mat
N
FBD
MAD
Assume impending sliding between the tires and the road.
= man =
Ft
3200 602
N = 4796 lb
32:2 200
= mat
3200 sin 20
0:6N = mat
3200
2
3200 sin 20
0:6(4796) =
at
at = 17:94 ft/s J
32:2
N
N
W cos 20 = m
v2
Fn
3200 cos 20 =
13.56
mv
R
2
mg
n
.
mv
t
µN
N
FBD
MAD
105
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(a)
v2
R
Fn
= man
N
mg = m
N
= m g+
v2
R
= 8 9:81 +
2:82
2:5
= 103:57 N J
(b)
Ft
v_
= mat
N
=
=
m
N = mv_
0:2(103:57)
=
8
2:59 m/s
2
J
13.57
T
FBD
=
mg
θ
• = 0, we have aR =
Since R_ = R
F
= ma
g sin
• =
R
FR
_2
_
maθ
MAD
maR
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 T
1
8:5
=
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
106
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13.59
ω
y
ρ
x
v
The bead moves on a horizonal circle of radius
v = ! = !R sin .
= R sin
with the speed
y
mg
mv2/ρ
x
N θ
FBD
Fx
= m
Fy
=
v2
MAD
N sin = m! 2 R sin
N = m! 2 R
0
mg N cos = 0
mg
g
9:81
cos =
= 2 =
= 0:8741
N
! R
(6:72 ) (0:25)
= 29:1 J
13.60
107
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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
N
y
F
n
W = 0.4 lb
FBD
v
=
an
=
30o
t
x
mat
ma n
ma
MAD
vx
8
=
= 9:238 ft/s
cos 30
cos 30
v2
9:2382
2
=
= 71:12 ft/s
1:2
Because vx is constant, we have ax = 0, so that a is vertical. Therefore (see
MAD),
2
at = an tan 30 = 71:12 tan 30 = 41:06 ft/s
108
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Ft
= mat
F cos 30 + W sin 30 = mat
0:4
F cos 30 + 0:4 sin 30 =
(41:06)
32:2
= 0:358 lb J
F
13.63
x
N
F
FBD
MAD
β R
man β
O
Fx
= max
+
kx0
x =
k
2
m_
=
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
(0
2 =
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
109
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13.65
When slider is at C:
mat
F 45o
man
N
mg
FBD
F
an
Ft
at
MAD
p
= k = 1:2(2:5 2 1:3) = 2:683 lb
v2
82
2
=
=
= 25:60 ft/s
R
2:5
= mat
F cos 45
mg = mat
2:683
F
cos 45
g=
cos 45
=
m
1:0=32:2
Fn
= man
N
= F sin 45
a=
F sin 45
a2n + a2t =
13.66
mg
1:0
(25:60) = 1:102 lb J
32:2
p
2
25:602 + 28:892 = 38:6 ft/s J
R
T
maR
.
θ
MAD
N FBD
FR = maR
2
N = man
man = 2:683 sin 45
q
32:2 = 28:89 ft/s
+ !
T =
• = 0. ) T =
(a) If string is intact, R
2:5 0
(b) After string breaks, T = 0. ) 0 =
• = 560 m/s2 J
)R
h
•
2:5 R
•
m(R
2
R_ )
1:4(20)2 = 1400 N J
1:4(20)2
i
110
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13.67
13.68
Assume impending sliding of the block up the cone.
mg
µsN
y
mv2 /ρ
x
β
β N
FBD
Fy
=
0
N cos
N=
Fx
MAD
cos
= max
mg
s N sin
mg = 0
mg
s sin
N sin
sin +
cos
+
s cos
s sin
s N cos
=m
=m
v2
v2
Substituting v = !:
)
!2
=
=
sin + s cos
= !2
cos
sin
s
g sin + s cos
32:2 sin 35 + 0:28 cos 35
=
cos
1:2 cos 35
0:28 sin 35
s sin
g
2
32:7 (rad/s)
! = 5:72 rad/s
111
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Assume impending sliding of the block down the cone. The direction of the
friction force s N must be reversed on the FBD. This is equivalent to reversing
the sign of s . The result is
!2
=
!
=
32:2 sin 35
0:28 cos 35
= 9:427
1:2 cos 35 + 0:28 sin 35
3:07 rad/s
) The block will not slide if 3:07 rad/s
!
5:72 rad/s J
13.69
13.70
112
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13.71
13.72
v
=
_ R = 1:2(2:2) = 2:640 m/s
an
=
v2
2:6402
2
=
= 3:168 m/s
R
2:2
Assume impending sliding when
35o
at = 0
= 35 .
9.81m
man
µs N
n
t
N
FBD
MAD
Fn
N
= man
N 9:81m cos 35 = m(3:168)
= m(9:81 cos 35 + 3:168) = 11:204m
Ft
= mat
9:81m sin 35 = 0
sN
9:81 sin 35
9:81 sin 35
=
=
= 0:502 J
N
11:204
s
113
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13.73
13.74
114
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
13.75
T
maθ
θ
maR
mg
FBD
MAD
• = 0, so that the acceleration components are aR =
Note that R_ = R
•
a =R .
d_ d
d_ _
=
d dt
d
But • =
d_ _
d
g
sin
R
)•=
mg sin = mR•
F = ma
2
R _ and
g
sin
R
=
_ d_ =
_2
g
sin d
R
2
=
g
cos + C
R
= 0 then _ = _ A
Initial condition: when
2
_2
A
2
=
)
_
g
g
+C
C= A
R
2
R
_2
_ 2 = 2g (1 cos )
A
R
(a) Rigid rod: pendulum will just reach position B if _ = 0 at
r
4g
) _A =
J
R
= 180
(b) Flexible string:
FR = maR
T + mg cos =
Pendulum will just reach position B if T = 0 at
mg
_A
2
mR _
r
5g
=
J
R
=
_2 = g
R
mR _
2
= 180
2
2
) _A = _ +
2g
5g
(2) =
R
R
115
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
13.76
aR
a
θ
aθ
R_ cos
R θ 2m
O
R_
=
4 m/s (given)
R _ sin
=
0
a
=
=
a =
R cos = 2 m
1
cos
=
R
2
m 1
R_ = R _ tan = 4 tan
1 d(4R)
1
1 d(R2 _ )
=
= 4 R_
R dt
R dt
R
cos
2
4
(4 tan ) = 8 sin m/s
2
a
8 sin
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.
R2 _
= constant ) R2 • + 2RR_ _ = 0
) a = R• + 2R_ _ = 0 Q.E.D
R• + 2R_ _ = 0
*13.78
(a)
θ
maR
mg
N
FR = maR
maθ
FBD
MAD
•
mg sin = m(R
R_ )
d
d d
d
=
= !0
dt
d dt
d
2
•
g sin = R
• = !2
)R
0
! 20 R
d2 R
d 2
116
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
= ! 20
g sin
d2 R
d 2
R
=
d2 R
d 2
g
sin
! 20
R
J
(a)
(b) Solution of the homogeneous equation
d2 R
d 2
A particular integral is
R = 0 is R = C1 sinh + C2 cosh
R=
g
sin
2! 20
Therefore, the solution of the di¤erential equation in Eq. (a) is
R = C1 sinh + C2 cosh
g
sin
2! 20
Initial conditions:
R
R_
=
0 when
=0
) C2 = 0
=
0 when
=0
C1
R=
g
(sinh
2! 20
g
=0
2! 20
C1 =
g
2! 20
sin ) J
13.79
117
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*13.80
13.81
118
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13.82
13.83
119
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
13.84
120
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
13.85
Initial conditions:
Letting x1 =
are
g
sin =
L
•=
Equation of motion:
= 60 =
3
9:81
sin = sin
9:81
rad _ = 0 at t = 0
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
[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
• = 0, we have aR =
(a) Since R_ = R
2
R _ and a = R•
θ mg
FBD
=
µN
maR
N
FR
F
= maR
= ma
+.
+&
mg sin
mg cos
maθ
MAD
N = mR _
N = mR•
2
121
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
• = 9:81 (cos
2
0:3 _
0:3 sin )
2
Letting x1 = and x2 = _ , the equivalent …rst-order equations and the initial
conditions are
x_ 1 = x2
x1 (0) = 0
x_ 2 = 4:905(cos
x2 (0) = 0
0:3 sin )
0:3x22
The corresponding MATLAB program is
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
N
MAD
122
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
!
=
0 cos !t
=
rad/s
FR
= maR
•
)R
= R_
•
R
R(0)
2
)_=
! 0 sin !t
+%
•
mg sin = m(R
g sin = R( ! 0 sin !t)2
12
=
12
rad
2
R_ )
g sin( 0 cos !t)
2
2
= R
0 = 15
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
= x2
x1 (0)
= 2 ft
x_ 2 = x1
12
x2 (0) = 0
sin t
32:2 sin
12
cos t
The corresponding MATLAB program is
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:
123
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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:
4:0718 3:9529
4 3:9529
=
3:4250 3:4
t 3:4
t = 3:410 s J
R_ 4:7323
4:8523 4:7323
=
3:410 3:4
3:4250 3:4
R_ = 4:78 ft/s J
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•
124
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6 tanβ
R
6 ft
s
β
z
0 = m(z • + 2z_ _ ) tan
FBD
β
mg N
=
maz
maθ
s
maR
β MAD
Kinetics:
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)
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
125
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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 ) hmin = 6
3:93 = 2:07 ft J
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_ )
126
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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
127
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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
128
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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
13.91
v = 150
mi
h
1h
3600 s
5280 ft
= 220 ft/s
1 mi
at = 0
129
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F1
F2
mat
man
MAD
W
FBD
Ft
= mat
F2 = 0
Fn
= man
F1 + W =
F1
= W
F
=
v2
g
28:1 lb # J
1+
Path
W v2
g
= 180
1+
2202
32:2(1300)
= 28:12 lb
13.92
130
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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
a =
a2R + a2 = ( 33:69)2 + 20:112 = 39:2 m/s J
aR
•
= R
131
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13.95
13.96
Kinematics:
• = 5 rad/s2
When
_ = 5t rad/s
= 2:5t2 rad
R = 0:18 m (constant)
= 30 = =6 rad:
t2 =
2
R_ = 0
2:5
=
=6
=
s2
2:5
15
aR
•
= R
a
2
= R• + 2R_ _ = 0:18(5) + 0 = 0:900 m/s
0:18(5t)2 =
0:18(25)
15
=
0:9425 m/s
2
Kinetics (assume impending sliding):
30o 9.81m
maθ
maR
µs N
FBD
F
FR
N
MAD
= ma
N 9:81m cos 30 = 0:900m
N = (0:900 + 9:81 cos 30 ) m
= maR
9:81m sin 30 = 0:9425m
sN
(0:900
+
9:81
cos
30 ) m 9:81m sin 30 =
s
9:396 s = 3:963
s = 0:422 J
0:9425m
132
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13.97
13.98
y = sinh x
dy
dx
d2 y
dx2
x = sinh 1 y = sinh 1 (3:8) = 2:045 m
=
cosh x = cosh(2:045) = 3:929
=
sinh x = sinh(2:045) = 3:800 m 1
h
i3=2
2
3=2
1 + (dy=dx)
1 + 3:9292
= 17:537 m
=
d2 y=dx2
3:800
=
an =
v2
=
82
2
= 3:649 m/s
17:537
133
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man
mg
FBD
N
θ
ma t
MAD
= tan 1 (dy=dx) = tan 1 (3:929) = 75:72
Fn
Ft
= man
N mg cos = man
N 12(9:81) cos 75:72 = 12(3:649)
N = 72:8 N J
= mat
mg sin = mat
2
v_ = at = g sin = 9:81 sin 75:72 = 9:51 m/s J
13.99
an =
v2
4:22
2
=
= 8:82 ft/s J
R
2
3 lb
mat
o
45
1.8 lb
ma n
N
FBD
Fn
= man
MAD
3 + 1:8 sin 45
N=
N = 3:451 lb J
Ft
= mat
at
=
1:8 cos 45 =
2
13:66 rad/s
3
(8:82)
32:2
3
at
32:2
J
13.100
mg
β man
µsN
z
N n
FBD
MAD
134
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Fz
=
N
=
0
cos
Fn
v
N cos
mg
s sin
s N sin
mg = 0
m(32:2)
= 37:30m lb
=
cos 10
0:7 sin 10
v2
= man
=
s N cos + N sin
r
N
=
( cos + sin )
m s
r
300(37:30m)
=
(0:7 cos 10 + sin 10 )
m
3600
= 98:27 ft/s = 98:27
= 67:0 mi/h J
5280
13.101
13.102
θ mg
maR
FBD
N
maθ
MAD
135
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F
= ma
mg sin = m(R• + 2R_ _ )
d_ _
g
• = g sin
= sin
R
d
R
1 _2
g
=
cos + C
2
R
+.
= R• + 0
g sin
_ d_
=
g
sin d
R
= 0. ) C = g=R.
Initial condition: _ = 0 when
g
=
(1 cos )
R
p
= R _ = 2gR(1
1 2
) _
2
v
FR
= maR
When N
=
cos
+-
0:
=
N
g cos = 0
2(1
r
2g
(1
R
cos ) J
_=
cos )
cos )
2
R_ )
•
mg cos = m(R
2
R_ =
= cos
1 2
3
2g(1
cos )
= 48:2
J
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
136
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13.104
13.105
137
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13.106
R
θ
2a
R = 2a cos
R_ = 2a sin
_ = 2a! 0 sin
vR = R_ = 2a! 0 sin
v = R _ = 2a! 0 cos
q
2 + v 2 = 2a! (constant) J
v = vR
0
13.107
138
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13.108
13.109
z
A
z
A
m
0.6
θ
T
R
B
FBD
ma n
MAD
mg
B
(a)
Fz
T
= maz
T cos
mg = 0
mg
1:4(9:81)
=
=
= 79:09 N J
cos
cos 80
(b)
Fn
v2
v
v2
R
T R sin
79:09(0:6 sin 80 ) sin 80
2
=
=
= 32:87 (m/s)
m
1:4
p
=
32:87 = 5:73 m/s J
= man
T sin = m
139
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Chapter 14
14.1
14.2
140
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14.3
14.4
14.5
(a) When L0 = b:
A
UA B
=
0
1
k
2
=
B =
p
2
B
2
A
2b
b = 0:4142b
1 h
2
=
k (0:4142b)
2
i
0 =
0:0858kb2 J
(b) When L0 = 0:8b:
A
UA B
= b
=
0:8b = 0:2b
1
k
2
2
B
2
A
B =
=
p
2b 0:8b = 0:6142b
i
1 h
2
k (0:6142b)
(0:2b)2 =
2
0:1686kb2 J
141
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14.6
14.7
180 lb
FBD
µkN
30 lb
20o
N
Fy = 0
N
180 + 30 sin 20 = 0
N = 169:74 lb
1
mv 2 0
2 2
1 180
[ 0:15 (169:74) + 30 cos 20 ] x =
4:02
2 32:2
x = 16:38 ft J
U1 2
= T2
UA B
= TB
T2
(
s N + 30 cos 20 )x =
14.8
d
=
TA
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)
142
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14.9
U1 2 = T2
T1
( k mg) x
1 2
k
2
=
0
1
mv 2
2 1
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 =
U1 2
v2
80 km
1h
1000 m
1 km
1h
= 22:22 m/s
3600 s
1
= T2 T1
mgh2 = m v22 v12
2
q
p
2
2
=
v1 2gh2 = 22:22
2(9:81)(10) = 17:249 m/s
=
17:249
m
s
1 km
1000 m
3600 s
= 62:1 km/h J
1h
14.11
R2 = 0.2 m
2
.
0.75 m
A
R1
1
.
The tension in the rope is a central force always passing through point A.
U1 2
T1
U1 2
=
=
F (R2
R1 ) mg h
p
=
10(0:2
0:22 + 0:752 )
= 1:3476 N m
0
T2 =
= T2
T1 :
1
2
mv 2 = 0:3vB
2 B
2
1:3476 = 0:3vB
0:6(9:81)(0:75)
vB = 2:12 m/s J
143
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14.12
14.13
h
T1
U1 2
h
= T2 = 0
= T2
=
T1 :
1:223 m J
1
2
U1 2 =
1
k( 22
2
1
(1200)(0
2
2
1)
0:12 )
mgh = 0
0:5(9:81)h = 0
144
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14.14
14.15
14.16
In the limiting case, package will arrive at D with zero velocity.
) TA = TD = 0.
2P
UA D = area under P -x diagram
10(9:81)(2) = 0
P = 98:1 N J
mghD = 0
145
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14.17
A
2R
25o
20o
B
2R cos 25
= 1:2
d
UA B
TA
UA B
p
= Fd = F
=
2:769 lb ft
0
TB =
= TB
TA :
=
p
2(1:8) cos 25
1
1 2 2
2
2
mvB
=
v = 0:03106vB
2
2 32:2 B
2
2:769 = 0:03106vB
vB = 9:44 ft/s J
14.18
U1 2
v22
= T2
=
1 2
k
2
T1
1
(2000) (0:122 )
2
2
89:43 (m/s)
30o
n
mg (1
cos 30 ) =
0:3(9:81)(2:5)(1
1
mv 2
2 2
0
cos 30 ) =
1
(0:3)v22
2
mg
man
t
FBD
mat
MAD
N
N
v2
mg cos 30 = m 2
Fn
= man
N
= m g cos 30 +
v22
= 0:3 9:81 cos 30 +
89:43
2:5
= 13:28 N J
146
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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 2
k
2
=
0
1
mv 2
2 1
1
1
(4000) 2 =
(5)(6)2
2
2
2000 2 + 18:089
144:27 = 0
The positive root is
= 0:2641 m
0:25(40:18)(3 + )
Fmax = k = 4000(0:2641) = 1056 N J
14.20
Release position (position 1)
Return position (position 2)
T1 = 0
U1 2
T2 =
1
2
= x1 = 1:5 ft
= x2 = 0
1 20 2
1
mv 2 =
v = 0:3106v22
2 2
2 32:2 2
1
2
k( 22
x1 )
k mg(x2
1)
2
1
=
(22:5)(0 1:52 ) 0:25(20)(0 + 1:5)
2
= 17:813 lb ft
=
U1 2
v2
= T2 T1
17:813 = 0:3106v22
= 7:57 ft/s J
147
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14.21
14.22
148
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14.23
14.24
Position 1 = position at entry (x = 0); Position 2 = position at exit (x = 54
in.).
U1 2
=
=
(area under the F -x diagram)
10 + 18
18 + 24
24 + 16
(24) +
(6) +
(24)
2
2
2
942 lb in. = 78:5 lb ft
T1
=
=
T2
=
U1 2
v2
1
1 3:2=16 2
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 v22
136:96
149
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14.25
14.26
Position 1 = release position ( = 0); choose as datum for gravitational potential
energy.
Position 2 = position where spring has maximum deformation ( = max ).
V1 = T 1 = 0
T1 + V1
Fmax
V2 =
1
W max + k 2max
2
0=
1
W max + k 2max
2
= T 2 + V2
= k max = 2W J
T2 = 0
max =
2W
k
14.27
Let B denote the lowest point of the drop and let A be the datum for gravitational potenetial energy.
TA
= TB = VA = 0
VB
=
=
mg(L +
120 2B
VB
h
B) +
150 B
= 0:
= L+
1 2
k =
2 B
24 000
150(160 +
B) +
1
(240) 2B
2
B = 14:78 ft
B = 160 + 14:78 = 174:8 ft
J
Pmax = k B = 240(14:78) = 3550 lb J
14.28
Position 1: position just before car hits the spring.
150
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Position 2: position where car come to a stop.
T 1 + V1
1
18
2
103
10 103
3600
1
1
mv 2 + 0 = 0 + k 2
2 1
2
= T 2 + V2
2
=
1
k(0:52 )
2
k = 556
103 N/m J
14.29
Position 1: position just before car hits the spring.
Position 2: position where car come to a stop.
2 δ2 1
δ1
0.3 m
2
0:3 = 0:2 m
Spring 1
Spring 2
2 = 0:5 m
T 1 + V1
1
18
2
103
10 103
3600
1 =
= T 2 + V2
2
=
1
1
mv12 + 0 = 0 + k
2
2
1
k(0:22 + 0:52 )
2
k = 479
2
1+
2
2
103 N/m J
14.30
151
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14.31
Choosing position A as the datum for gravitational potential energy:
VA
VB
TB
1 2
1
k = (2000) 2A = 1000 2A
TA = 0
2 A
2
= mgh = 0:5(9:81)(0:8) = 3:924 N m
1
1
=
mv 2 = (0:5)(1:52 ) = 0:5625 N m
2 B
2
=
VA + TA
A
= VB + TB :
1000 2A = 3:924 + 0:5625
= 0:0670 m J
14.32
152
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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
TA =
mv 2 = (0:6)(3)2 = 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 B = (200)(0:15)2 = 2:25 N m
2
2
1
1
2
2
2
mvB = (0:6)vB
= 0:3vB
2
2
2
2:70 + 9:235 = 0:3vB
+ 2:25
vB = 5:68 m/s J
14.34
153
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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=
T
man
FBD
MAD
mg
Fn
= man
+"
T
= m g+
v22
L
v2
mg = m 2
L
14:017
= 0:5 9:81 +
2
T
= 8:41 N J
14.37
The potential energy of a spring in terms of the spring force F is
Ve =
1 2
1
k = k
2
2
F
k
2
=
1 F2
2 k
154
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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 12002
1 W2
= 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 =
=
1
F2
2 660 103
T 1 + V1 = T 2 + V 2
7:576
1 F2
F
W
2 k
k
F
F (F 2400)
1200
=
3
660 10
1:32 106
)
10 7 F 2
1:8182
1192:5 + 1:0909
10 3 F
=
0+
F (F 2400)
1:32 106
1193:6 = 0
F = 40 900 lb J
14.38
(a) Choose the release position (x = 0) as the datum for gravitational potential
energy. Consequently, V1 = T1 = 0 in the release position. Noting that the
elongation of the spring is = 2x, we obtain for arbitrary x
V2
=
T2
=
1
1
k(2x)2 mgx = (50)(4x2 )
2
2
1 120 2
1
2
mv =
v = 1:8634v 2
2
2 32:2
V2 + T2
v2
v
(b)
120x = 100x2
120x
= V1 + T 1 :
100x2 120x + 1:8634v 2 = 0
120x 100x2
=
= 64:40x 53:67x2
1:8634
p
=
64:40x 53:67x2 ft/s J
xmax occurs when v
64:40xmax 53:67x2max
=
=
0:
0
xmax = 1:200 ft J
155
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
vmax occurs when d(v 2 )=dx = 0:
64:40 2(53:67)x = 0
x = 0:60 ft
p
vmax = 64:40(0:60) 53:67(0:60)2 = 4:40 ft/s J
14.39
1
2.5 in.
2
1
2
V1
V2
T2
10 in.
.
78 in
10.30
= 0:15 in. = 0:0125 ft
= (10:307 8 10) + 0:15 = 0:4578 in. = 0:0381 5 ft
1 2
= k 21 = 28(0:012 5)2 = 0:004 375 lb ft
= 2
k
2 1
1 2
2:5
=
Wy + 2
k 2 = 0:8
+ 28(0:038 15)2 =
2
12
1 0:8
1
mv22 =
v22 = 0:012 422v22
=
2
2 32:2
V1 + T1
v2
= V2 + T2
0:004 375 + 0 =
= 3:24 ft/s J
0:125 92 lb ft
0:125 92 + 0:012 422v22
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
156
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14.41
Choose line AC as the datum for gravitational potential.
V1
V2
T2
i2
1 h
(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
= mgR = 0:21(9:81)(0:5) = 1:0301 N m
1
1
mv 2 = (0:21)v22 = 0:105v22
=
2 2
2
=
T 1 + V1
v2
= T2 + V2
0 + 1:2004 = 0:105v22 + 1:0301
= 1:274 m/s J
14.42
157
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14.43
14.44
158
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14.45
14.46
159
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14.47
Choose point O as the datum for gravitational potential energy.
VA
VB
TA
1
1
= mgR + k(OA L0 )2 = 2:4(1:5) + (10)(1:5 1:0)2 = 4:850 lb ft
2
2
p
1
1
2
=
mgR + k(OB L0 ) = 2:4(1:5) + (10)(1:5 2 1:0)2
2
2
= 2:687 lb ft
1 2:4 2
1
2
2
=
v = 0:03727vB
= 0
TB = mvB
2
2 32:2 B
VA + T A = V B + T B
2
4:850 + 0 = 2:687 + 0:03727vB
vB = 7:62 ft/s J
160
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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
V1
=
T1
=
V2
=
(3:98 1014 )m
GMe m
= 6:633 107 m N m
=
R1
6 106
1
1
mv 2 = m(10 500)2 = 5:513 107 m N m
2 1
2
GMe m
(3:98 1014 )m
= 1:730 4 107 m N m
=
R2
23 106
V1 + T1
6:633
107 m + 5:513
= V2 + T 2
107 m =
v2
=
1
107 m + mv22
2
3490 m/s J
1:730 4
14.50
161
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14.51
mg
ma
F
N
FBD
F
P
MAD
4200
(12) = 1565:2 lb
32:2
5280
= 137 740 lb ft/s = 250 hp J
= F v = 1565:2 60
3600
= ma =
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 3
x= v +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
1 2
P
) 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
162
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14.54
P =
dW h
15
= 600
= 12 857 lb ft/s = 23:4 hp J
dt
0:7
14.55
Let P be the power supplied to driving wheels.
From Sample Problem 14.6:
Px =
When x =
)
1
mv 3 + C
3
5280
3600
0, v = 30 mi/h = 30
0=
When x =
)
P
=
Pengine
=
1 3200
(44:0)3 + C
3 32:2
= 44:0 ft/s
C=
320 ft, v = 50 mi/h = 50
2:822
5280
3600
= 73:33 ft/s
3200
(73:33)3 2:822
32:2
32 000
32 000 lb ft/s =
= 58:18 hp
550
P
58:18
=
= 71:0 hp J
0:82
P (320) =
1
3
106 lb ft2 /s
106
14.56
Pin
Pout
240 hp = 132 103 lb ft/s
dW
62:4 1600
=
h=
(58) = 96:51 103 lb ft/s
dt
60
Pout
96:51 103
=
100% =
100% = 73:1% J
Pin
132 103
=
14.57
v
y.
x.
100
2.8
From the velocity diagram:
y_ = p
2:8
v = 0:02799v
1002 + 2:82
163
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
P
v
= mg y_
4800
= 12:49 m/s J
103 = (1400
103 )(9:81)(0:02799v)
14.58
14.59
14.60
320(9.81) N
F
20o
y
0.35N
Fy
Fx
=
=
0
0
F
x
N
FBD
N 320(9:81) cos 20 = 0
N = 2950 N
F 0:35N 320(9:81) sin 20 = 0
0:35(2950) 320(9:81) sin 20 = 0
F = 2106 N
P = F v0
4000 = 2106v0
v0 = 1:899 m/s J
164
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14.61
(a)
Pout
=
=
(mg)v = 500(9:81)(0:68) = 3335 W
3335
Pout
100% =
100% = 79:4% J
Pin
4200
(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 + 0:14v 2
165
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
The power of F is
)a=
P = F v = mav + 0:14v 3
P
0:14v 3
mv
(a) When v = 50 km/h = 50(1000=3600) = 13:889 m/s:
a=
103 0:14(13:889)3
2
= 5:75 m/s J
2000(13:889)
160
(b) When v = 110 km/h = 110(1000=3600) = 30: 56 m/s:
a=
103 0:14(30:56)3
2
= 2:55 m/s J
2000(30:56)
160
14.65
F = F0 + cv 2
) P = F v = F0 v + cv 3
19:3 = F0 (50) + c(50)3
32:3 = F0 (60) + c(60)3
19:3 = 50F0 + 1:25
32:3 = 60F0 + 2:16
The solution is: c = 1:384 8
When v = 70 mi/h, we have
105
105
10 4 , F0 = 0:03979
P = 0:03979(70) + (1:3848
10 4 )(703 ) = 50:3 hp J
14.66
y
15 lb
x
F
35o
0.6N
vx
35o
3 ft/s
Velocity diagram
N
FBD
Fy
Fx
=
0
N cos 35 + 0:6N sin 35
15 cos 35 = 0
N = 10:562 lb
= 0
F 15 sin 35 + N sin 35
0:6N cos 35 = 0
F 15 sin 35 + 10:562 sin 35
0:6(10:562) cos 35 = 0
F = 7:737 lb
P = F vx = 7:737(3 cos 35 ) = 19:01 lb ft/s J
166
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
*14.67
14.68
T0
v1
L0 1
L0 1
14.69
L1 2 =
Z 5s
1
1
T1
v2
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
F dt =
0
p0
Z 5s
(2ti
0:6t2 j)dt = t2 i
0
mv1 + L1 2
v2
= mv2
= 22:5i
5s
0:2t3 j 0 = 25(i
2(10i) + 25(i
12:5j m/s J
j) N s
j) = 2v2
14.70
v1
v2
L1 2
= 60(cos 30 i + sin 30 j) = 51:96i + 30j ft/s
= 60j ft/s
= P t = 2:0P lb s
= m(v2 v1 )
3
2:0P =
[(0 51:96)i + (60
32:2
P =
2:42i + 1:398j lb J
L1 2
30)j]
167
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14.71
80 lb
40 lb
o
30
y
0.4N
N
FBD
(L1 2 )x
x
80 + 40 sin 30 = 0
Fy = 0
N
= m(v2
v1 )
[40 cos 30
N = 60 lb
0:4(60)] (4) =
v2 = 17:13 ft/s J
80
(v2
32:2
0)
14.72
W
FBD
F =µsW
N=W
The limiting force between the tires and the road is
F =
L0 1
t
s W = 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
168
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14.73
14.74
14.75
W
mv2
60o
y
x
N
FBD during
contact
68o
mv1
Momentum just
before impact
Momentum just
after impact
(a)
(L1 2 )x
v2
= m [(v2 )x (v1 )x ]
0 = m(v2 cos 60
cos 68
cos 68
= v1
= 28
= 20:98 ft/s J
cos 60
cos 60
v1 cos 68 )
169
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(b)
(L1 2 )y
= m [(v2 )y (v1 )y ] = m(v2 sin 60 + v1 sin 68 )
2:6=16
=
(20:98 sin 60 + 28 sin 68 ) = 0:223 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 2 i + sin 2 j)
mv2
Momentum just
after impact
20(cos 45 i
sin 45 j)]
Equating like components:
0 = v2 cos 2 20 cos 45
1:8
= v2 sin 2 + 20 sin 45
0:05
The solution is v2 cos 2 = 14:142, v2 sin 2 = 21:86.
p
) v2 =
14:1422 + 21:862 = 26:0 m/s J
21:86
= tan 1
= 57:1 J
2
14:142
14.77
Use principle of impulse and momentum to compute the velocity v2 immediately
after the initial impulse:
L1 2 = m (v2
v1 )
10 =
25
(v2
32:2
0)
v2 = 12:880 ft/s
25 lb
0.45(25) = 11.25 lb
N = 25 lb
FBD
170
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Use principle of work and kinetic energy to …nd the distance x3 traveled:
U2 3
= T3
T2
11:25x3 = 0
1
2
25
32:2
(12:8802 )
x3 = 5:72 ft J
Use principle of impulse and momentum to compute the time t3 of travel:
L2 3
= m(v3
t3 =
v2 )
11:25t3 =
25
(0
32:2
12:880)
12:880
= 0:889 s J
0:45(32:2)
14.78
14.79
171
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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 ) = 10 [3:6i
=
7:30i + 25j N s J
5(i cos 30
j sin 30 )]
14.82
F
) L1 2
(4)2
= pA = 4(1 e 0:5t )
= 50:27(1 e 0:5t ) lb
4
Z 10
Z 10
=
F dt = 50:27
(1 e 0:5t )dt = 402:8 lb s
L1 2 = m(v2
0
v1 )
0
402:8 =
500
(v2
32:2
0)
v2 = 25:9 ft/s J
14.83
172
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14.84
14.85
14.86
L1 2 =
Z t2
F (t) dt =
0
L1 2
Z t2
(600
0
= m(v2
t2
v1 )
600t2
240t + 4200 = 0
5t) dt = 600t2
5 2
t
2 2
5 2
t = 3500(3 0)
2 2
t2 = 19:0 s J
173
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14.87
Impulse of F
=
=
=
(L1 2 )F =
Z
F dt = area under F -t diagram
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 2 )W = W
(L1 2 )F
(L1 2 )W
3:425W
3W
= m(v2 v1 )
W
(v2 0)
=
9:81
t = 3W #
v2 = 4:17 m/s J
14.88
14.89
14.90
v
=
=
r =
hO
=
=
0:2i + 0:2j 0:3k
0:22 + 0:22 + 0:32
2:425i + 2:425j 3:638k m/s
!
OA = 0:6j m
i
j
k
2:425 2:425
3:638 (0:2)
(r v)m =
0
0:6
0
0:437i 0:291k N m s J
(5) p
174
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14.91
y
A
O
mv sin40o
mv
mv cos40o
x
R
40o
(a)
hO = mvR =
2=16
(12)(1:5) = 0:0699 lb ft s
32:2
J
(b)
hA
= mv cos 40 (R cos 40 ) mv sin 40 (R
2=16
(12)(1:5)(1
= mvR(1 sin 40 ) =
32:2
= 0:0250 lb ft s J
R sin 40 )
sin 40 )
14.92
2θ
v
y
m
r
O
R
2θ
θ
r = 2R cos (i cos + j sin )
h = r
=
(mv) = 2mvR cos
x
v = v( i sin 2 + j cos 2 )
i
cos
sin 2
j
sin
cos 2
k
0
0
2mvR cos (cos cos 2 + sin sin 2 )k = 2mvR cos2
kJ
175
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14.93
176
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14.94
14.95
14.96
.
θ
z
R
M
=
h1
=
Z t2
M dt
=
0
h2 = h1 +
Z t2
M
m 12 N.m
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
M dt
0 = [ 16:770 + (12t2
0
6)] k
t2 = 1:898 s J
177
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14.97
z
L
θ
R
ω
m
L
h = mRvk = m(L sin )(!L sin )k = mL2 ! sin k
h1 = 0:4(0:6)2 (12) sin2 60 k = 1:296k N m s
h2 = 0:4(0:6)2 ! 2 sin2 25 k = 0:02572! 2 k N m s
h1
!2
= h2
1:296k =0:02572! 2 k
= 50:4 rad/s J
14.98
x
A
mg
FBD
Because there are no horizontal forces acting on the projectile, the horizontal
component of velocity is constant. Therefore,
x = (v0 cos 0 ) t = (200 cos 30 ) t = 173:21t m
(AA )1 2
=
Z 10 s
MA dt =
0
=
2(9:81)(173:21)
Z 10 s
mgx dt =
0
102
2
Z 10 s
2(9:81)(173:21t) dt
0
= 169:9
103 N m s
178
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(hA )2
(hA )1
(hA )2
= (AA )1 2
(hA )2 0 = 169:9
3
= 169:9 10 N m s J
103 N m s
14.99
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
179
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14.102
14.103
Angular momentum about the z-axis is conserved: (hz )1 = (hz )2
(m! 1 R1 )R1 = (m! 2 R2 ) R2
!2 = !1
R12
= 10
R22
62
152
= 1:6 rad/s
(v )2 = ! 2 R2 = 1:6(15) = 24:0 in./s J
Kinetic energy is conserved: T1 = T2
1
1
m(! 1 R1 )2 =
m (v )22 + (vR )22
2
2
q
(vR )2
=
14.104
(! 1 R1 )2
(v )22
(102 )(62 )
24:02 = 55:0 in./s J
=
p
z
W
T
θ
N=W
R
180
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(a) Angular momentum about z-axis is conserved.
(hz )1
=
(hz )2
R1 (mv )1 = R2 (mv )2
R1 (mR1 _ 1 ) = R2 (mR2 _ 2 )
_2
=
hz
= R(mv ) = R(mR _ ) = mR2 _
h_ z
= m(2RR_ + R2 •) = 0
R1
R2
2
_1 =
1:5
0:8
2
(6) = 21:09 rad/s J
(b)
)•=
2R_ _
R
When R = 0:8 ft:
•2 =
2R_ _ 2
=
R2
2( 0:3)(21:09)
2
= 15:82 rad/s J
0:8
14.105
14.106
181
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14.107
h20
GMe (1 + e cos )
RA
10:5 106
1 + e cos 65
1 + e cos 70
)
=
=
RB
1 + e cos 0
16 106
1+e
Since e > 1, the trajectory is a hyperbola J
R=
10:5
RA
=
106
=
h0
=
h0 = RA v0
e = 1:4713
h20
GMe (1 + e cos 0)
(6:674
10:170
v0 =
h20
10 11 )(5:972
10
10
1024 )(1 + 1:4713)
2
m =s
h0
10:170 1010
= 9690 m/s J
=
RA
10:5 106
14.108
GMe
Re
= (6:674 10 11 )(5:972
= 6378 103 m
v1
1024 ) = 3:986
1014 m3 =s2
h2
h1
O
v2
R1
R2
182
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
R1 v1
) R2
= R2 v2 (angular momentum about O is conserved)
v1
6
=
R1 =
R1 = 0:8R1
v2
7:5
1 2 GMe
1 2 GMe
v1
=
v
(energy is conserved)
2
R1
2 2
R2
1
1
1 2
(v
v22 ) = GMe
2 1
R1
R2
1
1
1
(60002 75002 ) = 3:986 1014
2
R1
0:8R1
R1
h1
h2
= 9:842 106 m
R2 = 0:8(9:842 106 ) = 7:874 106 m
= R1 Re = (9:842 6:378) 106 = 3:464 106 m J
= R2 Re = (7:874 6:378) 106 = 1:496 106 m J
14.109
183
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14.110
From Prob.(14.109):
= 2:359
106 s
14.111
From Table 14.2 (for Mars):
GM = 6:674
10 11 (0:6417
1024 ) = 4:283
1013 m3 /s
2
14.112
(a)
Rmin
Rmax
GMe
= 6378 + 2500 = 8878 km
= 6378 + 4500 = 10 878 km
= (6:674 10 11 )(5:972 1024 ) = 3:986 1014 m3 =s2
Rmax Rmin
10 878 8878
e =
= 0:101 24
=
Rmax + Rmin
10 878 + 8878
184
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
vmax
vmin
=
=
s
s
GMe (1 + e)
=
Rmin
GMe (1 e)
=
Rmax
r
r
(3:986
1014 )(1 + 0:101 24)
= 7032 m/s J
8878 103
(3:986
1014 )(1 0:101 24)
= 5739 m/s J
10 878 103
(b)
h0
103 (5739) = 6:243
= Rmax vmin = 10 878
1010 m2 =s
2 h30
2 (6:243 1010 )3
=
(GMe )2 (1 e2 )3=2
(3:986 1014 )2 (1 0:101 242 )3=2
11 290 s J
=
=
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 )(6:674 10 11 )(5:972
= 5:163 1010 m2 =s
5:163 1010
h0
=
= 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
h20
GMe (1 + e)
Rmin
=
103
=
(6:674
h20
10 11 )(5:972 1024 )(1 + 0:02373)
h0
=
5:101
1010 m2 =s
h0 = Rv2
v2 =
h0
5:101
=
R
6688
L1 2 = m(v2
v1 ) = 6000(7627
1010
= 7627 m/s
103
7720) =
5:6
105 N s J
The answer is accurate to only 2 signi…cant …gures because 2 …gures were lost
in the subtraction.
185
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14.114
From Table 14.2 (for Venus):
GM
Rv
1016 ft3 /s
= (3:440 10 8 )(333:5 1021 ) = 1:1472
= 3761 mi = 19:858 106 ft
x1
2
F
Rv
y1
R1
vmax
Hmin
v1
(a)
q
x21 + y12 =
p
52 + 12
R1
=
h0
= v1 y1 = 8000(1:0
108 = 5:099
108 ) = 800
108 ft
109 ft2 /s
Equation (14.55):
E0
=
1 2
v
2 1
GM
1
= (8000)2
R1
2
=
9:501
106 (ft/s)
1016
108
1:1472
5:099
2
Since E > 0, the trajectory is a hyperbola J
(b) Equation (14.57):
s
2
h0
e =
1 + 2E0
GM
s
800 109
=
1 + 2(9:501 106 )
1:1472 1016
2
= 1:0452
Equation (14.62):
2
800 109
h20
=
GM (1 + e)
1:1472 1016 (1 + 1:0452)
Rmin
=
Hmin
= 27:278 106 ft
= Rmin Rv = (27:278 19:858)
= 7:42 106 ft = 1405 mi J
106
186
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(c) Angular momentum about F is conserved:
vmax Rmin = h0
vmax =
h0
800 109
= 29 300 ft/s J
=
Rmin
27:278 106
14.115
From Table 14.2 (for Venus):
GM
Rv
(3:440 10 8 )(333:5 1021 ) = 1:1472
3761 mi = 19:858 106 ft
=
=
2
Rv
x1
y1
1016 ft3 /s
Rmin
R1
v1
(a)
R1 =
Equation (14.55):
q
p
x21 + y12 = 52 + 12
E0 =
1 2
v
2 1
108 = 5:099
108 ft
GM
R1
For elliptical orbit E0 < 0. In the limit E0 = 0 (parabola), in which case Eq.
(14.55) yields
0
v1
1 2 GM
v
2 1
R1
r
r
2GM
2(1:1472 1016 )
=
=
= 6710 ft/s J
R1
5:099 108
=
(b) Equation (14.62):
Rmin =
h20
GM (1 + e)
Substituting h0 = v1 y1 and e = 1, we get
Rmin =
(6710)2 (1:0 108 )2
= 19:623
(1:1472 1016 )(2)
106 ft
Since Rmin < Rv , the spacecraft would crash. J
187
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14.116
(a) At insertion point:
= Re + H = (6:378 + 0:240) 106 = 6:618 106 m
= vR cos = 9600(6:618 106 ) cos 4 = 63:38 109 m2 /s
R
h0
Equation (14.55):
E0
1 2 GM
1
3:986 1014
v
= (9600)2
2
R
2
6:618 106
2
6
14:150 10 (m/s) < 0 ) Trajectory is an ellipse J
=
=
(b) Equation (14.57):
s
1 + 2E0
e =
s
h0
GM
1 + 2( 14:150
=
2
106 )
63:38
3:986
109
1014
2
= 0:5334
Solving Eq. (14.61) for :
=
cos 1
1
=
cos
=
11:45
"
1
e
h20
R GM
1
0:5334
1
63:38 109
(6:618 106 )(3:986
2
1014 )
!#
1
J
(c) Equation (14.62):
2
63:38 109
h20
=
= 6:572
GM (1 + e)
(3:986 1014 )(1 + 0:5334)
Rmin
=
Hmin
= Rmin
Re = (6:572
6:378)
106 = 1:94
106 m
105 m = 194 km J
14.117
(a) At point A:
R
E0
= Re + H = (6:378 + 0:240) 106 = 6:618 106 m
1
3:986 1014
1 2 GMe
v
= (11:4 103 )2
=
= 4:750
2
R
2
6:618 106
2
106 (m/s)
Since E0 > 0 , the trajectory is a hyperbola. Q.E.D.
188
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(b) When R ! 1, then E0 !
) v1 =
14.118
p
2E0 =
1 2
v
2 1
p
2 (4:750
106 ) = 3080 m/s J
For the circular orbits:
R1 = 3800 mi = 20:06
106 ft
R2 = 25 400 mi = 134:11
106 ft
Equation (14.66):
v1
v2
=
=
r
r
GMe
=
R1
GMe
=
R2
r
r
1:4076
20:06
1016
= 26:49
106
103 ft/s
1:4076
134:11
1016
= 10:245
106
103 ft/s
For the transfer orbit:
Rmin = R1
Rmax = R2
Equation (14.68):
e=
Rmax Rmin
134:11 20:06
= 0:7398
=
Rmax + Rmin
134:11 + 20:06
From Table 14.2:
vmax
=
s
GMe (1 + e)
=
Rmin
r
(1:4076
1016 ) (1 + 0:7398)
20:06 106
34:94 103 ft/s
s
r
GMe (1 e)
(1:4076 1016 ) (1 0:7398)
=
=
Rmax
134:11 106
=
vmin
At A
5:226
:
Lpark-tranfer = m(vmax v1 )
2000
(34:94 26:49) 103 = 525 103 lb s J
32:2
Ltransfer-geo = m(v2 vmin )
2000
(10:245 5:225) 103 = 312 103 lb s J
32:2
=
At B
103 ft/s
=
:
=
189
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14.119
14.120
GMe
Re
= (6:674 10 11 )(5:972 1024 ) = 3:986
= 6378 km
Rmax = Re + H
1014 m3 =s2
At point A:
(h0 )A
=
(vA cos ) Re = (2500 cos 45 ) (6378
(E0 )A
=
1 2
v
2 A
GMe
1
= (2500)2
Re
2
3:986
6378
103 ) = 1:1275
1014
=
103
5:937
1010 m2 =s
107 m2 =s2
At point B:
(h0 )B = vB Rmax
(E0 )B =
1 2
v
2 B
GMe
1 2
= vB
Rmax
2
3:986 1014
Rmax
190
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Angular momentum and mechanical energy are conserved:
(h0 )B
=
(h0 )A :
(E0 )B
=
(E0 )A :
vB Rmax = 1:1275 1010 m2 =s
1 2
3:986 1014
vB
= 5:937
2
Rmax
(a)
107 m2 =s2
(b)
2
Multiply Eq. (b) by Rmax
:
1 2 2
v R
2 B max
1014 Rmax + 5:937
3:986
2
107 Rmax
=0
Substitute for vB Rmax from Eq. (a):
1
2
(1:1275 1010 )2
3:986 1014 Rmax + 5:937 107 Rmax
2
2
5:937 107 Rmax
3:986 1014 Rmax + 6:356 1019
2
5:937Rmax
3:986
10
7
Rmax + 6:356
10
12
=
0
=
0
=
0
Solution is
Rmax
H
106 m = 6550 km
Re = 6550 6378 = 172:0 km J
= 6:550
= Rmax
14.121
(a)
maθ
R
FR
= maR
•
R
= R_
2
•
R
= R_
2
=
F
θ FBD
O
+%
O
•
F = m(R
3:440
2
GMe
= R_
R2
1:4076 1016
J
R2
F = ma
maR
+-
MAD
GmMe
•
= m(R
R2
10 8 (409:2 1021 )
R2
2
R_ )
0 = m(R• + 2R_ _ )
•=
2
R_ )
2R_ _
J
R
191
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
= 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
1:4076 1016
x21
x4
2x2 x4
x1
2:346
107
1:0899
10 3
2237
0
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)^2-1.4076e16/x(1)^2
x(4)
-2*x(2)*x(4)/x(1)];
end
end
192
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
7
3
x 10
2.9
2.8
R (ft)
2.7
2.6
2.5
2.4
2.3
2.2
0
6000
7000
(c) The following lines of printout span the time when
=2 :
t
7.0000e+003
7.0500e+003
1000
2000
3000
4000
t (s)
5000
8000
x1
x2
x3
x4
2.3402e+007 2.1687e+003 6.2548e+000 1.0953e-003
2.3513e+007 2.2835e+003 6.3093e+000 1.0850e-003
Use linear interpolation to …nd t at = 2 :
6:3093 6:2548
2
6:2548
=
7050 7000
t 7000
t = 7026 s J
(d) The partial printouts shown below span Rmax and Rmin :
t
2.7500e+003
2.8000e+003
2.8500e+003
x1
x2
x3
x4
2.9411e+007 1.0793e+002 2.2914e+000 6.9351e-004
2.9414e+007 1.5877e+000 2.3261e+000 6.9338e-004
2.9412e+007 -1.0476e+002 2.3608e+000 6.9350e-004
6.2500e+003
6.3000e+003
6.3500e+003
2.2616e+007 -2.3204e+002
2.2609e+007 -5.2497e+001
2.2611e+007 1.2733e+002
5.3933e+000
5.4520e+000
5.5107e+000
1.1728e-003
1.1735e-003
1.1733e-003
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
193
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14.122
(a) The equations of motion were derived in the solution of Prob. 14.121:
1:4076 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
1:4076 1016
x21
107
1917:4
0
x4
0:9342
2x2 x4
x1
10 3
T
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)^2-1.4076e16/x(1)^2
x(4)
-2*x(2)*x(4)/x(1)];
end
end
194
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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 = 2:0925
t
1.4400e+003
1.4500e+003
x1
x2
2.0942e+007 -5.0507e+003
2.0892e+007 -5.0835e+003
70
80
90
107 ft:
x3
1.3984e+000
1.4102e+000
x4
1.1723e-003
1.1780e-003
Linear interpolation:
2:0892
1:4102
2:0942
2:0925 2:0942
=
1:3984
1:3984
= 1:4024 rad = 80:4
J
14.123
y 24 lb
10 lb
FBD
o
0.2N
Fy
L1 2
30
x
N
= 0
N 24 + 10 sin 30 = 0
N = 19:0 lb
= m(v2 v1 ):
Fx t = m(v2 v1 )
24
[10 cos 30
0:2(19:0)] t =
(50 15)
32:2
t = 5:37 s J
195
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14.124
Since there is no impulse in the x-direction, vx does not change:
vx = 3 cos 35 = 2:458 m/s (constant)
(L1 2 )y
=
(py )2
Z 10 s
(py )1 :
P dt = m(vy )2
m(vy )1
0
(vy )2
v2
14.125
Z
1 10 s 0:3t
e
dt = 3 sin 35
m 0
=
2:239 m/s
p
2:4582 + 2:2392 = 3:32 m/s J
=
=
1
(3:1674)
0:8
(vy )1
(a)
T A + VA
= TB + VB = TC + VC
1
1
2
=
mv 2 + 0 = mvC
+ 8mg
2 B
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
v2
N =m C
=
2
mvC
mg N
Contact is lost when N = 0.
)
=
2
vC
17:722
=
= 32:0 m J
g
9:81
14.126
U1 2
1 2
k =0
2
1
1 35
0:2(35)(5 + ) + (280) 2 =
(12)2
2
2 32:2
= 0:5314 ft
= T2
T1 :
k W (5 +
)
1
mv 2
2 1
F = k = 280(0:5314) = 148:8 lb J
196
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14.127
Choose O as the datum for gravitational potential energy.
Between positions A and B energy is conserved.
TA + VA
2
vB
= T B + VB
=
2gR1 (1
0
1
mv 2
2 B
mgR1 cos 1 =
mgR1
cos 1 )
Between positions B and C angular momentum about point O is conserved.
(h0 )C
2
vC
=
(h0 )B
2
= vB
mvC R2 = mvB R1
R1
R2
2
= 2gR1 (1
cos 1 )
R1
R2
2
Between positions C and D energy is conserved.
1
mv 2
2 C
2
vC
= 2gR2 (1
mgR2 = 0
TC + VC = TD + VD
cos 2 = 1
cos 2 )
2gR1 (1
cos 1 )
R1
R2
2
cos 1 )
R1
R2
3
(1
(1
cos 1 )
mgR2 cos 2
= 2gR2 (1
cos 2 )
=1
cos 2
=1
(1
cos 45 )(1:05)3 = 0:6609
2 = 48:6
J
R1
R2
3
14.128
197
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14.129
y mg
P(N)
60
30
0
0
4
FBD Ps
µsmg
x
N = mg
8
12
t(s)
First determine if the crate will move. The smallest force that would move the
crate is
Ps = s mg = 0:3(16)(9:81) = 47:09 N
Since Ps < 60 N, the crate will move.
L1 2
= (area under P -t diagram)
t
k mg
1
= 60 8
(30 4) 0:25(16)(9:81)(8) = 106:08 N s
2
L1 2 = m(v2
v1 ):
106:08 = 16(v2
0)
v2 = 6:63 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
O
VO
TO
TA + VA
vO
p
= OB L0 = 42 + 32 2:5 = 2:50 ft
1 2
1
=
k O = (4:2)(2:50)2 = 13:125 lb ft
2
2
1
1 8 2
2
2
=
mvO =
v = 0:12422vO
lb ft
2
2 32:2 O
2
= TO + VO
0 + 83:88 = 0:12422vO
+ 13:125
= 23:9 ft/s J
14.131
198
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14.132
14.133
Angular momentum about AB is conserved:
m(R1 ! 1 )R1 = m(R2 ! 2 )R2
T
= T2
T
T1
=
R2 ! 2
R1 ! 1
% energy lost
=
T
T1
1
m(R2 ! 2 )2
2
2
R2
1=
R1
"
T1 =
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
14.134
Energy is conserved:
TA + VA
vB
1W 2
1
1W 2
1
= 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
=
vA +
( A
252 +
(0:52 1:02 )
B) =
W
1:2
= 23:97 ft/s J
199
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Angular momentum about O is conserved:
RA vA = RB (v )B
(v )B =
Rate of elongation
=
14.135
(px )A
vB
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
= (px )B :
mvA cos = mvB
= vA cos = 20 cos 35 = 16:383 m/s
(a)
mg
FBD
(LA B )y
t
=
(py )B
vA sin
=
g
(py )A :
mgt = 0 mvA sin
20 sin 30
=
= 1:019 s J
9:81
(b)
TA + VA
h
1
1
2
mv 2 + 0 = mvB
+ mgh
2 A
2
2
2
vB
vA
202 16:3832
=
= 6:71 m J
2g
2(9:81)
= T B + VB :
=
14.136
Dimensions in
108 feet
38.41o
y
3.960
5.0
vB
B
3.382
x
1.0
O
A
vA = 16 000 ft/s
200
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Angular momentum about point O is conserved.
m(16 000)(1:0) = mvB (3:960 cos 38:41
3:382 sin 38:41 )
16 000 = 1:0018vB
vB = 15 970 ft/s J
14.137
201
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14.138
14.139
LA
A
m
0.4
30o 0.2 m
Datum
L2A = 0:42 + 0:22
VA
=
=
=
VB
=
=
TB
=
2(0:4)(0:2) cos 30
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
2
mvB
= (0:6)vB
= 0:3vB
2
2
TA + VA = TB + VB
2
0 + 3:589 = 0:3vB
+ 3:146
vB = 1:215 m/s J
202
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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
v =
v = 0:6211v22 lb ft
2 g 2
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
203
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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
vA = 40 m/s
vB = 15 + 0:15t m/s
vB=A
= vB
vA =
xB=A
=
25 + 0:15t m/s
25t + 0:075t2 + C
Initial condition: xB=A = 3000 m when t = 0. ) C = 3000 m
xB=A =
25t + 0:075t2 + 3000
0:
25 + 0:15t = 0
(xB=A )min occurs when
vB=A
=
xB=A min
=
t = 166:67 s J
25(166:67) + 0:075(166:672 ) + 3000 = 917 m J
15.3
For aB=A to be zero, aB must be parallel to aA (vertical).
(at )B
o
60
(an )B
=
aB
=
aB=A = aB
2
vB
aB
(an)B
162
= 1:280 m/s
200
(an )B
1:280
2
=
= 1:478 m/s
sin 60
sin 60
aA = 0
=
aA = aB = 1:478 m/s # J
2
204
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15.4
Let P and W refer to the plane and the wind, respectively.
= vW + vP=W
vP
y
x
vP (sin i + cos j)
=
55(cos 28 i + sin 28 j) + 520j
Equating like components:
vP sin
vP cos
= 55 cos 28 = 48:56 mi/h
= 55 sin 28 + 520 = 545:82 mi/h
48:56
tan 1
= 5:084 J
545:82
p
=
48:562 + 545:822 = 548 mi/h J
=
vP
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
10
=
= 0:4167
vB=W
24
= 24:6
J
(b)
vB
=
t
=
q
2
vB=W
2 =
vW
p
242
102 = 21:82 km/h
d
4:8
=
= 0:220 h = 13:2 min J
vB
21:82
15.6
Balls collide if vA=B is directed from A toward B.
vA
= vB + vA=B
2j
=
y
x
3 (cos 30 i + sin 30 j) + vA=B ( sin i + cos j)
205
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Equating like components:
0 = 3 cos 30
vA=B sin
2 = 3 sin 30 + vA=B cos
= tan 1
vA=B sin = 2:598
vA=B cos = 0:5
2:598
= 79:11
0:5
J
15.7
15.8
vA
vB
vA=B
vA=B
15.9
= 50i mi/h
= (60 cos 25 ) i + (60 sin 25 ) j
= 54:38i + 25:36j mi/h
= vA vB = 4:38i 25:36j mi/h
p
=
4:382 + 25:362 = 25:74 mi/h J
206
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15.10
15.11
207
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15.12
15.13
vA
vB
vB=A
= 70j ft/s
= 40(i cos 30 + j sin 30 ) = 34:64i + 20:0j ft/s
= vB vA = 34:6i 50:0j ft/s J
6j ft/s
2
aA
=
aB
=
(aB )t + (aB )n = v_ B (i cos 30 + j sin 30 ) +
=
4(i cos 30 + j sin 30 ) +
402
(i sin 30
200
2
vB
(i sin 30
j cos 30 )
j cos 30 )
2
aB=A
= 7:464i 4:928j ft/s
= aB aA = 7:464i 4:928j
=
7:46i + 1:07j ft/s
2
( 6j)
J
15.14
Let W and B refer to the wind and the boat, respectively.
Position (a):
vW = (vB )a + (vW=B )a = 6j + va ( i cos 5 + j sin 5 )
208
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Position (b):
vW = (vB )b + (vW=B )b = 6i + vb ( i cos 28 + j sin 28 )
Equating like components:
va cos 5
6 + va sin 5
= 6 vb cos 28
= vb sin 28
Solution is
va = 6:349 mi/h
vb = 13:959 mi/h
6:32i + 6:55j mi/h J
vW = 6j + (6:349) ( i cos 5 + j sin 5 ) =
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) =
yB
xA
6 in./s = 6 in./s ! J
15.16
A
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
209
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Length of cable:
dL
dt
vA=B
vB
= vA
=
vB
1
(8) =
2
L = 2yB + yA + constant
=
0: 2vB + vA = 0
12 = vA
1
vA
2
vB =
1
vA
2
vA = 8 in./s # J
4 in./s = 4 in./s " J
15.18
15.19
210
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15.20
yA
yB
A
B
yC
C
Length of cable
= L = (yC yA ) + 2yC + (yC
= 4yC yA yB + constant
dL
dt
=
vC
=
vC
=
4vC
vA
yB ) + constant
vB = 0
vA + vB
3 + 1:2
=
=
4
4
0:45 ft/s " J
0:45 ft/s
15.21
1.2 m
yB
yA
B
A
Length of cable = L = 2yB +
dL
dt
=
vB
=
vB
=
q
2 + 22 + constant
yA
yA vA
2vB + p 2
=0
yA + 22
1
y
1
2
p A
p
vA =
(0:8) =
2
2
2
2 yA + 2
2 2 + 1:22
0:343 m/s
0:343 m/s " J
15.22
2.4 m
2.4 m
yB
yA
A
B
211
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Length of cable
dL
dt
vB
q
2 + 2:42 + constant
= L = y A + 2 yB
yB
vB = 0
= vA + 2 p 2
yB + 2:42
p
2 + 2:42
yB
=
vA
2yB
p
22 + 2:42
=
( 3:6) = 2:81 m/s " J
2(2)
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
x2 = L2
But L_ =
242
) 2xx_ = 2LL_
x_ =
LL_
x
10 in./s
) x_ =
10
L
x
Using x = 18 in.
L =
)
p
182 + 242 = 30 in.
30
x_ = 10
= 16:67 in./s = 16:67 in./s ! J
18
212
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15.25
15.26
213
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15.27
1
yC
yA
. C
D
.
yB
B
2
A
L1
L2
L_ 1
L_ 2
= yB + yC + (constant)
= 2(yA yC ) + yA + (constant) = 3yA
= vB + vC = 0
= 3vA 2vC = 0
2yC + (constant)
6 + vC = 0
vC = 6 in./s
3vA 2(6) = 0
vA = 4:00 in./s # J
15.28
214
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15.29
15.30
215
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
15.31
y
xB
xA
.
GA
B
A
x
(a) The coordinate x of the mass center of the system is given by
mA xA + mB xB = (mA + mB )x
Since there are no external forces acting on the system in the x-direction, x does
not change. Therefore, di¤erentiation with respect to time gives
mA vA + mB vB = 0
) vB =
vA
mA
=
mB
( 4:4)
18
= 1320 ft/s J
0:06
(b) The muzzle velocity is
vB=A = vB
( 4:4) = 1324 ft/s J
vA = 1320
15.32
20 ft
A
B
G
d
x
20 ft
A
B
x
The mass center G of the system is located at
x=
mi xi
(10)600
=
= 7:692 ft
mi
600 + 180
216
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
The mass center of the system does not move. Hence the displacement d of the
log is given by
) d + 2x = L
d=L
2x = 20
2(7:692) = 4:62 ft J
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
217
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
When t = 35 s:
x =
y =
120(35) = 4200 m
4:905(35)2 + 207:9(35) = 1267:9 m
(mA + mB )x = mA xA + mB xB
(20 + 40)(4200) = 20(3820) + 40xB
(mA + mB )y = mA yA + mB yB
(20 + 40)(1267:9) = 20(1960) + 40yB
xB = 4390 m J
yB = 922 m J
15.35
(a)
2000g
1400g
B
A
B
A
y
0.8(2000)g
1400g
(2000+1400)a
2000g
FBD
x
MAD
Fx
= max
0:8(2000)g = (2000 + 1400)a
0:8(2000)(9:81)
2
a =
= 4:616 m/s J
2000 + 1400
(b)
1400g
B
FBD
Fx = max + !
1400a
T
B
MAD
1400g
T = 1400a = 1400(4:616) = 6460 N = 6:46 kN J
218
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
15.36
15.37
System (crate + cart):
Fx = max
+ !
Crate:
100 + 300
a
32:2
P =
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
15.38
Check if the crate will slide on the cart:
s NA = 0:2(100) = 20 lb < 32 lb
) The crate will slide.
219
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100 lb
32 lb
A
0.18(100) = 18 lb
MAD
FBD NA= 100 lb
(100+300) = 400 lb
18 lb
300 a
32.2 B
B
B
FBD 400 lb
MAD
100
aA
32:2
Crate A :
Fx = mA aA + !
32
18 =
Cart B
Fx = mB aB + !
18 =
300
aB
32:2
:
100 a
32.2 A
A
aA = 4:51 ft/s
aB = 1:932 ft/s
2
2
J
J
15.39
System:
2850 N
A
80g N
A
80a
FBD
F
MAD
B
B
110g N
110a
= ma + "
2850
a = 5:190 m/s
2
(80 + 110)(9:81) = (80 + 110)a
J
220
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Block A:
2850/2 N
A
80g N
FBD
A
80a MAD
T
F
= ma + "
T
=
225 N J
2850
2
80(9:81)
T = 80(5:190)
15.40
Kinematics
xB
B
yA
A
• = 3aA + aB = 0
L
L = 3yA + xB + (constant)
aB =
3aA
Kinetics
3T
A
6aA
6(9.81) N
FBD
MAD
12(9.81) N
T
12aB
B
FBD N
MAD
221
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Block A :
Block B :
Solution is
Fy = mA aA + #
Fx = mB aB + !
T = 18:59 N J
6(9:81) 3T = 6aA
T = 12aB = 12( 3aA )
aA = 0:516 m/s # J
2
15.41
222
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15.42
15.43
Kinematic constraints: aA
= aB ! = aC # = a
223
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Kinetics:
24 a
32.2
24 lb
TAB
A
A
MAD
0.4(24) = 9.6 lb
24 lb
FBD
24 a
32.2
48 lb
TAB
TBC
B
0.4(48) = 19.2 lb
B
MAD
48 lb
TBC
FBD
FBD C
C
40 lb
MAD
40 a
32.2
Block A :
Fx = maA +
TAB
9:6 =
24
a
32:2
Block B
:
Fx = maB + !
TBC
TAB
19:2 =
Block C
:
Fy = maC + #
40
TBC =
24
a
32:2
40
a
32:2
The solution is
a = 4:10 ft/s
2
TAB = 12:66 lb J
TBC = 34:9 lb J
15.44
Kinematics: aB =
1
aA = 2 m/s.
2
WA
P
T
A
FBD's
N
y
T T
A
mA aA
MAD's
mBaB
x
B
B
WB
224
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
15.45
0.8 lb
P
A
N
x
B
FBD
0.2 lb
0.8 ag
y
T
T
40o
B
A
0.2 ag
MAD
Bob B:
Fy
Fx
=
0
T cos 40
0:2 = 0
T = 0:2611 lb
0:2
a
= ma
T sin 40 =
g
0:2
2
0:2611 sin 40 =
a
a = 27:02 ft/s
32:2
Block A:
Fx
= ma
P
0:8
a
32:2
0:8
0:2611 sin 40 =
(27:02)
32:2
P
T sin 40 =
P = 0:839 lb J
15.46
o
15o mg 45o
mg 15
o
T
A 15
B t
t
T
N
NA
B
FBD's
Ft = mat
+&
A
man
B
mat
man
MAD's
mg sin 15 + T cos 15
mg sin 45
T cos 15
= mat
= mat
mat
(Block A)
(Block B)
225
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
T
40ο
µANA
A
NA
y
x
µBNB
Α
WB
B
mAa
NB
T
FBD's
Β
mBa
MAD's
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
60 sin 40 =
a
32:2
T 24:78 =
1:8634a
(b)
a = 14:78 ft/s2 and T = 2:76 N J
15.48
Kinematics:
x2 + y 2 = 252
2xx_ + 2y y_ = 0
2
x•
x + x_ + y y• + y_ 2 = 0
Immediately after the release we have x_ = y_ = 0. Thus
x•
x + y y• = 0
y• =
x
x
•=
y
20
x
•=
15
4
x
•
3
(a)
226
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Kinetics:
4 ..
32.2 y
4 lb
NA
A
A
P
3
FBD
MAD
4
3 ..
x
32.2
3 lb
P
B
B
3
NB
Block A:
4
MAD
FBD
Fy = mA aA + "
Substituting for y• from Eq. (a), we get
Block B:
Fx = mB aB + !
3
4
P 4=
y•
5
32:2
3
4
P 4=
5
32:2
3
4
P =
x
•
5
32:2
4
x
•
3
(b)
(c)
The solution of Eqs. (b) and (c) yields
x
• = 16:99 ft/s
2
P = 1:978 lb J
15.49
yC
yA
C
1
2
yB
W B
W A
Kinematics:
Cable 1 :
Cable 2 :
yA + (yA yC ) = constant ) 2aA aC = 0
yB + 2yC = constant
) aB + 2aC = 0
aC = 2aA
aB = 4aA
227
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Kinetics:
F = ma + #
T1
T1
T2
B
T1
FBD's
W
W
:
T1
Block A :
W
Block B
W
:
T2
C
A
Pulley C
T2
2T2 = 0
T1 = 2T2
W
W
2T1 =
aA
W 4T2 =
aA
g
g
W
W
aB
W T2 = 4 aA
T2 =
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
*15.50
228
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15.51
229
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
15.52
(a)
2
mg
k(x1 − x2)
N
a1
=
a1
=
a2
=
1
F
=
2
N
FBD's
Block 1:
Block 2:
mg
ma2
1
ma1
MAD's
Fx = max
+ ! F k(x1 x2 ) = ma1
1
32:2
[F k(x1 x2 )] =
[1:2 (2 12) (x1 x2 )]
m
5
2
7:728 154:56(x1 x2 ) ft/s J
Fx = max
+ ! k(x1 x2 ) = ma2
2 12
k
(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
x_ 2
T
, 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
230
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
15.53
(a)
xB
xA
d
F F
A
FBD's
Particle A
:
aA
=
B
mAaA
=
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
+ !
F
0:005=(xB xA )2
=
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
x4
x(0)
=
0
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
231
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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
F
50(vB vA )
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
232
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(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
10 4 m = 0:567 mm J
xA = 5:67
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
233
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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
vB
T
, 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
234
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
mAaA
P
A
F
N
N F
=
MAD's
mBg
B
mBaB
NB
F =
k N sgn (vA
Block A
:
aA
=
vB )
P = mA aA
k
vB )
xA
mA
2000
0:3(9:81) sgn (vA vB )
xA
2
2
2:943 sgn (vA vB ) 1000xA m/s J
=
aB
k mA g sgn (vA
Fx = max
+ !
F P
=
k g sgn (vA
mA
=
Block B
vB ) =
F
:
Fx = max
+ ! F = mB aB
mA
F
= k
g sgn (vA vB )
=
mB
mB
2
= 0:3
(9:81) sgn (vA vB )
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
vB
2:943 sgn (x3
0:02 m
vA (0) = vB (0) = 0 J
0:02 m
0
T
, the equivalent …rst-order equations
x4 )
1000x1
1:4715 sgn (x3
x4 )
0
The ‡ollowing MATLAB program was used for the plot:
235
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
236
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
= 4:7124 rad/s
O
FR
FBD
F
maθ
R
= θ
MAD
maR
Fθ
k mg opposes the relative velocity vector v
vR
=
v
k mg
FR
= maR
•
R
= R_
2
= R_
2
F
!)2
R_
v
+&
where
v=
Initial conditions:
q
R_ 2 + R2 ( _
_
T
x1 x24
9:66
x2
v
x2
x(0)
=
1:0 ft 0
0
where
v=
9:66
!)
v
2
R_ )
R_
J
v
4:7124)2
_
R(0)
= (0) = _ (0) = 0 J
(b) Letting x = R R_
the initial conditions are
=
•
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
+%
2R_ _
R
2R_ _
R
=
v
=
v
2
FR
= R_
m
2
R_
0:3(32:2) = R _
v
= ma
• =
F =F
, the equivalent …rst-order equations and
x4
2x2 x4
x1
9:66
x4
4:7124
v
0
q
x22 + x21 (x4
4:7124)2
The corresponding MATLAB program is
237
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
R_ 3:6938
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
O
θ
400
A
Spring
B
L
400(1 - cosθ)
400 sinθ
350
= sin 1
0
40
θ
0
60
Dimensions
in mm
400
= 41:81
600
q
2
2
L =
(0:35 + 0:4 sin 41:81 ) + [0:4(1 cos 41:81 )] = 0:6250 m
= L L0 = 0:6250 0:35 = 0:2750 m
238
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Because the system rotates about O, we have
!=
vA
vB
=
0:6
0:4
vB =
2
vA
3
2
T2
=
=
1
2
2
1
1
2
2
=
mA vA
vA
mA + mB vA
+ mB
2
2
3
2
9
1
2
2
2
(4) + (2) vA
= 2:444vA
2
9
Choose the initial position as the datum for gravitational potential energy.
yA
=
V2
=
=
=
cos 41:81 ) = 0:101 86 m "
1 2
mA g yA + mA g yB + k
2
1
4(9:81)(0:4) + 2(9:81)(0:101 86) + (240)(0:2750)2
2
4:623 N m
0:4 m #
T1 + V 1 = T 2 + V 2
yB = 0:4(1
2
0 + 0 = 2:444vA
vA = 1:375 m/s J
4:623
15.59
vA = 22 ft/s
U1 2 = T2
vB = 11 ft/s "
)
xA = 2 yB
T1 :
1 2
1
1
2
2
k
xA WB yB = 0
mA vA
+ mB vB
k WA
2
2
2
1
1
5
9
1
k(5:42 ) 0:2(5)(5:4) 9(2:7) =
(222 )
2
2 32:2
2 32:2
k = 1:700 lb/ft J
(112 )
239
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
15.60
15.61
240
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
y
δ
x
The mass center of the system does not move:
mA xA + mB xB = mA + mB (xB + )
where xB is the centroidal coordinate of the cart in the initial position.
mA xA = (mA + mB )
mA xA
15(2:5)
= 0:798 m J
=
mA + mB
15 + 32
=
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
0 + k 2 = mA vA
+ mB vB
+0
2
2
2
1 0:75 2
1 0:5
v +
( 1:5vA )2
2 32:2 A 2 32:2
= T2 + V2
2
=
0:0675 =
vB =
vA = 1:523 ft/s
2
0:029 11vA
vA = 1:5228 ft/s
1:5( 1:5228) = 2:28 ft/s
J
vB = 2:28 ft/s ! J
241
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
15.65
15.66
) 2vA + vB = 0
Constraint: 2yA + xB = constant
T
vB =
2vA
T
T
B
A
7 lb
Block A: + # L1 2
(7
2T ) t
Block B: + ! L1 2
Tt
= p2
=
(7
2T ) t =
7
(6)
32:2
(7
2T ) t = 1:3043
= p2
=
7
vA
32:2
p1 :
p1 :
Tt =
12
( 12) = 4:472
32:2
12
vB
32:2
0
(a)
0
(b)
242
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Substituting (b) into (a):
7t
t = 1:464 s J
2(4:472) = 1:3043
15.67
Let vB = …nal velocity of block and vC = …nal velocity of cart, both positive to
the right.
Mometum is conserved:
p1
= p2 + !
mB
=
vB =
mC
vC
0 = mC vC + mB vB
8
vB = 0:4vB
20
Energy is conserved:
T 2 + V2 = T 1 + V1
vC
vB=C
vB=C
1
1
2
2
mC vC
+ mB vB
+0 =
2
2
1
1 2
(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
Constraint: 2yA + yB = constant
T
) 2vA + vB = 0
T
2vA
T
B
A
8 lb
+ " L1 2 = p2
vB =
6 lb
p1 :
Block A :
(2T
WA )t = mA vA
0
(2T
Block B
(T
WB ) t = mB vB
0
(T
:
8
vA
32:2
6
6)(7) =
( 2vA )
32:2
8)(7) =
The solution is T = 4:50 lb and
vA = 28:2 ft/s " J
243
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15.69
15.70
Since there are no external forces acting on the system in the horizontal direction, we have
p 1 = p2
0=
400
800
vA +
vB
g
g
vA =
2vB
400 lb
180 lb
(0.35)(400) lb
N =400 lb
FBD
(U1 2 )ext + (U1 2 )int
0 + [180
= T2 T1
1 800 2
1 400 2
vA +
v
(0:35)(400)] (8) =
2 32:2
2 32:2 B
1 400
1 800 2
2
2
320 =
( 2vB ) +
vB = 37:27vB
2
32:2
2
32:2
r
320
vB =
= 2:93 ft/s J
37:27
244
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15.71
15.72
245
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
15.73
246
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
15.74
15.75
247
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15.76
L
mωL
L
O
Momentum
diagram
L/3
mωL
G
2L/3
mωL
hG = 2(m!L)L + (m!L)
2L
8
= m!L2
3
3
J
15.77
z
8 m/s
2m
6 m/s
O
m
2 12 m 3
/s
y
x 1 4 3m
m
m
4
A
i
1
2
3
ri
4i
4i + 4j + 3k
3i + 2j
ri=A
4j
3k
7i 2j
=
X
=
340i
vi
8k
12j
6i
3
ri
vi
32j
36i + 48k
12k
36i 32j + 60k
(a) hO
= m
(b) hA
180i 160j + 300k N m s J
X
= m
ri=A vi = 5( 68i 12k)
ri
vi = 5( 36i
ri=A vi
32i
36i
12k
68i 12k
32j + 60k)
60k N m s J
248
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15.78
(a)
(AO )1 2
=
(hO )2
8 ( t)
=
140 4(0:62 ) + 7(0:32 )
(b)
+
!=
(hO )1
Z
+
C0 t = 0
_ ( mA r2
A
t = 36:23 s J
dt + C1 = t + C1
( is constant)
When t = 0, ! = 140 rad/s. ) C1 = 140 rad/s
When t = 36:23 s, ! = 0. ) 0 = (36:23) 140 )
+
d
When
= 0, ! =
When ! = 0,
=
2
mB r B
)
= 3:864 rad/s2
d! d
d!
d!
=
=
!
dt
d dt
d
1
= ! d!
3:864 = ! 2 + C2
2
=
1
( 140)2 = 9800 (rad/s)2
140 rad/s. ) C2 =
2
2536
9800
= 2536 rad. ) =
= 404 rev J
3:864
2
15.79
(a)
.
L
O
mv 2
L
mv1
.
O
Just after
impact
L
Momentum diagrams
Just before
impact
2mv2
Angular momentum of the system about O is conserved:
(hO )1
=
(hO )2
mv1 L = 3mv2 L
!2
=
v2
v1
=
J
L
3L
v2 =
1
v1
3
249
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(b)
2
1
1
1
1
1
mv 2
T2 = (3m)v22 = (3m)
v1
= mv12
2 1
2
2
3
6
(1=6) (1=2)
T2 T1
% energy change =
100% =
100% = 66:7%
T1
(1=2)
% energy lost = 66:7% J
T1
=
15.80
250
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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
3
v0
2
Final momenta
3mv0
mv0 = 3mvx + mvx
0 = m(vA )y + 3m(vB )y
1
v0
2
3(vB )y
vx =
(vA )y =
3L
L
3L
L
= (hG )2 +
mv0
+ 3mv0 = m(vA )y
+ 3m(vB )y
4
4
4
4
3
3
1
=
[(vB )y (vA )y ] = [(vB )y + 3(vB )y ]
(vB )y = v0
4
4
2
1
3
v0 =
v0
) (vA )y = 3
2
2
) vA =
1
i
2
1
1
i + j v0 J
2
2
3
j v0 J
2
vB =
0.3 m
0.3 m
15.82
yA
.
yB
A
B
q
2 + 0:32
y A + 2 yB
= L
) vA
=
When yB
=
yA = L
2y v
p B B
2 + 0:32
yB
0:4 m: vA =
q
2 + 0:32
2 yB
p
2(0:4)vB
=
0:42 + 0:32
1:60vB
251
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0.3 m
∆yA
A
m
A
m
0.5
m
.
m
0.5
B 2m
Position 2
Position 1 (Datum)
yA = yA jyB = 0
0.3 m
0.4 m
B 2m
.
yA jyB =0:25 m = [L
2(0:3)]
[L
2(0:5)] = 0:4 m
T1 + V1 = T2 + V2 :
1
1
2
mv 2 + (2m) vB
+ mg yA
2 A 2
1
2
0 =
( 1:60vB )2 + vB
+ 9:81(0:4)
2
vB = 1:312 m/s # J
0+0
=
2mg(0:4)
2(9:81)(0:4)
15.83
252
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15.84
2m(vB)y
B
A
mω0 L
L/2
.G L/2 B ω
0
2m (vB)x
L/2
G.
Position 2
L/2
Position 1
A
m(vA)y
m(vA)x
Because there are no external forces in the xy-plane, linear and angular momenta
about the mass center G are conserved.
Because the bar is rigid (vA )y = (vB )y
(px )1
=
)
(py )1
=
)
(px )2 : 0 = m(vA )x + 2m(vB )x
1
(vB )x =
(vA )x
2
(py )2 :
m! 0 L = m(vA )y + 2m(vB )y
!0 L
(vA )y = (vB )y =
3
!0 L =
3(vA )y
253
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
=
(hG )2 :
(m! 0 L)
L
L
= m(vA )x
2
2
!0 L =
(vA )x
2(vB )x
! 0 L = (vA )x
(vA )x =
!0 L
2
1
i
2
1
j
3
(hG )1
)
vA = ! 0 L
(vB )x =
J
2m(vB )x
2
L
2
1
(vA )x = 2(vA )x
2
!0 L
4
vB = ! 0 L
1
i
4
1
j
3
J
15.85
15.86
254
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#15.87
255
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15.88
15.89
+! (px )1 = (px )2 :
0 = 125 000v
106(1640) cos 35
v = 1:139 m/s J
256
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
15.90
p1
v…nal
= p2 + !
mA vA + mB vB = (mA + mB )v…nal
mA vA + mB vB
40(5) + 55(3)
=
=
= 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 = p 2 :
mA vA = (mA + mB ) v…nal
mA
0:008
v…nal =
vA =
vA = 0:004 975vA
mA + mB
1:6 + 0:008
1
1
2
(mA + mB )v…nal
+0 = 0+ k 2
2
2
1
1
2
(1:608) (0:004975vA ) =
(6500)(0:052)2
2
2
vA = 665 m/s J
T2 + V2 = T3 + V3 :
15.92
Position 1 = release position; Position 2 = just before impact; Position 3 = just
after impact; Position 4 = …nal (rest) position
U1 2
v2
U3 4
d
1 WA 2
= T2 T1
WA h =
v
2 g 2
p
p
=
2gh = 2(32:2)(8) = 22:70 ft/s
p3
= p3
v3
=
= T4
=
+
mA v2 = (mA + mB )v3
mA
8
v2 =
(22:70) = 6:985 ft/s
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)
257
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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 3
k (WA + WB )d
= T3
=
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
(8) cos 30 = 1:4846 ft/s J
vB cos 30 =
mA + mB
280
=
(b)
% energy lost =
1
=
1
T2
T1
100% = 1
!
2
280 (1:4846)
60 (8)
2
2
(mA + mB ) v…nal
2
mB vB
100%
100% = 83:9% J
15.95
Position 1 = just before collision; Position 2 = just after collision; Position 3 =
…nal (rest) position
p1
v…nal
= p2
=
+ !
mA vA = (mA + mB )v…nal
mA
5000
vA =
vA = 0:5814vA
mA + mB
8600
258
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
U2 3
= T3
0:8(3600)(22) =
T2
k WB d = 0
1 8600
2
(0:5814vA )
2 32:2
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:
1
T 1 + V 1 = T 2 + V2 :
0 + mgh1 = mv22 + 0
2
1
(8)v22
v2 = 3:388 m/s
8(9:81)(2:5)(1 cos 40 ) =
2
Horizontal momentum is conserved during impact:
+ ! p2 = p02 :
mv2 = mtot v20
8(3:388) = 11v20
v20 = 2:464 m/s
Energy beween Positions 2 and 3 is conserved:
1
2
mtot (v20 ) + 0 = 0 + mtot gh3
T 2 + V2 = T 3 + V3 :
2
1
(11)(2:464)2 + 0 = 0 + (11)(9:81)(2:5)(1 cos )
2
cos
= 0:8762
= 28:8 J
15.97
(a) Momentum is conserved during impact between carts (note that the parcel
keeps going at v = 15 ft/s):
p1
= p01
+ !
(mA + mC )v = (mA + mB )v 0 + mC v
mA
50
15 = 7:5 ft/s J
v=
mA + mB
100
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
0
100(7:5) + 35(15)
(mA + mB ) v + mC v
=
= 9:44 ft/s J
=
mA + mB + mC
135
15.98
(a) p1 = p2 :
[mA (vA )1
[4200(44)
mB (vB )1 sin 30 ] i + mB (vB )1 (cos 30 )j = (mA + mB )v2
3500(60) sin 30 ] i + 3500(60)(cos 30 )j = (4200 + 3500)v2
v2 = 10:36i + 23:6j ft/s J
259
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(b)
T1
=
=
T2
1
1
1 4200
1 3500
mA (vA )21 + mB (vB )21 =
(442 ) +
(602 )
2
2
2 32:2
2 32:2
321:9 103 lb ft
=
=
1 4200 + 3500
1
(mA + mB )v22 =
(10:362 + 23:62 )
2
2
32:2
79:4 103 lb ft
Energy absorbed = T1
T2 = (321:9
79:4)
103 = 243
103 lb ft J
15.99
p1
p2
= mA vA (i cos 25 + j sin 25 ) mB vB j
8
2:8=16
=
(130)(i cos 25 + j sin 25 )
(14)j
32:2
32:2
= 0:6403i 3:180j slug ft/s
(2:8=16) + 8
= (mA + mB )v2 =
v2 = 0:2539v2
32:2
p1
v2
= p2
0:6403i 3:180j = 0:2539v2
= 2:52i 12:52j ft/s J
15.100
A
Datum
4 ft
δ
B
A
B
Position 4
Position 1
Position 1: Release position.
Note that the springs are compressed by 0 = WB =k = 48=1200 = 0:04 ft
Position 2: Just before the impact.
Position 3: Just after the impact.
Position 4: Position of maximum displacement. Springs are compressed by
additional .
T1 + V1 = T2 + V2 :
p
p
1
v2 = 2gh = 2(32:2)(4) = 16:050 ft/s
0 + mA gh = mA v22 + 0
2
260
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
p 2 = p3 :
mA v2 = (mA + mB )v3
v3 =
mA
60
v2 =
(16:050) = 8:917 ft/s
mA + mB
60 + 48
T3 + V3 = T4 + V4 :
1
1
(mA + mB )v32 + 0 = 0 (mA + mB )g + k( 2
2
2
1 60 + 48
1
(8:9172 ) =
(60 + 48) + (1200) 2
2
32:2
2
2
0)
0:042
133:34 =
108 + 600 2 0:96
0 = 600 2 108
134:30
= 0:572 ft J
4
15.101
261
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15.102
262
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
15.103
15.104
15.105
263
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
15.106
264
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
*15.107
15.108
265
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
15.109
p 1 = p2 :
m(vA )1 + m(vB )1
(vA )1 + (vB )1
= m(vA )2 + m(vB )2
= (vA )2 + (vB )2
(a)
e = vsep =vapp :
(vB )2 (vA )2
(vA )1 (vB )1
e(vA )1 e(vB )1 = (vB )2
e =
(vA )2
(b)
Subtract Eq. (a) from Eq. (b):
(vA )1 (1
e) + (vB )1 (1 + e) = 2(vA )2
Q.E.D.
Add Eqs. (a) and (b):
(vA )1 (1 + e) + (vB )1 (1
e) = 2(vB )2
Q.E.D.
15.110
With 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
266
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
After impact of B and C:
(vB )3
=
(vC )3
=
1
1
(vB )2 =
4
4
3
3
(vB )2 =
4
4
3
v0
4
3
v0
4
3
v0
16
9
=
v0
16
=
After second impact of A and B:
(vA )4
=
(vB )4
=
3
1
1
(vA )2 + (vB )3 =
4
4
4
3
1
3
(vA )2 + (vB )3 =
4
4
4
1
v0
4
1
v0
4
3
4
1
+
4
+
3
v0
16
3
v0
16
13
v0
64
15
=
v0
64
=
Final velocities are:
vA =
13
v0 J
64
vB =
15
v0 J
64
vC =
9
v0 J
16
15.111
p 1 = p2 :
(vB )2
1:2
=
= 0:667
(vA )1
1:8
1:2 (vA )1 = 1:8 (vB )2
e = vsep =vapp :
e=
(vB )2
= 0:667 J
(vA )1
15.112
267
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15.113
+ ! p1 = p2 :
mA (vA )1 = mA (vA )2 + mB (vB )2
0:3(6) = 0:3( 2:5) + 1:5(vB )2
(vB )2 = 1:70 m/s !
e=
vsep
(vB )2 (vA )2
1:70 ( 2:5)
=
=
= 0:70 J
vapp
(vA )1
6
15.114
268
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15.115
15.116
θ2
θ1'
θ2'
θ3
θ1
Rebound formula
:
Geometry
:
Rebound formula
Geometry
tan 01 = e tan 1
1
tan 2 =
tan 01
e
tan 01
1
tan 01
: tan 3 =
=
= tan 1
e
tan 02
)
Q.E.D.
1 = 3
:
tan 02 = e tan 2 =
269
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
e=
(vBy )2 (vAy )2
vsep
=
vapp
(vAy )1
(vBy )2 (vAy )2
(vBy )2 (vAy )2
12 sin 70
9:585 m/s
(a)
0:85 =
=
(b)
Solution of (a) and (b) is
(vAy )2 =
) (vA )2 = 4:10i
2:631 m/s
(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
270
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mv 3mv
A
B
mv
3mv'
mv + 3mv = mv + 3mv 0
+
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
e=
15.120
Position 1 = release position
Position 2 = just before A hits the ground (choose as the datum position)
Position 3 = just after A hits the ground but before impacting with B
Position 4 = just after A and B impact
Position 5 = 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)
p
p2
(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
m(vB)3
10m(vA)3
A
p3 = p4
e=
+"
(vB )4 (vA )4
(vA )3 + (vB )3
B
m(vB)4
After
10m(vA)4 impact
A
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)
271
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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 )
*15.121
272
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15.122
A
18θ0 90
.
0.5(vA)y
y
.
6N s
4.8 N s
x
30o
B
Just before impact
0.6(vB)y
0.5(vA)x
0.6(vB)x
Just after impact
90
= 30
180
The x-component of the impulse of each disk is unchanged.
= sin1
Disk A :
Disk B :
6 sin 30 = 0:5(vA )x
4:8 sin 30 = 0:6(vB )x
(vA )x = 6:0 m/s
(vB )x = 4:0 m/s
Momentum of the system is unchanged in the y-direction.
4:8 cos 30
6 cos 30
0:5(vA )y + 0:6(vB )y
e=
vsep
vapp
0:6 =
=
=
0:6(vB )y + 0:5(vA )y
1:0392
(vA )y (vB )y
12 cos 30 + 8 cos 30
(vA )y
(vB )y = 10:392
(a)
(b)
Solution of Eqs. (a) and (b) is
(vA )y = 4:724 m/s
(vB )y =
5:668 m/s
The …nal speeds are
vA
vB
15.123
p
6:02 + 4:7242 = 7:64 m/s J
p
=
4:02 + 5:6682 = 6:94 m/s J
=
273
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15.124
15.125
274
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15.126
Mg
T/2
T/2
FBD
From FBD:
T = M g = 250 lb
Equation (15.49):
T = mu
_
P
250 = m(1100)
_
m
_ = 0:2273 slugs/s
1
1
mu
_ 2 = (0:2273)(11002 ) = 1:3752
2
2
1:3752 105
= 250:0 hp J
550
=
=
105 lb ft/s
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
(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:
P =
50) =
3142 N
F = 3140 N J
(b) Moving plate: vout = 4 m/s
m
_ = A(vin
F
P
vout ) = 1000(1:2566
10 3 )(50
= m(v
_ out vin ) = 57:80(4
=
F = 2660 N J
50) =
4) = 57:80 kg/s
2659 N
275
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15.128
y
vin
x
vout
28o
Control volume
m
_ =
vin
=
180(8:34)
= 0:7770 slugs/s
32:2(60)
45i ft/s
vout = 45(i cos 28 + j sin 28 ) = 39:73i + 21:13j ft/s
Force acting on control volume:
F = m(v
_ out
vin ) = 0:7770(39:73i + 21:13j 45i) =
4:095i + 16:418j lb
Force acting on vane:
P =
F = 4:10i
16:42j lb J
15.129
276
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15.130
Choosing the fan as the control volume, we have
vin
= v = 20 ft/s
m
_ =
Au = 2:33
vout = u = 50 ft/s
(6:42 )
10 3
(50) = 3:748 slugs/s
4
Equation (15.47):
T = F = m(v
_ out
vin ) = 3:784(50
20) = 113:5 lb J
15.131
277
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15.132
Pin
R
y
x
FBD of control vol. Pout
Let R = force exerted by the bend on water.
Equation (15.47):
F = m(v
_ out vin )
From FBD:
F = Pin Pout + R
Pin i + Pout j + R = m(
_ vout j vin i)
Substitute
m
_ =
Ain vin = 1000
vin
=
4:2 m/s (given)
vout
= vin
Ain
= 4:2
Aout
0:4
0:6
Pin
= pin Ain = 12
103
Pout
= pout Aout = (18
4
0:42 (4:2) = 527:8 kg/s
2
= 1:8667 m/s
(0:42 )
= 1508 N
4
(0:62 )
103 )
= 5089 N
4
) 1508i + 5089j + R = 527:8( 1:8667j
R = 3720i 6070j N
4:2i)
Force exerted by the water on the bend is
F=
R = 3720i + 6070j N J
278
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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 ) =
279
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15.135
d
a
5 in.
P
25.4 ft/s
a
d2
d2 62:4
vout =
(25:4) = 38:66d2 slugs/s
4
4 32:2
(5=12)2
D2
p=
(624) = 85:09 lb
= Aa a p =
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
280
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15.138
Mg
FBD
4.5 km
A
T
Control volume = rocket + wire above A.
M = 70 + 0:004(4500) = 88:0 kg
p_V
d(M v)
= M_ v + M a = ( v) v + M a
dt
0:004(220)2 + 88:0a = 193:6 + 88:0a N
=
=
+" F =
p_V
=
a =
F
+"
Mg
T =
88:0(9:81)
193:6 + 88:0a =
18 =
881:3
881:3 N
a=
12:21 m/s
2
12:21 m/s # J
2
15.139
Equation (15.48):
mu
_
Substituting m
_ =
M g = M v_
dM=dt and v_ = a (constant), we get
dM
u
dt
1
dM =
M
Mg = Ma
Integration yields
ln M =
a+g
dt
u
a+g
t+C
u
Initial condition:
M
= M0 when t = 0 ) C = ln M0
M
a+g
) ln
=
t
M0
u
When M = 0:25M0 :
ln (0:25) =
34:4 + 9:81
t
1300
t = 40:8 s J
281
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15.140
Mg
FBD
30 in.
P
m
_ =
"
0:075
vin Ain =
(80)
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
282
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15.142
15.143
30o
N2
A
y
60 a A
g A
x
40 lb
60 lb
FBD N1
N2
MAD
a B
B 40
g A
30o
40 a
g B/A
30o
FBD
MAD
Block A:
+
+
"
Fx = mA aA
N2 sin 30 =
Fy = 0
N2 cos 30
N1
60
aA
32:2
60 = 0
Block B:
40
aB=A cos 30
aA
32:2
40
aB=A sin 30
N2 cos 30 =
32:2
+ !
Fx = mB (aB )x
N2 sin 30 =
+
Fy = mB (aB )y
40
#
283
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After simplifying, the equations become
0:5N2 1:8634aA
N1 0:8660N2
0:5N2 + 1:2422aA 1:0758aB=A
0:8660N2 + 0:6211aB=A
= 0
= 60
= 0
= 40
The solution is
N1 = 85:7 lb
aA
aB
N2 = 29:7 lb
aA = 7:97 ft/s
2
=
7:97i ft/s J
= aA + aB=A = ( 7:97 + 23:0 cos 30 ) i
aB=A = 23:0 ft/s
2
2
=
11:95i
11:50j ft/s
2
(23:0 sin 30 ) j
J
15.144
15.145
p1 = p 2 +
4000(25) = 36 000v2
mcar v1 = mtot v2
v2 = 2:78 mi/h J
284
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15.146
p1 = p 2
+ !
0
0
mA vA + mB vB = mA vA
+ mB vB
0
8000(6) + 5000(2) = 8000vA
+ 5000(5:2)
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 = p2 :
mbullet v0 cos 20 = mtot v1
0:8
0:8
v0 cos 20
=
+ 5 v1
v1 = 9:304 10 3 v0
16
16
Energy is conserved after the impact :
T 2 + V2 = T 3 + V3 :
1
(9:304
2
1
mtot v12 + 0
2
10 3 )v0
2
v0
=
0 + mtot gd sin 20
=
32:2(1:6) sin 20
=
638 ft/s J
285
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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
= max + !
T = mrB ! 2
sW
=
2(0:4)! 2
4:905 + T = +0:8! 2
Block A:
Fy
0:25(2 9:81) T
= 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
286
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15.152
287
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
15.153
288
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15.154
15.155
L
Datum
L
L
L/2
y2
s1
y1
WA
Position 1
L
L
WA
WB
WB
s2
Position 2
289
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Length of rope is constant:
2s1 + y1
p
2 L2 + (0:5L)2 + y1
y1 y2
=
=
=
2s2 + y2
p
2 2L2 + y2
0:5924L
Energy is conserved: V1 = V2
WA (0:5L)
W B y1
0:5WA L
0:5WA L
WA
WB
=
WA L W B y 2
= WB (y1 y2 )
= 0:5924WB L
0:5924
=
= 1:185 J
0:5
15.156
290
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15.157
Position 1: release position.
Position 2: just before impact.
Position 3: just after impact.
T1 + V1 = T2 + V2 :
1
1
mA (vA )22 + k 22
2
2
1
1
1
(300) (0:32 ) =
(0:5)(vA )22 + (300)(0:052 )
2
2
2
(vA )2 = 7:246 m/s
1
0 + k 21
2
=
e = vsep =vapp :
e=
(vB )3 (vA )3
(vA )2
0:8 =
(vB )3 (vA )3
7:246
(a)
p 2 = p3 :
mA (vA )2 = mA (vA )3 + mB (vB )3
0:5(7:246) = 0:5(vA )3 + 0:2(vB )3
(b)
Solution of Eqs. (a) and (b) is
(vB )3 = 9:32 m/s J
(vA )3 = 3:52 m/s
15.158
Position 1 = just before shell is …red
Position 2 = just after the shell is …red
Position 3 = maximum compression of spring
System: + ! p1 = p2 :
0 = mA (vA )2
0 = 24(1800) 650 (vB )2
Barrel only: T2 + V2
1
2
650
32:2
(66:46)2
= T3
=
1
1
2
mB (vB )2 + 0 = 0 + k 23
2
2
V3 :
1
(26
2
mB (vB )2
(vB )2 = 66:46 ft/s
103 ) 23
3 = 1:852 ft
x
3 a
32.2
MAD
J
15.159
Block A:
3 lb
y
o
25
FBD NA
=
0.2NA
291
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Fy
=
Fx
= max
a =
0
+"
NA cos 25
+ !
0:2NA sin 25
3=0
NA sin 25 + 0:2NA cos 25 =
NA = 3:651 lb
3
a
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
0
MAD
Fy
=
+"
NB
9=0
Fx
= max
+ !
P
=
0:4(9) +
9
(23:66) = 10:21 lb J
32:2
mA (vA )1
mB(vB )1
P
NB = 9 lb
9
a
0:4NB =
32:2
15.160
(mA+ mB)v2
B
A
Before impact
A
B
After impact
(vA )1
=
40 mi/h = 40
(vB )1
=
26 mi/h = 26
5280
3600
5280
3600
= 58:67 ft/s
= 38:13 ft/s
(a) Momentum is conserved:
mA (vA )1 + mB (vB )1 = (mA + mB )v2
3800
3200 + 3800
3200
(58:67) +
(38:13) =
v2
g
g
g
v2 = 47:52 ft/s J
(b)
T1
=
=
1
1
mA (vA )21 + mB (vB )21
2
2
1 3200
1 3800
(58:67)2 +
2 32:2
2 32:2
(38:13)2 = 256:8
103 lb ft
292
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
T2 =
1
1
(mA + mB )v22 =
2
2
% energy loss =
T1
T2
T1
3200 + 3800
32:2
100% =
(47:52)2 = 245:5
256:8 245:5
256:8
103 lb ft
100% = 4:40% J
15.161
293
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Chapter 16
16.1
16.2
= 10e 0:5t
Initial conditions: ! =
)!=
20e 0:5t + C1
!=
= 40e 0:5t + C1 t + C2
= 0 when t = 0. ) C1 = 20 rad/s, C2 =
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 0:5t = 0:5
t = 1:3863 s
40(0:5) + 20(1:3863) 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: ! = 6000 rev/min when
)
When
1 2
(!
2
( is constant)
= 0. ) C1 = 60002 =2 (rev/min)
2
60002 ) =
= 3600 rev, ! = 1200 rev/min.
)
1
(12002
2
60002 ) = (3600)
=
4800 rev/min
2
d!
dt = d!
t = ! + C2
dt
Initial condition: ! = 6000 rev/min when t = 0. ) C2 = 6000 rev/min
When ! = 0:
=
t = C2
4800t =
6000
t = 1:250 min = 75:0 s J
294
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
295
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
296
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.9
!
Z
=
4t1=2
=
=
jt=6 s
jt=0 s =
! dt =
8 3=2
t +C
3
8 3=2
(6) + C
3
J
C = 39:2 rad
16.10
_=
= sin t
•=
cos t
2
sin t
= R• = R 2 sin t
2
an = R _ = R 2 cos2 t
q
q
a = a2t + a2n = R 2 sin2 t + cos4 t
at
The acceleration is maximized when
d
(sin2 t + cos4 t)
d( t)
=
0
2 sin t cos t + 4 cos3 t( sin t)
=
0
There are three solutions:
t
=
t
=
t
=
0 yielding a = R 2
yielding a = R 2
p
3
yielding a =
R 2
4
2
2
) amax = R 2 J
16.11
Pulley B:
v
=
(RB )o ! B
a =
(RB )o B
40
! B = 24 rad/s J
12 = 20! B
28
12 = 20 B
B =
16:8 rad/s
2
J
Belt between B and C:
v0
=
(RB )i ! B = 8(24) = 192 in./s
0
=
(RB )i
a
B = 8(
16:8) =
134:4 in./s
2
Pulley C:
v0
0
a
= RC ! C
= RC
C
192 = 24! C
134:4 = 24 C
! C = 8:0 rad/s J
C =
2
5:60 rad/s
J
297
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.12
Left pulley:
vA
=
(aA )n
=
a2A
=
(RA )i ! A
12 = 0:75! A
! A = 16 rad/s
2
2
2
(RA )o ! A = 2(16) = 512 ft/s
(aA )t = (RA )o A = 2 A
2
2
2
2
2
(aA )n + (aA )t
600 = 512 + 4 2A
A = 156:41 rad/s
Right pulley:
vB
= RB ! B
12 = 1:25! B
! B = 9:6 rad/s
2
2
2
RB ! B = 1:25(9:6) = 115:2 ft/s
(aB )n
=
(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
2
!B
!C
!D
rA
50
(320) = 457:1 rad/s
!A =
rB
35
= ! B = 457:1 rad/s
rC
20
=
!C =
(457:1) = 203 rad/s J
rD
45
=
16.14
298
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16.15
16.16
0:36i + 0:54j 0:3k
= 0:5035i + 0:7553j
0:362 + 0:542 + 0:32
AC = p
vB
= !
vB
=
rB=C = 4 AC
0:9063i
i
( 0:54j) = 4 0:5035
0
0:4196k
j
0:7553
0:54
k
0:4196
0
1:0876k m/s J
299
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
B
=
rB=C + ! (! rB=C ) =
rB=C + ! vB
= 6 AC ( 0:54j) + 2 AC vB
i
j
i
j
k
0:4196 + 4 0:5035 0:7553
= 6 0:5035 0:7553
0:9063
0
0
0:54
0
=
4:65i + 3:71j
1:107k m/s
2
k
0:4196
1:0876
J
16.17
Let w be the width of the tape.
dV1
dV2
dV1
!
= Vol. of tape leaving the reel during time dt is v0 hw dt
= Vol. change of tape on the reel (approx.) is
2 Rw dR
dR
vo h
= dV2 :
v0 h dt = 2 R dR
=
dt
2 R
=
=
v0
R
d!
d! dR
=
=
dt
dR dt
v0 h
2 R
v02 h
J
2 R3
v0 dR
=
R2 dt
v0
R2
15j + 9k
=
152 + 92
21:44j + 12:862k rad/s
i
0
12
k
12:862
9
=
16.18
!
rB=A
vB
= !
=
CA = 25 p
12i
9k in.
= !
rB=A =
j
21:44
0
= 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
vB = vA + vB=A
300
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
v)
v)
B
+
!
=
A
(vB )x = 8 cos 60 = 4 ft/s
q
p
2
(vB )y = vB
(vB )2x = 62
)
+
!
( B x
"
=
2ω
8 ft/s
60o + ω
A
( B y
2 ft
B
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!
vB = vC + vB=C
8 ft/s
B
+"
0.75ω
= 1.75ω +
B
0.75 ft
(vB)x C
ω
C
= 0:75!
! = 10:667 rad/s
J
= 1:75! = 1:75(10:667) = 18:67 ft/s
8
vC
! J
16.21
Wheel rolls without slipping: vC = R!
vA = vC + vB=C + vA=B
vA
A =
ω
Rω
C
+ C
Rω ω
AB
R
B +
B
2
2R
45o
A
2 2RωAB
+
+"
0
vA
vA
p
= R! 2 2R! AB sin 45
! AB = 0:5!
J
p
p
= R! + 2 2R! AB cos 45 = R! + 2 2R(0:5!) cos 45
= 2R!
J
301
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16.22
vO = vA + vO=A
6 ft/s
O = A
4 ft/s
2.5ω
+
O
2.5 ft
ω
+!
6=4
2:5!
.A
! = 4:0 rad/s
J
16.23
A
0.5 m/s
A
=
0.6 m/s
B
+
0.16 m
vA = vB + vA=B
B
+ ! 0:5 =
0:6 + 0:16!
0.16ω
ω
! = 6:875 rad/s
J
vC = vB + vC=B
C
=
0.6 m/s
B
+
0.6875 m/s
0.1 m
vC
C
6.875 rad/s
B
+ ! vC =
0:6 + 0:6875 = 0:0875 m/s ! J
302
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16.24
303
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16.25
16.26
vA = vB + vA=B
ω
12 in./s
=
+
A
in.
10
vA
.
B
B
A
10ω
4
3
304
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
+
+
"
3
(10!)
! = 2:0 rad/s
5
4
4
vA = (10!) = (10 2) = 16:0 in./s
5
5
0 = 12
J
16.27
16.28
D
B
y
A
2 ft/s
ω
θ
d
x
vB
2.5 ft
d = 2:5 csc
305
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Solution I (scalar notation)
vB = vA + vB=A
ωd
vB
θ
B
+ -
d
= A A + ωA
v
0 = vA sin + !d
! = 0:8 sin2 rad/s
0=
J
θ
B
2 sin + !(2:5 csc )
Solution II (vector notation)
vB
vB i
vB i
= vA + !
= 2 cos i
= 2 cos i
rB=A
2 sin j + !k 2:5 csc i
2 sin j + 2:5! csc j
Equating j-components:
!=
! = 0:8 sin2
2 sin + 2:5! csc
J
rad/s
16.29
= vB + vC=B
= ! AB rB=A + ! BC rC=B
=
2:8k ( 30i) + ! BC k (30i + 60j)
= 84j + ! BC (30j 60i)
vC
! CD rC=D
! CD k 60i
60! CD j
Equating like components:
60! BC
60! CD
=
=
0
84
) ! BC = 0 J
) ! CD = 1:40 rad/s
J
16.30
y
B
D 6
15
A
8
E
x
306
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
=
rB=A =
( 6i) =
36j =
vB
! AB
6k
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
307
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.32
B
A
Geometry:
rB=A
rC=B
18
30o x
27
y
30
φ
C
18 cos 30 + 27 cos = 30
= 57:74
= 18(i cos 30 + j sin 30 ) = 15:588i + 9j in.
= 27(i cos 57:74
j sin 57:74 ) = 14:412i 22:83j in.
= vB + vC=B
= ! AB rB=A + ! BC rC=B
= 20k (15:588i + 9j) + ! BC k (14:412i 22:83j)
= 311:8j 180i + ! BC (14:412i 22:83j) in./s
vC
vC j
Equating like components:
0
vC
=
180 + 22:83! BC
! BC = 7:884 rad/s
= 311:8 + 7:884(14:412) = 425 in./s " J
16.33
vB = vA + vB=A
30 o
vB
Β
+ ! vB sin 30
+ " vB cos 30
(0:3420!) cos 30
=
=
=
ω
Α
20o 0.5ω
+
2 m/s Α 0.5 m Β
=
0:5! sin 20
vB = 0:3420!
2 + 0:5! cos 20
2 + 0:5! cos 20
! = 11:516 rad/s
J
vC = vA + vC=A
(vC)y
Α
(v ) =
C
Cx
+
+
!
"
vC
=
2 m/s
11.516 rad/s
+
20o
Α 1.0 m
11.516 m/s
C
(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
308
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.34
A
m B
4
.
0 θ 0.25 m
0:25
= 38:68
0:4
vC = vB + vC=B
= sin 1
1.0 m/s
vC
C
=
2.5 rad/s
+
+
"
.
m
0.4
C
38.68o
o
0.4
38.68
m
B
0.4ωBC ωBC B
+
.
A
0 = (1:0 0:4! BC ) cos 38:68
! BC = 2:50 rad/s
vC = (1:0 + 0:4! BC ) sin 38:68 = [1:0 + (0:4 2:5)] sin 38:68
= 1:250 m/s
J
16.35
309
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.36
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
! 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
310
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.38
311
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.39
16.40
13 i
n.
12 in.
O 5 in. B ω
20 in./s
.
A
vA
Point B is the I.C. of the disk.
vO
vA
= ! OB
20 = 5!
! = 4 rad/s
= ! AB = 4(13) = 52 in./s J
312
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.41
12 in./s
6 in.
B
ω
O
in.
10
vA
8 in.
A
Point O is the I.C. of the disk.
vB
vA
= OB !
12 = 6!
! = 2 rad/s
= OA ! = 8(2) = 16 in./s
J
16.42
ω = 2.5 rad/s
300
vG
O vO= 500 mm/s
G
3
2
d = 200
e
C
Dimensions
in mm
vO
500
=
= 200 mm
!
2:5
The directions of vO and vG determine the location of the instant center C of
the wheel.
p
e =
3002 + 2002 = 360:6 mm
vG = e! = 360:6(2:5) = 902 mm/s %56:3 J
d=
313
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.43
16.44
314
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
315
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.47
0.5 m/s
0.5 m/s
A
a
I.C.
60 mm
C
160 − a
100 mm
B
0.6 m/s
0.6 m/s
Velocities of A and B establish the I.C. of the gear.
a
0:5
=
!
=
vC
=
160 a
a = 72:73 mm
0:6
vA
500
=
= 6:875 rad/s
J
a
72:73
(a AC)! = (72:73 60) (6:875) = 87:5 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
8
= 16:0 in.
cos 60
ωBC
13.856 in.
16
in.
E
C
4 in.
D
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
316
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.49
16.50
16.51
317
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.52
16.53
D
vC
60
Dimensions
in mm
C
60
ωCD
vB
B
30
A
2.8 rad/s
I.C. of link AB is point A. This determines the direction of vB .
318
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
I.C. of link CD is point D. This determines the direction of vC .
Since vB and vC are parallel, link BC is translating. Therefore,
= 0 J
= vC
! BC
vB
From motion of link AB:
vB = AB ! AB = 30(2:8) = 84:0 mm/s
From motion of link CD:
vC = CD ! CD
84:0 = 60! CD
! CD = 1:40 rad/s
J
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
6
12
=
= 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
=
319
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.55
A
27 in.
60o
12 rad/s
E
in.
15
vD
D
B
vB
ωBC F
Geometry: BF =
vB
! BC
AE
cos 60
AB =
27
cos 60
15 = 39 in.
= ! 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
F
BE
β
E
β
27 in.
Geometry:
ωBC
D
vD
BF
27 15
= tan 1
= 42:71
13
DF
= BF + DF cot = (27 15) + 13 cot 42:71 = 26:08 in.
= tan 1
vB
! BC
= ! AB AB = 12(15) = 180 in./s
vB
180
=
=
= 6:90 rad/s
J
26:08
BE
320
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
vC
C
24
16.57
E
18
F
ωBCD
B
24
A
20 o
35 vB
48 rad/s
18
D
BF
=
BE
=
DE
q
2
BC
2
CF =
Dimensions
in inches
p
242
vD
182 = 15:875 in.
BF
15:875
=
= 19:380 in.
cos 35
cos 35
= DF + BF tan 35 = 18 + 15:875 tan 35 = 29:12 in.
vB
! BCD
vD
= ! AB AB = 48(20) = 960 in./s
960
vB
=
= 49:54 rad/s
J
=
19:380
BE
= ! BCD DE = 49:54(29:12) = 1443 in./s ! J
321
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.58
ωAD
E
30o
22
.27
in.
in.
78
20.
24 in.
D
C
.
8 in
.
in.
12
18 in./s
.
vD
vC
60o
30o
A
.B
Point E is the I.C. of link AD. Its location is determined by the known directions
of the velocities at A and C.
AE
CE
DE
vA
vD
12
= 24 in.
sin 30
12
=
= 20:78 in.
tan
p 30
=
20:782 + 82 = 22:27 in.
=
18 = 24! AD
! AD = 0:75 rad/s
= AE ! AD
= DE ! AD = 22:27(0:75) = 16:70 in./s J
322
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.59
16.60
ωAB
8
E
vB
B
vA
O 16 in./s
A 3
5
4
C ωdisk
Geometry: OB
BE
AE
! disk
! AB
vB
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
vA
20
=
=
= 2:1575 rad/s
9:270
AE
= ! AB BE = 2:1575(8:562) = 18:47 in./s " J
=
323
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.61
y
B
o
30
B
8 m/s
2
=
x
+
o
6 m/s2
30
A
A
0.2ωAB2
m
0.2 o
30
0.2αAB
α
ωABAB(sense indeterminate)
aB = aA + aB=A
x+ %
y+ -
6 cos 30 = 8 sin 30
0:2! 2AB
! AB = 6:78 rad/s J
6 sin 30 = 8 cos 30
0:2 AB
AB = 49:6 rad/s
2
J
16.62
324
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.63
B
ω = 5 rad/s
rB/A
10
"
α = 12 rad/s2
.
A
4" r
v0
a0
D/A
y
D
x
No-slip condition gives
aA = R = 10(12) = 120 in./s
rD=A
=
=
2
2
12k rad/s
! = 5k rad/s
4j in.
rB=A = 10j in.
R = 10 in.
(a)
aD
= aA + aD=A = R i +
rD=A + !
= 10(12)i + ( 12k) ( 4j) + ( 5k)
=
120i
48i + 100j =72i + 100j in./s
(! rD=A )
[( 5k) ( 4j)]
2
J
(b)
aB
= aA + aB=A = R i +
rB=A + ! (! rB=A )
= 10(12)i + ( 12k) 10j + ( 5k) [( 5k) 10j]
=
120i + 120i
250j = 240i
250j in./s
2
J
(c) The acceleration of the end of the string is the same as the horizintal component of the acceleration of point D on the spool:
a0 = 72i in./s
2
J
325
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.64
B
ω = 5 rad/s
α = 12 rad/s2
10
"
.
A
4"
y
v0
a0
D
x
Note that the acceleration of point D on the spool is aD = (a0 !) + (aD )y "
aD = aA + aD=A
ω = 5 rad/s
α = 12 rad/s2
(aD)y
D
a
A
A
+
=
A
2
16 in./s
4"
4(12) = 48 in./s2
D
4(5)2 =100 in./s2
Equating like components:
+ !
+
"
16 = aA
48
aA = 64 in./s
(aD )y = 100 in./s
) aD = 16i + 100j in./s
J
2
2
2
aA = 64i in./s
2
J
aB = aA + aB=A
10(5)2 = 250 in./s2
aB
=
64 in./s2
A
+
10(12) = 120 in./s2
B
10"
A
aB = (64 + 120)i
ω = 5 rad/s
α = 12 rad/s2
250j = 184i
2
250j in./s
J
326
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.65
= B
A
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
Let A and B be points on the disk.
vA + vB=A = vB
3 m/s
A
0.4 m
+ω
A
+" 3
B
=
0.4ω
0:4! = 1:0
1.0 m/s
B
J
! = 5 rad/s
Note that the belt accelerations shown in Fig. P16.66 are the tangential components of the accelerations of points A and B on the disk.
aA + aB=A = aB
0.5 m/s2
A
0.2ω2
+A
+ " 0:5
0.4 m
α, ω
0:4 =
1:2
0.4ω2
B
0.4α
=
= 4:25 rad/s
0.2ω2
B
1.2 m/s2
2
J
327
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.67
vB
B
16
in.
vC
C
β
15 rad/s
A
8 in.
= sin 1
! BC = 0 (vB and vC are parallel)
6
= 30
12
aC = aB + aC=B
=
16αBC C
30o
+α
BC
16 i
n.
aC
8(15)2 in./s2
8 in.
15 rad/s
B
A
B
0 = 8(15)2
+ !
+
2
BC = 129:9 rad/s
2
16 BC cos 30
aC = 16 BC sin 30 = 16(129:9) sin 30 = 1039 in./s " J
"
16.68
3 rad/s
A
vC
! BC
sin
2
180
β
t
6f
6 ft
γ ft
4
C
vB
B
= ! AB = 3 rad/s (A is the I.C. for BC)
2
1
=
= 2 sin 1 = 38:94
6
3
38:94
=
+2
= 90
= 90
= 70:53
2
2
328
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
t
6f
aC
3 rad/s
A
12 rad/s2
β
C =
B
0 =
0 =
+
=
BC
aC =
=
+
+
γ
6(12) ft/s2 +
4(32) ft/s2
6(32) ft/s2
B
4 ft
C
αBC
3 rad/s
= aB + aC=B
aC
+"
4αBC
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
2
6(3 ) 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
) ! BC =
By inspection: b! = 2b! BC
aC
ω, α
C =
A
+
bα
ω/2, αBC
2
b
B
aC
= aB + aC=B
!
= b! 2 + 2b
2
aC
+"
bω +
0
2b(ω/2)2
2b
B
= b
1
2!
2
2b BC
=
3 2
b!
2
BC =
C
2bαBC
J
1
2
J
16.70
B
! BC =
4 m/s
30
6m
0.1
o
D
ωBC
C
vC
vB
4
=
= 28:87 rad/s (D is the I.C. for BC)
0:16 cos 30
BD
aC = aB + aC=B (note that aB = 0)
329
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C
28.87 rad/s
B
6m
0.1
αBC
aC =
30o
C
0.16(28.872) m/s2
0.16αBC
+
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
A
vA = 2 m/s
0.5
0.4
E 0.3
ωBC
vB
ωAB B
C
0.6
Point E is the I.C. of bar AB
! AB
vB
! BC
vA
2
=
= 5:0 rad/s
0:4
EA
= EB ! AB = 0:3(5) = 1:5 m/s
vB
1:5
= 2:5 rad/s
=
=
0:6
BC
=
aB = aA + aB=A
0.6αBC
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 m/s2
B
0.6
A
0.3
0.5αAB
+
!
+
"
0.5(5)2 m/s2
B
4
3
1:2 + (0:5 AB )
(0:5)(5)2
5
5
4
3
0:6 BC = (0:5)(5)2 + (0:5 AB )
5
5
0:6(2:5)2 =
330
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
2
AB = 31:125 rad/s and
The solution is
)
2
AB = 31:1 rad/s
J
BC =
32:23 rad/s2
J
2
BC = 32:2 rad/s
16.72
ωAB
vB
2 ft
C
B
1.5 ft
A
6 ft/s
Point C is the I.C. of bar AB.
1:5! AB
vB
2! BC
=
=
6
! AB = 4 rad/s
2! AB = 2(4) = 8 ft/s
8
vB
= = 4 rad/s
= vB
! BC =
2
2
aA = aB + aA=B
B
2αBC
2 ft/s2
A
=C
2 ft
αBC
ωBC= 4 rad/s
+ !
+
BC
+A
B
2(4)2 ft/s2
2.5(4)2 ft/s2
ft
2.5
3
αAB
ωAB= 4 rad/s
4
2.5αAB
4
3
2(4)2 + (2:5)(4)2 + (2:5) AB
5
5
2
J
AB = 1:3333 rad/s
2=
3
4
0 = 2 BC + (2:5)(4)2
(2:5) AB
5
5
3
4
0 = 2 BC + (2:5)(4)2
(2:5)(1:3333)
5
5
2
2
=
10:67 rad/s = 10:67 rad/s
J
"
331
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.73
vB = AB! AB = 1:2(10) = 12:0 m/s "
Note that vB and vD are parallel. Therefore, bar BD is translating in the
position shown.
) ! BD = 0
vD = vB = 12:0 m/s "
! DE =
vD
DE
12
= 5:0 rad/s
2:4
aD = aB + aD=B
3
ωDE= 5.0 rad/s 2.4α
DE
αDE
2.4 m
E
1.5αBD
D
2.4(5)2 m/s2
=
10.0 rad/s
1.2(10)2 m/s2
A
B 1.2 m
+
4
D
m
.5
αBD 1 3
4
B
+ !
+
"
2:4(5)2 = 1:2(10)2
4
(1:5 BD )
5
2
J
DE = 100 rad/s
2:4 DE =
3
(1:5 BD )
5
2:4 DE =
BD = 200 rad/s
2
J
4
(1:5)(200)
5
16.74
Acceleration Analysis
Assume
AB to be counterclockwise.
332
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.75
B
vB
0.8 m
ωBC
A 0.6 m
0.4 m
C
vC
. 8 rad/s
D
Point A is the I.C. of bar BC.
vC
0:6! BC
vB
0:8! AB
=
0:4(8) = 3:2 m/s
3:2
= vC
! BC =
= 5:333 rad/s
0:6
= 0:8! BC = 0:8(5:333) = 4:266 m/s
4:266
= vB
! AB =
= 5:333 rad/s
0:8
aC = aB + aC=B
333
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
0.8(5.333)2 m/s2
0.8αAB
B
=
0 = 0:8(5:333)2
BC = 0
AB
=
4
3
αBC
C
A
+ !
+
0.8 m
αAB
+ #
5.333 rad/s
αBC
m
1.0
0.4(8)2 m/s2 0.4 m
D
C
8 rad/s
B
J
5.333 rad/s
4
(5:333)2
5
3
5
(5.333)2 m/s2
BC
3
4
(5:333)2 +
BC
5
5
4
3
(5:333)2 + (0)
0:4(8)2 = 0:8 AB
5
5
2
2
53:3 rad/s = 53:3 rad/s
J
0:4(8)2 =
0:8 AB
16.76
n.
i
25
15 in.
B
C
vC
! AB
A
ωAB
D
vB
6
rad/s
.
in
5
1
vC
= vB = ! CD CD = 6(15) = 90 in./s
vB
90
=
=
= 3:6 rad/s
! BC = 0 (vB and vC are parallel)
25
AB
aB + aC=B = aC
αAB
20 rad/s2
D
A
αBC
45o 3.6 rad/s
45o 6 rad/s
+ 15
=
5
C 15
2
B
15(20) in./s2
C
2
2
15(6
)
in./s
α
25
15αBC
AB
25(3.62) in./s2
B
334
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
25(3:62 ) + 15 BC sin 45
+%
=
= 20:36 rad/s
= 15(20)
= 15(20)
BC
25 AB + 15 BC cos 45
25 AB + 15(20:36) cos 45
3:36 rad/s
ωBD
AB
J
2
J
2
=
8 in.
+&
15(62 )
16.77
D
in.
17 vD
B
vB
15 in.
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
! DE = ! BD = 0:8571 rad/s
aB = aD + aB=D
+ !
+"
A
10αDE
10 ω
=
8
10
2
6ωAB
B 6
ωAB
D
D
2
DE
6
ωDE
αDE
E
+
17
2
17ωBD
B
15
17αBD
ωBD
αBD
8
6
8
15
8
(10! 2DE )
(10 DE ) + (17! 2BD ) + (17 BD )
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
(a)
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
)
2
BD = 1:3775 rad/s and
2
BD = 1:378 rad/s
J
DE =
15
(17 BD )
17
(b)
3:444 rad/s2
DE = 3:44 rad/s
J
335
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.78
16.79
336
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.80
337
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.81
338
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.82
16.83
rA/O
y
ω
x
O
0.6 ft A, A'
! = 30
2
60
=
rad/s
339
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
P
30o
A
16
.97
1i
n.
in.
24
12 in.
45o
B
Let P 0 be …xed to bar BC.
vP = vP 0 + vP=BC
340
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.971ωBC
24(5) in./s
in.
24
30o
5 rad/s
16
.97
1i
n.
P
=
45o
ωBC
A
P'
vP/BC
+
45o
B
+ - 45
24(5) cos 15 = 16:971! BC
2
! BC = 6:83 rad/s
J
16.86
6 rad/s2
y
3 rad/s
rP/Ao
30
A
B
x
!
!
rP=A
rP=A
vP
aP
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
341
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.87
aP = aA + aP 0 =A + aP=disk + aC
Noting that A is a …xed point and that P 0 coincides with A, we have aA =
aP 0 =A = 0. Hence
aP = aP=disk + aC
82 = 8 in/s2
8
vP/disk = 8 in./s
P
aP
=
2(1.2)(8) = 19.2 in./s2
+
8 in.
vP/disk = 8 in./s
ω = 1.2 rad/s
=
11.2 in./s2
B
aP = 11:2 in./s " J
16.88
vP = vA + vP 0 =A + vP=AB = 0 + vP 0 =A + vP=AB
P
ft/s
45o 4(1.6971)
P'
45o
=
+
1
7
vP/AB
4 rad/s
.69
1
A
0
=
4(1:6971)
vP
vP
=
=
4(1:6971) + vP=AB cos 45
[4(1:6971) + 6:788] cos 45 = 9:60 ft/s J
vP
+
+"
vP=AB sin 45
vP=AB = 6:788 ft/s
vP2
9:602
2
=
= 76:8 ft/s
R
1:2
aP = aA + aP 0 =A + aP=AB + aC = 0 + aP 0 =A + aP=AB + aC
(aP )y =
1.6971(12) ft/s2
45o
P
(aP)x
76.8 ft/s2
=
P'
1
7
4 rad/s 69
1.
A
12 rad/s2
1.6971(42) ft/s2
+
ω
45o
45o
+ vP/AB
aP/AB
2(4)(6.788) ft/s2
342
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
+"
+ !
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
vP = vA + vP 0 =A + vP=AB = 0 + vP 0 =A + vP=AB
ωAB
0.2
309
m
0.2309ωAB
P'
vP/AB x
30o
1.2 m/s
+
=
60 o
P
y
A
+x +y .
1:2 cos 30
1:2 sin 30
= 0:2309! AB
! AB = 4:50 rad/s
= vP=AB
vP=AB = 0:6 m/s
J
aP = aA + aP 0 =A + aP=AB + aC
0 = 0 + aP 0 =A + aP=AB + aC
0.2309αΑΒ
0.2
309
m
30o
0 =
4.50 rad/s
A
+x -
0.2309 (4.52) m/s2
P'
ωAB
+ 60 o
+ v
P/AB
aP/AB
30o
2(4.50)(0.6) m/s2
αΑΒ
0 = 0:2309 AB
2(4:50)(0:6)
2
AB = 23:4 rad/s
J
16.90
aO
vP=disk
20
2
(2) = 3:333 ft/s
12
2
6 ft/s #
aP=disk = 28 ft/s #
= R =
=
aP = aO + aP 0 =O + aP=disk + aC
343
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
aP = 3.333 ft/s
O
2
+
O
+
+
!
"
)
1.0(52) ft/s2
P'
1.0 ft
1.0(2) ft/s
2
2(5)(6) ft/s2
ω
+
28 ft/s2
vP/disk
+
5 rad/s
2 rad/s2
(aP )x =
3:333
1:0(2)
2(5)(6) =
(aP )y =
1:0(52 )
28 =
53:0 ft/s
aP =
65:3i
53:0j ft/s
2
65:3 ft/s
2
2
J
*16.91
344
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.92
! = 5k rad/s
=0
rP 0 =A = 0:4(i cot 60 + j) = 0:2309i + 0:4j) m
(a)
vP 0 =A
vP=AB
= ! rP 0 =A = 5k (0:2309i + 0:4j) = 1:1547j 2i m/s
= vP=AB (i cos 60 + j sin 60 ) = vP=AB (0:5i + 0:8660j) m/s
vP
vP i
= vA + vP 0 =A + vP=AB
= 0 + 1:1547j 2i + vP=AB (0:5i + 0:8660j)
Equating y-components:
+
"
)
0 = 1:1547 + 0:8660vP=AB
vP=AB = 1:333 m/s
vP=AB = 1:333(i cos 60 + j sin 60 ) = 0:6665i 1:1544j m/s J
(b)
aP 0 =A
= !
(!
rP 0 =A ) = 5k
(1:1547j
aP=AB
aC
= aP=AB (0:5i + 0:8660j) m/s
= 2! vP=AB = 2(5k) ( 0:6665i
2i) =
5:776i
10j m/s
2
2
=
6:665j + 11:544i m/s
1:1544j)
2
= aA + aP 0 =A + aP=AB + aC
= 0 + ( 5:776i 10j) + aP=AB (0:5i + 0:8660j) + ( 6:665j + 11:544i)
aP
aP i
Equating y-components:
+
"
)
0=
10 + 0:8660aP=AB
6:665
aP=AB = 19:244 m/s
aP=AB = 19:244(0:5i + 0:8660j) = 9:62i + 16:67j m/s
2
2
J
16.93
vP = vP 0 =A + vP=AB
P
=
0.8
m
0.9 m
A
0.8(5) m/s
45o
5 rad/s
C
P'
ωAB
+
vP/AB
0.9ωAB
345
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
+ !
+ "
)
0:8(5) cos 45 = 0:9! AB
0:8(5) sin 45 = vP=AB
J
! AB = 3:143 rad/s
vP=AB = 2:828 m/s
aC = 2! AB vP=AB = 2(3:143)(2:828) = 17:777 m/s
2
Let P 0 be the point on AB that is coincident with P at this instant. Coordinates
are embedded in bar AB:
aP = aP 0 =A + aP=AB + aC
45o
P 0.8
m
ω = 5 rad/s
α = 3 rad/s2
=
A
0.9 m
0.8(5)2 m/s2 0.8(3) m/s2
αAB
ωAB = 3.143 rad/s
+
aP/AB
vP/AB = 2.828 m/s
+ 17.777 m/s2
0.9αAB
C
ωAB = 3.143 rad/s
0.9(3.143)2
+ !
0:8(52 ) sin 45 + 0:8(3) cos 45 = 0:9 AB
AB = 37:4 rad/s
2
17:777
J
16.94
346
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.95
16.96
347
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.97
B
P
390
A
150
2 rad/s
22.62o
360
D
vP = vA + vP 0 =A + vP=AB = 0 + vP 0 =A + vP=AB
348
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
P
0.3 m
0.3(2) m/s
2 rad/s
+x +y .
m
0.39 o
22.62
= ωAB
A
D
0:3(2) sin 22:62 = 0:39! AB
0:3(2) cos 22:62 = vP=AB
0.39ωAB
22.62o
P'
+ v
P/AB
x
y
! AB = 0:5917 rad/s
vP=AB = 0:5538 m/s
J
aP = aA + aP 0 =A + aP=AB + aC = 0 + aP 0 =A + aP=AB + aC
P
0.15 m
0.15(22) m/s2
A
22.62o
0.39(0.59172) m/s2
+
ωAB
22.62o
22.62 +
vP/AB
vP/AB
2(0.5917)(0.5538) m/s2
o
0.5917 rad/s
+x AB
αAB 0.39 m P'
=
2 rad/s
O
0.39αAB
=
0:15(22 ) cos 22:62 = 0:39 AB
0:260 rad/s
2
2(0:5917)(0:5538)
J
16.98
349
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
) vA =
vB
yA
) vA
=
0
(a)
=
vB tan =
1:4 tan 20 =
=
0:510 m/s # J
0:5096 m/s
(b) Di¤erentiating (a) with respect to time:
2
2
vB
+ xB aB + vA
+ yA aA
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
350
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.101
C
.
β
3 ft
.B
2.5 ft
t
2f
θ
0.8 rad/s
A
Geometry:
2 sin + 3 sin
= 2:5
(a)
_ + 3 cos
_ =0
(b)
Di¤erentiation yields:
2 cos
When
= 45 :
Eq. (a): 2 sin 45 + 3 sin = 2:5
= 21:22
Eq. (b): (2 cos 45 ) (0:8) + (3 cos 21:22 ) _ = 0
_ =
0:4046 rad/s
Di¤erentiate Eq. (b):
2 sin
When
_ 2 + 2 cos •
3 sin
_ 2 + 3 cos
•=0
= 45 :
(2 sin 45 ) (0:8)2 + 0
(3 sin 21:22 ) ( 0:4046)2 + (3 cos 21:22 ) • = 0
• = 0:3872 rad/s2
Therefore,
J
! BC = 0:405 rad/s
2
BC = 0:387 rad/s
J
16.102
B
0.25 m
φ
5m
0.7
0.5
m
C
x θ
D
351
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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 =
Substitute
0:75 _ sin
0:75 _ cos
0:5 _ sin
0:5 _ cos
(b)
(c)
= 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
(d)
(e)
1:302 rad/s and _ =
1:045
J
16.103
Geometry:
0:4 sin 1 + 0:3 sin 2
0:4 cos 1 _ 1 + 0:3 cos 2 _ 2
2
0:4 sin 1 _ 1 + 0:4 cos 1 •1
2
0:3 sin 2 _ 2 + 0:3 cos 2 •2
= 0:4
= 0
(a)
(b)
=
(c)
0
When 1 = 30 and _ 1 = 0:6 rad/s (note that •1 = 0):
Eqs. (a) and (b):
0:4 sin 30 + 0:3 sin 2
(0:4 cos 30 ) (0:6) + (0:3 cos 41:81 ) _ 2
=
=
0:4
0
2 = 41:81
_ 2 = 0:9295 rad/s J
Eq. (c):
( 0:4 sin 30 ) (0:6)2
(0:3 sin 41:81 )( 0:9295)2 + (0:3 cos 41:81 ) •2 = 0
•2 = 1:095 rad/s2 J
352
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.104
16.105
353
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.106
16.107
A
R
y
θ
e
O
_ = 800
2
60
= 83:78 rad/s
Law of cosines:
R2
802
= y 2 + e2 2ye cos
= y 2 + 402 2y(40) cos 50
(a)
y = 99:61 mm
354
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Di¤erentiating Eq. (a):
0
=
y_
=
=
2y y_
_
2ye
_ cos + 2ye sin
ye _ sin
99:61(40) (83:78) (sin 50 )
=
e cos
y
40 cos 50
99:61
3460 mm/s = 3:47 m/s # J
*16.108
A
θ
9 ft
C
φ
6 ft
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
L _ cos
L_ sin
L _ sin + L_ cos
= 20 and L_ =
0
0
(c)
(d)
2 ft/s, we get
From (a):
From (b):
6 sin 20
L sin = 0
6 cos 20 + L cos = 9 ft
The solution is L = 3:939 ft and
From (c): 6 _ cos 20
From (d):
6 _ sin 20
The solution is _ =
=
=
= 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 _ =
= 0
= 0
0:4265 rad/s. Therefore,
J
! AB = 0:427 rad/s
16.109
D
y
A
0.5
m
θ
θ
C
.5 m
0
m
0.5 B
θ
355
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
y
y_
= 3(0:5) sin = 1:5 sin m
= 1:5 _ cos m/s
y• =
Substitute
2
1:5 _ sin + 1:5• cos m/s
(a)
2
(b)
= 30 , y_ = 3 m/s and y• = 0:
From (a):
From (b):
3 = 1:5 _ cos 30
0=
_ = 2:309 rad/s
1:5(2:309) sin 30 + 1:5• cos 30
! ABC = 2:31 rad/s
2
J
• = 3:078 rad/s2
2
ABC = 3:08 rad/s
J
16.110
356
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.111
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
357
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.113
D
B
2 ft
θ
vA
A
x
Geometry:
2 cot
x
• =
_2
2 csc2 (2 cot
=
When
) x_ =
2 csc2 _
2
2 [2 csc ( csc cot )] _
2 csc2
x =
•
•)
= 30 :
x_ =
x
• =
2 csc2 30 _
2
•=0
(2 cot 30 ) (0:25)
2 ft/s
0
! AD
(aB )t
(aB )n
=
_ = 0:25 rad/s
2=
_ = 0:25 rad/s
2
• = 0:2165 rad/s2
J
= AB _ = 4(0:25)2 = 0:25 ft/s 30 . J
2
= AB • = 4(0:2165) = 0:866 ft/s 60 - J
2
358
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.114
359
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.115
16.116
360
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.117
1.5
ft
vO
O
A
25o
8 ft/s
25o
C
ω
The directions of the velocties of points O and A locate the instant center C of
the disk.
AC = 1:5 csc 25 = 3:549 ft
!=
OC = 1:5 cot 25 = 3:217 ft
vA
8
= 2:254 rad/s
=
3:549
AC
J
vO = OC ! = 3:217(2:254) = 7:25 ft/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.
vC
aC
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
361
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.119
vC = vB + vC=B
60 in./s
in.
10
17ωAB
Bω
BC
B
.
= ωAB 17 in
8 in. + 8 in.
10ωBC
C
15
in.
6
in.
A
C
+ !
0
=
+"
60
=
8
8
+ 10! BC
= 8! AB + 8! BC
17
10
15
6
17! AB
+ 10! BC
= 15! AB + 6! BC
17
10
17! AB
The solution is
! AB = ! BC = 2:86 rad/s
J
16.120
2
=
4 AB rad/s
= 20 AB rad/s
rC=B = 0:6j m
0:6i + 0:8j + 0:4k
= p
= 0:5571i + 0:7428j + 0:3714k
0:62 + 0:82 + 0:42
!
AB
= !
vC
aC
=
=
0:8914i
rC=B + !
(!
i
0:5571
0
=
20
=
0:484i
i
0:5571
0
rC=B = 4
j
0:7428
0:6
k
0:3714
0
rC=B + !
vC
1:337k m/s J
rC=B ) =
j
0:7428
0:6
k
0:3714
0
4:30j + 9:33k m/s
2
+4
i
0:5571
0:8914
j
0:7428
0
k
0:3714
1:337
J
16.121
AB
=
rBC
=
vC
aC
= !
=
3i + 5j + 4k
=
32 + 52 + 42
5j ft
p
0:4243i + 0:7071j + 0:5657k
= ! rBC = 5 ( 0:4243i + 0:7071j + 0:5657k)
= 14:143i + 10:608k ft/s J
vC = 5 ( 0:4243i + 0:7071j + 0:5657k)
37:5i + 62:5j
50:0k ft/s
2
( 5j)
(14:143i + 10:608k)
J
362
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.122
16.123
363
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.124
A
805
.0
E
4
3
15
0
A
201
.2
2
60
1
Geometry: AD
AE
vB
! BC
1
B
360
vA
D 7 rad/s
D 90
! AB
720
2
120
O
vA
ωAB
vB
180
C
ωBC
p
902 + 1802 = 201:2 mm
p
=
3602 + 7202 = 805:0 mm
=
= AD! OA = 201:2(7) = 1408:4 mm/s
1408:4
vA
=
= 1:7496 rad/s
=
805:0
AE
= BE! AB = 720(1:7496) = 1259:7 mm/s
1259:7
vB
=
=
= 7:00 rad/s
J
180
BC
16.125
(a)
vB
0
25
4 rad/s
5
37
O
300
A
200
B
150
225
200
C
vC
ωCD
D
364
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Point O is the I.C. for bar BC.
vB
! BC
vC
! CD
= AB! AB = 250(4) = 1000 mm/s
vB
1000
=
=
= 1:60 rad/s
Q.E.D
625
OB
= OC! BC = 500(1:60) = 800 mm/s
vC
800
=
= 4 rad/s
Q.E.D
=
200
CD
(b)
aB = aC + aB=C
375(1.6)2 mm/s2
375αBC B
250(4)2 mm/s 2
0
25 3
4 rad/s
B
=
200(4)2 mm/s2 200
C
4 rad/s
D
αCD
200αCD
4
+
A
375
αBC
1.6 rad/s
C
+ !
4
(250)(4)2 = 200(4)2 375 BC
5
2
J
BC = 17:07 rad/s
+ #
3
(250)(4)2 = 200 CD + 375(1:6)2
5
2
J
CD = 7:20 rad/s
C
F
ωBCD
B
E
vB
450
0
60
00
72 rad/s 5 o
35
vC
60
0
16.126
A
D
vD
365
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Geometry: BE
BF
DF
vB
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
! BCD
vD
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
( 36j)
432i + 180j + 48j + 432i = 90i + 444j in./s
2
J
16.128
vC = vB + vC=B
C
vC
=
0.1
2m
B 4
3 C
3
0.12(1.5) m/s
4
+ 0.09ωBC
. 1.5 rad/s
0.0
9m
ωBC
.
B
A
+ !
+
#
4
3
(0:12) (1:5)
(0:09)! BC
! BC = 2:667 rad/s
5
5
3
4
vC = (0:12)(1:5) + (0:09)! BC
5
5
3
4
= (0:12)(1:5) + (0:09)(2:667) = 0:300 m/s # J
5
5
0=
aC = aB + aC=B
366
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C
0.1
2m
0.12(1.5)2 m/s2 4
=
aC
B
+
. 1.5 rad/s
A
.
3
4
3
(0:12) (1:5)2 + (0:09)(2:667)2
(0:09) BC
5
5
5
2
BC = 6:484 rad/s
3
4
4
aC = (0:12)(1:5)2 + (0:09)(2:667)2 + (0:09) BC
5
5
5
4
3
4
2
2
= (0:12)(1:5) + (0:09)(2:667) + (0:09)(6:484)
5
5
5
2
= 1:067 m/s # J
+ !
+
C
0.09(2.667)2 m/s2
4 3
0.0
4
9m
0.09αBC
2.667 rad/s
αBC B
3
3
0=
#
16.129
(a) Because bars AB and CD are parallel, the velocities of points B and C are
also parallel. Therefore, the bar BC translates in the position shown.
) ! BC = 0
vC = vB = AB! AB = 75(12) = 900 mm/s
900
vC
=
! CD =
= 6 rad/s
J
150
CD
(b)
150(6)2 mm/s2
C
3
4
75(12)2 mm/s2
B
=
75
mm
αCD
150αCD
15
0m
m
aC = aB + aC=B
3
12 rad/s
6 rad/s
4
+
αBC
B
165αBC
165 mm
C
A
D
+
4
3
3
(150 CD ) + (150)(6)2 = (75)(12)2
5
5
5
2
=
27:0
rad/s
J
CD
367
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
+ "
3
4
4
(150 CD )
(150)(6)2 =
(75)(12)2 + 165 BC
5
5
5
3
4
4
(150)(27:0)
(150)(6)2 =
(75)(12)2 + 165 BC
5
5
5
2
J
BC = 40:9 rad/s
16.130
(a)
O
30o
B
o
vB 40
60 o 800
Geometry: BC
OC
OA
ωAB = 6 rad/s
y
C
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
(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
aA
rA=B
rA=B
aA
aA i
aA i
= 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
368
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.131
369
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.1
17.2
. R
2
R
2R
1
. R
1 3
1 2
R (R) =
R
3
3
1
5 3
R
R3 =
3
3
V2
1
m2 =
m= m
V
5
R2 (2R) = 2 R3
V1
=
V
= V1
m1
=
V1
12
m=
m
V
5
Iz
=
(Iz )1
=
1
2
V2 =
2
(Iz )2 =
12
m R2
5
V2 =
1
3
m1 R2
m2 R2
2
10
3 1
57
m R2 =
mR2 J
10 5
50
370
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.3
17.4
371
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.5
y
90 mm
90 mm
8 mm
.
2
144 mm
y- = 104.0 mm
1
x
10 mm
m1
=
4
d21 L1 =
4
(7850)(0:010)2 (0:144) = 0:08878 kg
d2 L2 = (7850)(0:008)2 (0:180) = 0:07103 kg
4 2
4
m = m1 + m2 = 0:08878 + 0:07103 = 0:159 81 kg
m2
y
=
0:08878(0:072) + 0:07103(0:144)
mi y i
=
= 0:1040 m = 104:0 mm
m
0:159 81
Coordinates of mass center are (0, 104.0 mm, 0) J
=
)
Iz
Iz
1
1
m1 L21 + m2 y22 = (0:08878)(0:144)2 + 0:07103(0:148)2
3
3
= 2:170 10 3 kg m2
= Iz my 2 = 2:170 10 3 0:159 81(0:1040)2
= 4:42 10 4 kg m2 J
=
17.6
"
Iz
=
1
3
mro d b2 + mro d
12
) Iz
=
1
1
mrod b2 = mb2 J
2
6
b
p
2 3
2
#
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 =
1 m 2 m
b +
12 4
4
b
2
2
=
1
mb2
12
372
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Bottom (mass = m=2): (Iz )2 =
1 m 2
m
(b + b2 ) +
12 2
2
Iz = 2(Iz )1 + (Iz )2 = 2
1
mb2
12
2
75
y
3
1
180
p b
2
2
2
=
1 2
mb
3
1
1
+ mb2 = mb2 J
3
2
17.8
150
x
(Ix )1
=
(Ix )2
=
(Ix )3
=
Ix
=
(Iz )1
=
(Iz )2
=
(Iz )3
=
Iz
=
m1
=
m2
=
m3
=
60
y
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
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
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
373
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
17.10
y
1
x
_
r
2 ft
G
2
1.5 ft
m1 =
1:2 2
= 21:30
32:2 3:5
10 3 slugs
m2 =
1:2 1:5
= 15:972
32:2 3:5
10 3 slugs
374
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(Iz )1
=
(Iz )2
=
=
Iz
=
1
1
m1 L21 =
21:30 10 3 (2)2 = 28:40 10 3 slug ft2
3
3 "
#
2
1
L2
1
2
2
m2 L2 + m2 L1 +
= m2 (3L21 + L22 )
12
2
3
1
(15:972 10 3 ) 3(2)2 + 1:52 = 75:87 10 3 slug ft2
3
(Iz )1 + (Iz )2 = (28:40 + 75:87) 10 3 = 104:27 10 3 slug ft2 J
x =
y
Iz
= Iz
=
24:4
=
2(0) + 1:5(0:75)
Li xi
=
= 0:3214 ft
Li
2 + 1:5
Li yi
2( 1:0) + 1:5( 2)
=
= 1:4286 ft
Li
2 + 1:5
mr2 = 104:27
(21:30 + 15:972)(0:32142 + 1:42862 )
10 3
10 3 slug ft2 J
17.11
375
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.12
I
1
msp oke R2
3
8
8
+
32:2
3
= Irim + 8(Isp oke ) = mrim R2 + 8
8
= R2 mrim + msp oke = 1:252
3
=
0:5
32:2
0:453 slug ft2 J
17.13
(a) Sphere:
Ix
=
=
2
mR2 + m(L + R)2
5
2
2 130
5
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 =
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
= 408:5 120(1:5479)2 = 121:0 kg m2 J
376
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.15
*17.16
17.17
377
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.18
17.19
B
20 lb
O
( MO )FBD
P
=
ma
O
2'
P
A
N
B
2'
A
MAD
FBD
= ( MO )M AD
= 23:1 lb J
60o
+
P (2 sin 60 )
20(4 cos 60 ) = 0
378
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.20
B
P
R mg
FBD
R
µs mg
A
ma
=
A
MAD
N = mg
Assume impending sliding.
MA
= maR +
2RP = maR
Fx
= ma + !
s mg + P = ma
a =
2 sg J
P =
ma
2
s mg +
ma
= ma
2
17.21
379
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.22
NA
y
4.5'
.2.25' =
B.
55 lb N
G
20 lb
B
FBD
Fx = max
O
20 =
.
G
55 a
g
MAD O
55
a
32:2
( MO )FBD
NA (4:5)
x
5.25'
A
a = 11:709 ft/s
2
J
=
( MO )M AD +
55
(11:709)(5:25)
55(2:25) =
32:2
NA = 50:8 lb J
17.23
380
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.24
The bar AB is translating. Therfore, all points on the bar have the same velocity
and acceleration.
TA
35o
A
.
2 ft
y
TB
.G
x
ma n
.B = A
2 ft
mg
FBD
G
ma t
.
B
MAD
an = 3 _
2
at = 3•
( MG )FBD = ( MG )M AD + :
(TA cos 35 ) (2)
(TB cos 35 )(2) = 0
TA = TB
( Fx )FBD = ( Fx )M AD + .:
mg sin 35 =
mat
48 sin 35 =
48
3•
32:2
•=
2TA
48 cos 35 =
32 )
TA = TB = 39:8 lb J
6:16 rad/s
2
J
( Fy )FBD = ( Fy )M AD + -:
TA + TB
mg cos 35
= man
2TA
48 cos 35
=
48
(3
32:2
2
48
(3 _ )
32:2
17.25
y
80 lb
FBD
F
80 a
32.2 n
o
= 30
MAD
x
N
The box translates along a circular path of 5-ft radius
) an = R! 2 = 5(2)2 = 20 ft/s
Fy
= may
+"
Fx
= max
+ !
N
2
80
(20) sin 30
32:2
80
(20) cos 30
F =
32:2
80 =
N = 104:8 lb J
F = 43:0 lb J
381
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.26
L/2
A
B = A
mg FBD
N
( MA )FBD
L
+
mg
2
aB
L/2
mL2α
12
B
MAD
m Lα
2
=
( MA )M AD
L
L mL2
=
m
+
2
2
12
3
= L = g #J
2
=
3g
2L
17.27
A
N2
N1
=
7a
y
x
MAD
FBD
A
Ax
=
35
3a
3(9.81) N
2m
35 o
35 o
T
B
o
B
FBD
1.0 m
Ay
7(9.81) N
MAD
System:
Fx = max
+ . 7(9:81) sin 35 = 7a
a = 5:627 m/s
2
Bar:
( MA )FBD
T
= ( MA )M AD +
(T sin 35 )(2) = (3a cos 35 )(1:0)
= 2:142a = 2:142 (5:627) = 12:05 N J
17.28
1 2
B 12 mLαAC
mg
B
C
C = A
L/2
L/2
L/2 R L/2
T
FBD's
1 mL2α MAD's
mg
DE
T
12
F
D
=
D
E
E
L/2
L/2
L/2
L/2
F Lα
m 2 ( AC + αDE)
A
382
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Kinematics:
# aE = aC =
L
2
# aF = aE +
AC
L
2
DE =
L
( AC +
2
DE )
Kinetics— bar AC:
( MB )FBD = ( MB )M AD
1
L
T =
mL2 AC
2
12
+
AC =
6T
mL
Kinetics— barDE:
( MF )FBD
=
( MF )M AD
Fy
= may
T
=
1
6T
L
T =
mL2 DE
DE =
2
12
mL
L
L
6T
T = m ( AC + DE ) = m
2
2
2
mL
+
+#
mg
1
mg J
7
17.29
.G
NB
=
mg
o
60
.
A
.A
Ax
Ay
60o
MAD
FBD
( MA )FBD = ( MA )M AD +
mg/2
8f
t
8f
t
10
ft
.G
B
:
mg (8 cos 60 )
10NB
=
60(8 cos 60 )
10NB
=
mg
(8 sin 60 )
2
60
(8 sin 60 )
2
NB = 3:215 lb J
( Fx )FBD = ( Fx )M AD +!:
Ax
NB sin 60
=
Ax
=
mg
Ax
2
32:78 lb
3:215 sin 60 =
60
2
383
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
( Fy )FBD = ( Fy )M AD + ":
Ay + NB cos 60
17.30
mg = 0
Ay + 3:215 cos 60
Ay = 58:39 lb
p
A = 32:782 + 58:392 = 67:0 lb J
60 = 0
384
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.31
17.32
40(0.252)α
40(9.81) N
G 0.6 m
0.6 m 0.6 m
=
G
T
800 N
R
T
FBD's
A
60(9.81) N
60(0.6α)
A
MAD's
System:
800(0:6)
( MG )FBD = ( MG )M AD + :
60(9:81)(0:6) = 40(0:25)2 + 60(0:6 )(0:6)
=
2
5:263 rad/s
J
385
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Block A:
Fy
T
= may + "
T 60(9:81) = 60(0:6)
= 60(9:81) + 60(0:6)(5:263) = 778 N J
17.33
FBD
mg
2 ft
=
A
B
2 ft
µs NA
( MB )FBD
ma
4:5mg 7:5
= max
4.5 ft 3 ft
+ !
=
( MB )M AD
=
2ma
ma A
B
NB 4.5 ft 3 ft NA
Fx
MAD
G
s NA = 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
Ay
.A A
x
.G
R
I
2R
2R
b
= IC
π
b
MAD
= R2 +
=
. 2R
mbα
FBD
sin
A
θ
.
R
=
mg
b2
C
I-α
G
2
=
cos =
m
1+
4
2
R2
R
b
2
2R
= mR2
m
2R
2
=
4
1
2
mR2
( MA )FBD = ( MA )M AD + :
mgR
g
= I + (mb )b =
=
2R
=
g
2R
1
4
2
mR2 + 1 +
4
2
mR2
J
386
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
( Fx )FBD = (Fx )M AD +!:
Ax = mb sin = mb
g
2R
2R
b
=
mg
! J
( Fy )FBD = (Fy )M AD +":
Ay
mg =
mb cos =
mb
g
2R
R
b
Ay =
mg
" J
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
Kinetics— racks:
FA = 2(aC + 3 )
FB = 4( aC + 2 )
Kinetics— gear:
2
IC = mkC
= 5(2)2 = 20 slug ft2
( 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
Solution of (a) and (b) is
= 22:0 rad/s2
(b)
and aC = 4:0 ft/s2 ! J
387
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.36
x
aA
A
6' T
B
aB
y
α 8'
G
B
B
G
8'
mg
=
A
6'
1 m(102)α
12
3mα
4mα
A
MAD
N FBD
Kinematics:
aG
=
k
rG=A = rG=B = 3i + 4j ft
rG=A = ( 3j + 4i)
= aA + aG=A = aA i +
rG=A = aA i + ( 3j + 4i)
)
(aG )y = 3
aG
= aB + aG=B =
) (aG )x = 4
aB j +
rG=B =
aB j + (3j
4i)
Kinetics:
Fx = max +
( MA )FBD
T = 4m
=
( MA )M AD +
1
m(102 ) + 3m (3)
3mg 8T =
12
3mg 8(4m ) = 1:3333m
=
T =4
4m (4)
2
0:090g = 0:090(32:2) = 2:898 rad/s
50
(2:898) = 18:0 lb
32:2
J
J
388
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.37
389
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.38
390
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
391
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.41
4
= 0:124 22 slugs
32:2
1
1
mL2 =
(0:124 22)(32 ) = 0:093 17 slug ft2
=
12
12
= mr = 0:124 22(0:75) = 0:093 17 "
= mr! 2 = 0:124 22(0:75)(42 ) = 1:4906 lb
m =
I
may
max
y
Cy
Cx
G
1.5 lb ft C
.
..
0.75'
=
FBD
x
MAD
0.09317α
G 1.4906 lb
.C . 0.09317α
Only the horizontal forces are shown on the FBD.
( MC )FBD
=
=
Fx
Fy
( MC )M AD +
2
9:20 rad/s
= max +
= may + "
C=
17.42
1:5 = 0:093 17 + 0:093 17 (0:75)
J
Cx = 1:4906 lb
Cy = 0:093 17 = (0:09317) (9:20) = 0:8572 lb
p
1:49062 + 0:85722 = 1:720 lb J
392
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.43
Ay
A
d
W
L
=
.C
P
A
Ax = 0
Iα
G
C
mat
d − L/2
.
FBD
MAD
B
B
( MC )FBD = ( MC )M AD + :
0
= I
d
=
mat d
L
2
=
mL2
12
m
L
2
d
L
2
2
L J
3
393
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.44
17.45
394
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.46
30
= 0:9317 slugs
32:2
2
1
1
16
=
mR2 = (0:9317)
= 0:8282 slug ft2
2
2
12
!
p
302 + 162
= 2:640
= mL = 0:9317
12
m =
I
mat
System:
16"
A
.
By
34"
30 lb
.
B
Bx
=
y
x
FBD
0.8282α
34"
"
16 A
2.640α
.
.
B
MAD
395
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
( MB )FBD
=
( MB )M AD +
=
10:231 rad/s
30
34
12
= 0:8282 + 2:640
34
12
J
2
Disk:
30 lb
Ax
.A
16"
Ay
FBD
( MA )FBD
=
T
=
. 0.8282α
=
A
T
2.640α
C
MAD
16
T = 0:8282
12
0:6212 = 0:6212(10:231) = 6:36 lb J
( MA )M AD +
17.47
A
.G
Iα
4 ft
=
θ
40 lb
O
C
θ
4 ft
θ
A
C
NA
B
FBD
NB
.G
ma n
ma t
B
O
MAD
The path of the mass center G is a circle of radius r = 4 ft centered at O.
Let the angular acceleration of line OG be = • clockwise. Then the angular
396
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
acceleration of the bar is also , but it is directed counterclockwise (see MAD).
m =
man
mat
=
=
I
=
=
W
40
=
= 1:2422 slugs
g
32:2
mr! 2 = 1:2422(4)(2)2 = 19:875 lb
mr = 1:2422(4) = 4:969
1:2422(8)2
mL2
=
= 6:625 slug ft2
12
12
30
( MC )FBD = ( MC )M AD +
:
W (4 sin 30 )
= I + mat (4)
40(4 sin 30 )
=
6:625 + 4(4:969 )
= 3:019 rad/s
2
J
( Fx )FBD = ( Fx )M AD + !:
NA
=
=
man sin 30 + mat cos 30
19:875 sin 30 + 4:969(3:019) cos 30 = 3:05 lb ! J
( Fy )FBD = ( Fy )M AD +":
NB
NB
W =
man cos 30
mat sin 30
40 =
19:875 cos 30
4:969(3:019) sin 30
NB = 15:29 lb " J
17.48
397
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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
398
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17.51
399
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.52
400
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.53
401
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.54
17.55
Kinematics:
y
1.2 m
ω, α
O G
x
0.4 m
402
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
C F
Ma-y
O G
0.4 m
=
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
403
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.57
404
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.58
aC = aO + aC=O
Kinematics (C is the midpoint of bar AB):
aC
aO
=O
+
0.15 α
0.15
C
2
0.15ω
ω, α
O
Kinetics:
NB
0.3
0.2
4
0.15
3
C
mg
0.2
Ay
Iα
=
C
m(a0 + 0.15ω2)
A
Ax
A
B
m(0.15α)
B
FBD
MAD
2
! = 5 rad/s
= 8 rad/s
a0 = R = 0:45(8) = 3:60 m/s
m = 25 kg
mg = 25(9:81) = 245:3 N
mL2
25(0:5)2
I =
=
= 0:5208 kg m2
12
12
2
( MA )FBD = ( MA )M AD :
0:15mg
0:5NB
NB
= I + m(a0 + 0:15! 2 )(0:2) (0:15m ) (0:15)
= 0:5208(8) + 25 3:60 + 0:15(5)2 (0:2) 0:15(25)(8)(0:15)
= 0:757 N & J
17.59
1
1
mL2 =
(4:2) (1:22 ) = 0:5040 kg m2
12
12
Kinematics (the system has 2 DOF):
Bar AB:
I=
aA + aG=A = aG
405
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
aA
A
α
+
A
= G
aA− 0.6α
0.6 m
0.6α
G
Kinetics:
WA
A .
NA
G
A
.
14 N
2.4aA
0.6 m
= 0.5040Gα . 4.2(aA − 0.6α)
.
WAB
y
MAD
FBD
Fx = max + !
x
14 = 2:4aA + 4:2(aA
( MA )FBD = ( MA )M AD +
0 = 0:5040
4:2(aA
Solution is
=
5:07 rad/s
2
J
0:6 )
0:6 )(0:6)
aA = 4:06 m/s ! J
2
406
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.60
407
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.61
17.62
For the assembly:
MB
= IB
=
40 =
2
6:822 rad/s
1
2
40
32:2
(1:2)2 +
40
(2)2
32:2
J
408
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
For the disk:
Ay
.
A
NC
0.8 ft
IAα
=
Ax C
.A man
mat
FBD
MAD
0:8NC =
1
2
40
32:2
MA
= IA
NC
=
W
= mg = 150(9:81) = 1471:5 N
(1:2)2 (6:822)
7:63 lb " J
17.63
= m(AG! 2 ) = 150(0:6)
man
v0
1:8
!=
2
= 27:78v02
y
Ax
v0
v0
=
R
1:8
y
0.6 G
θ
A
1.8 Ay 1471.5 N
G
x
=
27.78v02
P
FBD
MAD
Fy
Ay
27:78v02 sin
= may + "
Ay 1471:5 =
= 1471:5 27:78v02 sin
θ
x
409
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Disk stays in bearing if Ay
this condition is
0 for all . The largest v0 that does not violate
r
1471:5
v0 =
= 7:28 m/s J
27:78
17.64
410
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.65
*17.66
411
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.67
412
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.68
(a)
MA = IA
k =
mL2
3
=
3k
3k
=
2
mL
mL2
J
(b)
3k
d!
!
! d! =
d
mL2
Initial condition: ! = 0 when = 0
=
0
=
)
1 2
! =
2
d
3k 2
+C
2 0
2mL
r
3k
( 2
!=
mL2 0
(c)
! max = !j =0 =
r
C=
2
3k 2
+C
2mL2
3k 2
2mL2 0
) J
3k
0 J
mL2
17.69
(a)
1.2 m mg
F
FBD D N d
360
Fx
30v
P
0.45 m
=
C y
ma
0.6 m
C
x
MAD
= max + !
P FD = ma
2
= 60a
a = 6:0 0:5v m/s Q.E.D
(a)
413
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(b)
dv
Zdt
a =
t
=
dv
= dt
a
dv
=
6:0 0:5v
dv
= dt
6:0 0:5v
2:0 ln(6:0
0:5v) + C
Initial condition: v = 0 when t = 0. ) C = 2 ln 6:0
)t=
2:0 ln
6:0
0:5v
=
6:0
v
12
2:0 ln 1
(b)
When tipping impends, we have d = 0 on the FBD.
( MC )FBD = ( MC )M AD +
1:2P 0:45mg = 0:6ma
2
1:2(360) 0:45(60)(9:81) = 0:6(60)a
a = 4:643 m/s
Eq. (a):
4:643 = 6:0
0:5v
v = 2:714 m/s
2:714
2:0 ln 1
= 0:513 s J
12
Eq. (b):
t=
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 45
o
0.18 m
0.03888α
=
Fy
NC
= may
+%
= 20:47 N
C
y
NO
C 0.15NC
FBD
NC
O
x
MAD
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
414
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(b)
=
d!
!
d
)
d = ! d!
14:215 d = ! d!
14:215 =
1 2
! +C
2
= 0: ) C = 0
1
) 14:215 = ! 2
) = 0:03517! 2
2
Initial condition: ! = 0 when
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
40o
0.75N B FBD
N
2.236α
20o
2' 18 lb
C Cx
Cy
=
C
MAD
FBD
Bar AB :
+
N =
MA = 0
6(1:0 sin 20 ) + 0:75N (2 cos 40 )
15:033 lb
Wheel
+
( 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
N (2 sin 40 ) = 0
2:236
J
(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
415
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.72
(a)
Wheel: I = mR2 =
Ay
N
A
Ax
1.0'
6 lb
40
1.0' 20o
18
(2)2 = 2:236
32:2
0.75N
o
0.75N B FBD
N
2.236α
20o
2' 18 lb
C Cx
Cy
=
C
MAD
FBD
Bar AB :
+
N =
MA = 0
6(1:0 sin 20 )
0:8429 lb
Wheel
+
( MC )FBD = ( MC )M AD
0:75N (2) = 2:236
0:75(0:8429)(2) =
:
0:75N (2 cos 40 )
2
=
0:5654 rad/s = 0:5654 rad/s
2
N (2 sin 40 ) = 0
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
17.73
(a)
IA
=
IB
=
1
2
1
2
18
32:2
30
32:2
(1:52 ) = 0:6289 slug ft2
(2:02 ) = 1:8634 slug ft2
416
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18 lb
Ax
'
1.5
A
=
0.6289αA
0.3N = 5.4 lb
N = 18 lb
N = 18 lb
0.3N = 5.4 lb
30 lb
2'
B
=
Bx
1.8634αB
By
FBD's
( MA )FBD
MAD's
=
( MA )M AD +
( MB )FBD
=
=
12:880 rad/s = 12:880 rad/s
J
( MB )M AD +
5:4(2:0) = 1:8634 B
B
=
5:796 rad/s
A
5:4(1:5) =
2
2
0:6289 A
2
J
(b)
d!
!=
dt
During sliding:
=
!A
=
!B
=
dt
! = t + ! 0 ( constant, ! 0 = initial speed)
2
= 12:880t + 18:850 rad/s
60
5:796t + 0 = 5:796t rad/s
12:880t + 180
Sliding stops when
RA ! A
t
= RB ! B
= 0:9147 s
1:5( 12:880t + 18:850) = 2:0(5:796t)
Final speeds:
!A
!B
=
=
12:880(0:9147) + 18:850 = 7:07 rad/s J
5:796(0:9147) = 5:30 rad/s J
417
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.74
(a) Let _ be the angular velocity of the disk. Noting that A is the I.C. of the
disk, we have
vA = vG + vA=G
0
. G
.
O
θ
=
R
β
r
+
.
rβ
.
G
0 = R_
r
) _ =
r_
R•
r
G
G
..
Iβ r
ma t
A
at = R•
an = R _
( MA )FBD = ( MA )M AD +
A
MAD
2
(point G rotates about O)
:
= I • + (mat ) r
3 •
R
2
=
N
F
=
)•=
ma n
FBD
mg(r sin )
R_
r
mg
θ
g sin
. Rθ
.A
•=
mgr sin =
2g
sin
3R
mr2
2
R•
+ mR• r
r
J
(b) The di¤erential equation is equivalent to
d_ _
d
=
)
2g
_ d_ =
sin
3R
1 _2
2g
=
cos + C
2
3R
2g
sin d
3R
418
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Initial condition: _ = 0 when
0
=
1 _2
2
=
= 60
2g
cos 60 + C
3R
2g
1g
cos
3R
3R
C=
_=
1g
3R
r
2g
(2 cos
3R
1) J
17.75
419
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.76
420
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.77
421
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.78
Assume impending sliding.
B
Iα
G
30o
L/2
B
30o
mg
F = µsN
A
A
MAD
N FBD
( MA )FBD = ( MA )M AD +
L
mg sin 30
2
=
:
L
2
3g
3g
sin 30 =
2L
4L
= I +
mL α
2
mL
2
L
mg sin 30 =
2
mL2
mL2
+
12
4
( Fx )FBD = ( Fx )M AD + !:
sN =
mL
mL
cos 30 =
2
2
3g
sin 30
2L
cos 30 = 0:3248mg
( Fy )FBD = ( Fy )M AD + ":
N
mg
N
mL
sin 30
2
mL 3g
= mg
sin 30
2
2L
=
s =
sN
N
=
sin 30 = 0:8125mg
0:3248
= 0:400 J
0:8125
422
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.79
A
θ
L/2
θ
L/2
mg
2
=
( MA )FBD
=
m L
g( + L) cos
2 2
=
d!
!
d
m L
2 2
18 g
cos
17 L
) ! d! =
Initial condition: ! = 0 when
mLα
2
( MA )M AD
=
=
m Lω2
22
mLα
22
mLω2
2
1m 2
12 2 L α
MAD
mg
2
FBD
+
1m 2
12 2 L α
A
= 0:
r
)!=
2
+
m 2
L +2
2
1 m 2
L
12 2
J
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
A
An
6 ft
At
t
Bt
n
750 lb . ft B
FBD
Bn
mat
man
100 lb
2 ft 2 ft
θ A
C
Cn
An
At
ma t
= A
θ
C
man
MAD
FBD
100
(6) = 18:634 lb
32:2
= mL! 2 = 18:634! 2 lb
= mL =
(a)
Bar AB:
MB = 0 +
6At
750 = 0
At = 125 lb
423
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Bar AC:
125
Ft
100 cos
= mat + &
= 18:634
At 100 cos = 18:634
2
= 6:708 5:367 cos rad/s J
(b)
=
d!
!
d
! d! =
d = (6:708
5:367 cos ) d
1 2
! = 6:708
5:367 sin + C
2
Initial condition: ! = 0 when = 0: ) C = 0
p
) ! = 13:412
10:734 sin rad/s J
17.81
424
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.82
17.83
425
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
*17.84
426
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
=
d!
!
d
! d! =
1 2
! =
2
d = 4:905 sin
4:905 cos + C
Initial condition: ! = 0 when = 30
) C = 4:905 cos 30
p
2(4:905)(cos 30
cos )
! =
p
= 3:132 0:8660 cos rad/s J
(b) Substituting the expressions for ! and
Eq. (a), we get
NA
=
=
, and the values of M and L into
25(3)
[(4:905 sin ) cos
2(4:905)(0:8660
2
183:94(3 cos
1:7320) sin N J
(c)
NA = 0 when 3 cos
1:7320 = 0
)
= 54:7
cos ) sin ]
J
17.85
(a) Kinematics of bar AB:
aA
L/2
A
L
+ 2α
G
ω, α
Lω2
2
B
a = aA + aG=A
427
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Kinetics:
m2g
Ax
P0
FBD's
Collar:
Bar:
MAD's
Ay
A
G θ
m1g
Fx = max
Ax
m Lα
= 12
L/2
L/2
B
m2aA
=
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
+ !
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
P0
m1
+
L
2
(m1 aA )
1
m1 L(! 2 sin +
2
m1 + m2
m2 aA
L
cos
2
m1
L
cos
2
cos ) L
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
cos
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
428
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
t
8.6000e-001
8.8000e-001
is maximized. Note
x1
x2
2.4173e+001 1.8418e+000
2.4179e+001 -1.3238e+000
By inspection we see that
max = 24:2
J
429
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(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
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)
(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:
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)’)
430
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
17.87
(a) Geometry:
B
A
φ
L
R
θ C
431
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
L sin
= R sin
tan
=
sin
p
1
Kinetics:
) sin
φ
θ
R
sin
L
sin
=q
(L=R)2
sin2
sin2
Iα
P B
P tanφ
=
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
(L=R)2 sin2
0
cos
gP0 R sin @
sin + q
=
2
Wk
(L=R)2
2
sin
1
A Q.E.D.
(b) Substituting the given data, the equation of motion becomes
0
1
cos
32:2(24)(0:75) sin @
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
The corresponding MATLAB program is
function problem17_87
[t,x] = ode45(@f,[0:0.02:2.4],[pi/2,0]);
432
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
t
2:24
=
9:8875
10
0
0
0.5
2:24
9:8975
t = 2:25 s J
(c)
12
omega (rad/s)
10
8
6
4
2
1
1.5
2
2.5
t (s)
433
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
L/2
θ β x
mg
N
mg sin
G
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
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
434
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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)
435
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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 _
32:2 cos (3 + 2r) + 4rr_ _
3 + 2r2
2
The initial conditions are: = =3, _ = 0, r = 1:5 ft and r_ = 0 when t = 0
_ r r_ T , the …rst-order equations and the initial conditions
Letting x =
are
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)
436
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
aθ
ar
2
17.90
(a) Kinematics:
Top view
..
y
r
ω, α
D
C + C
aD = aC + aD=C
437
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Kinetics:
B
D
L/2 C Cx N
r
mABy..
..
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 r0 ) + r _ + ap2 sin pt sin Q.E.D
r• =
m
r0 )
(b) Substituting the given data, we get
• =
r
(312:5
10 6 ) =0:125 + r2
r sin 10t cos
=
r• =
=
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
438
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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 =
x3 (sin 10t cos x1 2x4 x2 )
0:0025 + x23
x_ 3
= x4
x_ 4 =
25(x3
=
x(0)
0
0
0:05 m
0
0:05) + x3 x22 + sin 10t sin x1
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)
439
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.91
17.92
Consider the disk as a composite of the uniform disk 1 and the half-disk 2.
m
2m
+ O x- 2 G2
O
2
1
(a)
x2
=
x =
4
R
3
m1 x1 + m2 x2
0 + mx2
4
=
=
R J
m1 + m2
3m
9
440
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(b)
IO = (IO )1 + (IO )2 =
I = IO
3mx2 =
m1 R 2
m2 R 2
2mR2
mR2
5
+
=
+
= mR2
2
4
2
4
4
5
mR2
4
3m
4
R
9
2
= 1:190mR2 J
17.93
30o
mg
x
G
A =
0.4
0.2N N
0.4
0.4
P
y
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
17.94
441
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.95
Assume impending loss of contact at B.
N
0.9
'
0.9
'
A
B
Ay
A
Ax
0.9
'
0.2N
FBD's
W + 12 a
g
18 lb =
W + 12 lb
=
W
MAD's
Wa
g
60o
60o
Bar AB:
( MA )FBD
=
a =
( MA )M AD +
W (0:9 cos 60 ) =
W
a(0:9 sin 60 )
g
0:5774g
System:
Fy
Fx
=
0+"
N
(W + 12) = 0
N = W + 12 lb
W + 12
a
= max + !
18 0:2N =
g
W + 12
18 0:2(W + 12) =
(0:5774g)
W = 11:15 lb J
g
442
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.96
D
T
G
A
B FBD
y
L/2
L/2 o
45 mg
1 mL2α
aG = 45o B + G L/2 B
D
12
α
Lα
aB
L/2
L/2
2
A
G
aG = aB + aG/B
maB o
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
T
m
B
x
B MAD
T + mg sin 45 =
L
4
1
mL sin 45
2
T = 0:283mg J
sin 45
17.97
ma = mCG! 2 = 6
( MA )FBD = ( MA )M AD
=
C
45o
_
ma
45o
G
m
B
MAD
0.2
2 4
Ax
45o
0.24 m
A
2
FBD
Ay
0.2
4
ω
A
0:24
p
2
(10)2 = 101:82 N
+
0:24T
=
T
=
0:24
p
2
72:0 N J
101:82
17.98
y
279.5α
1800 lb
o
62.11aA MAD
α
419.3
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
443
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
I
=
max
=
1 2000 2
1
mL2 =
9 = 419:3 slug ft2
12
12 32:2
2000
L
2000
aA = 62:11aA
may = m
=
32:2
2
32:2
9
2
= 279:5
( MA )FBD = ( MA )M AD
(1800 sin 40 )(9) 2000(4:5) = 419:3 + 279:5 (4:5)
=
Fy = may
+"
N
N
Fx = max
+ ! 0:8N
0:8(1078:5)
J
2
0:8427 rad/s
2000 + 1800 sin 40
2000 + 1800 sin 40
N
1800 cos 40
1800 cos 40
=
=
aA
=
aA
=
=
=
=
279:5
279:5(0:8427)
1078:5 lb
62:11aA
62:11aA
8:309 ft/s
8:31 ft/s
2
2
J
17.99
444
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.100
aG
I
= aA + aG=A = 50 -aA + # 2:25
=
mL2
80(4:5)2
=
= 135:0 kg m2
12
12
y'
40o
T
A
40o
x'
G
80(9.81) N
( MG )FBD
T
Fy 0
B
2.25 m
2.25 m
80aA
=
2.25 m
135.0α
2.25 m
o
40
80(2.25)α
FBD
= ( MG )M AD +
= 93:34
MAD
(T sin 40 ) (2:25) = 135:0
T 80(9:81) sin 40 = 80(2:25) sin 40
= may0 + %
93:34
80(9:81) sin 40 = 80(2:25) sin 40
= 2:413 rad/s
J
T = 93:34(2:413) = 225 N J
2
17.101
445
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.102
Assume impending sliding
B 60 lb
Iα
G
=
120 lb
8 ft
b
F = µsN
A
ma t
FBD
A
MAD
N
:
120 162
mL2
=
= 79:50 slug ft2
12
32:2 12
L
120 16
= m
=
= 29:81 lb
2
32:2 2
I
=
mat
Fy = 0: N = 120 lb
( Fx )FBD = ( Fx )M AD + !:
60
s N = mat
60
( MA )FBD = ( MA )M AD +
60b
b
0:3(120) = 29:81
2
= 0:8051 rad/s
:
= 79:50 + 29:81 (8)
= 5:230 = 5:230(0:8051) = 4:21 ft J
446
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17.103
17.104
110 lb y
G
200 lb
2 ft
1.0 ft
D
T
FBD
x
N
=
0.3N
G
20.12α
D 12.422α
MAD
447
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The spool "rolls" on point D. ) a = DG = 2 ft/s2
I
mk
2
200
(1:8)2 = 20:12 slug ft2
32:2
200
(2 ) = 12:422 lb
32:2
=
ma =
Fy = 0 + "
N
200 = 0
N = 200 lb
( MD )FBD = ( MD )M AD +
110(5) 0:3N (1:0) = 20:12 + 12:422 (2)
110(5) 0:3(200)(1:0) = 20:12 + 12:422 (2)
=
10:898 rad/s
a =
Fx
2
2(10:898) = 21:8 ft/s
2
J
= ma + !
110 T + 0:3N = 12:422
110 T + 0:3(200) = 12:422(10:898)
T = 34:6 lb J
17.105
n
A An
FBD
R
At
=
θ mg
t mRα
A
mRω2
MAD
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
448
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(c)
Fn
An
mg cos
Ft
&
At + mg sin
=
A=
s
= man %
An
g
= mR
cos
R
At + mg sin =
g
mR
sin
2R
(2mg cos ) +
1
mg sin
2
20(9.81) N
B y
3m
2
2
= mg
mg cos = mR! 2
An = 2mg cos
mR
1
mg sin
2
At =
r
4 cos2 +
1
sin2
4
J
17.106
P
A
3m
FBD
NA
Fx
( MA )FBD
20(9:81)(3) + 1:5P
20(9:81)(3) + 1:5P
+
B
x
=
1.5 m
A
20a
2.25 m
MAD
= max
+ ! P = 20a
) a = 0:05P
= ( MA )M AD
= 20a(2:25)
= 20(0:05P )(2:25)
P = 785 N J
17.107
(a)
θC
1.0'
G
. .
F
N
mg
FBD
( MC )FBD = ( MC )M AD +
mg(1:0) cos
40 cos
=
Iα
. .
C
man
1.0' G
mat
MAD
:
= I + mat (1:0)
40
82
40
=
+
(1:02 )
32:2 12
32:2
= 5:084 cos rad/s
2
J
449
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(b)
d!
! = 5:084 cos
d
1 2
! =
2
! d! = 5:084 cos d
Initial condition: ! = 0 when
5:084 sin + C
= 0: ) C = 0
p
! = 3:189 sin rad/s
) ! 2 = 10:168 sin
J
(c)
( Ft )FBD = ( Ft )M AD + %:
N
mg cos
=
mat
N
=
40 cos
N
40
(1:0)
32:2
40 cos =
40
(5:084 cos ) = 33:68 cos lb J
32:2
( Fn )FBD = ( Fn )M AD + -:
F
mg sin
F
F
=
40
(10:168 sin ) = 52:63 sin lb J
32:2
40 sin +
(d) Sliding impends when F=N =
40 sin =
40
(1:0)! 2
32:2
= man
s:
52:63
tan = 0:6
33:68
= 21:0
J
17.108
(a)
P
A
y
L/2
mg θ
FBD
+
P (L cos )
=
)
L/2
By
x
=
B
Bx
( 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
1 2
3 2P
! =
g sin + C
2
2L m
2P
m
g cos d
450
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Initial condition: ! = 0 when = 0: ) C = 0
s
3 2P
)!=
g sin
L m
J
(b)
Fx
Bx
L
L
Bx = m ! 2 cos + m
sin
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
O
1 mR2α
2
MAD
µkmg
N = mg
Final angular speed of disk: ! 0 =
( MO )FBD
=
)
v
5
=
= 20 rad/s
R
0:25
( 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
451
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Chapter 18
18.1
18.2
18.3
452
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18.4
B
2.4
ft
4
.
10 ft
2 lb
lb
10
k = 12 lb/ft
θ
L0 =2.5 ft
40 lb.ft
A
Work of gravity:
Work of couple:
(U1 2 )c =
Work of spring:
(U1 2 )s =
=
=
Total work: U1 2 =
C
(U1 2 )g =
2W
h=
C
35
180
= 24:43 lb ft
= 40
2(10)(1:2 sin 35 ) =
1
k (L2 L0 )2 (L1 L0 )2
2
1
(12) (4:8 cos 35
2:5)2 (4:8
2
19:437 lb ft
13:766 lb
ft
2:5)2
13:766 + 24:43 + 19:473 = 30:1 lb ft J
18.5
U1 2
=
12(9:81)(0:375)
=
78:9 J J
8(9:81)
p
0:752 + 0:32
0:45 + 40
2
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
453
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18.7
De…nition of power:
Pout = C!
De…nition of e¢ ciency:
Pout = Pin
(a) C
=
(b) C
=
)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
M =I =I
d!
dt
But P = M !, so that
P
d!
=I
!
dt
P dt = I! d!
Pt =
1 2
I! + C
2
Initial condition: ! = 0 when t = 0: ) C = 0
r
2P t
!=
J
I
18.9
454
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18.10
18.11
455
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18.12
18.13
456
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18.14
18.15
I
= mk 2 = 60(0:24)2 = 3:456 kg m2
IC
= 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
T
2
457
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18.16
C ω
d
A
vA = 12 ft/s
3 ft
ft
10
8 ft
3 ft
G
B
3
vB
4
Instant center of zero velocity is at C. From geometry
AC
=
d2
=
IC = I + md2 =
T =
vA
12
=
= 1:5 rad/s
8
AC
82 + 32 = 73 ft2
8 ft
!=
62
+ 73
12
24
1 W 2 W 2
L +
d =
12 g
g
32:2
= 56:65 slug ft2
1
1
IC ! 2 = (56:65)(1:5)2 = 63:7 lb ft J
2
2
18.17
0.3 rad/s C
d
30o
G
4 ft
6 ft
1.8 ft/s
A
Point C is the I.C. of the bar. From geometry:
d2 = (4 sin 30 )2 + (6
4 cos 30 )2 = 10:431 ft2
458
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1
1
mL2 =
12
12
20
32:2
I
=
IC
= I + md2 = 3:313 +
T
=
(8)2 = 3:313 slug ft2
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
0.8 ft
3.2 ft/s
C
2.4 f
t
10 lb
3.2 ft/s
B 2 lb
15 lb
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
1
2
10
2
(3:2)2 +
(3:2)2
0:14907(4) +
2
32:2
32:2
= 3:10 lb ft J
18.19
1.5 m/s
1.4 kg
110 mm
B
220
mm
2.8
kg
1.5 m/s
A
13.64 rad/s
C
459
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Bar BC is translating with the velocity v = 1:5 m/s.
! AB
=
TAB
=
=
TBC
=
1:5
= 13:64 rad/s
0:110
1
1 1
2
IA ! 2AB =
mAB AB ! 2AB
2
2 3
1
(1:4)(0:110)2 (13:64)2 = 0:5253 J
6
1
1
mBC v 2 = (2:8)(1:5)2 = 3:150 J
2
2
T = 0:5253 + 3:150 = 3:68 J J
18.20
5
= 0:155 28 slugs
32:2
2
= mR = 0:155 28(1:5)2 = 0:3494 slug ft2
1
1
1
1
=
mv 2 + I! 2 = (0:155 28)(30)2 + (0:3494)(102 ) = 87:3 lb ft J
2
2
2
2
m =
I
T
18.21
vB
B
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
=
=
= ! AB AB = 7:5(2) = 15 m/s
15
vB
=
=
= 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
460
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TBC
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
A
=
=
+# 3=8
+# v=8
I=
T
B
G
v
vB = vA + vB=A
v = vA + vG=A
ω 1.2 m
1.2 m
2:4!
1:2! = 8
! = 2:083 rad/s
1:2(2:083) = 5:500 m/s
2
18(2:4)
= 8:640 kg m2
12
1 2 1
1
1
I! + mv 2 = (8:640)(2:083)2 + (18)(5:500)2
2
2
2
2
291 J J
18.23
ωDE
E
vD
.D
1.6 m
m
1.0
vB
B
.
0.8 m
A
ωAB
Kinematics
vB
vD
! ED
= ! AB LAB = 12(0:8) = 9:6 m/s
= vB = 9:6 m/s
! BD = 0
v BD = 9:6 m/s
vD
9:6
=
=
= 6:0 rad/s
LED
1:6
Mass properties
(IAB )A
=
(IED )E
=
mBD
=
1
1
mAB L2AB = (2
3
3
1
1
2
mED LED = (2
3
3
2 1:0 = 2:0 kg
0:8)(0:8)2 = 0:3413 kg m2
1:6)(1:6)2 = 2:731 kg m2
461
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Kinetic energy
T
1
1
1
2
(IAB )A ! 2AB + (IED )E ! 2ED + mBD vBD
2
2
2
1
1
1
(0:3413)(12)2 + (2:731)(6:0)2 + (2:0)(9:6)2
2
2
2
165:9 J J
=
=
=
18.24
.
B
5 m/s
25 rad/s
.A
.C8.333 rad/s
5 m/s
Disk A:
IA
=
TA
=
mA R2
8(0:2)2
=
= 0:160 kg m2
2
2
1
1
1
1
2
IA ! 2A + mA vA
= (0:160)(25)2 + (8)(5)2
2
2
2
2
150:0 J
=
Bar AB (translates):
TAB =
1
1
2
mAB vAB
= (6)(5)2 = 75:0 J
2
2
Bar BC (rotates about C):
TBC =
1
1 mBC L2BC 2
1 4(0:6)2
(IBC )C ! 2BC =
! BC =
(8:333)2 = 16:67 J
2
2
3
2 3
System:
T = TA + TAB + TBC = 150:0 + 75:0 + 16:67 = 242 J J
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
462
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
0.25 m
0.15 m
m = 4 kg
k- = 0.18 m
C
6 kg A
vA
12 kg
B
vB
The displacements of the blocks during a 90 clockwise turn of the pulley are
U1 2
T2
yA
= 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
mA vA
+ mB vB
+ mC k 2 ! 2
=
2
2
2
1
2
=
6(0:25!) + 12(0:15!)2 + 4(0:18)2 ! 2 = 0:3873! 2
2
U1 2 = T2
T1
4:621 = 0:3873! 2
0
! = 3:45 rad/s J
463
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18.27
.
.
C
C
3 ft
B.
6 ft
.A 5 rad/s
(IA )AB =
V1
T1
V2
T2
9 ft
2
t
7f
.81
10
1
Datum
.Aω 6 ft
B
mAB L2AB
(20=32:2) (6)2
=
= 7:453 slug ft2
3
3
= WAB (3) = 20(3) = 60 lb ft
1
1
=
(IA )AB ! 20 = (7:453) (5)2 = 93:16 lb ft
2
2
1 2
1
=
k = (5)(10:817 3)2 = 152:76 lb ft
2
2
1
1
=
(IA )AB ! 2 = (7:453) ! 2 = 3:727! 2
2
2
V 1 + T 1 = V 2 + T2
60 + 93:16 = 152:76 + 3:727! 2
! = 0:328 rad/s J
18.28
464
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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 2:22 10 3 (3)
3
32:2
=
1:7236
U1 2
=
area under M - diagram =
T2
=
1 2
1
I! = (1:7236
2
2
U1 2
!
= T2
=
3
12
+
(2:22
1
mAB L2AB
3
+ mBC L2AB
2
10 3 )(2)
32:2
3
12
0:88 + 0:50
2
= 2:168 lb ft
10 5 slug ft2
T1
501:6 rad/s
2:168 =
10 5 )! 2
1
(1:7236
2
10 5 )! 2
) vBC = LAB ! =
0
3
(501:6) = 125:4 ft/s J
12
*18.30
465
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18.31
466
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18.32
18.33
467
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18.34
vB
v
1.5
ft
0.5 ft
ω
Kinematics: Let vB be the velocity of the block. From the velocity diagram of
the spool we see that
!=
vB
= 0:5vB
2:0
v = 1:5! = 0:75vB
It follows that
sA = 0:75sB
Conservation of energy: Let the release position (position 1) be the datum for
potential energy.
V1 = 0
T1 = 0
468
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V2
T2
=
=
WB sB + WA sA sin 30 = 38(6) + 64(0:75 6)(0:5)
84:0 lb ft
1
1
1
2
=
mA v 2 + mA k 2 ! 2 + mB vB
2
2
2
1 64
1 38 2
1 64
(0:75vB )2 +
(1:25)2 (0:5vB )2 +
v
=
2 32:2
2 32:2
2 32:2 B
2
= 1:5373vB
V1 + T1
0+0
= V2 + T2
2
=
84:0 + 1:5373vB
vB = 7:39 ft/s J
18.35
M = 0.15P
G
C
P
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
2
=
IC ! =
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
469
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18.37
18.38
(a)
A
0.3
m
0.075 m
o
30
Datum
B
A
vA
0.3 m
B
ωAB
Position 2
Position 1
In position 2 point B is stationary. ) Point B is the I.C. of bar AB.
V1
T2
= mAB g(0:15 sin 30 ) = 3:2(9:81)(0:15 sin 30 ) = 2:354 J
1
1 1
1
=
(IAB )B ! 2AB =
mAB L2AB ! 2AB = (3:2)(0:3)2 ! 2AB
2
2 3
6
=
0:0480! 2AB
T 1 + V1 = T 2 + V2
0 + 2:354 = 0:0480! 2AB + 0
! AB = 7:003 rad/s
470
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
vA = LAB ! AB = 0:3(7:003) = 2:10 m/s #J
(b)
Datum
δ
B
0.3 m
A
Position 3
V3
1
+ k 2
2
1
3:2(9:81) + (1200) 2 = 600 2
2 2
=
mAB g
=
T1 + V1
2
15:696
= T 3 + V3
0 + 2:354 = 0 + 600 2
= 0:0771 m J
15:696
G 2.7 ft/s
0.25'
ω2
18.39
3 rad/s
0.65'
L0
C
Position 1
I
v1
V1
T1
V2
T2
Datum
0.25'
0.4ω2
0.4' G
C
L0 + 0.65π
Position 2
16
(0:25)2 = 0:198 94 slug ft2
32:2
= CG ! 1 = 0:9(3) = 2:7 ft/s
v2 = CG ! 2 = 0:4! 2
= IO
md2 = 0:23
= mge = 16(0:25) = 4:0 lb ft
1 2 1
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 2 1
1
1 16
=
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
471
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.40
L/2
30o
L/2
B
L/2
B
Position 1
ω2
A
A
P
C
L/2
C
Position 2
P
In Position2, point A is the I.C. of the bar
U1 2
U1 2
= P (L sin 30 )
= T2
T2 =
1
1
IA ! 22 =
2
2
1
P L sin 30 = mL2 ! 22
6
T1
1
mL2 ! 22
3
0
!2 =
r
3P
J
mL
18.41
IC
T
2
2
2
= I + mCG = m(k 2 + CG )! 2 = 320(0:42 + CG ) = 51:20 + 320CG
1
1
2
2
IC ! 2 =
51:20 + 320CG ! 2 = 25:60 + 160CG ! 2
=
2
2
2
Dimensions in meters
0.3
Datum
1.4
G
G
0.3
ω3
ω2
G
0.3
1.7
1.1
C
Position 2
C
Position 1
C
Position 3
(a) Position 1 (initial position): V1 = 0
2
CG
T1
=
=
1:42 + 0:32 = 2:050 m2
[25:60 + 160(2:050)] (2:8)2 = 2772 J
Position 2 (position of maximum T ):
2
CG
T2
V2
=
=
=
1:12 = 1:21 m2
[(25:60 + 160(1:21)] ! 22 = 219:2! 22
mg(0:3) = 320(9:81)(0:3) = 941:8 J
472
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
T1 + V1
2772 + 0
= T 2 + V2
= 219:2! 22
941:8
! 2 = 4:12 rad/s J
(b) Position 3 (position of minimum T ):
2
CG
T3
V3
T1 + V 1
2772 + 0
= 1:72 = 2:89 m2
= [25:60 + 160(2:89)] ! 23 = 488:0! 23
= mg(0:3) = 320(9:81)(0:3) = 941:8 J
= T3 + V3
= 488:0! 23 + 941:8
! 3 = 1:937 rad/s J
473
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.42
G
G
a
b
1
Datum
ω
A
2
In position 2:
IA
= I + mb2 = m
= m
V1
V2
T2
2
+ b2
a2 + 5b2
+ mb2 = m
4
4
1:02 + 5(0:75)2
= 0:9531m
4
= mga = m(32:2)(1:0) = 32:2m
T1 = 0
= mgb = m(32:2)(0:75) = 24:15m
1
1
IA ! 2 = (0:9531m)! 2 = 0:4766m! 2
=
2
2
V 1 + T 1 = V 2 + T2
32:2m + 0 = 24:15m + 0:4766m! 2
! = 4:11 rad/s J
18.43
474
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18.44
18.45
475
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.46
18.47
.
A
.B
2 ft
ft
2.5
.
ω2 A
1.5 ft
. Datum
D
.
C
Position 1
2 ft
vB
.
2 ft
0.5 ft
ωCD
.
D
B
2 ft
2.5
ft
Position 2
vC
.C
Position 1 (release position):
T1 = 0
V1 = 2w0 (1:5) + 2:5w0 (0:75) = 4:875w0
476
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Position 2: Note that BC is translating (! BC = 0)
V2 = 2w0 (0:5)
vB
! CD
mBC
(IAB )A
T2
=
=
=
2:5w0 (1:25)
2w0 (1:0) =
4:125w0
= vC = vBC = LAB ! 2 = 2! 2
vC 2! 2
= !2
=
LCD 2
2:5w0
= 0:07764w0
=
32:2
1 2w0
= (ICD )D =
(2)2 = 0:08282w0
3 32:2
1
1
1
2
(IAB )A ! 22 + (ICD )D ! 2CD + mBC vBC
2
2
2
1
1
1
(0:08282w0 )! 22 + (0:08282w0 )! 22 + (0:07764w0 )(2! 2 )2
2
2
2
0:2381w0 ! 22
T 1 + V1
0 + 4:875w0
!2
= T2 + V2
= 0:2381w0 ! 22 4:125w0
= 6:15 rad/s J
18.48
Kinematics ( = 45 ):
C
2.121 ft
1.5
ft
A
45o
ωAB
. v
vA
G
G
B
vB
Point C is the I.C. of bar AB.
vG
=
!B
=
1:5! AB
vB = 2:121! AB
vB
2:121! AB
=
= 1:414! AB
R
1:5
477
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Conservation of energy:
.
1.5'
IAB
=
IB
=
1
45o
Datum
. 1.5'
1
1 12
mAB L2AB =
(3)2 = 0:2795 slug ft2
12
12 32:2
1 15
1
mB R2 =
(1:5)2 = 0:5241 slug ft2
2
2 32:2
V1 = WAB (1:5) = 12(1:5) = 18 lb ft
V2
(T2 )AB
(T2 )B
2
T1 = 0
= WAB (1:5 sin 45 ) = 12(1:5 sin 45 ) = 12:728 lb ft
1
1
2
=
IAB ! 2AB + mAB vG
2
2
1 12
1
(0:2795)! 2AB +
(1:5! AB )2 = 0:5590! 2AB
=
2
2 32:2
1
1
2
IB ! 2B + mB vB
=
2
2
1
1 15
=
(0:5241)(1:414! AB )2 +
(2:121! AB )2
2
2 32:2
= 1:5718! 2AB
V1 + T 1
18 + 0
! AB
= V2 + T 2
= 12:728 + (0:5590 + 1:5718)! 2AB
= 1:5730 rad/s
vB = 2:121! AB = 2:121(1:5730) = 3:34 ft/s J
18.49
478
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.50
60 lb
A v2
60 lb
C
C
3 ft
3 ft
3 ft
3 ft
A
o
240 lb.ft
60 240 lb.ft
Position 1
D
B
Position 2
B
D
Bar AC undergoes curvilinear translation
U1 2
T2
= C
=
mg
h = 240
30
180
60(3
3 sin 60 ) = 101:55 lb ft
1 60 2
1
mv 2 =
v
2 2
2 32:2 2
U1 2 = T2
T1
101:55 =
1 60 2
v +0
2 32:2 2
v2 = 10:44 ft/s J
479
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.51
30o
80(9.81) N
µkN =
FBD
T
Fy
kN
x
y
MAD
80a
N
= 0
+ - N 80(9:81) cos 30 = 0
= 0:4(679:7) = 271:9 N
U1 2
=
2mgy
T2
=
2
=
k N x = 2(80)(9:81)(x sin 30 )
N = 679:7 N
271:9x = 512:9x
1 1
1
1
2
mv22 + IA ! 2A = mv22 +
mRA
2
2
2 2
1
1
mv22 = 1 +
80(5)2 = 2500 J
1+
4
4
U1 2 = T2
T1
512:9x = 2500
0
v2
RA
2
x = 4:87 m J
18.52
C
A
0.5
m
0.3 m
P
0.4 m B
Position 1
ωC
Datum A
2 m/s
C
0.08 m
0.5 m ωAB B
Position 2
P
Position 1:
V1
1
= mC g(0:3) + mAB g(0:15) + k 21
2
1
= 2(9:81)(0:3) + 3(9:81)(0:15) + (72)(0:4
2
= 13:541 J
0:1)2
480
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Position 2 (B is the I.C. of bar AB):
IC
=
(IAB )B
=
! AB
=
!C
=
V2
T2
1
1
2
mC RC
= (2)(0:08)2 = 0:0064 kg m2
2
2
1
1
2
mAB LAB = (3)(0:5)2 = 0:25 kg m2
3
3
vA
2
=
= 4 rad/s
LAB
0:5
vA
2
=
= 25 rad/s
RC
0:08
1
1 2
k 2 P x = (72)(0:7 0:1)2 P (0:5 0:4)
2
2
= 12:96 0:1P
1
1
1
2
IC ! 2C + mC vA
+ (IAB )B ! 2AB
=
2
2
2
1
1
1
(0:0064)(25)2 + (2)(2)2 + (0:25)(4)2 = 8:0 J
=
2
2
2
=
T1 + V 1 = T 2 + V 2
0 + 13:541 = 8:0 + 12:96
0:1P
P = 74:2 N J
18.53
Kinematics: ! A = vB =0:25 = 4vB
Choose the ground as the datum for potential energy.
V1 = mB gh = 5(9:81)(6) = 294:3 J
V2
T2
T1 = 0
=
0
1
1
1
1
2
2
=
mA R! 2A + mB vB
= mA R2 (4vB )2 + mB vB
2
2
2
2
1
1
2
2
2
=
(8)(0:25)2 (16vB
) + (5)vB
= 6:50vB
2
2
V1 + T1 = V2 + T2
2
294:3 + 0 = 0 + 6:50vB
vB = 6:73 m/s J
481
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
*18.54
482
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.55
vC = 2vA
15"
vA
ωA
vA
!A
= vB = 0:5vC
) sA = sB = 0:5
vC
vC
=
= 0:4vC
= !B =
2R
30=12
IA = IB = mR2 =
250
32:2
15
12
sC
2
= 12:131 slug ft2
Cylinder A reaches corner of the slab when
sC
T2
U1 2
sA = 20 in.
sC
0:5 sC = 20 in.
sC = 40 in.
1
1
1
2
2
mC vC
+2
IA ! 2A + mA vA
2
2
2
1 1500
250
2
2
=
vC
+ 12:131(0:4vC )2 +
(0:5vC )2 = 27:174vC
2 32:2
32:2
40
= P sC = 180
= 600 lb ft
12
=
U1 2 = T2
T1
2
600 = 27:174vC
0
vC = 4:70 ft/s J
18.56
483
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.57
D
180
180
A
B
G
120
x- = 380
100
x=
IAB
=
ID
=
8(180) + 16(480)
mi xi
=
= 380 mm
mi
8 + 16
1
1
mAB L2AB =
(8)(0:36)2 = 0:0864 kg m2
12
12
1
1
mD R2 = (16)(0:120)2 = 0:1152 kg m2
2
2
(a)
I
h
= IAB + mAB (0:2)2 + ID + mD (0:1)2
= 0:0864 + 8(0:2)2 + 0:1152 + 16(0:1)2 = 0:6816 kg m2
= I! = 0:6816(5) = 3:408 N m s J
(b)
IA
hA
= I + (mAB + mD )x2 = 0:6816 + (8 + 16)(0:38)2
= 4:147 kg m2
= IA ! = 4:147(5) = 20:7 N m s J
18.58
484
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.59
mABv-AB
IABω
m v- BC BC
IBCω
B
A 1.2'
d
C
3.2'
(a)
mv-
=
B
A
C
(b)
Momentum diagram (a):
IAB !
=
IBC !
=
mAB vAB
mBC vBC
hA
=
=
1
1 3
mAB L2AB ! =
(2:4)2 ! = 0:04472!
12
12 32:2
1
1 5:5
mBC L2BC ! =
(1:6)2 ! = 0:03644!
12
12 32:2
3
(1:2!) = 0:11180!
32:2
5:5
= mBC (rBC !) =
(3:2!) = 0:5466!
32:2
= mAB (rAB !) =
[0:04472 + 0:03644 + 0:11180(1:2) + 0:5466(3:2)] !
1:9644!
Momentum diagram (b):
mv =
8:5
(!d) = 0:2640!d
32:2
hA = mvd = 0:2640!d2
Diagrams (a) and (b) are equivalent:
1:9644! = 0:2640!d2
d = 2:73 ft J
485
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.60
18.61
18.62
(AO )1 2
=
Z 60 s
C(t)dt = 12
0
(hO )2 = mk 2 ! 2 =
0
=
(hO )1
=
60 s
1
0
714:0 lb ft s
(AO )1 2 = (hO )2
Z 60 s
(hO )1 :
40
32:2
1
e 2t dt = 12 t + e 2t
2
0
9
12
714:0 = 0:6988! 2
2
! 2 = 0:6988! 2
0
! 2 = 1022 rad/s J
486
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.63
20
in. Ay
Ax
A
120 lb
20
in.
. I-ω
.
1
A
mBω1R
B
50 lb
FBD
IA =
Momentum diagram
1
1 120
mA R2 =
2
2 32:2
2
20
12
= 5:176 slug ft2
Z t
(hA )1
IA ! 1 + (mB ! 1 R) R
0
Z t
MA dt
=
mB gR dt =
(hA )2
0
0
5:176(12) +
50
(12)
32:2
20
12
2
50
20
12
t
=
0
t = 1:367 s J
18.64
h1
=
A1 2
=
0
Z t
0
A1 2
!1
= h2
=
h1
h2 = I!
Z t
M (t)dt =
M0 e t=t0 dt = M0 t0 1
e t=t0
e t=t0 = I!
M0 t 0
1
I
M 0 t0 1
0
0
!=
e t=t0
M 0 t0
4:6(3:8)
=
= 24:3 rad/s J
0:72
I
487
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.65
488
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
mk 2 ! 1 =
(hO )1 =
(AO )1 2 = (hO )2
(hO )1
20(0:16)2 400
1:08P0 = 0
2
60
=
21:45 N m s
P0 = 19:86 N J
( 21:45)
18.67
mrodvG
C Irodω
G
.
A
mrodvG
C Irodω
G
mdiskvB
.
.B R
L/2
L/2
Momentum diagram (a)
A
mdiskvB
.B R
Idiskω
L/2
L/2
Momentum diagram (b)
(a) Angular speed of disk A will remain zero.
hC
=
(hC )disk + (hC )ro d
=
(mdisk vB )L + Iro d ! + (mro d vG )
=
(mdisk !L)L +
=
L
2
1
L
mro d L2 ! + mro d !
12
2
1
mdisk + mro d
3
L2 ! =
44 + (18=3)
32:2
L
2
16
12
2
! = 2:761!
(AC )1 2 = M0 t = 6(3) = 18 lb ft s
(AC )1 2 = (hC )2
(hC )1
18 = 2:761!
0
! = 6:52 rad/s J
489
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(b) Disk and rod have the same angular speed. We must add the following term
to hC :
Idisk !
1
1 44
mdisk R2 ! =
(1:0)2 ! = 0:6832!
2
2 32:2
hC = 2:761! + 0:6832! = 3:444!
=
)
(AC )1 2 = (hC )2
(hC )1
18 = 3:444!
0
! = 5:23 rad/s J
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
v
2
=
= 0:3398 s
g
0:6(9:81)
k
(AO )1 2 = (hO )2
k mgRt
=
!
=
(hO )1
mv
k mgRt = 0
1
mR2 !
2
2 k gt
2(0:6)(9:81)(0:3398)
=
= 13:33 rad/s
R
0:3
I!
J
18.69
490
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.70
P
R
Iω
R
mωR
mg
F
C
FBD
C
Momentum diagram
N = mg
(a)
Z t
+
(AC )1 2 = (hC )2
P (2R) dt
= I! + m!R(R)
(hC )1
0
0
2P Rt =
t
=
mR2
! + m!R2
2
3 m!R
3 60(10)(0:4)
=
= 0:900 s J
4 P
4
200
(b)
Z t
+ ! L1 2 = p2
(P + F )dt
= m!R
0
p1
(P + F )t = m!R
0
0
F
=
Pt
m!R
=
t
200(0:9)
(60)(10)(0:4)
= 66:67 N
0:9
491
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Smallest
s that would prevent sliding is
s =
F
66:67
=
= 0:1133 J
N
60(9:81)
18.71
B
A
IAω1
mB Lω2
2
I-
IAω2
ω
B 2
10" 10"
Final momenta
Initial momenta
IA
=
1
1 6
mA R2 =
2
2 32:2
IB
=
1 4
1
mB L2 =
12
12 32:2
(hO )1
=
0:052 41(16)
=
0:8386
=
+
mALω2
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
(hO )2
"
0:052 41 + 0:028 76 +
6
32:2
20
12
2
+
4
32:2
10
12
2
L
2
#
2
!2
!2
! 2 = 1:224 rad/s J
0:6850! 2
18.72
L/2
x
C
A
1 m L2ω
12 AB
B
mC xω
mAB(L/2)ω
Momentum diagram
(a) Angular momentum of the system about A is conserved.
hA
=
=
1
L
mAB L2 ! + mAB !
12
2
1
mAB L2 + mC x2 !
3
L
+ (mC x!) x
2
492
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
When x =
When x =
1 2
0:3
(1:6)2 +
(0:5)2 (4) = 0:2213 lb ft s
3 32:2
32:2
1 2
0:3
1:6 ft: (hA )2 =
(1:6)2 +
(1:6)2 ! = 0:07685!
3 32:2
32:2
0:5 ft: (hA )1 =
(hA )1 = (hA )2
0:2213 = 0:07685!
! = 2:88 rad/s J
(b) The act of the mass leaving the rod does not change the angular momentum
of the rod.
) ! = 2:88 rad/s J
*18.73
18.74
493
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.75
18.76
18.77
O
0.125mω1
O
Iω1
0.125 m
Initial momenta
I=
0.375 m
0.375mω2
Iω2
Final momenta
1
1
mL2 =
(18)(0:75)2 = 0:8438 kg m2
12
12
(hO )1
=
(hO )2
2
I! 1 + (0:125) m! 1 = I! 2 + (0:375)2 m! 2
0:8438 + (0:125)2 (18) (16) = 0:8438 + (0:375)2 (18) ! 2
! 2 = 5:33 rad/s J
494
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.78
With rods vertical:
(Iz )1
=
2
"
1
mR (a2 + b2 ) + mR d21 + mB k 2
12
9
12
2
1
mR (a2 + L2 ) + mR d22 + mB k 2
12
"
#
2
1 18 1:52 + 182
18
15
24
2
+
+
12 32:2
122
32:2 12
32:2
9
12
2
=
0:7043 slug ft2
6
12
2
#
24
32:2
=
18
1 18 1:52 + 2:52
+
2
2
12 32:2
12
32:2
+
With rods horizontal:
(Iz )2
=
=
=
2
2:377 slug ft2
Angular momentum about z-axis is conserved:
(Iz )1 ! 1 = (Iz )2 ! 2
0:7043(40) = 2:377! 2
! 2 = 11:85 rad/s J
495
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
*18.79
496
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.80
18.81
497
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.82
Ι-ABω1 I-1ω1 m(2.5ω1)
I1ω1
m(2.5ω1)
2.5' O 2.5'
Initial momenta
m(2.5ω2)
I2ω2
m(2.5ω2)
Ι ω
I2ω2
AB 2
A
O B
Final momenta
One plate:
1
1 12
mb2 =
(2)2 = 0:124 22 slug ft2
12
12 32:2
1
m(2b2 ) = 2I1 = 0:2484 slug ft2
I2 =
12
In position 1:
I1 =
In position 2:
IAB =
1
1 8
mAB L2 =
(3)2 = 0:186 34 slug ft2
12
12 32:2
+
(hO )1
2 I1 ! 1 + m(2:5) ! 1 + IAB ! 1
= (hO )2
= 2 I2 ! 2 + m(2:5)2 ! 2 + IAB ! 2
2
12
(2:5)2 + 0:186 34 (12)
32:2
12
(2:5)2 + 0:186 34 ! 2
2 0:2484 +
32:2
5:342! 2
! 2 = 11:44 rad/s J
2 0:124 22 +
=
61:12
=
18.83
498
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18.84
A
Iω1
.
0.25mL ω2
=
+
F dt
C
C
Momenta
before impact
Impulse
during impact
G
C
.
.
Iω2
0.25L
Momenta
after impact
Angular momentum about C is conserved:
I! 1
2
mL
!1
12
!2
= I! 2 + 0:25mL! 2 (0:25L)
mL2
=
! 2 + 0:25mL! 2 (0:25L)
12
= 0:5714! 1 = 0:5714(25) = 14:29 rad/s J
18.85
499
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.86
(a)
ω2 Datum
ω2d
Velocities
after impact d
θ
C
G
Position of
maximum
displacement
0:12
(1600)(0:60) = 3:578 lb ft s
32:2
+
(hC )2 = (mAB + mD )! 2 d2 + I! 2
1 24
24 + 0:12
! 2 (0:6)2 +
(6)2 ! 2 = 2:506! 2
=
32:2
12 32:2
(hC )1 = (hC )2
3:578 = 2:506! 2
! 2 = 1:4278 rad/s J
+
(hC )1 = mD v1 d =
(b)
A
C
D mv
D1
^
F
B
Momenta before
impact
T2
V2
V3
L/2
^
R
^
F
d
Iω2
(mAB + mD)ω2d
L/2
FBD during
impact
Momenta after
impact
1 2 1
I! + (mD + mAB )(! 2 d)2
2 2 2
1 1 24
1 0:12 + 24
=
(6)2 (1:4278)2 +
(1:4278 0:6)2
2 12 32:2
2 32:2
= 2:554 lb ft
=
(WD + WAB )d = (0:12 + 24)(0:6) = 14:472 lb ft
=
(WD + WAB )d cos = (0:12 + 24)(0:6 cos ) = 14:472 cos
=
2:554
T2 + V2 = T3 + V3
14:472 = 0 14:472 cos
= 34:6
J
500
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.87
P^
A
L/2
A
Iω
mv-
A
ω
L/2
FBD during
impact
+
(hA )1 = (h2 )2 :
vC = v
L
2
!
Momenta after
impact
L
2
0 = I!
mv
y
6
v
L
=v
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
0.25 m
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
A Datum
3 Imminent
tipping
(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
=
=
501
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.89
(a)
+
!2
(hA )1 = (hA )2
=
mv1
L
= IA ! 2
2
mv1
3
3
v1 =
(5:5) = 9:167 rad/s
2L
2(0:9)
L
1
= mL2 ! 2
2
3
J
(b) Choose the horizontal plane through A as the datum for potential energy.
V2 + T 2
L 1
mg + IA ! 22
2
2
L 1 1
mL2 ! 22
mg +
2
2 3
s
r
g
! 3 = ! 22 6 = 9:1672
L
= V3 + T3
L 1
= mg + IA ! 23
2
2
L 1 1
= mg +
mL2 ! 23
2
2 3
6
9:81
0:9
J
= 4:32 rad/s
18.90
1 mR2ω
1
2
1 mR2ω2
2
mRω1
20o R
A
Momenta before
impact
mRω2
R
A
^
^
N
F
FBD during impact
A
Momenta after
impact
+
(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
502
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.91
18.92
503
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.93
2
1
G mv1
.
+
.
G I-ω
2
F dt
Momenta A
after impact
ft
2.5
=
2 ft
Momenta
before impact
A
A
mv2
Impulse
during impact
Angular momentum about A is conserved during impact:
mv1 (2)
2mv1
!2
= I! 2 + mv2 (2:5)
m 2
=
(3 + 42 )! 2 + m! 2 (2:5)2
12
= 0:240v1
3
2
.G
.G
2 ft
2.5 ft
ω2
Datum
A
A
Energy is conserved after the impact:
For the box to ‡ip, it has to reach position 3.
V2
=
T2
=
=
2mg = 64:4m
V3 = 2:5mg = 2:5(32:2m) = 80:50m
1 2 1
1m 2
1
I! + mv 2 (2:5) =
3 + 42 ! 22 + m! 22 (2:5)2
2 2 2 2
2 12
2
2
4:167m! 2
V2 + T2
64:4m + 4:167m! 22
!2
v1 =
T3 = 0
= V3 + T 3
= 80:50m + 0
= 1:9656 rad/s
1:9656
= 8:19 ft/s J
0:240
504
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.94
18.95
(a)
R/2
R
G
Momenta before
impact
FBD during
impact
F^
mv-2
G
Iω2
Momenta after
impact
505
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Z
F^ dt = mv2 = 2:8m
Z
R
F^ dt = I! 2 0
(AG )1 2 = (hG )2 (hG )1
+
2
R
2
3:5
3:5
(2:8m) =
mR2 ! 2
!2 =
=
= 33:60 rad/s J
2
5
R
1:25=12
L1 2
= p2
p1
+
(b)
mg
mv-2
Iω2
G
R
C
Momenta after
impact
(hC )1
mv2 R
=
G
R
m(Rω3)
F
N
FBD during
slipping
C
G
R
C
Iω3
Momenta during
rolling
+
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
2
mR2 ! 2
5
!3
(hC )1
18.96
Angular momentum about the z-axis is conserved:
mA (0:8)2
!2
4
12(0:8)2
!2
0:009(850)(0:6) = 0:009(310)(0:6) +
4
! 2 = 1:519 rad/s J
mB v1 (0:6)
= mB v2 (0:6) +
18.97
IA
=
=
=
1
2
2
2
2
mAB AB + mC RC
+ mc AB + RC
3
5
2
2
20
2 5
3
5
23
1 2
+
+
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)
506
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Since impact is elastic, energy is conserved
T1
1
(0:6318) ! 21
2
1
1
IA ! 21 = mD v22
2
2
= T2 :
=
1
2
W
32:2
p
! 1 = 0:2217 W v2
v22
(b)
Equating (a) and (b):
p
0:094 21W = 0:2217 W
W = 5:54 lb J
18.98
(a)
mCv1
30o
mCω2b
b
b
IBω2
C
B
A
A
Momentum before impact
Bar AB: IB =
L/2
mAB L ω
2 2
Momenta after impact
B
1 60
1
mAB L2 =
(8)2 = 39:75 slug ft2
3
3 32:2
(hB )1 = (hB )2
(mC v1 sin 30 ) b = IB ! 2 + (mC ! 2 b)b
28
28
(15 sin 30 )(6) = 39:75! 2 +
! 2 (6)2
32:2
32:2
! 2 = 0:5507 rad/s J
(b) Initial deformation of the spring is
1 =
0:5WAB
0:5(60)
=
= 1:5 in.
k
20
L
b
L/2
Lθ
A
Datum
bθ Lθ/2 θ
G
C
B
507
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
T2
=
=
V3
=
=
=
1
IB ! 22 + mC (! 2 b)2
2
1
28
39:75(0:5507)2 +
(0:5507 6)2 = 10:774 lb ft
2
32:2
1
2
k(L
(mAB g) (b )
1)
2
2
1:5
1
(20 12) 8
60(4 ) 28(6 )
2
12
7680 2
648 + 1:875
V 2 + T 2 = V 3 + T3
0 + 10:774 = 7680 2 648 + 1:875 + 0
0 = 7680:0 2 648:0
8:90
= 0:09639 rad = 5:52 J
18.99
508
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.100
18.101
(a)
B
B
L/2
L
mb v1
A
Momentum before impact
+ !
mb v1
=
+
(mb v1 ) L
=
mb v1
=
!2
=
G mv2
1
2
12 mL ω2
A
Momenta after impact
p 1 = p2
mv 2
v2 =
mb
0:04
v1 =
(2800) = 6:222 ft/s
m
18
(hB )1 = (hB )2
1
L
mL2 ! 2 + (mv 2 )
12
2
1
1
mb
mL! 2 + m
v1
12
2
m
6mb v1
6(0:04)(2800)
=
= 7:467 rad/s J
mL
18(5)
509
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(b) Due to absence of horizontal forces, v remains constant after impact (v3 =
v2 ).
B
B
Datum
θ
L/2
v-2
G
G
ω3 = 0
ω2
L cosθ
2
v-3 = v-2
A
A
Immediately after impact
V2
=
T2
=
=
Position of maximum θ
L
= 18(2:5) = 45:0 lb ft
2
1
mL2 ! 22 + mv 22
12
1 18
18
(5)2 (7:467)2 +
(6:222)2
12 32:2
32:2
mg
1
2
1
2
V3
=
T3
=
= 43:29 lb ft
L
cos = 18(2:5 cos ) = 45:0 cos lb ft
2
1
1
18
mv 22 =
(6:222)2 = 10:820 lb ft
2
2 32:2
mg
V2 + T 2 = V 3 + T 3
45:0 + 43:29 =
45:0 cos + 10:820
45:0 + 43:29 10:820
= cos 1
45:0
= 73:8
J
18.102
Kinematics
When = 0, the rods are vertical and the collars at B and D are at the limit
of their travel. Therefore,
vB = vD = 0
510
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Also note that due to symmetry the angular velocities of the rods are equal in
magnitude but opposite sense:
! AC =
! CE = !
vA and ! are related by the relative velocity equation
vB = vA + vB=A
.
A ω
0
vA
=
+ L/2
ωL/2
B
0=
vA +
!L
2
!=
2
vA
L
Conservation of energy
Choose the horizontal rod that guides collar A as the datum for potential energy.
V1
=
V2
=
T2
=
0
2
T1 = 0
L 3L
mg
+
2
2
1 2
I!
2
=
V1 + T 1
0+0
vA
18.103
=
2mgL
mL2 2
mL2
! =
12
12
2vA
L
2
=
1
mv 2
3 A
= V2 + T2
1
2
2mgL + mvA
3
p
=
6Lg J
=
511
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.104
18.105
Iω
G
G
R
mg
F = µkmg
FBD
mv-
C
Momentum diagram
N = mg
Sliding stops when
vC = v0
v + !R = v0
(a)
(a) During sliding:
p1 +
Z t
F dt = p2
0+
k mgt = mv
v=
k gt
(b)
0
512
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(hC )1 +
Z t
F R dt =
(hC )2
0+
k mgRt = I!
k mgRt
mR2
!
2
!=
2 k gt
R
(c)
v0
J
3 kg
(d)
0
=
Substituting Eqs. (b) and (c) into Eq. (a) yields
k gt +
2 k gt
R = v0
R
t=
(b) Subsitution of Eq. (d) into Eqs (b) and (c) results in
v=
kg
v0
3 kg
=
v0
J
3
!=
2 kg
R
v0
3 kg
=
2 v0
J
3R
18.106
513
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.107
18.108
3 ft
3 ft
3 ft
B
L − 6.708'
3 ft
C
8'
6.70
A
B
A
Position 1
3 ft
L − 3'
ωAB
Position 2
C
vC
vC = AB! AB = 3! AB
Let L be the length of the rope.
U1 2
=
=
T2
=
=
=
AB
+ WC [(L 3) (L 6:708)]
2
W (1:5) + W (3:708) = 2:208W
1
1
2
(IAB )A ! 2AB + mC vC
2
2
11 W
1 W
(3)2 ! 2AB +
(3! AB )2
2 3 32:2
2 32:2
0:18634W ! 2AB
WAB
514
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
U1 2
2:208W
vC
= T2 T1
= 0:18634W ! 2AB 0
! AB = 3:442 rad/s
= 3(3:442) = 10:33 ft/s # J
18.109
^
N
T^A
C
A
G
^
N
o
45o 45
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
515
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.111
18.112
V1
T2
= mgh
T1 = 0
V2 = 0
1 2 1
1 2
=
mR2 ! 2
I! + mv 2 =
2
2
2 5
T 1 + V1
!
1
9
+ m(!R)2 =
mR2 ! 2
2
10
9
= T 2 + V2
0 + mgh =
mR2 ! 2
10
p p
p
10 gh
gh
=
= 1:054
J
3 R
R
516
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18.113
18.114
^y
O
L
L
mv1
F^
A
Momentum
before impact
O
Datum
Lm ω
2 OA 2
Iω2
F^
FBD during
impact
I=
O
^
O
x
L/2
O
θ
L/2
L/2
A
mv2
A
Momenta after Position of max.
displacement
impact
1
1 30
mOA L2 =
(3)2 = 0:6988 slug ft2
12
12 32:2
System
+
(hO )1 = (hO )1
mv1 L = I! 2 + mOA
L
2
2
! 2 + mv2 L
2
5
30
3
5
(1800)(3) = 0:6988! 2 +
!2 +
(1400)(3)
16 32:2
32:2 2
16 32:2
52:41 = 2: 795 ! 2 + 40: 76
! 2 = 4:1667 rad/s
517
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Bar only
T 2 + V2
= T3 + V3
2
1 2 1
L
L
! 22 WOA
I! + mOA
2 2 2
2
2
"
#
2
1
L
I + mOA
! 22
2
2
1
2
0:6988 +
30
2
(1:5)2 (4:167)
32:2
=
0
WOA
L
cos
2
L
= WOA (1
2
cos )
=
30(1:5)(1
cos )
=
62:6
J
18.115
IAω1
Α^y Α^x
Α 1.5'
IAωA2
IA
=
IB
=
Momenta before
impact
F^
Β^y Β^
x
1.2' Β
IBωB2
F^
FBD's during
impact
Momenta after
impact
1
1 8
mA L2A =
(3)2 = 0:18634 slug ft2
12
12 32:2
1
1 6
mB LB =
(2:4)2 = 0:08944 slug ft2
12
12 32:2
Kinematics (plastic impact):
1:5! A2 = 1:2! B2
! B2 = 1:25! A2
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
Z
A1 2
=
(hA )2
F^ dt = IB (! B2
F^ dt =
(hA )1
! B1 ) = 0:08944 (1:25! A2 )
0
0:09317! A2
518
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0:12423 (! A2
J
! A2 = 2:857 rad/s
5) = 0:09317! A2
J
! B2 = 1:25(2:857) = 3:57 rad/s
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
18.117
θ
Position 1
1.8 m
1.2
m
Position 2
1.5 m/s
!1 =
V2
T1
0.6 m
Datum
2.5 rad/s
v1
1:5
=
= 2:5 rad/s
R
0:6
= W h = (9:81m) [1:2(1 cos )] = 11:772(1
1
1
1 1 2
=
mv 2 + I! 21 = mv12 +
mr ! 21
2 1
2
2 2
1
1
2
2
=
m (1:5) + (0:6) (2:5)2 = 1:6875m
2
2
cos )m
519
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
V1 + T1
cos
= 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
C C
O
B
(mΒ + mC)vI- ω
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
20
10
2
IB ! B = mB kB
!B =
(! A 25) = 0:4313! A
32:2 12
(mB + mC )v =
2
IC ! C = mC kC
!A =
12
32:2
3
12
10:783
2
! A = 0:023 29! A
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
520
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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
521
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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
=
4:99 rad/s
2
!
= !
J
J
2 =
(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 = ! p
rB =
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!
vB
160
) !=
=
= 8:46 rad/s J
18:916
18:916
522
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19.5
19.6
523
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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
524
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19.9
19.10
z
ω1 = 36 rad/s
x
ω2 = 14 rad/s
23.2o
(a) Let the xyz-axes be attached to the frame B.
!1
!2
!
= 36(i cos 23:2 + k sin 23:2 ) = 33:09i + 14:182k rad/s
= 14k rad/s
= ! 1 + ! 2 = 33:09i + 28:18k rad/s (valid only at the instant) J
525
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(b) Since the xyz-axes are attached to the frame, we have
)
=
=
d!
dt
320k + 14k
+ !2
!=!
_ 2 + !2
= !2 .
!
=B
2
(33:09i + 28:18k) = 463j + 320k rad/s
J
19.11
19.12
526
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19.13
rA=B = 0:18i
vA=B
= !
=
rA=B =
i
!x
0:18
0:2j + 0:1k m
j
!y
0:2
k
!z
0:1
(0:1! y + 0:2! z )i + ( 0:1! x + 0:18! z ) j + ( 0:2! x
vA = vA i
0:18! y )k
vB = 2k m/s
vA = vB + vA=B
Equating like components, we get
vA = 2 + 0:1! y + 0:2! z
0 =
0:1! x + 0:18! z
0 = 2 0:2! x 0:18! y
(a)
(b)
(c)
Assuming the spin velocity to be zero, we have ! rA=B = 0:
0 = 0:18! x
0:2! y + 0:1! z
(d)
Solution of Eqs. (a)-(d) is
!x
vA
= 4:85 rad/s
= 3:11 m/s J
! y = 5:72 rad/s
! z = 2:70 rad/s
) ! = 4:85i + 5:72j + 2:70k rad/s
527
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19.14
(a) Let the xyz-coordinate system be attached to the arm AB.
!1
=
8k rad/s
!2
=
2i rad/s
6
! 3 = p (j + k) = 4:243(j + k) rad/s
2
Angular velocity of the disk is
!
= ! 1 + ! 2 + ! 3 = 8k + 2i + 4:243(j + k)
= 2i + 4:243j + 12:243k rad/s J
(b)
= ! 2 + ! 3 = 2i + 4:243(j + k) rad/s
Angular acceleration of the disk is
= !
_ = (!)
_ =xyz +
=
33:9i
! =0+
i
2
2
j
4:243
4:243
k
4:243
12:243
J
2
16:0j rad/s
19.15
z
P
2.4
ft
5
4
4 ft
3.2
ft
3
A
ωPQ/OA
O
Q
y
(a) Let the xyz-axes be attached to OA.
! OA
! P Q=OA
!P Q
rP=O
= !2 =
16k rad/s
3j + 4k
= ! 1 = 25
= 15j + 20k rad/s
5
= ! OA + ! P Q=OA = 15j + 4k rad/s
= 4k ft
528
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Noting that point O is …xed, we have
vP = ! P Q
(b) Note that
PQ =
aP
rP=O = (15j + 4k)
4k = 60i ft/s J
= ! OA .
d! P Q
dt
+ ! OA
! P Q = 0 + ( 16k)
=OA
=
rP=O + ! P Q (! P Q rP=O ) = P Q rP=O + ! P Q
PQ
= 240i 4k+ (15j + 4k) 60i = 960j + ( 900k) + 240j
=
720j
2
(15j + 4k) = 240i rad/s
900k ft/s
2
vP
J
19.16
19.17
rB=A
vA
=
=
0:75i + 0:9j + 0:27k m
2:1i m/s
vB = vB j
529
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
= !
vB=A
=
rB=A =
(0:27! y
The equations vB
0
vB
0
0
i
!x
0:75
j
!y
0:9
k
!z
0:27
0:9! z )i + ( 0:27! x
0:75! z )j + (0:9! x + 0:75! y )k
= vA + vB=A and ! rB=A = 0 (no spin velocity) yield
=
2:1 + 0:27! y 0:9! z
=
0:27! x 0:75! z
= 0:9! x + 0:75! y
=
0:75! x + 0:9! y + 0:27! z
The solution is
!x
vB
=
=
0:327 rad/s
1:750 m/s J
! y = 0:392 rad/s
!z =
2:22 rad/s
19.18
z
A
0.8
m
θ
B
y
cos
=
)
When
=
y
0:8
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
)
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
530
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19.19
19.20
531
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19.21
532
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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)
!
= (10i + 3k) ( 10k) = 100j in./s
= ! (! rQ=B ) +
rQ=B
= (10i + 3k) 100j + 30j ( 10k)
rQ=B
aQ=B
2
= (1000k 300i) 300i = 1000k 600i in./s
= aB + aQ=B = AB! 21 j + aQ=B = 30(3)2 j + 1000k
aQ
=
600i
270j + 1000k in./s
2
600i
J
19.23
(a) Let the xyz-axes be attached to arm OA, so that
!1
!
= !1 .
= 3k rad/s
! 2 = 5(j cos 35 + k sin 35 ) = 4:096j + 2:868k rad/s
= ! 1 + ! 2 = 4:096j + 5:868k rad/s
= !
_ = (!)
_ =xyz +
=
12k
!=
12:288i rad/s
2
12k + 3k
(4:096j + 5:868k)
J
(b)
rA=O
vA
= 0:480((j cos 35 + k sin 35 ) = 0:3932j + 0:2753k m
= ! 1 rA=O = 3k (0:3932j + 0:2753k) = 1:1796i m/s
rP=A
vP=A
=
=
=
=
vP
0:2( j sin 35 + k cos 35 ) = 0:11472j + 0:16383k m
! rP=A = (4:096j + 5:868k) ( 0:11472j + 0:16383k)
1:3442i m/s
vA + vP=A = ( 1:1796 + 1:3442)i = 0:1646i m/s J
533
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
!
79:79i
rB=A
=
108j
aB
= aB=A = ! (! rB=A ) +
rB=A
= (79:79i 212:75j 59:84k) + 79:79i
=
212:75j
59:84k m/s
2
=
(0:9456j + 0:7388k) =79:79i m/s
159:6i
212:8j
59:8k m/s
2
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
534
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(b)
T
=
=
1
1
mL2
! h0 = (!k)
! (k cos
2
2
12
1
mL2 ! 2 cos2 J
24
+ j sin ) cos
19.26
y
5m
0.1
O
O
x
G
G
^
P
0.1
5m
z
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
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
(AO )1 2 = (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
= 4:0 rad/s
) ! 2 = 4:0i
T2
=
=
! 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
535
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
19.27
z
m
0.4 kg
2
0.
4 k8 m
g
O
g
4k
0.8
4k m
g
x
m
0.4 kg
2
y
(a)
Iy
=
Ixy
=
4(0:8)2
(2)(0:4)2
+
+ 2(4)(0:4)2 = 1:7067 kg m2
3
12
2 [2(0:2)( 0:8)] + 2 [4(0:4)( 0:4)] = 1:920 kg m2
2
!=
(hO )x
(hO )y
(hO )z
18j rad/s
=
Ixy ! y = ( 1:920)( 18) = 34:56 N m s
= Iy ! y = 1:7067( 18) = 30:72 N m s
= 0
hO =
34:56i
30:72j N m s J
(b)
T =
1
1
! hO = ( 18)( 30:72) = 276 J J
2
2
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
536
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(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
z
1
. .
b/2 O b
b 3 b/2
.
y
2
With ! x = ! y = 0, ! z = ! Eqs. (19.17a) are equivalent to
hO =
Ixz !i
Iyz !j + Iz !k
b
2
b
2
Ixz
=
0
Iyz
= m( b)
Iz
= m ( b) + mb2 +
+ m(b)
2
+0=
mb2
8
1
(2m)(2b)2 = mb2
12
3
8
hO = mb2 ! j + k
3
J
From Eq. (19.25) we get
T =
4
1
Iz ! 2 = mb2 ! 2 J
2
3
19.30
0.3 m
Position 2
A
x
0.4 m
ω y
m
0.5
Position 1
B
δ
ωrod
537
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In Position 2, the spring is undeformed and the rod is rotating about B with
the angular velocity
vB
0:15!
! ro d =
=
= 0:3!
0:5
AB
.
T2
=
=
=
1
1
(Idisk )O ! 2 + (Iro d )B ! 2ro d
2
2
1 1
1 1
mdisk R2 ! 2 +
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
T1 + V1
!
= T 2 + V2
0 + 10 = 0:04125! 2 + 0
= 15:57 rad/s J
19.31
538
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19.32
539
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
19.33
19.34
O
z
20o
L=1
.5'
Instant axis
ω1
3
1
y
R=
10
0.5'
ω0
C
!0
!1
!
ω
= 4( j sin 20 + k cos 20 ) = 1:3681j + 3:759k rad/s
=
!1 j
3j + k
p
=
! = ( 0:9487j + 0:3162k) !
10
!
( 0:9487j + 0:3162k) !
= !0 + !1
= ( 1:3681
! 1 ) j + 3:759k
540
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Equating like terms:
0:9487!
0:3162!
=
=
1:3681
3:759
!1
The solution is
! = 11: 888 rad/s
! 1 = 9: 910 rad/s
) ! = 11:888 ( 0:9487j + 0:3162k) =
Iy
Iz
11:278j + 3:759k rad/s
1W 2
1
6
R =
(0:5)2 = 0:02329 slug ft2
2 g
2 32:2
W 2
W 1 2
6
0:52
= Iz +
L =
R + L2 =
+ 1:52
g
g 4
32:2
4
=
= 0:4309 slug ft2
Since the xyz-axes are the principal axes at O, we have
hO
= Ix ! x i + Iy ! y j + Iz ! z k = 0:02329( 11:278)j + 0:4309(3:759)k
=
0:263j + 1:620k lb ft s J
19.35
(a)
OB
rB=A
vA
vB
= vA + !
) vB =
0 =
0 =
=
=
=
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
0:9! y + 1:2! z
0:9! x + 0:9950! z 1:4
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)
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! =
=
541
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(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
19.37
z
ω1
A
R
O
2R
ω
ω2
B
ω2 = 2ω1
y
ω
2
1
ω1
! = ( 2j + k)! 1
For the disk:
v
Iy
Ix
=
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
542
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T2
U1 2
1
1
Iy ! 2y + Iz ! 2z + mv 2
2
2
1
1 14 2
2
=
0:05435( 2! 1 ) + 0:02717! 21 +
! = 0:3397! 21
2
2 32:2 1
= C0 (4 ) = 0:35(4 ) = 4:398 lb ft
=
U1 2 = T2
4:398 = 0:3397! 21
T1
! 1 = 3:60 rad/s J
0
19.38
z
A
x
4.8 m
vA
4m
Position 2
vB = 0
B
m
53
2.6
y
Kinematics (position 2)
Note that in position 2 collar B has reached the limit of its travel. Therefore,
vB = 0.
rA=B = 2:653i 4j m
vA = vB + !
rA=B
vA k = 0 +
i
!x
2:653
j
k
!y !z
4 0
Expanding the determinant and equating like terms:
4! x
4! z
2:653! z
2:653! y
= 0
= 0
=
vA
Setting the spin velocity of the bar to zero gives the fourth equation:
! rA=B = 0
2:653! x
4! y = 0
The above equations yield
!x
= 0:173 62vA
! y = 0:115 16vA
!z = 0
2
2
) ! 2 = 0:173 622 + 0:115 162 vA
= 0:043 406vA
Conservation of energy
543
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Using the xy-plane as the datum for potential energy:
V1
=
V2
=
T2
=
=
1
(1:2mg) = 0:6(9:81)m = 5:886m
T1 = 0
2
0
1
1
mv 2 + I! 2
2
2
1
vA 2 1 m(4:8)2
2
2
+
m
(0:043 406vA
) = 0:166 67mvA
2
2
2 12
V1 + T1 = V2 + T2
2
5:886 + 0 = 0 + 0:166 67vA
vA = 5:94 m/s J
19.39
544
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19.40
Ix
=
Iy
=
Iz
=
1
1 300
mb2 =
(4)2 = 12:422 slug ft2
12
12 32:2
1 300
1
ma2 =
(3)2 = 6:988 slug ft2
12
12 32:2
1 300 2
1
m(a2 + b2 ) =
(3 + 42 ) = 19:410 slug ft2
12
12 32:2
(hO )2
(AO )1 2 = rAO
= Ix ! x i + Iy ! y j + Iz ! z k
= 12:422! x i + 6:988! y j + 19:410! z k
Z
k P^ dt = ( 1:5i + 2j) ( 20k) = 40i
(AO )1 2 = (hO )2
(hO )1 = (hO )2
30j lb ft s
0
Equating like components, we get
40 = 12:422! x
30 = 6:988! y
0 = 19:410! z
!=
3:22i
) ! x = 3:22rad/s
) !y
4:29 rad/s
) !z = 0
4:29j rad/s J
19.41
(a)
L1 2 = mv2 mv1
0:18i = 2v2 0
v2 =
0:09i m/s J
(b)
Ix
Iy
h2
A1 2
1
1
mR2 = (2)(0:09)2 = 0:0081 kg m2
2
2
1
= Iz = mR2 = 0:00405 kg m2
4
= Ix ! x i + Iy ! y j + Iz ! z k = 0:0045(2! x i + ! y j + ! z k)
=
= rA=G L1 2 = 0:09(j cos 30
= 0:0081j + 0:01403k N m s
k sin 30 )
( 0:18i)
A1 2 = h2 h1 = h2 0
0:0081j + 0:01403k = 0:0045(2! x i + ! y j + ! z k)
!x = 0
! y = 1:8 rad/s
! z = 3:118 rad/s
545
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T
=
=
=
1
1
1
1
h2 ! 2 + mv 22 = A1 2 ! 2 + mv 22
2
2
2
2
1
1
[0:0081(1:8) + 0:01403(3:118)] + (2)(0:09)2
2
2
0:0373 N m J
19.42
19.43
546
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19.44
547
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
19.45
548
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
19.46
549
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
0.7
2 ft
19.47
C
1.6 ft
RC
B
ωx
60o
0.7
2 ft
y
x
A
D
1.6 ft
60o
−ωy
ω
RD
FBD
Let the xyz-axes be attached to bar AB.
!x
!y
!z
= ! cos 60 = 5 cos 60 = 2:5 rad/s
=
! sin 60 = 5 sin 60 = 4:330 rad/s
= 0
For bar AB (CD does not contribute to dynamic reactions):
Ix
=
0
Iy
= Iz =
1
1
mL2 =
12
12
14
32:2
(1:44)2 = 0:07513 slug ft2
Euler’s equation (the other two equations are trivially satis…ed):
Mz = ! x ! y (Iy Ix )
RC (1:6) RD (1:6) = 2:5( 4:330)(0:07513)
RC + RD = 0:5083
F = 0:
RC RD = 0
) RC = 0:254 lb " J
RD = 0:254 lb # J
19.48
Let the xyz-axes be embedded in the plate.
!x = 0
Ix =
mb2
12
!y =
Iy =
! 0 sin
ma2
12
! z = ! 0 cos
Iz =
m(a2 + b2 )
12
550
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C
O
R
FBD
Since the xyz-axes are the principal axes of inertia, we apply Eqs. (19.33). With
!_ x = !_ y = !_ z = ! x = 0, these equations become
Mx
= ! y ! z (Iz
My
Mz
= 0
= 0
Iy )
) Cx = ( ! 0 sin )(! 0 cos )
) Cy = 0
) Cz = 0
mb2
12
1
mb2 ! 20 sin cos i J
12
F = ma
)R=0 J
C=
19.49
Let the xyz-axes be embedded in the plate.
!x = 0
!y =
! 0 sin
! z = ! 0 cos
2
Iz
Ix
=
Iy
=
Iz
=
Iy
=
mb2
m b
5
+
mb2
=
12
12 2
48
ma2
12
2
m(a2 + b2 )
m b
ma2
5
+
=
+ mb2
12
12 2
12
48
5
mb2
48
551
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C
O
R
FBD
Since the xyz-axes are the principal axes of inertia, we can apply Eqs. (19.33).
With !_ x = !_ y = !_ z = ! x = 0, these become
Mx
= ! y ! z (Iz
My
Mz
=
=
) Cx = ( ! 0 sin )(! 0 cos )
Iy )
5
mb2
48
) Cy = 0
) Cz = 0
0
0
5
mb2 ! 20 sin cos
48
C=
i J
ω0
z
βm
aA
A
y
b cosβ
2
aA =
b
cos ! 20 (j cos
2
+ k sin )
F = mA aA
R
= mA aA =
=
m
12
b
cos ! 20 (j cos
2
1
mb! 20 cos (j cos
24
+ k sin )
+ k sin ) J
552
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19.50
B
FBD
L mg
R
Ay z ω
A
C
Az = mg
x
O
y
Rotation about …xed axis:
Mx
= Iyz ! 2z
C
=
C = myz! 2 = m( R)
L
2
!2
1
mRL! 2 J
2
19.51
y
x
RA A
0.012 m
G
0.4
8m
B
0.3
2m
RB
z
FBD (only horizontal forces are shown)
We have rotation about a …xed axis. Equations (19.39) apply.
Mx = Iyz ! 2z
My =
Ixz ! 2z
With origin of xyz-axes at A:
Iyz
Ixz
My
= myz = 20(0)(0:48) = 0
= mxz = 20(0:012)(0:48) = 0:1152 kg m2
=
Ixz ! 2z
0:8RB = 0:1152(18)2
RB = 46:7 N J
With origin of xyz-axes at B:
Iyz
Ixz
My
= myz = 20(0)( 0:32) = 0
= mxz = 20(0:012)( 0:32) = 0:0768 kg m2
=
Ixz ! 2z
0:8RA = ( 0:0768)(18)2
RA = 31:1 N J
553
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19.52
The xyz-axes are attached to the rod. Because we have rotation about a …xed
axis, Eqs. (19.39) apply.
Iyz = 0
Ixz = m
y
A
Ax
ma2
a
(a) =
2
2
a
O
1
a
a
2
x
B
Bx
a
z
FBD (only the dynamic reactions are shown)
Ixz ! 20
My =
max =
mx1 ! 20
Ax a
Ax =
ma2 2
!
2 0
a
m ! 20
2
ma! 20 =
Ax + Bx =
3
ma! 20
2
Bx =
1
ma! 20 J
2
mx2 ! 20 =
Fx = max
Bx a =
ma! 20 J
3
ma! 20
2
19.53
5m
0.0
W1
y
z
x 0
.1 m R
B
W2
0
RA
B .1 m
A
0.8 N. m
FBD
m
0.1
W1
5m
0.0
554
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m1
=
m2
=
W1
=
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
m
1:8
=
= 0:3 kg (0.1 m 0.05 m plate)
6
6
4m
4(1:8)
=
= 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
=
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
Rotation about …xed axis:
Mz = Iz !_ z
My =
0:8 = 0:012!_ z
!_ z =
Iyz !_ z
(2W1 + W2 )(0:05)
(2 2:943 + 11:772) (0:05)
0:1RB
0:1RB
RB
Fx = 0
(2
2
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
y
RA
.
z
RB
B
m
0.1
A
FBD (only horizontal forces are shown)
From solution of Problem 19.53:
Iyz = 0:0015 kg m2
Izx = 0
555
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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
2 lb
D
0.7
5'
RB
C
'
.
5
0.7 ω z
'
5
0.7 3 lb
Use Eqs. (19.37) with ! z = 0 , !_ z = !_
Ixz
=
0
Iyz = myz =
Iz
=
1
1
mL2 =
3
3
Mz = Iz !_
3(0:75)
My =
1:25RB 2 =
RA + 2:80
) RA =
3
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
3
(0:75)(2) = 0:13975 slug ft2
32:2
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
C
z
R
FBD
O
x
y
mg
556
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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 !x
Iy
Iz
z !y
x !z
Cx
Cy
= 0 0=0
= (9:6 10 4 )(40)( 180)
=
6:91 N m
= 0 0=0
x !y
Ix
y !x
Cz
)C=
6:91k N m J
0
19.57
Az
Ay
A
Ax
0.7
5 ft
z
G
_
an = 0.75ω12
FBD
25 lb
x
y
Let the xyz-axes be attached to the shaft AG.
! = 30j + ! 1 k
Iy
Ix
= !1 k
1W 2
1 25
R =
(1)2 = 0:3882 slug ft2
2 g
2 32:2
1
= Iz = Iy = 0:1941 slug ft2
2
=
First equation of Eqs. (19.34) with G as the moment center:
Mx
0:75Az
= Iz y ! z Iy z ! y
= 0 (0:3882) ! 1 (30)
(a)
557
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Fz = maz
Az
25 = 0
Az = 25 lb
Equation (a) becomes
! 1 = 1:610 rad/s J
0:75(25) = (0:3882) ! 1 (30)
19.58
ωz
m
0.2
y
z
Oz
ωy
m
0.1
z
O
ω0
25o
A
Oy y
O
9.81m
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)
19.59
558
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19.60
ωy
y
A
x
50o
T
B
FBD
Ay
Ax A
0.9
m
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
Euler equation:
+
T (1:8 sin 50 )
Mz = ! x ! y (Iy Ix )
20(9:81)(0:9 cos 50 ) = (4:596)(3:857)(21:60
T = 360 N J
0)
559
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
19.61
A
L/4
H
ω
A
=
G
mrω2
x'
r
MAD
L/
2
V
L/
2
45o
G
mg y'
FBD
B
From FBD-MAD:
r=
Fx0
Fy 0
B
L
sin 45
2
y
ω x
o
45
G
A
B
L
= 0:103 55L
4
H = mr! 2 = 0:103 55mL! 2
mg = 0
V = mg
= max0
= 0
V
Euler equation (xyz-axes embedded in bar AB):
Mz
= ! x ! y (Iy Ix )
L
sin 45
+
V
2
=
(! sin 45 ) (! cos 45 )
mg
=
L
cos 45
2
1
mL2 0
12
H
L
sin 45
2
0:103 55mL! 2
L
cos 45
2
1
mL2 ! 2 sin 45 cos 45
12
The solution is
! = 2:126
r
g
= 2:126
L
r
32:2
= 8:53 rad/s J
2
19.62
CA
RA
A x
b/2
b
FBD
b
G
mg
z
y
560
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
z !y :
x ! z + Ix
x !y
CA =
(CA )x = 0
z !x :
Ix
y !x :
2 2
m! 0 b
8
m(b=2)2
(! 0 )(2! 0 ) =
2
(CA )y =
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
! z = 4 cos rad/s
!_ y = 0
!_ z = ( 4 sin ) _
!_ y = 0
!_ z =
Substituting _ = 12 rad/s, we get
!_ x = 48 cos rad/s
2
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
2
48 sin rad/s
= Ix !_ x + ! y ! z (Iz Iy )
= 7:680(48 cos ) + ( 12)( 4 cos )(0
Iz = 0
7:680) = 0
561
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
My
Cy
= 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
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
19.65
The no precession condition is
) Iz = I
= Iz =I = 1
1
1
mR2 =
m(3R2 + h2 )
2
12
p
h
= 3 J
R
562
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19.66
Z,y
.
φ
z
R
6 ft
O
x
ψ.
mg
A
FBD
_ =
F
_ = _R =
L
16 rad/s
0.8 ft
0:8
6
16
=
2:133 rad/s
Since the precession axis is perpendicular to the spin axis, Eq. (19.48) applies:
(F
Mx
= Iz _ _
4)(6)
=
1
2
(F
4
32:2
mg)L =
1
mR2 _ _
2
F = 4:23 lb J
(0:8)2 ( 16)( 2:133)
19.67
z
x
Z,y .
φ
mg
B
ψ.
O
0.5
m
0.5
m
F
A
mAg
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
563
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19.68
z Z
5o
β ω
y
1
Iz
1
=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
v
600(5280)
1
=
= 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
= Iz _ _
C
= mk 2 _ _ =
500
(1:2)2 (0:083 33)(1570:8) = 2930 lb ft J
32:2
564
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19.70
z
mg Z
FBD
L/2
L/2
RO = 2mg
y
O mg
θ
Iz
Iz
=
2
"
1
mR2
2
R
= mR2
I
=
1
2 mR2 + m
4
I
=
1
mR2 1
2
L
2
2
#
=
1
L2
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
565
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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
= tan 1
_ = ! = 3 rad/s
_ = _ =0
R
3
= tan 1 = 26:57
L
6
= 180
Iz
=
1
1
mR2 =
2
2
6
32:2
3
12
I
=
1
1
mR2 + mL2 =
4
4
6
32:2
=
49:50
mg(L sin )
6
N
=
=
2
= 5: 823
3
12
2
+
10 3 slug ft2
6
32:2
6
12
2
10 3 slug ft2
Mx
+
= 153:4
N
6
sin 26:57
12
5: 823 10 3
2:12 lb J
L
cos
N
49:50
2
=
(Iz
I) _ sin cos + Iz _ _ sin
=
(Iz
I) _ sin cos + Iz _ _ sin
2
6=12
cos 26:57
10 3 (3)2 sin 153:4 cos 153:4 + 0
19.73
The xyz coordinate system is embedded in the sphere-rod assembly with the
origin at the …xed point O.
Iz =
2
mR2
5
I=
2
47
mR2 + m(3R)2 =
mR2
5
5
566
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Oz
Oy
O
2R
θ
R
mg
FBD
Equation (19.46):
2
(Iz I) _ sin cos + Iz _ _ sin
45
2
mR2 _ sin cos + mR2 _ _ sin
mg(3R sin ) =
5
5
2
3g =
9R _ cos + 0:4R _ _
3(9:81) + 0:4(0:08)(60)(960)
3g + 0:4R _ _
=
cos
=
= 0:7225
2
9(0:08)(60)2
9R _
= 43:7 J
Mx
=
567
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#19.74
568
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19.75
Z, y
1.25 ft
1.25 ft
A
RA
Bz
G
RB
380 lb
'_
=
1:2 rad/s
_ = 160 rad/s
Iz
=
1
1
mR2 =
2
2
380
32:2
Fy = 0
+"
Mx
1:25RA 1:25RB
R A RB
5
12
2
RA + RB
= 1:0244 slug ft2
380 = 0
(a)
= Iz '_ _
= 1:0244(1:2)(160)
= 157:35
(b)
Solution of (a) and (b) is
RA = 269 lb J
RB = 111:3 lb J
569
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19.76
19.77
Z
z
mm
48
G
30o
y
0.5(9.81) N
Oy
O
Oz
FBD
We have steady precession. Equation (19.46) is
(0:5
Mx
=
9:81) (0:048 sin 30 )
=
0:11772
=
(Iz
2
I) _ sin cos + Iz _ _ sin
2
(5 10 4 20 10 4 ) _ sin 30 cos 30
+(5 10 4 )(120) _ sin 30
6:495
2
10 4 _ + 0:03 _
The two solutions are
_ = 4:33 and 41:9 rad/s J
570
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19.78
(a) The third equation of Eqs. (19.45) is
Mz = Iz !_ z
! z = Iz !_ z
!_ z =
Iz
!z
10 4 kg m2 (see Prob. 19.77). The solution is
where Iz = 5
! z = Ce ( =Iz )t
The initial condition ! z = (! z )0 at t = 0 yields C = (! z )0 .
) ! z = (! z )0 e ( =Iz )t J
(b) When ! z = 0:5(! z )0 , then
0:5
t
= e ( =Iz )t
=
(ln 0:5)
Iz
Iz
=
t = ln 0:5
(ln 0:5)
5 10 4
= 2:77 s J
1:25 10 4
19.79
19.80
571
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#19.81
572
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19.82
573
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19.83
19.84
574
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19.85
z
Z
25 o
80 rad/s
β
.
(a) We have torque-free motion with
tan
=
)
ω
y
= Iz =I = 2:105. From Eq. (19.52):
tan = 2:105 tan 25 = 0:9816
= 44:47 J
Equation (19.53c):
_
=
_
=
(1
(1
) _ cos
_
) cos
=
2:105( 80)
= 168:2 rad/s J
(1 2:105) cos 25
=
( 80)
cos 25 = 65:6 rad/s J
(1 2:105)
(b) Equation (19.53a):
_
= !(1
!
=
) cos
_
(1
) cos
19.86
575
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19.87
576
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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
)
f
=
E=
s
x20 +
v0
p
2
1:5
v0
=p
= 153:90 rad/s
0:0122 0:0072
E 2 x20
153:90
p
=
= 24:5 Hz J
2
2
p= p
(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
577
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20.4
20.5
578
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20.6
20.7
579
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20.8
580
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20.9
θ
ft
1.5
1.8 lb
8(1.0)θ
Oy
Ox
.
ft
1.0
ft
1.5
FBD
2.4 lb
= IO •
2:4
1:8
(2:5)2 +
(1:5)2 •
8(1:0 )(1:0) =
32:2
32:2
7:10 = 0:5171•
MO
1:8(2:5 )
2:4(1:5 )
This equation has the form m• + k = 0. Therfore,
r
r
k
7:10
p=
=
= 3:71 rad/s J
m
0:5171
20.10
Lθ T
x
FBD
Lθ T
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.
581
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20.11
..
m Lθ
FBD
m g kbθ
L
.2
m Lθ
=
MAD
θ
bOx
Oy
Use small angle approximations: sin
O
and cos
1
( MO )FBD = ( MO )M AD
+
mg(L )
kb (b)
• + kb
If
O
= mL•(L)
2
mgL
mL2
=
0
= 1:2 s, then p = 2 = = 2 =1:2 = 5: 236 rad/s
kb2 mgL
mL2
2
2
p mL + mgL kb2 = 0
2
(5: 236) (0:3)L2 + 0:3(9:81)L 300(0:08)2 = 0
8:225L2 + 2:943L 1:92 = 0
) p2 =
The positive root is L = 0:336 m J
20.12
a
mg
θ
O
b
R
k(bθ +∆) =
FBD
Use small angle approximations: sin
( MO )FBD = ( MO )M AD
where
and cos
+
O
a
.2
maθ
..
maθ MAD
1
mga + k(b +
)b =
ma2 •
is the static elongation of the spring. Substituting the static equilibrium
582
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
equation
mga + k
kb2
)p
b = 0, we get
2
• + kb
ma2 •
=0
ma2
s
r
k b
20
12
=
=
= 7:583 rad/s
ma
2:8=32:2 24
=
=
2
2
=
= 0:829 s J
p
7:583
20.13
#20.14
583
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#20.15
W
Water level after displacement x
Water level at equilibrium
∆
ρ
ρw
Area = A
h =
Ww
FBD
..
(W/g)x
MAD
584
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20.16
20.17
12(2)(θ + θs)
Ax
A
Ay
2 ft
θ
3 − 8(4)(θ + θs)
B
2 ft
C
FBD
= IA •
3
12(2)( + s )(2) + [3 8(4)( + s )] 4 =
(4)2 •
32:2
12 176( + s ) = 1:4907•
MA
The static displacement s is obtained by setting
ft . Therefore, the equation of motion becomes
= • = 0 yielding
s = 12=176
1:4907• + 176 = 0
585
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
p
1
) f=
=
2
2
r
k
1
=
m
2
r
176
= 1:729 cps J
1:4907
20.18
(a)
s
p
=
r
ccr
=
2mp = 2
=
6
c
=
= 0:3643 < 1
ccr
16:472
k
=
m
14 12
= 20:40 rad/s
13=32:2
13
32:2
(20:40) = 16:472 lb s/ft
) Underdamped.
(b)
ln
xn+1
xn
xn+1
xn
=
2
p
1
2
=
2 (0:3643)
p
=
1 0:36432
2:458
= e 2:458 = 0:0856 J
20.19
Damping is critical if
c
=1
= p
2 km
s
p
c = 2 km = 2 (16
20.20
c
= p
=1
2 km
12)
20
32:2
= 21:8 lb s/ft J
p
p
) c = 2 km = 2 80(3) = 31:0 N s/m J
20.21
=
)
20.22
q
2
2
2
= 0:5 p
) 1
= 0:5
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
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
586
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
0:14836(0:1)] e 5:164(0:1) =
2:89
10 3 m J
20.23
(a) Equivalent spring constant is
k=
1
1
= 90 lb/ft
=
1=k1 + 1=k2
1=225 + 1=150
m=
p
=
1:2
= 0:03727 slugs
32:2
r
r
k
90
=
= 49:14 rad/s
m
0:03727
2(2:8)
c
=
= 1:5288 overdamped J
2mp
2(0:03727)(49:14)
=
(b)
+
q
q
2
1 p
=
1:5288 +
2
1 p
=
1:5288
p
1:52882
p
1:52882
1 49:14 =
18:30 s 1
1 49:14 =
132:0 s 1
) x(t) = A1 e 18:30t + A2 e 132:0t
Initial conditions:
1:5
= 0:125 ft
A1 + A2 = 0:125 ft
12
x(0)
_
=0
18:30A1 + 132:0A2 = 0
x(0) =
The solution is A1 = 0:14511 ft, A2 =
0:02011 ft
) x(t) = 0:14511e 18:30t
0:02011e 132:0t
ft J
587
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20.24
2x
.
2cx
bx
FBD
Ox
kx
b Oy
O
b
x
MAD
=
O
b
..
mx
mg
(MO )FBD =
(MO )M AD
(kx)b + 2cx(2b)
_
+ mgx =
mxb
•
mg
m•
x + 4cx_ + k +
x = 0
b
p
=
=
d
20.25
=
r
k + mg=b
1800 + 3(9:81=0:2)
=
= 25:48 rad/s
m
3
4c
4(30)
=
= 0:7849
2mp
2(3)(25:48)
2
2
p
p
= 0:398 s J
=
2
25:48
1
0:78492
p 1
r
588
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20.26
589
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20.27
20.28
590
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20.29
h
L
From Eq. (20.31) the amplitude the relative motion (displacement of trailer
relative to the wheels) is
(!=p)2
Z=Y
1 (!=p)2
where
Y
=
p2
=
1:5=12
2 v
h
=
= 0:0625 ft
!=
2
2
L
2k
2(240 12)
2
=
= 231:8 (rad/s)
m
800=32:2
The amplitude of the absolute motion is
X =Z +Y =Y
(!=p)2
+Y =
1 (!=p)2
1
Y
(!=p)2
The displacement of the trailer is
x(t) = X sin !t =
1
Y
sin !t
(!=p)2
x
•(t) =
Y !2
sin !t
1 (!=p)2
The wheels are about to leave the road if j•
xjmax = g, i.e.,
Y !2
=g
1 (!=p)2
If !=p < 1:
Y !2
1 (!=p)2
= g
!
=
r
v
=
!L
12:664(6)
=
= 12:07 ft/s
2
2
g
=
Y + g=p2
s
32:2
= 12:644 rad/s
0:0625 + 32:2=231:8
591
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
If !=p > 1:
Y !2
(!=p)2 1
= g
!
=
r
v
=
!L
20:53(6)
=
= 19:60 ft/s
2
2
g
g=p2
Y
=
s
32:2
= 20:53 rad/s
32:2=231:8 0:0625
Wheels lose contact with the road if 12:07 ft/s < v < 19:60 ft/s J
20.30
h
L
From Eq. (20.31) the maximum relative displacement (displacement of trailer
relative to the wheels) is
zmax = jZj = Y
(!=p)2
1 (!=p)2
where
Y
=
!
=
p2
=
h
1:5=12
=
= 0:0625 ft
2
2
v
12(5280=3600)
2
=2
= 18:431 rad/s
L
6
2k
2(240 12)
2
=
= 231:8 (rad/s)
m
800=32:2
18:4312
= 1:4655
231:8
1:4655
zmax = 0:0625
= 0:196 76 ft
1 1:4655
) (!=p)2 =
But jZj is also the deformation of each spring, measured from the static equilibrium position. Thus the maximum force in each spring is
Fmax
=
=
W
+ kzmax
2
800
+ (240
2
12)(0:196 76) = 967 lb J
592
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#20.31
593
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20.32
594
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20.33
20.34
r
p
=
Z
= Y
k
=
m
s
36
= 43:95 rad/s
0:6=32:2
(!=p)2
= 0:6
1 (!=p)2
0:8283
1 0:8283
!
p
2
40
43:95
=
2
= 0:8283
= 2:89 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
595
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(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
596
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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
597
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20.40
20.41
598
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#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 =
599
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20.44
20.45
20.46
600
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20.47
X
=
!
= 0:5
p
c
8
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 2
0:25 =
2
0:5 ) + (2(0:5350)(0:5))
P0 =60
0:9213
13:82 lb J
#20.48
601
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602
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20.49
20.50
603
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20.51
0:8
= 0:024 84 slugs
k = 12(12) = 144 lb/ft
32:2
r
r
k
144
=
=
= 76:14 rad/s
m
0:024 84
c
0:4
!
50
=
=
= 0:105 75
=
= 0:6567
2mp
2(0:024 84)(76:14)
p
76:14
m =
p
X
=
Y
q
[1
=
tan
=
x(t)
1
2
0:2=144
q
(1
=
2
(!=p)2 ] + (2 !=p)
2: 372 3
2 2
2
0:6567 ) + [2(0:105 75)(0:6567)]
10 3 ft = 0:02847 in.
2 !=p
2(0:105 75)(0:6567)
=
= 0:2442
(!=p)2
1 0:65672
= X sin(!t
) = 0:02847 sin(50t
x(t) lags P (t) by 13:72 J
= 13:72
0:2442) in. J
604
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20.52
.
cx
x
FBD
mg
k(y − x − ∆)
F = m•
x
k(y
x
)
cx_ + mg = m•
x
Substituting the static equilibrium condition k
= mg, we get
k(y x) cx_ = m•
x
m•
x + cx_ + kx = kY sin !t
Letting P0 =k = Y , the last equation can be written as
m•
x + cx_ + kx = P0 sin !t
p
=
=
X
=
P0 =k
q
[1
=
=
1
x(t)
2
(!=p)2 ] + (2 !=p)2
12
q
(1
tan
r
k
90 103
=
= 600 rad/s
m
0:25
c
60
!
900
=
= 0:2
=
= 1:5
2mp
2(0:25)(600)
p
600
r
= 8:65 mm
2
2
1:52 ) + [2(0:2)(1:5)]
2(0:2)(1:5)
2 !=p
=
=
(!=p)2
1 1:52
0:48
=
0:448 rad = 25:6
= X sin(!t
) = 8:65 sin(900t + 0:448) mm J
x(t) leads y(t) by 25:6 J
605
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20.53
20.54
606
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20.55
20.56
607
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20.57
P(t)
R
2 ft
θ
4 lb
. 1.5 ft O. 2 ft
.
c(2θ)
k(1.5θ + ∆)
FBD
Assume that the angular displacement
P (t)(3:5)
k(1:5 +
)(1:5)
of the bar is small.
c(2 _ )(2)
MO = IO •
4(2) = m(2)2 •
(a)
where
is the deformation of the spring at static equilibrium. Its value is
obtained by setting P (t) = = • = 0 which yields 1:5k
8 = 0: Substituting
this into Eq. (a) gives us
3:5P (t)
2:25k
4c _
3:5(0:4 sin 4t) 2:25(6)
4(2:2) _
0:4969• + 8:80 _ + 13:50
4
(4)•
32:2
= 0:4969•
= 1:40 sin 4t
=
which has the form
M • + C _ + K = P0 sin !t
608
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
p
r
r
K
13:50
=
= 5:212 rad/s
M
0:4969
C
8:8
=
= 1:6989
2M p
2(0:4969)(5:212)
4=5:212 = 0:7675
=
=
!=p
P0 =k
=q
=
max
=
[(1
=
q
(1
2
2
(!=p)2 ] + (2 !=p)
1:40=13:5
= 0:03928 rad
2
0:76752 )2 + [2(1:6989)(0:7675)]
( block )max = 24( max ) = 24(0:03928) = 0:943 in.
20.58
(a)
mg
6x.
80(0.004 sinωt − x)
FBD
60x
N
Fx = max
+
!
80(0:004 sin !t
x)
6x_
60x = 12•
x
12•
x + 6x_ + 140x = 0:32 sin !t J
(b) Comparing with M x + Cx + Kx = P0 sin !t, we obtain
r
r
K
140
p =
=
= 3:416 rad/s
M
12
C
6
=
=
= 0:07319
2M p
2(12)(3:416)
With ! = p Eqs. (20.26) and (20.27) yield
X
=
0:32=140
P0 =K
=
= 0:015 615 m = 15:62 mm
2
2(0:07319)
=
tan 1 1 = =2 rad
x(t) = 15:62 sin(3:42t + =2) mm J
609
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20.59
(a)
P0 sinωt
Oy
Ox O
θ
b
θ
b
.
mg − cbθ
FBD
kb(θ + θs)
(mg
cb _ )b
kb(
MO
s )b + (P0 sin !t)b
= IO •
= mb2 •
Substituting the static equilibrium equation
mgb + kb2 s = 0
the equation of motion becomes
mb2 • + cb2 _ + kb2
= P0 b sin !t
P0
sin !t J
=
b
m• + c _ + k
(b)
p
=
=
r
r
k
30
!
6
=
= 8:660 rad/s
=
= 0:6928
m
0:4
p
8:660
c
3
=
= 0:4330
2mp
2(0:4)(8:660)
=
(P0 =b)=k
q
[1
=
(1:5=0:4)=30
q
(1
=
tan
=
=
2
(!=p)2 ] + (2 !=p)2
2
2
0:69282 ) + (2(0:4330)(0:6928))
0:157 44 rad = 9:02
J
2 !=p
2(0:4330)(0:6928)
=
= 1:1537
2
1 (!=p)
1 0:69282
49:1 J
610
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#20.60
#20.61
611
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612
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20.62
mg
kx
R
FBD
F
C
N
Fx
Fy
= m•
x
= 0
MC
= IC •
F
N
kx = m•
x
F = kx + m•
x
mg = 0
N = mg
1
x
•
(kx) R =
mR2 + mR2
2
R
) F = kx
m
2 k
x
3m
=
1
kx
3
Fmax =
x
•=
2 k
x
3m
1
kx0
3
At impending sliding we have
Fmax
=
N
s
kx0
=
3mg
s
k=
3mg s
3(8)(9:81)(0:2)
=
= 314 N/m J
x0
0:15
20.63
613
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20.64
20.65
Use formula given in Prob. 20.64:
p
2
=
)
r
gy
k2 + y2
gy
k2 = 2 y2
p
(a)
(b)
Pin at A:
p=
2
=
y
=
0:28 m
=2s
From Eq. (b) k 2
=
9:81(0:28)
3:14162
0:282 = 0:199 91 m2
2
= 3:1416 rad/s
2
Pin at B:
y
=
From Eq. (a) p2
=
)
0:21 m
9:81(0:21)
2
= 24:51 (rad/s)
0:199 912 + 0:212
2
2
=p
=
= 1:269 s J
p
24:51
614
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20.66
(a) Let
s be the static angular displacement
mg
+
mg
Upon substituting
becomes
Ox O L/2
L/2
Oy
.
c2Lθ/2
c2 L _
2
h
L
2
θ
FBD
.
kL(θs + θ) + c1Lθ
MO = IO •
i
1
kL( s + ) + c1 L _ L =
mL2 •
3
2
s = mg=(2kL) and cancelling L , the equation of motion
c2
1 •
m + c1 +
3
4
_
k =0 J
(b) Comparing with M • + C _ + K = 0, we obtain
s
s
r
K
k
80
=
=
= 3:464 rad/s
p =
M
m=3
20=3
Ccr
=
=
m
20
p=2
3:464 = 46:19 N s/m
3
3
C
c1 + c2 =4
25 + 16=4
= 0:628 J
=
=
Ccr
Ccr
46:19
2M p = 2
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
p =
=
= 0:5941
M
2:833R
R
r
r
p
0:5941 g
g
f =
=
= 0:0946
Hz J
2
2
R
R
615
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20.68
F2
Ox
IO
2
= I + m OG =
=
Letting
θ
8(9.81) N
G
5m
2
1
0.
O 0.15 m
Oy
0.2 m
F1
FBD
1
m(a2 + b2 ) + m
12
b2
a2
+
4
4
=
1
m(a2 + b2 )
3
1
(8)(0:152 + 0:22 ) = 0:166 67 kg m2
3
s be the static angular displacement, the spring forces are
F1
F2
= k1 1 = 300 [0:2( s + )] = 60( s + )
= k2 2 = 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•
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
K
1
25:5
)f =
=
=
= 1:969 Hz J
2
2
M
2
0:166 67
616
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20.69
x
A
L/4
kx
By
B
Bx
L/4
θ G
L/2
mg
FBD
k(3x − x0 sinωt)
C
3x
m=
IB =
mL2
+m
12
12x
12
= 0:3727 slugs
32:2
L
4
2
=
=
4
2
x= x
L
3
7
7
mL2 =
(0:3727)(6)2 = 1:9567 slug ft2
48
48
MB
L
3L
mgx kx
k (3x x0 sin !t)
4
4
3
6
150 (3x x0 sin 8t)
(6)
150x
4
4
2262x + 675x0 sin 8t
1:3045•
x + 2262x
K
p
= IB •
= IB
2
3
= 1:9567
=
=
x
•
2
x
•
3
1:3045•
x
675x0 sin 8t
=
2262 lb
M = 1:3045 slug ft
r
r
K
2262
=
=
= 41:64 rad/s
M
1:3045
2
(!=p)
2
xmax
8
= 0:03691
41:64
(!=p)2
0:03691
= x0
= 0:0796 in. J
2 = 2 (1
2
0:03691)2
[1 (!=p) ]
=
617
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20.70
20.71
x
mg
kx
R
F
FBD
C
N
(a)
m=
8
= 0:2484 slugs
32:2
MC = IC •
p=
(kx) R =
r
2 k
=
3m
r
1
3
mR2 + mR2 = mR2
2
2
3
x
•
2
k
mR2
x
•+
x=0
2
R
3m
IC =
2 110
= 17:182 rad/s J
3 0:2484
(b)
Fx = m•
x
F
kx = m•
x
F = m•
x + kx
618
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
p2 X sin pt:
Substitute x = X sin pt , x
•=
F
Fmax
= X( mp2 + k) sin pt
1:2
= X( mp2 + k) =
12
s =
(0:2484)(17:182)2 + 110 = 3:667 lb
Fmax
3:667
=
= 0:458 J
mg
8
20.72
kLθ
.
cLθ
L
O
L
L/2
θ
L/2
.
mg
FBD (only forces contributing
to MOare shown)
MO
L
(kL )L (cL _ )L mg
2
mg
m• + c _ + k +
2L
= IO •
mL2
= 3
3
=
•
0
Comparing with M • + C _ + K = 0, we obtain
M
K
p
!d
1:8
= 0:05590 slugs
C=c
32:2
mg
1:8
= k+
= 20 +
= 20:60 lb/ft
2L
2(1:5)
= m=
r
r
K
20:60
=
=
= 19:197 rad/s
M
0:05590
q
p
2
= p 1
= 19:1971 1 0:52 = 16:625 rad/s J
619
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20.73
(a)
Oy
O
Ox
b
mg
IO =
mg
θ FBD
.
2cbθ
b/2
kb(θs + θ)
+
2b
1
m(3b)2 + m
12
b
2
2cb _ (2b)
The static angular displacement
b
2
2
= mb2
MO
= IO •
kb( s + )b
= mb2 •
s is given by
(mg)
b
2
kb2 s = 0
(a)
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
3200
p =
=
= 24:12 rad/s
m
5:5
4c
mp
5:5(24:12)
=
ccr =
=
= 66:3 N s/m J
2mp
2
2(1)
20.74
.
chθ
A
mg
x-
h
Bx
MB = IB
B
+
By
FBD
b
θ
C
kb(θs + θ)
mgx
ch2 _
kb2 ( s + ) = IB •
Substituting the static angular displacement s = mgx=(kb2 ), the equation of
motion becomes
IB • + ch2 _ + kb2 = 0
620
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Comparison with M • + C _ + K = 0 yields
p2
K
kb2
=
M
IB
=
C
ch2
c(1:0)2
=
=
= 0:004 650c
2M p
2IB p
2(7:709)(2
2:22)
Damping is critical when
) IB =
(20
kb2
=
2
p
(2
12)(2:5)2
=
2
2:22)
= 7:709 slug ft2
=1
) ccr =
1
= 215 lb s/ft J
0:004 650
20.75
621
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
#20.76
622
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20.77
623
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20.78
*20.79
624
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
*20.80
20.81
k1Rθ
Oy
θ
R
k2Rθ
Vg
=
Ve
=
V
W
O
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
625
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
)K=
WR
44(2:4)
+ (k1 + k2 )R2 =
+ (25 + 35)(2:4)2 = 398:4 lb ft
2
2
T
)M
1 _2
I
2
=
= I = mk 2 =
)p=
r
K
=
M
r
44
(1:8)2 = 4:427 lb ft s2
32:2
398:4
= 9:49 rad/s J
4:427
20.82
Datum
A
L
V
=
T
=
=
θ
B
Lθ
C
L
L
1
1
k(L )2 2mg
= kL2 2 mgL
) K = kL2
2
2
2
2
2
1
1
2
1
(IA + IC ) • =
2
mL2 •
) M = mL2
2
2
3
3
r
r
r
2
M
2mL2 =3
m
=
5:13
=2
=2
J
p
K
kL2
k
20.83
(a)
m
θa
Datum
V
T
b
θ
1
1 2
1
= mga cos + k(b )2 mga 1
+ kb2 2
2
2
2
1
= mga +
kb2 mga 2
) K = kb2 mga
2
1
=
m(a _ )2
) M = ma2
2
r
r
K
kb2 mga
p=
=
J
M
ma2
626
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(b)
kb2
mga > 0
k>
mga
J
b2
20.84
.
G
h
.G θ
Datum
max
R
θmax
Position of Vmax
Position of Tmax
Use Rayleigh’s principle:
h
=
R
cos max
R=R
1
cos max
cos max
R
1
(1
1
2
max =2)
2
max =2
2
R max
2
2
Vmax
Tmax
= mgh = mgR max
2
1 _2
1 mL2 _ 2
1
=
IG max =
=
mL2 p2 2max
2
2 12 max
24
Tmax
= Vmax
1
mL2 p2 2max
24
= mgR
p2
=
12gR
L2
V0
2
max
2
0
p=
2p
3gR J
L
20.85
627
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20.86
20.87
628
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20.88
Datum
L
θ
m
V
=
T
=
mg(L cos )
R
1 2
) K = mgL
2
2
_2
_ 2 = 1 m L2 + R
2
2
mgL 1
1
1
m(L _ )2 +
2
2
1
mR2
2
R2
) M = m L2 +
2
s
s
r
m L2 + R2 =2
2
M
L2 + R2 =2
=
=2
=2
=2
p
K
mgL
gL
The condition
= 1:2 s yields
s
L2 + R2 =2
2
=
gL
1:22 g
R2
L+
= 0
2
4
2
1:1745L + 0:03125 = 0
L2
L2
4 2
1:2
L2
L2 + R2 =2
= 1:22
gL
1:22 (32:2)
1
2
L + (0:25) = 0
4 2
2
The two solutions are
L1 = 0:02724 ft = 0:327 in. J
L2 = 1:1473 ft = 13:76 in. J
629
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20.89
Use Rayleigh’s method.
.
A.
L/2
L/2
mg
b
.
.
.
.
C
B
3mg
Position of maximum V
Vmax
V0
Tmax
p2
=
4mgL cos max
4mgL
=
= Vmax
=
L
cos max )
2
2(mg
=
12 g
11 L
Lθmax
.
b
D
Position of maximum T
=
=
Tmax
L
L
L/2
mg
E . θmax
L/2
L/2
θmax
.E
Datum
.
A . θmax
Vmax
3mL cos max
4mgL 1
V0 = 2mgL
2
max
2
2
max
2
1
mL2 _ 2
_
2
max + 3m L max
2
3
2
11
11
mL2 _ max =
mL2 p2 2max
6
6
11
mL2 p2 2max = 2mgL 2max
6
r
3g
p=2
J
11L
V0
0
630
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20.90
20.91
Datum position (spring undeformed)
β
R
x+
Let
∆
be the static elongation of the spring
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 1
x_
1
1 3
T =
mR2
+ mx_ 2 =
m x_ 2
2 2
R
2
2 2
3
) M= m
2
r
r
r
p
1
K
1
2k
k
f=
=
=
= 0:1299
J
2
2
M
2
3m
m
V
=
631
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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 water
V
=
T
=
p
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 water
=
h wo o d
32:2(0:036)
= 10:266 rad/s
0:5(0:022)
20.93
632
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20.94
633
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
#20.95
634
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20.96
Datum
ω -
θ
R−r
R
v
r
v = (R
V
=
T
=
mg(R
r) cos
v
R r_
=
r
r
r) _
!=
mg(R
r) 1
1 2
2
) K = mg(R
r)
2
=
12 2 R r
1 2 1
_ 2 + 1 m(R r)2 _ 2
I! + mv 2 =
mr
2
2
25
r
2
2
1 7
7
m(R r)2 _
) M = m(R r)2
2 5
5
s
r
r
1
1
p
K
5g
g
f=
=
=
= 0:1345
J
2
2
M
2
7(R r)
R r
20.97
Use Rayleigh’s method
.
θmax
Datum θmax
h
Position
of Vmax
R−r
r
R
C
R
Position
of Tmax
635
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
h
IC
Vmax
=
mgh =
= r + (R r) cos max
= I + mR2 = mR2 + mR2 = 2mR2
mg [r + (R
mg r + (R
V0
=
Tmax
=
r) 1
r) cos max ]
1 2
1
= mgR + mg(R
max
2
2
r) 2max
mgR
2
1 _2
1
2mR2 _ max = mR2 p2 2max
IC max =
2
2
Tmax
p
1
= Vmax V0
mR2 p2 2max = mg(R
2
r
g(R r)
J
=
2R2
r) 2max
20.98
(a)
x
Fx = max
(b)
kx
mg
FBD
N
+ !
2kx = m•
x
kx
2(150)x = 6•
x
x
• + 50x = 0 J
p
50
p
=
= 1:125 Hz J
f=
2
2
(c)
x = E sin(pt + )
s
s
2
2
v
( 120)
E =
x20 + 02 = 252 +
= 30:22 mm
p
50
p !
x
p
25
50
0
= tan 1
= tan 1
= 0:9744 rad
v0
120
)
x = 30:22 sin(7:071t
0:9744) mm J
636
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20.99
20.100
637
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20.101
L/2
C(t)
θ
L/2
O
k1 L θ
2
k2 L θ
2
FBD (only forces contributing
to MO are shown)
C0 sin !t
(k1 + k2 )
L
2
MO
L
2
m• + 3(k1 + k2 )
= IO
mL2 •
=
12
C0
= 12 2 sin !t
L
This equation has the form
M • + K = P0 sin !t
where
p
=
=
X
=
18
= 0:5590 slugs
32:2
3(k1 + k2 ) = 3(20 + 30) = 150:0 lb=ft
C0
0:5
12 2 = 12
= 0:16667 lb/ft
L
62
M
= m=
K
=
P0
=
r
r
K
150
=
= 16:381 rad/s
M
0:5590
P0 =K
0:16667=150:0
=
2
2 = 0:02582 rad
[1 (!=p)2 ]
[1 (18=16:381)2 ]
L
6
= 0:02582 = 0:07746 ft = 0:930 in. J
2
2
20.102
Momentum is conserved during impact. Letting v0 be the velocity of the block
(and the embedded bullet) immediately after the impact, we have
0:012(1200) = (5:012)v0
v0 = 2:873 m/s
638
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Now consider the ensuing damped free vibration with the inital conditions x = 0
and x_ = 2:873 m/s at t = 0.
r
r
k
1800
p =
=
= 18:951 rad/s
m
5:012
ccr = 2mp = 2(5:012)(18:951) = 189:96 N s/m
c
50
=
=
= 0:2632
Since < 1, the system is underdamped
ccr
189:96
q
p
2
= 18:951 1 0:26322 = 18:283 rad/s
!d = p 1
x(t) = Ee pt sin(! d t + )
x(t)
_
=
E pe pt sin(! d t + ) + E! d e
p = 0:2632(18:951) = 4:988
x =
x_ =
)
pt
cos(! d + )
0 at t = 0
E sin = 0
2:873 m/s at t = 0
E p sin + E! d cos
2:873
2:873
= 0:157 14 m
=
=0
E=
!d
18:283
= 2:873
) x (t) = 0:1571e 0:499t sin 18:28t m J
20.103
639
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20.104
Since
= 0:6 < 1, the system is underdamped.
s
r
k
300
p =
=
= 34:75 rad/s
m
8=32:2
p
!d
=
0:6(34:75) = 20:85 rad/s
q
p
2
= p 1
= 34:75 1 0:62 = 27:80 rad/s
x(t) = Ee
pt
sin(! d t + ) = Ee 20:85t sin(27:80t + )
Initial conditions:
x = 3 in. when t = 0. ) E sin = 3 in.
x_ = 0 when t = 0. )
20:85E sin + 27:80E cos = 0
27:80
= 1:3333
= 53:13 (0.9273 rad)
) tan =
20:85
3
) E=
= 3:750 in.
sin 53:13
x(t) = 3:750e 20:85t sin(27:80t + 0:9273) in. J
20.105
640
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20.106
20.107
641
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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 = 8 + 14 = 22 kN/m
= k2 Y = 14(0:024) = 0:336 m
=
=
!
p
=
r
r
k
22 000
=
= 52:44 rad/s
m
8
c
50
=
= 0:05959
2mp
2(8)(52:44)
18
= 0:3432
52:44
642
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
The steady-state displacement is x(t) = X sin(!t
X
=
=
=
)
q
q
), where
P0 =k
[1
2
2
(!=p)2 ] + (2 !=p)
0:336=22
(1
2
= 0:017 293 m
2
0:34322 ) + [2(0:05959)(0:3432)]
2 !=p
2(0:05959)(0:3432)
= tan 1
= 0:04633 rad
2
1 (!=p)
1 0:34322
x(t) = 0:01729(sin 18t 0:0463) m J
x(t) lags y(t) by 0:0463 rad (2:65 ) J
tan 1
20.110
643
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20.111
b
L/2
Ox O
.
Oy
FBD
..
θ
kb(θ + θs)
mg
L
1
kb2 ( + s ) = mL2 •
2
3
The static angular displacement s of the bar is oblained by setting
MO = I0 •
mg
Substituting
L
2
mg
kb2 s = 0
)
s =
(a)
= 0:
mgL
2kb2
s into Eq. (a) yields
mg
L
2
mgL
1
=
mL2 •
2kb2
3
2
• + 3kb
= 0
mL2
r
b 3k
) p=
J
L m
kb2
+
20.112
644
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20.113
b
az
F
θ
z
a
y,z
b
FBD
Oy
x
O
Ox
=
may
mg
F =k =k
kb
But kb
is
b
+ z
a
ax = a _
( MO )FBD
+
F b (mg) a
b
k
+ z b mga
a
b2
mga + k z + ma•
z
a
MAD
a
max
O
2
ay = a• + y• = z• + y•
=
=
( MO )M AD
(may ) a
=
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
645
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20.114
From Eq. (20.31):
Y
= Z
!
p
=
)Y
=
=
rh
1
(!=p)
2
i2
+ (2 !=p)2
(!=p)2
18
= 0:9549
2 (3)
q
2
2
(1 0:95492 ) + [2(0:25)(0:9549)]
10
0:95492
5:32 mm J
646
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
647
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
An Instructor’s Solutions Manual to Accompany
ENGINEERING MECHANICS: DYNAMICS,
4TH EDITION
ANDREW PYTEL
JAAN KIUSALAAS
ISBN: 978-1-305-88500-4
© 2017, 2010 Cengage Learning
WCN: 01-100-101
ALL RIGHTS RESERVED. No part of this work covered by the
copyright herein may be reproduced, transmitted, stored, or
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For product information and technology assistance, contact us at
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READ IMPORTANT LICENSE INFORMATION
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Instructor's Solutions Manual
to Accompany
Engineering Mechanics:
Dynamics
4th EDITION
ANDREW PYTEL
JAAN KIUSALAAS
Contents
Chapter 11:.......................................................................................................................... 1
Chapter 12:........................................................................................................................ 83
Chapter 13:...................................................................................................................... 140
Chapter 14:...................................................................................................................... 204
Chapter 15:...................................................................................................................... 294
Chapter 16:..................................................................................................................... .370
Chapter 17:...................................................................................................................... 452
Chapter 18:...................................................................................................................... 521
Chapter 19:...................................................................................................................... 577
Chapter 20:...................................................................................................................... 647
Appendix F
F.1
F.2
647
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.3
648
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.4
F.5
649
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.6
F.7
650
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.8
F.9
651
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.10
F.11
652
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.12
653
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.13
F.14
654
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.15
F.16
655
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.17
656
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.18
F.19
657
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.20
F.21
658
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.22
659
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.23
660
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.24
F.25
661
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.26
662
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.27
663
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.28
664
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.29
665
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.30
Iy0
Iz0
Iy0 z0
m b2
3
m (2b)2
+
= mb2
2 3
4 3
4
= (Iz0 )1 + (Iz0 )2 + (Iz0 )3
m (2b)2
m
m
m b2
3
=
+ (2b2 ) + b2 +
= mb2
2 12
2
4
4 3
2
b
m
=
= (Iy0 z0 )2 = (Iyz )2 + m2 y 0 z 0 = 0 + ( b)
4
2
=
(Iy )1 + (Iy )2 =
IAB = mb2
3
4
1
2
+
3
2
1
2
2
1
8
1
2
1 2
mb
8
= mb2 J
F.31
666
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.32
667
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.33
F.34
668
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.35
669
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.36
F.37
670
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.38
671
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.39
672
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F.40
F.41
673
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.