solution_manual_vme_9e_chapter_16

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CHAPTER 16CHAPTER 16
3
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PROBLEM 16.1
A conveyor system is fitted with vertical panels, and a 300-mm rod AB 
of mass 2.5 kg is lodged between two panels as shown. Knowing that the 
acceleration of the system is 1.5 m/s2 to the left, determine (a) the force 
exerted on the rod at C, (b) the reaction at B.
SOLUTION
Geometry d
b
=
∞
=
= =
200
20
212 84
0 15 51 3
 mm
 mm
 mm) 20 mm
cos
.
( . sin .∞
 
Weight W mg= = =( . ) . .2 5 9 81 24 525 kg m/s N2 
+
S SM M C W maB B= - = -( ) : ( . ( . ) ( . )eff mm) mm mm212 84 51 3 140 95
 
C W ma
C a
C
= -
= -
=
0 241 0 6622
0 241 24 525 0 6622 2 5
5 911
. .
. ( . ) . ( . )( )
.
 N kg
 N -1 656. a (1)
The given acceleration is a = 1 5 2. m/s 
(a) Force at C.
From Eq. (1): C = -5 911 1 656 1 5. . ( . ); C = 3.43 N 20° �
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PROBLEM 16.1 (Continued)
(b) Reaction at B.
 
S SF F B W C
B
y y y
y
= - + ∞ =
= - ∞ =
( ) : sin
. ( . )sin .
eff
 N N
20 0
24 525 3 43 20 23 355 N ≠
 S SF Fx x= ( ) :eff B C max - ∞ =cos20
 Bx - =( . )cos ( . )( . )3 43 20 2 5 1 5
2N kg m/s∞
 Bx = + = ¨3 22 3 75 6 97. . . N B = 24 4. N 73.4∞ �
+
+
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PROBLEM 16.2
A conveyor system is fitted with vertical panels, and a 300-mm rod AB of 
mass 2.5 kg is lodged between two panels as shown. If the rod is to remain 
in the position shown, determine the maximum allowable acceleration of the 
system.
SOLUTION
Geometry
Weight W mg= = =( . ) . .2 5 9 81 24 525 kg m/s N2 
 
+ S SM M C W ma
C
B B= - = -
=
( ) : ( . ( . ) ( . )
.
eff mm) mm mm212 84 51 3 140 95
0 241WW ma
C a
C
-
= -
= -
0 6622
0 241 24 525 0 6622 2 5
5 911 1 6
.
. ( . ) . ( . )( )
. .
 N kg
 N 556a (1)
6
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PROBLEM 16.2 (Continued)
Maximum allowable acceleration.
For amax , C = 0 
From Eq. (1), C a
a
= -
= -
5 911 1 656
0 5 911 1 656
. .
. . max
 N
 N
 
 amax .= 3 57
2 m/s amax .= 3 57
2 m/s ¨ �
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SOLUTION
For the board to remain in the position shown, we need B ≥ 0. When the board is just about to start rotating 
about A, we have B = 0.
+ S SM MA A= ( ) :eff W
W
g
a( )cos ( )sin1 78 1 78 m m∞ = ∞
 a g= ∞ = ∞cot ( . )cot78 9 81 78 m/s2 a = 2 09. m/s2 �
PROBLEM 16.3
A 2-m board is placed in a truck with one end resting against a 
block secured to the floor and the other leaning against a vertical 
partition. Determine the maximum allowable acceleration of the 
truck if the board is to remain in the position shown.
8
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PROBLEM 16.4
A uniform rod BC weighing 4 kg is connected to a collar A by a 250 mm 
cord AB. Neglecting the mass of the collar and cord, determine (a) the 
smallest constant acceleration aA for which the cord and the rod lie in a 
straight line (b) the corresponding tension in the cord.
SOLUTION
Geometry and kinematics:
Distance between collar and floor = AD = + =250 350 600 mm mm mm
When cord and rod lie in a straight line:
AC AB BC
AD
AC
= + = + =
= =
= ∞
250 400 650
600
650
22 62
 mm mm mm
cos
.
q
q
Kinetics
(a) Acceleration at A.
 
+
S SM MC C= ( ) :eff W CG ma CG( )sin ( )cosq q=
 ma mg= tanq
 a g= = ∞tan ( . ) tan .q 9 81 22 62 m/s2
 a aA = = 4 09. m/s
2 Æ �
(b) Tension in the cord.
 
+ S SF F T ma mg
T
mg
x x= = =
= =
∞
( ) : sin tan
cos
( )( . )
cos
eff
22.62
q q
q
4 9 81
 T = 42 5. N 67.4∞ �
9
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PROBLEM 16.5
Knowing that the coefficient of static friction between the 
tires and the road is 0.80 for the automobile shown, determine 
the maximum possible acceleration on a level road, assuming 
(a) four-wheel drive, (b) rear-wheel drive, (c) front-wheel drive.
SOLUTION
(a) Four-wheel drive:
+SF N N W N N W mgy A B A B= + - = + = =0 0:
 Thus: F F N N N N W mgA B k A k B k A B k+ = + = + = =m m m m( ) .0 80 
+ S SF F F F max x A B= + =( ) :eff
 0 80. mg ma=
 a g= =0 80 0 80 9 81 2. . ( . ) m/s a = Æ7 85. m/s2 �
(b) Rear-wheel drive:
+ S SM M W N maB B A= - = -( ) : ( ) ( ) ( )eff mm mm mm1000 2500 500
N W maA = +0 4 0 2. .
 Thus: F N W ma mg maA k B= = + = +m 0 80 0 4 0 2 0 32 0 16. ( . . ) . . 
+ S SF F F max x A= =( ) :eff
0 32 0 16
0 32 0 84
. .
. .
mg ma ma
g a
+ =
=
 a =
0 32
0 84
9 81
.
.
( . ) m/s2 a = Æ3 74 2. m/s �
10
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PROBLEM 16.5 (Continued)
(c) Front-wheel drive:
+ S SM M N W maA A B= - = -( ) : ( ) ( ) ( )eff mm mm mm2500 1500 500
N W maB = -0 6 0 2. .
 Thus: F N W ma mg maB k B= = - = -m 0 80 0 6 0 2 0 48 0 16. ( . . ) . . 
+ S SF F F max x B= =( ) :eff
 
0 48 0 16
048 1 16
0 48
1 16
9 81 2
. .
. .
.
.
( . )
mg ma ma
g a
a
- =
=
= m/s a = Æ4 06. m/s2 �
11
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PROBLEM 16.6
For the truck of Sample Problem 16.1, determine the distance through which the truck will skid if (a) the rear-
wheel brakes fail to operate, (b) the front-wheel brakes fail to operate.
SOLUTION
(a) If rear-wheel brakes fail to operate
+ S SM M N W ma
N W
A A B
B
= - =
= +
( ) : ( . ) ( . ) ( . )
.
.
.
.
eff m m m3 6 1 5 1 2
1 5
3 6
1 2
3 6
WW
g
a W
W
g
a= 5
12
1
3
+
 
S SF F F ma
N
W
g
a
W
W
g
a
W
g
a
a
x x B
k B
= =
=
+
Ê
ËÁ
ˆ
¯̃
=
=
( ) :
.
.
eff
m
0 728
5
12
1
3
0 728 5512
29 81
1 0 24267
( )
-
( . )
.
 m/s
 a = ¨3 9292 2. m/s
 Uniformly accelerated motion
 v v ax x2 0
2 22 0 10 2 3 9292= + = -( ) ( . m/s m/s )2 x = 12 73. m �
(b) If front-wheel brakes fail to operate
+ S SM M W N ma
N W
W
g
a
B B A
A
= - =
= -
( ) : ( . ) ( . ) ( . )eff m m m1 5 3 6 1 2
7
12
1
3
+
12
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PROBLEM 16.6 (Continued)
 
+ S SF F F ma
N
W
g
a
W
W
g
a
W
g
a
a
x x A
k A
= =
=
-
Ê
ËÁ
ˆ
¯̃
=
=
( ) :
.
.
eff
m
0 728
7
12
1
3
0 728 7712
29 81
1 0 24267
( )
+
( . )
.
 m/s
 a = 3 35244 2. m/s ¨
Uniformly accelerated motion
 v v ax x2 0
2 22 0 10 2 3 35244= + = -( ) ( . m/s m/s )2 x = 14 91. m �
13
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PROBLEM 16.7
A 20-kg cabinet is mounted on casters that allow it to move freely 
(m = 0) on the floor. If a 100-N force is applied as shown, determine 
(a) the acceleration of the cabinet, (b) the range of values of h for which 
the cabinet will not tip.
SOLUTION
(a) Acceleration
+ S SF F ma
a
x x= =
=
( ) :
(
eff N
 N kg)
100
100 20
 a = Æ5 00 2. m/s �
(b) For tipping to impend : A = 0
+ S SM M h mg ma
h
B B= - =
-
( ) : ( ) ( . ) ( . )
( ) ( )(
eff N m m
 N kg
100 0 3 0 9
100 20 9.. )( . ) ( )( . )
.
81 0 3 100 0 9
1 489
 m/s m N m
 m
2 =
=h
 For tipping to impend : B = 0
+ S SM M h mg ma
h
A A= + =
+
( ) : ( ) ( . ) ( . )
( ) ( )(
eff N m m
 N kg
100 0 3 0 9
100 20 99 81 0 3 100 0 9
0 311
. )( . ) ( )( . )
.
 m/s m N m
 m
2 =
=h
 Cabinet will not tip: 0 311 1 489. . m m£ £h �
14
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PROBLEM 16.8
Solve Problem 16.7, assuming that the casters are locked and slide on 
the rough floor (mk = 0.25).
PROBLEM 16.7 A 20-kg cabinet is mounted on casters that allow it to 
move freely (m = 0) on the floor. If a 100-N force is applied as shown, 
determine (a) the acceleration of the cabinet, (b) the range of values 
of h for which the cabinet will not tip.
SOLUTION
(a) Acceleration.
+SF N N W
N W mg
y A B
A B
= + - =
+ =
0 0:
 But, F N= m , thus F F mgA B+ = m( )
 
+ S SF F F F ma
mg ma
x x A B= - + =
- =
-
( ) : ( )
. ( (
eff N
 N
 N kg)
100
100
100 0 25 20
m
99 81 20
2 548 2
. ) ( )
.
 m/s kg
 m/s
2 =
=
a
a a = 2 55 2. m/s Æ �
(b) Tipping of cabinet.
W mg
W
= =
=
( )( . )
.
20 9 81
196 2
2 kg m/s
 N
 For tipping to impend : N A = 0.
+ S SM M h W ma
h
B B= - =
-
( ) : ( ( . ( . )
( ) ( . )
eff N) m) m
 N N
100 0 3 0 9
100 196 2 (( . ) ( )( . )
.
0 3 20 2 548
1 047
 m kg m/s )(0.9 m
 m
2=
=h
15
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PROBLEM 16.8 (Continued)
 For tipping to impend : N B = 0
+ S SM M h W ma
h
A A= + =
+
( ) : ( ) ( . ) ( . )
( ) ( . )
eff N m m
 N N
100 0 3 0 9
100 196 2 (( . ) ( )( . )( . )
.
0 3 20 9 81 0 9
0 130
 m kg m/s m
 m (impossible)
2=
= -h
 Cabinet will not tip: h £ 1 047. m �
16
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PROBLEM 16.9
The forklift truck shown weighs 1125 kg and is used to 
lift a crate of weight W = 1250 kg. Knowing that the 
truck is at rest, determine (a) the upward acceleration 
of the crate for which the reactions at the rear wheels 
B are zero, (b) the corresponding reaction at each of 
the front wheels A.
SOLUTION
(a) Acceleration of crate for B = 0.
 
+ S SM M aA A= - = -( ) : ( ( . ( . ( )( . ( . ( ) (eff ) ) ) ) )1250 9 81 0 9 1125 9 81 1 2 1250 00 9
7500 9000
2500
3
1
5
9 81
5
. )
( )
.
- =
= =
g
a
a g a = ≠1 962. m/s2 �
(b) Reaction at A:
+S SF F A
A
g
y y= - - =
Ê
ËÁ
ˆ
¯̃
- = ÊËÁ
( ) :eff g g
g
g
2 1250 1125 1250
5
2 2375 1250
5
ˆ̂
¯̃
=2 2625A g = 25751.25 N
 For one wheel: A = ≠12880 N �
17
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PROBLEM 16.10
The forklift truck shown weighs 1125 kg and is 
used to lift a crate of mass m = 1250 kg. The truck 
is moving to the left at a speed of 3 m/s when the 
brakes are applied on all four wheels. Knowing that 
the coefficient of static friction between the crate and 
the fork lift is 0.30, determine the smallest distance in 
which the truck can be brought to a stop if the crate is 
not to slide and if the truck is not to tip forward.
SOLUTION
Assume crate does not slide and that tipping impends about A. ( )B = 0
+ S SM MA A= ( ) :eff
 
( ( . ( ) ( . ) ( . ( .
;
1250 0 9 1125 1 2 1250 1 2 1125 0 9) ) ) )
 
g g a a- = - -
=a
g
a
6
67
== =6 9.8167
¥
0 87851. m/s2 a = Æ0 879 2. m/s
Uniformly accelerated motion
 v v ax x2 0
2 2 22 0 3 2 0 87851= + = -; ( ) ( . ) m/s m/s x = 5 12. m �
Check whether crate slides
N W
F ma
F
N
a
g
=
=
= = = =mreq
6
67
0 09.
 mreq = 0 09 0 30. . .< The crate does not slide. �
18
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PROBLEM 16.11
The support bracket shown is used to transport a cylindrical can from one 
elevation to another. Knowing that ms = 0 25. between the can and the 
bracket, determine (a) the magnitude of the upward acceleration a for which 
the can will slide on the bracket, (b) the smallest ratio h/d for which the can 
will tip before it slides.
SOLUTION
(a) Sliding impends
 + S SF Fy x= ( ) :eff F ma= ∞cos30
+S SF F N mg may y= - = ∞( ) : sineff 30
N m g a= + ∞( sin )30
ms
F
N
ma
m g a
g a a
=
= ∞
+ ∞
+ ∞ = ∞
0 25
30
30
30 4 30
.
cos
( sin )
sin cos
 
a
g
=
∞ - ∞
1
4 30 30cos sin
 a = 0 337. g 30∞ �
(b) Tipping impends
S SM M F h N dG G=
Ê
ËÁ
ˆ
¯̃
- ÊËÁ
ˆ
¯̃
=( ) :eff 2 2
0
F
N
d
h
=
 m = =F
N
d
h
; 0 25. ; 
h
d
= 4 00. �
19
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PROBLEM 16.12
Solve Problem 16.11, assuming that the acceleration a of the bracket is 
directed downward.
PROBLEM 16.11 The support bracket shown is used to transport a cylindrical 
can from one elevation to another. Knowing that ms = 0 25. between the can 
and the bracket, determine (a) the magnitude of the upward acceleration a for 
which the can will slide on the bracket, (b) the smallest ratio h/d for which the 
can will tip before it slides.
SOLUTION
(a) Sliding impends:
 + S SF Fx x= ( ) :eff F ma= ∞cos30 
+S SF F N mg may y= - = - ∞( ) : sineff 30
N m g a= - ∞( sin )30
ms
F
N
ma
m g a
g a a
=
= ∞
- ∞
- ∞ = ∞
0 25
30
30
30 4 30
.
cos
( sin )
sin cos
 
a
g
=
∞ + ∞
=1
4 30 30
0 25226
cos sin
. a = 0 252. g �
(b) Tipping impends:
+
S SM M F h W d F
N
d
hG G
= ÊËÁ
ˆ
¯̃
= ÊËÁ
ˆ
¯̃
=( ) :eff 2 2
 m = =F
N
d
h
; . ;0 25 
h
d
= 4 00. �
20
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PROBLEM 16.13
A completely filled barrel and its contents have a combined 
weight of 100 kg. A cylinder C is connected to the barrel 
at a height h = 550 mm as shown. Knowing ms = 0 40. and 
mk = 0 35. , determine the maximum weight of C so the 
barrel will not tip.
SOLUTION
Kinematics: Assume that the barrel is sliding, but not tipping.
a a= =0 G aÆ
Since the cord is inextensible, aC a= Ø
Kinetics: Draw the free body diagrams of the barrel and the cylinder.
The barrel is sliding. F N NF k= =m 0 35.
Assume that tipping is impending, so that the line of action of the reaction on the bottom of the barrel passes 
through Point B.
e = 250 mm
For the barrel. 
+
 SF N W N Wy B B= - = =0 0:
+ SM Ne T N
T
e
N
G = - - =
= - = -
0 100 450 0 35 0
450 0 35
100
250 157 5
1
: ( )( . )
( )( . ) .
000
0 925W WB B= .
+ SF
W
g
a T N
W
g
a
a
g
T
W
N
W
x
B B
B B
= - =
= - = - =
: .
. . . .
0 35
0 35 0 925 0 35 0 575
21
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PROBLEM 16.13 (Continued)
For the cylinder: +SF
W
g
a W T W
a
g
C
C C= - =:
 W
T
a
g
W
C
B=
-
=
-
=
1
0 925
1 0 575
2 1765 100 9 81
.
.
( . )( )( . ) WC = 2140 N �
22
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PROBLEM 16.14
A uniform rectangular plate has a mass of 5 kg and is held in 
position by three ropes as shown. Knowing that q = 30∞, determine, 
immediately after rope CF has been cut, (a) the acceleration of the 
plate, (b) the tension in ropes AD and BE.
SOLUTION
(a) Acceleration + 30 30∞ = ∞ =S SF F mg maeff : sin
 a g= =0 5 4 905. . m/s2 a = 4 91. m/s2 30∞ �
(b) Tension in ropes
+
 
S SM M T mg maA A B= ∞ - = - ∞( ) : ( cos )( . ) ( . ) (cos )( . )eff m m m30 0 3 0 15 30 0 12 -- ∞
- = -
ma
TB
(sin )( . )
. ( )( . )( . ) ( )(
30 0 15
0 2598 5 9 81 0 15 5
m
kg m/s m kg2 44 905 0 1039 0 07
0 2598 7 3575 4 388
. )( . . )
. . .
m/s2 +
- =TB
 TB = +11 43. N TBE = 11 43. N �
+ 10 11 43 30 0
11 43 5 9 81 30
∞ = + - ∞ =
+ -
S SF F T mg
T
A
A
eff N
N kg
: . cos
. ( )( . )cos ∞∞ =
+ - =
0
11 43 42 48 0TA . .N N
 TA = 31 04. N TAD = 31 0. N �
23
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PROBLEM 16.15
A uniform rectangular plate has a mass of 5 kg and is held 
in position by three ropes as shown. Determine the largest 
value of q for which both ropes AD and BE remain taut 
immediately after rope CF has been cut.
SOLUTION
S SF F mg ma
a g
= =
=
eff : sin
sin
q
q
 
+
S SM M mg ma ma
mg
B B= = +( ) : ( . ) cos ( . ) sin ( . )
( . )
eff m m m0 15 0 12 0 15
0 15
q q
== +
= -
- =
m g( sin )( . cos . sin )
. sin cos sin
sin .
q q q
q q q
q
0 12 0 15
1 0 8
1 02
2
88
0 8
1 0 8
1 25
2
sin cos
cos . sin cos
.
sin
cos
tan .
q q
q q q
q
q
q
=
=
= q = ∞51 3. �
24
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PROBLEM 16.16
A uniform circular plate of mass 3 kg is attached to two links AC and BD of the 
same length. Knowing that the plate is released from rest in the position shown, in 
which lines joining G to A and B are, respectively, horizontal and vertical, determine 
(a) the acceleration of theplate, (b) the tension in each link.
SOLUTION
(a) Acceleration.
+
S SF F mg ma= ∞ =eff : cos75
 a g= ∞ =cos .75 2 5390 2m/s a = 2 54 2. m/s 15∞ �
(b) Tension in each link.
+ S SM M F r F r ma r
F
B B A A
A
= ∞ + ∞ = ∞
∞ +
( ) : ( cos ) ( sin ) ( sin )
(cos
eff 75 75 75
75 ssin ) ( )( . )sin75 3 2 539 752∞ = ∞kg m/s
 FA = 6 01. FA = 6 01. N Tension �
 + 75 eff∞ =S SF F : F F mgA B+ - ∞ =sin 75 0
3 3 9 81 75 02kg kg m/s+ - ∞ =FB ( )( . )sin
 FB = 22 42. N FBD = 22 4. N Tension �
25
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PROBLEM 16.17
Three bars, each of 4 kg, are welded together and are pin-connected 
to two links BE and CF. Neglecting the weight of the links, 
determine the force in each link immediately after the system is 
released from rest.
SOLUTION
Mass center of ABCD is at G
W mg=
40 50 6 3057∞ = ∞ = =S SF F mg maeff
2m/s: cos .a 40°
 ( )( . )cos12 9 81 50∞ = ma ma = 75 669. N
+ S SM MB B= ( ) :eff
( sin )( ) ( )( . )( ) sin ( ) cosF ma maCF 50 400 12 9 81 200 40 200 4∞ - = - ∞ -mm mm 00
400
3
∞ÊËÁ
ˆ
¯̃
mm
306 418 23544 128 558 102 139
306 418 23544 75
. ( . . )
. ( .
F ma
F
CF
CF
- = - +
= - 6669 230 697)( . )
 FCF = +19 866. N FCF = 19 87. N compression �
50°S SF F FBE= + - ∞ =eff : . ( )( . )sin19 866 12 9 81 50 0
 FBF = +70 313. N FBE = 70 3. N compression �
26
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PROBLEM 16.18
At the instant shown, the angular velocity of links BE and CF is 6 rad/s counterclockwise and is decreasing at 
the rate of 12 rad/s2. Knowing that the length of each link is 300 mm and neglecting the weight of the links, 
determine (a) the force P, (b) the corresponding force in each link. The mass of rod AD is 6 kg.
SOLUTION
Links: a rn = =w
2 20 3 6( . )( )m rad/s
 an = 10 80. m/s
2 60∞
 a rt = =a ( . )( )0 3 12m rad/s
2
 at = 3 6. m/s
2 30∞
Bar AD is in translation. a a aB D= =
(b) Force in each link.
+ S SM M F FG G CF BE= ∞ - ∞( ) : cos ( . ) cos ( . )eff m m60 0 3 60 0 3
F FCF BE=
+ S SM M F mgB B CF= ∞ -( ) : sin ( . ) ( . )eff m m60 0 6 0 3
= + ∞ + ∞
-
ma ma Ft n CFsin ( . ) sin ( . ) .
( )( . )
30 0 3 60 0 3 0 5196
6 9 81
m m
kg m/s2 (( . )
( )( . )sin ( . )
( )( . )(sin
0 3
6 3 6 30 0 3
6 10 80 60
m
kg m/s m
kg m/s
2
2
= + ∞
+ ∞∞ -
= + +
)( . ) . .
. .
0 3 0 5196 17 658
3 24 16 836
m FCF
 FCF = +72 62. N F FCF BE= = 73 6. N tension �
27
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PROBLEM 16.18 (Continued)
(a) Force P.
+ S SF F F F F ma max x BE CF P t n= + ∞ + = - ∞ + ∞( ) : ( )cos cos cos
( .
eff 60 30 60
2 72 622 60 6 3 6 30 6 10 8 60
7
N kg m/s kg m/s2 2)cos ( )( . )cos ( )( . )cos∞ + = - ∞ + ∞FP
22 62 18 706 37 40. . .+ = - +FP
 FP = -58 9. N P = ¨58 9. N �
28
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PROBLEM 16.19
The 7.5-kg rod BC connects a disk centered at A to crank CD. 
Knowing that the disk is made to rotate at the constant speed of 
180 rpm, determine for the position shown the vertical components 
of the forces exerted on rod BC by the pins at B and C.
SOLUTION
We first determine the acceleration of Point B of disk.
w
w
w
= =
=
= =
=
180 18 85
0 2 18
2
rpm rad/s
Since constant
m) 
.
( )
( . (
a a rB B n
.. )85 2rad/s
aB = 71 0645. m/s
2 60∞
Since rod BC is in translation.
a = =a m/s271 0645. 60∞
Vertical components of forces at B and C.
 + S SM M C W maB B y= - = - ∞( ) : ( . ) ( . ) sin ( . )eff 0 75 0 375 60 0 375
Since W ma= = = =( . )( . ) . ( . )( . ) .7 5 9 81 73 575 7 5 71 0645 532 984N and N
 
0 75 0 375 73 575 532 984 60 0 375
2 461 5
. ( . )( . ) ( . )(sin )( . )
.
C
C
y
y
- = - ∞
= - 778 73 575 388 003+ = - = -. . N, 194.001 NCy Cy = -194 0. N Ø �
 +S SF Fy y= ( ) :eff B W C may y- + = - ∞sin60 
B W C may y= - - ∞ = + -sin . .60 73 575 194 461 578
 By = -194 003. N B y = Ø194 0. N �
29
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PROBLEM 16.20
The triangular weldment ABC is guided by two pins that slide 
freely in parallel curved slots of radius 150 mm cut in a vertical 
plate. The weldment weighs 8 kg and its mass center is located 
at Point G. Knowing that at the instant shown the velocity of 
each pin is 750 mm/s downward along the slots, determine 
(a) the acceleration of the weldment, (b) the reactions at A and B.
SOLUTION
Slot: v = 750 mm/s = 0.75 m/s
a
v
rn
= = =
2 20 75
0 15
3 75
( . )
( .
.
m/s
m)
m/s2
an = 3 75. m/s
2 30°
at ta= 60°
Weldment is in translation an = 3 75. m/s
2
60° S SF F mg mat= ∞ =eff : cos30
at = 8 4957. m/s
2 60∞
(a) Acceleration
b = = = ∞
= +
= +
- -tan tan
.
.
.
( . ) ( .
1 1
2 2 2
2
3 75
8 4957
23 82
8 4957 3
a
a
a
n
t
t na a
775 2)
 a = 9 2865. m/s2 83.8° a = 9 29. m/s2 83.8∞ �
30
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PROBLEM 16.20 (Continued)
(b) Reactions
ma = =( . ) .8 9 2865 74 292kg) ( m/s N2
 + S SM MA A= ( ) :eff
B(cos )( . ) ( )( . )( . ) ( . )cos( . )( .30 0 225 8 9 81 0 15 74 292 83 82 0 075∞ - = ∞ )) - (74.292)sin(83.82 )(0.15)
0 19486 11 772 0 59983 11 079. . . .B - = -
 B = +6 6347. N B = 6 63. N 30° �
+ S SF F A B max x= ∞ + ∞ = ∞( ) : cos cos coseff 3.8230 30 8
A
A
cos ( . )cos ( . )cos .
cos .
30 6 6347 30 74 292 83 82
30 5 7458 7
∞ + ∞ = ∞
∞ + =
N N
..9977
 A = 2.6003 N A = 2 60. N 30° �
31
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PROBLEM 16.21*
Draw the shear and bending-moment diagrams for the vertical rod 
AB of Problem 16.17.
PROBLEM 16.17 Three bars, each of mass 4 kg, are welded 
together and are pin-connected to two links BE and CF. Neglecting 
the weight of the links, determine the force in each link immediately 
after the system is released from rest
SOLUTION
Distributed weight per unit length:
W = =
( )( . )
( .
.
4 9 81
0 4
98 1
N
m)
N/m
Horizontal component of effective forces
W
g
ax = ∞
=
98 1
9 81
6 3057 40
48 3045
2
2.
.
( . )cos
.
N/m
m/s
m/s
N/m
S SM M MB B B= =( ) : ( . )
( . )
eff N/m
m
48 3045
0 4
2
2
MB = ◊3 86436. N m
 + S SF F= ( ) :eff VB = ( . )( . )48 3045 0 4N/m m 
VB = 19 3218. N
We note that V MA A= = 0 and sketch the V and M diagrams.
 | | .maxV = 19 32 N �
 | | .maxM = ◊3 86 N m �
32
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PROBLEM 16.22*
Draw the shear and bending-moment diagrams for the connecting 
rod BC of Problem 16.19.
PROBLEM 16.19 The 7.5-kg rod BC connects a disk centered 
at A to crank CD. Knowing that the disk is made to rotate at the 
constant speed of 180 rpm, determine for the position shown the 
vertical components of the forces exerted on rod BC by the pins 
at B and C.
SOLUTION
We first determine the acceleration of Point B of disk.
w
w
w
= =
=
= =
=
180 18 85
0 2 18
2
rpm rad/s
Since constant
m) 
.
( )
( . (
a a rB B n
.. )85 2rad/s
aB = 71 0645. m/s
2 60∞
Since rod BC is in translation.
a = =a m/s271 0645. 60∞
Vertical components of forces at B and C.
 + S SM M C W maB B y= - = - ∞( ) : ( . ) ( . ) sin ( . )eff 0 75 0 375 60 0 375
Since W ma= = = =( . )( . ) . ( . )( . ) .7 5 9 81 73 575 7 5 71 0645 532 984N and N
 
0 75 0 375 73 575 532 984 60 0 375
2 461 5
. ( . )( . ) ( . )(sin )( . )
.
C
C
y
y
- = - ∞
= - 778 73 575 388 003+ = - = -. . N, 194.001 NCy Cy = -194 0. N Ø �
 +S SF Fy y= ( ) :eff B W C may y- + = - ∞sin60 
B W C may y= - - ∞ = + -sin . .60 73 575 194 461 578
 By = -194 003. N B y = Ø194 0. N
 ay = ∞ =71 0645 60 61 544. sin . m/s
2 a y = 61 544. m/s
2 �
33
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PROBLEM 16.22* (Continued)
Distributed weight per unit length = = =W
( . )( .
( .
.
7 5 9 81
0 75
9
N)
m)
8 1 N/m
Distributed mass per unit length = = =
W
g
98 1
9 81
10
.
.
kg/m
Vertical component of effective forces = =W
g
ay ( ( . )10 61 544kg/m) m/s
2
= 615 44. N/m
+ S SF F B Cy y y y= + + =( ) : ( . )( . ) ( . )( . )eff m N/m m m0 75 98 1 0 75 615 44 N/
B Cy y+ = 388 005. N
From symmetry. B Cy y= = 194 003. N B Cy y= = Ø194 0. N
Maximum value of BM occurs at G, where V = 0:
| |maxM = Area under V-diagram from B to G =
Ê
ËÁ
ˆ
¯̃
◊
1
2
194 003
0 75
2
( . )
.
N
m
N m= 36.376
 | | .maxM = ◊36 4 N m �
 VB = -194 0. N �
34
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PROBLEM 16.23
For a rigid slab in translation, show that the system of the effective forces consists 
of vectors ( )Dmi a attached to the various particles of the slab, where a is the 
acceleration of the mass center G of the slab. Further show, by computing their sum 
and the sum of their moments about G, that the effective forces reduce to a single 
vector ma attached at G.
SOLUTION
Since slab is in translation, each particle has same acceleration as G, namely a. The effective forces consist of 
( ) .Dmi a
The sum of these vectors is: S S( ) ( )D = Dm mi ia a
or since SD =m mi ,
S( )D =m mi a a
The sum of the moments about G is: S S¢¥ D = D ¢ ¥r m m ri i i i( ) ( )a a (1)
But, SD ¢= ¢ =m r mri i 0, because G is the mass center. It follows that the right-hand member of Eq. (1) is zero. 
Thus, the moment about G of ma must also be zero, which means that its line of action passes through G and 
that it may be attached at G.
35
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PROBLEM 16.24
For a rigid slab in centroidal rotation, show that the system of the effective forces 
consists of vectors -( ) ’Dmi iw
2r and ( )Dmi ( ’ )a ¥ r i attached to the various particles 
Pi of the slab, where w and a are the angular velocity and angular acceleration of the 
slab, and where r ’i denotes the position vector of the particle Pi relative to the mass 
center G of the slab. Further show, by computing their sum and the sum of their 
moments about G, that the effective forces reduce to a couple I a.
SOLUTION
For centroidal rotation: a a a r ri i t i n i i= + = ¥ ¢ - ¢( ) ( ) a w
2
Effective forces are: ( ) ( )( ) ( ) ( )( )D = D ¥ - D ¢ D ¥m a m m r mi i i i i i i i� r rw a
2
S S( ) ( )( ) ( )D D S Dm m r m ri i i i i ia = ¥ ¢ - ¢a w
2
= ¥ D ¢ - D ¢a wS S( ) ( )m m ri i i ir
2
Since G is the mass center, S( )D ¢ =m ri i 0
effective forces reduce to a couple, Summiml moments about G
S S S( ) [ ( )( )] ( )¢ ¥ D = ¢ ¥ D ¥ ¢ - ¢ ¥ D ¢r a r r r1
2m m m ri i i i i i i ia w
But, ¢ ¥ D ¢ = D ¢ ¥ ¢ =r r r ri i i i i im m( ) ( )( )w w
2 2 0
and, ¢ ¥ D ¥ ¢ = D ¢ri i i im r m r( )( ) ( )a a1
2
Thus, S S( ) ( )¢ ¥ D = D = D( ) ¢ÈÎ ˘̊r ai i i i i im m r m r12 2a aS
Since S( ) ,D ¢ =m r Ii
2
the moment of the couple is I a
36
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PROBLEM 16.25
It takes 10 min for a 3000-kg flywheel to coast to rest from an angular velocity of 300 rpm. Knowing that the 
radius of gyration of the flywheel is 900 mm, determine the average magnitude of the couple due to kinetic 
friction in the bearings.
SOLUTION
m k= = =3000 900 0 9kg mm m.
I mk= = =2 2 23000 0 9 2430( )( . ) kg m◊
w p p0 300
2
60
10= ÊËÁ
ˆ
¯̃
= rpm rad/s
w w a= +0 t
0 10 600= +p a rad/s s( )
a = -0 05236. rad/s2
M I= = ◊ - = ◊a ( )( . ) .2430 0 05236 127 235 kg m rad/s N m2 2
or M = ◊127 2. N m �
37
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PROBLEM 16.26
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
250 0 180
1 62
( )( . )
. kg m2
M I= ◊ = ◊a a: N m kg m3 5 1 62 2. ( . )
a
w p
p
=
= ÊËÁ
ˆ
¯̃
=
2 1605
3600
120
0
. rad/s (deceleration)
 rpm
2
60
 r
2
aad/s
rad/s rad/s
w w aq
p q
q
2
0
2
2 2
2
0 120 2 2 1605
32 891 10
= +
= + -
= ¥
( ) ( . )
. 33
5235 26
rad
rev= .
or q = 5230 rev �
38
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SOLUTION
Belt: F Nk= m
Disk:
 + S SF Fx x= ( )eff : N FAB- =cosq 0
 F NAB cosq = (1)
+ S SF F N F mgy y k AB= + - =( ) sineff : m q 0
 F mg NAB ksinq m= - (2)
Eq.
Eq.
:
( )
( )
tan
2
1
q
m
=
-mg N
N
k
N mg N
N
mg
k
F N
mg
k
k
k
k
tan
tan
tan
q m
q m
m
m
q m
= -
=
+
= =
+
PROBLEM 16.27
The 180-mm disk is at rest when it is placed in contact with a belt moving 
at a constant speed. Neglecting the weight of the link AB and knowing 
that the coefficient of kinetic friction between the disk and the belt is 
0.40, determine the angular acceleration of the disk while slipping occurs.
39
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PROBLEM 16.27 (Continued)
+ S SM M Fr IA A= =( )eff : a
a
m
q m
m
q m
=
= ◊
+
= ◊
+
r
I
F
r
mr
mg
g
r
k
k
k
k
1
2
2
2 tan
tan
Data: r
k
=
= ∞
=
0 18
60
0 40
.
.
m
q
m
 a = ◊
∞ +
2 9 81
0 18
0 40
60 0 40
2( . )
.
.
tan .
m/s
m
 a = 20 4 2. rad/s �
40
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PROBLEM 16.28
Solve Problem 16.27, assuming that the direction of motion of the belt is 
reversed.
PROBLEM 16.27 The 180-mm disk is at rest when it is placed in contact 
with a belt moving at a constant speed. Neglecting the weight of the link 
AB and knowing that the coefficient of kinetic friction between the disk 
and the belt is 0.40, determine the angular acceleration of the disk while 
slipping occurs.
SOLUTION
Belt: F Nk= m
Disk:
 + S SF F N F F Nx x AB AB= - =( ) cos ; coseff : q q (1)
+ S SF F F mg Ny y AB k= - - =( ) sineff : q m 0
 F mg NAB ksinq m= + (2)
Eq.
Eq.
:
( )
( )
tan
2
1
q
m
=
+mg N
N
k
N mg N N
mg
k
k
tan ;
tan
q m
q m
= + =
-
+
 S SM M Fr IA A= =( )eff : a
a
m m
q m
m
q m
= = = ◊
-
= ◊
-
Fr
r
Nr
I
r
mr
mg g
r
k k
k
k
k
1
2
2
2 tan tan
Data: r k= = ∞ =0 18 60 0 40. , , .m q m
 a = ◊
∞ -
2 9 81
0 18
0 40
60 0 40
2( . )
.
.
tan .
m/s
m
 a = 32 7. rad/s2 �
41
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PROBLEM 16.29
The 150-mm-radius brake drum is attached to a larger flywheel that 
is not shown. The total mass moment of inertia of the drum and the 
flywheel is 75 2kg m◊ . A band brake is used to control the motion of 
the system and the coefficient of kinetic friction between the belt and the 
drum is 0.25. Knowing that the 100-N force P is applied when the initial 
angular velocity of the system is 240 rpm clockwise, determine the time 
required for the system to stop. Show that the same result is obtained if 
the initial angular velocity of the system is 240 rpm counterclockwise.
SOLUTION
Lever ABCD
Static Equilibrium:
+ SM T TA B0 0 80 80 100 320 0= + - =: ( ) ( ) ( )( )mm mm N mm
 T TA B+ = 400 N (1)
w p p0 240
2
60
8= ÊËÁ
ˆ
¯̃
=rpm rad/s 
Thus a will be 
r = 0 15. m
+ S SM M T r T r IE E B A= - =( )eff : a
 T T
I
rB A
- = a (2)
42
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PROBLEM 16.29 (Continued)
Belt friction: b p
m b p
= ∞ =
= = = =
270
3
2
3 2480 25 1 178
3
2
rad
T
T
e e eB
A
k ( . ) . .
 T TB A= 3 248. (3)
Eq. (1): T T
T T
T
T
A B
A A
A
B
+ =
+ =
=
=
=
400
3 248 400
94 16
3 248 94 16
305 9
N
N
N
N
N
.
.
. ( . )
.
Eq. (2): T T
I
rB A
- =
- =
◊
=
a
a
a
305 9 94 16
75
0 15
0 423
2
2
. .
.
.
N N
kg m
m
rad/s
Uniformly accelerated motion.
 w a p0
28 0 423= =t t: rad/s rad/s( . ) t = 59 4. s �
Note: If a is reversed than TA and TB are interchanged, thus causing no change in Eq. (1) and Eq. (2). Thus from 
Eq. (3), a is not changed.
43
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PROBLEM 16.30
The 200-mm-radius brake drum is attached is a larger flywheel that is not 
shown. The total mass moment of inertia of the drum and the flywheel is 
20 2kg m◊ and the coefficient of kinetic friction between the drum and 
the brake shoe is 0.35. Knowing that the angular velocity of the flywheel 
is 360 rpm counterclockwise when a force P of magnitude 400 N is 
applied to the pedal C, determine the number of revolutions executed by 
the flywheel before it comes to rest.
SOLUTION
Lever ABC: Static equilibrium (friction force Ø)
F N Nk= =m 0 35.
+
 SM N FA = - - =0 250 50 400 250 0: mm mm N mm( ) ( ) ( )( )
250 50 0 35 100000 0
430 108
0 35 430 108
150
N N
N
F Nk
- - =
=
=
=
=
( . )
.
. ( . )
.
N
N
m
5538 N
Drum:
r = =
= ÊËÁ
ˆ
¯̃
=
200 0 2
360
2
60
12
0
0
mm m
rpm
rad/s
.
w p
w p
+
 S SM M Fr ID D= =( )eff : a
( . )( . ) ( )150 538 02 20 2 N m kg m= ◊ a
a = 1 50538 2. rad/s (deceleration)
w w aq p q2 0
2 2 22 0 12 2 1 50538= + = + -: ( ) ( . )rad/s rad/s
 
q
q
p
=
= ÊËÁ
ˆ
¯̃
=
472 048
472 048
1
2
75 129
.
. .rad rev q = 75 1. rev �
44
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PROBLEM 16.31
Solve Problem 16.30, assuming that the initial angular velocity of the 
flywheel is 360 rpm clockwise.
PROBLEM 16.30 The 200 mm-radius brake drum is attached is a larger 
flywheel that is not shown. The total mass moment of inertia of the 
drum and the flywheel is 20 kg.m2 and the coefficient of kinetic friction 
between the drum and the brake shoe is 0.35. Knowing that the angular 
velocity of the flywheel is 360 rpm counterclockwise when a force P of 
magnitude 400 N is applied to the pedal C, determine the number of 
revolutions executed by the flywheel before it comes to rest.
SOLUTION
Lever ABC: Static equilibrium (friction force ≠)
F N Nk= =m 0 35.
+
 SM N FA = + - =0 250 50 400 250 0: ( ) ( ) ( )( ) mm mm N mm
250 50 0 35 100000 0
373 832
0 35 373 832
130
N N
N
F Nk
+ - =
=
=
=
=
( . )
.
. ( . )
.
N
N
m
8841 N
Drum:
r = =
= ÊËÁ
ˆ
¯̃
=
200 0 2
360
120
 mm m
 rpm
2
60
 rad/s
0
.
w p
w p
+ S SM M Fr ID D= = =( ) :eff 0 a
( . )( . ( )
. ( )
130 84 0 2 20
1 30841 2
1N m) kg m
 rad/s deceleration
2= ◊
=
a
a
w w aq p q2 0
2 2 22 0 12 2 1 30841= + = + -: ( ) ( . ) rad/s rad/s
 
q
q
=
= =
543 1107
543 1107
2
86 439
.
.
.
 rad
rev rev
p
 q = 86 4. rev �
45
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PROBLEM 16.32
The flywheel shown has a radius of 500 mm, a mass of 120 kg, and a radius 
of gyration of 375 mm. A 15-kg block A is attached to a wire that is wrapped 
around the flywheel, and the system is released from rest. Neglecting the effect 
of friction, determine (a) the acceleration of block A, (b) the speed of block A 
after it has moved 1.5 m.
SOLUTION
 Kinematics: Kinetics:
+
 S SM M m g r I m a rB B A A= = +( ) ( ) ( )eff : a
m gr m k
a
r
m ar
a
m g
m m
A F A
A
A F
k
r
= ÊËÁ
ˆ
¯̃
+
=
+ ( )
2
2
(a) Acceleration of A a =
+ ( )
=( )( . )
( )
.
15 9 81
15 120
1 7836
375 2
 kg m/s
 kg kg
 m
2
 mm
500 mm
//s2
or aA = 1 784. m/s
2 Ø �
(b) Velocity of A v v a sA A A
2
0
2 2= +( )
 For s = 1 8. m v
v
A
A
2 0 2 1 7836 1 5
5 3509
2 3132
= +
=
=
( . )( . )
.
.
 m/s m
 m /s
 m/s
2
2 2
or vA = 2 31. m/s �
46
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PROBLEM 16.33
In order to determine the mass moment of inertia of a flywheel of radius 
600 mm, a 12-kg block is attached to a wire that is wrapped around the 
flywheel. The block is released and is observed to fall 3 m in 4.6 s. To eliminate 
bearing friction from the computation, a second block of mass 24 kg is used 
and is observed to fall 3 m in 3.1 s. Assuming that the moment of the couple 
due to friction remains constant, determine the mass moment of inertia of the 
flywheel.
SOLUTION
 Kinematics: Kinetics:
+
 S SM M m y r M I m a rB B A f A= - = +( ) : ( ) ( )eff a
 m gr M I
a
r
m arA f A- = + (1)
Case 1: y
t
=
=
3
4 6
 m
 s.
y at
a
a
mA
=
=
=
=
1
2
3
1
2
4 6
0 2836
12
2
2 m s
 m/s
 kg
2
( . )
.
Substitute into Eq. (1)
( )( . )( . )
.
.
(12 9 81 0 6
0 2836
0 6
12 kg m/s m
 m/s
 m
 k2
2
- =
Ê
ËÁ
ˆ
¯̃
+M If gg m/s m)( . )( . )0 2836 0 6
2
 20 632 0 4727 2 0419. ( . ) .- = +M If (2)
47
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PROBLEM 16.33 (Continued)
Case 2: y
t
y at
a
a
mA
=
=
=
=
=
=
3
3 1
1
2
3
1
2
3 1
0 6243
24
2
2
 m
s
 m s
 m/s
 kg
2
.
( . )
.
Substitute into Eq. (1):
( )( . )( . )
.
(24 9 81 0 6
0 6243
24 kg m/s m
 m/s
0.6 m
 k2
2
- =
Ê
ËÁ
ˆ
¯̃
+M If gg m/s m
2)( . )( . )0 6243 0 6
 141 264 1 0406 8 9899. ( . ) .- = +M If (3)
Subtract Eq. (1) from Eq. (2), to eliminate Mf
 
20 632 1 0406 0 4727 6 948
13 684 0 5679
112 14
. ( . . ) .
. ( . )
.
= - +
=
= ◊
I
I
I kg m22 I = ◊112 1 2. kg m �
48
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PROBLEM 16.34
Each of the double pulleys shown has a mass moment of inertia of 20 2 kg m◊ and is initially at rest. The 
outside radius is 500 mm and the inner radius is 250 mm. Determine (a) the angular acceleration of each pulley, 
(b) the angular velocity of each pulley after Point A on the cord has moved 3 m.
SOLUTION
Case 1: (a) + S SM M0 0 800 0 25 20= = ◊( ) : ( )( . (eff
2 N m) kg m )a
 a = 10 2 rad/s �
 (b) q = =(
. )
3
0 250
12
 m)
( m
rad
 w aq2 2 2 10 12= = ( )( ) rad/s rad2 w = 15 49. rad/s �
Case 2: (a) + S SM M ma0 0 800 0 25 20 0 25= = +( ) : ( )( . ) ( . )eff a
200 20
800
9 81
0 25 0 25
200 20 5 09684
7 9691
= +
= +
=
a a
a
a
.
( . )( . )
( . )
. rad/ss2
 a = 7 97 2. rad/s �
 (b) q = =3
0 250
12
 m
m
 rad
.
 
w aq2 2
2 7 9691 12
=
= ( . )( ) rad/s rad2 w = 13 83. rad/s �
49
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PROBLEM 16.34 (Continued)
Case 3: (a) + S SM M0 0= ( ) :eff
2300 0 25 1500 0 25 20
2300
9 81
0 25
1500
9 81
0 252( . ) ( . )
.
( . )
.
( . )- = + +a a a 22
2
200 20 14 6534 9 5566
4 5239
= + +
=
a a a
a
. .
. rad/s
 a = 4 52 2. rad/s �
 (b) q = =3
0 25
12
 m
m
 rad
.
 w aq2 2 2 4 5239 12= = ( . )( ) w = 10 42. rad/s �
Case 4: (a) + S SM M a0 0 400 0 5 20
400
9 81
0 5= = +( ) : ( . )
.
( . )eff a
200 20
400
9 81
0 5
200 20 10 19368
6 6239
2
2
= +
= +
=
a a
a a
a
.
( . )
.
. rad/s
 a = 6 62 2. rad/s �
 (b) q = =3
0 25
12
 m
 m
 rad
.
 w aq2 2 2 6 6239 12= = ( . )( ) rad/srad2 w = 12 61. rad/s �
50
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PROBLEM 16.35
Each of the gears A and B weighs 10 kg and has a radius of gyration of 
150 mm; gear C weighs 2.5 kg and has a radius of gyration of 60 mm 
If a couple M of constant magnitude 6 N.m is applied to gear C, 
determine (a) the angular acceleration of gear A, (b) the tangential 
force which gear C exerts on gear A.
SOLUTION
Kinematics:
We express that the tangential components of the accelerations of the gear teeth are equal:
 
at A
B
C
B A
C A
=
=
=
=
=
0 2
0 2
0 08
2 5
.
.
.
.
a
a
a
a a
a a (1)
Kinetics:
Gear A: I m kA A A= =
= ◊
2 210 0 15
0 225
( . )
. kg m2
+ S SM M FA A A= =( ) : ( . ) .eff 0 2 0 225a
 F A= 1 125. a (2)
Because of symmetry, gear C exerts an equal force F on gear B.
Gear C: I m kC C C=
=
= ◊
2
22 5 0 06
0 009
. ( . )
. kg m2
+
 S SM M M Fr IC C C C C= - =( ) :eff 2 a
6 2 0 08 0 009- =F C( . ) . a
51
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PROBLEM 16.35 (Continued)
(a) Angular acceleration of gear A.
 Substituting for aC from (1) and for F from (2):
 
6 2 1 125 0 08 0 009 2 5
6 0 18 0 0225
6 0 2025
- =
- =
=
( . )( . ) . ( . )
. .
.
a a
a a
a
A A
A A
A
aa A = 29 6296
2. rad/s a A = 29 6. rad/s
2 �
(b) Tangential force F.
Substituting for a A into (2): F = 1 125 29 6296. ( . ) FA = 33 3. N �
52
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PROBLEM 16.36
Solve Problem 16.35, assuming that the couple M is applied to disk A.
PROBLEM 16.35 Each of the gears A and B weighs 10 kg and has a 
radius of gyration of 150 mm gear C weighs 2.5 kg and has a radius 
of gyration of 60 mm. If a couple M of constant magnitude 6 N· m is 
applied to gear C, determine (a) the angular acceleration of gear A, 
(b) the tangential force which gear C exerts on gear A.
SOLUTION
Kinematics: a A A= a a A A= a aB B= a 
At the contact point between gears A and C,
a r r
r
r
t A A C C
C
A
C
A A A
= =
= = =
a a
a a a a0 2
0 08
2 5
.
.
.
At the contact point between gears B and C,
a r r
r
r
t B B C C
B
C
A
C A A
= =
= = ◊ =
a a
a a a a0 08
0 2
2 5
.
.
( . )
Kinetics:
Gear B: 
+
 S SM MB B= ( ) :eff F r I IBC B B B B A= =a a
F
I
rBC
B
B
A= a
Gear C: + S SM M F r F r IC C BC C AC C C C= + =( ) :eff a
r
r
I F r I
F
r
I I
C
B
B A AC C C A
AC
C
B C A
a a
a
+ =
= +
2 5
1
0 4 2 5
.
( . . )
Gear A: + M M M r F IA A A A AC A A= - =S( ) :eff a
M
r
r
I I I
M I I I
A
A
C
B C A A A
A A B C A
- + =
= + + ( )( )
( . . )
.
0 4 2 5
2 5 2
a a
a
53
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PROBLEM 16.36 (Continued)
Data: m m
k k
I I m k
A B
A B
A B A A
= =
= =
= = =
= ◊
10
0 15
10 0 15
0 225
2 2
 kg
 m
 kg m2
.
( )( . )
.
m
k
I
M
C
C
C
A
=
=
= = ◊
= ◊
=
2 5
0 06
2 5 0 06 0 009
6
6 0 22
2 2
.
.
( . )( . ) .
.
 kg
 m
kg m
 N m
55 0 225 2 5 0 0092+ +ÈÎ ˘̊. ( . ) ( . ) a A
(a) Angular acceleration. a A = 11 8519
2. rad/s a A = 11 85
2. rad/s �
(b) Tangential gear force. FAC = +[ ]
1
0 08
0 4 0 225 2 5 0 009 11 8519
.
( . )( . ) ( . )( . ) ( . )
 FAC = 16 67. N �
54
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PROBLEM 16.37
Two uniform disks and two cylinders are assembled as indicated. Disk A 
weighs 10 kg and disk B weighs 6 kg. Knowing that the system is released 
from rest, determine the acceleration (a) of cylinder C, (b) of cylinder D.
Disks A and B are bolted together and the cylinders are attached to separate 
cords wrapped on the disks.
SOLUTION
Total I : I m r m rA A B B= +
= +
= + =
1
2
1
2
1
2
10 0 2
1
2
6 0 15
0 2 0 0675 0 2
2 2
2 2( )( . ) ( )( . )
. . . 6675 kg m2
Kinematics:
Kinetics: + S SM M0 0= ( ) :eff
( . )( . ) ( )( . )
( . )( . )( . )
7 5 9 81 9 9 81
7 5 9 81 0 2
r r I m a r m a rA B C C A D D B- = + +a
-- = +
+
9 9 81 0 15 0 2675 7 5 0 2 0 2
9 0 15 0 15
( . )( . ) . ( . )( . )( . )
( . )( . )
a a
a
14 715 13 2435 0 2675 0 3 0 2025
1 4715 0 77
1 91104
. . ( . . . )
. .
.
- = + +
=
=
a
a
a raad/s2
(a) Acceleration of C. aC = =0 2 0 2 1 91104
2. ( . ( . )a m) rad/s aC = 0 382
2. m/s Ø �
(b) Acceleration of D. aD = =( . ) ( . )( . )0 15 0 15 1 91104
2a m rad/s aD = 0 287
2. m/s ≠ �
55
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PROBLEM 16.38
Two uniform disks and two cylinders are assembled as 
indicated. Disk A weighs 10 kg and disk B weighs 6 kg. 
Knowing that the system is released from rest, determine the 
acceleration (a) of cylinder C, (b) of cylinder D.
The cylinders are attached to a single cord that passes over 
the disks. Assume that no slipping occurs between the cord 
and the disks.
SOLUTION
Moments of inertia: I m r
I m r
A A A
B B B
= = = ◊
= = =
1
2
1
2
10 0 2 0 2
1
2
1
2
6 0 15 0
2 2 2
2 2
( )( . ) .
( )( . )
 kg m
..0675 2 kg m◊
Kinematics: a
a
A
A
B
B
a
r
a
a
a
r
a a
= = =
= = =
0 2
5
0 15
20
3
.
.
m
m
Kinetics:
Disk A: FAB = Tension in cord between disks
+
 S SM MA A= ( ) :eff
 
F I m a
F
AB A A C
AB
( . ) ( . ) ( . )
. . ( . )
0 2 7 5 0 2
0 2 14 715 0 2
- = +
- =
(9.81)(0.2) a
(( ) ( . ) ( . )
. . .
. . .
5 7 5 0 2
0 2 14 715 1 5
0 2 14 715 2 5
a a
F a a
F a
F
AB
AB
AB
+
- = +
- =
== +73 575 12 5. . a (1)
56
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PROBLEM 16.38 (Continued)
Disk B: S SM MB B= ( ) :eff
( ( . ( . ( . ) ( . )
. . (
9 9 81 0 15 0 15 0 15
13 2435 0 15 0
) ) ) - = +
- =
F I m a
F
AB B B D
AB
a
.. ) ( . )
. . . .
0675
20
3
9 0 15
13 2435 0 15 0 45 1 35
a
a
F a aAB
Ê
ËÁ
ˆ
¯̃
+
- = +
Substitute for FAB from Eq. (1)
13 2435 0 15 73 575 12 5 1 8
13 2435 11 03625 1 875 1 8
. . ( . . ) .
. . . .
- + =
- - =
a a
a aa
a
a
2 20725 3 675
0 6006 2
. .
.
=
= m/s
Both aC and aD have the same magnitude.
(a) Acceleration of C. aC = 0 601
2. m/s ≠ �
(b) Acceleration of D. aD = 0 601
2. m/s Ø �
57
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PROBLEM 16.39
Disk A has a mass of 6 kg and an initial angular velocity of 360 rpm 
clockwise; disk B has a mass of 3 kg and is initially at rest. The disks 
are brought together by applying a horizontal force of magnitude 20 N 
to the axle of disk A. Knowing that mk = 0 15. between the disks and 
neglecting bearing friction, determine (a) the angular acceleration of 
each disk, (b) the final angular velocity of each disk.
SOLUTION
While slipping occurs, a friction force F ≠ is applied to disk A, and F Ø to disk B
Disk A:
I m rA A A=
=
= ◊
1
2
1
2
6 0 08
0 0192
2
2( )( . )
.
 kg m
 kg m2
SF N P: = = 20 N
F N= = =m 0 15 20 3. ( ) N
+ S SM M Fr IA A A A A= =( ) :eff a
 
( )( . ) ( . )
.
3 0 08 0 0192
10 227
 N m kg m2= ◊
=
a
a
A
A a A = 12 50. rad/s
2 �
Disk B:
I m rB B B=
=
= ×
1
2
1
2
0 06
0 0054
2
2
2
( )( . )
.
3 kg m
 kg m
+ S SM M Fr IB B B B B= =( ) :eff a
 
( )( . ) ( . )
.
3 0 06 0 0054
33 333
 N m kg m
 rad/s
2
2
= ◊
=
a
a
B
B aB = 33 3. rad/s
2 �
( )w p pA 0 360
2
60
12= ÊËÁ
ˆ
¯̃
= rpm rad/s 
58
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PROBLEM 16.39 (Continued)
Sliding stops when V VC C= ¢ or w wA A B Br r=
r t r t
t
A A A B B[( ) ]
( . )[ ( . ) ] ( .
w a a
p
0
0 08 12 12 5 0 0
- =
- = m rad/s rad/s2 66 33 333
1 00531
 m rad/s
 s
2)( . )
.
t
t =
+ w w a
p
A A At= -
= -
=
( )
( . )( . )
.
0
12 12 5 1 00531
25 132
 rad/s rad/s s
 rad
2
//s
w
pA
= ÊËÁ
ˆ
¯̃
=
( . )
.
25 132
60
2
240 00
 rad/s
 rpm
or w A = 320 rpm �
+ w aB Bt=
=
=
33 333 1 00531
33 510
. ( . )
.
 rad/s s
 rad/s
2
w
pB
= ÊËÁ
ˆ
¯̃
=
( . )
.
33 510
60
2
320 00
 rad/s
 rpm
or w B = 320 rpm �
59
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PROBLEM 16.40
Solve Problem 16.39, assuming that initially disk A is at rest and disk B 
has an angular velocity of 360 rpm clockwise.
PROBLEM 16.39 Disk A has a mass of 6 kg and an initial angular velocity 
of 360 rpm clockwise; disk B has a mass of 3 kg and is initially at rest. The 
disks are brought together by applying a horizontal force of magnitude 
20 N to the axle of disk A. Knowing that mk = 0 15. between the disks and 
neglecting bearing friction, determine (a) the angular acceleration of each 
disk, (b) the final angular velocity of each disk.
SOLUTION
While slipping occurs, a friction force F ≠ is applied to disk A, and F Ø to disk B.
Disk A: I m rA A A=
=
= ×
1
2
1
2
6 0 08
0 0192
2
2( )( . )
.
 kg m
 kg m2
SF N P: = = 20 N
F N= = =m 0 15 20 3. ( ) N
+ S SM M Fr IA A A A A= =( ) :eff a
 
( )( . ) ( . )
.
3 0 08 0 0192
10 227
 N m kg m2= ◊
=
a
a
A
A a A = 12 50. rad/s
2 �
Disk B: I m rB B B=
=
= ◊
1
2
1
2
0 06
0 0054
2
2
2
( )( . )
.
3 kg m
 kg m
+ S SM M Fr IA B B B B= =( ) :eff a
 
( )( . ) ( . )
.
3 0 06 0 0054
33 333
 N m kg m
 rad/s
2
2
= ◊
=
a
a
B
B aB = 33 3. rad/s
2 �
60
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PROBLEM 16.40 (Continued)
Sliding starts when V VC C= ¢ . That is when
w w
a w a
A A B B
A A B B B
r r
t r t r
t
=
= -
=
( ) [( ) ]
[( . ) ]( . ) [
0
212 5 0 08 12 rad/s m pp rad/s rad/s m
 s
-
= fi =
( . )( )]( . )
. .
33 333 0 06
4 02124 0 75398
2 t
t t
 w aA At= = =( . )( . ) .12 5 0 75398 9 4248 rad/s s rad/s
2 
+ w
pA
= ÊËÁ
ˆ
¯̃
=( . ) .9 4248 60
2
90 00 rad/s rpm
or w A = 90 0. rpm �
+ w w a
p
B B Bt= -
= -
( )
( . )( . )
0
12 33 333 0 75398 rad/s rad/s s2
= 12 566. rad/s 
w
pB
= ÊËÁ
ˆ
¯̃
=
( . )
.
12 566
60
2
120 00
 rad/s
 rpm
or w B = 120 0. rpm �
61
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PROBLEM 16.41
A belt of negligible mass passes between cylinders A and B and is pulled to 
the right with a force P. Cylinders A and B weigh, respectively, 2.5 and 10 kg. 
The shaft of cylinder A is free to slide in a vertical slot and the coefficients 
of friction between the belt and each of the cylinders are ms = 0 50. and 
mk = 0 40. . For P = 18 N, determine (a) whether slipping occurs between the 
belt and either cylinder, (b) the angular acceleration of each cylinder.
SOLUTION
Assume that no slipping occurs.
Then: a A Bbelt m m= =( . ) ( . )0 1 0 2a a a aB A=
1
2
 (1)
Cylinder A. + S SM M F IG G e A A A= =( ) : ( .ff m)0 1 a
 
F
F
A A
A
A
( . ) ( . )( . )0 1
1
2
2 5 0 1
8
2=
=
a
a
 (2)
Cylinder B. 
+
S SM M F IG G B B B= =( ) : ( .eff m)0 2 a
 
F
F
B
A
B
A
( . ) ( )( . )0 2
1
2
10 0 2
2
2
2=
=
a
a
 (3)
Belt + SF P F FA A B= - - =0 0: (4)
18
8 2
0- - =
a aA A
a A =
Ê
ËÁ
ˆ
¯̃
8
5
18( )
a A = 28 8. rad/s
2
62
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PROBLEM 16.41 (Continued)
Check that belt does not slip.
From (2): FA = =
1
8
28 8 3 6( . ) . N
From (4):F P FB A= - = - =18 3 6 14 4. . N
But F Nm s= = =m ( . )( . )( . ) .0 5 2 5 9 81 12 2625 N
Since F FB m� ,assumption is wrong.
 Slipping occurs between disk B and the belt. �
We redo analysis of B, assuming slipping. a aB Aπ
Ê
ËÁ
ˆ
¯̃
1
2
 
F NB k= = =m 0 4 2 5 9 81 9 81. ( . )( . ) . N
+ S SM MG G B= =( ) ( . )( . ) ( )( . )eff : 9 81 0 2
1
2
10 0 2 2a
 aB = 9 81. aB = 9 81. rad/s
2 �
Belt Eq. (4): P F F
F P F
A B
A B
- - =
= -
= -
=
0
18 9 81
8 19
.
. N
 
Since F FA m� , There is no slipping between A and the belt. �
Our analysis of disk A, therefore is valid. Using Eq. (2),
We have 8 19
8
. N =
a A
 a A = =( )( . ) .8 8 19 65 52 a A = 65 5. rad/s
2 �
63
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PROBLEM 16.42
Solve Problem 16.41 for P = 10 N
PROBLEM 16.41 A belt of negligible mass passes between cylinders A and B 
and is pulled to the right with a force P. Cylinders A and B weigh, respectively, 
2.5 and 10 kg. The shaft of cylinder A is free to slide in a vertical slot and the 
coefficients of friction between the belt and each of the cylinders are ms = 0 50. 
and mk = 0 40. . For P = 18 N , determine (a) whether slipping occurs between 
the belt and either cylinder, (b) the angular acceleration of each cylinder.
SOLUTION
Assume that no slipping occurs.
Then: a A Bbelt m m= =( . ) ( . )0 1 0 2a a a aB A=
1
2
 (1)
Cylinder A + S SM M F IG G A A A= =( ) : ( .eff m)0 1 a
 
F
F
A A
A
A
( . ) ( . )( . )0 1
1
2
2 5 0 1
8
2=
=
a
a
 (2)
Cylinder B 
+
S SM M F IG G B B B= =( ) : ( .eff m)0 2 a
 
F
F
B
A
B
A
( . ) ( )( . )0 2
1
2
10 0 2
2
2
2=
=
a
a
 (3)
Belt + SF P F FA A B= - - =0 0:
10
8 2
0- - =
a aA A
a A =
Ê
ËÁ
ˆ
¯̃
( )10
8
5
a A = 16 rad/s
2
64
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PROBLEM 16.42 (Continued)
From (2): FA = =
16
8
2 N
From (4): 
FB = =
16
2
8 N
But F NM s= = =m ( . )( . )( . ) .0 5 2 5 9 81 12 2625 N
Thus, FA and FB are both � Fm . Our assumption was right:
There is no slipping between cylinders and belt a A = 16 00. rad/s
2 �
From (1): aB = 8 00. rad/s
2 �
65
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PROBLEM 16.43
The 3-kg disk A has a radius rA = 75 mm and an initial angular velocity 
w0 375= rpm clockwise. The 7.5 kg disk B has a radius rB = 125 mm 
and is at rest. A force P of magnitude 12.5 N is then applied to bring 
the disks into contact. Knowing that mk = 0 25. N between the disks and 
neglecting bearing friction, determine (a) the angular acceleration of 
each disk, (b) the final angular velocity of each disk.
SOLUTION
While slipping occurs:
(a) Angular accelerations. F N
P
k
k
=
=
=
=
m
m
0 25 12 5
3 125
. ( . )
.
N
N
 Disk A:
+ S SM M Fr I m rA A A A A A A= = =( ) :eff a a
1
2
2
F m rA A
A
=
=
1
2
3 125
1
2
3
a
a. (N kg)(0.075 m)
 a A =
250
9
 rad/s2 a A = 27 8. rad/s
2 �
 Disk B:
+ S SM M Fr I m rB B B B B B B B= = =( ) :eff a a
1
2
2
F m rB B B
B
=
=
1
2
3 125
1
2
7 5
a
a. ( .N kg)(0.125 m)
 aB =
20
3
rad/s2 aB = 6 67. rad/s
2 �
(b) Final angular velocities. ( )w pA 0 375
2
60
25
2
= ÊËÁ
ˆ
¯̃
=rpm rad/sp
( )wB 0 0=
66
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PROBLEM 16.43 (Continued)
When disks stop sliding
 v v r rP P A A B B= =: w w (1)
[( ) ] ( )
( ) (
w a aA B A B Bt r t r
t
0
25
2
25
9
75
20
3
125
- =
-ÊËÁ
ˆ
¯̃
= ÊËÁ
ˆ
¯̃
p 0
mm t mmm
s
)
.
( )
( . )
t
tA A A
=
= -
= -
1 0098
25
2
25
9
1 0098
0w w a
p 0
 w
pA
= ÊËÁ
ˆ
¯̃
=11 22 60
2
107 1422. .rad/s rpm w A = 107 1. rpm �
Eq. (1) ( . )( ) ( )107 1422 75 125rpm mm mm= wB w B = 64 3. rpm �
67
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PROBLEM 16.44
Solve Problem 16.43, assuming that disk A is initially at rest and that 
disk B has an angular velocity of 375 rpm clockwise.
SOLUTION
While slipping occurs:
(a) Angular accelerations. F N
P
k
k
=
=
=
=
m
m
0 25 12 5
3 125
. ( . )
.
N
N
 Disk A:
+ S SM M Fr I m rA A A A A A A= = =( ) :eff a a
1
2
2
F m rA A
A
=
=
1
2
3 125
1
2
3
a
a. ( N kg)(0.075 m)
 a A =
250
9
rad/s2 a A = 27 8. rad/s
2 �
 Disk B:
+ S SM M Fr I m rB B B B B B B B= = =( ) :eff a a
1
2
2
F m rB B B
B
=
=
1
2
3 125
1
2
7 5 0 125
a
a. ( . )( . N kg m)
 aB =
20
3
rad/s2 aB = 6 67. rad/s
2 �
(b) Final angular velocities
( )
( )
w
w p
A
B
0
0
0
375
=
= ÊËÁ
ˆ
¯̃
=rpm 2
60
25
2
rad/s
p
68
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PROBLEM 16.44 (Continued)
Eq. (1): w w
a w a
p
A A B B
A A B B B
r r
tr t r
=
= -
Ê
ËÁ
ˆ
¯̃
= -ÊËÁ
[( ) ]
( )
0
250
9
75t tmm
25
2
20
3
ˆ̂
¯̃
=
( )
. sec
125
1 683
mm
t
 
w a
p
A At=
= ÊËÁ
ˆ
¯̃
=
= ÊËÁ
ˆ
¯̃
250
9
rad/s
rad/s
( . ) .
.
1 683 46 75
46 75
60
2
 w A = 446 rpm �
Eq. (1): ( . )( ) ( )446 428 75 125rpm mm mm= wB w B = 268 rpm �
69
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PROBLEM 16.45
Disk B has an angular velocity w0 when it is brought into contact with 
disk A, which is at rest. Show that (a) the final angular velocities of the 
disks are independent of the coefficient of friction mk between the disks 
as long as mk π 0, (b) the final angular velocity of disk B dependsonly 
upon w0 and the ratio of the masses mA and mB of the two disks.
SOLUTION
While slipping occurs: F N Pk k= =m m
Disk A:
S SM M Fr IA A A A A= =( ) :eff a
 
m a
a
m
k A A A A
A
k
A A
Pr m r
P
m r
=
=
1
2
2
2
 (1)
Disk B: + = =SM M Fr IB B B B B( ) :eff a
m ak B B B BPr m r=
1
2
2
 a
m
B
k
B B
P
m r
=
2
 (2)
At any time t
 + w a
m
A A
k
A A
t
P
m r
= + =0
2
 (3)
 
+ w w a w
m
B B
k
B B
t
P
m r
t= - = -0 0
2
 (4)
Slipping ends when w w
w a
a a w
A A B B
A A B B
A A B B B
r r
a tr t r
r r t r
=
= -
+ =
( )
( )
0
0
70
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PROBLEM 16.45 (Continued)
Substitute from Eqs. (1) + (2): 
2 2
0
P
m r
r
P
m r
r t rk
A A
A
k
B B
B B
m m
w+
È
Î
Í
˘
˚
˙ =
2
1 1
2
1
2
0
0
1 1
0
P
m m
t r
t
r
P
t
r
P
m m
k
A B
B
B
k m m
B
k
A B
A B
m w
w
m
w
m
+
Ê
ËÁ
ˆ
¯̃
=
= ◊
+
= ◊
&
mm mA B+
Eq. (3): w
m w
m
w w
A A
k
A A
B
k
A B
A B
A
B
A
B
A B
a t
P
m r
r
P
m m
m m
r
r
m
m m
= = ◊ ◊
+
È
Î
Í
˘
˚
˙
= ◊
+
2
2
0
0
(wA is independent of mk,QED)
Eq. (4): w w a w
m w
m
w w
B B
k
B B
B
k
A B
A B
B
A
A
t
P
m r
r
P
m m
m m
m
m
= - = - ◊ ◊
+
È
Î
Í
˘
˚
˙
= -
+
0 0
0
0
2
2
1
mm
m m m
m m
m
m m
B
B
A B A
A B
B
A B
B m
m
A
B
Ï
Ì
Ó
¸
˝
˛
=
+ -
+
=
+
=
+
w w w
w
w
0 0
0
1
wB depends only upon w0 and 
m
m
A
B
 (QED)
71
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PROBLEM 16.46
Show that the system of the effective forces for a rigid slab in plane motion reduces to a single vector, and express 
the distance from the mass center G of the slab to the line of action of this vector in terms of the centroidal 
radius of gyration k of the slab, the magnitude a of the acceleration of G, and the angular acceleration a.
SOLUTION
We know that the system of effective forces can be reduced to the vector ma at G and the couple I a. We 
further know from Chapter 3 on statics that a force-couple system in a plane can be further reduced to a single 
force.
The perpendicular distance d from G to the line of action of the single vector ma is expressed by writing
+ S SM M I ma dG G= =( ) : ( )eff a
 d
I
ma
mk
ma
= =a a
2
 d
k
a
=
2a
 �
72
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PROBLEM 16.47
For a rigid slab in plane motion, show that the system of the effective forces 
consists of vectors ( ) ,Dmi a - ¢( ) ,Dm ri iw
2 and ( )( )Dm ri ia ¥ ¢ attached to the 
various particles Pi of the slab, where a is the acceleration of the mass center G 
of the slab, w is the angular velocity of the slab, a is its angular acceleration, 
and ri¢ denotes the position vector of the particle Pi relative to G. Further show, 
by computing their sum and the sum of their moments about G, that the effective 
forces reduce to a vector ma attached at G and a couple I a.
SOLUTION
Kinematics: The acceleration of PL is
a a a
a a r r
a r r
i P G
i i i
i i
i
= +
= + ¥ ¢ + ¥ ¥ ¢
= + ¥ ¢ - ¢
/
a w w
a
( )
w2
Note: that a ¥ ¢ ^ ¢r ri i is to 
Thus, the effective forces are as shown in Figure P16.47 (also shown above). We write
( ) ( ) ( )( ) ( )D D D Dm m m m ri i i i i i ia a r= + ¥ ¢ - ¢a w
2
The sum of the effective forces is
S D S D S D S D
S D S D
( ) ( ) ( )( ) ( )
( ) ( )
m m m m
m m
i i i i i i i
i i i
a a r r
a a
= + ¥ ¢ - ¢
=
a w2
++ ¥ ¢ - ¢a S D S D( ) ( )m mi i i ir rw
2
We note that S D( ) .m mi = And since G is the mass center,
S D( )m mri i i¢ = =r ¢ 0
Thus S D( )m mi ia a= (1)
The sum of the moments about G of the effective forces is:
S D S D S D S D
S
( ) ( )( ) ( )
(
¢ ¥ = ¢ ¥ + ¢ ¥ ¥ ¢ - ¢ ¥ ¢r a r a r r r ri i i i i i i i i i im m m ma w
2
¢¢ ¥ = ¢ + ¢ ¥ ¥ ¢ - ¢ ¥ ¢r a r a r r r ri i i i i i i i i i im m m mD S D S D S D) ( ) [ ( ) ] ( )a w
2
Since G is the mass center, S D¢ =ri im 0
Also, for each particle, ¢ ¥ ¢ =r ri i 0
Thus, S D S D( ) [ ( ) ]¢ ¥ = ¢ ¥ ¥ ¢r a r ri i i i i i im ma
73
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PROBLEM 16.47 (Continued)
Since a ^ ¢ri , we have ¢ ¥ ¥ ¢ =r ri i ir( )a
2a and
S D S D
S D
( ) ( )¢ ¥ = ¢
= ¢( )
r aii i i i i
i i
m r m
r m
2
2
a
a
Since S D¢ =r m Ii i
2 S D( )¢ ¥ =r ai i im I a (2)
From Eqs. (1) and (2) we conclude that system of effective forces reduce to ma attached at G and a couple I a.
74
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PROBLEM 16.48
A uniform slender rod AB rests on a frictionless horizontal 
surface, and a force P of magnitude 1 N is applied at A in a 
direction perpendicular to the rod. Knowing that the rod weighs 
9 N, determine the acceleration of (a) Point A, (b) Point B.
SOLUTION
m
W
g
I
W
g
L
=
= 1
12
2
+ S SF F P ma W
g
ax x= = =( ) :eff
 a
P
W
g g
g= = =
1
9 9
N
N
 a = 1
9
g Æ
+ S SM M P L I W
g
LG G= = =( ) :eff 2
1
12
2a a
 a = =6 6
1
9
P
W
g
L
g
L
N
N
 a = 2
3
g
L
(a) Acceleration of Point A.
 + a aA
L g L g
L
g= + = + ◊ = ÊËÁ
ˆ
¯̃
=
2 9 2
2
3
4
9
4 36a . )m/s2 aA = 4 36. m/s
2 Æ �
(b) Acceleration of Point B.
 + a aB
L g L g
L
g= - = - ◊ = - = -
2 9 2
2
3
2
9
2
9
9 81a ( . )m/s2 aB = 2 18. m/s
2 ¨ �
75
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PROBLEM 16.49
(a) In Problem 16.48, determine the point of the rod AB at which 
the force P should be applied if the acceleration of Point B is to 
be zero. (b) Knowing that P = 1 N, determine the corresponding 
acceleration of Point A.
SOLUTION
+ S SF F P ma W
g
ax X= = =( ) :eff
a = ÆP
W
g
+ S SM M Ph I W
g
LG G= = =( ) :eff a a
1
12
2
a = 12
2
Ph
WL
g
(a) Position of forceP. + a aB
L= -
2
a
0
2
12
6
1
6
2
= - ◊
= =
P
W
g
L Ph
WL
g
h
L
m
 Thus, P is located 
1
3
m from end A. �
 For h
L P
WL
g
P
W
g
L
L
= =
( )
= ◊
6
12
26
2
: a
(b) Acceleration of Point A.
+ a a
L P
W
g
L P
W
g
L
P
W
gA = + = + ◊ =2 2
2 2a
aA =
Ê
ËÁ
ˆ
¯̃
2
1
9
9 81( . ) aA = 2 18. m/s
2 Æ �
76
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PROBLEM 16.50
A force P of magnitude 3 N is applied to a tape wrapped 
around the body indicated. Knowing that the body 
rests on a frictionless horizontal surface, determine the 
acceleration of (a) Point A, (b) Point B.
A thin hoop of mass 2.4 kg.
SOLUTION
Hoop: I mr= 2
+ S SF F P max x= =( ) :eff
a = P
m
 Æ
+ S SM M Pr I mrG G= = =( ) :eff a a
2
a = P
mr
(a) Acceleration of Point A.
+ a a r
P
m
r
P
mr
P
mA
= + = + ÊËÁ
ˆ
¯̃
=a 2
 aA = =2
3
2 4
2 5
N
kg
m/s2
.
. aA = 2 50. m/s
2 Æ �
(b) Acceleration of Point B.
 + a a r
P
m
r
P
mrB
= - = - ÊËÁ
ˆ
¯̃
=a 0 aB = 0 �
77
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PROBLEM 16.51
A force P of magnitude 3 N is applied to a tape wrapped 
around the body indicated. Knowing that the body rests on a 
frictionless horizontal surface, determine the acceleration of 
(a) Point A, (b) Point B.
A uniform disk of mass 2.4 kg.
SOLUTION
Disk: I mr= 1
2
2
 + S SF F P maX X= =( ) :eff
 a = P
m
Æ
 + S SM MG G= =( ) :eff Pr Ia
 Pr = 1
2
2mr a
 a = 2P
mr
 
(a) Acceleration of Point A.
+ a a r
P
m
r
P
mr
P
mA
= + = + ◊ =a 2 3
 aA = =3
3
2 4
3 75
N
kg
m/s2
.
. aA = 3 75. m/s
2 Æ �
(b) Acceleration of Point B.
+ a a r
P
m
r
P
mr
P
mB
= - = - ◊ = -a 2
 aB = - = -
3
2 4
1 25
N
kg
m/s2
.
. aB = 1 25
2. /m s ̈ �
78
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PROBLEM 16.52
A force P is applied to a tape wrapped around a uniform 
disk that rests on a frictionless horizontal surface. Show 
that for each 360∞ rotation of the disk the center of the 
disk will move a distance pr.
SOLUTION
Disk: I mr= 1
2
2
 + S SF F P max x= =( )eff :
 a = P
m
Æ
 + S SM M IG G= =( ) :eff Pr a
 Pr =
1
2
2mr a
 a = 2P
mr
 
Let t1 be time required for 360∞ rotation.
 
q a
p
p
=
= ÊËÁ
ˆ
¯̃
=
1
2
2
1
2
2
2
1
2
1
2
1
2
t
P
mr
t
t
mr
P
rad
Let x1 = distance G moves during 360∞ rotation.
 x at
P
m
mr
P1 1
21
2
1
2
2= = ÊËÁ
ˆ
¯̃
p
 x r1 = p Q E.D. . �
79
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PROBLEM 16.53
A 120-kg satellite has a radius of gyration of 600 mm with respect to 
the y axis and is symmetrical with respect to the zx plane. Its orientation 
is changed by firing four small rockets A, B, C, and D, each of which 
produces a 16.20-N thrust T directed as shown. Determine the angular 
acceleration of the satellite and the acceleration of its mass center G 
(a) when all four rockets are fired, (b) when all rockets except D are fired.
SOLUTION
m I mky= = = = ◊120 120 43 2
2 kg, kg)(0.600 m) kg m2 2( .
(a) With all four rockets fired:
 S SF F= eff : 0 = ma a = 0 �
 + S SM M Tr IG G= =( )eff : 4 a
 4 16 20 43 2( . .N)(0.8 m) = a a = 1 200. rad/s2
 a = -( . )1 200 rad/s2 j �
(b) With all rockets except D:
+ S SF F a ax x x x= - = = -( ) . .eff
2: N 120 m/s16 20 0 1350
 + S SF Fz z= ( )eff : 0 120= =a az 0 a = -( . )0 1350
2 m/s i �
 + S SM MG G= ( ) :eff 3Tr I= a
 3 16 20 43 2( . .N)(0.8 m) = a a = -( . )0 900 rad/s2 j �
80
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PROBLEM 16.54
A rectangular plate of mass 5 kg is suspended from four 
vertical wires, and a force P of magnitude 6 N is applied to 
corner C a shown. Immediately after P is applied, determine 
the acceleration of (a) the midpoint of edge BC, (b) corner B.
SOLUTION
 I m b h= + = + = ◊1
12
1
12
5 0 3 0 104172 2( ) ( ( . ] . kg)[(0.4 m) m) kg m2 2 2
 + S SF F P ma= =eff :
 6 5 1 2N a a= = +( ( . ) kg) m/s2 k
 + S SM M P IG G= =( ) ( .eff : m)0 2 a
 ( ( . )6 0 10417 N)(0.2 m) kg m2= ◊ a a = -( . )11 52 rad/s2 j
(a) a a r
k j i
E E G= + ¥
= + - ¥
= +
a /
( . ( . ( .
( .
1 2 11 52 0 2
1 2
 m/s rad/s ) m)
 m
2 2) )
//s m/s2 2)k )k+ ( .2 304 a )kE = ( .3 50 m/s
2 �
81
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PROBLEM 16.54 (Continued)
(b) a a r
k j i
B B G= + ¥
= + - ¥
a /
( . ) ( . [( .1 2 11 52 0 2 m/s rad/s) m) +(0.15 m2 )) ]
 m/s m/s m/s2 2 2
k
k k i= + + +( . ) ( . ) ( . )1 2 2 304 1 728
 a i kB = +( . ) ( . )1 728 3 5 m/s m/s
2 2 �
82
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PROBLEM 16.55
A 3-kg sprocket wheel has a centroidal radius of gyration of 70 mm and is 
suspended from a chain as shown. Determine the acceleration of Points A and B 
of the chain, knowing that TA = 14 N and TB = 18 N.
SOLUTION
 
I mk
r
W mg
=
=
= ¥ ◊
=
=
=
-
2
3
3
14 7 10
0 08
3
(
.
.
(
 kg)(0.07 m)
 kg m
m
 kg)(9.
2
2
881 m/s
 N
2 )
.W = 29 43
 + S SF F T T W may y A B= + - =( )eff :
 T T aA B+ - =29 43 3. ( N kg)
 + a T TA B= + -
1
3
29 43( . ) (1)
 + S SM M T T IG G B A= - =( ) ( . ( .eff : m) m)0 08 0 08 a
 ( )( . ( . )T TB A- = ¥ ◊
-0 08 14 7 10 3 m) kg m2 a
 + a = -5 442. ) (T TB A (2)
 T TA B= =14 18 N, N
Eq. (1): + a = + - =
1
3
14 18 29 43 0 8567(. ) . m/s2
Eq. (2): + a = - =5 442 18 14 21 769. ( ) . rad/s2
83
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PROBLEM 16.55 (Continued)
+aA A ta
a r
=
= +
= -
( )
. ( . )( . )
a
0 8567 0 08 21 769
 = -0 885. m/s2 aA = 0 885. m/s
2 Ø �
+aB B ta
a r
=
= +
= +
¢( )
. ( . )( . )
a
0 8567 0 08 21 769
 = +2 60. m/s2 aB = 2 60. m/s
2 ≠ �
84
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PROBLEM 16.56
Solve Problem 16.55, assuming that TA = 14 N and TB = 12 N.
PROBLEM 16.55 A 3-kg sprocket wheel has a centroidal radius of gyration of 
70 mm and is suspended from a chain as shown. Determine the acceleration of 
Points A and B of the chain, knowing that TA =14 N and TB = 18 N.
SOLUTION
 
I mk
r
W mg
=
=
= ¥ ◊
=
=
=
-
2
3
3
14 7 10
0 08
3
(
.
.
(
 kg)(0.07 m)
kg m
m
 kg)(9.8
2
2
11 m/s
 N
2 )
.W = 29 43
 + S SF F T T W may y A B= + - =( )eff :
 T T aA B+ - =29 43 3. ( N kg)
 + a T TA B= + -
1
3
29 43( . ) (1)
 + S SM M T T IG G B A= - =( ) ( . ( .eff : m) m)0 08 0 08 a
 ( )( . ( . )T TB A- = ¥ ◊
-0 08 14 7 10 3 m) kg m2 a
 + a = -5 442. ) (T TB A (2)
 T TA B= =14 12 N, N
Eq. (1): + a = + - = -
1
3
14 12 29 43 1 1433( . ) . m/s2 a = 1 1433. m/s2 Ø
Eq. (2): + a = - = -5 442 12 14 10 884. ( ) . rad/s2 a = 10 884. rad/s2 
85
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PROBLEM 16.56 (Continued)
+ aA A ta
a r
=
= +
= - +
( )
. ( . )( . )
a
1 1433 0 08 10 884
 = -0 273. m/s2 aA = 0 273. m/s
2Ø �
 + aB B ta
a r
=
= +
= - -
¢( )
. ( . )( . )
a
1 1433 0 08 10 884
 = -2 01. m/s2 aB = 2 01. m/s
2 Ø �
86
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SOLUTION
Kinematics: a aB A= + (5 m)a
 1≠ = 6≠ + 5a
 a = 1 rad/s2 
 a a aA B= + = +
1
2
1
2
1 6( ) ( )
 a = 3 5. m/s2≠
Kinetics:
 
I mL=
=
= ◊
1
12
1
12
2500
9 81
5
530 921
2
( )
( . )
(
.
)
 kg m
2
2
 + S SM M T W ma IB B A= - = +( ) ( ( . ( .eff : m) m) m)5 2 5 2 5 a
 
TA( ( ( . )( .
( .
5 2500
2500
3 5 2 5
530 921
 m) N)(2.5 m)
9.81
 m/s m)2- =
+ )) )
. .
.
(1 rad/s
 N
2
5 6250 2229 87 530 921
1802 16
T
T
A
A
- = +
= TA = 1802 N �
PROBLEM 16.57
A 5 m beam weighing 2500 N is lowered by means of two cables 
unwinding from overhead cranes. As the beam approaches the 
ground, the crane operators apply brakes to slow the unwinding 
motion. Knowing that the deceleration of cable A is 6 m/s2 and 
the deceleration of cable B is 1 m/s2, determine the tension in each 
cable.
87
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PROBLEM 16.57 (Continued)
+ S SF F T T W maA B= + - =eff :
 
1802 16 2500
9 81
3 5
1589 79
.
.
( . )
.
 N N
2500
N
+ - =
=
T
T
B
B TB = 1590 N �
88
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PROBLEM 16.58
A 5-m beam weighing 2500 N is lowered by means of two cables 
unwinding from overhead cranes. As the beam approaches the 
ground, the crane operators apply brakes to slow the unwinding 
motion. Knowing that the deceleration of cable A is and the 
deceleration of cable B is 1 m/s2, determine the tension in each 
cable.
SOLUTION
Kinematics: a aB A= + 4a
 1≠ = 6≠ + 4a
 a = 1 2. rad/s2 
 a = +aA 2 5. a
 = 6 ≠ + ( . )( . )2 5 1 25 Ø
 a = 2 875. m/s2≠
 I mL= = = ◊1
2
1
12
2500
9 81
5 530 9212
( )
( . )
( .) kg m2 2
Kinetics:
 + S SM M T W ma IB B A= + = +( ) ( ( . ( .eff : m) m) m)4 1 5 1 5 a
 
TA( ( .
( . )( .
( . )
4 2500
2500
9 81
2 875 1 5
530 921
 m) N)(1.5 m) )
 (1.
- =
+ 225
N
)
. .
.
4 3750 1099 01 663 65
1378 165
T
T
A
A
- = +
fi = TA = 1378 N �
 + S SF F T T W maA B= + - =eff :
 1378.165
2500
9.81
+ - =TB 2500 2 875( . ) TB = 1855 N �
89
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PROBLEM 16.59
The steel roll shown has a mass of 1200 kg, a centroidal radius of gyration 
of 150 mm, and is lifted by two cables looped around its shaft. Knowing 
that or each cable TA = 3100 N and TB = 3300 N, determine (a) the angular 
acceleration of the roll, (b) the acceleration of its mass center.
SOLUTION
Data: m
I mk
r d
=
= = = ◊
= = =
1200
1200 0 150 27
1
2
1
2
0 100 0 05
2 2
 kg
 kg m2( )( . )
( . ) . 00
3100
3300
m
 N
N
T
T
A
B
=
=
(a) Angular acceleration.
 + S SM M T r T r IG G B A= - =( )eff : 2 2 a
 
a =
-
= -
2
2 3300 3100 0 050
27
( )
( )( )( . )
T T r
I
B A
 a = 0 741. rad/s2 �
(b) Acceleration of mass center.
 + S SF F T T mg may y A B= + - =( )eff : 2 2
 
a
T T
m
gA B=
+
-
= + -
2
2 3100 3300
1200
9 81
( )
( )
. a = 0 857. m/s2 ≠ �
90
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PROBLEM 16.60
The steel roll shown has a mass of 1200 kg, has a centroidal radius of 
gyration of 150 mm, and is being lowered by two cables looped around 
its shaft. Knowing that at the instant shown the acceleration ofthe roll 
is 150 mm/s2 downward and that for each cable TA = 3000 N, determine 
(a) the corresponding tension TB , (b) the angular acceleration of the roll.
SOLUTION
Data: m
I mk
r d
=
= = = ◊
= = =
1200
1200 0 150 27
1
2
1
2
0 100 0 05
2 2
 kg
 kg m2( )( . )
( . ) . 00
3000
m
 NTA =
 a = 0 150. m/s2 Ø
(a) Tension in cable B.
 + S SF F T T mg may y A B= + - = -( ) :eff 2 2
 
T
mg ma
T
m g a
T
B A
A
= - -
= - -
= - -
=
2
2
1200 9 81 0 150
2
3000
2796
( )
( )( . . )
 N TB = 2800 N �
(b) Angular acceleration.
 + S SM M T r T r IG G B A= - =( )eff : 2 2 a
 
a =
-
= -
= -
2
2 2796 3000
27
15 11
( )
( )( )
.
T T r
I
B A
 rad/s2 a = 15 11. rad/s2 �
91
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PROBLEM 16.61
By pulling on the string of a yo-yo, a person manages to make the yo-yo spin, while 
remaining at the same elevation above the floor. Denoting the mass of the yo-yo by m, 
the radius of the inner drum on which the string is wound by r, and the centroidal radius 
of gyration of the yo-yo by k ,determine the angular acceleration of the yo-yo.
SOLUTION
+ S SF F T mg T mgy y= - = =( ) : ;eff 0
 + S SM MG G= ( ) :eff Tr I
mgr mk
=
=
a
a2
 a =
r
k
B
2
 a = r
k
B
2
 �
92
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PROBLEM 16.62
The 100 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
m = 100g = 0.1 kg
 + S SF F T mg may y= - =( ) :eff
 T - =0 1 0 1 1. . ( ) (9.81) 
 T = 1 081. N T = 1 081. N �
 + S SM M Tr IG G= =( ) :eff a
 ( . ) ( . )1 081
6
1000
0 1
30
1000
2Ê
ËÁ
ˆ
¯̃
= ÊËÁ
ˆ
¯̃
a
 a = 1081
15
rad/s2 a = 72 1. rad/s2 �
93
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PROBLEM 16.63
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 bar, (b) the acceleration of Point A, (c) the 
acceleration of Point B.
SOLUTION
Statics: T T W mg1 2
1
2
1
2
= = =
(a) Angular acceleration:
+ S SM M T
L
G G=
Ê
ËÁ
ˆ
¯̃
=( ) :eff 2
Ia
 
1
2 2
1
12
2mg
L
mLÊËÁ
ˆ
¯̃
= a
 a = 39
L
 a = 39
L
 �
 + S SF F W T may y= - =( ) :eff 1
 mg mg ma- =1
2
 a g g= =1
2
1
2
a Ø
(b) Acceleration of A:
a a aA G A G= + /
 + aA g
L= -1
2 2
a
 = - ÊËÁ
ˆ
¯̃
1
2 2
39
g
L
L
 a gA = - aA g= ≠ �
94
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PROBLEM 16.63 (Continued)
(c) Acceleration of B:
a a aB G B G= + /
 + a a
L
g
L
L
gB = + = +
Ê
ËÁ
ˆ
¯̃
= +
2
1
2 2
39
2a aB g= 2 Ø �
95
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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 bar, (b) the acceleration of Point A, (c) the 
acceleration of Point B.
SOLUTION
Statics: T T W mg1 2
1
2
1
2
= = =
(a) Angular acceleration:
 + S SM M T
L
B
IG G=
Ê
ËÁ
ˆ
¯̃
=( ) :eff a
 
1
2
1
12
2mg
L
B
mLÊËÁ
ˆ
¯̃
= a
 a = g
L
 a = g
L
 �
 + S SF F W T may y= - =( ) :eff 1
 mg mg ma- =1
2
 a g g= =1
2
1
2
a Ø
(b) Acceleration of A:
 a a aA G A G= + /
 + aA g
L= -1
2 2
a
 = - ÊËÁ
ˆ
¯̃
=1
2 2
0g
L g
L
 aA = 0�
96
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PROBLEM 16.64 (Continued)
(c) Acceleration of B:
a a aB G B G= + /
 + aB a
L
g
L g
L
g= + = + ÊËÁ
ˆ
¯̃
=
2
1
2 2
4a aB g= Ø �
97
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PROBLEM 16.65
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 bar, (b) the acceleration of Point A, (c) the 
acceleration of Point B.
SOLUTION
Before spring 1 breaks:
 + SF T T Wy = ∞ + ∞ - =0 30 30 01 2: sin sin
Since T1 = T2 by symmetry,
 2 301T W mgsin ∞ = =
 T1 = mg 30∞
Immediately after spring 2 breaks, elongation of spring 1 is unchanged. Thus, we still have
 T1 = mg 30∞
(a) Angular acceleration:
 + S SM M T
L
IG G= ∞ =( ) : ( sin )eff 1 30 6
a
 ( sin )mg
L
mL30
6
1
12
2∞ = a a = g
L
 �
 + S SF Fx x= ( ) :eff T max1 30cos ∞ =
 mg ma gxcos , .30 0 866∞ = =a ¨
 + S SF Fy y= ( ) :eff W T may- ∞ =1 30sin
 mg mg ma gy y- ∞ = =sin .30 0 5a Ø
98
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PROBLEM 16.65 (Continued)
 Accelerations of A and B
Translation + Rotation about G
(b) Acceleration of A:
a a aA G A G g= + =/ [ .0 866 ¨ + 0 5. g Ø] +
Ê
ËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
g
L
L
2
≠
 = 0 866.g ¨ + 0 aA g= 0 866. ¨ �
(c) Acceleration of B:
a a aB G B G g= + =/ [ .0 866 ¨ + 0 5. g Ø] +
Ê
ËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
g
L
L
2
Ø
 = 0 866. g ¨ + g Ø aB g= 1 323. 49.1∞ �
99
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PROBLEM 16.66
A thin plate of the shape indicated and of mass m is suspended from two 
springs as shown. If spring 2 breaks, determine the acceleration at that instant 
(a) of Point A, (b) of Point B.
A circular plate of diameter b.
SOLUTION
 I m
b
mb= ÊËÁ
ˆ
¯̃
=1
2 2
1
8
2
2
Statics: T T W mg1 2
1
2
1
2
= = =
Kinetics: + S SM MG G= ( ) :eff T
b
I1 2
= a
 
1
2 2
1
8
2mg
b
mbÊËÁ
ˆ
¯̃
= a
 a = 2 g
b
 
 + S SF F W T may y= - =( )eff : 1
 mg mg ma g- = =1
2
1
2
a Ø
Kinematics:
Plane motion = Translation + Rotation
(a) a a a aA G A G= + =/ Ø+ 
b
2
a ≠= 1
2
g Ø+ ÊËÁ
ˆ
¯̃
 
b g
b2
2 ≠ aA g=
1
2
≠ �
(b) a a a aB G B G= + =/ Ø+ 
b
2
a Ø= 1
2
g Ø+ ÊËÁ
ˆ
¯̃
 
b g
b2
2 Ø aB g=
3
2
Ø �
100
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PROBLEM 16.67
A thin plate of the shape indicated and of mass m is suspended from two 
springs as shown. If spring 2 breaks, determine the acceleration at that instant 
(a) of Point A, (b) of Point B.
A thin hoop of diameter b.
SOLUTION
 I m
b
mb= ÊËÁ
ˆ
¯̃
=
2
1
4
2
2
Statics: T T W mg1 2
1
2
1
2
= = =
Kinetics: + S SM M T
b
IG G=
Ê
ËÁ
ˆ
¯̃
=( ) :eff 1 2
a
 
1
2 2
1
4
2mg
b
mbÊËÁ
ˆ
¯̃
=
 a = g
b
 
 
+ S SF F W T may y= - =( ) :eff
 mg mg ma g- = =1
2
1
2
a Ø
Kinematics:
Plane motion = Translation + Rotation
(a) a a a aA G A G= + =/ Ø+ 
b
2
a ≠= 1
2
g Ø+ ÊËÁ
ˆ
¯̃
 
b g
b2
≠ aA = 0 �
(b) a a a aB G B G= + =/ Ø+ 
b
2
a Ø= 1
2
g Ø+ ÊËÁ
ˆ
¯̃
 
b g
b2
Ø aB g= Ø �
101
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PROBLEM 16.68
A thin plate of the shape indicated and of mass m is suspended from two 
springs as shown. If spring 2 breaks, determine the acceleration at that 
instant (a) of Point A, (b) of Point B.
A square plate of side b.
SOLUTION
 I m b b= +1
12
2 2( )
 I mb= 1
6
2
Statics: T T W mg1 2
1
2
1
2
= = =
Kinetics: + S SM MG G= ( ) :eff T
b
I
2
= a
 
1
2 2
1
6
2mg
b
mbÊËÁ
ˆ
¯̃
= a
 a = 3
2
g
b
 
 
+
 S SF F W T may y= - =( ) :eff 1
 mg mg ma g- = =1
2
1
2
a Ø
Kinematics:
Plane motion = Translation + Rotation
102
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PROBLEM 16.68 (Continued)
(a) a a a aA G A G= + =/ Ø + 
b
2
a ≠
 aA
g=
2
 Ø + 3ÊËÁ
ˆ
¯̃
 
b g
b2 2
 ≠ = g
4
≠ aA g=
1
4
≠ �
(b) a a a aB G B G= + =/ Ø + 
b
2
a Ø
 aB g=
1
2
 Ø + ÊËÁ
ˆ
¯̃
 
b g
b2
3
2
 Ø = 5
4
g Ø aB g=
5
4
Ø �
103
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PROBLEM 16.69
A bowler projects an 200-mm-diameter ball weighing 6 kg along an alley with a forward 
velocity v0 of 5 m/s and a backspin w0 of 9 rad/s. Knowing that the coefficient of kinetic 
friction between the ball and the alley is 0.10, determine (a) the time t1 at which the ball 
will start rolling without sliding, (b) the speed of the ball at time t1, (c) the distance the 
ball will have traveled at time t1.
SOLUTION
Kinetics:
 + S SF F mg max x k= =( ) :eff m
 a = mk g ¨
 + S SM MG G= ( ) :eff Fr I= a
 ( )mk mg r mr=
2
5
2
 a = 5
2
mk g
r
 
Kinematics: When the ball rolls, the instant center of rotation is at C, and when
 t t v r= =1 w (1)
 v v at v gk= - = -0 0 m t (2)
 w w a w
m
= - + = - +0 0
5
2
t
g
r
tk
When t = t1:
Eq. ( ) : :1 v r= w v gt
g
r
t rk
k
0 1 0 1
5
2
- = - +Ê
ËÁ
ˆ
¯̃
m w
m
 v gt r gtk k0 1 0 1
5
2
- = - +m w m
 t
v r
gk
1
0 02
7
=
+( )w
m
 (3)
 v r0 5 9 100 0 1= = = =m/s, rad/s, mm m0w .
(a) t1
2 5 0 1 9
7 0 1 9 81
1 71836= + =( . ( ))
( . )( . )
. s t1 1 718= . s �
104
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PROBLEM 16.69 (Continued)
(b) Eq. (2): v v gtk1 0 1
5 0 1 9 81 1 71836
= -
= -
m
. ( . )( . )
 = 3 31429. m/s v1 3 31= . m/s �
(c) a gk= = =m 0 1 9 81 0 981. ( . ) .m/s m/s
2 2
 + s v t at1 0 1 1
2
2
1
2
5 1 71836
1
2
0 981 1 71836
= -
= -( )( . ) ( . )( . )
 = - =8 5918 1 44833 7 14347. . . m s1 7 14= . m �
105
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PROBLEM 16.70
Solve Problem 16.69, assuming that the bowler projects the ball with the same forward 
velocity but with a backspin of 18 rad/s.
PROBLEM 16.69 A bowler projects a 200-mm-diameter ball of mass 6 kg along an 
alley with a forward velocity v0 of 5 m/s and a backspin w0 of 9 rad/s. Knowing that 
the coefficient of kinetic friction between the ball and the alley is 0.10, determine 
(a) the time t1 at which the ball will start rolling without sliding, (b) the speed of the ball 
at time t1, (c) the distance the ball will have traveled at time t1.
SOLUTION
Kinetics:
 + S SF F mg max x k= =( ) :eff m
 a = mk g ¨
 
+
 
S SM M Fr I
mg r mr
g
r
G G
k
k
= =
=
=
( ) :
( )
eff a
m
m
2
5
5
2
2
a
Kinematics: When the ball rolls, the instant center of rotation is at C, and when
 t t v r= =1 w (1)
 v v at v gk= - = -0 0 m t (2)
 
w w a w
m
= - + = - +0 0
5
2
t
g
r
tk
When t = t1:
Eq. (1) n = rw: v gt
g
r
t r
v gt r gt
t
v
k
k
k k
0 1 0 1
0 1 0 1
1
0
5
2
5
2
2
7
- = - +Ê
ËÁ
ˆ
¯̃
- = - +
=
+
m wm
m w m
( rr
g
v r
k
w
m
w
0
0 05 18 0 1
)
.= = =m/s, rad/s, m
 (3)
106
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PROBLEM 16.70 (Continued)
(a) Eq. (3): t1
2
7
5 0 1 18
0 1 9 81
1 98049=
+( )
=
( . )( )
. ( . )
. s t1 1 980= . s �
(b) Eq. (2): v v g
v
k1 0
1
5 0 1 9 81 1 98049
3 0571
= -
= -
=
m t
. ( . )( . )
. m/s v1 3 06= . m/s �
(c) a g
s v t a t
k= = =
= -
=
m ( . )( . ) .
( )( .
0 1 9 81 0 981
1
2
5 1 980
1 0 1 1
2
m/s m/s
m/s
2 2
449
1
2
0 981 1 98049
9 90245 1 92391
7 9785
2s m/s s
m
2) ( . )( . )
. .
.
-
= -
= s1 7 98= . m �
+
107
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PROBLEM 16.71
A sphere of radius r and mass m is projected along a rough horizontal surface with the 
initial velocities indicated. If the final velocity of the sphere is to be zero, express, in 
terms of v0, r, and mk, (a) the required magnitude of w0 (b) the time t1 required for the 
sphere to come to rest, (c) the distance the sphere will move before coming to rest.
SOLUTION
Kinetics: I mk
F F F ma
mg ma g
M M Fr I
mg
x x
k k
G G
k
=
= =
= = ¨
= =
2
S S
S S
( )
( )
(
eff
eff
:
:
m m
a
m
a
))r mk
gr
k
k
=
=
2
2
a
ma
Kinematics: + v v at
v v gtk
= -
= -
0
0 m
For n = 0 when t = t1 0 0 1 1
0= - =v gt t
v
gk k
m
m
; (1)
 + w w a
w w
m
= -
= -
0
0 2
t
gr
k
tk
For w = 0 when t t= 1 0 0 2 1 1
2
0= - =w
m
m
wk
k
gr
k
t t
k
gr
; (2)
Set Eq. (1) Eq. (2)= 
v
g
k
gr
r
k
v
k k
0
2
0 0 2 0m m
w w= =; (3)
Distance traveled: s v t at1 0 1 1
21
2
= -
 s v
v
g
g
v
g
s
v
gk
k
k k
1 0
0 0
2
1
0
21
2 2
=
Ê
ËÁ
ˆ
¯̃
-
Ê
ËÁ
ˆ
¯̃
=
m
m
m m
( ) ; (4)
+
+
108
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PROBLEM 16.71 (Continued)
For a solid sphere k r2 2
2
5
=
(a) Eq. (3): w0 2
5
2 0
05
2
= =r
r
v
v
r
 w0 0
5
2
=
v
r
 �
(b) Eq. (1) t
v
gk
1
0=
m
 �
(c) Eq. (4) s
v
gk
1
0
2
2
=
m
 �
109
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PROBLEM 16.72
Solve Problem 16.71, assuming that the sphere is replaced by a uniform thin hoop of 
radius r and mass m.
PROBLEM 16.71 A sphere of radius r and mass m is projected along a rough horizontal 
surface with the initial velocities indicated. If the final velocity of the sphere is to be zero, 
express, in terms of v0, r, and mk , (a) the required magnitude of w 0, (b) the time t1 required 
for the sphere to come to rest, (c) the distance the sphere will move before coming to rest.
SOLUTION
Kinetics: I mk= 2
 + S SF Fx x= ( )eff : F ma=
 m mk kmg ma g= =a ¨
 + S SM M Fr IG G= =( )eff : a
 ( )m ak mg r mk=
2
 a = mk gr
k 2
 
Kinematics: + v v at
v v gtk
= -
= -
0
0 m
For v = 0 when t t= 1 0 0 1 1
0= - =v gt t
v
gk k
m
m
; (1)
 + w w a
w w
m
= -
= -
0
0 2
t
gr
k
tk
For w = 0 when t t= 1 0 0 2 1 1
2
0= - =w
m
m
wk
k
gr
k
t t
k
gr
; (2)
Set Eq. (1) Eq. (2)= 
v
g
k
gr
r
k
v
k k
0
2
0 0 2 0m m
w w= =; (3)
Distance traveled: s v t at1 0 1 1
21
2
= -
 s v
v
g
g
v
g
s
v
gk
k
k k
1 0
0 0
2
1
0
21
2 2
=
Ê
ËÁ
ˆ
¯̃
-
Ê
ËÁ
ˆ
¯̃
=
m
m
m m
( ) ; (4)
110
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PROBLEM 16.72 (Continued)
For a hoop, k r=
(a) Eq. (3): w0 2 0
0= =r
r
r
v
r
 w0 0=
v
r
 �
(b) Eq. (1): t
v
gk
1
0=
m
 �
(c) Eq. (4): s
v
gk
1
0
2
2
=
m
 �
111
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PROBLEM 16.73
A uniform sphere of radius r and mass m is placed with no initial velocity 
on a belt that moves to the right with a constant velocity v1. Denoting by mk 
the coefficient of kinetic friction between the sphere and the belt, determine 
(a) the time t1 at which the sphere will start rolling without sliding, (b) the 
linear and angular velocities of the sphere at time t1.
SOLUTION
Kinetics: + S SF F F max x= =( )eff :
 mk mg ma=
 a = mk g Æ
 + S SM M Fr IG G= =( )eff : a
 ( )m ak mg r mr=
2
5
2
 a = 5
2
mk g
r
 
Kinematics: + v at gtk= = m (1)
 + w a
m
= =t
g
r
tk
5
2
 (2)
C = Point of contact with belt
 + v v r gt
g
r
t r
v gt
C k
k
C k
= + = + Ê
ËÁ
ˆ
¯̃
=
w m
m
m
5
2
7
2
(a) When sphere starts rolling ( ),t t= 1 we have
 
v v
v gt
C
k
=
=
1
1 1
7
2
m t
v
gk
1
12
7
=
m
 �
(b) Velocities when t t= 1
 Eq. (1): v g
v
gk
=
Ê
ËÁ
ˆ
¯̃
m
m
2
7
1 v =
2
7 1
v Æ �
 Eq. (2): w
m
m
= Ê
ËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
5
2
2
7
1k
k
g
r
v
g
 w = 5
7
1v
r
 �
112
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PROBLEM 16.74
A sphere of radius r and m has a linear velocity v0 directed to the left 
and no angular velocity as it is placed on a belt moving to the right 
with a constant velocity v1. If after first sliding on the belt the sphere 
is to have no linear velocity relative to the ground as it starts rolling on 
the belt without sliding, determine in terms of v1 and the coefficient of 
kinetic friction mk between the sphere and the belt (a) the required value 
of v0, (b) time t1 at which the sphere will start rolling on the belt, (c) the 
distance the sphere will have moved relative to the ground at time t1.
SOLUTION
Kinetics: + S SF F F maxx= =( ) :eff
 mk mg ma=
 a = mk g Æ
 + S SM M F IG G r= =( )eff : a
 ( )m ak mg r mr=
5
2
2
 a = 5
2
mk g
r
 
Kinematics: + v v at v gtk= - = -0 0 m (1)
 + w a
m
= =t
g
r
tk
5
2
 (2)
C = Point of contact with belt
 + v v r
v v r
g
r
t
C
C
k
= - +
= - +
w
m5
2
 v v
g
tC
k= - +
5
2
m
 (3)
But, when t t v= =1 0, and v vc = 1
Eq. (3): v
g
tk1 1
5
2
=
m
 t
v
gk
1
12
5
=
m
 �
Eq. (1): v v gtk= -0 m
When t t v= =1 0, , 0
2
50
1= -
Ê
ËÁ
ˆ
¯̃
v g
v
gk k
m
m
 v v0 1
2
5
= �
113
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PROBLEM 16.74 (Continued)
Distance when t t= 1: + s v t at= -0 1 1
21
2
 s v
v
g
g
v
g
s
v
g
k
k
k
k
= ÊËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
-
Ê
ËÁ
ˆ
¯̃
=
2
5
2
5
1
2
2
5
4
25
1
1 1
2
1
2
m
m
m
m
( )
--ÊËÁ
ˆ
¯̃
2
25
; s =
2
25
1
2v
gkm
¨ �
114
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PROBLEM 16.75
Show that the couple I a of Figure 16.15 can be eliminated by attaching the vectors mat 
and man at a Point P called the center of percussion, located on line OG at a distance 
GP k r= 2 / from the mass center of the body.
SOLUTION
OG r rt= = a a
We first observe that the sum of the vectors is the same in both figures. To have the same sum of moments 
about G, we must have
+ S SM M I ma GPG G t= =: a ( )( )
 mk mr GP2a a= ( ) GP k
r
=
2
( . . .)Q E D �
Note: The center of rotation and the center of percussion are interchangeable. Indeed, since OG r= , we may 
write
GP
k
GO
GO
k
GP
= =
2 2
or
Thus, if Point P is selected as center of rotation, then Point O is the center of percussion.
115
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PROBLEM 16.76
A uniform slender rod of length L = 1 m and weight W = 2 kg hangs freely from a hinge at A. 
If a force P of magnitude 8 N is applied at B horizontally to the left (h = L), determine (a) the 
angular acceleration of the rod, (b) the components of the reaction at A.
SOLUTION
a I mL= =1
2
1
12
2a
+ S SM M PL ma L IA A= = +( ) : ( )eff 2
a
= ÊËÁ
ˆ
¯̃
+
=
m
L L
mL
PL mL
2 2
1
12
1
3
2
2
a a
a
(a) Angular acceleration. a =
=
3
3 8
2
P
mL
(
(
 N)
 kg)(1 m)
= 12 rad/s2 a = 12 00. rad/s2 �
(b) Components of the reaction at A.
+ S SF F A Wy y y= - =( ) :eff 0
A Wy = = 2g )= 2(9.81 = 19.62 N A y = 19 62. N ≠ �
+ S SF F A P max x x= - = -( ) :eff
A P m
L
P m
L P
mL
P
A
P
x
x
= - ÊËÁ
ˆ
¯̃
= - ÊËÁ
ˆ
¯̃
= -
= - = - = -
2 2
3
2
2
8
2
4
a
 N A x = 4 00. N ¨ �
116
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PROBLEM 16.77
In Problem 16.76, determine (a) the distance h for which the horizontal component of the 
reaction at A is zero, (b) the corresponding angular acceleration of the rod.
PROBLEM 16.76 A uniform slender rod of length L = 1 m and mass m = 2 kg hangs freely 
from a hinge at A. If a force P of magnitude 8 N is applied at B horizontally to the left (h = L), 
determine (a) the angular acceleration of the rod, (b) the components of the reaction at A.
SOLUTION
a a=
=
L
I mL
2
1
12
2 
+ S SF Fx x= ( ) :eff P ma=
P m
L= ÊËÁ
ˆ
¯̃2
a
(b) Angular acceleration. a = 2P
mL
a = 2 8
2
(
(
 N)
 kg)(1 m)
 a = 8 00. rad/s2 �
+
 S SM MG G= ( ) :eff P h
L
I P h L
P h
L
mL
P
mL
PL
h
L
-ÊËÁ
ˆ
¯̃
= -
-ÊËÁ
ˆ
¯̃
= ÊËÁ
ˆ
¯̃
=
-Ê
2
2
1
12
2
6
2
2
a : ( )
ËËÁ
ˆ
¯̃
= = + = = =L h L L L
6 2 6
2
3
2
3
1
2
3
; ( ) m
(a) Distance h. h = 0 667. m �
117
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PROBLEM 16.78
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 r I mL= =a 1
12
2
+
 S SM M P r L ma r IC C= +
Ê
ËÁ
ˆ
¯̃
= +( ) : ( )eff 2
a
= +
+ÊËÁ
ˆ
¯̃
= +ÊËÁ
ˆ
¯̃
( )mr r mL
P r
L
m r L
a a
a
1
12
2
1
12
2
2 2
Substitute data: ( .
.
( ( .7 5
0 9
4
1
12
0 9 N) 0.225 m
 m
2
 kg) (0.225 m) m)2 2+È
ÎÍ
˘
˚̇
= +È
ÎÍÍ
˘
˚̇
a
50 125 0 4725
107 14
. .
.
=
=
a
a rad/s2 a = 107 1. rad/s2 �
(b) Components of reaction at C.
+ S SF F C Wy y y= - =( ) :eff 0
C W mgy = = = ( )4 kg)(9.81 m/s
2 Cy = 39 2. N ≠ �
+ S SF F C P max x x= - = -( ) :eff
C P ma P m Px = - = -
= -
( )
( )
a
75 4 N kg)(0.275 m)(107.14 rad/s2
Cx = -
= -
75 96 4 N N
21.4 N
.
 Cx = 21 4. N ¨ �
118
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PROBLEM 16.79
In Problem 16.78, determine (a) the distance r for which the horizontal component of the reaction at C is zero, 
(b) the corresponding angular acceleration of the rod.
SOLUTION
(a) Distance r . a r
I mL
=
=
a
1
12
2
+ S SF F P max x= =( ) :eff
P m r
P
mr
=
=
( )a
a (1)
+
 S SM M P L IG G= =( ) :eff 2
a
P
L
mL
P
L
mL
P
mr
2
1
12
2
1
12
2
2
=
= ÊËÁ
ˆ
¯̃
a
L L
r
r L r
2 12
1
6
900
2
=
= = mm
6
 r = 150 mm �
(b) Angular acceleration.
 Eq. (1): a = = ( ) =
P
mr
P
m
P
mLL
6
6
a = =6 75
4
125
(
(
 N)
 kg)(0.9 m)
 rad/s2 a = 125 rad/s2 �
119
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PROBLEM 16.80
Auniform slender rod of length l and mass m rotates about a vertical axis AA' 
with a constant angular velocity w. Determine the tension in the rod at a distance 
x from the axis of rotation.
SOLUTION
Consider the portion of rod of length ( )l x- and mass
¢ =
-
m m
l x
l
:
We have a = 0
a a= =n rw
2 ¨
+ S SF F T m a m rn n= = ¢ = ¢( )eff : w
2
= -ÊËÁ
ˆ
¯̃
+
m
l x
l
l xw2
2
 T
m
l
l x= -1
2
2 2 2w ( ) �
120
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PROBLEM 16.81
A large flywheel is mounted on a horizontal shaft and rotates at a constant rate of 1200 rpm. Experimental 
data indicate that the total force exerted by the flywheel on the shaft varies from 55 kN upward to 85 kN 
downward. Determine (a) the mass of the flywheel, (b) the distance from the center of the shaft to the 
mass center of the flywheel.
SOLUTION
w p
p
w
= ÊËÁ
ˆ
¯̃
=
=
1200
2
60
40
2
rpm
rad/s
a rm
 + S SF F mg man= - =eff kN: 85
 85 2- =mg mrw (1)
 + S SF F mg man= + =eff kN: 55
 55 2+ =mg mrw (2)
(a) Mass of the flywheel.
 Eq. (2) – Eq. (1): 30 2 0
30 10 2 9 813
kN
N m/s2
- =
¥ =
mg
m( . )
m = 1529 kg m = 1529 kg �
(b) Distance r .
 Eq. (1) + Eq. (2): 140 2
140 10 2 1529 40
2
3 2
kN
N kg
=
¥ =
mr
r
w
p( ) ( )
r = ¥ -2 90 10 3. m r = 2 90. mm �
121
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PROBLEM 16.82
A turbine disk of mass 26 kg rotates at a constant rate of 9600 rpm. 
Knowing that the mass center of the disk coincides with the center of 
rotation O, determine the reaction at O immediately after a single blade 
at A, of mass 45 g, becomes loose and is thrown off.
SOLUTION
w p
w p
= ÊËÁ
ˆ
¯̃
=
9600
2
60
320
rpm
rad/s
Consider before it is thrown off.
+ S SF F R ma mFn= = =eff : w
2
= ¥
=
-( )( . )( )
.
45 10 0 3 320
13 64
3 2kg m
kN
p
R
Before blade was thrown off, the disk was balanced ( ).R = 0 Removing vane at A also removes its reaction, so 
disk is unbalanced and reaction is
 R = 13 64. kN Æ �
122
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PROBLEM 16.83
The shutter shown was formed by removing one quarter of a disk of 20 mm 
radius and is used to interrupt a beam of light emanating from a lens at C. 
Knowing that the shutter has a mass of 50 g and rotates at the constant rate 
of 24 cycles per second, determine the magnitude of the force exerted by 
the shutter on the shaft at A.
SOLUTION
See inside front cover for centroid of a circular sector.
r
r
r
r
a rn
=
=
=
=
=
2
3
2 20
3
4 0014
24
3
4
3
4
2
sin
( )sin ( )
( )
.
a
a
p
p
w
w
mm
mm
cycless
rad/s
rad/s
eff
=
=
= = =
24 2
150 8
2
( )
.
:
p
w
wS SF F R ma mrn
+
= ÊËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
50
1000
4 0014
1000
150 8 2
.
( . )
R = 4 5495. N
Force on shaft is R = 4 5495. N
Magnitude: R = 4 55. N �
123
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PROBLEM 16.84
A uniform rod of length L and mass m is supported as shown. If 
the cable attached at end B suddenly breaks, determine (a) the 
acceleration of end B, (b) the reaction at the pin support.
SOLUTION
w a
L= =0
2
a
+
 S SM M W L I ma LA A= = +( ) :eff 2 2
a
mg
L
mL m
L L
2
1
12 2 2
2= + ÊËÁ
ˆ
¯̃
a a
mg
L
mL
g
L2
1
3
3
2
2= =a a 
(b) Reaction at A.
+ S SF F A mg ma m Ly y= - = - = -( ) :eff 2
a
A mg m
L g
L
A mg mg
- = - ÊËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
- = -
2
3
2
3
4
A mg= 1
4
 A = 1
4
mg≠ �
(a) Acceleration of B.
a a aB n B A L= + = +/ 0 a Ø
aB L
g
L
g= ÊËÁ
ˆ
¯̃
=3
2
3
2
Ø aB g=
3
2
Ø �
124
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PROBLEM 16.85
A uniform rod of length L and mass m is supported as shown. If 
the cable attached at end B suddenly breaks, determine (a) the 
acceleration of end B, (b) the reaction at the pin support.
SOLUTION
w a= =0
4
a
L
+
 S SM M W L I ma LC C= = +( ) :eff 4 4
a
mg
L
mL m
L L
mg
L
mL
4
1
12 4 4
4
7
48
2
2
= + ÊËÁ
ˆ
¯̃
=
a a
a a = 12
7
g
L
(b) Reaction at C.
+ S SF F C mg ma m Ly y= - = - = -( ) :eff 4
a
C mg m
L g
L
C mg mg
- = - ÊËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
- = -
4
12
7
3
7
C mg= 4
7
 C = 4
7
mg ≠ �
(a) Acceleration of B.
a a a
L
a
L g
L
g
B C C B
B
= + = +
= ÊËÁ
ˆ
¯̃
=
/ 0
3
4
3
4
12
7
9
7
a
 aB g=
9
7
Ø �
125
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PROBLEM 16.86
A slender uniform cone of mass m can swing freely about the horizontal 
rod AB. If the cone is released from rest in the position shown, determine 
(a) the acceleration of the tip D, (b) the reaction at C.
SOLUTION
From inside cover I m a hD = +
Ê
ËÁ
ˆ
¯̃
3
5
1
4
2 2
For slender cone, neglect ‘a’ I mhD =
3
5
2
Parallel axis theorem I I m h
mh I mh
I mh I mh
D = +
Ê
ËÁ
ˆ
¯̃
= +
= -ÊËÁ
ˆ
¯̃
=
3
4
3
5
9
16
3
5
9
16
3
80
2
2 2
2 2
+
 S SM M W h I ma hC C= = +( ) :eff 4 4
a
mg
h
mh m
h h
mg
h
mh
mg
h
mh
4
3
80 4 4
4
3
80
1
16
4
1
10
2
2
2
= + ÊËÁ
ˆ
¯̃
= +ÊËÁ
ˆ
¯̃
=
a a
a
aa
a = 5
2
g
h
(a) Acceleration of D. a h h
g
hD
= = ÊËÁ
ˆ
¯̃
a 5
2
 aD g= 2 50. Ø �
126
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PROBLEM 16.86 (Continued)
(b) Reaction at C. + S SF F C mg may y= - = -( ) :eff
C mg m
h
C mg m
h g
h
C mg mg
- = -
- = - ÊËÁ
ˆ
¯̃
- = -
4
4
5
25
8
a
 C = 3
8
mg ≠ �
127
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PROBLEM 16.87
The object ABC consists of two slender rods welded together at Point B. Rod AB has 
a mass of 1 kg and bar BC has a mass of 2 kg. Knowing the magnitude of the angular 
velocity of ABC is 10 rad/s when q = 0, determine the components of the reaction at 
Point C when q = 0.
SOLUTION
Masses and lengths: m L m LAB AB BC BC= = = =1 0 6 2 0 3kg m, kg, m, . .
Moments of inertia: I m L
I m L
AB AB AB
BC BC BC
= = = ¥ ◊
= =
-1
12
1
12
1 0 3 7 5 10
1
12
1
1
2 2 3
2
( )( . ) . kg m2
22
2 0 6 60 102 3( )( . ) = ¥ ◊- kg m2
Geometry: r
r L
AB
BC BC
= + =
= =
=
= ∞
0 6 0 15 0 61847
1
2
0 3
0 15
0 6
14 036
2 2. . .
.
tan
.
.
.
m
m
b
b
Kinematics: Let a = a be the angular acceleration of object ABC.
( )aAB t ABr= a b
( )aB n ABr= w
2 b
( )aBC BCt r= a Æ
( )aBC n BCr= w
2≠
Kinetics: + SMC = + S( ) :MC eff m g
L
I r m a tAB
AB
AB AB AB AB2
= +a ( )
+ +
= + + +( )
I r m a t
I m r I m r
BC BC BC BC
AB AB AB BC BC BC
a
a
( )
2 2
( )( . )( . ) [( . ) ( )( . )
( ) ( )( .
1 9 81 0 15 7 5 10 1 0 61847
60 10 2 0
3 2
3
= ¥ +
+ ¥ +
-
- 33 2) ]a a = 2 3357. rad/s2
128
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PROBLEM 16.87 (Continued)
+ SFX = + S( ) :FX eff C m a m a m ax AB AB t AB AB n BC BC t= + +( ) cos ( ) sin ( )b b
= + +
=
m r m rAB AB BC BC( cos sin )
( )( . )( . cos .
a b w b a2
1 0 61847 2 3357 14 035∞∞ + ∞ +
=
10 14 035 2 0 3 2 3357
17 802
2 sin . ) ( )( . )( . )
. N
+ SFy = 
+ S( ) :Fy eff C m g m g m a m a m ay AB BC AB AB t AB AB n BC BC n- - = - + +( ) sin ( ) cos ( )b b
C m m g m r m r
C
y AB BC AB AB BC BC
y
= + + - +
= +
( ) ( cos sin )
( )( . ) ( )
w b a b w2
3 9 81 1 (( . )( cos . . sin . )
( )( . )( )
0 61847 10 14 035 2 3357 14 035
2 0 3 10
2
2
∞ - ∞
+
Cy == 149 08. N
Reaction at C. C = +
=
17 802 149 08
150 1
2 2. .
. N
tan
.
.
.
b
b
=
= ∞
17 802
149 08
83 2 C = 150 1. N 83 2. ∞ �
129
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PROBLEM 16.88
An 4-kg slender rod AB and a 2.5-kg slender rod BC are connected 
by a pin at B and by the cord AC. The assembly can rotate in a 
vertical plane under the combined effect of gravity and a couple M 
applied to rod BC. Knowing that in the position shown the angular 
velocity of the assembly is zero and the tension in cord AC is equal 
to 30 N, determine (a) the angular acceleration of the assembly, 
(b) the magnitude of the couple M.
SOLUTION
(a) Angular acceleration.
 Rod AB: I m lAB AB AB=
=
= ◊
1
12
1
12
4 0 6
0 12
2
2( ( . )
.
kg) m
kg m2
+
 S SM M IB B AB= - + =( ) : ( )( . ) ( )( . )( .eff N m m)
3
5
30 0 6 4 9 81 0 3 a
I ABa = 0 972. Nm (1)
( . ) .0 12 0 972a = Nm a = 8 10. rad/s2 �
 Entire assembly: Since AC is taut, assembly rotater about C as a rigid body.
 Kinematics:
 CB
CG CBBC
= + =
= =
0 3 0 225 0 375
1
2
0 1875
2 2. . .
.
m
m
 aBC = ( . )0 1875 m a 
 aAB = ( . )0 225 m a Æ
130
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PROBLEM 16.88 (Continued)
 Kinetics: I m CBBC BC=
= -
= ◊
1
12
1
12
2 5 0 375
15
512
2
2
( )
( . ( . )kg) m
kg m2
(b) Couple M.
+ S SM MC C= ( ) :eff
M W m I m I
M
BC BC BC BC AB AB AB- = + + +
-
( . ) ( . ) ( . )
( .
0 15 0 1875 0 225
2
m m ma aa a
55 9 81 0 3 2 5 0 1875
15
512
4 0 2252 2)( . )( . ) ( . )( . ) ( ) ( )( . )= + + +a a a aI AB
 Substitute a = 8 1. rad/s2 and from Eq. (1), IABa = 0 972.
M
M
- =
= ◊
7 3575 3 56146
10 919
. .
. N m M = ◊10 92. N m �
131
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PROBLEM 16.89
Two uniform rods, ABC of mass 3 kg and DCE of mass 4 kg, are 
connected by a pin at C and by two cords BD and BE. The T-shaped 
assembly rotates in a vertical plane under the combined effect of gravity 
and of a couple M which is applied to rod ABC. Knowing that at the 
instant shown the tension is 8 N in cord BD determine (a) the angular 
acceleration of the assembly, (b) the couple M.
SOLUTION
We first consider the entire system and express that the external forces are equivalent to the effective forces of 
both rods.
( ) .
( ) .
( )( . )
.
a
a
I
I
AC t
DE t
AC
=
=
=
= ¥ ◊-
0 15
0 3
1
12
3 0 3
22 5 10
2
3
a
a
kg m
kg m2
DDE =
= ¥ ◊-
1
12
4 0 4
53 333 10
2
3
( )( . )
.
kg m
kg m2
+ S SM M M I a I a
M
A A AC AC t DE DE t= = + + +
= ¥
( ) : ( ) ( . ) ( ) ( . )
.
eff a a3 0 15 4 0 3
22 5 110 3 0 15 0 15 53 333 10 4 0 3 0 33 3- -+ + ¥ +a a a a( . )( . ) . ( . )( . )
M = 0 50333. a (1)
We now consider rod DE alone:
+ S SM M T T IC C BE BD DE=
Ê
ËÁ
ˆ
¯̃
- ÊËÁ
ˆ
¯̃
=( ) : ( . ) ( . )eff m m
3
5
0 2
3
5
0 2 a
0 12 53 333 10
0 44444
3. ( ) .
.
T T
T T
BE BD
BE BD
- = ¥
- =
- a
a (2)
132
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PROBLEM 16.89 (Continued)
Given data: T
T
BD
BE
=
=
2
8
N
N
(a) Angular acceleration.
 Substitute in (2): 8 2 0 44444- = . a a = 13 50. rad/s2 �
(b) Couple M.
 Carry value of a into (1): M = 0 50333 13 50. ( . ) M = 6.79 N · m �
133
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PROBLEM 16.90
A 1.5-kg slender rod is welded to a 5-kg uniform disk asshown. The 
assembly swings freely about C in a vertical plane. Knowing that in the 
position shown the assembly has an angular velocity of 10 rad/s clockwise, 
determine (a) the angular acceleration of the assembly, (b) the components 
of the reaction at C.
SOLUTION
Kinematics:
a CGn = =( ) ( . )( )w
2 0 14 10m rad/s2
an = 14 m/s
2¨
at CG= =( ) ( . )a a0 14 m Ø
Kinetics: I m CGdisk disk
2
kg m
kg m
=
=
= ¥ ◊-
1
2
1
2
5 0 08
16 10
2
2
3
( )
( )( . )
I m ABAB AB=
=
= ¥ ◊-
1
12
1
12
1 5 0 12
1 8 10
2
3
( )
( . )( . )
.
kg m
kg m2
(a) Angular acceleration.
+ S SM MC C= ( ) :eff
W I m a IAB AB t AB( . ) ( . )
( . )( . )( .
0 14 0 14
1 5 9 81 0 14
m m
kg m/s m
disk
2
= + +a a
)) ( . )( . )
. ( .
= + +
◊ = ¥ + ¥- -
I I ABdisk kg m
N m
a a a1 5 0 14
2 060 16 10 29 4 10
2
3 3 ++ ¥
◊ = ¥ ◊
-
-
1 8 10
2 060 47 2 10
3
6
. )
. ( . )
a
aN m kg m2
a = 43 64. rad/s2 a = 43 6. rad/s2 �
134
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PROBLEM 15.90 (Continued)
(b) Components of reaction of C.
+ S SF F C m ax x x AB n= = - = -( ) : ( . )( )eff
2kg m/s1 5 14
Cx = -21 0. N Cx = 21 0. N¨ �
+ S SF Fy y= ( ) :eff at = ( . )( )0 14 m a
C m g m g m a
C
y AB AB t
y
- - = -
- - = -
disk
kg kg kg( ) . ( . ) . ( . )( .5 9 81 1 5 9 81 1 5 0 144 43 64
49 05 14 715 9 164
m rad/s
N N N
2)( . )
. . .Cy - - = -
Cy = +54 6. N Cy = 54 6. N ≠ �
135
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PROBLEM 16.91
A 5-kg uniform disk is attached to the 3-kg uniform rod BC by 
means of a frictionless pin AB. An elastic cord is wound around 
the edge of the disk and is attached to a ring at E. Both ring E and 
rod BC can rotate freely about the vertical shaft. Knowing that the 
system is released from rest when the tension in the elastic cord 
is 15 N, determine (a) the angular acceleration of the disk, (b) the 
acceleration of the center of the disk.
SOLUTION
(a) Angular acceleration of the disk.
 Disk: I m rdisk disk kg m= =
1
2
1
2
5 0 0752 2( )( . )
Idisk kg m= ¥ ◊
-14 06 10 3 2.
+ S SM M IA A= =( ) ( )( . )eff disk disk: N N15 0 075 a
1 125 14 06 103 2. ( . )N m kg m disk◊ = ¥ ◊ a
adisk rad/s= 80 0
2. adisk rad/s= 80 0
2. �
(b) Acceleration of center of disk.
 Entire assembly
I m BCBC BC= = = ¥ ◊
1
12
1
12
3 0 15 5 625 102 2 3 2( ) ( )( . ) .kg m kg m
136
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PROBLEM 16.91 (Continued)
 Assume aBC is aA BC= ( . )0 15 m a ≠
a BC= ( . )0 075 m a
+ S SM M I m a m a IC C A BC BC= = - - -( ) ( . ) ( . )eff disk disk disk: m m0 0 15 0 075a aBBC
0 14 06 10 80 5 0 153 2 2 2= ¥ ◊ --( . )( ) ( )( . )kg m rad/s kg m aBC
- - ¥ ◊( )( . ) ( . )3 0 075 5 625 102 3 2kg m kg ma aBC BC
0 1 125 0 1125 16 375 10 5 625 103 3= - - ¥ - ¥- -. . . .a a aBC BC BC
0 1 125 0 135= -. . aBC
aBC = +8 333
2. rad/s aBC = 8 33
2. rad/s
a ACA BC= =( ) ( . )( . )a 0 15 8 333
2m rad/s
aA = +1 25
2. m/s aA = 1 250
2. m/s ≠ �
 Note: Answers can also be written:
adisk rad/s m/s= = -( ) ( . )80 1 25
2 2j a iA
137
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PROBLEM 16.92
Derive the equation SM IC C= a for the rolling disk of Figure 16.17, where SMC represents the sum of the 
moments of the external forces about the instantaneous center C, and IC is the moment of inertia of the disk 
about C.
SOLUTION
+
 S S SM M M ma r I mr r IC C C= = + = +( ) ( ) ( )eff : a a a
= +( )mr r Ia a
But, we know that I mr IC = +
2
Thus: SM IC C= a (Q.E.D.) �
138
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PROBLEM 16.93
Show that in the case of an unbalanced disk, the equation derived in Problem 16.92 is valid only when the mass 
center G, the geometric center O, and the instantaneous center C happen to lie in a straight line.
SOLUTION
Kinematics:
a a a= +C G C/
= + ¥ + ¥ ¥a r rC G C G Ca w w/ /( )
or, since w ^ rG C/
a a r= + ¥ -C G C G Cra / /a
2 (1)
Kinetics:
S S SM M I mC C C G C= = + ¥( ) /eff : M a r a
Recall Eq. (1): SM r a rC G C C G C G CI m= + ¥ + ¥ -a / / /( )a w
2r
SM r r rC G C C G C G C G C G CI ma mr r m= + ¥ + ¥ ¥ - ¥a / / / / /( )a w
2
But: r rG C G C/ /¥ = 0 and a ^ rG C/
rG C G C G Cm mr/ / /( )¥ ¥ =a ar
2
Thus: SM r aC G C G C CI mr m= +( ) + ¥/ /2 a
Since I I mrC G C= + /
2
SM r aC C G C CI m= + ¥a / (2)
Eq. (2) reduces to SM IC C= a when r aG C Cm/ ;¥ = 0 that is, when rG C/ and aC are colliear.
Referring to the first diagram, we note that this will occur only when Points G, O, and C lie in a straight line.
 (Q.E.D.) �
139
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PROBLEM 16.94
A wheel of radius r and centroidal radius of gyration k is released from rest on 
the incline and rolls without sliding. Derive an expression for the acceleration of 
the center of the wheel in terms of r, k , b , and g.
SOLUTION
I mka a= 2
ma mr= a
+
 S SM M W r ma r IC C= = +( ) ( sin ) ( )eff : b a
( sin ) ( )mg r mr r mkb a a= + 2
rg r ksin ( )b a= +2 2
a b=
+
rg
r k
sin
2 2
a r r
rg
r k
= =
+
a bsin
2 2
 a
r
r k
g=
+
2
2 2
sin b �
140
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PROBLEM 16.95
A flywheel is rigidly attached to a shaft of 40-mm radius that can roll along 
parallel rails asshown. When released from rest, the system rolls 5 m in 40 s. 
Determine the centroidal radius of gyration of the system.
SOLUTION
Kinematics: S v t a t= +0
21
2
5 0
1
2
40 2m s= + a( )
a = 1
160
2m/s 15∞
Since r = =40 0 04mm m.
a r= =a a; ( . )1
160
0 04
a = 5
32
2rad/s
Kinetics:
+ S SM M mg r I ma rC C= ∞ = +( ) ( sin ) ( )eff : 15 a
( sin ) ( )mg r mk mr r15 2∞ = +a a
gr k rsin ( )15 2 2∞ = + a
Data: r = 0 04. m a = 5
32
2rad/s
( . )( . )sin ( )9 81 0 04 15
5
32
2 2 2 2m/s m rad/s∞ = +k r
k r2 2 22 51136+ = . m
k 2 2 20 04 2 51136+ =( . ) .m m
k 2 2 50976= . k = 1 584. m �
141
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PROBLEM 16.96
A flywheel of centroidal radius of gyration k is rigidly attached to a shaft that can 
roll along parallel rails. Denoting by ms the coefficient of static friction between 
the shaft and the rails, derive an expression for the largest angle of inclination b 
for which no slipping will occur.
SOLUTION
I mk= 2
m r= a
+ S SM M mg r I ma rC C= = +( ) : ( sin ) ( )eff b a
mg r mk mrsin b a a= +2 2
a b=
+
gr
r k2 2
sin (1)
+
 S SF F= eff : F mg ma- = -sin b
F mg mr- = -sin b a
F mg mr= -sin b a
+ S SF F= eff : N mg- =cosb 0
N mg= cosb
If slipping impends F N
F
Ns s
= =m mor
m b a
b
b a
bs
F
N
mg mr
mg
r
g
= = - =
-
sin
cos
sin
cos
Substitute for a from Eq. (1) m
b b
bs
r
g
gr
r k=
- ◊
+
sin sin
cos
2 2
m b bs
r
r k
k
r k
= -
+
È
Î
Í
˘
˚
˙ =
+
È
Î
Í
˘
˚
˙tan tan1
2
2 2
2
2 2
tan b m= +s
r k
k
2 2
2
 tan b m= + ÊËÁ
ˆ
¯̃
È
Î
Í
Í
˘
˚
˙
˙
s
r
k
1
2
 �
142
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PROBLEM 16.97
A homogeneous sphere S, a uniform cylinder C, and a thin pipe 
P are in contact when they are released from rest on the incline 
shown. Knowing that all three objects roll without slipping, 
determine, after 4 s of motion, the clear distance between (a) the 
pipe and the cylinder, (b) the cylinder and the sphere.
SOLUTION
General case: I mk a r= =2 a
+
S SM M w r I marC C= = +( ) : ( sin )eff b a
mg r mk mrsin b a a= +2 2
a q b=
+
r
r k
sin
2 2
a r r
rg
r k
= =
+
a bsin
2 2
a
r
r k
g=
+
2
2 2
sin b (1)
For pipe: k r= a r
r r
g gP = +
=
2
2 2
1
2
sin sinb b
For cylinder: k 2
1
2
= a r
r
r
g gC =
+
=
2
2
2
2
2
3
sin sinb b
For sphere: k 2
2
5
= a r
r r
g gS =
+
=
2
2 22
5
5
7
sin sinb b
(a) Between pipe and cylinder.
a a a g gC P C P/ = - = -
Ê
ËÁ
ˆ
¯̃
=2
3
1
2
1
6
sin sinb b
x a t g tC P C P/ /= =
Ê
ËÁ
ˆ
¯̃
1
2
1
2
1
6
2 2sin b
 SI units: xC P/ m/s s m=
Ê
ËÁ
ˆ
¯̃
∞ =1
2
1
6
9 81 10 4 2 272 2. sin ( ) . �
143
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PROBLEM 16.97 (Continued)
(b) Between sphere and cylinder.
a a a g gS C S C/ = - = -
Ê
ËÁ
ˆ
¯̃
=5
7
2
3
1
21
sin sinb b
x a t g tS C S C/ /= =
Ê
ËÁ
ˆ
¯̃
1
2
1
2
1
21
2 2sin b
 SI units: xS C/ m/s s m=
Ê
ËÁ
ˆ
¯̃
∞ =1
2
1
21
9 81 10 4 0 6492 2. sin ( ) . �
144
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PROBLEM 16.98
A drum of 100-mm radius is attached to a disk of 200-mm radius. The 
disk and drum have a combined weight of 50 N and a combined radius 
of gyration of 150 mm. A cord is attached as shown and pulled with 
a force P of magnitude 25 N. Knowing that the coefficients of static 
and kinetic friction are ms = 0 25. and mk = 0 20. , respectively, determine 
(a) whether or not the disk slides, (b) the angular acceleration of the disk 
and the acceleration of G.
SOLUTION
Assume disk rolls: a r= =a a( .0 2 m)
I mk= =2 250
9 81
0 15
( . )
( . )
I = ◊0 114679 2. kg m
+
 S SM M ma r IC C= = +( ) ( )( . ( )eff : N m)25 0 2 a
5
50
9 81
0 2 0 1146792N m m◊ = +
.
( . ) .a a
5 0 31885= . a
a = 15 681 2. rad/s a = 15 68 2. rad/s
a r= =
=
a ( . ( . )
.
0 2 15 681
3 1362
2
2
m) rad/s
m/s a = 3 14 2. m/s Æ
+ S SF F F max x= - + =( )eff : N25
- + =
=
F
F
25
50
3 1362
9 0153
9.81
N
( . )
.
+ S SF Fy y= ( )eff : N - =50 0N N = 50 N
F Nm s= = =m 0 25 50 12 5. ( ) .N N
(a) Since F Fm� , Disk rolls without sliding �
(b) Angular acceleration of the disk. a = 15 68 2. rad/s �
 Acceleration of G. a = 3 14 2. m/s Æ �
145
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PROBLEM 16.99
A drum of 100-mm radius is attached to a disk of 200-mm radius. The disk and 
drum have a combined weight of 50 N and a combined radius of gyration of 
150 mm. A cord is attached as shown and pulled with a force P of magnitude 
25 N. Knowing that the coefficients of static and kinetic friction are ms = 0 25. 
and mk = 0 20. , respectively, determine (a) whether or not the disk slides, 
(b) the angular acceleration of the disk and the acceleration of G.
SOLUTION
Assume disk rolls: a r= =a a( .0 2 m)
I mk= 2
= 50
9 81
0 15 2
( . )
( . )
I = ◊0 114679. kg m2
+
 S SM M ma r IC C= = +( ) ( )( . ) ( )eff : N m25 0 3 a
7 5
50
9 81
0 2 0 1146792.
.
( . ) .N m◊ = +a a
7 5 0 31855. .= a
a = 23 544 2. rad/s a = 23 5 2. rad/s
a r= = =a ( . ( . ) .0 2 23 544 4 70882m) rad/s m/s2 a = 4 71 2. m/s Æ
+ S SF Fx x= ( )eff : - + =F ma25 N
- + = =F F25 50
9 81
4 7088 1 002N m/s N
( . )
( . ); .
 + S SF Fy y= ( )eff : N - =50 0N N = 50 N
F Nm s= = =m 0 25 50 12 5. ( ) .N N
(a) Since F Fm� , Disk rolls without sliding �
(b) Angular acceleration of the disk. a = 23 5 2. rad/s �
 Acceleration of G. a = 4 71 2. m/s Æ �
146
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PROBLEM 16.100
A drum of 100-mm radius is attached to a disk of 200 mm radius. The 
disk and drum have a combined weight of 50 N and a combined radius 
of gyration of 150mm. A cord is attached as shown and pulled with a 
force P of magnitude 25 N. Knowing that the coefficients of static and 
kinetic friction are ms = 0 25. and mk = 0 20. , respectively, determine 
(a) whether or not the disk slides, (b) the angular acceleration of the 
disk and the acceleration of G.
SOLUTION
Assume disk rolls: a r= =a a( .0 2 m)
I mk= =2 250
9 81
0 15
( )
( . )
( . )
I = ◊0 114679. kg m2
+
 S SM M ma r IC C= = +( ) ( ( . ( )eff : N) m)25 0 1 a
2 5
50
9 81
0 2 0 1146792.
( . )
( . ) .N m◊ = +a a
2 5 0 31885. .= a
a = 7 8407 2. rad/s a = 7 84 2. rad/s
a r= = =a ( . . ) .0 2 7 8407 1 568142 2m)( rad/s m/s a = 1 568 2. m/s Æ
+ S SF Fx x= ( )eff : - + =F ma25 N
- + =F 25 50
9 81
1 56814N
( . )
( . )
F = 17 007. N
 + S SF Fy y= ( )eff : N - =50 0N N = 50 N
F Nm s= = =m 0 25 50 12 5. ( ) .N N
(a) Since F Fm� , Disk slides �
 Knowing that disk slides
F Nk= = =m 0 20 50 10. ( )N N
147
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PROBLEM 16.100 (Continued)
(b) Angular acceleration.
+
 S SM M F IG G= - =( ) ( . ( )( .eff : m) N m)0 2 25 0 1 a
( )( . . .10 0 2 2 5 0 114679N m) N m- ◊ = a
- =0 5 0 114679. . a
a = -4 36 2. rad/s a = 4 36 2. rad/s �
(c) Acceleration of G.
+ S SF F F max x= - + =( )eff : N25
- + =10 25 50
9.81
a
a = 2 943 2. m/s a = 2 94 2. m/s Æ �
148
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PROBLEM 16.101
A drum of 100-mm radius is attached to a disk of 200 mm radius. The disk and drum 
have a combined weight of 50 N and a combined radius of gyration of 150 mm. 
A cord is attached as shown and pulled with a force P of magnitude 25 N. Knowing 
that the coefficients of static and kinetic friction are ms = 0 25. and mk = 0 20. , 
respectively, determine (a) whether or not the disk slides, (b) the angular acceleration 
of the disk and the acceleration of G.
SOLUTION
Assume disk rolls: a r= =a a( .0 2 m)
 I mk= =2 250
9 81
0 15
( . )
( . )
 I = ◊0 114679. kg m2
 + SM MC C= S( )eff : ( )( . ) ( )25 0 1N m = +ma r Ia
 2 5
50
9 81
0 2 0 1146792.
.
( . ) .N m◊ = +a a
 2 5 0 31885. .= a
 a = 7 8407 2. rad/s a = 7 84 2. rad/s
 a r= =a ( . ( . )0 2 7 8407 2m) rad/s a = 1 56814 2. m/s ¨
 + S SF Fx x= ( )eff : F ma=
 F F= =
50
9 81
1 56814 7 9926
( . )
( . ); . N
 + S SF Fy y= ( )eff : N - + =50 25 0N N N = 25 N
 F Nm s= = =m 0 25 25 6 25. ( ) .N N
(a) Since F Fm� , Disk slides �
 Knowing that disk slides
 F Nk= =m ( . )( )0 2 25
 F = 5 N
149
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PROBLEM 16.101 (Continued)
(b) Angular acceleration.
 + S SM M F IS S= - =( ) ( )( . ( .eff : N m) m)25 0 1 0 2 a
 ( )( . ( )( . .25 0 1 5 0 2 0 114679N m) N m)- = a
 1 5 0 114679. .= a
 a = 13 08 2. rad/s a = 13 08 2. rad/s �
(c) Acceleration of G.
 + S SF F F max x= =( )eff :
 5
50
9 81
N =
.
a
 a = 0 981 2. m/s a = 0 981 2. m/s ¨ �
150
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PROBLEM 16.102
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 r
I mk
I
= =
=
=
= ¥ ◊-
a a( . )
( )( . )
.
0 12
6 0 09
40 5 10
2
2
3
m
kg m
kg m2
 + S SM M ma r IC C= = +( ) : ( )( . ) ( )eff N m20 0 12 a
 2 4 6 0 12 40 5 10
2 4 135 0 10
2 3
3
. ( )( . ) .
. .
N m kg m kg m2◊ = + ¥ ◊
= ¥
-
-
a
a
(a) a = 17 778. rad/s2 a = 17 78. rad/s2 �
 
a r= =
=
a ( . )( . )
.
0 12 17 778
2 133
m rad/s
m/s
2
2 a = 2 13. m/s2 �
(b) + S SF Fy y= ( ) :eff N mg- = 0
 N = ( )( . )6 9 81kg m/s2 N = 58 86. N ≠
 + S SF F F max x= - =( ) :eff N20
 20 6 2 133 7 20N kg m/s N2- = =F ( )( . ) .F ¨
 ( )
.
.min
ms
F
N
= =
7 20
58 86
N
N
 ( ) .minms = 0 122 �
151
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PROBLEM 16.103
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 r
I mk
I
= =
=
=
= ¥ ◊-
a a( . )
( )( . )
.
0 12
6 0 09
40 5 10
2
2
3
m
kg m
kg m2
 + S SM M ma r IC C= = +( ) : ( )( . ) ( )eff N m20 0 18 a
 
3 6 6 0 12 40 5
3 6 135 10
2
3
. ( )( . ) .
.
N m kg m kg m2◊ = + ◊
= ¥ -
a
a
(a) a = 26 667. rad/s2 a = 26 7. rad/s2 �
 
a r= =
=
a ( . )( . )
.
0 12 26 667
3 2
m rad/s
m/s
2
2 a = 3 20. m/s2 Æ �
(b) + S SF F N mgy y= - =( ) :eff 0
 N = ( )( . )6 9 81kg m/s2 ≠ N = 58 86. N
 + S SF F F max x= - =( ) :eff N20
 20 6 3 2N kg m/s2- =F ( )( . ) F = 0 8. N ¨
 ( )
.
.min
ms
F
N
= =
0 8
58 86
N
N
 ( ) .minms = 0 0136 �
152
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PROBLEM 16.104
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 coefficientof static friction compatible with this motion.
SOLUTION
a r
I mk
I
= =
=
=
= ◊
a a( . )
( )( . )
.
0 12
6 0 09
40 5
2
2
m
kg m
kg m2
 + S SM M ma r IC C= = +( ) : ( )( . ) ( )eff N m20 0 06 a
 
1 2 6 0 12 40 5 10
1 2 135 10
2 3
3
. ( )( . ) .
.
N m kg m kg m2◊ = + ¥ ◊
= ¥
-
-
a
a
(a) a = 8 889. rad/s2 a = 8 89. rad/s2 �
 
a r= =
=
a ( . )( . )
.
0 12 8 889
1 0667
m rad/s
m/s
2
2 a = 1 067. m/s2 Æ �
(b) + S SF Fy y= ( ) :eff N mg- = 0
 N = ( )( . )6 9 81kg m/s2 N = 58 86. N≠
 + S SF Fk x= ( ) :eff 20 N - =F ma
 ( ) ( )( . )20 6 1 0667N kg m/s2- =F F =13 6. N ¨
 ( )
.
.min
ms
F
N
= =
13 6
58 86
N
N
 ( ) .minms = 0 231 �
153
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PROBLEM 16.105
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 r
I mk
I
= =
=
=
= ◊
a a( . )
( )( . )
.
0 12
6 0 09
40 5
2
2
m
kg m
kg m2
 + S SM M ma r IC C= = +( ) : ( )( . ) ( )eff N m20 0 06 a
 
1 2 6 0 12 40 5 10
1 2 135 10
2 3
3
. ( )( . ) .
.
N m kg m kg m2◊ = + ¥ ◊
= ¥
-
-
a
a
(a) a = 8 889. rad/s2 a = 8 89. rad/s2 �
 
a r= =
=
a ( . )( . )
.
0 12 8 889
1 0667
m rad/s
m/s
2
2 a = 1 067. m/s2 ¨ �
(b) + S SF F N mgy y= + - =( ) :eff N20 0
 N + -20 6 9 81N kg m/s2( )( . ) N = 38 86. N ≠
 + S SF F F max x= =( ) :eff
 F = =( )( . ) .6 1 0667 6 4kg m/s N2 F ¨
 ( )
.
.min
ms
F
N
= =
6 4
38 86
N
N
 ( ) .minms = 0 165 �
154
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PROBLEM 16.106
A bar of mass m is held as shown between four disks, each of mass m¢ and radius 
r = 75 mm. Determine the acceleration of the bar immediately after it has been released 
from rest, knowing that the normal forces on the disks are sufficient to prevent any 
slipping and assuming that (a) m = 5 kg and ¢ =m 2 kg, (b) the mass m¢ of the disks is 
negligible, (c) the mass m of the bar is negligible.
SOLUTION
Kinematics:
a r
a
r
= =a a
Kinetics of bar Kinetics of one disk
 
S SF F
W F ma
y y=
- =
( ) :eff
4 (1) 
S SM M Fr I
Fr m r
a
r
F m a
C C= =
= ¢ ÊËÁ
ˆ
¯̃
= ¢
( ) :eff a
1
2
1
2
2
(2)
Substitute for F from (2) into (1).
mg m a ma- ¢ÊËÁ
ˆ
¯̃
=4 1
2
mg m m a a
mg
m m
= + ¢ =
+ ¢
( )2
2
(a) m m a g= ¢ = =
+
5 2
5
5 2 2
kg, kg:
( )
 a = 5
9
g Ø �
(b) ¢ = =m a g0: a = g Ø �
(c) m a= =0 0 a = 0 �
+
+
155
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PROBLEM 16.107
A bar of mass m is held as shown between four disks, each of mass m¢ and radius r = 75 mm. 
Determine the acceleration of the bar immediately after it has been released from rest, knowing 
that the normal forces on the disks are sufficient to prevent any slipping and assuming that 
(a) m = 5 kg and ¢ =m 2 kg, (b) the mass m¢ of the disks is negligible, (c) the mass m of the 
bar is negligible.
SOLUTION
Kinematics:
Disk is rolling on vertical wall
¢ = =
= =
a a r
a a r
C
r
a
a2
Therefore:
a
a
=
¢ = =
a
r
a r a
2
1
2
Kinetics of bar Kinetics of one disk
 
S SF F
W F ma
mg F ma
y y=
- =
- =
( ) :
( )
eff
4
2 2 (1)
I m r= ¢
1
2
2
+ S SM MD D= ( ) :eff
F r m gr I m a r
Fr m gr m r
a
r
m
a
r
F m a
( )2
2
1
2 2 2
2
1
4
2
+ ¢ = + ¢ ¢
= ¢ = ¢ ÊËÁ
ˆ
¯̃
+ ¢
= ¢ +
a
11
2
a m gÊËÁ
ˆ
¯̃
- ¢ (2)
Substitute for 2F from (2) into (1):
mg m a m g ma
m m a m m g a
m m
m m
- ¢ ÊËÁ
ˆ
¯̃
+ ¢ =
+ ¢ÊËÁ
ˆ
¯̃
= + ¢ =
+ ¢
+ ¢
2
3
4
2
3
2
2
2
3
2
( ) gg
+
156
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PROBLEM 16.107 (Continued)
(a) m m a g= ¢ = =
+
+
5 2
5 4
5 3
kg, kg : a = 9
8
g Ø �
(b) ¢ = =m a g0 : a = g Ø �
(c) m a g= =0 2
3 2
:
/
 a = 4
3
g Ø �
Note: In parts (a) and (c), a g> , from Eq. (1) it follows that F < 0. The disks push the bar down.
157
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PROBLEM 16.108
A bar of mass m is held as shown between four disks, each of mass m¢ and radius r = 75 mm. 
Determine the acceleration of the bar immediately after it has been released from rest, knowing 
that the normal forces on the disks are sufficient to prevent any slipping and assuming that 
(a) m = 5 kg and ¢ =m 2 kg, (b) the mass m¢ of the disks is negligible, (c) the mass m of 
the bar is negligible.
SOLUTION
Kinematics:
Disk is rolling on vertical wall
a a r= ¢ = a
Kinetics of bar Kinetics of one disk
 
S SF F
W F ma
mg F ma
y y=
- =
- =
( ) :eff
4
4 (1)
I m r
a
r
a = ¢ ÊËÁ
ˆ
¯̃
1
2
2
S SM M
F w r I m a r
F m g r m r
a
r
m at
F
D D=
+ ¢ = + ¢ ¢
+ ¢ = ¢ ¢ ÊËÁ
ˆ
¯̃
+
( ) :
( )
( ) ’
eff
a
1
2
== ¢ - ¢
3
2
m a m g (2)
Substitute for F from (2) into (1):
mg m a m g ma
m m a m m g a
m m
m m
g
- ¢ + ¢ =
+ ¢ = + ¢ =
+ ¢
+ ¢
6 4
6 4
4
6
( ) ( )
(a) m m a= ¢ = =
+
+
5 2
5 8
5 12
kg, kg : a = 13
17
g Ø �
(b) ¢ = =m a g0 : a = g Ø �
(c) m a g= =0 4
6
: a = 2
3
g Ø �
+
+
158
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PROBLEM 16.109
Two uniform disks A and B, each of mass 2 kg, are connected 
by a 1.5-kg rod CD as shown. A counterclockwise couple M of 
moment 2.25 N · m is applied to disk A. Knowing that the disks 
roll without sliding, determine (a) the acceleration of the center 
of each disk, (b) the horizontal component of the force exerted 
on diskB by pin D.
SOLUTION
Since the rod CD is translating, its acceleration is only in the x direction and aCD = 0. So it exerts forces only 
in x direction at pins C and D.
a a aA B A B r= = = = =a a a a
Disk A:
W = 2g = 19.62 N
 I mr= 1
2
2
 + S SM M M C ma r Is s= - = +( ) : ( . ( )eff m)0 2 a
 
2 25 0 2
1
2
3
2
2 25 0 2
3
2
2 0 15
2
2
2
. . ( )
. . ( )( . )
- = +
=
- =
C mr r mr
mr r
C
a a
a
 2 25 0 2 0 0675. . .- =C a (1)
Disk B: I mr= 1
2
2
 + S SM M D ma r I mr mrH H= = + = +( ) : ( . ( )eff
2m)0 2
1
2
2 2a a a
 
D
D
( . ) ( )( . )
. .
0 2
3
2
2 0 15
0 2 0 0675
2=
=
a
a (2)
Rod CD:
+ a a a aR C A C A= = + /
 aR = + =0 15 0 05 0 2. . .a a a
159
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PROBLEM 16.109 (Continued)
 + S SF Fx x= ( ) :eff C D m a
C D
R R- =
- = ( . )( .1 5 0 2a)
Multiply
By 0.2 0 2 0 2 0 06. . .C D- = a (3)
Add (1), (2), (3): 2 25 0 2 0 2 0 2 0 2 0 0675 0 0675 0 06. . . . . ( . . . )- + + - = + +
fi
C D C D a
a = 11.5385 raad/s2 a = 11 538. rad/s2
(a) a aA B r= = =
=
a ( . )0 15 m)(11.5385 rad/s
m/s
2
1.7308 2
 a aA B= = ¨1 731. m/s
2 �
(b) Substitute for a in (2)
 0 2 0 0675 11 5385. ( . )( . )D = D = ¨3 89. N �
160
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PROBLEM 16.110
Gear C has a weight of 5 kg and a centroidal radius of gyration of 75 mm. The uniform 
bar AB has a weight of 3 kg gear D is stationary. If the system is released from rest in the 
position shown, determine (a) the angular acceleration of gear C, (b) the acceleration of 
Point B.
SOLUTION
Kinematics: Let a AB AB= a be the angular acceleration of the rod AB and aC C= a be that of gear D. Let 
Point G be the mass center of rod AB. l lAG AB=
1
2
. Normal accelerations vanish since w = 0.
Rod AB rotates about the fixed Point A.
 aG AG ABr= Øa aB AB ABl= Øa (1)
Gear C is effectively rolling on the stationary gear D. At the contact Point P, a p = 0, as wgear = 0.
 aB C Cr= Øa
Matching the two expressions for aB gives
l r
r
lAB AB C C AB
C
AB
Ca a a a= =or
Kinetics: Let FD be the tangential force between gears C and D.
Gear C. + SMB = + S( ) :M F r IG D D C Ceff = a
 F
I
rD
C C
C
=
a
Rod AB and gear C.
 SM A = S( ) : ( )M l r F W l W lA AB C D C AB AB AGeff + - -
= - - -
+ - -
=
I m a l I m a l
l r
I
r
W l W l
I
C C C B AB AB AB D B AB
AB C
C
C
C C AB AB AG
C
a a
a( )
aa a a aC C AB C C
C
A
AB C D AB C C
AB
C
C C AB C
C
AB
AB
m l r
r
l
I m l r
l
r
I m l r
r
l
I
- - -
+ + ( ++
È
Î
Í
˘
˚
˙
= +
m l
W l W l
AB AG C
C AB AB AG
2 ) a
161
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PROBLEM 16.110 (Continued)
Data: W W l rC AB AB C= = = = =5 3 250 0 25g g mm m 125 mm 0.125 m. =
 m m
I m k
I
C AB
C C
AB
= =
= = = ◊
=
5 3
5 0 028125
1
12
2
kg kg
kg) (0.075 m) kg m2 2( .
mm l
l
AB AB
AB
C
2 21
12
3 0 25 0 015625
2
= = ◊
=
( )( . ) . kg m2
r
( )( . ) ( )( . )( . ) ( . )( . ( )( . ) )2 0 028125 5 0 25 0 125 0 5 0 015625 3 0 125 2+ + +{{ }
= +
aC
( )( . )( . ) ( )( . )( . )5 9 81 0 25 3 9 81 0 125
0 24375 15 94125 65 4. . .a aC C= = rad/s
2
(a) Angular acceleration. aC = 65 4. rad/s
2 �
(b) Acceleration of B.
 From Eq. (1), aB = ( . )( . )0 125 65 4m rad/s
2 aB = 8 175. m/s
2 Ø �
162
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PROBLEM 16.111
A half section of a uniform cylinder of mass m is at rest when a force P is applied as 
shown. Assuming that the section rolls without sliding, determine (a) its angular 
acceleration, (b) the minimum value of ms compatible with the motion.
SOLUTION
The analysis will be restricted to very small rotation, i.e. position change is neglected.
Kinematics: Assume a , a0 = Æra
From inside cover of text OG
r= 4
3p
 a a a a= + = +O G O O OG/ [ ] [( ) ]Æ ¨a
 a = Æ +
È
ÎÍ
˘
˚̇
= -Ê
ËÁ
ˆ
¯̃
[ ]r
r
ra
p
a
p
a4
3
1
4
3
¨ Æ
Kinetics: m = mass of half cylinder:
I mr
I I m OG
mr I
r
m
I mr
O
O
=
= +
= + ÊËÁ
ˆ
¯̃
= -Ê
ËÁ
ˆ
¯̃
1
2
1
2
4
3
1
2
16
9
2
2
2
2
2
2
( )
p
p
(a) Angular acceleration.
+
 S SM M ma rC O= = + -
Ê
ËÁ
ˆ
¯̃( ) ( )eff : Pr Ia 1
4
3p
Pr mr
Pr mr
= -ÊËÁ
ˆ
¯̃ + -
Ê
ËÁ
ˆ
¯̃ -
Ê
ËÁ
ˆ
¯̃
= -
2
2
2
1
2
16
9
1
4
3
1
4
3
1
2
1
p
a
p
a
p
mr r
66
9
1
8
3
16
92
2
2p
a
p p
aÊËÁ
ˆ
¯̃ + - +
Ê
ËÁ
ˆ
¯̃mr
163
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PROBLEM 16.111 (Continued)
Pr mr
Pr mr
= -ÊËÁ
ˆ
¯̃
=
2
2
3
2
8
3
0 65717
p
a
a( . )
 a = 1 5357. P
mr
 a = 1 536. P
mr
 �
(b) Minimum value of ms.
 + S SF F F max x= =( )eff :
F mr mr
F mr
P
mr
= -ÊËÁ
ˆ
¯̃
=
= ÊËÁ
ˆ
¯̃
=
1
4
3
0 57559
0 57559 1 5357
p
a a( . )
( . ) . 00 8839. P
 + S SF F N P mgy y= - - =( ) :eff 0
 
N mg P
F
N
P
mg Ps
= +
= =
+
( )
.
minm
0 8839
 ( ) .minms
P
mg P
=
+
0 884 �
164
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PROBLEM 16.112
Solve Problem 16.111, assuming that the force P applied at B is directed 
horizontally to the right.
PROBLEM 16.111 A half section of a uniform cylinder of mass m is at rest 
when a force P is applied as shown. Assuming that the section rolls without 
sliding, determine (a) its angular acceleration, (b) the minimum value of ms 
compatible with the motion.
SOLUTION
The analysis will be restricted to very small rotation, i.e., position change is neglected.
Kinematics: Assume a , a0 = ra Æ
From inside cover of text OG
r= 4
3p
 a a= + = Æ - ¨O G O O OGa / [ ] [( ) ]a a
 
a = Æ + ¨È
ÎÍ
˘
˚̇
= -ÊËÁ
ˆ
¯̃ Æ[ ]r
r
ra 4
3
1
4
3p
a
p
a
Kinetics: m = mass of half cylinder:
I mr
I I m OG
mr I
r
m
I mr
O
O=
= +
= + ÊËÁ
ˆ
¯̃
= -Ê
ËÁ
ˆ
¯̃
1
2
1
2
4
3
1
2
16
9
2
2
2
2
2
2
( )
p
p
(a) Angular acceleration:
 + S SM M rC C= = + -
Ê
ËÁ
ˆ
¯̃( ) ( )eff : Pr I maa 1
4
3p
 
Pr mr= -ÊËÁ
ˆ
¯̃ + -
Ê
ËÁ
ˆ
¯̃ -
Ê
ËÁ
ˆ
¯̃
2
2
1
2
16
9
1
4
3
1
4
3p
a
p
a
p
mr r
165
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PROBLEM 16.112 (Continued)
 
Pr mr mr= -ÊËÁ
ˆ
¯̃ + - +
Ê
ËÁ
ˆ
¯̃
2
2
2
2
1
2
16
9
1
8
3
16
9p p p
aa
 
Pr mr
Pr mr
= -ÊËÁ
ˆ
¯̃
=
=
2
2
3
2
8
3
1 65113
1 5367
p
a
a
a
( . )
.
P
mr
 a = 1 536.
P
mr
 �
(b) Minimum value of ms.
 
S SF F P F ma mr
P F mr
P
x x= - = = -
Ê
ËÁ
ˆ
¯̃
- = -ÊËÁ
ˆ
¯̃
( )
.
eff : 1
4
3
1
4
3
1 5357
p
a
p mmr
P
F P P P
Ê
ËÁ
ˆ
¯̃
=
= - =
0 8839
0 8839 0 1161
.
. .
 + S SF F N mgy y= - =( ) :eff 0
 
N ma
F
N
P
mgS
=
= =( ) .minm
0 1161
 ( ) .minmS
P
mg
= 0 116 �
+
166
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PROBLEM 16.113
A small clamp of mass mB is attached at B to a hoop of mass mh . The system is 
released from rest when q = ∞90 and rolls without sliding. Knowing that m mh B= 3 , 
determine (a) the angular acceleration of the hoop, (b) the horizontal and vertical 
components of the acceleration of B.
SOLUTION
Kinematics: aA B Ar a r= Æ = Øa a, /
 a aB A B Aa= Æ + Ø/
 aB r r= Æ+ Øa a
 ( ) ( )aB x B yr a r= Æ = Øa a
Kinetics:
m m
I m r m r
h B
h B
=
= =
3
32 2
(a) Angular acceleration.
 
S SM M W r I m a r m a r m a r
m gr m r m
C C B h A B B x B B y
B B
= = + + +
= +
( ) : ( ) ( )
(
eff a
a3 32 BB B Br m r m r
gr r
) 2 2 2
28
a a a
a
+ +
= a =
1
8
g
r
 �
(b) Components of acceleration of B. ( )aB x g=
1
8
 Æ �
 ( )aB y g=
1
8
 Ø �
 ( )aB x r g= = Æa
1
8
 ( )aB y r g= =a
1
8
 Ø �
+
167
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PROBLEM 16.114
A small clamp of mass mB is attached at B to a hoop of mass mh. Knowing that the 
system is released from rest and rolls without sliding, derive an expression for the 
angular acceleration of the hoop in terms of mB, mh, r, and q.
SOLUTION
Kinematics: aA B Ar a r= Æ =a a/ q
 aB A B Aa a= Æ + / q
 aB r r= Æ +a a q
Kinetics: I m rh=
2
+ S SM M W r I m a r m r r r m r rC C B h A B B= = + + + +( ) sin ( cos ) sin ( sineff : q a a q a q q ))
cos ( cos )+ +m r r rB a q q
 
m gr m r m r r m r r r
m r r
B h h B
B
sin ( ) ( cos )( cos )
sin ( sin
q a a a q q
a q
= + + + +
+
2 1
qq
q a a q q
a a
)
sin [( cos ) sin ]
[
m gr m r m r
m r m r
B h B
h B
= + + +
= + +
2 1
2 1 2
2 2 2 2
2 2 ccos cos sin ]q q q+ +2 2
 m gr r m mB h Bsin [ ( cos )]q a q= + +
2 2 2 2 a
q
q
=
+ +
g
r
m
m m
B
h B2 1
sin
( cos )
 �
168
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PROBLEM 16.115
The center of gravity G of a 1.5-kg unbalanced tracking wheel 
is located at a distance r = 18 mm from its geometric center B. 
The radius of the wheel is R = 60 mm and its centroidal radius of 
gyration is 44 mm. At the instant shown the center B of the wheel 
has a velocity of 0.35 m/s and an acceleration of 1.2 m/s2, both 
directed to the left. Knowing that the wheel rolls without sliding 
and neglecting the mass of the driving yoke AB, determine the 
horizontal force P applied to the yoke.
SOLUTION
Kinematics: Choose positive vB and aB to left.
 Trans. with B + Rotation abt. B = Rolling motion
a = + ¨[ ]a rB w
2 +
r
R
aB
È
ÎÍ
˘
˚̇
≠
Kinetics:
 
S SM M PR Wr ma r ma R I
PR mgr m
r
R
a r m a
C C y x
B
= - = + +
- = ÊËÁ
ˆ
¯̃
+
( ) ( ) ( )
(
eff : a
BB
B
B
B
r R mk
a
R
ma
r
R
R
k
R
mr
v
r
R
P mg
r
R
+ +
= + +
Ê
ËÁ
ˆ
¯̃
+ Ê
ËÁ
ˆ
¯̃
= ÊËÁ
ˆ
w2 2
2 2 2
)
¯̃̄
+ + +
Ê
ËÁ
ˆ
¯̃
+ma r k
R
m
r
R
vB B1
2 2
2 2
2 (1)
+
169
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PROBLEM 16.115 (Continued)
Substitute: m
r
R
k g
=
=
=
= =
1 5
0 018
0 06
0 044 9 81
.
.
.
. .
 kg
m
m
 mm and m/s in Eq. (1)2
 
P aB= + +
+Ê
ËÁ
ˆ
¯̃
+1 5 9 81 0 018
0 06
1 5 1
0 018 0 044
0 06
1 5
2 2
2
. ( . )
.
.
. ( )
. .
.
.
00 018
0 06
4 4145 2 4417 7 5
2
2
2
.
.
. . .
v
P a v
B
B B= + + (2)
 vB Bv= ¨ = +0 35 0 35. ; . m/s m/s
 aB Ba= ¨ = +1 2 1 2. ; . m/s m/s
2 2
Substitute in Eq. (2): P = + + + +4 4145 2 4417 1 2 7 5 0 35 2. . ( . ) . ( . )
 
= + +
= +
4 4145 2 9300 0 9188
8 263
. . .
. N P = ¨8 26. N �
170
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PROBLEM 16.116
A 2-kg bar is attached to a 5-kg uniform cylinder by a square pin, P, as shown. 
Knowing that r = 0.4 m, h = 0.2 m, q = 20∞, L = 0.5 m and w = 2 rad/s at the 
instant shown, determine the reactions at P at this instant assuming that the 
cylinder rolls without sliding down the incline.
SOLUTION
Masses and moments of inertia.
Bar: m
I m L
B
B B
=
= = = ◊
2
1
12
1
12
2 0 5 0 041672 2
 kg
 kg m2( )( . ) .
Disk: m
I m r
D
D C
=
= = = ◊
5
1
2
1
2
5 0 4 0 42 2
 kg
 kg m2( )( . ) .
Kinematics. Let Point C be the point of contact between the cylinder and the incline and Point G be the mass 
center of the cylinder without the bar. Assume that the mass center of the bar lies at Point P.
a = a . aG Ga= 20∞
For rolling without slipping, ( )
( )
( ) ( )
( )
a a r a r
a a h
a r h
a h h
C t G G
P t G
P t
P h
= - = =
= +
= +
= + =
a a
a
a
w w
0
0 2 2
Kinetics. Using the cylinder plus the bar as a free body,
 + S SM M m gr m r hC C D B= + + +( ) sin ( )sineff : q q
 
= + + + +
= + + + +
=
+
I m a r I m a r h
I m r I m r h
m r m
D D G B B P t
D D B B
D
a a
a
a
( ) ( )
[ ( ) ]
[
2 2
BB
D D B B
r h g
I m r I m r h( )] sin
( )
[( )( . ) ( )( . )]( . )
+
+ + + +
= +
q
2 2
5 0 4 2 0 6 9 81 ssin
. ( )( . ) ( . ) ( )( . )
.
( )
2 ∞
+ + +
=
0
0 4 5 0 4 0 04167 2 0 6
5 4732
2 2
 rad/s2
aP tt
P na
= =
= =
( . )( . ) .
( ) ( . )( ) .
0 6 5 4732 3 2840
0 2 2 0 82
m
m
/s
/s
2
2
171
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PROBLEM 16.116 (Continued)
Using the bar alone as a free body,
 + S SF Fx x= ( )eff : P m a m ax B P t B P n= ∞ - ∞( ) cos ( ) sin20 20
= ∞ - ∞
=
( )( . )cos ( )( . )sin
.
2 3 2840 20 2 0 8 20
5 625Px N
 ± ≠ = - = - ∞ - ∞S SF F P m g m a m ay y y B B P t B P n( ) ( ) sin ( ) coseff : 20 20
 
P
P
P
y
y
=
= - ∞ - ∞
=
=
( )( . )
( )( . )sin ( )( . )cos
.
2 9 81
2 3 2840 20 2 0 8 20
15 87 N
55 625 15 87 16 84
15 87
5 625
70 5
2 2. . .
tan
.
.
.
+ =
=
= ∞
 N
b
b P =16 84. N 70 5. ∞ �
 ± S SM MP P= ( )eff : M IP B= a
 = ( . )( . )0 04167 5 4732 MP = ◊0 228. N m �
172
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PROBLEM 16.117
The ends of the 10-kg 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 q = 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 a w = 0
 a a aB A B A AaØ = + = Æ +/ [ ] [ .1 2a 25∞]
 aB = ∞ =( . )cos .1 2 25 1 0876a a
 aA = ∞ =( . )sin .1 2 25 0 5071a a
 a aG A G A A= + = Æ +a a/ [ ] [ .0 6a 25∞]
 aG = Æ +[ . ] [ .0 5071 0 6a a 25∞]
 ax G xa= = Æ + ¨( ) [ . ] [ . ]0 5071 0 2536a a
 ax = Æ0 2535. a
 ay = ∞ ¨ = Ø[ . ] .0 6 25 0 5438a acos
We have found for a ax = Æ0 2535. a
 ay = Ø0 5438. a
173
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PROBLEM 16.117 (Continued)
Kinetics: I mL m= =1
12
1
12
1 22 2( . m)
(a) Angular acceleration.
 + S SM ME E= ( )eff : mg I ma max y( . ( . ( .0 5438 0 2535 0 5438 m) m) m)= + +a
 mg m m m( . ) ( . ) ( . ) ( . )0 5438
1
12
1 2 0 2535 0 54382 2 2+ +a a
 g g( . ) . .0 5438 0 48 1 133= =a a a = 11 11. rad/s2 �
(b) + S SF F A mg ma my y y= - = - =( ) ( . )eff : 0 5438a
 
A
A
- = - ( )
= -
=
10 9 81 10 0 5438 11 115
98 1 60 44
37 66
( . ) ( )( . ) .
. .
. N A = ≠37 7. N �
(c) + S SF Fx x= ( )eff : B ma mx= = ( . )0 2535a
 
B
B
=
=
10 0 2535 11 115
28 18
( . )( . )
. N B = Æ28 2. N �
174
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PROBLEM 16.118
The ends of the 10-kg uniform rod AB are attached to collars of negligible 
mass that slide without friction along fixed rods. A vertical force P is 
applied to collar B when q = ∞25 , causing the collar to start from rest 
with an upward acceleration of 12 m/s2. Determine (a) the force P,
(b) the reaction at A.
SOLUTION
Kinematics: w = 0
 a a aB A B A Aa= + = ¨/ ; ] [ ][ m/s
212 ≠ + [ .1 2a 25 ]°
 a aB B A= ∞/ cos25
 12 1 2 25 m/s2 = ∞( . )cosa
 a = 11 034. rad/s2
 aA = ∞ = ¨12 25 5 596tan . m/s
2
 a a a aG A G A G= + = ¨/ [ . ]: 5 596 + [ .0 6a 25∞]
 aG = ¨[ . ]5 596 + [ . ( . )0 6 11 034 25∞]
 a ax G x= = ¨( ) .2 798 m/s
2
 a ay G y= = ≠( ) .6 00 m/s
2
175
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PROBLEM 16.118 (Continued)
Kinetics: I mL m= =1
12
1
12
1 22 2( . )
(a) 
+
 S SM M P W I ma maE E x y= - = + +( ) ( . ) ( . ) ( . ) ( .eff : 1 0876 0 5438 0 2535 0 5438a ))
 
W mg
I mL
= = =
=
=
=
10 9 81 98 1
1
12
1
12
10 1 2 11 034
13 24
2
2
( . ) .
( )( . ) ( . )
.
 N
a a
 N m
 N
 N
◊
= =
= =
-
ma
ma
P
x
y
( )( . ) .
( )( )
( . ) ( .
10 2 798 27 98
10 6 60
1 0876 98 11 0 5438 13 24 27 98 0 2535 60 0 5438
1 0876 53
)( . ) . ( . )( . ) ( )( . )
( . )
= + +
-P .. . . .347 13 24 7 096 32 628= + +
 
P
P
( . ) .
.
1 0876 106 311
97 8
=
= N P = ≠97 8. N �
(b) + S SF F A W P may y y= - + =( )eff : 
 A - + =98 1 97 8 60. . N A = ≠60 3. N �
176
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PROBLEM 16.119
The motion of the 4-kg uniform rod AB is guided by small wheels of negligible 
weight that roll along without friction in the slots shown. If the rod is released 
from rest in the position shown, determine immediately after release (a) the 
angular acceleration of the rod, (b) the reaction at B.
SOLUTION
Kinematics:
 aG A/ m)= Æ( .0 375 a
 = Æ0 375. a
 aB A/ m)= Æ( .0 75 a
 = Æ0 75. a
 a a aB A B A Ba= + Ø/ : [ ] = ∞ + Æ[ ] [ . ]aA 30 0 75a
 aA = ∞
=0 75
30
0 86603
.
cos
.
a a 30∞
 aB = ∞ = Ø( . ) tan .0 75 30 0 43301a a
 a a a a a= = + = ∞ + ÆG A G A/ ] ]; [ . [ .0 86603 30 0 375a a
 ax = ¨ + Æ = ¨[ . ] [ . ] .0 75 0 375 0 375a a a
 a y = Ø = Ø[ . ] .0 433015 0 433015a a
177
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PROBLEM 16.119 (Continued)
We have:
a ax y= ¨ = Ø0 375 0 43301. ; .a a
 
I mL
I
= ◊ = ◊
= ◊
1
12
1
12
4 0 75
0 1875
2 ( )( .
.
)
 kg m
2
2
(a) Angular acceleration.
+ S SM M W I ma maE E x y= = + +( ) ( . ( . ( .eff : ) m) m)0 43301 0 375 0 43301a
 
( )( . )( . . ( . )
( )
.
4 9 81 0 43301 0 1875 4 0 375
4
16 9913 1
2
2
= +
+
=
a a
a0.43301
..
.
5
11 328
a
a = rad/s2 a = 11 33. rad/s2 �
(b) Reaction at B.
 + S SM M B I maA A x= = - +( ) ( .eff: (0.75 m) m)a 0 375
 
0 75 0 1875 11 328
4 0 375 11 328 0 375
0 75 2
. ( . )( . )
( )( . )( . )( . )
. .
B
B
= -
+
= - 1124 6 372
5 664
+
=
.
.B N B = ¨5 66. N �
178
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PROBLEM 16.120
The 2-kg uniform rod AB is attached to collars of negligible mass 
which may slide without friction along the fixed rods shown. 
Rod AB is at rest in the position q = ∞25 when a horizontal force 
P is applied to collar A, causing it to start moving to the left with 
an acceleration of 4 m/s2. Determine (a) the force P, (b) the 
reaction at B.
SOLUTION
Kinematics aA = ¨4 m/s
2 ( )w = 0
 a a aB A B A= + /
 aB 20 4 0 8 25∞ = ¨ + ∞( . )a
Law of sines: 
( . )
sin sin
0 8
70
4
45
a
∞
=
∞
 a = 6 64463. rad/s2
 a a a a= = + = ¨ ∞G A G A/ 4 0 4 6 64463 65( . )( . )
 + ax x= - + ∞ = - = ¨4 0 4 6 64463 65 2 4088 2 8767( . )( . )cos . , .a m/s
2
 + ay y= + ∞ = = ≠( . )( . )sin . , .0 4 6 64463 65 4088 2 4088+2 m/s
2a
Kinetics m
I ml
=
= = =
2
1
12
1
12
2 0 8
8
75
2 2
kg
kg m2( )( . ) ◊
179
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PROBLEM 16.120 (Continued)
From triangle ABE: 
AE AB
AE
GD
sin sin
( .
.
( .
45 70
0 8
0 60199
0 4
∞
=
∞
=
=
=
 m)
sin 45
sin 70
 m
 
°
°
mm)sin 65 0.36252 m
 m
° =
= - = - ∞
=
DE AE AD 0 60199 0 4 65
0 43294
. ( . )cos
.
(a) Force P.
 + S SEM M P AE W GD I ma DE ma GDE E x y= - = + +( ) ( ) ( ) ( ) ( )eff : a
 
P( . ) ( )( . )( . ) ( . ) ( . )( .0 60199 2 9 81 0 36252
8
75
6 64463 2 2 8767 0 432- = + 994
2 2 4088 0 36252
0 60199 7 1126 0 70876 2 4909 1 74
)
( . )( . )
. . . . .
+
- = + +P 6648)
 P = 20 031. N P = ¨20 0. N �
(b) Reaction at B.
 + S SF F B P max x x= ∞ - = -( ) coseff : 20
 
B
B
cos . ( . )
.
.
20 20 031 2 2 8767
14 278
15 194
∞ = -
=
= N B = 15 19. N 20∞ �
180
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PROBLEM 16.121
The 2-kg uniform rod AB is attached to collars of negligible mass which may slide without friction along the 
fixed rods shown. If rod AB is released from rest in the position q = ∞25 , determine immediately after release 
(a) the angular acceleration of the rod, (b) the reaction at B.
SOLUTION
Kinematics: Assume a ( )w = 0
 a a aB A B A= + /
 a aB A70 0 8 65∞ = Æ + ∞( . m)a 
Law of sines: 
a aA B
sin sin
( . )
sin45 65
0 8
70∞
=
∞
=
∞
a
 a aA B= Æ = ∞0 60199 0 77158 70. .a a 
 a a a a= = + = Æ + ∞G A B
1
2
0 3010 0 38579 70( ) . .a a
 ax = + ∞ = Æ0 3010 0 38579 70 0 43295. . cos .a a a
 ay = ∞ = Ø0 38579 70 0 36252. sin .a a
Kinetics: m I ml= = = =2 1
12
1
12
2 0 8
8
75
2 2 2kg kg m( )( . ) ◊
From triangle ABE:
 
AE AB
sin sin45 70∞
=
∞
 
AE
AE
= ∞
∞
=
( .
sin
sin
.
0 8
45
70
0 60199
m)
m
 
GD
DE AE AD
= ∞ =
= - = - ∞ =
( . sin .
. ( . )cos .
0 4 65 0 36252
0 60199 0 4 65 0 43
m) m
m 2294 m
181
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PROBLEM 16.121 (Continued)
(a) Angular acceleration.
 
+ S SM ME E= ( ) :eff W GD I ma DE ma GDx y( ) ( ) ( )= + +a
 
( )( . )( . ) ( . ) ( . ) ( . ) (2 9 81 0 36252
8
75
2 0 43295 0 43294 2 0 36252 0= + +a a a .. )
. ( . . . )
36252
7 11264 0 10667 0 37488 0 26284= + +a
 a = 9 555. rad/s2 a = 9 56. rad/s2 �
(b) Reaction at B.
 + S SF FX X= ( ) :eff B ma mxcos ( . )20 0 43295∞ = = a
 B cos ( . )( . )20 2 0 43295 9 555∞ = B = 8 80. N 20∞ �
182
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PROBLEM 16.122
The motion of the uniform rod AB of mass 5 kg and length 
L = 750 mm is guided by small wheels of negligible mass that 
roll on the surface shown. If the rod is released from rest when 
q = ∞20 , determine immediately (a) the angular acceleration of 
the rod, (b) the reaction at A.
SOLUTION
Kinematics:
a aB A B Aa= + /
 a a LB A[ ¨ ] = [ ∞] + [ ∞]60 20a
Law of sines:
 
a a LA B
sin sin sin70 50 60∞
=
∞
=
∞
a
 aA L= 1 0851. a 60∞
 aB L= ¨0 88455. a
 a
L
G A/ = 2
a 20∞
 a a a a LG A G A= = + = [/ .1 0851 a 60 2∞] +
È
ÎÍ
L a 20∞˘
˚̇
 + a L Lx = ∞ + ∞( . )cos ( . )sin1 0851 60 0 5 20a a
 = + = ¨0 54254 0 17101 0 71355. . ; .L L Lxa a a a
 + a L Ly = ∞ - ∞( . )sin ( . )cos1 0851 60 0 5 20a a
 = - = Ø0 93972 0 46985 0 46985. . ; .L L Lya a a a
183
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PROBLEM 16.122 (Continued)
We have: ax L= ¨0 71355. a
 ay L= Ø0 46985. a
Kinetics:
Triangle ABE: – = ∞ABE 70
 – = ∞ – = ∞ABE BAE70 50, 
Law of Sines: AB L=
 
BE L
BE L
sin sin
; .
50 60
0 88455
∞
=
∞
= 
(a) Angular acceleration.
+ S SM M mg L I ma L ma LE E x y= = + +( ) : ( . ) ( . ) ( . )eff 0 46985 0 71355 0 46985a
 
0 46985
1
12
0 71355 0 71355
0 46985 0 4
2. ( . )( . )
( . )( .
mgL mL m L L
m L
= +
+
a a
a 66985L)
 0 46985 0 81325. ( . )mgL mL= 2 a
 
a = =
=
0 57775 0 57775
9 81
0 75
7 557
. .
.
.
.
g
L
 m/s
 m
 rad/s
2
2 a = 7 56. rad/s2 �
(b) Reaction at A.
 + S SF Fx x= ( ) :eff A ma m Lxsin ( . )60 0 71355∞ = = a
 Asin ( )( . )( . )( . ) .60 0 71355 0 75 7 557 32 354∞ = =5 kg m rad/s2
 A = 23 349. N or A = 23 3. N 30∞ �
184
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PROBLEM 16.123
End A of the 8-kg uniform rod AB is attached to a collar that can 
slidewithout friction on a vertical rod. End B of the rod is attached 
to a vertical cable BC. If the rod is released from rest in the 
position shown, determine immediately after release (a) the angular 
acceleration of the rod, (b) the reaction at A.
SOLUTION
Kinematics: w = 0
a a aB A B A Ba= + / ; [ Æ] = [aA Ø ] +[La q ]
+ 0 = -a LA a qcos
aA L= a qcos Ø
a a a= +A G A/
a = [La qcos Ø ] + È
ÎÍ
L
2
a q ]
ax
L=
2
a qsin Æ; a y
L=
2
a qcos ≠
Kinetics:
+ S SM M mg L I ma L ma LE E x y= = +
Ê
ËÁ
ˆ
¯̃
+ ÊËÁ
ˆ
¯̃
( ) : cos sin coseff 2 2 2
q a q q
mg
L
mL m
L
m
L
2
1
12 2 2
2
2 2
cos sin cosq a q a q a= + ÊËÁ
ˆ
¯̃
+ ÊËÁ
ˆ
¯̃
 mg
L
mL
2
1
3
2cosq a= a = 3
2
g
L
cosq 
185
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PROBLEM 16.123 (Continued)
+ S SF F A ma m Lx x x= = =( ) : sineff 2
a q
A m
L g
L
= ÊËÁ
ˆ
¯̃2
3
2
cos sinq q
A = 3
4
mg sin cosq q Æ
Data: m L= = ∞ =8 30 0 75kg m, , .q
(a) Angular acceleration. a = ∞3
2
9 81
0 75
30
.
.
cos
m/s
m
2
 a = 16 99. rad/s2 �
(b) Reaction at A. A = ∞ ∞3
4
8 9 81 30 30( )( . )sin coskg m/s2 A = 25 5. N Æ �
186
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PROBLEM 16.124
The 4-kg uniform rod ABD is attached to the crank BC and is fitted with a 
small wheel that can roll without friction along a vertical slot. Knowing that 
at the instant shown crank BC rotates with an angular velocity of 6 rad/s 
clockwise and an angular acceleration of 15 rad/s2 counterclockwise, 
determine the reaction at A.
SOLUTION
Crank BC:
BC = 0 1. m
 ( ) ( ) ( . )( ) .a BCB t = = =a 0 1 15 1 5m rad/s m/s
2 2 ( ) .aB t = 1 5 rad/s
2 Ø
 ( ) ( ) ( . )( ) .a BCB n = = =w
2 20 1 6 3 6 m rad/s m/s2 ( ) .aB n = 3 6 m/s
2 Æ
Rod ABD: q = = = ∞- -sin sin .
.
1 1 0 1
0 2
30
BC
AB
 m
 m
a a aA B A G= + /
[aA≠] = [ .1 5Ø+ 3 6. Æ] [ .+ 0 2a b]
+ 0 3 6 0 2= -. ( . )cosa b
 a
b
= =
∞
=3 6
0 2
18
30
20 78
.
. cos cos
. rad/s2 a = 20 78. rad/s2 
Kinetics:
+
 S SM M A I m LG G= = =( ) : ( . )eff m0 1732
1
12
2a a
= 1
12
4 0 4 20 782( )( . ) ( . )kg m rad/s2
 A = 6 399. N A = 6 40. N ¨ �
187
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PROBLEM 16.125
The 250-mm uniform rod BD, of mass 5 kg, is connected as shown to disk A and to a 
collar of negligible mass, which may slide freely along a vertical rod. Knowing that 
disk A rotates counterclockwise at the constant rate of 500 rpm, determine the reaction 
at D when q = 0.
SOLUTION
Kinematics: For disk A: ( )a A = 0
w A - =500 52 36rpm rad/s.
 
v r
a r
B A
B A
= =
=
= =
w
w
( . )( . )
.
( . )( .
0 05 52 36
2 618
0 05 52 362
m rad/s
m/s
m radd/s
m/s
2 )
.= 137 08
For rod (velocities)
 
BC = -
=
( . ) ( . )
.
0 25 0 15
0 20
2 2
m
 
cos
.
.
sin
.
.
b
b
= =
= =
0 15
0 25
3
5
0 20
0 25
3
5
 w = =
v
BC
B 2 618
0 20
.
.
m/s
m
 w = 13 09. rad/s
Kinematics of rod (accelerations)
 a a aB D B D+ =/
 a aB D B nØ + ( )/ 5 4
3
 + ( )aD B t/ 53 4 = aD b
 137 08. Ø + 0 25 13 09 2. ( . ) 5 4
3
+ 0 25. a 53
4
 = rD b
188
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PROBLEM 16.125 (Continued)
x components:
 
3
5
42 84
4
5
0 25 0
25 704 0 2 0
( . ) ( . )
. .
- =
- =
a
a a = 128 52. rad/s2
a a a a= = +G B G B/
 = Ø + +137 08 0 125 13 09 0 125 128 522. . ( . ) . ( . )Z ^
 = Ø137.08 21.42+ b +16 065. b
+ ax = - =
3
5
21 42
4
5
16 065 0( . ) ( . )
 +ay = - - =137 08
4
5
21 42
3
5
16 065 110 31. ( . ) ( . ) . m/s2
Kinetics
 
I ml= =
=
1
12
1
12
5 0 25
0 026042
2 2( )( . )
.
 kg m
 S SM M D W ma IB B= + = -( ) : ( . ) ( . ) ( . )eff m m0 20 0 075 0 075 a
0 2 5 9 81 0 075 551 6 0 075 3 347. ( . )( . ) . ( . ) .D + = -
 
0 2 41 370 3 347 3 679 34 344
171 7
. . . . .
.
D
D
= - - =
= N
 D = Æ171 7. N �
+
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PROBLEM 16.126
Solve Problem 16.125 when q = ∞90 .
PROBLEM 16.125 The 250-mm uniform rod BD, of mass 5 kg, is connected as 
shown to disk A and to a collar of negligible mass, which may slide freely along 
a vertical rod. Knowing that disk A rotates counterclockwise at the constant rate 
of 500 rpm, determine the reaction at D when q = 0.
SOLUTION
Kinematics: For disk A: ( )a A = 0
w
w
A
B A
B
v r
a
= =
= =
=
500 52 36
0 05 52 36
2 618
rpm rad/s
 m rad/s
m/s
.
( . )( . )
.
== =
=
r Aw
2 20 05 52 36
137 08
( . )( . )
.
 m rad/s
m/s2
For rod (velocities)
Since v vD B//
We have w = 0
We also note that cos
.
.
.
f
f
=
= ∞
0 10
0 25
66 42
For rod (accelerations)
( )aD B n/ = 0
Since w = 0
a a aB D B D+ =/ _
137 08 0 25. .Æ + =a_ baD
( . )sin . .0 25 66 42 137 08a ∞ =
a = 598 3. rad/s2
a a a a= = +G B G B/
+ ax = - ∞ = +137 08 0 125 598 3 66 42 68 54. . ( . )sin . . m/s
2
+ ay = ∞ = +0 125 598 3 66 42 29 92. ( . )cos . . m/s
2
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PROBLEM 16.126 (Continued)
Summary of kinematics: a = 598 3 2. rad/s , ax = 68 54. m/s
2 Æ, ay = 29 92. m/s
2 ≠
I ml= =
=
1
12
1
12
5 0 25
0 026042
2 2( )( . )
.
kg m
Kinetics:
We recall that f = ∞66 42. . Thus: h = =( . )sin .0 25 0 2291m mf
+ S SM M Dh W ma h ma IB B x y= + =
Ê
ËÁ
ˆ
¯̃
- -( ) : ( . ) ( . )eff 0 05 2
0 05 a
D( . ) ( . )( . ) ( . )
.
( . )( . ) .0 2291 5 9 81 0 05 342 7
0 2291
2
149 6 0 05 15 58+ = - - 11
 0 2291 37 256 7 480 15 581 2 453 13 742. . . . . .D = - - - = D = 60 0. NÆ �
191
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PROBLEM 16.127
The 300-mm uniform rod BD weighs 4 kg and is connected as 
shown to crank AB and to a collar D of negligible weight, 
which can slide freely along a horizontal rod. Knowing that 
crank AB rotates counterclockwise at the constant rate of 
300 rpm, determine the reaction at D when q = 0.
SOLUTION
Crank AB: AB = =60 0 06mm m.
w p pAB =
Ê
ËÁ
ˆ
¯̃
=300 2
60
10rpm rad/s
v ABB AB= = =( ) ( . )( ) .w p0 06 10 1 88496 m/s≠
aB ABAB= = =( ) ( . )( ) .w p
2 20 06 10 59 218 m/s2 ¨
Geometry sin cosb b= =3
5
4
5
Rod BD:
Velocity: Instantaneous center at C
v BCB BD
BD
=
=
( )
. ( . )
w
w1 88496 0 24m/s
wBD = 7 854. rad/s
Acceleration
BD = =300 0 3mm m.
( ) ( ) .a BDD B t BD BD/ = =a a0 3 4
3
( ) ( ) ( . )( . ) .a BDD B n BD/
2m/s= = =w2 20 3 7 854 18 5056 4
3
a a a a a aD B D B B D B t D B n= + = + +/ / /( ) ( )
[aD ¨] [ .= 59 218 m/s
2 ¨] [ .+ 0 3a b ] [ .+ 18 5056 m/s2 b]
192
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PROBLEM 16.127 (Continued)
+
0 0 3
4
5
18 5056
3
5
= -( . ) ( . )a aBD = 46 264. rad/s
2
( ) ( )
.
( . ) .aG B t BDBG/
2 2rad/s m/s= = ÊËÁ
ˆ
¯̃
=a 0 3
2
46 264 6 9396 5 4
3
( ) ( )
.
( . ) .a BGG S n GD/
2rad/s m/s= = ÊËÁ
ˆ
¯̃
=w2 20 3
2
7 854 9 2528 
5
4
3
a a a a a a= + = + +B G B B G B t G B n/ / /( ) ( )
a = [ .59 218 m/s2 ] [ .+ 6 9396 m/s2 b] [ .+ 9 2528 m/s2 b ]
+ ax = +
Ê
ËÁ
ˆ
¯̃
+ ÊËÁ
ˆ
¯̃
59 218 6 9396
3
5
9 2528
4
5
. ( . ) ( . )
ax = 70 784. m/s
2 ¨
+
 a ay y=
Ê
ËÁ
ˆ
¯̃
- ÊËÁ
ˆ
¯̃
= =( . ) ( . ) ;6 9396 4
5
9 2528
3
5
0 0
Kinetcs:
I L
I
= = ÊËÁ
ˆ
¯̃
= ◊
= ◊
1
12
1
12
4 0 03
0 03
2 2m kg)(0.3 m) kg m
kg m
2
2
( .
.
+ S SM M D W I maB B BD x= - = +( ) : ( . ( . ( .eff m) m) m)0 24 0 12 0 09a
0 24 4 9 81 0 12 0 03 46 264 4 70 784 0 09
0 24
. ( )( . )( . ) ( . )( . ) ( )( . )( . )
.
D - = +
DD - = +4 7088 1 38792 25 48224. . . )
 D = 131 579. N D = 131 6. N≠ �
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PROBLEM 16.128
Solve Problem 16.127 when q = ∞90 .
PROBLEM 16.127 The 300-mm uniform rod BD has a mass of 
4 kg and is connected as shown to crank AB and to a collar D of 
negligible weight, which can slide freely along a horizontal rod. 
Knowing that crank AB rotates counterclockwise at the constant 
rate of 300 rpm, determine the reaction at D when q = 0.
SOLUTION
Crank AB: AB = =60 0 06mm m.
w p p= ÊËÁ
ˆ
¯̃
=300 2
60
10rpm rad/s
v ABB AB= = =( ) ( . )( ) .w p0 06 10 1 88496 m/s ̈
aB ABAB= = =( ) ( . )( ) .w p
2 20 06 10 59 218 m/s2 Ø
Rod BD:
Instantaneous center at 
wBD = 0
b = = ∞-sin
.
.
.1
0 12
0 3
23 58
m
m
Acceleration:
BD = =300 0 3mm m.
( ) ( ) ( . )a BDD B t BD/ m= =a 0 3 b
a a a a a aD B D B B D B t D B n= + = + +/ / /( ) ( )
[aD ¨] [ .= 59 218 m/s
2 Ø] [ .+ 0 3aBD b] + 0
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PROBLEM 16.128 (Continued)
+ 0 59 218 0 3= - +. ( . )cosa bBD
aBD = ∞
59 218
0 3 23 57
.
. cos .
 aBD = 215 36. rad/s
2
( ) ( )
.
( . ) .a BGG B t BD/
2m m/s= = ÊËÁ
ˆ
¯̃
=a 0 3
2
215 36 32 304
a a a a a a= + = + +B G B B G B t G B n/ / /( ) ( )
a = [ .59 218 m/s2 Ø] [ .+ 32 304 m/s2 b] + 0
+ a
a
x x
y
= ∞ = =
= -
[ . ]sin . . , .
. [
32 304 23 57 12 9174 12 9174
59 218
m/s m/s2 2a
332 304 23 57 59 218 29 609. ]cos . . .∞ = -
¨
+
 ay = 29 609. m/s
2 ay = 29 609. m/s
2Ø
Kinetics
I BD= = = ◊1
12
1
12
4 0 3 0 032 2m kg m2( ) ( )( . ) .
+ S SM M D W I ma maB B x y= - = + -( ) : ( . ) ( . ) ( . ) ( .eff 0 27495 0 13748 0 06 0 1374a 88)
0 27495 4 9 81 0 13748 0 03 215 36 4 12 9174 0 06. ( )( . )( . ) ( . )( . ) ( . )( .D - = + )) ( . )( . )
. . . . .
-
- = + -
4 29 609 0 13748
0 27495 5 3947 6 4608 3 1002 16 2826D
 D = -4 82597. N D = 4 83. N Ø �
195
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PROBLEM 16.129
The 3-kg 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: w
a
BD B
BD
= = =
=
15 0 08 15 1 2
60
rad/s, m rad/s m/s
rad/s2
v ( . )( ) . ¨
( ) ( . )( ) .aB x = =0 08 60 4 8m rad/s m/s
2 2¨
( ) ( . )( )aB y = =0 08 15 18m rad/s m/s
2 2 ≠
Rod AB:
Velocity: Instantaneous center at C.
CB = ∞ =( . ) tan .0 5 30 0 86603m / m
wAB
Bv
CB
= = =
1 2
0 86603
1 3856
.
.
.
m/s
m
rad/s
Acceleration:
( ) ( ) .aA B t AB ABAB/ = =a a0 5 Ø
( ) ( ) ( . )( . ) .aA B n ABAB/
2m/s= = =w2 20 5 1 3856 0 96 Æ
( ) ( ) .aA B t AB ABGB/ = =a a0 25 Ø
( ) ( ) ( . )( . ) .aA B n ABGB/
2m/s= = =w2 20 25 1 3856 0 40 Æ
a a a a a aA B B A B B A t G A n= + = + +/ / /( ) ( )
[aA 30 4 8∞ =] [ . ¨] [+ 18≠] [ .+ 0 5a ABØ] [ .+ 0 96Æ]
196
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PROBLEM 16.129 (Continued)
+ aA Acos . . ; .30 4 8 0 96 4 434∞ = - =a m/s
2 30∞
+ ( . )sin . ; .4 434 30 18 0 5 31 566∞ = - =a AB ABa rad/s
2
a a a a a a= + = + +B G B B G B t G B n/ / /( ) ( )
a = [ .4 8Æ] [+ 18≠] [ . ( . )+ 0 25 32 566 Ø] [ .+ 0 48Æ]
+ a ax x= - = =4 8 0 48 4 32 4 32. . . ; . m/s
2 ¨
+ a ay y= - = =18 2 892 10 108 10 108. . ; . m/s
2 ≠
Kinetics: I m AB= = = ◊1
12
3
12
0 5 0 06252 2( ) ( . ) .
kg
m kg m2
I m AB= = =1
12
3
12
0 5 02 2( ) ( . )
 kg
m
+
S SM M A mg Ia maB B AB B= ∞ - = - +( ) : ( sin )( . ) ( . ) ( . )eff m m m60 0 5 0 25 0 25
0 433 3 9 81 0 25 0 0625 31 566. ( )( . )( . ) ( . )( . )A - = - ◊kg m/s m kg m rad/s2 2 2
++ ( )( . )( . )3 10 108 0 25kg m/s m2
0 433 7 358 1 973 7 581. . . .A - = - +
 A = 29 94. N A = 29 9. N 60° �
197
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PROBLEM 16.130
In Problem 16.129, determine the reaction at A, knowing that 
in the position shown crank BD rotates with an angular velocity 
of 15 rad/s clockwise and an angular acceleration of 60 rad/s2 
counterclockwise.
PROBLEM 16.129 The 3-kg 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: w BD = 15 rad/s
vB = =( . )( ) .0 08 15 1 2m rad/s m/s¨
aBD = 60 rad/s
2
( ) ( . )( ) .aB x = =0 08 60 4 8m rad/s m/s
2 2Æ
( ) ( . )( )aB y = =0 08 15 18
2m rad/s m/s2 ≠
Rod AB:
Velocity: Instantaneous center at C.
CB = ∞ =( . ) tan .0 5 30 0 86603m / m
w AB B
v
CB
= = =
1 2
0 86603
1 3856
.
.
.
m/s
m
rad/s
Acceleration:
( ) ( ) .aA B t AB ABAB/ = =a a0 5 Ø
( ) ( ) ( . )( . ) .aA B n ABAB/
2m/s= = =w2 20 5 1 3856 0 96 Æ
( ) ( ) .aA B t AB ABGB/ = =a a0 25 Ø
( ) ( ) ( . )( . ) .aA B n ABGB/
2m/s= = =w2 20 25 1 3856 0 48 Æ
a a a a a aA B B A B B A t B A n= + = + +/ / /( ) ( )
[aA 30 4 8∞ =] [ . Æ] [+ 18≠] [ .+ 0 5a ABØ] [ .+ 0 96Æ]
198
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PROBLEM 16.130 (Continued)
+ aA A= ∞ = + =cos . . ; .30 4 8 0 96 6 651a m/s
2 30°
+ ( . )sin . ; .6 651 30 18 0 5 42 65∞ = - + =a AB ABa rad/s
2
a a a a a a= + = + +B G B B G B t B A n/ / /( ) ( )
a = [ .4 8Æ] +[18 ≠] [ . ( . )+ 0 25 42 65 Ø] [ .+ 0 48Æ]
+ a ax x= + = =4 8 0 48 5 58 5 28. . . ; . m/s
2 Æ
+ a y y= - = =18 10 663 7 337 7 337. . ; .a m/s
2≠
Kinetics: I m AB= = = ◊1
12
3
12
0 5 0 06252 2( ) ( . ) .
kg
m kg m2
+
S SM M A mg Ia maB B AB y= ∞ - = - +( ) : ( sin )( . ) ( . ) ( . )
.
eff m m m60 0 5 0 25 0 25
0 4333 3 9 81 0 25 0 0625 42 65
3
A - = - ◊
+
( )( . )( . ) ( . )( . )
(
kg m/s m kg m rad/s
k
2 2
gg m/s m2)( . )( . )7 337 0 25
0 433 7 358 2 666 5 503. . . .A - = - +
 A = 23 55. N A = 23 5. N 60° �
199
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PROBLEM 16.131
A driver starts his car with the door on the passenger’s side wide 
open ( ).q = 0 The 40-kg door has a centroidal radius of gyration 
k = 300 mm and its mass center is located at a distance 
r = 500 mm from its vertical axis of rotation. Knowing that the 
driver maintains a constant acceleration of 2 m/s2, determine the 
angular velocity of the door as it slams shut ( ).q = ∞90
SOLUTION
Kinematics: a = aA
where ( )/aG A t r= a q
Kinetics:
+ S SM M I mr r ma r
mk mr ma r
a r
A A A
A
A
= = + -
+ =
=
( ) : ( ) ( cos )
cos
eff 0
2 2
a a q
a a q
a
kk r2 2+
cosq
Setting a w wq=
d
d , and using r k= =0 5 0 3. , .m m
200
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PROBLEM 16.131 (Continued)
w
w
q
q
q
w w
d
d
a
a
d
A
A
=
( )
+ÈÎ ˘̊
=
=
0 5
0 3 0 5
1 4706
1 47
2 2
.
( . ) ( . )
cos
. cos
.
m
m m
006a dA cosq q
 
w w q q
w q
w
w p
w
p
d a d
a
f
f
A
A
f
0 0
2
2
0
0
2
2
1 4706
1
2
1 4706
Ú Ú=
=
( . )cos
. sin
/
/| |
== 2 9412. aA (1)
Given data: aA = 2 m/s
2 Æ
 
w f
2 2 9412 2
5 8824
=
=
. ( )
. w f = 2 43. rad/s �
201
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PROBLEM 16.132
For the car of Problem 16.131, determine the smallest constant 
acceleration that the driver can maintain if the door is to close and 
latch, knowing that as the door hits the frame its angular velocity 
must be at least 2 rad/s for the latching mechanisms to operate.
SOLUTION
Kinematics: a = aA
where ( )/aG A t r= a q
Kinetics:
+ S SM M I mr r ma r
mk mr ma r
a r
A A A
A
A
= = + -
+ =
=
( ) : ( ) ( cos )
cos
eff 0
2 2
a a q
a a q
a
kk r2 2+
cosq
Setting a w wq=
d
d , and using r k= =0 5 0 3. , .m m
w w
q
q
q
w w
d
d
a
a
d
A
A
=
( )
+ÈÎ ˘̊
=
=
0 5
0 3 0 5
1 4706
1 47
2 2
.
( . ( .
cos
. cos
.
m
m) m)
006a dA cosq q
202
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PROBLEM 16.132 (Continued)
 
w w q q
w q
w
w p
w
p
d a d
a
f
f
A
A
f
0 0
2
2
0
0
2
2
1 4706
1
2
1 4706
Ú Ú=
=
( . )cos
. | sin |
/
/
== 2 9412. aA (1)
Given data: w f = 2 rad/s
Eq. (1): w f A
A
a
a
2
2
2 9412
2 2 9412
=
=
.
( ) .
 aA = 1 3600. m/s
2 aA = 1 360. m/s
2 Æ �
203
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PROBLEM 16.133
Two 4-kg 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
a BC= =( ) ( .a a0 3 m)
a = 0 3. a Ø
sinq q= = ∞300 30 mm
600 mm
Bar BC: aD B L/ = a
Must be zero since aD Ø aBD BDa a= =0 and
Kinetics:
Bar BD
+ S SF F B W may y y= - = -( ) :eff
 
B
B
y
y
- = -
= -
( )( . ) ( . )
. .
4 9 81 4 0 3
39 24 1 2
a
a (1)
204
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PROBLEM 16.133 (Continued)
+ S SM M D W maB B= - = -( ) : ( . ( . ( .eff m) m) m)0 5196 0 15 0 15
 
D
D
( . ( ( . )( .
. .
0 5196 4 4 0 3 0 15
11 3279 0 3
 m) )(9.81)(0.15m) m)- = -
= -
a
44642a (2)
Bar AC: I m AC=
=
= ◊
1
12
1
12
4 0 6
0 12
2( )
( ( .
.
 kg) m)
kg m
2
2
+ S SM M W B I mC C y= + = +( ) : ( . ( . ( . )( . )eff m) m) m0 3 0 3 0 3 0 3a a
Substitute from Eq. (1) for By
( )( . )( . ) ( . . )( . ) . ( . )( . )
.
4 9 81 0 3 39 24 1 2 0 3 0 12 4 0 3 0 3
11 772
+ - = +
+
a a a
111 772 0 36 0 12 0 36
23 544 0 84
28 0286
. . . .
. .
.
- = +
=
=
a a a
a
a rad/s2
Eq. (2), D = -
= -
= -
11 3279 0 34642
11 3279 0 34642 28 0286
11 3279 9 7097
. .
. . ( . )
. .
a
 D = 1 6182. N D = 1 618. N ¨ �
205
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PROBLEM 16.134
The linkage ABCD is formed by connecting the 3-kg bar BC to the 
1.5-kg bars AB and CD. The motion of the linkage is controlled by 
the couple M applied to bar AB. Knowing that at the instant shown 
bar AB has an angular velocity of 24 rad/s clockwise and no angular 
acceleration, determine (a) the couple M, (b) the components of the 
force exerted at B on rod BC.
SOLUTION
Kinematics:
Bar AB: Bar CD:
Bar BC:
b
b
=
= ∞
-tan
.
1 5
12
22 62
206
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PROBLEM 16.134 (Continued)
( ) ( )
.
./aG B t BC BC BCBG= =
Ê
ËÁ
ˆ
¯̃
=a a a0 325
2
0 1625m 
5
13 12
( ) ( ) ( . ./aC B t BC BC BCBC= = =a a a0 325 0 325 m)
a a aC B C B t C x C y B C B ta a a a= + + = +( ) ; ) ( ) ( )/ /(
[ .0 125aCDÆ] [+ 72 m/s
2Ø] [= 72 m/s2≠] .+ ÈÎ0 325aBC 12
5
13
˘
˚̇
+ 72 72 0 325
12
13
480= - + =( . ) ;aBC BCa rad/s
2
+ 0 125 0 325 480
5
13
480. ( . )( ) ;aCD CD=
Ê
ËÁ
ˆ
¯̃
=a rad/s2
a a a a a= + = +B G B B G B t/ /( )
= [72 m/s2 ≠] .+ ÈÎ0 1625aBC
5
12 ˘
˚̇
= [72 m/s2 ≠] . ( )+ ÈÎ0 1625 480 12
5
13
˘
˚̇
a = [72 m/s2≠] + È
ÎÍ
78 m/s2 12
5
13
˘
˚̇
+ ax =
Ê
ËÁ
ˆ
¯̃
=78 5
13
30 m/s2 Æ
+ ay = -
Ê
ËÁ
ˆ
¯̃
= - =72 78 12
13
72 72 0
Total acceleration of G is 30 m/s2Æ
Bar CD:
Kinematics:
( ) ( . )aCD x = =0 0625 480 30 m)( rad/s m/s
2 2 Æ
207
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PROBLEM 16.134 (Continued)
Kinetics: I m CD
I
CD
CD
=
=
= ¥ ◊-
1
12
1
12
1 5 0 125
1 953 10
2
2
3
( )
( . .
.
 kg)( m)
 kg m2
+
S SM M C I m a
C
D D x CD CD x
x
= = +
=
( ) ( . ( ) ( .
. ( .
eff : m) m)0 125 0 0625
0 125 1
a
9953 10 480
1 5 30 0 0625
0
3¥ ◊
+
- kg m rad/s
 kg)( m/s m)
2 2
2
)( )
( . )( .
.1125 3 75
30
C
C
x
x
=
=
.
 N
Bar BC: I m BCBC = = = ¥ ◊
-1
12
1
12
3 0 325 26 406 102 2 3( ) ( . . kg)( m) kg m2
m aBC = =( )3 30 90 kg)( m/s N
2 Æ
+ S SF F B C m ax x x x BC= - =( )eff :
 
B
B
x
x
- =
= +
30 90
120
 N N
 N Bx = 120 N Æ �
+
S SM M B B W
I m a
C C y x
BC BC BC
= - -
= -
( ) ( . ( . ( .
( )
eff : m) m) m)0 3 0 125 0 15
a (( .0 0625 m)
0 3 120 0 125 3 9 81 0 15 26 406 10 4803. ( )( . ) ( )( . )( . ) ( . )(By - - = ¥ ◊
- kg m2 rad/s
 N)(0.0625)
2 )
(- 90
 
0 3 15 4 4145 12 675 5 625
0 3 26 465
88 22
. . . .
. .
.
B
B
B
y
y
y
- - = -
=
= + N By = 88 2. N ≠ �
208
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PROBLEM 16.134 (Continued)
Bar AB:
Kinetics:
+
S SM M M BA A x= + =( ) : ( .eff m)0 125 0
 
M
M
+ =
= - ◊
( .120 0 125 0
15
 N)( m)
 N m M = ◊15 N m �
209
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PROBLEM 16.135
Solve Problem 16.134, assuming that at the instant shown bar AB 
has an angular velocity of 24 rad/s clockwise and an angular 
acceleration of 160 rad/s2 counterclockwise.
PROBLEM 16.134 The linkage ABCD is formed by connecting 
the 3-kg bar BC to the 1.5-kg bars AB and CD. The motion of the 
linkage is controlled by the couple M applied to bar AB. Knowing 
that at the instant shown bar AB has an angular velocity of 24 rad/s 
clockwise and no angular acceleration, determine (a) the couple M, 
(b) the components of the force exerted at B on rod BC.
SOLUTION
Kinematics:
Bar AB: ( ) ( ) ( . )( )a ABB x AB= =a 0 125 160
( )aB x = 20 m/s
2 Æ
( ) ( ) ( . )( )a ABB y AB= =w
2 20 125 24
( )aB y = 72 m/s
2 ≠
Bar CD:
Bar BC:
 
b
b
=
= ∞
-tan
.
1 5
12
22 62
210
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PROBLEM 16.135 (Continued)
( ) ( ) ( . ) .aC B t BC BC BCBC/ m= = =a a a0 325 0 325
5
12
( ) ( )
.
.aG B BC BC BCBG/
 m= = ÊËÁ
ˆ
¯̃
=a a a0 325
2
0 1625 12513
a a a a aC B C B C x C y B x B y C B ta a a= + + = + +/ /: ( ) ( ) ( ) ( ) ( )
[ .0 125aCDÆ] [+ 72 m/s
2Ø] [= 20 m/s2Æ] [+ 72 m/s2≠] [ .+ 0 325aBC 12
5
13 ]
+ 72 72 0 325
12
13
480= - + =( . ) ;aBC BCa rad/s
2
+ 0 125 20 0 325 480
5
13
640. ( . )( ) ;aCD CD= +
Ê
ËÁ
ˆ
¯̃
=a rad/s2
a a a a a a= + = + +B G B B z B y G B t/ /( ) ( ) ( )
a = [20 m/s2Æ] [+ 72 m/s2≠] [ .+ 0 1625aBC 12
5
13 ]
= [20 m/s2Æ] [+ 72 m/s2≠] [ . ( )+ 0 1625 480 5 12 ]
a = [20 m/s2 Æ] [+ 72 m/s2≠] [+ 78 m/s2 12513 ]
+ ax = +
Ê
ËÁ
ˆ
¯̃
= + =20 78 5
13
20 30 50 m/s2
+ a y = -
Ê
ËÁ
ˆ
¯̃
=72 78 12
13
0
Total acceleration of G is 50 m/s2Æ
Bar CD:
Kinematics:
( ) ( . )( )aCD x = =0 0625 640 40m rad/s m/s
2 2 Æ
211
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PROBLEM 16.135 (Continued)
Kinetics: I m CD
I
CD
CD
=
=
= ¥ ◊-
1
12
1 5
12
0 125
1 953 10
2
3
( )
.
( .
.
 kg
 m)
 kg m
2
2
+ S SM M C I m a
C
D D x CD CD CD z
x
= = +
=
( ) ( . ( ) ( . )
. ( .
eff : m) m0 125 0 0625
0 126 1
a
9953 10 640
1 5 0 0625
0
3¥ ◊
+
- kg m rad/s
 kg)(40 m/s m)
2 2
2
)( )
( . )( .
.1126 5 00
40
C
C
x
x
=
=
.
 N
Bar BC: I m BCBC =
=
= ¥ ◊-
1
12
3
12
0 325
26 406 10
2
2
3
( )
( . )
.
 kg
 m
 kg m2
+
m a
F F B C m a
BC
x x x y BC
= = Æ
= - =
( )
( )
3 150 kg)(50 m/s N
:
2
effS S
 
B
B
x
x
- =
=
40 150
190
 N N
 N By = 190 N Æ �
+
 S SM M B B W I m aC C y x BC BC BC= - - = -( ) ( . ( . ( . ( )eff : m) m) m)0 3 0 125 0 15 a (( .0 0625 m)
0 3 190 3 26 406 10 3. ( ( ( . )By - - = ¥ ◊
-)(0.125) kg)(9.81)(0.15 m) kg m2 (( )
( )( . )
. . . .
480
150 0 0625
0 3 23 75 4 4145 12 675 9
 rad/s
 N m
2
-
- - = -By ..
. .
375
0 3 31 465By =
 By = 104 88. N By = 104 9. N ≠ �
212
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PROBLEM 16.135 (Continued)
Bar AB:
Kinetics:
+
S SM M M B m aA A x AB AB AB AB x= - = - -( ) : ( . ( ) ( .eff m) I m)0 125 0 0625a
M + = - -( )( . ) ( . )( . ) ( ) ( . )( )( . )190 0 125 1
12
1 5 0 125 160 1 5 10 0 06252
 
M
M
+ = - -
= - ◊
23 75 0 3125 0 9375
25 0
. . .
. N m M = ◊25 0. N m �
213
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PROBLEM 16.136
The 2-kg rod AB and the 3-kg rod BC are connected as shown to a disk 
that is made to rotate in a vertical plane at a constant angular velocity of 
6 rad/s clockwise. For the position shown, determine the forces exerted 
at A and B on rod AB.
SOLUTION
Kinematics:
Velocity w
w
AB
B A
BC
B
v v
v
=
=
=
=
= =
0
0 06 6
0 36
0 18
0 36
2
( .
.
.
.
 m)( rad/s)
 m/s
 m
 m/s22
 m0 18.
wBC = 2 rad/s
Acceleration: Note that aBC = 0 for the given configuration, i.e. w AB = 0.
aA = ( . )0 06
2 m)(6 rad/s
aA = 2 16. m/s
2 Ø
aB = =( . )( ) .0 18 2 0 72
2 rad/s m/s2 2Ø
a
a
BC
BC
=
=
( . )
.
0 09
0 36
2 m)(2 rad/s
 m/s2 Ø
214
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PROBLEM 16.136 (Continued)
a a aAB A B= + = +
1
2
1
2
2 16 0 72( ) ( . . )
aAB = 1 44. m/s
2 Ø
a aA B AB
AB
= +
= +
( .
. . ( .
0 12
2 16 0 72 0 12
 m)
 m/s m/s m)2 2
a
a
a AB = 12 rad/s
2
Kinetics: I m AB
I
AB AB
AB
= =
= ¥ ◊-
1
12
1
12
2 0 12
2 4 10
2
3
( ) ( ( .
.
kg) m)
kg m
2
2
Rod BC:
Since aBC xa= =0 0,
SM BC x= =0 0yields
Rod AB:
+ S SF F Ax x x= =( ) :eff 0
+ S SM M B W I m aA A y AB AB AB AB AB= - = -( ) : ( . ( . ( .eff m) m) m)0 12 0 06 0 06a
 
0 12 2 2 4 10 12
2 1 44
0 12
3. ( ( . )( )
( . )( )
.
B
B
y
y
- = ¥
-
=
-)(9.81)(0.06)
0.06
11 1772 0 0288 0 1728
0 12 1 0332
8 61
. . .
. .
.
= -
=
=
B
B
y
y N B = 8 61. N ≠ �
+ S SF F A W B m ay y y AB y AB AB= - + = -( ) :eff
 
A
A
y
y
- = -
=
( )( . ) ( . )
.
2 9 81 2 1 44
8 13
+ 8.61 
 N A y = 8 13. N ≠ �
215
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PROBLEM 16.137
The 2-N rod AB and the 3-N rod BC are connected as shown to a disk 
that is made to rotate in a vertical plane. Knowing that at the instant 
shown the disk has an angular acceleration of 18 2rad/s clockwise and 
no angular velocity, determine the components of the forces exerted at A 
and B on rod AB.
SOLUTION
Kinematics: Velocity of all elements = 0
Acceleration: For given configuration aAB = 0
a aB A= = =( . ) .0 06 18 1 08
2 m)( rad/s m/s2Æ
aBC
Ba= =
0 18
1 08
0 18.
.
. m
 m/s
 m
2
aBC = 6
2rad/s
aBC = =( . ) .0 09 6 0 54
2 m)( rad/s m/s2 Æ
a a aAB A B= = = 0 54. m/s
2 Æ
Kinetics: I m BCBC BC= =
1
12
1
12
3 0 182( ) ( )( . )2
Rod BC: IBC = ¥ ◊
-8 1 10 3. kg m2
+ S SM M B I m aA A x BC BC BC BC= = +( ) : ( . ( .eff m) m)0 18 0 09a
0 18 8 1 10 6
3 0 54 0 09
0 18 0 0486 0 1458
3. ( . )( )
( )( . )( .
. . .
B
B
B
x
x
x
= ¥
+
= +
-
)
== 1 08. N
( ) .on NAB xB = 1 08 ¨
216
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PROBLEM 16.137 (Continued)
Rod AB:
+ S SF F A B m ax x x x AB AB= - =( ) :eff
A
A
A
x
x
x
- =
- =
=
1 08 2 1 08
1 08
3 24
. ( ( . )
.
.
 N kg) m/s
 N 2.16 N
 N
2
 Ax = 3 24. N Æ
SNA y: NB = 9 81. ≠
SMB y: NA = 9 81. ≠
 �
217
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PROBLEM 16.138
In the engine system shown l = 250 mm and b = 100 mm. The 
connecting rod BD is assumed to be a 1.2-kg uniform slender rod and 
is attached to the 1.8-kg piston P. During a test of the system, crank 
AB is made to rotate with a constant angular velocity of 600 rpm 
clockwise with no force applied to the face of the piston. Determine 
the forces exerted on the connecting rod at B and D when q = ∞180 . 
(Neglect the effect of the weight of the rod.)
SOLUTION
Kinematics: Crank AB:
w pAB =
Ê
ËÁ
ˆ
¯̃
=600 2
60
62 832 rpm rad/s.
a ABB AB= =( ) ( . )w
2 0 1 m)(62.832 rad/s2
aB = 394 78. m/s
2 ¨
Also: v ABB AB= = =( ) ( .w 0 1 m)(62.832 rad/s) 6.2832 m/sØ
Connecting rod BD:
Velocity Instantaneous center at D.
wBD
Bv
BD
= = =6 2832
0 25
25 133
.
.
.
 m/s
 m
 rad/s
Acceleration:
a a aD B D B Ba= + =/ [ ¨] [( )+ BD BDw
2 Æ]
aD = [ .394 78 m/s
2 ¨] [( . )+ 0 25 2 m)(25.133 rad/s Æ]
aD = [ .397 78 m/s
2 ¨] [ .+ 157 92 m/s2 ¨] .= 236 86 m/s2¨
aBD B D= + =
1
2
1
2
394 78( ) ( .a a ¨ + 236 86. ¨) .= 315 82 m/s2 ¨
Kinetics of piston
 D = 426 34. N ¨
218
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PROBLEM 16.138 (Continued)
Force exerted on connecting rod at D is:
 D = 416 35. Æ
Kinetics of connecting rod: (neglect weight)
+ S SF F B D m ax x BD BD= - =( )eff :
B
B
- =
= +
=
426 34
426 35 378 48
. )
. .
 N (1.2 kg)(315.82 m/s
 N N
805.33 N
2
Forces exerted on connecting rod. B = 805 N ¨ �
 D = 426 N Æ �
219
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PROBLEM 16.139
Solve Problem 16.138 when q = ∞90 .
PROBLEM 16.138 In the engine system shown l = 250 mm and 
b = 100 mm. The connecting rod BD is assumed to be a 1.2-kg 
uniform slender rod and is attached to the 1.8-kg piston P. During a 
test of the system, crank AB is made to rotate with a constant angular 
velocity of 600 rpm clockwise with no force applied to the face of 
the piston. Determine the forces exerted on the connecting rod at B 
and D when q = ∞180 . (Neglect the effect of the weight of the rod.)
SOLUTION
Geometry: l b x l b
B A D B
= = = - =
= = -
0 25 0 1 0 2291
0 1 0 22
2 2. . .
( . ( .
 m, m, m
 m) ,/ /r j r 991 0 1 m) m)i j- ( .
Kinematics: w AB = =600 62 832 rev/min rad/s.
Velocity: w AB
B AB B A
= -
= ¥
= - ¥
=
( .
( . ) ( . )
( .
62 832
62 832 0 1
6 2832
 rad/s)
 
/
k
v r
k i
w
mm/s)i
v v
i i k i j
i
D B BD D B
D BD
r= + ¥
= + ¥ - -
= +
w
w
/
v 6 2832 0 2291 0 1
6 2832 0 1
. ( . . )
. . ww wBD BDj i- 0 2291.
Equating components gives
vD BD= =6 2832 0. . m/s, w
Acceleration: a a
a
AB BD BD D D
B AB B A AB B A
a
r
= = =
= ¥ - = -
0
0 62 832 0 12 2
,
( . ) ( . )
a
w
k a i
a r j/ / == -
= + ¥ -
= - + ¥
( . )
.
394 78
394 78
2
 m/s2
/ /
j
a a r r
i j k
D B BD B D BD D B
D ABa
a w
a (( . . )
. . .
- - -
= - - +
0 2291 0 1 0
394 78 0 2291 0 1
i j
j j ia aAB AB
Equate like components.
 j: 0 394 78 0 2291 1723= - - = -. . a aAB AB rad/s
2
 i: aD = - = -( . )( ) .0 1 1723 172 3 m/s
2
 a AB = -( )1723 rad/s
2 k a iD = -( . )172 3 m/s
2
220
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PROBLEM 16.139 (Continued)
Acceleration of mass center G of bar BD
rG D
G D BD G D
/
/
 m) m)
= +
= +
= + ¥
1
2
0 2291 0 1
0 11455 0 05
( . . )
( . ( .
i j
i j
a a ra --
= - + - ¥ + -
= - -
wBD G D
2
172 3 1723 0 11455 0 05 0
172 3 197
r
i k i j
i
/
. ( ) ( . . )
. .. .
( . ) ( . )
34 86 15
86 15 197 34
j i
i j
+
= - - m/s m/s2 2
Force on bar BD at P in B.
Use piston + bar BD as a free body
m
m
D D
BD G
a i
i
a i j
= -
= -
= - -
( . )( . )
( . N)
( . )( . .
1 8 172 3
310 14
1 2 86 15 197 34 ))
( . ) ( . )
( . )( . ) ( )
= - -
= - = -
103 38 236 88
1
12
1 2 0 25 1723 102
N Ni j
IBD BDa ..769 N m◊
+ S SF F B m a m ax x x D D BD G x= = +( ) : ( )eff
B Bx x= - - = -310 14 103 38 413 52. . . N
+ = - = + ¥ + ¥S SM M x I m m aB B BD BD D B D D G B BD G( ) : ( ) ( )eff / /Nk k r a ra
- = - + - + - -
¥ -
0 2291 10 769 0 1 310 14 0 11455 0 05
103
. . ( . )( . ) ( . . )
(
Nk k k i j
.. . )
. . . .
.
38 236 38
10 769 31 014 27 038 5 169
86 923
i j
k k k k
N
-
= - - + -
= N
+ = + =S SF F N B m ay y y BD G y( ) : ( )eff
 
86 923 236 88 323 80
413 52 323 80 525 22 2
. . .
. . .
tan
+ = - = -
= + =
=
B B
B
y y N
N
b 3323 80
413 52
38 1
.
.
.b = ∞ B = 525 N 38 1. ∞ �
221
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PROBLEM 16.139 (Continued)
Force exerted by bar BD as piston D.
Use piston D as a free body S SF
D j i
D i j
i j
=
+ =
= -
= - +
F
N m a
m a N
D D
D D
eff
 N) N)( . ( .310 14 86 923
Force exerted by the piston on bar BD. By Newton’s third law,
¢ = - = -D D i j( . ( .310 14 86 923 N) N) ¢ =D 322 N 15 7. ∞ �
222
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PROBLEM 16.140
Two identical rods AC and CE, each of weight W, are attached to 
form the linkage shown. Knowing that at the instant shown the force 
P causes the roller attached at D to move to the left with a constant 
velocity vD, determine the magnitude of the force P in terms of L, 
W, vD, and q.
SOLUTION
Kinematics: Velocity:
Horizontal component of vC equals vD Z/ ¨
vC Dvcosq =
1
2
vC
Dv=
2cosq
 q
Rod AC: v
v
v
C AC
D
AC
AC
D
L
L
L
=
=
=
2
2 2
w
q
w
w
q
cos
cos
By symmetry: | | | |w wCD AC=
Also since aD = 0, horizontal component of aC is zero. Thus aC is vertical
Acceleration: Rod AC
223
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PROBLEM 16.140 (Continued)
( ) ( ) tan
tan
a a
L L
C t C n
AC AC
=
=
q
a w q
2 2
2
a AC AC D D
v
L
v
L
= =
Ê
ËÁ
ˆ
¯̃
=w q
q
q
q
q
2
2 2
2 2
tan
cos
tan
tan
cos
By symmetry: | | | |a aAC CD=
aCE D
v
L
=
2
2 2
tan
cos
q
q
Kinetics: Entire system
S SM M W D BD IB B y CE AC= - = -( ) : ( )( ) ( )eff a a
Since a aCG AC y W= =, we find D ≠
Kinetics: Rod CE
+
S SM M P L IC C CE= = -( ) coseff : 2
q a
P
L m
L
v
L
D
2 12
2
2
2 2
cos
tan
cos
q
q
q
= - ◊
Ê
ËÁ
ˆ
¯̃
 P
mv
L
D= ◊
2
36
tan
cos
q
q
 �
224
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PROBLEM 16.141
At the instant shown, the 6 m long, uniform 50-kg pole ABC has an angular 
velocity of 1 rad/s counterclockwise and Point C is sliding to the right. 
A 500 N horizontal force P acts at B. Knowing the coefficient of kinetic 
friction between the pole and the ground is 0.3, determine at this instant 
(a) the acceleration of the center of gravity, (b)the normal force between 
the pole and the ground.
SOLUTION
Data: l m P
I ml
k= = = =
= = = ◊
6 50 500 0 3
1
12
1
12
50 6 1502 2 2
 m, kg, N,
 kg m
m .
( )( )
Kinematics: a = a aC Ca= Æ
a a a aG C G C t C G n= + +( ) ( )/ /
= [aC Æ] +
È
ÎÍ
l
2
a 10
2
2
∞˘˚̇
+ l w 80° (1)
Kinetics: Sliding to the right: F Nf k= m
+ SMG = + S( ) : sin cos cosM N
l
F
l
P
l
h IG feff 2
10
2
10
2
10∞ - ∞ + ∞ -ÊËÁ
ˆ
¯̃
= a
I N
l
N
l
P
l
hka m
a
- ∞ + ∞ = ∞ -ÊËÁ
ˆ
¯̃
- ∞ -
2
10
2
10
2
10
150 3 10
cos sin cos
( )(sin 00 3 10 500 3 10 2. cos ) ( cos )∞ = ∞ -N (2)
225
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PROBLEM 16.141 (Continued)
+ SFy =
+ S( ) : sin cosF N mg ml mly eff - = -
Ê
ËÁ
ˆ
¯̃
∞ - ∞
2
10
2
102a w
ml
N mg
ml
N
2
10
2
10
50 3 10 50 9
2sin cos
[( )( )sin ] ( )(
∞ÊËÁ
ˆ
¯̃
+ = - ∞
∞ + =
a w
a .. ) ( )( )( ) cos81 50 3 1 102- ∞ (3)
Solving Eqs. (2) and (3) simultaneously,
a = =2 5054 277 52. . rad/s NN
+ SFx = + S( ) : cos sinF P F ma
ml ml
x f Ceff - = - ∞ + ∞2
10
2
102a w
ma P N
ml ml
a
C k
C
= - + ∞ - ∞
= - +
m a w
2
10
2
10
50 500 0 3 277 52 50
2cos sin
( . )( . ) ( ))( )( . )cos
( )( )( ) sin
.
3 2 5054 10
50 3 1 10
15 2159
2
∞
- ∞
=aC m/s
Using Eq. (1), aG = +15 2159 3 2 5054. ( )( . ) 10 3 1
2∞ + ( )( ) 80∞
( ) . . cos cos
.
( ) .
a
a
G x
G y
= - ∞ + ∞
=
= -
15 2159 7 5162 10 3 80
8 3348
7 516
2 m/s
22 10 3 80
4 2596
8 3348 4 2596
9 36
2
2 2
sin sin
.
( . ) ( . )
.
∞ - ∞
= -
= +
=
 m/s
 m
aG
//s2
tan
.
.
.
b
b
=
= ∞
4 2586
8 3348
27 1 aG = 9 36. m/s
2 27.1° �
(a) Acceleration at Point G.
(b) Normal force. N = 278 N ≠ �
226
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PROBLEM 16.142*
A uniform disk of mass m = 4 kg and radius r = 150 mm 
is supported by a belt ABCD that is bolted to the disk at 
B and C. If the belt suddenly breaks at a point located 
between A and B, determine, (a) the acceleration of the 
center of the disk, (b) the tension in portion CD of the 
belt.
SOLUTION
Kinematics: w =
= = +
0
a a a aG C G C/
where a rG C/ = a
Kinetics:
+ S SF F W may y C= - ∞ = -( ) : coseff 30
aC g= 0 866. 60°
+ S SM M W r I ma rC C G C= ∞ = +( ) : ( sin ) ( )eff /30 a
mgr mr mr r
g r
g
r
sin ( )30
1
2
1
2
3
2
1
3
2∞ = ÊËÁ
ˆ
¯̃
+
= =
a a
a a
a r g gG C G C/ /= = =a
1
3
1
3
a 30°
(a) b
b
=
= ∞
=
=
-tan
.
.
.
cos
.
cos .
1 1 3
0 866
21 052
0 866
0 866
21 052
/ g
g
a
g
g
a g= =0 9279 0 9279 9 81. . ( . ) a = 9 10. m/s2 81.1° �
(b) 
+
S SF F T W max x G C= - ∞ = -( ) : sineff /30
T m g s mg
T
= - =
=
=
0 5
1
6
1
6
4 9 81
6 54
2
. ( )
( )( . )
.
 mg /
 kg m/s
 N T = 6 54. N �
227
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PROBLEM 16.143*
Two disks, each of mass m and radius r are connected as shown by a continuous chain 
belt of negligible mass. If a pin at Point C of the chain belt is suddenly removed, 
determine (a) the angular acceleration of each disk, (b) the tension in the left-hand 
portion of the belt, (c) the acceleration of the center of disk B.
SOLUTION
Kinematics: Assume a A and aB
w wA B= = 0
aD Ar= a Ø
aE D Aa r= = a Ø
aB E B E A Ba a r r= + = +/ ( )a a
aB A Br= +( )a a Ø
Kinetics: Disk A: + S SM M Tr IA A A= =( )eff : a
Tr mr A=
1
2
2a
a A
T
mr
= 2 (1)
Disk B: 
+
S SM M Tr IB B B= =( )eff : a
Tr mr B=
1
2
2a
aB
T
mr
= 2 (2)
228
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PROBLEM 16.143* (Continued)
From (1) and (2) we note that a aA B=
+
S SM M Wr I ma rE E B B= = +( ) ( )eff : a
Wr mr mr rB A B= + +
1
2
2a a a( )
a a aA B AWr mr= =:
5
2
2 a A
g
r
= 2
5
 �
 aB
g
r
= 2
5
 �
Substitute for a A into (1): 
2
5
2g
r
T
mr
= T mg= 1
5
 �
a r
r
r
g
r
B A B
A
= +
=
= ÊËÁ
ˆ
¯̃
( )
( )
a a
a2
2
2
5
 aB g=
4
5
Ø �
229
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PROBLEM 16.144
A uniform rod AB, of weight 15 kg and length 1 m, is attached to the 
20-kg cart C. Neglecting friction, determine immediately after the system 
has been released from rest, (a) the acceleration of the cart, (b) the angular 
acceleration of the rod.
SOLUTION
Let G be centre of mass of rod AB
Kinematics: We resolve the acceleration of G into the accel. of the cart and the accel. of G relative to A:
a a a a
a a a
R G A G A
R C G A
= = +
= +
/
/
where a LG A/ =
1
2
a
Kinetics: Cart and rod
W
W
L
I m L
R
C
R R
= =
= =
=
= =
( )( . ) .
( )( . ) .
15 9 81 147 15
20 9 81 196 2
1
1
12
2
N
 N
 m
11
12
15 1 1 25
1
2
1 0 5
2( )( ) .
( ) .
=
= =
kg m2
/
◊
aG A a a
+
S SF F W W m m m ax x C R C R C R G A= + ∞ = + - ∞( ) : ( )sin ( ) coseff /25 25a
( . . )sin [( ) ( . )cos ]
.
147 15 196 2 25 20 15 15 0 3 25
145 106 35
+ ∞ = + - ∞
=
a
a
C a
CC
Ca
-
= +
4 0784
4 1459 0 11653
.
( . ) .
a
a (1)
230
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PROBLEM 16.144 (Continued)
Rod
+
S SM M I m a L m a LA A R G R R C= = + - ∞( ) : ( ) ( cos )eff /0 2
25
2
a
1 25 15 0 5 0 5 15 25 0 5 0
1 25 3 75 6 797
. ( . )( . ) ( )( cos )( . )
. . .
a a
a a
+ - ∞ =
+ =
aC
33
1 3595
a
a
C
Ca = . (2)
(a) Acceleration of the cart.
 Substitute for a from (2) into (1):
a aC C= +4 1459 0 11653 1 3595. . ( . )
aC = 4 9263
2. m/s aC = 4 93. m/s
2 25° �
(b) Angular acceleration.
 From (2): a = ( . )( . )1 3595 4 9263
= 6 6973 2. rad/s a = 6 70 2. rad/s �
231
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PROBLEM 16.145*
A uniform slender bar AB of mass m is suspended as shown from a uniform 
disk of the same mass m. Neglecting the effect of friction, determine the 
accelerations of Points A and B immediately after a horizontal force P has 
been applied at B.
SOLUTION
Kinematics: w = 0
Cylinder: Rolling without sliding ( )aC x = 0
+ a a aA C x A C Ar= + = +( ) / 0 a
aA Ar= a Æ
a A
Aa
r
=
+ a
L
aA AB AB= -2
a
+
S SM M A r ma r IC C x A A= = +( )eff : a
Cylinder: A r ma r mr
a
r
A ma
A m
L
a
x A
A
x A
x AB AB
= + Ê
ËÁ
ˆ
¯̃
=
= -ÊËÁ
ˆ
¯̃
1
2
3
2
3
2 2
2
a (1)
+
S SM MA A= ( )eff : PL ma
L
IAB AB= +2
a
 PL ma
L m
LAB AB= +2 12
2a (2)
+ S SF Fx x= ( )eff : P A max AB+ =
Rod AB:
Rod AB:
Kinetics:
232
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PROBLEM 16.145* (Continued)
Substitute from (1):
P m
L
a ma
P ma mL
AB AB AB
AB AB
+ -ÊËÁ
ˆ
¯̃
=
= -
3
2 2
5
2
3
4
a
a (3)
Multiply by 
L
9
: 
1
9
5
18
1
12
2PL
L
ma mLAB AB= - a (4)
(4) + (2): 10
9
1
2
5
18
7
9
PL mLa mLaAB AB= +
Ê
ËÁ
ˆ
¯̃
=
a
P
mAB
= 10
7
¨ (5)
(5) Æ (3) P m P
m
mL
P P mL
AB
AB
= ÊËÁ
ˆ
¯̃
-
= -
5
2
10
7
3
4
25
7
3
4
a
a
- =18
7
3
4
P mL ABa a AB
P
mL
= 24
7
+ a
L
a
L P
mL
P
m
A AB AB= -
= ÊËÁ
ˆ
¯̃
-
2
2
24
7
10
7
a
a
P
mA
= -ÊËÁ
ˆ
¯̃
12
7
10
7
 aA
P
m
= Æ2
7
 �
+ a
L
a
L P
mL
P
m
B AB AB= +
= ÊËÁ
ˆ
¯̃
+
2
2
24
7
10
7
a
a
P
mB
= +ÊËÁ
ˆ
¯̃
12
7
10
7
 aB
P
m
= ¨22
7
 �
233
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PROBLEM 16.146*
The 5-kg slender rod AB is pin-connected to an 8-kg 
uniform disk as shown. Immediately after the system 
is released from rest, determine the acceleration of 
(a) Point A, (b) Point B.
SOLUTION
Kinematics: w = 0
a rB C= a Ø
+ a a a= +B G B/
a = +r LC ABa a2
Ø
Kinetics: Disk + S SM M Br IC C C= =( ) :eff a
Br m r
B m r
C C
C C
=
=
1
2
1
2
2a
a
Rod AB:
+ S SM M B L IG G AB= =( ) :eff 2
a
1
2 2
1
12
1
3
2m r
L
m L
m
m
L
r
C C AB AB
C
AB
C
AB
a a
a a
Ê
ËÁ
ˆ
¯̃
=
= ◊ (1)
+ S SF F m g B m ay y AB AB= - =( ) :eff
m g m r m r
L
g
L m
m
r
AB C C AB C AB
AB
C
AB
C
- = +ÊËÁ
ˆ
¯̃
= + +
Ê
ËÁ
ˆ
¯̃
1
2 2
2
1
2
1
a a a
a a
234
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PROBLEM 16.146* (Continued)
g
L
m
m
r
L
m
m
L
r
g
L
m
AB
C
AB
AB
C
AB= + +
Ê
ËÁ
ˆ
¯̃
◊ ◊
Ê
ËÁ
ˆ
¯̃
= + +
1
2
1
2
1
1
3
1
2
1
6
1
3
a a
AAB
C
AB
AB
C
AB
AB m
m
m
m
m
g
L AB
C
Ê
ËÁ
ˆ
¯̃
= +
Ê
ËÁ
ˆ
¯̃
=
+( )
a a
a
1
3
2
3 1
2
 (2)
m m r LAB C= = = =5 8 0 1 0 25 kg kg m, m, , . .
Eq. (1): a AB = ◊
+
=3 9 81
0 25
1
2
44 846
2
5
2( . )
.
.
m/s
 m
 rad/s
 kg
8 kg
Eq. (2): aC = ◊ ◊ =
1
3
5
8
0 25
0 1
44 846 23 3572 2
 kg
 kg
 m
 m
 rad/s rad/s
.
.
( . ) .
(b) Acceleration of B. a rB C=
=
a
( . )( . )0 1 23 357 2 m rad/s
= 2 336 2. m/s aB = 2 34. m/s
2 Ø �
(a) Acceleration of A. + a a aA B B A= + /
a aA B ABL= + [ a Ø]
a
a
A
A
=
+
= +
2 336
0 25 44 846
2 336 11 212
2
2
.
( . )( . )
. .
 m/s
 m rad/s
 aA = 13 55. m/s
2 Ø �
235
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PROBLEM 16.147*
The 3-kg cylinder B and the 2-kg wedge A are held at rest in the position 
shown by cord C. Assuming that the cylinder rolls without sliding on 
the wedge and neglecting friction between the wedge and the ground, 
determine, immediately after cord C has been cut, (a) the acceleration 
of the wedge, (b) the angular acceleration of the cylinder.
SOLUTION
Kinematics: We resolve aB into aA and a component parallel to the incline
a a aB A B A= + /
where a rB A/ = a , since the cylinder rolls on wedge A and wcyl. = 0
aB A/ m= ( . )0 1 a
Kinetics: Cylinder and wedge
+ S SF F m a m a m ax x A A B A B B A= = + - ∞( ) : coseff /0 20
0 3 2 3 0 1 20
0 06 20
= + - ∞
= ∞
( ) ( . )cos
( . cos )
a
a
A
A
a
a (1)
Cylinder
I mr
I
= =
=
1
2
1
2
3 0 1
0 015
2 2( )( . )
. kg ◊ m2
236
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PROBLEM 16.147* (Continued)
(b) Angular acceleration of the cylinder.
+ S SM M I m a m aE E B B A B A= ∞ = + -( ) : ( )sin ( . ) ( . ) coeff / m m3 20 0 1 0 1¥ 9.81 a ss ( . )20 0 1∞ m
2 943 20 0 015 3 0 1 0 1
3 20 0 1
1 0066 0 015
. sin . ( . )( . )
cos ( . )
. .
∞ = +
- ∞
=
a a
aA
aa a+ -0 03 0 28191. . aA
 Substitute from (1): 1 0066 0 015 0 03 0 28191 0 06 20
1 0066 0 029105
. . . . ( . cos )
. .
= + - ∞
=
a a a
a
a = 34 585 2. rad/s a = 34 6 2. rad/s �
(a) Acceleration of the wedge.
 Eq. (1): aA = ∞
= ∞
( . cos )
( . cos )( . )
0 06 20
0 06 20 34 585
a
aA = 1 9499
2. m/s aA = 1 950
2. m/s Æ �
237
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PROBLEM 16.148*
The 3-kg cylinder B and the 2-kg wedge A are held at rest in the 
position shown by cord C. Assuming that the cylinder rolls without 
sliding on the wedge and neglecting friction between the wedge and 
the ground, determine, immediately after cord C has been cut, (a) the 
acceleration of the wedge, (b) the angular acceleration of the cylinder.
SOLUTION
Kinematics: We resolve aB into aA and a horizontal component aB A/
a a aB A B A= + /
where a rB A/ = a , since the cylinder B rolls on wedge A, and wcyl. = 0
aB A/ m= ( . )0 1 a
Kinetics: Cylinder and wedge:
+
S SF F W Wm m a m ax x A B A B A B B A= + ∞ = + - ∞( ) : ( )sin ( ) coseff /20 20
( )( . )sin ( ) ( )( . )cos
. sin ( .
3 2 9 81 20 5 3 0 1 20
9 81 20
3
5
0
+ ∞ = - ∞
= ∞ +
a
a
A
A
a
11 20
9 81 20 0 06 20
)cos
. sin . cos
∞
= ∞ + ∞
a
aaA (1)
Cylinder: + S SM M I m a m aE E B B A B A= = + - ∞( ) : ( )( . ) ( cos )( . )eff / m m0 0 1 20 0 1a
0
1
2
3 0 1 3 0 1 0 1
3 20 0 1
0 0
2= +
- ∞
=
( ( . ) ( ( . )( . )
( cos ( . )
.
kg) m kg)
kg)
a a
aA
0015 0 03 0 2819
0 0 045 0 2819
6 2646
a a
a
a
+ -
= -
=
. .
. .
.
a
a
a
A
A
A (2)
238
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PROBLEM 16.148* (Continued)
(a) Acceleration of the wedge.
 Substitute for a from (2) into (1):
a a
a a
A A
A A
= ∞ + ∞
= +
-
9 81 20 0 06 20 6 2646
3 3552 0 375876
1 0
. sin . cos ( . )
. .
( .. ) .
.
375876 3 3552
5 3759 2
a
a
A
A
=
= m/s aA = 5 38
2. m/s 20° �
(b) Angular acceleration of the cylinder.
 Eq. (2): a =
=
6 2646
6 62646 5 3759
.
. ( . )
aA
a = 35 623. rad/s2 a = 35 6 2. rad/s �
239
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PROBLEM 16.149*
Each of the 3-kg bars AB and BC is of length L = 500 mm. A horizontal force P of 
magnitude 20 N is applied to bar BC as shown. Knowing that b = L (P is applied at C ), 
determine the angular acceleration each bar.
SOLUTION
Kinematics: Assume a AB aBC and w wAB BC= = 0
Kinetics: Bar BC
+ S SM M PL I ma LB B BC BC= = +( ) ( )eff : a 2
= + +ÊËÁ
ˆ
¯̃
= +
m
L m L
L L
P mL mL
BC AB BC
AB BC
12 2 2
1
2
1
3
2a a a
a a (1)
+ S SF F P B max x x BC= - =( )eff :
P B m L Lx AB BC- = +
Ê
ËÁ
ˆ
¯̃
a a1
2
 (2)
240
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PROBLEM 16.149* (Continued)
Bar AB:
+ S SM M B L I ma LA A x AB AB= = +( ) ( )eff : a 2
= + ÊËÁ
ˆ
¯̃
=
m
L m
L L
B mL
AB AB
x AB
12 2 2
1
3
2a a
a (3)
Add (2) and (3): P mL mLAB BC= +
4
3
1
2
a a (4)
Subtract (1) from (4) 0
5
6
1
6
= +mL mLAB BCa a
a aBC AB= -5 (5)
Substitute for aBC in (1): P mL mL mLAB AB AB= + - = -
1
2
1
3
5
7
6
a a a( )
a AB
P
mL
= - 6
7
 (6)
Eq. (5) a aBC BC
P
mL
P
mL
= - -ÊËÁ
ˆ
¯̃
=5 6
7
30
7
 (7)
Data: L m P= = = =500 0 5 3 20 mm m, kg, N.
a AB
P
mL
= - = -
= -
6
7
6
7
20
3 0 5
11 249 2
 N
 kg m
 rad/s
( )( . )
. a AB = 11 43. rad/s
2 �
aBC
P
mL
= =
=
30
7
30
7
20
3 0 5
57 143 2
 N
 kg m
 rad/s
( )( . )
. aBC = 57 1. rad/s
2 �
241
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PROBLEM 16.150*
Each of the 3-kg bars AB and BC is of length L = 500 mm. A horizontal force P of 
magnitude 20 N is applied to bar BC. For the position shown, determine (a) the distance 
b for which the bars move as if they formed a single rigid body, (b) the corresponding 
angular acceleration of the bars.
SOLUTION
Kinematics: We choose a a a= =AB BC
Kinetics: Bars AB and BC (acting as rigid body)
m m
I m L
I mL
ABC =
=
=
2
1
12
2 2
2
3
2
2
( )( )
+ S SM M P L b I m a LA A ABC ABC B= + = +( ) ( )eff : a
P L b mL m L L
P L b mL
( ) ( )( )
( )
+ = +
+ =
2
3
2
8
3
2
2
a a
a (1)
242
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PROBLEM 16.150* (Continued)
Bar BC:
+ S SM M Pb I ma LB B BC BC= = +( ) ( )eff : a 2
= + ÊËÁ
ˆ
¯̃
=
=
m
L m L
L
Pb mL
Pb
mL
12
3
2 2
5
6
6
5
2
2
2
a a
a
a (2)
Substitute for a into (1): P L b mL Pb
mL
PL Pb Pb
L b b
b
( )+ = ÊËÁ
ˆ
¯̃
+ =
= -ÊËÁ
ˆ
¯̃
=
=
8
3
6
5
16
5
16
5
1
11
5
5
2
2
111
L
Eq. (2) a a= ÊËÁ
ˆ
¯̃
=6
5
5
11
6
112
P
mL
L
P
mL
Data: L
m
P
= =
=
=
500 0 5
3
20
 mm m
 kg
 N
.
(a) b L= = =5
11
5
11
0 5 0 22727( . ) . m b = 227 mm �
(b) a = = =6
11
6
11
20
3 0 5
7 2727 2
P
mL
( )
( )( . )
.
 N
 kg m
 rad/s a = 7 27. rad/s2 �
243
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PROBLEM 16.151*
(a) Determine the magnitude and the location of the maximum bending moment in the rod of Problem 16.76. 
(b) Show that the answer to Part a is independent of the weight of the rod.
SOLUTION
Rod AB: a
L=
2
a
+
S SM M PL ma L IA A= = +( ) : ( )eff 2
a
= ÊËÁ
ˆ
¯̃
+m L m L1
2 2
1
12
2a a
a = 3P
mL
 (1)
+ S SF F A P max x x= - = -( ) :eff
A P m
L
P m
L P
mL
P
x = - = -
Ê
ËÁ
ˆ
¯̃
= -
2 2
3
2
a A x P=
1
2
¨
Portion AJ of Rod:
External forces: A Wx AJ, , axial force FJ, shear VJ, and bending moment MJ.
Effective forces: Since acceleration at any point is proportional to distance from A, effective forces are linearly 
distributed. Since mass per unit length is m/L, at Point J we find
m
L
a
m
L
xJ
Ê
ËÁ
ˆ
¯̃
= ( )a
Using (1): 
m
L
a
m
L
P
mL
m
L
a
Px
L
J
J
= ÊËÁ
ˆ
¯̃
=
3
3
2
+ S S(M M M A x Px
L
x
M Px
P
L
x
J J J x
J
= - = - ÊËÁ
ˆ
¯̃
¥ ÊËÁ
ˆ
¯̃
= -
) :eff
1
2
3 2
3
1
2
1
2
2
2
33
 
(2)
For Mmax : 
dM
dx
P P
L
xJ = - =
2
3
2
0
2
2
x
L=
3
 (3)
244
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PROBLEM 16.151* (Continued)
Substituting into (2)
( )
( )
max
max
M
PL P
L
L PL
M
PL
J
J
= - Ê
ËÁ
ˆ
¯̃
= ÊËÁ
ˆ
¯̃
=
1
2 3
1
2 3
1
2 3
2
3
3 3
2
3
 (4)
Note: Eqs. (3) and (4) are independent of W.
Data: L P= =1 8m N,
Eq. (3): x
L= = =
3
1
3
0 5773m m.
Eq. (4):( )
( )( )
.
maxMJ =
= ◊
8 1
3 3
1 5396
N m
N m
 M Amax . .= ◊1 540 0 577N m located m below �
245
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PROBLEM 16.152*
Draw the shear and bending-moment diagrams for the beam of Problem 16.84 immediately after the cable at 
B breaks.
SOLUTION
From answers to Problem 16.84:
a g A mgB = =
3
2
1
4
 ≠
We now find
a = =
a
L
g
L
B 3
2
a x
g
L
xJ = =a
3
2
Ø
Portion AJ of rod:
External forces: Reaction A, distributed load per unit length mg/L, shear VJ , bending moment MJ .
Effective forces: Since a x: , the effective forces are linearly distributed. The effective force per unit length at J is:
m
L
a
m
L
g
L
x
mg
L
xJ = ◊ =
3
2
3
2 2
+S SF F mg
L
x
mg
V
mg
L
x xy y J= - + =
Ê
ËÁ
ˆ
¯̃
( )eff : 4
1
2
3
2 2
V
mg mg
L
x
mg
L
xJ = - +4
3
4 2
2
+ S SM M mg
L
x
x mg
x M
mg
L
x x
x
J J J=
Ê
ËÁ
ˆ
¯̃
- + = ÊËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
( )eff : 2 4
1
2
3
2 32
M
mg
x
mg
L
x
mg
L
xJ = - +4
1
2
1
4
2
2
3
Find Vmin: 
dV
dx
mg
L
mg
L
x x LJ = - + = =3
2
0
2
32
; 
V
mg mg
L
L
mg
L
L V
mg
min min;= -
Ê
ËÁ
ˆ
¯̃
+ ÊËÁ
ˆ
¯̃
= -
4
2
3
3
4
2
3 122
2
 
246
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PROBLEM 16.152* (Continued)
Find Mmax where V V
mg mg
L
x
mg
L
xJ J= = - + =0 4
3
4
02:
3 4 02 2x Lx L- + =
( )( )3 0
3
x L x L x
L
x L- - = = = and 
M
mg L mg
L
L mg
L
L mgL
max =
Ê
ËÁ
ˆ
¯̃
- ÊËÁ
ˆ
¯̃
+ ÊËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
=
4 3
1
2 3
1
4 3 27
2 3
M
mg
L
mg
L
L
mg
L
Lmin = - + =4
1
2
1
4
02 3
 M
mgL L
Amax = 27 3
 at from �
247
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PROBLEM 16.153
The axis of a 100-mm-radius disk is fitted into a slot that forms an angle 
q = 30∞ with the vertical. The disk is at rest when it is placed in contact with 
a conveyor belt moving at constant speed. Knowing that the coefficient of 
kinetic friction between the disk and the belt is 0.20 and neglecting bearing 
friction, determine the angular acceleration of the disk while slipping occurs.
SOLUTION
+ S SF Fx x= ( ) :eff m qk N R- =cos 0
 R Nkcosq m= (1)
+S SF F R N mgy y= + - =( ) : sineff q 0
 R mg Nsinq = - (2)
Divide (2) by (1): tan .
.
q
m
= - = -mg N
N
mg N
Nk
; 0 5774
0 2
0 1155
1 1155
0 8965.
.
.N mg N N
mg= - = =; mg
+ S SM M Nr Ik0 0= =( )eff : m a
( . )( . )0 2 0 8965
1
2
2mg r mr= a
a =
=
=
0 35858
0 35858
9 81
0 1
35 177
.
.
.
( .
.
g
r
 m/s
m)
 rad/s
2
2 a = 35 2. rad/s2 �
248
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PROBLEM 16.154
Solve Problem 16.153, assuming that the direction of motion of the conveyor 
belt is reversed.
PROBLEM 16.153 The axis of a 100-mm-radius disk is fitted into a slot that 
forms an angle of q = 30° with the vertical. The disk is at rest when it is 
placed in contact with a conveyor belt moving at constant speed. Knowing 
that the coefficient of kinetic friction between the disk and the belt is 0.20 and 
neglecting bearing friction, determine the angular acceleration of the disk of 
change while slipping occurs.
SOLUTION
+ S SF Fx x= ( )eff : R Nkcosq m- = 0
R Nkcosq m= (1)
+S SF F R mg Ny y= + - =( ) sineff : q 0
R N mgsinq = - (2)
Divide (2) by (1): tan .
.
q
m
= - = -N mg
N
N mg
Nk
; 0 5774
0 2
0 1155
0 8845
1 1306.
.
.N N mg N
mg
mg= - = =:
+
S SM M Nr Ik0 0= =( )eff : m a
( . )( . )
.
.
.
( . )
0 2 1 1306
1
2
0 4522
0 4522
9 81
0 1
2mg r mr
g
r
=
=
=
a
a
 m/s
 m
2
a = 44 361. rad/s2 a = 44 4. rad/s2 �
249
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PROBLEM 16.155
Identical cylinders of mass m and radius r are pushed by a series of moving arms. Assuming the coefficient 
of friction between all surfaces to be m �1 and denoting by a the magnitude of the acceleration of the arms, 
derive an expression for (a) the maximum allowable value of a if each cylinder is to roll without sliding, (b) the 
minimum allowable value of a if each cylinder is to move to the right without rotating.
SOLUTION
(a) Cylinder rolls without sliding a r= a or a = a
r
 P is Horizontal component of force arm exerts on cylinder.
+
S SM M P r I ma rA A k= - = +( ) : ( ) ( )eff Pr m a
P r mr
a
r
ma r
P
ma
( ) ( )
( )
1
1
2
3
2 1
2- = ÊËÁ
ˆ
¯̃
+
=
-
m
m
 (1)
+SFy = 0: N P mg- - =m 0 (2)
+ S SF Fx x= ( ) :eff P N ma- =m (3)
 Solve (2) for N and substitute for N into (3).
P P mg ma- - =m m2
 Substitute P from (1): ( )
( )
1
3
2 1
2-
-
- =m
m
mma mg ma
3 1 2 2
1 3 2 0
( )
( )
+ - =
+ - =
m m
m m
a g a
a g a g=
+
2
1 3
m
m
 �
250
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PROBLEM 16.155 (Continued)
(b) Cylinder translates: a = 0
 Sliding occurs at A: A Nx = m
 Assume sliding impends at B: B Py = m
+ S SM MA A= ( ) :eff Pr Pr ma r- =m ( )
P r ma r( )1- =m
P
ma
=
-1 m
 (4)
+ S SF Fx x= ( ) :eff P N ma- =m (5)
+S SF Fy y= ( ) :eff N P mg- - =m 0 (6)
 Solve (5) for N and substitute for N into (6).
P P mg ma- - =m m2
 Substitute for P from (4):
ma
mg ma
a g a
a g
1
1
1
0
2
-
- - =
+ - =
- =
m
m m
m m
m m
( )
( )
 a g= �
Summary: a g�
2
1 3
m
m+
: rolling
3
1 3
m
m+
g a g� � : Rotating and sliding
a g� : Translation
251
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PROBLEM 16.156
A cyclist is riding a bicycle at a speed of 30 kmph on a horizontal road. The distance between the axles is 
1050 mm, and the mass center of the cyclist and the bicycle is located 650 mm behind the front axle and 
1000 mm above the ground. If the cyclist applies the brakes only on the front wheel, determine the shortest 
distance in which he can stop without being thrown over the front wheel.
SOLUTION
v0 30= kmph
When cyclist is about to be thrown over the front wheel, N A = 0
+ S SM M mg maB B= =( ) : ( . ) ( )eff m m0 65 1
a g= = =0 65 0 65 9 81 6 3765. . ( . ) .m/s m/s2 2
Uniformly accelerated motion:
v
v v as s
s
0
2
0
2
30
25
3
2 0
25
3
2 6 3765
= =
- = - ÊËÁ
ˆ
¯̃
= -
=
kmph m/s
m/s
2
2
2: ( . )
55 4453. m s = 5 45. m �
252
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PROBLEM 16.157
The uniform rod AB of weight W is released from rest when b = ∞70 . 
Assuming that the friction force between end A and the surface is large 
enough to prevent sliding, determine immediately after release (a) the 
angular acceleration of the rod, (b) the normal reaction at A, (c) the 
friction force at A.
SOLUTION
We note that rod rotates about A. w = 0
I mL
a
L
=
=
1
12
2
2
a
+
S SM MA A= ( ) :eff mg
L
I ma
L
2 2
cos ( )b aÊËÁ
ˆ
¯̃
= +
1
2
1
12 2 2
1
3
2
2
mgL mL m
L L
mL
cosb a a
a
= + ÊËÁ
ˆ
¯̃
=
a b= 3
2
g
L
cos
 (1)
+ S SF Fx x= ( ) :eff F maA = sin b
F m
L
m
L g
LS
= = ÊËÁ
ˆ
¯̃2 2
3
2
a b b bsin cos sin
F mgA =
3
4
sin cosb b (2)
+S SF Fy y= ( ) :eff N mg ma m
L
A - = - = -
Ê
ËÁ
ˆ
¯̃
cos cosb a b
2
N mg m
L g
L
N mg
A
A
- = - ÊËÁ
ˆ
¯̃
= -ÊËÁ
ˆ
¯̃
2
3
2
1
3
4
2
cos
cos
cos
b b
b
 (3)
253
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PROBLEM 16.157 (Continued)
For b = ∞70 :
(a) Eq. (1): a = ∞3
2
70g
L
cos
 a = 0 513. g
L
 �
(b) Eq. (3): N mgA = - ∞
Ê
ËÁ
ˆ
¯̃
1
3
4
702cos NA mg= 0 912. ≠ �
(c) Eq. (2): F mgA = ∞ ∞
3
4
70 70sin cos FA mg= 0 241. Æ �
254
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PROBLEM 16.158
The uniform rod AB of weight W is released from rest when b = ∞70 . 
Assuming that the friction force is zero between end A and the surface, 
determine immediately after released (a) the angular acceleration of the 
rod, (b) the acceleration of the mass center of the rod, (c) the reaction at A.
SOLUTION
 + S SF Fx x= ( ) :eff a m x x= =a a 0
a aA y
L= +
2
a b
+ 0
2
= -a Ly a bcos
a y
L=
2
a bcos
 +S SF Fy y= ( ) :eff mg A ma m
L
y- = =
Ê
ËÁ
ˆ
¯̃2
a bcos (1)
+ S SM M A L I mLG G=
Ê
ËÁ
ˆ
¯̃
= =( ) : coseff 2
1
12
2b a a
A
mL=
6
a
bcos
 (2)
Substitute (2) into (1):
mg
mL
m
L
g
L L
g
L
- =
= +
Ê
ËÁ
ˆ
¯̃
= +
Ê
6 2
2 6
6
3
1
a
b
a b
b
b
a
b
b
cos
cos
cos
cos
cos
cosËËÁ
ˆ
¯̃
= +
Ê
ËÁ
ˆ
¯̃
a
b
b
ag L
6
3 12cos
cos
a b
b
=
+
Ê
ËÁ
ˆ
¯̃
6
1 3 2
g
L
cos
cos
255
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PROBLEM 16.158 (Continued)
a = = ◊
+
Ê
ËÁ
ˆ
¯̃
=
+
Ê
ËÁ
ˆL L g
L
g
2 2
6
1 3
3
1 32
2
2
a b b
b
b b
b
cos
cos
cos
cos
cos
cos ¯̃̄
¨
A = ◊ = ◊ ◊
+
Ê
ËÁ
ˆ
¯̃
=
+
mL mL g
L
mg
6 6
6
1 3
1 1
1 32 2
a
b
b
b b bcos
cos
cos cos cos
≠
For b = ∞70 :
(a) a = ∞
+ ∞
6 70
1 3 702
g
L
cos
cos
 a = 1 519. g
L
 �
(b) a g=
+ ∞
3
70
1 3 70
2
2
cos
cos
 a = 0 260. g Ø �
(c) A mg=
+ ∞
1
1 3 702cos
 A = 0 740. mg ≠ �
256
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PROBLEM 16.159
A uniform plate of mass m is suspended in each of the ways shown. For each case determine immediately 
after the connection at B has been released (a) the angular acceleration of the plate, (b) the acceleration 
of its mass center.
SOLUTION
(1) Plate attached to pins
 Kinematics: Assume a w( )= 0
a = ra q
 + x comp: a r rx = =a q q asin ( sin )
 + y comp: a r ry = =a q q acos ( cos )
 Thus: a = 1
4
Ca ¨; ay C=
1
2
a Ø (1)
 Kinetics: I m C
C
mC= + ÊËÁ
ˆ
¯̃
È
Î
Í
Í
˘
˚
˙
˙
=1
12 2
5
48
2
2
2
(a) 
+
S SM M W C I ma C ma CA A x y=
Ê
ËÁ
ˆ
¯̃
= + ÊËÁ
ˆ
¯̃
+ ÊËÁ
ˆ
¯̃
( ) : ( ) ( )eff 2 4 2
a
1
2
5
48
1
4 4
1
2 2
2mgC mC m C
C
m C
C= + ÊËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
+ ÊËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
a a a
1
2
20
48
1 22mgC mC
g
C
= =a a . a = 1 2. g
C
 �
257
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PROBLEM 16.159 (Continued)
(b) From (1): a C g gx x= = =
1
4
1
4
1 2 0 3a ( . ) .a ¨
a C g gy y= = =
1
2
1
2
1 2 0 6a ( . ) .a Ø a = 0 671. g 63.4° �
(2) Plate suspended from wires.
 Kinematics: Assume a ( )w = 0
a a a a= = +C A G A/
= ´ +a rA a q
 + y comp.
a r
r
y = -
= -
0 a q
q a
cos
( cos )
r Ccosq = 1
2
 Thus: a C a Cy y= - =
1
2
1
2
a a Ø (2)
 Kinetics: I m C
C I
mC= + ÊËÁ
ˆ
¯̃
È
Î
Í
Í
˘
˚
˙
˙
=1
12 2 48
2
2
2
+ S SF Fx x= ( ) :eff 0
0
2=
=
ma
ax
+
S SM M W C I ma CA A y=
Ê
ËÁ
ˆ
¯̃
= + ( )ÊËÁ
ˆ
¯̃
( ) :eff
1
2
1
2
a
 Recalling (1): 
1
2
5
48
1
2
1
2
2mgC mC mC C= + ÊËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
a a
1
2
17
48
24
17
2mgC mC
g
C
= =a a a = 24
17
g
C
 �
a C C
g
C
gy = =
Ê
ËÁ
ˆ
¯̃
=1
2
1
2
24
17
12
17
a a = 12
17
g Ø �
258
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it without permission.PROBLEM 16.159 (Continued)
(3) Plate suspended from springs. Immediately after spring B is released, the tension in spring A is still 12 mg
 since its elongation is unchanged.
(a) Angular acceleration.
+
S SM M mg C IG G=
Ê
ËÁ
ˆ
¯̃
Ê
ËÁ
ˆ
¯̃
=( ) :eff
1
2
1
2
a
1
4
5
48
2mgC mC= a a = 2 4. g
C
 �
(b) Acceleration at mass center.
+ S SF F ma ax x x x= = =( ) :eff 0 0
+S SF Fy y= ( ) :eff mg mg may- =
1
2
a gy =
1
2
 a = 0 5. g Ø �
259
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PROBLEM 16.160
The slender bar AB of weight W is held in equilibrium by 
two counterweights each weighing 12W . If the wire at B is 
cut, determine the acceleration at that instant (a) of Point A, 
(b) of Point B.
SOLUTION
Kinematics:
w =
=
0
a aC A
[a Ø] [= aA ≠ ] + Ø
È
ÎÍ
˘
˚̇
L
2
a [ ] [a aB AØ = ≠] [+ La Ø]
a = -ÊËÁ
ˆ
¯̃
1
2
L aAa Ø a L aB A= -( )a Ø
Kinetics: Counterweight m = mass of bar AB
+S SF F W T m a may y C C A= - = =( ) :eff
1
2
1
2
1
2
1
2
1
2
mg T ma
T m g a
A
A
- =
= -( ) (1)
260
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PROBLEM 16.160 (Continued)
Kinetics: Bar AB
+ S SF F mg T my y= - =( ) :eff a
mg m g a m L aA A- - = -
Ê
ËÁ
ˆ
¯̃
1
2
1
2
( ) a
2 2 3g g a L a g a LA A A- + = - + =a a (2)
+
S SM M T L IG G= =( ) :eff 2
a
1
2 2
1
12
2m g a
L
mLA( )- = a
3 3g a LA- = a (3)
Add Eqs. (2) and (3): 4 2
2
g L
g
L
= =a a
(a) Acceleration at Point A.
 Substitute into (2): g a L
g
LA
+ = ÊËÁ
ˆ
¯̃
3
2
 aA g=
1
3
≠ �
a L a L
g
L
gA= -
Ê
ËÁ
ˆ
¯̃
= ÊËÁ
ˆ
¯̃
-1
2
1
2
2 1
3
a a = 2
3
g Ø
(b) Acceleration at Point B. a L a L
g
L
gB A= - = -
Ê
ËÁ
ˆ
¯̃
( )a 2 1
3
 aB g=
5
3
Ø �
261
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PROBLEM 16.161
The mass center G of a 5-kg wheel of radius R = 300 mm is located at a distance 
r= 100 mm from its geometric center C. The centroidal radius of gyration is 
k = 150 mm. As the wheel rolls without sliding, its angular velocity varies and it is 
observed that w = 8 rad/s in the position shown. Determine the corresponding angular 
acceleration of the wheel.
SOLUTION
Kinematics: (Rolling motion) a RC = a
Translation with C + Rotation about C = Rolling motion
a = +( )a rC w
2 Æ+ra ≠
Kinetics:
I mk
m
k
R
r
a RC
=
=
=
=
=
=
=
-2
5
0 15
0 3
0 1
8
kg
m
m
m
rad/s
.
.
.
a
w
+
S SM M Wr I ma R ma rE E x y= - = + +( ) : ( ) ( )eff a
- = + + +
= + + +
- = + +
mgr mk m a r R mr r
mk mR mrR mr
gr k R r
C
2 2
2 2 2 2
2 2
a w a
a a w a
( )
( 22 2
2 2 29 81 0 1 0 15 0 3 0 1 0 1 0 3 8
)
. ( . ) [( . ) ( . ) ( . ) ] ( . )( . )( )
a w
a
+
- = + + +
rR
22
a = - 23 68. rad/s2 a = 23 7. rad/s2 �
262
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PROBLEM 16.162
Two slender rods, each of length l and mass m, are released from 
rest in the position shown. Knowing that a small knob at end B of 
rod AB bears on rod CD, determine immediately after release (a) the 
acceleration of end C of rod CD, (b) the force exerted on the knob.
SOLUTION
Rod AB: ( )w = 0
+
S SM M mg l Bl I ma lA A AB=
Ê
ËÁ
ˆ
¯̃
- = + ÊËÁ
ˆ
¯̃
( ) :eff 2 2
a
1
2
1
12 2 2
2mgl Bl mL m
l l
AB AB- = +
Ê
ËÁ
ˆ
¯̃
a a
1
2
1
3
2mgl Bl ml AB- = a (1)
Rod CD: ( )w = 0
+ S SM M mg l B l I ma lD D CD CD=
Ê
ËÁ
ˆ
¯̃
+ ÊËÁ
ˆ
¯̃
= +( ) :eff 2 2 2
a
1
2
1
2
1
12 2 2
2mgl Bl ml m
l l
CD CD+ = +
Ê
ËÁ
ˆ
¯̃
a a
Multiply by 2: mgl Bl ml CD+ =
2
3
2a (2)
Add (1) and (2): 
3
2
1
3
2
3
2mgl ml AB CD= +
Ê
ËÁ
ˆ
¯̃
a a (3)
Multiply by 3: a aAB CD
g
l
+ =2 9
2
 (4)
263
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PROBLEM 16.162 (Continued)
Kinematics: We must have
(aB Ø) (= ¢aB Ø)
l
l
AB CDa a= 2
l AB CDa a=
1
2
 (5)
Substitute for a AB from (5) into (4)
1
2
2
9
2
a aCD CD
g
l
+ =
5
2
9
2
1 8a aCD CD
g
l
g
l
= =; . (6)
(a) Acceleration of C: a l l
g
lC CD
= = ÊËÁ
ˆ
¯̃
a 1 8. aC g= 1 8. Ø �
(b) Force on knob B:
 Substitute for aCD from (6) into (2)
mgl Bl ml
g
l
B mg mg
+ = ÊËÁ
ˆ
¯̃
= -
2
3
1 8
1 2
2 .
.
 (on rod AB): B = 0 2. mg≠ �
264
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PROBLEM 16.163
The motion of a square plate of side 150 mm and mass 2.5 kg is guided 
by pins at corners A and B that slide in slots cut in a vertical wall. 
Immediately after the plate is released from rest in the position shown, 
determine (a) the angular acceleration of the plate, (b) the reaction at 
corner A.
SOLUTION
Kinematics: AG
L L
a AB
L
G A= = = =2
2
2 2
/ ( )a
a
Plane motion = Translation + Rotation
aBØ = aA ¨ + aB A/ 60°
aBØ = aA ¨+ La 60°
a = aA¨+ aG A/ 15°
a = ∞La sin 30 ¨+
La
2
 15 0 5∞ = . La ¨+ 0 707. La 15∞
Law of cosines
a a a a a
a L L
L
A G A A G A
2 2 2
2 2 2
2 15
0 5 0 707
2 0 5
= ◊ - ∞
= +
-
/ / cos
( . ) ( . )
( . )(
a a
a 00 707 15
0 25 0 5 0 68301
0 06699
2 2 2
2 2 2
. )cos
( . . . )
( . )
L
a L
a L
a
a
a
a
∞
= + -
=
= 00 25882. La
265
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PROBLEM 16.163 (Continued)
Law of sines
a a a
a
L
L
G A G A
sin sin
; sin sin
.
.
sin
/ /
15
15
0 707
0 25950
15
∞
= = ∞ = ∞
b
b a
a
sin . ;b b= = ∞0 707 135
a = 0 2583. La 45°
Kinetics( )w = 0
We find the location of Point E where lines of action of A and B intersect.
MG
L
LEAG
=
= ∞
2
15
( ) . . .
( ) ( . ) .
EG L L L
EG L L
y = - =
= =
0 6829 0 5 0 183
0 183 2 0 2588
I mL= 1
6
2
266
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PROBLEM 16.163 (Continued)
(a) Angular acceleration.
+
S SM M mg L I ma LA A= = +( ) : ( . ) ( )( . )eff 0 183 0 2588a
0 183
1
6
0 2588 0 2588
0 183
1
6
0 06698
2
2
. ( . )( . )
. .
mgL mL m L L
gL L
= +
= +Ê
a a
a
ËËÁ
ˆ
¯̃
= =0 183 0 2336 0 7834. . ; .g
L
g
L
a a
a = 0 7834 9 81
6 15
2
.
.
.
 m/s
 m
 a = 51 2. rad/s2 �
(b) Reaction at corner A.
+S SF F A mg may y= - = - ∞( ) : sineff 45
= - ∞
= - ÊËÁ
ˆ
¯̃
∞
- =
m L
m L
g
L
A mg
( . )sin
( . ) . sin
.
0 2588 45
0 2588 0 7834 45
0
a
11434
0 8566
0 8566 2 5 9 81
21 01
mg
A mg=
=
=
.
. ( . )( . )
.
kg m/s
N
2
 A = 21 0. N ≠ �
267
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PROBLEM 16.164
Solve Problem 16.163, assuming that the plate is fitted with a single pin 
at corner A.
Problem 16.163 The motion of a square plate of side 150 mm and mass 
2.5 kg is guided by pins at corners A and B that slide in slots cut in 
a vertical wall. Immediately after the plate is released from rest in the 
position shown, determine (a) the angular acceleration of the plate, 
(b) the reaction of corner A.
SOLUTION
Since both A and mg are vertical, ax = 0 and a is Ø
Kinematics
AG
L=
2
 15∞ =aG A AG/ ( )a 15∞
a Ø = aA ¨ +aG A/ 15∞
a = 0 183. La 15∞
Kinetics I mL= 1
6
2
268
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PROBLEM 16.164 (Continued)
(a) Angular acceleration.
+
S SM M mg AG I ma AGA A= ∞ = + ∞( ) : ( )sin ( )sineff 15 15a
mg
L
mL m L
L
g
L
2
15
1
6
0 183
2
15
0 183
1
6
2Ê
ËÁ
ˆ
¯̃
∞ = + Ê
ËÁ
ˆ
¯̃
∞
=
sin ( . ) sin
.
a a
++ÊËÁ
ˆ
¯̃
=
=
=
0 033494
0 183 0 2002
0 9143
0 9143
9 81
0
.
. .
.
.
.
.
g
L
g
L
a
a
m/s2
115 m
 a = 59 8. rad/s2 �
(b) Reaction at corner A.
+S SF F A mg may y= - = -( ) :eff
A mg m L
m L
g
L
A mg mg
A
- = -
= - ÊËÁ
ˆ
¯̃
- = -
=
( . )
( . ) .
.
0 183
0 183 0 9143
0 1673
0
a
..
. ( . )( . )
8326
0 8326 2 5 9 81
mg
A = kg m/s2 A = 20 4. N ≠ �