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CHAPTER 16CHAPTER 16

3

PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

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

20212 84

0 15 51 3

mm mm

mm) 20 mmcos

.

( . 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° �

4


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∞ �

+

+

5


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

CB 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



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 ¨ �

7


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 WW

ga( )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


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

65022 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/s2 Æ �

(b) Tension in the cord.

+ S SF F T ma mg

Tmg

x x= = =

= =∞

( ) : sin tan

cos

( )( . )

cos

eff

22.62

q q

q4 9 81

T = 42 5. N 67.4∞ �

9


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

0 849 81

.

.( . ) m/s2 a = Æ3 74 2. m/s �

10



(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

0 48 1 16

0 48

1 169 81 2

. .

. .

.

.( . )

mg ma ma

g a

a

- ==

= m/s a = Æ4 06. m/s2 �

11


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

ga W

W

ga= 5

12

1

3+

S SF F F ma

NW

ga

WW

ga

W

ga

a

x x B

k B

= =

=

+ÊËÁ

ˆ¯̃

=

=

( ) :

.

.

eff

m

0 7285

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 x202 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 WW

ga

B B A

A

= - =

= -

( ) : ( . ) ( . ) ( . )eff m m m1 5 3 6 1 2

7

12

1

3

+

12



+ S SF F F ma

NW

ga

WW

ga

W

ga

a

x x A

k A

= =

=

-ÊËÁ

ˆ¯̃

=

=

( ) :

.

.

eff

m

0 7287

12

1

3

0 728 7712

29 81

1 0 24267

( )+

( . )

.

m/s a = 3 35244 2. m/s ¨


v v ax x202 22 0 10 2 3 35244= + = -( ) ( . m/s m/s )2 x = 14 91. m �

13


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

ax 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


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 max 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



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


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 90002500

3

1

5

9 81

5

. )

( )

.

- =

= =

ga

a g a = ≠1 962. m/s2 �

(b) Reaction at A:

+S SF F A

Ag

y y= - - = ÊËÁ

ˆ¯̃

- = ÊËÁ

( ) :eff g gg

g

2 1250 1125 12505

2 2375 12505

ˆ̂¯̃

=2 2625A g = 25751.25 N

For one wheel: A = ≠12880 N �

17


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

ga

6

67== =6 9.81

67

¥0 87851. m/s2 a = Æ0 879 2. m/s


v v ax x202 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

==

= = = =mreq6

670 09.

mreq = 0 09 0 30. . .< The crate does not slide. �

18


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

msF

Nma

m g a

g a a

=

= ∞+ ∞

+ ∞ = ∞

0 2530

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 Fh

Nd

G G= ÊËÁ

ˆ¯̃

- ÊËÁ

ˆ¯̃

=( ) :eff 2 20

F

N

d

h=

m = =F

N

d

h; 0 25. ;

h

d= 4 00. �

19


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

msF

Nma

m g a

g a a

=

= ∞- ∞

- ∞ = ∞

0 2530

30

30 4 30

.cos

( sin )

sin cos

a

g=

∞ + ∞=1

4 30 300 25226

cos sin. a = 0 252. g �

(b) Tipping impends:

+S SM M F

hW

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


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

Te

N

G = - - =

= - = -0 100 450 0 35 0

450 0 35

100

250 157 5

1

: ( )( . )

( )( . ) .

0000 925W WB B= .

+ SFW

ga T N

W

ga

a

g

T

W

N

W

xB B

B B

= - =

= - = - =

: .

. . . .

0 35

0 35 0 925 0 35 0 575

21



For the cylinder: +SFW

ga W T W

a

gC

C C= - =:

WT

ag

WC

B=-

=-

=1

0 925

1 0 5752 1765 100 9 81

.

.( . )( )( . ) WC = 2140 N �

22


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

TA

A

eff N

N kg

: . cos

. ( )( . )cos ∞∞ =+ - =

0

11 43 42 48 0TA . .N N

TA = 31 04. N TAD = 31 0. N �

23


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

qq

+S SM M mg ma ma

mgB 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

costan .

q q

q q qqq

q

=

=

= q = ∞51 3. �

24


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 the plate, (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


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 maeff2m/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 00400

3∞ÊËÁ

ˆ¯̃

mm

306 418 23544 128 558 102 139

306 418 23544 75

. ( . . )

. ( .

F ma

FCF

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


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 = =w2 20 3 6( . )( )m rad/s

an = 10 80. m/s2 60∞

a rt = =a ( . )( )0 3 12m rad/s2

at = 3 6. m/s2 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



(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


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.

ww

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/s2 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


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

av

rn = = =2 20 75

0 153 75

( . )

( ..

m/s

m)m/s2

an = 3 75. m/s2 30°

at ta= 60°

Weldment is in translation an = 3 75. m/s2

60° S SF F mg mat= ∞ =eff : cos30

at = 8 4957. m/s2 60∞

(a) Acceleration

b = = = ∞

= +

= +

- -tan tan.

..

( . ) ( .

1 1

2 2 2

2

3 75

8 495723 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



(b) Reactions

ma = =( . ) .8 9 2865 74 292kg) ( m/s N2


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


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

0 498 1

N

m)N/m

Horizontal component of effective forces

W

gax = ∞

=

98 1

9 816 3057 40

48 3045

22.

.( . )cos

.

N/m

m/sm/s

N/m

S SM M MB B B= =( ) : ( . )( . )

eff N/mm

48 30450 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


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.

ww

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/s2 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/s2 a y = 61 544. m/s2 �

33


PROBLEM 16.22* (Continued)

Distributed weight per unit length = = =W( . )( .

( ..

7 5 9 81

0 759

N)

m)8 1 N/m

Distributed mass per unit length = = =W

g

98 1

9 8110

.

.kg/m

Vertical component of effective forces = =W

gay ( ( . )10 61 544kg/m) m/s2

= 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 = ÊËÁ

ˆ¯̃

◊

1

2194 003

0 75

2( . )

.N

m

N m= 36.376

| | .maxM = ◊36 4 N m �

VB = -194 0. N �

34


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


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 w2

Effective forces are: ( ) ( )( ) ( ) ( )( )D = D ¥ - D ¢ D ¥m a m m r mi i i i i i i i� r rw a2

S S( ) ( )( ) ( )D D S Dm m r m ri i i i i ia = ¥ ¢ - ¢a w2

= ¥ 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 r12m 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 w2 2 0

and, ¢ ¥ D ¥ ¢ = D ¢ri i i im r m r( )( ) ( )a a12

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 Ii2

the moment of the couple is I a

36


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 3002

6010= Ê

ËÁˆ¯̃

= 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


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)

rpm2

60

r

2

aad/s

rad/s rad/s

w w aq

p q

q

202

2 2

2

0 120 2 2 1605

32 891 10

= +

= + -

= ¥

( ) ( . )

. 33

5235 26

rad

rev= .

or q = 5230 rev �

38


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

1q

m=

-mg N

Nk

N mg N

Nmg

k

F Nmg

k

k

kk

tan

tan

tan

q m

q m

mm

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



+ S SM M Fr IA A= =( )eff : a

a

mq m

mq m

=

= ◊+

= ◊+

r

IF

r

mr

mg

g

r

k

k

k

k

122

2 tan

tan

Data: r

k

== ∞=

0 18

60

0 40

.

.

m

qm

a = ◊∞ +

2 9 81

0 18

0 40

60 0 40

2( . )

.

.

tan .

m/s

m a = 20 4 2. rad/s �

40


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

1q

m=

+mg N

Nk

N mg N Nmg

kk

tan ;tan

q mq m

= + =-

+ S SM M Fr IA A= =( )eff : a

am m

q mmq m

= = = ◊-

= ◊-

Fr

r

Nr

I

r

mr

mg g

rk k

k

k

k12

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


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 2402

608= Ê

ËÁˆ¯̃

=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 TI

rB A- = a (2)

42



Belt friction: b p

m b p

= ∞ =

= = = =

2703

2

3 2480 25 1 17832

rad

T

Te 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 TI

rB A- =

- =◊

=

a

a

a

305 9 94 1675

0 15

0 423

2

2

. ..

.

N Nkg m

m

rad/s

Uniformly accelerated motion.

w a p028 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


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

3602

60

12

0

0

mm m

rpm

rad/s

.

w p

w p+

S SM M Fr ID D= =( )eff : a

( . )( . ) ( )150 538 0 2 20 2 N m kg m= ◊ a

a = 1 50538 2. rad/s (deceleration)

w w aq p q202 2 22 0 12 2 1 50538= + = + -: ( ) ( . )rad/s rad/s

q

qp

=

= ÊËÁ

ˆ¯̃

=

472 048

472 0481

275 129

.

. .rad rev q = 75 1. rev �

44


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

rpm2

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 q202 2 22 0 12 2 1 30841= + = + -: ( ) ( . ) rad/s rad/s

q

q

=

= =

543 1107

543 1107

286 439

.

..

rad

rev revp

q = 86 4. rev �

45


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 ka

rm ar

am g

m m

A F A

A

A Fkr

= ÊËÁ

ˆ¯̃

+

=+ ( )

2

2

(a) Acceleration of A a =+ ( )

=( )( . )

( ).

15 9 81

15 1201 7836

375 2

kg m/s

kg kg m

2

mm500 mm

//s2

or aA = 1 784. m/s2 Ø �

(b) Velocity of A v v a sA A A2

02 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


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 Ia

rm arA f A- = + (1)

Case 1: y

t

==

3

4 6

m

s.

y at

a

a

mA

=

=

==

1

2

31

24 6

0 2836

12

2

2 m s

m/s

kg

2

( . )

.

Substitute into Eq. (1)

( )( . )( . ).

.(12 9 81 0 6

0 2836

0 612 kg m/s m

m/s

m k2

2

- =Ê

ËÁˆ

¯̃+M If gg m/s m)( . )( . )0 2836 0 62

20 632 0 4727 2 0419. ( . ) .- = +M If (2)

47



Case 2: y

t

y at

a

a

mA

==

=

=

==

3

3 1

1

2

31

23 1

0 6243

24

2

2

m

s

m s

m/s

kg

2

.

( . )

.

Substitute into Eq. (1):

( )( . )( . ).

(24 9 81 0 60 6243

24 kg m/s m m/s

0.6 m k2

2

- =Ê

ËÁˆ

¯̃+M If gg m/s m2)( . )( . )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


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= = ◊( ) : ( )( . (eff2 N m) kg m )a

a = 10 2 rad/s �

(b) q = =(

. )

3

0 25012

m)

( mrad

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 20800

9 810 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 25012

m

m rad

.

w aq2 2

2 7 9691 12

=

= ( . )( ) rad/s rad2 w = 13 83. rad/s �

49



Case 3: (a) + S SM M0 0= ( ) :eff

2300 0 25 1500 0 25 202300

9 810 25

1500

9 810 252( . ) ( . )

.( . )

.( . )- = + +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 2512

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 20400

9 810 5= = +( ) : ( . )

.( . )eff a

200 20400

9 810 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 2512

m

m rad

.

w aq2 2 2 6 6239 12= = ( . )( ) rad/s rad2 w = 12 61. rad/s �

50


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

.

.

.

.

aaa

a aa 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



(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 aa a

a

A A

A A

A

aa A = 29 6296 2. rad/s a A = 29 6. rad/s2 �

(b) Tangential force F.

Substituting for a A into (2): F = 1 125 29 6296. ( . ) FA = 33 3. N �

52


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

CA

CA A A

= =

= = =

a a

a a a a0 2

0 082 5

.

..

At the contact point between gears B and C,

a r r

r

r

t B B C C

BC

AC A A

= =

= = ◊ =

a a

a a a a0 08

0 22 5

.

.( . )

Kinetics:

Gear B: +

S SM MB B= ( ) :eff F r I IBC B B B B A= =a a

FI

rBCB

BA= a

Gear C: + S SM M F r F r IC C BC C AC C C C= + =( ) :eff a

r

rI F r I

Fr

I I

C

BB A AC C C A

ACC

B C A

a a

a

+ =

= +

2 5

10 4 2 5

.

( . . )

Gear A: + M M M r F IA A A A AC A A= - =S( ) :eff a

Mr

rI I I

M I I I

AA

CB C A A A

A A B C A

- + =

= + + ( )( )( . . )

.

0 4 2 5

2 5 2

a a

a

53



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

0 080 4 0 225 2 5 0 009 11 8519

.( . )( . ) ( . )( . ) ( . )

FAC = 16 67. N �

54


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

21

210 0 2

1

26 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 aa

14 715 13 2435 0 2675 0 3 0 2025

1 4715 0 77

1 91104

. . ( . . . )

. .

.

- = + +=

=

aa

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


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

210 0 2 0 2

1

2

1

26 0 15 0

2 2 2

2 2

( )( . ) .

( )( . )

kg m

..0675 2 kg m◊

Kinematics: a

a

AA

BB

a

r

aa

a

r

a a

= = =

= = =

0 25

0 15

20

3

.

.

m

m

Kinetics:

Disk A: FAB = Tension in cord between disks


F I m a

FAB 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



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

.. ) ( . )

. . . .

067520

39 0 15

13 2435 0 15 0 45 1 35

aa

F a aAB

ÊËÁ

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


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

21

26 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= ◊=

aa

A

A a A = 12 50. rad/s2 �

Disk B:

I m rB B B=

=

= ×

1

21

20 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

aB

B aB = 33 3. rad/s2 �

( )w p pA 0 3602

6012= Ê

ËÁˆ¯̃

= rpm rad/s

58



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

p0

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

pA A At= -

= -=

( )

( . )( . )

.

0

12 12 5 1 00531

25 132

rad/s rad/s s

rad

2

//s

wpA = Ê

ËÁˆ¯̃

=

( . )

.

25 13260

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

wpB = Ê

ËÁˆ¯̃

=

( . )

.

33 51060

2

320 00

rad/s

rpm

or w B = 320 rpm �

59


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

21

26 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= ◊=

aa

A

A a A = 12 50. rad/s2 �

Disk B: I m rB B B=

=

= ◊

1

21

20 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

aB

B aB = 33 3. rad/s2 �

60



Sliding starts when V VC C= ¢ . That is when

w wa 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/s2

+ wpA = Ê

ËÁˆ¯̃

=( . ) .9 424860

290 00 rad/s rpm

or w A = 90 0. rpm �

+ w w a

pB B Bt= -

= -

( )

( . )( . )

0

12 33 333 0 75398 rad/s rad/s s2

= 12 566. rad/s

wpB = Ê

ËÁˆ¯̃

=

( . )

.

12 56660

2

120 00

rad/s

rpm

or w B = 120 0. rpm �

61


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

AA

( . ) ( . )( . )0 11

22 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

BA

BA

( . ) ( )( . )0 21

210 0 2

2

2

2=

=

a

a (3)

Belt + SF P F FA A B= - - =0 0: (4)

188 2

0- - =a aA A

a A = ÊËÁ

ˆ¯̃

8

518( )

a A = 28 8. rad/s2

62



Check that belt does not slip.

From (2): FA = =1

828 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 AπÊËÁ

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 21

210 0 2 2a

aB = 9 81. aB = 9 81. rad/s2 �

Belt Eq. (4): P F F

F P FA 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 8 198

. N =a A

a A = =( )( . ) .8 8 19 65 52 a A = 65 5. rad/s2 �

63


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

AA

( . ) ( . )( . )0 11

22 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

BA

BA

( . ) ( )( . )0 21

210 0 2

2

2

2=

=

a

a (3)

Belt + SF P F FA A B= - - =0 0:

108 2

0- - =a aA A

a A = ÊËÁ

ˆ¯̃

( )108

5

a A = 16 rad/s2

64



From (2): FA = =16

82 N

From (4): FB = =16

28 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/s2 �

From (1): aB = 8 00. rad/s2 �

65


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

Pk

k

====

mm0 25 12 5

3 125

. ( . )

.

N

N

Disk A:

+ S SM M Fr I m rA A A A A A A= = =( ) :eff a a1

22

F m rA A

A

=

=

1

2

3 1251

23

a

a. (N kg)(0.075 m)

a A = 250

9 rad/s2 a A = 27 8. rad/s2 �

Disk B:

+ S SM M Fr I m rB B B B B B B B= = =( ) :eff a a1

22

F m rB B B

B

=

=

1

2

3 1251

27 5

a

a. ( .N kg)(0.125 m)

aB = 20

3rad/s2 aB = 6 67. rad/s2 �

(b) Final angular velocities. ( )w pA 0 375

2

60

25

2= Ê

ËÁˆ¯̃

=rpm rad/sp

( )wB 0 0=

66



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

975

20

3125

- =

-ÊËÁ

ˆ¯̃

= ÊËÁ

ˆ¯̃

p 0mm t mmm

s

)

.

( )

( . )

t

tA A A

== -

= -

1 0098

25

2

25

91 0098

0w w ap 0

wpA = Ê

ËÁˆ¯̃

=11 2260

2107 1422. .rad/s rpm w A = 107 1. rpm �

Eq. (1) ( . )( ) ( )107 1422 75 125rpm mm mm= wB w B = 64 3. rpm �

67


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

Pk

k

====

mm0 25 12 5

3 125

. ( . )

.

N

N

Disk A:

+ S SM M Fr I m rA A A A A A A= = =( ) :eff a a1

22

F m rA A

A

=

=

1

2

3 1251

23

a

a. ( N kg)(0.075 m)

a A = 250

9rad/s2 a A = 27 8. rad/s2 �

Disk B:

+ S SM M Fr I m rB B B B B B B B= = =( ) :eff a a1

22

F m rB B B

B

=

=

1

2

3 1251

27 5 0 125

a

a. ( . )( . N kg m)

aB = 20

3rad/s2 aB = 6 67. rad/s2 �

(b) Final angular velocities

( )

( )

w

w pA

B

0

0

0

375

=

= ÊËÁ

ˆ¯̃

=rpm2

60

25

2rad/s

p

68



Eq. (1): w wa w a

p

A A B B

A A B B B

r r

tr t r

== -

ÊËÁ

ˆ¯̃

= -ÊËÁ

[( ) ]

( )

0

250

975t tmm

25

2

20

3ˆ̂¯̃

=

( )

. sec

125

1 683

mm

t

w a

p

A At=

= ÊËÁ

ˆ¯̃

=

= ÊËÁ

ˆ¯̃

250

9rad/s

rad/s

( . ) .

.

1 683 46 75

46 7560

2 w A = 446 rpm �

Eq. (1): ( . )( ) ( )446 428 75 125rpm mm mm= wB w B = 268 rpm �

69


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 depends only 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

am

k A A A A

Ak

A A

Pr m r

P

m r

=

=

1

22

2

(1)

Disk B: + = =SM M Fr IB B B B B( ) :eff a

m ak B B B BPr m r= 1

22

am

Bk

B B

P

m r=

2 (2)

At any time t

+ w am

A Ak

A A

tP

m r= + =0

2 (3)

+ w w a w

mB B

k

B B

tP

m rt= - = -0 0

2 (4)

Slipping ends when w ww 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



Substitute from Eqs. (1) + (2): 2 2

0P

m rr

P

m rr t rk

A AA

k

B BB B

m mw+

È

ÎÍ

˘

˚˙ =

21 1

2

1

2

0

01 1

0

Pm m

t r

tr

P

tr

P

m m

kA B

B

B

k m m

B

k

A B

A B

m w

wm

wm

+ÊËÁ

ˆ¯̃

=

= ◊+

= ◊&

mm mA B+

Eq. (3): wm w

m

w w

A Ak

A A

B

k

A B

A B

AB

A

B

A B

a tP

m r

r

P

m m

m m

r

r

m

m m

= = ◊ ◊+

È

ÎÍ

˘

˚˙

= ◊+

2

20

0

(wA is independent of mk,QED)

Eq. (4): w w a wm w

m

w w

B Bk

B B

B

k

A B

A B

BA

A

tP

m r

r

P

m m

m m

m

m

= - = - ◊ ◊+

È

ÎÍ

˘

˚˙

= -+

0 00

0

2

2

1mm

m m m

m m

m

m m

B

BA B A

A B

B

A B

B mm

A

B

ÏÌÓ

¸˝˛

=+ -

+=

+

=+

w w w

ww

0 0

0

1

wB depends only upon w0 and m

mA

B

(QED)

71


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

dI

ma

mk

ma= =a a2

dk

a=

2a �

72


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 iw2 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 w2

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 rw2

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 w2

¢¢ ¥ = ¢ + ¢ ¥ ¥ ¢ - ¢ ¥ ¢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 w2

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



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 i2 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


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

mW

g

IW

gL

=

= 1

122

+ S SF F P maW

gax x= = =( ) :eff

aP

Wg g

g= = =1

9 9

N

N a = 1

9g Æ

+ S SM M PL

IW

gLG G= = =( ) :eff 2

1

122a a

a = =6 61

9

P

W

g

L

g

L

N

N a = 2

3

g

L

(a) Acceleration of Point A.

+ a aAL g L g

Lg= + = + ◊ = Ê

ËÁˆ¯̃

=2 9 2

2

3

4

94 36a . )m/s2 aA = 4 36. m/s2 Æ �

(b) Acceleration of Point B.

+ a aBL g L g

L

g= - = - ◊ = - = -2 9 2

2

3

2

9

2

99 81a ( . )m/s2 aB = 2 18. m/s2 ¨ �

75


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 maW

gax X= = =( ) :eff

a = ÆP

Wg

+ S SM M Ph IW

gLG G= = =( ) :eff a a1

122

a = 122

Ph

WLg

(a) Position of force P. + a aBL= -2

a

02

12

6

1

6

2= - ◊

= =

P

Wg

L Ph

WLg

hL

m

Thus, P is located 1

3m from end A. �

For hL P

WLg

P

W

g

L

L

= =( )

= ◊6

1226

2: a

(b) Acceleration of Point A.

+ a aL P

Wg

L P

W

g

L

P

WgA = + = + ◊ =

2 22 2a

aA = ÊËÁ

ˆ¯̃

21

99 81( . ) aA = 2 18. m/s2 Æ �

76


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 a2

a = P

mr


+ a a rP

mr

P

mr

P

mA = + = + ÊËÁ

ˆ¯̃

=a 2

aA = =23

2 42 5

N

kgm/s2

.. aA = 2 50. m/s2 Æ �


+ a a rP

mr

P

mrB = - = - ÊËÁ

ˆ¯̃

=a 0 aB = 0 �

77


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

22

+ S SF F P maX X= =( ) :eff

a = P

mÆ

+ S SM MG G= =( ) :eff Pr Ia

Pr = 1

22mr a

a = 2P

mr


+ a a rP

mr

P

mr

P

mA = + = + ◊ =a 23

aA = =33

2 43 75

N

kgm/s2

.. aA = 3 75. m/s2 Æ �


+ a a rP

mr

P

mr

P

mB = - = - ◊ = -a 2

aB = - = -3

2 41 25

N

kgm/s2

.. aB = 1 25 2. /m s ̈ �

78


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

22

+ S SF F P max x= =( )eff :

a = P

mÆ

+ S SM M IG G= =( ) :eff Pr a

Pr = 1

22mr a

a = 2P

mr

Let t1 be time required for 360∞ rotation.

q a

p

p

=

= ÊËÁ

ˆ¯̃

=

1

2

21

2

2

2

12

12

12

t

P

mrt

tmr

P

rad

Let x1 = distance G moves during 360∞ rotation.

x atP

m

mr

P1 121

2

1

2

2= = ÊËÁ

ˆ¯̃

p

x r1 = p Q E.D. . �

79


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 22 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= - = = -( ) . .eff2: 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


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

125 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/s2 �

81



(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/s2 2 �

82


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

329 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

314 18 29 43 0 8567( . ) . m/s2

Eq. (2): + a = - =5 442 18 14 21 769. ( ) . rad/s2

83



+aA A ta

a r

== += -

( )

. ( . )( . )

a0 8567 0 08 21 769

= -0 885. m/s2 aA = 0 885. m/s2 Ø �

+aB B ta

a r

== += +

¢( )

. ( . )( . )

a0 8567 0 08 21 769

= +2 60. m/s2 aB = 2 60. m/s2 ≠ �

84


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

329 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

314 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



+ aA A ta

a r

== += - +

( )

. ( . )( . )

a1 1433 0 08 10 884

= -0 273. m/s2 aA = 0 273. m/s2Ø �

+ aB B ta

a r

== += - -

¢( )

. ( . )( . )

a1 1433 0 08 10 884

= -2 01. m/s2 aB = 2 01. m/s2 Ø �

86


SOLUTION

Kinematics: a aB A= + (5 m)a

1≠ = 6≠ + 5a

a = 1 rad/s2

a a aA B= + = +1

2

1

21 6( ) ( )

a = 3 5. m/s2≠

Kinetics:

I mL=

=

= ◊

1

121

12

2500

9 815

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 25002500

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

TA

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



+ S SF F T T W maA B= + - =eff :

1802 16 25009 81

3 5

1589 79

..

( . )

.

N N2500

N

+ - =

=

T

T

B

B TB = 1590 N �

88


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

9 815 530 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 25002500

9 812 875 1 5

530 921

m) N)(1.5 m) )

(1.

- =

+ 225

N

)

. .

.

4 3750 1099 01 663 65

1378 165

T

TA

A

- = +fi = TA = 1378 N �

+ S SF F T T W maA B= + - =eff :

1378.1652500

9.81+ - =TB 2500 2 875( . ) TB = 1855 N �

89


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

20 100 0 05

2 2

kg

kg m2( )( . )

( . ) . 00

3100

3300

m

N

N

T

TA

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

IB A

a = 0 741. rad/s2 �

(b) Acceleration of mass center.

+ S SF F T T mg may y A B= + - =( )eff : 2 2

aT T

mgA B=

+-

= + -

2

2 3100 3300

12009 81

( )

( ). a = 0 857. m/s2 ≠ �

90


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 of the 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

20 100 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

Tmg ma

T

m g aT

B A

A

= - -

= - -

= - -

=

2

21200 9 81 0 150

23000

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

IB A

rad/s2 a = 15 11. rad/s2 �

91


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

kB2

a =r

kB2

�

92


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 0816

10000 1

30

1000

2ÊËÁ

ˆ¯̃

= ÊËÁ

ˆ¯̃

a

a = 1081

15rad/s2 a = 72 1. rad/s2 �

93


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 21

2

1

2= = =

(a) Angular acceleration:

+ S SM M TL

G G= ÊËÁ

ˆ¯̃

=( ) :eff 2Ia

1

2 2

1

122mg

LmLÊ

ËÁˆ¯̃

= 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

2a Ø

(b) Acceleration of A:

a a aA G A G= + /

+ aA gL= -1

2 2a

= - ÊËÁ

ˆ¯̃

1

2 2

39g

L

L

a gA = - aA g= ≠ �

94



(c) Acceleration of B:a a aB G B G= + /

+ a aL

gL

LgB = + = + Ê

ËÁˆ¯̃

= +2

1

2 2

392a aB g= 2 Ø �

95


PROBLEM 16.64


SOLUTION


2

1

2= = =


+ S SM M TL

BIG G= Ê

ËÁˆ¯̃

=( ) :eff a

1

2

1

122mg

L

BmLÊ

ËÁˆ¯̃

= 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

2a Ø


a a aA G A G= + /

+ aA gL= -1

2 2a

= - ÊËÁ

ˆ¯̃

=1

2 20g

L g

L aA = 0�

96



(c) Acceleration of B:a a aB G B G= + /

+ aB aL

gL g

Lg= + = + Ê

ËÁˆ¯̃

=2

1

2 24a aB g= Ø �

97


PROBLEM 16.65


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∞


+ S SM M TL

IG G= ∞ =( ) : ( sin )eff 1 306

a

( sin )mgL

mL306

1

122∞ = 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



Accelerations of A and B

Translation + Rotation about G


a a aA G A G g= + =/ [ .0 866 ¨ + 0 5. g Ø] + ÊËÁ

ˆ¯̃

ÊËÁ

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] + ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

g

L

L

2Ø

= 0 866. g ¨ + g Ø aB g= 1 323. 49.1∞ �

99


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 mb

mb= ÊËÁ

ˆ¯̃

=1

2 2

1

8

22


2

1

2= = =

Kinetics: + S SM MG G= ( ) :eff Tb

I1 2= a

1

2 2

1

82mg

bmbÊ

ËÁˆ¯̃

= a

a = 2g

b

+ S SF F W T may y= - =( )eff : 1

mg mg ma g- = =1

2

1

2a Ø

Kinematics:

Plane motion = Translation + Rotation

(a) a a a aA G A G= + =/ Ø+ b

2a ≠= 1

2g Ø+ Ê

ËÁˆ¯̃

b g

b22 ≠ aA g= 1

2≠ �

(b) a a a aB G B G= + =/ Ø+ b

2a Ø= 1

2g Ø+ Ê

ËÁˆ¯̃

b g

b22 Ø aB g= 3

2Ø �

100


PROBLEM 16.67


A thin hoop of diameter b.

SOLUTION

I mb

mb= ÊËÁ

ˆ¯̃

=2

1

4

22


2

1

2= = =

Kinetics: + S SM M Tb

IG G= ÊËÁ

ˆ¯̃

=( ) :eff 1 2a

1

2 2

1

42mg

bmbÊ

ËÁˆ¯̃

=

a = g

b

+ S SF F W T may y= - =( ) :eff

mg mg ma g- = =1

2

1

2a Ø

Kinematics:


(a) a a a aA G A G= + =/ Ø+ b

2a ≠= 1

2g Ø+ Ê

ËÁˆ¯̃

b g

b2≠ aA = 0 �

(b) a a a aB G B G= + =/ Ø+ b

2a Ø= 1

2g Ø+ Ê

ËÁˆ¯̃

b g

b2Ø aB g= Ø �

101


PROBLEM 16.68


A square plate of side b.

SOLUTION

I m b b= +1

122 2( )

I mb= 1

62


2

1

2= = =

Kinetics: + S SM MG G= ( ) :eff Tb

I2

= a

1

2 2

1

62mg

bmbÊ

ËÁˆ¯̃

= a

a = 3

2

g

b

+

S SF F W T may y= - =( ) :eff 1

mg mg ma g- = =1

2

1

2a Ø

Kinematics:


102



(a) a a a aA G A G= + =/ Ø + b

2a ≠

aAg=2

Ø + 3ÊËÁ

ˆ¯̃

b g

b2 2 ≠ = g

4≠ aA g= 1

4≠ �

(b) a a a aB G B G= + =/ Ø + b

2a Ø

aB g= 1

2 Ø + Ê

ËÁˆ¯̃

b g

b2

3

2 Ø = 5

4g Ø aB g= 5

4Ø �

103


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

52

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 wm

= - + = - +0 05

2t

g

rtk

When t = t1:

Eq. ( ) : :1 v r= w v gtg

rt rk

k0 1 0 1

5

2- = - +Ê

ËÁˆ¯̃

m wm

v gt r gtk k0 1 0 15

2- = - +m w m

tv r

gk1

0 02

7=

+( )wm

(3)

v r0 5 9 100 0 1= = = =m/s, rad/s, mm m0w .

(a) t12 5 0 1 9

7 0 1 9 811 71836= + =( . ( ))

( . )( . ). s t1 1 718= . s �

104



(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/s2 2

+ s v t at1 0 1 12

2

1

2

5 1 718361

20 981 1 71836

= -

= -( )( . ) ( . )( . )

= - =8 5918 1 44833 7 14347. . . m s1 7 14= . m �

105


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

55

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

2t

g

rtk

When t = t1:

Eq. (1) n = rw: v gtg

rt r

v gt r gt

tv

kk

k k

0 1 0 1

0 1 0 1

10

5

2

5

22

7

- = - +ÊËÁ

ˆ¯̃

- = - +

=+

m wm

m w m

( rr

g

v rk

wm

w

0

0 05 18 0 1

)

.= = =m/s, rad/s, m

(3)

106



(a) Eq. (3): t12

7

5 0 1 18

0 1 9 811 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 12

m/s m/s

m/s

2 2

4491

20 981 1 98049

9 90245 1 92391

7 9785

2s m/s s

m

2) ( . )( . )

. .

.

-

= -= s1 7 98= . m �

+

107


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 ma

m

a

))r mk

gr

kk

=

=

2

2

ama

Kinematics: + v v at

v v gtk

= -= -

0

0 m

For n = 0 when t = t1 0 0 1 10= - =v gt t

v

gkk

mm

; (1)

+ w w a

w wm

= -

= -

0

0 2

t

gr

ktk

For w = 0 when t t= 1 0 0 2 1 1

2

0= - =wm

mwk

k

gr

kt t

k

gr; (2)

Set Eq. (1) Eq. (2)= v

g

k

gr

r

kv

k k

02

0 0 2 0m mw w= =; (3)

Distance traveled: s v t at1 0 1 121

2= -

s vv

gg

v

gs

v

gkk

k k1 0

0 02

1021

2 2=

ÊËÁ

ˆ¯̃

-ÊËÁ

ˆ¯̃

=m

mm m

( ) ; (4)

+

+

108



For a solid sphere k r2 22

5=

(a) Eq. (3): w0 25

2 005

2= =r

rv

v

r w0

05

2=

v

r �

(b) Eq. (1) tv

gk1

0=m

�

(c) Eq. (4) sv

gk1

02

2=

m �

109


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 10= - =v gt t

v

gkk

mm

; (1)

+ w w a

w wm

= -

= -

0

0 2

t

gr

ktk

For w = 0 when t t= 1 0 0 2 1 1

2

0= - =wm

mwk

k

gr

kt t

k

gr; (2)

Set Eq. (1) Eq. (2)= v

g

k

gr

r

kv

k k

02

0 0 2 0m mw w= =; (3)

Distance traveled: s v t at1 0 1 121

2= -

s vv

gg

v

gs

v

gkk

k k1 0

0 02

1021

2 2=

ÊËÁ

ˆ¯̃

-ÊËÁ

ˆ¯̃

=m

mm m

( ) ; (4)

110



For a hoop, k r=

(a) Eq. (3): w0 2 00= =r

rr

v

r w0

0=v

r �

(b) Eq. (1): tv

gk1

0=m

�

(c) Eq. (4): sv

gk1

02

2=

m �

111


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

52

a = 5

2

mk g

r

Kinematics: + v at gtk= = m (1)

+ w am

= =tg

rtk5

2 (2)

C = Point of contact with belt

+ v v r gtg

rt r

v gt

C kk

C k

= + = + ÊËÁ

ˆ¯̃

=

w mm

m

5

2

7

2

(a) When sphere starts rolling ( ),t t= 1 we have

v v

v gt

C

k

=

=

1

1 17

2m t

v

gk1

12

7=

m �

(b) Velocities when t t= 1

Eq. (1): v gv

gk

=ÊËÁ

ˆ¯̃

mm

2

71 v = 2

7 1v Æ �

Eq. (2): wm

m= Ê

ËÁˆ¯̃

ÊËÁ

ˆ¯̃

5

2

2

71k

k

g

r

v

g w = 5

71v

r �

112


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 max x= =( ) :eff

mk mg ma=

a = mk g Æ

+ S SM M F IG G r= =( )eff : a

( )m ak mg r mr= 5

22

a = 5

2

mk g

r

Kinematics: + v v at v gtk= - = -0 0 m (1)

+ w am

= =tg

rtk5

2 (2)

C = Point of contact with belt

+ v v r

v v rg

rt

C

Ck

= - +

= - +

wm5

2

v vg

tCk= - +

5

2

m (3)

But, when t t v= =1 0, and v vc = 1

Eq. (3): vg

tk1 1

5

2=

m t

v

gk1

12

5=

m �

Eq. (1): v v gtk= -0 m

When t t v= =1 0, , 02

501= -

ÊËÁ

ˆ¯̃

v gv

gkk

mm

v v0 12

5= �

113



Distance when t t= 1: + s v t at= -0 1 121

2

s vv

gg

v

g

sv

g

kk

k

k

= ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

-ÊËÁ

ˆ¯̃

=

2

5

2

5

1

2

2

5

4

25

11 1

2

12

mm

m

m

( )

--ÊËÁ

ˆ¯̃

2

25; s = 2

2512v

gkm¨ �

114


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= ( ) GPk

r=

2

( . . .)Q E D �

Note: The center of rotation and the center of percussion are interchangeable. Indeed, since OG r= , we may write

GPk

GOGO

k

GP= =

2 2

or

Thus, if Point P is selected as center of rotation, then Point O is the center of percussion.

115


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

122a

+ S SM M PL maL

IA A= = +( ) : ( )eff 2a

= ÊËÁ

ˆ¯̃

+

=

mL 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 mL

P mL P

mL

P

AP

x

x

= - ÊËÁ

ˆ¯̃

= - ÊËÁ

ˆ¯̃

= -

= - = - = -

2 2

3

2

2

8

24

a

N A x = 4 00. N ¨ �

116


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

21

122

+ S SF Fx x= ( ) :eff P ma=

P mL= Ê

ËÁˆ¯̃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

LI P h L

P hL

mLP

mL

PL

hL

-ÊËÁ

ˆ¯̃

= -

-ÊËÁ

ˆ¯̃

= ÊËÁ

ˆ¯̃

=

-Ê

2

2

1

12

2

6

2

2

a : ( )

ËËÁˆ¯̃

= = + = = =Lh

L LL

6 2 6

2

3

2

31

2

3; ( ) m

(a) Distance h. h = 0 667. m �

117


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 r I mL= =a 1

122

+ S SM M P r

Lma r IC C= +Ê

ËÁˆ¯̃

= +( ) : ( )eff 2a

= +

+ÊËÁ

ˆ¯̃

= +ÊËÁ

ˆ¯̃

( )mr r mL

P rL

m r L

a a

a

1

12

2

1

12

2

2 2

Substitute data: ( ..

( ( .7 50 9

41

120 9 N) 0.225 m

m

2 kg) (0.225 m) m)2 2+È

ÎÍ˘˚̇

= +ÈÎÍÍ

˘˚̇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/s2 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


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

=

=

a1

122

+ S SF F P max x= =( ) :eff

P m r

P

mr

=

=

( )a

a (1)

+ S SM M P

LIG G= =( ) :eff 2a

PL

mL

PL

mLP

mr

2

1

12

2

1

12

2

2

=

= ÊËÁ

ˆ¯̃

a

L L

r

r L r

2 121

6

900

2

=

= = mm

6 r = 150 mm �


Eq. (1): a = = ( ) =P

mr

P

m

P

mLL6

6

a = =6 75

4125

(

(

N)

kg)(0.9 m) rad/s2 a = 125 rad/s2 �

119


PROBLEM 16.80

A uniform 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 rw2 ¨

+ S SF F T m a m rn n= = ¢ = ¢( )eff : w2

= -ÊËÁ

ˆ¯̃

+m

l x

l

l xw2

2 T

m

ll x= -1

22 2 2w ( ) �

120


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¯̃

=

=

12002

60

402

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


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¯̃

=

96002

60

320

rpm

rad/s

Consider before it is thrown off.

+ S SF F R ma mFn= = =eff : w2

= ¥=

-( )( . )( )

.

45 10 0 3 320

13 64

3 2kg m

kN

pR

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


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.

rr

r

r

a rn

=

=

=

==

2

32 20

3

4 0014

24

34

34

2

sin

( )sin ( )

( )

.

aa

pp

ww

mm

mm

cycless

rad/s

rad/s

eff

==

= = =

24 2

150 82

( )

.

:

pw

wS SF F R ma mrn

+

= ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

50

1000

4 0014

1000150 8 2.

( . )

R = 4 5495. N

Force on shaft is R = 4 5495. N

Magnitude: R = 4 55. N �

123


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 aL= =02

a

+ S SM M W

LI ma

LA A= = +( ) :eff 2 2

a

mgL

mL mL L

2

1

12 2 22= + Ê

ËÁˆ¯̃

a a

mgL

mLg

L2

1

3

3

22= =a a

(b) Reaction at A.

+ S SF F A mg ma mL

y y= - = - = -( ) :eff 2a

A mg mL g

L

A mg mg

- = - ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

- = -

2

3

2

3

4

A mg= 1

4 A = 1

4mg≠ �

(a) Acceleration of B.

a a aB n B A L= + = +/ 0 a Ø

aB Lg

Lg= Ê

ËÁˆ¯̃

=3

2

3

2Ø aB g= 3

2Ø �

124


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= =04

aL

+ S SM M W

LI ma

LC C= = +( ) :eff 4 4

a

mgL

mL mL L

mgL

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 mL

y y= - = - = -( ) :eff 4a

C mg mL g

L

C mg mg

- = - ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

- = -

4

12

7

3

7

C mg= 4

7 C = 4

7mg ≠ �

(a) Acceleration of B.

a a aL

aL g

Lg

B C C B

B

= + = +

= ÊËÁ

ˆ¯̃

=

/ 03

43

4

12

7

9

7

a

aB g= 9

7Ø �

125


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 = +ÊËÁ

ˆ¯̃

3

5

1

42 2

For slender cone, neglect ‘a’ I mhD = 3

52

Parallel axis theorem I I m h

mh I mh

I mh I mh

D = + ÊËÁ

ˆ¯̃

= +

= -ÊËÁ

ˆ¯̃

=

3

4

3

5

9

163

5

9

16

3

80

2

2 2

2 2

+ S SM M W

hI ma

hC C= = +( ) :eff 4 4

a

mgh

mh mh h

mgh

mh

mgh

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 hg

hD = = ÊËÁ

ˆ¯̃

a 5

2 aD g= 2 50. Ø �

126



(b) Reaction at C. + S SF F C mg may y= - = -( ) :eff

C mg mh

C mg mh g

h

C mg mg

- = -

- = - ÊËÁ

ˆ¯̃

- = -

4

4

5

2

5

8

a

C = 3

8mg ≠ �

127


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

121 0 3 7 5 10

1

12

1

1

2 2 3

2

( )( . ) . kg m2

222 0 6 60 102 3( )( . ) = ¥ ◊- kg m2

Geometry: r

r L

AB

BC BC

= + =

= =

=

= ∞

0 6 0 15 0 61847

1

20 3

0 15

0 614 036

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= w2 b

( )aBC BCt r= a Æ

( )aBC n BCr= w2≠

Kinetics: + SMC = + S( ) :MC eff m gL

I r m a tABAB

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



+ 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 0883 2 C = 150 1. N 83 2. ∞ �

129


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


Rod AB: I m lAB AB AB=

=

= ◊

1

121

124 0 6

0 12

2

2( ( . )

.

kg) m

kg m2

+ S SM M IB B AB= - + =( ) : ( )( . ) ( )( . )( .eff N m m)

3

530 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

20 1875

2 2. . .

.

m

m

aBC = ( . )0 1875 m a

aAB = ( . )0 225 m a Æ

130



Kinetics: I m CBBC BC=

= -

= ◊

1

121

122 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 187515

5124 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


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

123 0 3

22 5 10

2

3

aa

kg m

kg m2

DDE =

= ¥ ◊-

1

124 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= ÊËÁ

ˆ¯̃

- ÊËÁ

ˆ¯̃

=( ) : ( . ) ( . )eff m m3

50 2

3

50 2 a

0 12 53 333 10

0 44444

3. ( ) .

.

T T

T TBE BD

BE BD

- = ¥- =

- aa (2)

132



Given data: T

TBD

BE

==

2

8

N

N


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


PROBLEM 16.90

A 1.5-kg slender rod is welded to a 5-kg uniform disk as shown. 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 = =( ) ( . )( )w2 0 14 10m rad/s2

an = 14 m/s2¨

at CG= =( ) ( . )a a0 14 m Ø

Kinetics: I m CGdisk disk

2

kg m

kg m

=

=

= ¥ ◊-

1

21

25 0 08

16 10

2

2

3

( )

( )( . )

I m ABAB AB=

=

= ¥ ◊-

1

121

121 5 0 12

1 8 10

2

3

( )

( . )( . )

.

kg m

kg m2


+ 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



(b) Components of reaction of C.

+ S SF F C m ax x x AB n= = - = -( ) : ( . )( )eff2kg 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


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

25 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

123 0 15 5 625 102 2 3 2( ) ( )( . ) .kg m kg m

136



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 1 252 2j a iA

137


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


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 / /a2 (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 w2r

SM r r rC G C C G C G C G C G CI ma mr r m= + ¥ + ¥ ¥ - ¥a / / / / /( )a w2

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


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

sin2 2

a r rrg

r k= =

+a bsin

2 2 a

r

r kg=

+

2

2 2sin b �

140


PROBLEM 16.95

A flywheel is rigidly attached to a shaft of 40-mm radius that can roll along parallel rails as shown. 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= +021

2

5 01

240 2m s= + a( )

a = 1

1602m/s 15∞

Since r = =40 0 04mm m.

a r= =a a; ( . )1

1600 04

a = 5

322rad/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

322rad/s

( . )( . )sin ( )9 81 0 04 155

322 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


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 2sin (1)

+

S SF F= eff : F mg ma- = -sin bF 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 NF

Ns s= =m mor

m b ab

b a

bsF

N

mg mr

mg

rg

= = - =-

sin

cos

sin

cos

Substitute for a from Eq. (1) mb b

bs

rg

gr

r k=- ◊

+sin sin

cos

2 2

m b bsr

r k

k

r k= -

+

È

ÎÍ

˘

˚˙ =

+

È

ÎÍ

˘

˚˙tan tan1

2

2 2

2

2 2

tan b m= +s

r k

k

2 2

2 tan b m= + Ê

ËÁˆ¯̃

È

ÎÍÍ

˘

˚˙˙

sr

k1

2

�

142


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

sin2 2

a r rrg

r k= =

+a bsin

2 2a

r

r kg=

+

2

2 2sin b (1)

For pipe: k r= ar

r rg gP =

+=

2

2 2

1

2sin sinb b

For cylinder: k 2 1

2= a

r

rr

g gC =+

=2

22

2

2

3sin sinb b

For sphere: k 2 2

5= a

r

r rg gS =

+=

2

2 225

5

7sin sinb b

(a) Between pipe and cylinder.

a a a g gC P C P/ = - = -ÊËÁ

ˆ¯̃

=2

3

1

2

1

6sin sinb b

x a t g tC P C P/ /= = ÊËÁ

ˆ¯̃

1

2

1

2

1

62 2sin b

SI units: xC P/ m/s s m= ÊËÁ

ˆ¯̃

∞ =1

2

1

69 81 10 4 2 272 2. sin ( ) . �

143



(b) Between sphere and cylinder.

a a a g gS C S C/ = - = -ÊËÁ

ˆ¯̃

=5

7

2

3

1

21sin sinb b

x a t g tS C S C/ /= = ÊËÁ

ˆ¯̃

1

2

1

2

1

212 2sin b

SI units: xS C/ m/s s m= ÊËÁ

ˆ¯̃

∞ =1

2

1

219 81 10 4 0 6492 2. sin ( ) . �

144


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

9 810 15

( . )( . )

I = ◊0 114679 2. kg m+

S SM M ma r IC C= = +( ) ( )( . ( )eff : N m)25 0 2 a

550

9 810 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

2550

3 1362

9 01539.81

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


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


I mk= 2

= 50

9 810 15 2

( . )( . )

I = ◊0 114679. kg m2

+ S SM M ma r IC C= = +( ) ( )( . ) ( )eff : N m25 0 3 a

7 550

9 810 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 F2550

9 814 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


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 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


I mk= =2 250

9 810 15

( )

( . )( . )

I = ◊0 114679. kg m2

+ S SM M ma r IC C= = +( ) ( ( . ( )eff : N) m)25 0 1 a

2 550

9 810 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 2550

9 811 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



(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 2550

9.81a

a = 2 943 2. m/s a = 2 94 2. m/s Æ �

148


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


I mk= =2 250

9 810 15

( . )( . )

I = ◊0 114679. kg m2

+ SM MC C= S( )eff : ( )( . ) ( )25 0 1N m = +ma r Ia

2 550

9 810 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

9 811 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




+ 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 :

550

9 81N =

.a

a = 0 981 2. m/s a = 0 981 2. m/s ¨ �

150


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.

SOLUTIONa 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 ¨

( ).

.minmsF

N= =

7 20

58 86

N

N ( ) .minms = 0 122 �

151


PROBLEM 16.103


SOLUTION

a r

I mk

I

= =

=

=

= ¥ ◊-

a a( . )

( )( . )

.

0 12

6 0 09

40 5 10

2

2

3

m

kg m

kg m2


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 ¨

( ).

.minmsF

N= =

0 8

58 86

N

N ( ) .minms = 0 0136 �

152


PROBLEM 16.104


SOLUTIONa r

I mk

I

= =

=

=

= ◊

a a( . )

( )( . )

.

0 12

6 0 09

40 5

2

2

m

kg m

kg m2


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 ¨

( ).

.minmsF

N= =

13 6

58 86

N

N ( ) .minms = 0 231 �

153


PROBLEM 16.105


SOLUTION

a r

I mk

I

= =

=

=

= ◊

a a( . )

( )( . )

.

0 12

6 0 09

40 5

2

2

m

kg m

kg m2


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 ¨

( ).

.minmsF

N= =

6 4

38 86

N

N ( ) .minms = 0 165 �

154


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 ra

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 ra

r

F m a

C C= =

= ¢ ÊËÁ

ˆ¯̃

= ¢

( ) :eff a12

12

2

(2)

Substitute for F from (2) into (1).

mg m a ma- ¢ÊËÁ

ˆ¯̃

=41

2

mg m m a amg

m m= + ¢ =

+ ¢( )2

2

(a) m m a g= ¢ = =+

5 25

5 2 2kg, kg:

( ) a = 5

9g Ø �

(b) ¢ = =m a g0: a = g Ø �

(c) m a= =0 0 a = 0 �

+

+

155


PROBLEM 16.107


SOLUTION

Kinematics:

Disk is rolling on vertical wall¢ = =

= =a a r

a a rC

r

aa2

Therefore:

a

a

=

¢ = =

a

r

a r a

21

2


S SF F

W F ma

mg F ma

y y=

- =- =

( ) :

( )

eff

4

2 2 (1)

I m r= ¢1

22

+ S SM MD D= ( ) :eff

F r m gr I m a r

Fr m gr m ra

rm

ar

F m a

( )2

21

2 2 2

21

4

2

+ ¢ = + ¢ ¢

= ¢ = ¢ ÊËÁ

ˆ¯̃

+ ¢

= ¢ +

a

11

2a m gÊ

ËÁˆ¯̃

- ¢ (2)

Substitute for 2F from (2) into (1):

mg m a m g ma

m m a m m g am m

m m

- ¢ ÊËÁˆ¯̃

+ ¢ =

+ ¢ÊËÁ

ˆ¯̃

= + ¢ = + ¢

+ ¢

23

42

3

22

232

( ) gg

+

156



(a) m m a g= ¢ = = ++

5 25 4

5 3kg, kg : a = 9

8g Ø �

(b) ¢ = =m a g0 : a = g Ø �

(c) m a g= =02

3 2:

/ a = 4

3g Ø �

Note: In parts (a) and (c), a g> , from Eq. (1) it follows that F < 0. The disks push the bar down.

157


PROBLEM 16.108


SOLUTION

Kinematics:

Disk is rolling on vertical wall

a a r= ¢ = a


S SF F

W F ma

mg F ma

y y=

- =- =

( ) :eff

4

4 (1)

I m ra

ra = ¢ Ê

ËÁˆ¯̃

1

22

S SM M

F w r I m a r

F m g r m ra

rm at

F

D D=+ ¢ = + ¢ ¢

+ ¢ = ¢ ¢ ÊËÁˆ¯̃

+

( ) :

( )

( ) ’

eff

a1

2

== ¢ - ¢3

2m a m g (2)

Substitute for F from (2) into (1):

mg m a m g ma

m m a m m g am m

m mg

- ¢ + ¢ =

+ ¢ = + ¢ = + ¢+ ¢

6 4

6 44

6( ) ( )

(a) m m a= ¢ = = ++

5 25 8

5 12kg, kg : a = 13

17g Ø �

(b) ¢ = =m a g0 : a = g Ø �

(c) m a g= =04

6: a = 2

3g Ø �

+

+

158


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 disk B 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

22

+ S SM M M C ma r Is s= - = +( ) : ( . ( )eff m)0 2 a

2 25 0 21

23

2

2 25 0 23

22 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

22

+ S SM M D ma r I mr mrH H= = + = +( ) : ( . ( )eff2m)0 2

1

22 2a a a

D

D

( . ) ( )( . )

. .

0 23

22 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



+ S SF Fx x= ( ) :eff C D m a

C DR 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/s2 �

(b) Substitute for a in (2)

0 2 0 0675 11 5385. ( . )( . )D = D = ¨3 89. N �

160


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 rr

lAB AB C C ABC

ABCa 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

FI

rDC 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 rI

rW l W l

I

C C C B AB AB AB D B AB

AB CC

CC C AB AB AG

C

a a

a( )

aa a a aC C AB C CC

AAB C D AB C C

AB

CC C AB C

C

ABAB

m l rr

lI m l r

l

rI m l r

r

lI

- - -

+ + ( ++È

ÎÍ

˘

˚˙

= +

m l

W l W l

AB AG C

C AB AB AG

2 ) a

161



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

123 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/s2

(a) Angular acceleration. aC = 65 4. rad/s2 �

(b) Acceleration of B.

From Eq. (1), aB = ( . )( . )0 125 65 4m rad/s2 aB = 8 175. m/s2 Ø �

162


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 OGr= 4

3p

a a a a= + = +O G O O OG/ [ ] [( ) ]Æ ¨a

a = Æ + ÈÎÍ

˘˚̇

= -ÊËÁ

ˆ¯̃

[ ]rr

rap

ap

a4

31

4

3¨ Æ

Kinetics: m = mass of half cylinder:

I mr

I I m OG

mr Ir

m

I mr

O

O

=

= +

= + ÊËÁ

ˆ¯̃

= -ÊËÁ

ˆ¯̃

1

2

1

2

4

3

1

2

16

9

2

2

22

22

( )

p

p(a) Angular acceleration.

+ S SM M ma rC O= = + -Ê

ËÁˆ¯̃( ) ( )eff : Pr Ia 1

43p

Pr mr

Pr mr

= -ÊËÁ

ˆ¯̃ + -Ê

ËÁˆ¯̃ -Ê

ËÁˆ¯̃

= -

22

2

12

16

91

43

14

3

12

1

pa

pa

pmr r

66

91

83

16

922

2pa

p paÊ

ËÁˆ¯̃ + - +Ê

ËÁˆ¯̃mr

163



Pr mr

Pr mr

= -ÊËÁ

ˆ¯̃

=

2

2

32

83

0 65717

pa

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 mrP

mr

= -ÊËÁ

ˆ¯̃

=

= ÊËÁ

ˆ¯̃

=

14

30 57559

0 57559 1 5357

pa 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


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 OGr= 4

3p

a a= + = Æ - ¨O G O O OGa / [ ] [( ) ]a a

a = Æ + ¨È

ÎÍ˘˚̇

= -ÊËÁ

ˆ¯̃ Æ[ ]r

rra 4

31

43p

ap

a

Kinetics: m = mass of half cylinder:

I mr

I I m OG

mr Ir

m

I mr

O

O

=

= +

= + ÊËÁ

ˆ¯̃

= -ÊËÁ

ˆ¯̃

1

2

1

2

4

3

1

2

16

9

2

2

22

22

( )

p

p(a) Angular acceleration:

+ S SM M rC C= = + -ÊËÁ

ˆ¯̃( ) ( )eff : Pr I maa 1

43p

Pr mr= -Ê

ËÁˆ¯̃ + -Ê

ËÁˆ¯̃ -Ê

ËÁˆ¯̃

22

12

16

91

43

14

3pa

pa

pmr r

165



Pr mr mr= -Ê

ËÁˆ¯̃ + - +Ê

ËÁˆ¯̃

22

22

12

16

91

83

16

9p p paa

Pr mr

Pr mr

= -ÊËÁ

ˆ¯̃

=

=

2

2

32

83

1 65113

1 5367

pa

a

a

( . )

.P

mr a = 1 536.

P

mr �

(b) Minimum value of ms.

S SF F P F ma mr

P F mrP

x x= - = = -ÊËÁ

ˆ¯̃

- = -ÊËÁ

ˆ¯̃

( )

.

eff : 14

3

14

31 5357

pa

p mmrP

F P P P

ÊËÁ

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


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


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 32BB 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


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 rB h h B

B

sin ( ) ( cos )( cos )

sin ( sin

q a a a q qa 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 aq

q=

+ +g

r

m

m mB

h B2 1

sin

( cos ) �

168


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 w2 +r

RaB

ÈÎÍ

˘˚̇

≠

Kinetics:

S SM M PR Wr ma r ma R I

PR mgr mr

Ra r m a

C C y x

B

= - = + +

- = ÊËÁ

ˆ¯̃

+

( ) ( ) ( )

(

eff : a

BBB

BB

r R mka

R

mar

RR

k

Rmr

v

rR

P mgr

R

+ +

= + +Ê

ËÁˆ

¯̃+ Ê

ËÁˆ¯̃

= ÊËÁ

ˆ

w2 2

2 2 2

)

¯̃̄+ + +Ê

ËÁˆ

¯̃+ma

r k

Rm

r

RvB B1

2 2

2 22 (1)

+

169



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

ËÁˆ

¯̃+1 5 9 81

0 018

0 061 5 1

0 018 0 044

0 061 5

2 2

2. ( . )

.

.. ( )

. .

..

00 018

0 06

4 4145 2 4417 7 5

22

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/s2 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


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

122 0 5 0 041672 2

kg

kg m2( )( . ) .

Disk: m

I m r

D

D C

=

= = = ◊

5

1

2

1

25 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 aaa

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

( )

[( )( . ) ( )( . )]( . )

++ + + +

= +

q2 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



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 62570 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


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



Kinetics: I mL m= =1

12

1

121 22 2( . m)


+ 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 54381

121 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


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/s212 ≠ + [ .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/s2

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/s2

a ay G y= = ≠( ) .6 00 m/s2

175



Kinetics: I mL m= =1

12

1

121 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

121

1210 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


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

300 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



We have:

a ax y= ¨ = Ø0 375 0 43301. ; .a a

I mL

I

= ◊ = ◊

= ◊

1

12

1

124 0 75

0 1875

2 ( )( .

.

)

kg m

2

2


+ 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


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/s2 ( )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/s2

+ ay y= + ∞ = = ≠( . )( . )sin . , .0 4 6 64463 65 4088 2 4088+2 m/s2a

Kinetics m

I ml

=

= = =

2

1

12

1

122 0 8

8

752 2

kg

kg m2( )( . ) ◊

179



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 362528

756 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


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 B1

20 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= = = =21

12

1

122 0 8

8

752 2 2kg kg m( )( . ) ◊

From triangle ABE:

AE AB

sin sin45 70∞=

∞

AE

AE

= ∞∞

=

( .sin

sin.

0 845

700 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




+ S SM ME E= ( ) :eff W GD I ma DE ma GDx y( ) ( ) ( )= + +a

( )( . )( . ) ( . ) ( . ) ( . ) (2 9 81 0 36252

8

752 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


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

aL

G A/ =2

a 20∞

a a a a LG A G A= = + = [/ .1 0851 a 602

∞] + ÈÎÍ

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



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 Lsin sin

; .50 60

0 88455∞

=∞

=


+ S SM M mg L I ma L ma LE E x y= = + +( ) : ( . ) ( . ) ( . )eff 0 46985 0 71355 0 46985a

0 46985

1

120 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 577759 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


PROBLEM 16.123

End A of the 8-kg uniform rod AB is attached to a collar that can slide without 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 ] + ÈÎÍ

L

2a q ]

axL=2

a qsin Æ; a yL=2

a qcos ≠

Kinetics:

+ S SM M mgL

I maL

maL

E E x y= = + ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

( ) : cos sin coseff 2 2 2q a q q

mgL

mL mL

mL

2

1

12 2 22

2 2

cos sin cosq a q a q a= + ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

mgL

mL2

1

32cosq a= a = 3

2

g

Lcosq

185



+ S SF F A ma mL

x x x= = =( ) : sineff 2a q

A mL g

L= Ê

ËÁˆ¯̃2

3

2cos sinq q

A = 3

4mg sin cosq q Æ

Data: m L= = ∞ =8 30 0 75kg m, , .q

(a) Angular acceleration. a = ∞3

2

9 81

0 7530

.

.cos

m/s

m

2

a = 16 99. rad/s2 �

(b) Reaction at A. A = ∞ ∞3

48 9 81 30 30( )( . )sin coskg m/s2 A = 25 5. N Æ �

186


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/s2 2 ( ) .aB t = 1 5 rad/s2 Ø

( ) ( ) ( . )( ) .a BCB n = = =w 2 20 1 6 3 6 m rad/s m/s2 ( ) .aB n = 3 6 m/s2 Æ

Rod ABD: q = = = ∞- -sin sin.

.1 1 0 1

0 230

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

ab

= =∞

=3 6

0 2

18

3020 78

.

. cos cos. rad/s2 a = 20 78. rad/s2

Kinetics:+

S SM M A I m LG G= = =( ) : ( . )eff m0 17321

122a a

= 1

124 0 4 20 782( )( . ) ( . )kg m rad/s2

A = 6 399. N A = 6 40. N ¨ �

187


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

50 20

0 25

3

5

w = =v

BCB 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 43

+ ( )aD B t/ 534

= aD b 137 08. Ø + 0 25 13 09 2. ( . ) 5

43

+ 0 25. a 534

= rD b

188



x components:

3

542 84

4

50 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

521 42

4

516 065 0( . ) ( . )

+ay = - - =137 084

521 42

3

516 065 110 31. ( . ) ( . ) . m/s2

Kinetics

I ml= =

=

1

12

1

125 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 �

+

189


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

ww

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 Aw2 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 2566 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/s2

+ ay = ∞ = +0 125 598 3 66 42 29 92. ( . )cos . . m/s2

190



Summary of kinematics: a = 598 3 2. rad/s , ax = 68 54. m/s2 Æ, ay = 29 92. m/s2 ≠

I ml= =

=

1

12

1

125 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 mah

ma IB B x y= + = ÊËÁ

ˆ¯̃

- -( ) : ( . ) ( . )eff 0 052

0 05 a

D( . ) ( . )( . ) ( . ).

( . )( . ) .0 2291 5 9 81 0 05 342 70 2291

2149 6 0 05 15 58+ = - - 11

0 2291 37 256 7 480 15 581 2 453 13 742. . . . . .D = - - - = D = 60 0. NÆ �

191


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 = ÊËÁ

ˆ¯̃

=3002

6010rpm rad/s

v ABB AB= = =( ) ( . )( ) .w p0 06 10 1 88496 m/s≠

aB ABAB= = =( ) ( . )( ) .w p2 20 06 10 59 218 m/s2 ¨

Geometry sin cosb b= =3

5

4

5Rod BD:

Velocity: Instantaneous center at C

v BCB BD

BD

==

( )

. ( . )

ww1 88496 0 24m/s

wBD = 7 854. rad/s

Acceleration

BD = =300 0 3mm m.

( ) ( ) .a BDD B t BD BD/ = =a a0 3 43

( ) ( ) ( . )( . ) .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/s2 ¨] [ .+ 0 3a b ] [ .+ 18 5056 m/s2 b]

192



+0 0 3

4

518 5056

3

5= -( . ) ( . )a aBD = 46 264. rad/s2

( ) ( ).

( . ) .aG B t BDBG/2 2rad/s m/s= = Ê

ËÁˆ¯̃

=a 0 3

246 264 6 9396 5 4

3

( ) ( ).

( . ) .a BGG S n GD/2rad/s m/s= = Ê

ËÁˆ¯̃

=w2 20 3

27 854 9 2528

54

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 = + ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

59 218 6 93963

59 2528

4

5. ( . ) ( . )

ax = 70 784. m/s2 ¨

+ a ay y= Ê

ËÁˆ¯̃

- ÊËÁ

ˆ¯̃

= =( . ) ( . ) ;6 93964

59 2528

3

50 0

Kinetcs:

I L

I

= = ÊËÁ

ˆ¯̃

= ◊

= ◊

1

12

1

124 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 09a0 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≠ �

193


PROBLEM 16.128


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¯̃

=3002

6010rpm rad/s

v ABB AB= = =( ) ( . )( ) .w p0 06 10 1 88496 m/s ̈

aB ABAB= = =( ) ( . )( ) .w p2 20 06 10 59 218 m/s2 Ø

Rod BD:

Instantaneous center at

wBD = 0

b = = ∞-sin.

..1 0 12

0 323 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/s2 Ø] [ .+ 0 3aBD b] + 0

194



+ 0 59 218 0 3= - +. ( . )cosa bBD

aBD =∞

59 218

0 3 23 57

.

. cos . aBD = 215 36. rad/s2

( ) ( ).

( . ) .a BGG B t BD/2m m/s= = Ê

ËÁˆ¯̃

=a 0 3

2215 36 32 304


a = [ .59 218 m/s2 Ø] [ .+ 32 304 m/s2 b] + 0

+ a

ax 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/s2 ay = 29 609. m/s2Ø

Kinetics

I BD= = = ◊1

12

1

124 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


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

aBD 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/s2 2¨

( ) ( . )( )aB y = =0 08 15 18m rad/s m/s2 2 ≠

Rod AB:

Velocity: Instantaneous center at C.

CB = ∞ =( . ) tan .0 5 30 0 86603m / m

wABBv

CB= = =

1 2

0 866031 3856

.

..

m/s

mrad/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



+ aA Acos . . ; .30 4 8 0 96 4 434∞ = - =a m/s2 30∞+ ( . )sin . ; .4 434 30 18 0 5 31 566∞ = - =a AB ABa rad/s2


a = [ .4 8Æ] [+ 18≠] [ . ( . )+ 0 25 32 566 Ø] [ .+ 0 48Æ]

+ a ax x= - = =4 8 0 48 4 32 4 32. . . ; . m/s2 ¨+ a ay y= - = =18 2 892 10 108 10 108. . ; . m/s2 ≠

Kinetics: I m AB= = = ◊1

12

3

120 5 0 06252 2( ) ( . ) .

kgm kg m2

I m AB= = =1

12

3

120 5 02 2( ) ( . )

kgm

+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


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/s2

( ) ( . )( ) .aB x = =0 08 60 4 8m rad/s m/s2 2Æ

( ) ( . )( )aB y = =0 08 15 182m rad/s m/s2 ≠

Rod AB:

Velocity: Instantaneous center at C.

CB = ∞ =( . ) tan .0 5 30 0 86603m / m

w ABBv

CB= = =

1 2

0 866031 3856

.

..

m/s

mrad/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



+ aA A= ∞ = + =cos . . ; .30 4 8 0 96 6 651a m/s2 30°

+ ( . )sin . ; .6 651 30 18 0 5 42 65∞ = - + =a AB ABa rad/s2

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/s2 Æ+ a y y= - = =18 10 663 7 337 7 337. . ; .a m/s2≠

Kinetics: I m AB= = = ◊1

12

3

120 5 0 06252 2( ) ( . ) .

kgm 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


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 02 2

a a q

a a q

akk r2 2+

cosq

Setting a w wq= d

d , and using r k= =0 5 0 3. , .m m

200



wwq

q

qw 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

wp

d a d

a

f

f

A

A

f

0 0

2

2

00

2

2

1 4706

1

21 4706

Ú Ú=

=

( . )cos

. sin

/

/| |

== 2 9412. aA (1)

Given data: aA = 2 m/s2 Æ

w f2 2 9412 2

5 8824

=

=

. ( )

. w f = 2 43. rad/s �

201


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 02 2

a a q

a a q

akk r2 2+

cosq

Setting a w wq= d

d , and using r k= =0 5 0 3. , .m m

w wq

q

qw 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



w w q q

w q

w

w p

wp

d a d

a

f

f

A

A

f

0 0

2

2

00

2

2

1 4706

1

21 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/s2 aA = 1 360. m/s2 Æ �

203


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= = ∞30030

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



+ 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)- = -= -

a44642a (2)

Bar AC: I m AC=

=

= ◊

1

121

124 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 a111 772 0 36 0 12 0 36

23 544 0 84

28 0286

. . . .

. .

.

- = +=

=

a a aa

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


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

1222 62

206



( ) ( ).

./aG B t BC BC BCBG= = ÊËÁ

ˆ¯̃

=a a a0 325

20 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/s2Ø] [= 72 m/s2≠] .+ ÈÎ0 325aBC 12

513

˘˚̇

+ 72 72 0 32512

13480= - + =( . ) ;aBC BCa rad/s2

+ 0 125 0 325 4805

13480. ( . )( ) ;aCD CD= Ê

ËÁˆ¯̃

=a rad/s2

a a a a a= + = +B G B B G B t/ /( )

= [72 m/s2 ≠] .+ ÈÎ0 1625aBC

512 ˘

˚̇

= [72 m/s2 ≠] . ( )+ ÈÎ0 1625 480 12

513

˘˚̇

a = [72 m/s2≠] + ÈÎÍ78 m/s2 12

513

˘˚̇

+ ax = ÊËÁ

ˆ¯̃

=785

1330 m/s2 Æ

+ ay = - ÊËÁ

ˆ¯̃

= - =72 7812

1372 72 0

Total acceleration of G is 30 m/s2Æ

Bar CD:

Kinematics:

( ) ( . )aCD x = =0 0625 480 30 m)( rad/s m/s2 2 Æ

207



Kinetics: I m CD

I

CD

CD

=

=

= ¥ ◊-

1

121

121 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

Cx

x

==

.

N

Bar BC: I m BCBC = = = ¥ ◊-1

12

1

123 0 325 26 406 102 2 3( ) ( . . kg)( m) kg m2

m aBC = =( )3 30 90 kg)( m/s N2 Æ+ S SF F B C m ax x x x BC= - =( )eff :

B

Bx

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



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


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/s2 Æ

( ) ( ) ( . )( )a ABB y AB= =w2 20 125 24

( )aB y = 72 m/s2 ≠

Bar CD:

Bar BC:

b

b

=

= ∞

-tan

.

1 5

1222 62

210



( ) ( ) ( . ) .aC B t BC BC BCBC/ m= = =a a a0 325 0 325 512

( ) ( ).

.aG B BC BC BCBG/ m= = Ê

ËÁˆ¯̃

=a a a0 325

20 1625 12

513

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/s2Ø] [= 20 m/s2Æ] [+ 72 m/s2≠] [ .+ 0 325aBC 125

13 ]

+ 72 72 0 32512

13480= - + =( . ) ;aBC BCa rad/s2

+ 0 125 20 0 325 4805

13640. ( . )( ) ;aCD CD= + Ê

ËÁˆ¯̃

=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 125

13 ]

= [20 m/s2Æ] [+ 72 m/s2≠] [ . ( )+ 0 1625 4805

12 ]

a = [20 m/s2 Æ] [+ 72 m/s2≠] [+ 78 m/s212

513 ]

+ ax = + ÊËÁ

ˆ¯̃

= + =20 785

1320 30 50 m/s2

+ a y = - ÊËÁ

ˆ¯̃

=72 7812

130

Total acceleration of G is 50 m/s2Æ

Bar CD:

Kinematics:

( ) ( . )( )aCD x = =0 0625 640 40m rad/s m/s2 2 Æ

211



Kinetics: I m CD

I

CD

CD

=

=

= ¥ ◊-

1

121 5

120 125

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

Cx

x

==

.

N

Bar BC: I m BCBC =

=

= ¥ ◊-

1

123

120 325

26 406 10

2

2

3

( )

( . )

.

kg m

kg m2

+

m a

F F B C m aBC

x x x y BC

= = Æ= - =

( )

( )

3 150 kg)(50 m/s N

:

2

effS S

B

Bx

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



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 1251

121 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


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

BCB

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/s2 Ø

aB = =( . )( ) .0 18 2 0 722 rad/s m/s2 2Ø

a

a

BC

BC

=

=

( . )

.

0 09

0 36

2 m)(2 rad/s

m/s2 Ø

214



a a aAB A B= + = +1

2

1

22 16 0 72( ) ( . . )

aAB = 1 44. m/s2 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/s2

Kinetics: I m AB

I

AB AB

AB

= =

= ¥ ◊-

1

12

1

122 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


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 1 082 m)( rad/s m/s2Æ

aBCBa

= =0 18

1 08

0 18.

.

. m

m/s

m

2

aBC = 6 2rad/s

aBC = =( . ) .0 09 6 0 542 m)( rad/s m/s2 Æ

a a aAB A B= = = 0 54. m/s2 Æ

Kinetics: I m BCBC BC= =1

12

1

123 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



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


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

ËÁˆ¯̃

=6002

6062 832 rpm rad/s.

a ABB AB= =( ) ( . )w2 0 1 m)(62.832 rad/s2

aB = 394 78. m/s2 ¨

Also: v ABB AB= = =( ) ( .w 0 1 m)(62.832 rad/s) 6.2832 m/sØ

Connecting rod BD:

Velocity Instantaneous center at D.

wBDBv

BD= = =6 2832

0 2525 133

.

..

m/s

m rad/s

Acceleration:

a a aD B D B Ba= + =/ [ ¨] [( )+ BD BDw2 Æ]

aD = [ .394 78 m/s2 ¨] [( . )+ 0 25 2 m)(25.133 rad/s Æ]

aD = [ .397 78 m/s2 ¨] [ .+ 157 92 m/s2 ¨] .= 236 86 m/s2¨

aBD B D= + =1

2

1

2394 78( ) ( .a a ¨ + 236 86. ¨) .= 315 82 m/s2 ¨

Kinetics of piston

D = 426 34. N ¨

218



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


PROBLEM 16.139


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= + ¥= + ¥ - -= +

ww

/

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

aAB 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 kD B BD B D BD D B

D ABa

a wa (( . . )

. . .

- - -= - - +

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/s2

i: aD = - = -( . )( ) .0 1 1723 172 3 m/s2

a AB = -( )1723 rad/s2 k a iD = -( . )172 3 m/s2

220



Acceleration of mass center G of bar BD

rG D

G D BD G D

/

/

m) m)

= +

= +

= + ¥

1

20 2291 0 1

0 11455 0 05

( . . )

( . ( .

i j

i j

a a ra --= - + - ¥ + -= - -

wBD G D2

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

121 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 5238 1

.

..b = ∞ B = 525 N 38 1. ∞ �

221



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 ND 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


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

vCDv

=2cosq

q

Rod AC: v

v

v

C AC

DAC

ACD

L

L

L

=

=

=

2

2 2

w

qw

wq

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



( ) ( ) tan

tan

a a

L LC t C n

AC AC

=

=

q

a w q2 2

2

a AC ACD Dv

L

v

L= =

ÊËÁ

ˆ¯̃

=w qq

qqq

22 2

2 2tan

costan

tan

cos

By symmetry: | | | |a aAC CD=

aCEDv

L=

2

2 2

tan

cos

qq

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

LIC C CE= = -( ) coseff :

2q a

PL m

Lv

LD

2 122

2

2 2cos

tan

cosq

qq

= - ◊Ê

ËÁˆ

¯̃ P

mv

LD= ◊2

36

tan

cos

qq

�

224


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

1250 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] + ÈÎÍ

l

2a 10

22

∞˘˚̇

+ l w 80° (1)

Kinetics: Sliding to the right: F Nf k= m

+ SMG = + S( ) : sin cos cosM Nl

Fl

Pl

h IG feff 210

210

210∞ - ∞ + ∞ -Ê

ËÁˆ¯̃

= a

I Nl

Nl

Pl

hka m

a

- ∞ + ∞ = ∞ -ÊËÁ

ˆ¯̃

- ∞ -2

102

102

10

150 3 10

cos sin cos

( )(sin 00 3 10 500 3 10 2. cos ) ( cos )∞ = ∞ -N (2)

225



+ SFy = + S( ) : sin cosF N mgml ml

y eff - = -ÊËÁ

ˆ¯̃

∞ - ∞2

102

102a w

mlN mg

ml

N

210

210

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 maml ml

x f Ceff - = - ∞ + ∞2

102

102a w

ma P Nml ml

a

C k

C

= - + ∞ - ∞

= - +

m a w2

102

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 334827 1 aG = 9 36. m/s2 27.1° �

(a) Acceleration at Point G.

(b) Normal force. N = 278 N ≠ �

226


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 rg

r

sin ( )301

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

3a 30°

(a) b

b

=

= ∞

=

=

-tan.

.

.

cos

.

cos .

1 1 3

0 866

21 052

0 866

0 866

21 052

/ g

g

ag

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 51

61

64 9 81

6 54

2

. ( )

( )( . )

.

mg /

kg m/s

N T = 6 54. N �

227


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

22a

a AT

mr= 2

(1)

Disk B: +

S SM M Tr IB B B= =( )eff : a

Tr mr B= 1

22a

aBT

mr= 2

(2)

228



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

22a a a( )

a a aA B AWr mr= =:5

22 a A

g

r= 2

5 �

aBg

r= 2

5 �

Substitute for a A into (1): 2

5

2g

r

T

mr= T mg= 1

5 �

a r

r

rg

r

B A B

A

= +=

= ÊËÁ

ˆ¯̃

( )

( )

a aa2

22

5 aB g= 4

5Ø �

229


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 aR G A G A

R C G A

= = += +

/

/

where a LG A/ = 1

2a

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

122

N

N

m

1112

15 1 1 25

1

21 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

aC a

CC

Ca

-= +

4 0784

4 1459 0 11653

.

( . ) .

aa (1)

230



Rod

+S SM M I m a

Lm a

LA A R G R R C= = + - ∞( ) : ( ) ( cos )eff /0

225

2a

1 25 15 0 5 0 5 15 25 0 5 0

1 25 3 75 6 797

. ( . )( . ) ( )( cos )( . )

. . .

a aa a

+ - ∞ =+ =

aC

33

1 3595

a

aC

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/s2 25° �


From (2): a = ( . )( . )1 3595 4 9263

= 6 6973 2. rad/s a = 6 70 2. rad/s �

231


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 AAa

r=

+ aL

aA AB AB= -2

a

+S SM M A r ma r IC C x A A= = +( )eff : a

Cylinder: A r ma r mra

r

A ma

A mL

a

x AA

x A

x AB AB

= + ÊËÁ

ˆ¯̃

=

= -ÊËÁ

ˆ¯̃

1

2

3

23

2 2

2

a (1)

+S SM MA A= ( )eff : PL ma

LIAB AB= +

2a

PL maL m

LAB AB= +2 12

2a (2)

+ S SF Fx x= ( )eff : P A max AB+ =

Rod AB:

Rod AB:

Kinetics:

232



Substitute from (1):

P mL

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

122PL

Lma mLAB AB= - a (4)

(4) + (2): 10

9

1

2

5

18

7

9PL mLa mLaAB AB= +Ê

ËÁˆ¯̃

=

aP

mAB = 10

7¨ (5)

(5) Æ (3) P mP

mmL

P P mL

AB

AB

= ÊËÁ

ˆ¯̃

-

= -

5

2

10

7

3

4

25

7

3

4

a

a

- =18

7

3

4P mL ABa a AB

P

mL= 24

7

+ aL

a

L P

mL

P

m

A AB AB= -

= ÊËÁ

ˆ¯̃

-

2

2

24

7

10

7

a

aP

mA = -ÊËÁ

ˆ¯̃

12

7

10

7 aA

P

m= Æ2

7 �

+ aL

a

L P

mL

P

m

B AB AB= +

= ÊËÁ

ˆ¯̃

+

2

2

24

7

10

7

a

aP

mB = +ÊËÁ

ˆ¯̃

12

7

10

7 aB

P

m= ¨22

7 �

233


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 = +rL

C 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

21

2

2a

a

Rod AB:

+ S SM M BL

IG G AB= =( ) :eff 2a

1

2 2

1

12

1

3

2m rL

m L

m

m

L

r

C C AB AB

CAB

CAB

a a

a a

ÊËÁ

ˆ¯̃

=

= ◊ (1)

+ S SF F m g B m ay y AB AB= - =( ) :eff

m g m r m rL

gL m

mr

AB C C AB C AB

ABC

ABC

- = +ÊËÁ

ˆ¯̃

= + +ÊËÁ

ˆ¯̃

1

2 2

2

1

21

a a a

a a

234



g

L

m

m

r

L

m

m

L

r

g

L

m

ABC

AB

AB

CAB= + +

ÊËÁ

ˆ¯̃

◊ ◊ÊËÁ

ˆ¯̃

= + +

1

2

1

21

1

3

1

2

1

6

1

3

a a

AAB

CAB

AB

CAB

AB mm

m

m

m

g

L AB

C

ÊËÁ

ˆ¯̃

= +ÊËÁ

ˆ¯̃

=+( )

a a

a

1

32

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

244 846

2

52( . )

..

m/s

m rad/s

kg8 kg

Eq. (2): aC = ◊ ◊ =1

3

5

8

0 25

0 144 846 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/s2 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/s2 Ø �

235


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

aA

A

aa (1)

Cylinder

I mr

I

= =

=

1

2

1

23 0 1

0 015

2 2( )( . )

. kg ◊ m2

236



(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 aaA

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 aa

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


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 W m 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 81 203

50

+ ∞ = - ∞

= ∞ +

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

01

23 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 aa

a

+ -= -=

. .

. .

.

a

a

a

A

A

A (2)

238



(a) Acceleration of the wedge.

Substitute for a from (2) into (1):

a a

a aA 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


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 maL

B B BC BC= = +( ) ( )eff : a2

= + +ÊËÁ

ˆ¯̃

= +

mL 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- = +ÊËÁ

ˆ¯̃

a a1

2 (2)

240



Bar AB:

+ S SM M B L I maL

A A x AB AB= = +( ) ( )eff : a2

= + ÊËÁ

ˆ¯̃

=

mL 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

2a a (4)

Subtract (1) from (4) 05

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

35

7

6a a a( )

a ABP

mL= - 6

7 (6)

Eq. (5) a aBC BCP

mL

P

mL= - -Ê

ËÁˆ¯̃

=56

7

30

7 (7)

Data: L m P= = = =500 0 5 3 20 mm m, kg, N.

a ABP

mL= - = -

= -

6

7

6

7

20

3 0 5

11 249 2

N

kg m

rad/s

( )( . )

. a AB = 11 43. rad/s2 �

aBCP

mL= =

=

30

7

30

7

20

3 0 5

57 143 2

N

kg m

rad/s

( )( . )

. aBC = 57 1. rad/s2 �

241


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

122 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

32

8

3

2

2

a a

a (1)

242



Bar BC:

+ S SM M Pb I maL

B B BC BC= = +( ) ( )eff : a2

= + ÊËÁ

ˆ¯̃

=

=

mL m L

L

Pb mL

Pb

mL

12

3

2 2

5

66

5

2

2

2

a a

a

a (2)

Substitute for a into (1): P L b mLPb

mL

PL Pb Pb

L b b

b

( )+ = ÊËÁ

ˆ¯̃

+ =

= -ÊËÁ

ˆ¯̃

=

=

8

3

6

5

16

516

51

11

5

5

22

111L

Eq. (2) a a= ÊËÁ

ˆ¯̃

=6

5

5

11

6

112

P

mLL

P

mL

Data: L

m

P

= ===

500 0 5

3

20

mm m

kg

N

.

(a) b L= = =5

11

5

110 5 0 22727( . ) . m b = 227 mm �

(b) a = = =6

11

6

11

20

3 0 57 2727 2P

mL

( )

( )( . ).

N

kg m rad/s a = 7 27. rad/s2 �

243


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: aL=2

a

+S SM M PL ma

LIA A= = +( ) : ( )eff 2a

= ÊËÁ

ˆ¯̃

+mL

m L1

2 2

1

122a a

a = 3P

mL (1)

+ S SF F A P max x x= - = -( ) :eff

A P mL

P mL P

mL

Px = - = - Ê

ËÁˆ¯̃

= -2 2

3

2a 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

La

m

LxJ

ÊËÁ

ˆ¯̃

= ( )a

Using (1): m

La

m

L

P

mL

m

La

Px

L

J

J

= ÊËÁ

ˆ¯̃

=

3

32

+ S S(M M M A xPx

L

x

M PxP

Lx

J J J x

J

= - = - ÊËÁ

ˆ¯̃

¥ ÊËÁ

ˆ¯̃

= -

) :eff1

2

3 2

3

1

2

1

2

2

233

(2)

For Mmax : dM

dx

P P

LxJ = - =

2

3

20

22

xL=3

(3)

244



Substituting into (2)

( )

( )

max

max

MPL P

L

L PL

MPL

J

J

= - ÊËÁ

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): xL= = =3

1

30 5773m m.

Eq. (4): ( )( )( )

.

maxMJ =

= ◊

8 1

3 31 5396

N m

N m

M Amax . .= ◊1 540 0 577N m located m below �

245


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

LB 3

2

a xg

LxJ = =a 3

2Ø

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

La

m

L

g

Lx

mg

LxJ = ◊ =3

2

3

2 2

+S SF Fmg

Lx

mgV

mg

Lx xy y J= - + = Ê

ËÁˆ¯̃

( )eff :4

1

2

3

2 2

Vmg mg

Lx

mg

LxJ = - +

4

3

4 22

+ S SM Mmg

Lx

x mgx M

mg

Lx x

xJ J J= Ê

ËÁˆ¯̃

- + = ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

( )eff :2 4

1

2

3

2 32

Mmg

xmg

Lx

mg

LxJ = - +

4

1

2

1

42

23

Find Vmin: dV

dx

mg

L

mg

Lx x LJ = - + = =3

20

2

32;

Vmg mg

LL

mg

LL V

mgmin min;= - Ê

ËÁˆ¯̃

+ ÊËÁ

ˆ¯̃

= -4

2

3

3

4

2

3 122

2

246



Find Mmax where V Vmg mg

Lx

mg

LxJ J= = - + =0

4

3

402:

3 4 02 2x Lx L- + =

( )( )3 03

x L x L xL

x L- - = = = and

Mmg L mg

L

L mg

L

L mgLmax = Ê

ËÁˆ¯̃

- ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

=4 3

1

2 3

1

4 3 27

2 3

Mmg

Lmg

LL

mg

LLmin = - + =

4

1

2

1

402 3

MmgL L

Amax =27 3

at from �

247


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 ..

qm

= - = -mg N

N

mg N

Nk

; 0 57740 2

0 11551 1155

0 8965..

.N mg N Nmg= - = =; mg

+ S SM M Nr Ik0 0= =( )eff : m a

( . )( . )0 2 0 89651

22mg r mr= a

a =

=

=

0 35858

0 358589 81

0 1

35 177

.

..

( .

.

g

r

m/s

m)

rad/s

2

2 a = 35 2. rad/s2 �

248


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 ..

qm

= - = -N mg

N

N mg

Nk

; 0 57740 2

0 11550 8845

1 1306..

.N N mg Nmg

mg= - = =:

+S SM M Nr Ik0 0= =( )eff : m a

( . )( . )

.

..

( . )

0 2 1 13061

2

0 4522

0 45229 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


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 mra

rma r

Pma

( ) ( )

( )

11

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): ( )( )

13

2 12-

-- =m

mmma

mg ma

3 1 2 2

1 3 2 0

( )

( )

+ - =+ - =

m mm ma g a

a g a g=+2

1 3

mm

�

250



(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

Pma

=-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):

mamg ma

a g a

a g

11

1

0

2

-- - =

+ - =- =

mm m

m mm m

( )

( )

a g= �

Summary: a g�2

1 3

mm+

: rolling

3

1 3

mm+

g a g� � : Rotating and sliding

a g� : Translation

251


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

202

3025

3

2 025

32 6 3765

= =

- = - ÊËÁ

ˆ¯̃

= -

=

kmph m/s

m/s

2

22: ( . )

55 4453. m s = 5 45. m �

252


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

aL

=

=

1

12

2

2

a

+S SM MA A= ( ) :eff mg

LI ma

L

2 2cos ( )b aÊ

ËÁˆ¯̃

= +

1

2

1

12 2 2

1

3

2

2

mgL mL mL 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 mL

mL g

LS = = ÊËÁ

ˆ¯̃2 2

3

2a b b bsin

cossin

F mgA = 3

4sin cosb b (2)

+S SF Fy y= ( ) :eff N mg ma mL

A - = - = - ÊËÁ

ˆ¯̃

cos cosb a b2

N mg mL g

L

N mg

A

A

- = - ÊËÁ

ˆ¯̃

= -ÊËÁ

ˆ¯̃

2

3

2

13

42

coscos

cos

b b

b (3)

253



For b = ∞70 :

(a) Eq. (1): a = ∞3

2

70g

L

cos a = 0 513.

g

L �

(b) Eq. (3): N mgA = - ∞ÊËÁ

ˆ¯̃

13

4702cos NA mg= 0 912. ≠ �

(c) Eq. (2): F mgA = ∞ ∞3

470 70sin cos FA mg= 0 241. Æ �

254


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 yL= +2

a b

+ 02

= -aL

y a bcos

a yL=2

a bcos

+S SF Fy y= ( ) :eff mg A ma mL

y- = = ÊËÁ

ˆ¯̃2

a bcos (1)

+ S SM M AL

I mLG G= ÊËÁ

ˆ¯̃

= =( ) : coseff 2

1

122b a a

AmL=6

abcos

(2)

Substitute (2) into (1):

mgmL

mL

gL L

gL

- =

= +ÊËÁ

ˆ¯̃

= +Ê

6 2

2 6

63

1

ab

a b

bb

a

bb

coscos

coscos

coscosËËÁ

ˆ¯̃

= +Ê

ËÁˆ

¯̃

a

bb

agL

6

3 12cos

cos

a bb

=+

Ê

ËÁˆ

¯̃6

1 3 2

g

L

cos

cos

255



a = = ◊+

Ê

ËÁˆ

¯̃=

+

Ê

ËÁˆL L g

Lg

2 2

6

1 33

1 32

2

2a b b

bb b

bcos

cos

coscos

cos

cos ¯̃̄¨

A = ◊ = ◊ ◊+

Ê

ËÁˆ

¯̃=

+mL mL g

Lmg

6 6

6

1 3

1 1

1 32 2

ab

bb 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=+ ∞

370

1 3 70

2

2

cos

cos a = 0 260. g Ø �

(c) A mg=+ ∞

1

1 3 702cos A = 0 740. mg ≠ �

256


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

4Ca ¨; ay C= 1

2a Ø (1)

Kinetics: I m CC

mC= + ÊËÁ

ˆ¯̃

È

ÎÍÍ

˘

˚˙˙

=1

12 2

5

482

22

(a) +

S SM M WC

I maC

maC

A A x y= ÊËÁ

ˆ¯̃

= + ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

( ) : ( ) ( )eff 2 4 2a

1

2

5

48

1

4 4

1

2 22mgC mC m C

Cm C

C= + ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

ÊËÁ

ˆ¯̃

a a a

1

2

20

481 22mgC mC

g

C= =a a . a = 1 2.

g

C �

257



(b) From (1): a C g gx x= = =1

4

1

41 2 0 3a ( . ) .a ¨

a C g gy y= = =1

2

1

21 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

2a a Ø (2)

Kinetics: I m CC I

mC= + ÊËÁ

ˆ¯̃

È

ÎÍÍ

˘

˚˙˙

=1

12 2 482

22

+ S SF Fx x= ( ) :eff 0

02=

=ma

ax

+S SM M W C I ma CA A y= Ê

ËÁˆ¯̃

= + ( )ÊËÁ

ˆ¯̃

( ) :eff1

2

1

2a

Recalling (1): 1

2

5

48

1

2

1

22mgC mC mC C= + Ê

ËÁˆ¯̃

ÊËÁ

ˆ¯̃

a a

1

2

17

48

24

172mgC mC

g

C= =a a a = 24

17

g

C �

a C Cg

Cgy = = Ê

ËÁˆ¯̃

=1

2

1

2

24

17

12

17a a = 12

17g Ø �

258



(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.


+S SM M mg C IG G= Ê

ËÁˆ¯̃

ÊËÁ

ˆ¯̃

=( ) :eff1

2

1

2a

1

4

5

482mgC 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


PROBLEM 16.160

The slender bar AB of weight W is held in equilibrium by two counterweights each weighing 1

2W . 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 ≠ ] + ØÈÎÍ

˘˚̇

L

2a [ ] [a aB AØ = ≠] [+ La Ø]

a = -ÊËÁ

ˆ¯̃

1

2L 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= - = =( ) :eff1

2

1

2

1

2

1

21

2

mg T ma

T m g a

A

A

- =

= -( ) (1)

260



Kinetics: Bar AB

+ S SF F mg T my y= - =( ) :eff a

mg m g a m L aA A- - = -ÊËÁ

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

LIG G= =( ) :eff 2a

1

2 2

1

122m g a

LmLA( )- = a

3 3g a LA- = a (3)

Add Eqs. (2) and (3): 4 22

g Lg

L= =a a

(a) Acceleration at Point A.

Substitute into (2): g a Lg

LA+ = ÊËÁ

ˆ¯̃

32

aA g= 1

3≠ �

a L a Lg

LgA= -Ê

ËÁˆ¯̃

= ÊËÁ

ˆ¯̃

-1

2

1

2

2 1

3a a = 2

3g Ø

(b) Acceleration at Point B. a L a Lg

LgB A= - = -Ê

ËÁˆ¯̃

( )a 2 1

3 aB g= 5

3Ø �

261


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 w2 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

.

.

.

aw

+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

C2 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

+

- = + + +

rR22

a = - 23 68. rad/s2 a = 23 7. rad/s2 �

262


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

lBl I ma

lA A AB= Ê

ËÁˆ¯̃

- = + ÊËÁ

ˆ¯̃

( ) :eff 2 2a

1

2

1

12 2 22mgl Bl mL m

l lAB AB- = + Ê

ËÁˆ¯̃

a a

1

2

1

32mgl Bl ml AB- = a (1)

Rod CD: ( )w = 0

+ S SM M mgl

Bl

I mal

D D CD CD= ÊËÁ

ˆ¯̃

+ ÊËÁ

ˆ¯̃

= +( ) :eff 2 2 2a

1

2

1

2

1

12 2 22mgl Bl ml m

l lCD CD+ = + Ê

ËÁˆ¯̃

a a

Multiply by 2: mgl Bl ml CD+ = 2

32a (2)

Add (1) and (2): 3

2

1

3

2

32mgl ml AB CD= +Ê

ËÁˆ¯̃

a a (3)

Multiply by 3: a aAB CDg

l+ =2

9

2 (4)

263



Kinematics: We must have

(aB Ø) (= ¢aB Ø)

ll

AB CDa a=2

l AB CDa a= 1

2 (5)

Substitute for a AB from (5) into (4)

1

22

9

2a aCD CD

g

l+ =

5

2

9

21 8a aCD CD

g

l

g

l= =; . (6)

(a) Acceleration of C: a l lg

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 mlg

l

B mg mg

+ = ÊËÁ

ˆ¯̃

= -

2

31 8

1 2

2 .

.

(on rod AB): B = 0 2. mg≠ �

264


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: AGL L

a ABL

G A= = = =2

22 2

/ ( )a a


aBØ = aA ¨ + aB A/ 60°

aBØ = aA ¨+ La 60°

a = aA¨+ aG A/ 15°

a = ∞La sin 30 ¨+ La2

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 A2 2 2

2 2 2

2 15

0 5 0 707

2 0 5

= ◊ - ∞

= +-

/ / cos

( . ) ( . )

( . )(

a aa 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



Law of sines

a a a

a

L

LG A G A

sin sin; sin sin

.

.sin

/ /

1515

0 707

0 2595015

∞= = ∞ = ∞

bb 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.

MGL

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

62

266




+S SM M mg L I ma LA A= = +( ) : ( . ) ( )( . )eff 0 183 0 2588a

0 1831

60 2588 0 2588

0 1831

60 06698

2

2

. ( . )( . )

. .

mgL mL m L L

gL L

= +

= +Ê

a a

aËËÁ

ˆ¯̃

= =0 183 0 2336 0 7834. . ; .g

L

g

La a

a = 0 78349 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 Lg

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


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

AGL=2

15∞ =aG A AG/ ( )a 15∞

a Ø = aA ¨ +aG A/ 15∞

a = 0 183. La 15∞

Kinetics I mL= 1

62

268



(a) Angular acceleration.+

S SM M mg AG I ma AGA A= ∞ = + ∞( ) : ( )sin ( )sineff 15 15a

mgL

mL m LL

g

L

215

1

60 183

215

0 1831

6

2ÊËÁ

ˆ¯̃

∞ = + ÊËÁ

ˆ¯̃

∞

=

sin ( . ) sin

.

a a

++ÊËÁ

ˆ¯̃

=

=

=

0 033494

0 183 0 2002

0 9143

0 91439 81

0

.

. .

.

..

.

g

Lg

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 Lg

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 ≠ �

vector mechanics ch 16 v.2011.06.11 02 - wordpress.com · chapter 16. 3 proprietary material. ... 8...

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