1. No category

PROBLEM 16.4
The motion of the 2.5-kg rod AB is guided by two small wheels which roll
freely in horizontal slots. If a force P of magnitude 8 N is applied at B,
determine (a) the acceleration of the rod, (b) the reactions at A and B.
SOLUTION
ΣFx = Σ( Fx )eff : P = ma
(a)
a=
P
8N
=
= 3.20 m/s 2
m 2.5 kg
a = 3.20 m/s 2
2r 

 2r 
ΣM B = Σ( M B )eff : W  r −  − Ar = ma  
π 

π 
(b)
2
2

2

2
A = W 1 −  − ma   = mg 1 −  − P  
 π
π 
 π
π 
2

2
= (2.5 kg)(9.81 m/s 2 ) 1 −  − (8 N)  
 π
π 
= 8.912 N − 5.093 N = 3.819 N



A = 3.82 N 
ΣFy = 0: A + B − W = 0
B = W − A = (2.5)(9.81) − 3.819,
B = 20.71 N 
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1407
PROBLEM 16.19
The triangular weldment ABC is guided by two pins that slide
freely in parallel curved slots of radius 6 in. cut in a vertical plate.
The weldment weighs 16 lb and its mass center is located at Point G.
Knowing that at the instant shown the velocity of each pin is
30 in./s downward along the slots, determine (a) the acceleration of
the weldment, (b) the reactions at A and B.
SOLUTION
v = 30 in./s
Slot:
an =
v 2 (30 in./s)2
=
= 150 in./s 2
r
6 in.
a n = 12.5 ft/s 2
at = at
Weldment is in translation
30°
60°
an = 12.5 ft/s 2
60° ΣF = ΣFeff : mg cos 30° = mat
at = 27.886 ft/s 2
(a)
60°
Acceleration
β = tan −1
an
12.5
= tan −1
= 24.14°
27.886
at
a 2 = at2 + an2
= (27.886) 2 + (12.5) 2
a = 30.56 ft/s 2
84.1°
a = 30.6 ft/s 2
84.1° 


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1428
PROBLEM 16.19 (Continued)
(b)
Reactions
ma =
16 lb
(30.56 ft/s 2 ) = 15.185 lb
2
32.2 ft/s
ΣM A = Σ( M A )eff :
B cos 30°(9 in.) − (16 lb)(6 in.) = (15.185 lb)(cos84.14°)(3 in.) − (15.185 lb)(sin 84.14°)(6 in.)
7.794 B − 96 = + 4.651 − 90.634
B = +1.285 lb
B = 1.285 lb
30° 
ΣFx = Σ( Fx )eff : A cos 30° + B cos 30° = ma cos84.14°
A cos 30° + (1.285 lb) cos 30° = (15.185 lb) cos84.14°
A cos 30° + 1.113 lb = 1.550 lb
A = +0.505 lb
A = 0.505 lb
30° 
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1429
PROBLEM 16.25
The rotor of an electric motor has an angular velocity of 3600 rpm when the load and power are cut off. The
50-kg rotor, which has a centroidal radius of gyration of 180 mm, then coasts to rest. Knowing that kinetic
friction results in a couple of magnitude 3.5 N ⋅ m exerted on the rotor, determine the number of revolutions
that the rotor executes before coming to rest.
SOLUTION
I = mk 2
= (50)(0.180) 2
= 1.62 kg ⋅ m 2
M = I α:
3.5 N ⋅ m = (1.62 kg ⋅ m 2 )α
α = 2.1605 rad/s 2 (deceleration)
 2π 

 60 
= 120π rad/s
ω0 = 3600 rpm 
ω 2 = ω02 + 2αθ
0 = (120π rad/s) 2 + 2(−2.1605 rad/s 2 )θ
θ = 32.891 × 103 rad
= 5235.26 rev
θ = 5230 rev 
or
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1439
PROBLEM 16.48
A uniform slender rod AB rests on a frictionless horizontal
surface, and a force P of magnitude 0.25 lb is applied at A
in a direction perpendicular to the rod. Knowing that the rod
weighs 1.75 lb, determine (a) the acceleration of Point A,
(b) the acceleration of Point B, (c) the location of the point
on the bar that has zero acceleration.
SOLUTION
W
g
1 W 2
I=
L
12 g
m=
ΣFx = Σ( Fx )eff : P = ma =
a=
ΣM G = Σ( M G )eff : P
W
a
g
P
0.25 lb
1
g=
g= g
W
1.75 lb
7
a=
1
g
7
L
1 W 2
Lα
= Iα =
2
12 g
α =6
P 5
0.25 lb g 6 g
=6
⋅ =
W L
1.75 lb L 7 L
α=
6g
7L
We calculate the accelerations immediately after the force is applied. After the rod acquires angular velocity,
there will be additional normal accelerations.
(a)
Acceleration of Point A.
aA = a +
(b)
L
1
L 6
4
4
α = g + ⋅ g = g = (32.2 ft/s2 )
2
7
2 7
7
7
a A = 18.40 ft/s 2

a B = 9.20 ft/s 2

Acceleration of Point B.
aB = a −
L
1
L 6
2
2
α = g − ⋅ g = − g = − (32.2 ft/s2 )
2
7
2 7
7
7
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1478
PROBLEM 16.48 (Continued)
(c)
Point of zero acceleration.
aP = 0
a − ( z P − zG )α = 0
z P − zG =
Since zG =
a
α
=
1 g
7
6 g
⋅
7 L
=
1
L
6
1
L
2
1
1
2
L+ L= L
2
6
3
2
z P = (36 in.)
3
zP =
z P = 24.0 in. 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1479
PROBLEM 16.56
The 80-g yo-yo shown has a centroidal radius of gyration of 30 mm. The radius
of the inner drum on which a string is wound is 6 mm. Knowing that at the instant
shown the acceleration of the center of the yo-yo is 1 m/s2 upward, determine
(a) the required tension T in the string, (b) the corresponding angular acceleration
of the yo-yo.
SOLUTION
W = mg
W = 0.080 kg (9.81 m/s 2 ) = 0.7848 N
ΣFy = Σ( Fy )eff : T − W =
W
a
g
T − (0.08 kg)(9.81 m/s 2 ) = (0.08 kg)(1 m/s 2 )
T = 0.8648 N
(a)
T = 0.865 N 
Tension in the string.
ΣM G = Σ( M G )eff : Tr = Iα
(0.8648 N)(0.006 m) = mk 2α
5.1888 × 10−3 N ⋅ m = (0.08 kg)(0.03 m)2 α
(b)
Angular acceleration.
α = 72.067 rad/s 2
α = 72.1 rad/s 2

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1488
PROBLEM 16.64
A beam AB of mass m and of uniform cross section is suspended
from two springs as shown. If spring 2 breaks, determine at that
instant (a) the angular acceleration of the beam, (b) the acceleration
of Point A, (c) the acceleration of Point B.
SOLUTION
ΣFy =
2mg
= maG
3
1
 mg  L
mL2α
ΣM G = 
=

 3  2 12
aG =
(a)
(b)
aA =
2g 2 g L
+
3
L 2
=
2g
3
,
α =
2g
L
g
3 ,
g
3

5g
3

aA =
(c)
aB =
2g 2 g L
+
3
L 2
=

5g
3
aB =
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1499
PROBLEM 16.76
A uniform slender rod of length L = 900 mm and mass m = 4 kg is suspended
from a hinge at C. A horizontal force P of magnitude 75 N is applied at end B.
Knowing that r = 225 mm, determine (a) the angular acceleration of the rod,
(b) the components of the reaction at C.
SOLUTION
(a)
Angular acceleration.
a = rα
I =
1
mL2
12
L

ΣM C = Σ( M C )eff : P  r +  = (ma )r + Iα
2

1
mL2α
12
L

 2 1 2
P  r +  = m  r + L α
2
12 


= (mrα )r +
Substitute data:
0.9 m 
1



= (4 kg) (0.225 m) 2 + (0.9 m)2  α
(75 N) 0.225 m +

2 
12



50.625 = 0.4725α
α = 107.14 rad/s 2
(b)
α = 107.1 rad/s 2

Components of reaction at C.
ΣFy = Σ( Fy )eff : C y − W = 0
C y = W = mg = (4 kg)(9.81 m/s 2 )
C y = 39.2 N 
ΣFx = Σ( Fx )eff : C x − P = −ma
Cx = P − ma = P − m(r α )
= 75 N − (4 kg)(0.225 m)(107.14 rad/s 2 )
Cx = 75 N − 96.4 N

= −21.4 N

C x = 21.4 N

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1527
PROBLEM 16.86
A 12-lb uniform plate rotates about A in a vertical plane under the
combined effect of gravity and of the vertical force P. Knowing that
at the instant shown the plate has an angular velocity of 20 rad/s and
an angular acceleration of 30 rad/s2 both counterclockwise, determine
(a) the force P, (b) the components of the reaction at A.
SOLUTION
 6 
at = rα =  ft  (30 rad/s 2 ) = 15 ft/s 2
 12 
 6 
an = rω 2 =  ft  (20 rad/s) 2 = 200 ft/s 2
 12 
Kinematics.
Mass and moment of inertia.
I=
m=
W
12 lb
=
= 0.37267 lb ⋅ s 2 /ft
g 32.2 ft/s 2
2
2
m  10   20  
  +    = (0.37267 lb ⋅ s 2 /ft)(0.28935 ft 2 ) = 0.10783 lb ⋅ s 2 ⋅ ft
12  12   12  
Kinetics.
(a)
Force P.
 16 
 6 
 6 
ΣM A = Σ( M A )eff : P  ft  − W  ft  = mat  ft  + I α
 12 
 12 
 12 
4
1
1
P = (12)   + (0.37267)(15)   + (0.10783)(30)
3
2
2
P = 9.0224 lb.
P = 9.02 lb 
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1537
PROBLEM 16.86 (Continued)
(b)
Reaction at A.
ΣFy = Σ( Fx )eff : Ax = − man = −
12
(200)
32.2
Ax = −74.53 lb
A x = 74.5 lb

ΣFy = Σ( Fy )eff : Ay + P − W = mat
Ay = W + mat − P = 12 +
12
(15) − 9.02 = 8.57 lb
32.2
A y = 8.57 lb 
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1538
PROBLEM 16.98
A drum of 60-mm radius is attached to a disk of 120-mm radius.
The disk and drum have a total mass of 6 kg and a combined radius
of gyration of 90 mm. A cord is attached as shown and pulled with
a force P of magnitude 20 N. Knowing that the disk rolls without
sliding, determine (a) the angular acceleration of the disk and the
acceleration of G, (b) the minimum value of the coefficient of static
friction compatible with this motion.
SOLUTION
a = rα = (0.12 m)α
I = mk 2
= (6 kg)(0.09 m) 2
I = 48.6 × 10−3 kg ⋅ m 2
ΣM C = Σ( M C )eff : (20 N)(0.12 m) = (ma ) r + I α
2.4 N ⋅ m = (6 kg)(0.12 m) 2 α + 48.6 × 10−3 kg ⋅ m 2
2.4 = 135.0 × 10−3 α
α = 17.778 rad/s 2
(a)
α = 17.78 rad/s 2

a = 2.13 m/s 2

a = rα = (0.12 m)(17.778 rad/s 2 )
= 2.133 m/s 2
(b)
ΣFy = Σ( Fy )eff :
N − mg = 0
N = (6 kg)(9.81 m/s 2 )
N = 58.86 N
ΣFx = Σ( Fx )eff : 20 N − F = ma
20 N − F = (6 kg)(2.133 m/s 2 ) F = 7.20 N
( μ s ) min =
F
7.20 N
=
N 58.86 N
( μ s ) min = 0.122 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1558
PROBLEM 16.117
The ends of the 20-lb uniform rod AB are attached to collars of
negligible mass that slide without friction along fixed rods. If the rod is
released from rest when θ = 25°, determine immediately after release
(a) the angular acceleration of the rod, (b) the reaction at A, (c) the
reaction at B.
SOLUTION
Kinematics: Assume α
ω=0
] + [4α
a B = a A + a B/A = [ a A
25°]
aB = (4α ) cos 25° = 3.6252α
a A = (4α )sin 25° = 1.6905α
] + [2α
aG = a A + aG/ A = [ a A
aG = [1.6905α
] + [2α
ax = (aG ) x = [1.6905α
25°]
25°]
] + [0.84524α
]
ax = 0.84524α
a y = [2α cos 25° ] = 1.8126α
We have found for α
ax = 0.84524α
a y = 1.8126α
Kinetics:
I =
1
1
mL2 = m(4 ft) 2
12
12
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1588
PROBLEM 16.117 (Continued)
(a)
Angular acceleration.
ΣM E = Σ( M E )eff :
mg (1.8126 ft) = I α + max (0.84524 ft) + ma y (1.8126 ft)
mg (1.8126) =
1
m(4) 2 α + m(0.84524)2 α + m(1.8126) 2α
12
g (1.8126) = 5.3333α
(b)
α = 0.33988g

ΣFy = Σ( Fy )eff : A − mg = −ma y = m(1.8126α )
 20 
A − 20 = − 
 (1.8126)(10.944)
 32.2 
A = 20 − 12.321
= 7.6791 lb
(c)
α = 10.944 rad/s 2
ΣFx = Σ( Fx )eff :
A = 7.68 lb 
B = max = m(0.84524α )
20
(0.84524)(10.944)
32.2
B = 5.7453 lb
B=

B = 5.75 lb

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1589
PROBLEM 16.125
The 7-lb uniform rod AB is connected to crank BD
and to a collar of negligible weight, which can slide
freely along rod EF. Knowing that in the position
shown crank BD rotates with an angular velocity of
15 rad/s and an angular acceleration of 60 rad/s2,
both clockwise, determine the reaction at A.
SOLUTION
Crank BD:
 4 
ω BD = 15 rad/s, v B =  ft  (15 rad/s) = 5 ft/s
 12 
α BD = 60 rad/s 2
 4 
(a B ) x = 
ft  (60 rad/s 2 ) = 20 ft/s 2
 12 
 4 
(a B ) y = 
ft  (15 rad/s) 2 = 75 ft/s 2
12


Rod AB:
Velocity: Instantaneous center at C.
 25 
CB = 
ft  / tan 30° = 3.6084 ft
 12 
ω AB =
vB
5 ft/s
=
= 1.3856 rad/s
CB 3.6084 ft
Acceleration:
(a A/B )t = ( AB )α AB =
25
α AB
12
 25 
2
(a A /B ) n = ( AB)ω AB
=   (1.3856)2 = 4 ft/s 2
 12 
(aG/B )t = (GB)α AB =
12.5
α AB
12
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1602
PROBLEM 16.125 (Continued)
(aG /B ) n
 12.5 
2
2
2
= (GB)ω AB
=
 (1.3856) = 2 ft/s
12


a A = a B + a B/A = a B + (a B/A )t + (aG/A ) n
[a A
30°] = [20
a A cos 30° = 20 − 4;
(18.475) sin 30° = 75 −
 25
] + [75 ] +  α AB
 12
a A = 18.475 ft/s 2

 + [4

]
30°
25
α AB ; α AB = 31.566 rad/s 2
12
a = a B + aG/B = a B + (aG/B )t + (aG/B )n
a = [20
] + [75 ] + [ 12.5
(31.566) ] + [2
12
ax = 20 − 2 = 18;
]
ax = 18 ft/s 2
a y = 75 − 32.881 = 42.119; a y = 42.119 ft/s 2
2
Kinetics:
I =
1
7 lb  25 
m( AB )2 =
ft = 0.078628 slug ⋅ ft 2
12
12(32.2)  12 
 25 
 12.5 
 12.5 
ft  − mg 
ft  = − I α AB + ma y 
ft 
ΣM B = Σ( M B )eff : ( A sin 60°) 
 12 
 12

 12

 12.5 
 7

 12.5 
1.8042 A − (7 lb) 
ft  = −(0.078628 slug ⋅ ft 2 )(31.566 rad/s 2 ) + 
slug  (42.119 ft/s 2 ) 
ft 
 12

 32.2

 12

1.8042 A − 7.2917 = −2.4820 + 9.5378
A = 7.9522 lb
A = 7.95 lb
60° 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1603
PROBLEM 16.134
Two 8-lb uniform bars are connected to form the linkage shown.
Neglecting the effect of friction, determine the reaction at D
immediately after the linkage is released from rest in the position
shown.
SOLUTION
Kinematics:
Bar AC:
Rotation about C
 15 
a = ( BC )α =  ft  α
 12 
a = 1.25α
sin θ =
15 in.
θ = 30°
30 in.
Bar BC:
a D/B = Lα
Must be zero since aD
α BD = 0 and aBD = a
Kinetics:
Bar BD
ΣFy = Σ( Fy )eff : By − W = − ma
8 lb
(1.25α )
32.2
By = 8 − 0.3105α
By − 8 lb = −
(1)
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1621
PROBLEM 16.134 (Continued)
ΣM B = Σ( M B )eff : D (2.165 ft) − W (0.625 ft) = − ma (0.625 ft)
8 lb
(1.25α )(0.625 ft)
32.2
D = 2.309 − 0.08965α
D (2.165 ft) − (8 lb)(0.625 ft) = −
(2)
1
m( AC ) 2
12
1 8 lb
=
(2.5 ft) 2
12 32.2
= 0.1294 lb ⋅ ft ⋅ s 2
I =
Bar AC:
ΣM C = Σ( M C )eff : W (1.25 ft) + By (1.25 ft) = I α + m(1.25α )(1.25)
Substitute from Eq. (1) for By
8
(1.25) 2 α
32.2
10 + 10 − 0.3881α = 0.1294α + 0.3882α
20 = 0.9057α
8(1.25) + (8 − 0.3105α )(1.25) = (0.1294)α +
α = 22.08 rad/s 2
Eq. (2),
D = 2.309 − 0.08965α
= 2.309 − 0.08965(22.08)
= 2.309 − 1.979

D = 0.330 lb 
D = 0.330 lb

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1622