Capitulo 15 by Govata

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Transcript of Capitulo 15 by Govata

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 1.

Angular coordinate: Angular velocity:

= 8t 3 6 ( t 2 ) radians= =d = 24t 2 12 ( t 2 ) rad/s dt d = 48t 12 rad/s 2 dt

2

Angular acceleration: (a) When the angular acceleration is zero.

48 t 12 = 0(b) Angular coordinate and angular velocity at t = 0.250 s.

t = 0.250 s

= ( 8 )( 0.250 ) ( 6 )( 0.250 2 )2

3

2

= 18.25 radians = 22.5 rad/s

= ( 24 )( 0.250 ) (12 )( 0.250 2 )

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 2.

= 0.5 e3 t cos 4 t = =d = 0.5 3 e 3 t cos 4 t 4 e3 t sin 4 t dt

(

)

d = 0.5 9 2e3 t cos 4 t + 12 2e3 t sin 4 t + 12 2e3 t sin 4 t 16 2e 3 t cos 4 t dt

( = 0.5 ( 24 e

)

2 3 t

sin 4 t 7 2e 3 t cos 4 t

)

(a)

t = 0,

= ( 0.5 ) = ( 0.5 )( 3 ) = 4.71 = ( 0.5 ) 7 2 = 34.5

= 0.500 rad = 4.71 rad/s = 34.5 rad/s 2

(

)

(b) t = 0.125 s,

cos 4 t = cos

2

= 0,

sin 4 t = sin

2

=1

e3 t = 0.30786

= ( 0.5 )( 0.30786 )( 0 ) = 0 = ( 0.5 )( 0.30786 )( 4 ) = 1.93437 = ( 0.5 )( 0.30786 ) 24 2 = 36.461

=0 = 1.934 rad/s = 36.5 rad/s 2

(

)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 3.

= 0e7 t/6 sin 4 t

=

d 7 7 t/6 = 0 e sin 4 t + 4 e7 t/6 cos 4 t dt 6 49 2 7 t/6 d 28 2 7 t/6 28 2 7 t/6 = 0 e sin 4 t e cos 4 t e cos 4 t 16 2e7 t/6 sin 4 t 36 dt 6 6 49 2 28 2 cos 4 t = 0e 7 t/6 16 sin 4 t + 36 3

=

(a) 0 = 0.4 rad,

t = 0.125 se7 (0.125)/6 = 0.63245, 4 t =

2

, sin

2

= 1, cos

2

=0

= ( 0.4 )( 0.63245 )(1) = 0.25298 radians = ( 0.4 )( 0.63245) 7 (1) = 0.92722 rad/s 6

= 0.253 rad = 0.927 rad/s

= ( 0.4 )( 0.63245 ) 16 (b) t = ,e7 t/6 = 0

49 2 2 (1) = 36.551 rad/s 36

= 36.6 rad/s 2 =0 =0 =0

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 4.

Angular coordinate: Initial angular velocity: Angular acceleration:

= 1800 rev = 3600 radians 0 = 6000 rpm = 200 rad/s =d d = = constant dt d

d = d

0 d = d 02 = 0

0

1 2

( 200 ) = 17.4533 rad/s2 2 = 0 = 2 ( 2 )( 3600 )(a) Time required to coast to rest.

2

= 0 + tt=

0 0 200 = 17.4533

t = 36.0 s

(b) Time to execute the first 900 revolutions.

= 900 rev = 1800 radians = 0t + t 21800 = 200 t 1 (17.4533) t 2 2 1 2

t 2 72t + 648 = 0t= 72

( 72 )2 ( 4 )( 648)2

t = 10.54 s

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 5.

1 = 2400 rpm =(a )

( 2400 )( 2 )60

= 80 rad/s,

0 = 0, t1 = 4 s80 = 20 rad/s 2 4

1 = 0 + t = t , 1 = 0t +

=

1t

=

1 2 1 160 t = 0 + ( 20 ) 42 = 160 rad = = 80 rev 2 2 2

( )

1 = 80 rev(b)

1 = 80 rad/s,

2 = 0, =1 2

t2 t1 = 40 s

2 = 1 + ( t2 t1 ) ,

2 1t2 t12

=

0 80 = 2 rad/s 2 40

2 1 = 1 ( t2 t1 ) + ( t2 t1 ) = ( 80 )( 40 ) += 1600 radians =1600 = 800 rev 2

1 ( 2 )( 40 )2 2

2 1 = 800 rev

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 6.

Angular acceleration: Angular velocity:

= 30 e 0.2t =t

d dt

= 0 + 0 dt= 0 + 30 0 e0.2t dt =30 0.2t t e 0 0.2 t

= 150 1 e 0.2t =When t = 0.5 s,

(

)

d dt

= 150 1 e( 0.2 )( 0.5)

(

) = 14.27 rad/s

Angular coordinate:

= 0 + 0 dt= 0 + 150 0 1 e0.2t dt = 150t 150 0.2t t e 0 0.2 t

t

(

)

= 150t 750 1 e 0.2tWhen t = 0.5 s,

(

)

= (150 )( 0.5 ) 750 1 e ( 0.2 )( 0.5)

(

) = 3.63 radians

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 7.

(a )

= 0.5 ,Integrating,0 30 d = 0.5 0 d

d = 0.5 d

d = 0.5d

30 = 0.560 = 9.55 rev 2

= 60 radians =(b)Integrating,

= 9.55 rev

d = 0.5 dtt 0 0 dt = 2 30

dt = 2d 0 = 30

d

t = 2 ln

t =

(c)

= ( 0.02 )( 30 ) = 0.6 rad/st 0.6 0 dt = 2 30

d

t = 2 ln

0.6 = 2 ln 50 30

t = 7.82 s

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 8.

= kIntegrating,

d = k d

d = k d

0 6 12 d = k 0 d

0

62 122 = k 2 0 2

(a )

k =

122 = 9 s 2 62

k = 9.00 s 2

3 12 d = k 0 d

22

32 122 = 9 0 2 2 2

(b)

2 = 122 ( 9 )( 3) = 63 rad 2/s 2

= 7.94 rad/s

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 9.

rA/O = ( 5 in.) i + ( 31.2 in.) j + (12 in.) k rB/O = ( 5 in.) i + (15.6 in.) jlOA =Angular velocity.

( 5)2 + ( 31.2 )2 + (12 )2lOA rA/O =

= 33.8 in.

=

6.76 ( 5i + 31.2 j + 12k ) 33.8

= (1.0 rad/s ) i + ( 6.24 rad/s ) j + ( 2.4 rad/s ) kVelocity of point B.

v B = rB/Oi j k v B = 1.0 6.24 2.4 = 37.44i + 12 j 15.6k 5 15.6 0

v B = ( 37.4 in./s ) i + (12.00 in./s ) j (15.60 in./s ) kAcceleration of point B.

aB = vB

i j k 2.4 = 126.1i 74.26 j + 245.6k a B = 1.0 6.24 37.4 12 15.60

a B = 126.1 in./s 2 i 74.3 in./s 2 j + 246 in./s 2 k

(

) (

) (

)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 10.

rA/O = ( 5 in.) i + ( 31.2 in.) j + (12 in.) k rB/O = ( 5 in.) i + (15.6 in.) jlOA =

( 5)2 + ( 31.2 )2 + (12 )2lOA rA/O =

= 33.8 in.

Angular velocity.

=

3.38 ( 5i + 31.2j + 12k ) 33.8

= ( 0.5 rad/s ) i + ( 3.12 rad/s ) j + (1.2 rad/s ) kVelocity of point B.

v B = rB/Oi j k v B = 0.5 3.12 1.2 = 18.72i + 6 j 7.80k 5 15.6 0

v B = (18.72 in./s ) i + ( 6.00 in./s ) j ( 7.80 in./s ) kAngular Acceleration.

=

lOA

rA/O =

5.07 ( 5i + 31.2 j + 12k ) 33.8

= 0.75 rad/s 2 i 4.68 rad/s 2 j 1.8 rad/s 2 kAcceleration of point B.

(

) (

) (

)

a B = rB/O + v Bj k i a B = 0.75 4.68 1.8 5 15.6 0 + i j k 0.5 3.12 1.2 18.72 6 7.8

= 28.08i 9 j + 11.7k 31.536i 18.564 j + 61.406k

a B = 3.46 in./s 2 i 27.6 in./s 2 j + 73.1 in./s 2 k

(

) (

) (

)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 11.

rB/ A = ( 500 mm ) i ( 225 mm ) j + ( 300 mm ) k = ( 0.5 m ) i ( 0.225 m ) j + ( 0.3 m ) k

l AB =Angular velocity vector.

0.52 + 0.2252 + 0.32 = 0.625 m

=

l AB

rB/ A =

10 ( 0.5i 0.225j + 0.3k ) 0.625

= ( 8 rad/s ) i ( 3.6 rad/s ) j + ( 4.8 rad/s ) krE/B = ( 300 mm ) k = ( 0.3 m ) kVelocity of E.

v E = rE/B

i j k = 8 3.6 4.8 = 1.08i + 2.4 j 0 0 0.3

v E = (1.080 m/s ) i + ( 2.40 m/s ) jAcceleration of E.

aE = vE =

i j k 8 3.6 4.8 = 11.52i + 5.184 j + 23.088k 1.08 2.4 0

a E = 11.52 m/s 2 i + 5.18 m/s 2 j + 23.1 m/s 2 k

(

) (

) (

)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 12.

rB/ A = ( 500 mm ) i ( 225 mm ) j + ( 300 mm ) k = ( 0.5 m ) i ( 0.225 m ) j + ( 0.3 m ) k

l AB =Angular velocity vector.=

0.52 + 0.2252 + 0.32 = 0.625 m10 ( 0.5i 0.225j + 0.3k ) 0.625

l AB

rB/ A =

= ( 8 rad/s ) i ( 3.6 rad/s ) j + ( 4.8 rad/s ) kAngular acceleration vector.

=

l AB

rB/ A =

20 ( 0.5i 0.225j + 0.3k ) 0.625

= 16 rad/s 2 i + 7.2 rad/s 2 j 9.6 rad/s 2 k

(

) (

) (

)

Velocity of C.

vC = rC/BrC/B = ( 500 mm ) i = ( 0.5 m ) i

i j k vC = 8 3.6 4.8 = 2.4 j 1.8k 0.5 0 0

vC = ( 2.40 m/s ) j (1.800 m/s ) k Acceleration of C. aC = rC/B + vCj k i = 16 7.2 9.6 0.5 0 0 = 18i + 19.2 j 15.6k + i j k 8 3.6 4.8 0 2.4 1.8

= 4.8j + 3.6k + 18i + 14.4 j 19.2k

aC = 18.00 m/s 2 i + 19.20 m/s 2 j 15.60 m/s 2 k

(

) (

) (

)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 13.

rA/D = ( 200 mm ) i + (120 mm ) j + ( 90 mm ) k

d DA = DA =

( 200 )2 + (120 )2 + ( 90 )2

= 250 mm

rA/D = 0.8 i + 0.48j + 0.36k d DA

= DA = ( 75 )( 0.8i + 0.48 j + 0.36k ) = ( 60 rad/s ) i + ( 36 rad/s ) j + ( 27 rad/s ) k =

d DA = 0 dt

rB/A = ( 200 mm ) i = ( 0.2 m ) iVelocity of corner B.

v B = rB/ A

i j k = 60 36 27 = 5.4 j 7.2k 0.2 0 0

v B = ( 5.40 m/s ) j ( 7.20 m/s ) k Acceleration of corner B. a B = rB/ A + v Bi j k = 0 + 60 36 27 = 405i 432 j + 324k 0 5.4 7.2

a B = 405 m/s 2 i 432 m/s 2 j + 324 m/s 2 k

(

) (

) (

)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 14.

rA/D = ( 200 mm ) i + (120 mm ) j + ( 90 mm ) k

d AD = DA =

( 200 )2 + (120 )2 + ( 90 )2

= 250 mm

rA/D = 0.8 i + 0.48j + 0.36k d AD

= DA = ( 75 )( 0.8i + 0.48 j + 0.36k ) = ( 60 rad/s ) i + ( 36 rad/s ) j + ( 27 rad/s ) k =

d DA = ( 600 )( 0.8i + 0.48j + 0.36k ) dt= 480 rad/s 2 i 288 rad/s 2 j 216 rad/s 2 k

(

) (

) (

)

rB/A = ( 200 mm ) i = ( 0.2 m ) iVelocity of corner B.

v B = rB/ A

i j k = 60 36 27 = 5.4 j 7.2k 0.2 0 0

v B = ( 5.40 m/s ) j ( 7.20 m/s ) k Acceleration of corner B. a B = rB/ A + v Bj k i j k i = 480 288 216 + 60 36 27 0.2 0 0 0 5.4 7.2 = 43.2 j 57.6k 405i 432 j 324k

a B = 405 m/s 2 i 389 m/s 2 j 266 m/s 2 k

(

) (

) (

)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 15.

93,000,000 mi = 491.04 109 ft 365.24 days = 31.557 106 s, 1 rev = 2 radAngular velocity.

=Velocity of the earth.

2 = 199.11 109 rad/s 31.557 106

v = r = 491.04 109 199.11 109 = 97.77 103 ft/s

(

)(

)

v = 66.7 103 mi/hAcceleration of the earth.

a = r 2 = 97.77 103 199.11 109

(

)(

)

2

= 19.47 103 ft/s 2

a = 19.47 103 ft/s 2

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 16.

23 h 56 min = 23.933 h = 86.16 103 s,

1 rev = 2 rad

=

2 = 72.925 106 rad/s 86.16 103

R = 6370 km = 6.37 106 m = j,

r = R cos i + R sin j

v = r = R cos k

= 72.925 106 6.37 106 cos k = ( 464.53cos m/s ) k

(

)(

)

a = v p = j ( R cos ) i = 2 R cos i = 33.876 103 cos m/s 2 i

(

)

(a) Equator.

(

= 0 )

cos = 1.000v = 465 m/s

v = ( 465 m/s ) ka = 33.9 103 m/s 2 i

(

)

a = 0.0339 m/s 2 cos = 0.76604v = 356 m/s

(b) Philadelphia.

(

= 40 )

v = ( 464.52 )( 0.76604 ) k = ( 356 m/s ) ka = 33.876 103 ( 0.76604 ) i3 2

( ) = ( 0.273 10 m/s ) i(

a = 0.0259 m/s 2 cos = 0v=0a=0

(c) North Pole.

= 90 )

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 17.

vB = v A = 300 mm/s

rB = 120 mm

( aB )t(a )vB = rB ,

= a A = 180 mm/s

=

vB 300 = = 2.5 rad/s rB 120 =

= 2.50 rad/s

( aB )t(b)

= rB ,

=

( aB )trB

180 = 1.5 rad/s 2 1202

= 1.500 rad/s 2

( aB ) n

= rB 2 = (120 )( 2.5 ) = 750 mm/s 22 ( aB )t2 + ( aB )n

aB =

=

(180 )2 + ( 750 )2

= 771 mm/s 2

tan =

750 , 180

= 76.5

a B = 771 mm/s 2

76.5

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 18.

B = 4 rad/s ,

rB = 120 mm2

( aB ) n

2 = rB B = (120 )( 4 ) = 1920 mm/s 2

aB = 2400 mm/s 2

( aB )t

=

2 aB ( aB ) n =

2

24002 19202 = 1440 mm/s 2

( aB )t

= rB ,

=

( aB )trB

=

1440 = 12 rad/s 2 120

12.00 rad/s 2

or

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 19.

Let vB and aB be the belt speed and acceleration. These are given as vB = 12 ft/s (a) Angular velocity and angular acceleration of each pulley. Pulley A.

and

aB = 96 ft/s 2.

These are also the speed and tangential acceleration of periphery of each pulley provided no slipping occurs.

rA = 8 in. = 0.6667 ft

A = A =Pulley C.

v A vB 12 = = = 18 rad/s rA rA 0.6667 a A aB 96 = = = 144 rad/s 2 0.6667 rA rArC = 5 in. = 0.41667 ft

A = 18 rad/s A = 144 rad/s 2

C = C =

vC vB 12 = = = 28.8 rad/s rC rC 0.41667 aC aB 96 = = = 230.4 rad/s 2 rC rC 0.41667

C = 28.8 rad/s C = 230 rad/s 2

(b) Acceleration of point P on pulley C.

C = 5 in. = 0.41667 ft

( aP )t( aP ) n =aP =

= aB = 96 ft/s 2=

C

2 vP

=

C

2 vB

(12 )20.41667=

= 345.6 ft/s 2= 358.7 ft/s 2

2 ( aP )t2 + ( aP )n

(96)2 + ( 345.6)2

tan =

96 345.6

= 15.52a P = 359 ft/s 215.52

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 20.1 rev = radians, 2

rA = 8 in. = 0.6667 ft, d , d

rC = 5 in. = 0.41667 ft=

= 120 0.002 2 =

d120 0.002

2

500 d = d 60000 2

Integrating and applying initial condition = 0 at = 0 and noting that = radians at the final state, 2 0 60000 2 = 250 ln ( 60000 ) 0 = 0 d =

500 d

60000 250 ln 60000 2 ln 60000 = 250 ln 60000

(

)

2

=

60000 2 = e /250 60000

2 = 60000 1 e /250 = 749.26 rad 2 /s 2 = 27.373 rad/s = 120 0.002 2 = 120 ( 0.002 )( 749.26 ) = 118.50 rad/s(a) Tangential velocity and acceleration of point B on the belt.

vB = v A = rA = ( 0.6667 )( 27.373) = 18.249 ft/saB = a A = rA = ( 0.6667 )(118.50 ) = 79.0 ft/s 2

aB = 79.0 ft/s 2(b) Acceleration of point P on pulley C.

C = 5 in. = 0.41667 ftvP = vB = 18.249 ft/s

( a P )n

=

C

2 vB

=

18.2492 = 799.3 ft/s 2 0.41667

( a P )taP =

= aB = 79.0 ft/s 2 = 803.2 ft/s 2

( 799.3)2 + ( 79.0 )279 , 799.3

tan =

= 5.64a P = 803 ft/s 25.64

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 21.

Left pulley. Inner radius Outer radius r1 = 50 mm r2 = 100 mm

v A = 0.6 m/s = 600 mm/s

1 =

v A 600 = = 6 rad/s r2 100

Speed of intermediate belt.

v1 = r11 = ( 50 )( 6 ) = 300 mm/sRight pulley.

Inner radius Outer radius

r3 = 50 mm r4 = 100 mm

2 =

v1 300 = = 3 rad/s r4 100

(a) Velocity of C.

vC = r3 2 = ( 50 )( 3) = 150 mm/svC = 0.1500 m/s(b) Acceleration of point B.2 aB = r4 2 = (100 )( 3) = 900 mm/s 2 2

a B = 0.900 m/s 2

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 22.

Left pulley. Inner radius Outer radius r1 = 50 mm r2 = 100 mm

v A = 0.6 m/s = 600 mm/s

( a A )t

= 1.8 m/s = 1800 mm/sv A 600 = = 6 rad/s r2 100

1 = 1 =Intermediate belt.

( a A )tr2

=

1800 = 18 rad/s 2 100

v1 = r11 = ( 50 )( 6 ) = 300 mm/s

( a1 )tRight pulley.

= r11 = ( 50 )(18 ) = 900 mm/s 2

Inner radius Outer radius

r3 = 50 mm r4 = 100 mm

2 =

v1 300 = = 3 rad/s r4 100

2 =(a) Velocity and acceleration of point C.

( a1 )tr4

=

900 = 9 rad/s 2 100

vC = r3 2 = ( 50 )( 3) = 150 mm/svC = 0.150 m/s

( aC )t

= r3 2 = ( 50 )( 9 ) = 450 mm/saC = 0.450 m/s

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

(b) Acceleration of point B.2 ( aB )n = r4 2 = (100 )( 3)2 = 900 mm/s2

( a B )n = 0.900 m/s2 ( aB )t= r4 2 = (100 )( 9 ) = 900 mm/s 2

( a B )t

= 0.900 m/s 2

a B = 1.273 m/s 2

45

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 15, Solution 23.

(a) Let point C be the point of contact between the shaft and the ring.

vC = r1 A

B =

vC r = 1 A r2 r2

B =(b) On shaft A:2 a A = r1 A

r1 A r2

2 a A = r1 A

On ring B :

aB =

2 r2 B

r = r2 1 A r2

2

aB =

2 r12 A r2

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Chapter 15, Solution 24.

(a) Let point C be the point of contact between the shaft and the ring.

vC = r1 A = ( 0.5 )( 25 ) = 12.5 in./s

B =(b) On shaft A:

vC 12.5 = = 5.0 rad/s 2.5 r22

B = 5.00 rad/s

2 a A = r1 A = ( 0.5 )( 25 )

= 312.5 in./s 2 ,

a A = 26.0 ft/s 22

On ring B :

2 aB = r2 B = ( 2.5 )( 5.0 )

= 62.5 in./s 2 ,

a B = 5.21 ft/s 2r = r3 = 3.5 in.

(c) At a point on the outside of the ring,2 a = r B = ( 3.5 )( 5.0 ) = 87.5 in./s 2 2

a = 7.29 ft/s 2

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Chapter 15, Solution 25.

(a )

A = 600 rpm =

( 600 )( 2 )60

= 20 rad/s.

Let points A, B, and C lie at the axles of gears A, B, and C, respectively. Let D be the contact point between gears A and B.

vD = rD/ A A = ( 2 )( 20 ) = 40 in./s

B =

vD 40 60 = = 10 rad/s = 10 = 300 rpm rD/B 4 2

B = 300 rpmLet E be the contact point between gears B and C.vE = rE/B B = ( 2 )(10 ) = 20 in./s

C =

vE 20 60 = = 3.333 rad/s = ( 3.333 ) = 100 rpm rE/C 6 2

C = 100 rpm(b) Accelerations at point E.2 ( 20 ) = 1973.9 in./s2 vE = 2 rE/B 2

On gear B :

aB =

a B = 1974 in./s 2On gear C :

( 20 ) = 658 in./s 2 v2 aC = E = 6 rE/CaC = 658 in./s 2

2

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Chapter 15, Solution 26.

(a) At time t = 2 s,

A = 600 rpm =

( 600 )( 2 )60

= 20 rad/s

A = At ,

A =

At

= 10 rad/s 2

Let D be the contact point between gears A and B.

( aD )tB =

= rD/ A A = ( 2 )(10 ) = 20 in./s 2=

( aD )trD/B

20 = 5 rad/s 2 4

B = 15.71 rad/s 2

Let E be the contact point between gears B and C.

( aE )tC =(b) At t = 0.5 s.

= rE/B B = ( 2 )( 5 ) = 10 in./s 210 = 1.6667 rad/s 2 6

( aE )trE/C

=

C = 5.24 rad/s 2

For gear B, B = Bt = ( 5 )( 0.5 ) = 2.5 rad/s

( a E )n

2 = rE/B B = ( 2 )( 2.5 ) = 123.37 in./s 2

2

( a E )t

= 10 in./s 2

= 31.416 in./s 2=

aE =

( aE )2 + ( aE )t2 n31.416 , 123.37

(123.37)2 + ( 31.416)2

= 127.3 in./s 2

tan =

= 14.29

a E = 127.3 in./s 2

14.29

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For gear C,

C = C t = (1.6667 )( 0.5 ) = 0.83333 rad/s

( aE ) n

2 = rE/CC = ( 6 )( 0.83333 ) = 41.123 in./s 2

2

( aE )t

= 31.416 in./s 2

aE =

( aE )2 + ( aE )t2 n

=

( 41.123)2 + (31.416)2

= 51.75 in./s 2

tan =

31.416 = 37.4, 41.123

a E = 51.8 in./s 2

37.4

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Chapter 15, Solution 27.

(a) For the pulley,

r =

1 d, 2

rA =

1 ( 0.3) = 0.15 m 2

rB =

1 ( 0.2 ) = 0.1 m 2

v A = rA ,

vB = rB

v A/B = v A vB = ( rA rB ) v A/B rA rB0.8 = 16 rad/s 0.15 0.1 0.4 = 8 rad/s 0.15 0.1

=At t = 0, At t = 0.25 s,

0 = 1 = = 1 2t

=

8 16 = 32 rad/s 2 0.25

= 32 rad/s 2

a A = rA a B = rB

= ( 0.15 )( 32 ) = 4.80 m/s 2 = ( 0.1)( 32 ) = 3.2 m/s 2

a A/B = a A a B = 1.6 m/s 2

a A/B = 1.600 m/s 2

(b)

x A/B = v A/B

(

)0 t + 1 (a A/B ) t 2 21 (1.6 )( 0.25)2 = 0.15 m 2

= ( 0.8 )( 0.25 ) +

x A/B = 150.0 mm

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Chapter 15, Solution 28.

For the pulley,

r =

1 d, 2

rA = rB =

1 ( 0.3) = 0.15 m 2 1 ( 0.2 ) = 0.1 m 2vB = rB v A/B = ( rA rB )

v A = rA ,

=At = 0,

v A/B rA rB

0 = = =d d = dt d

0.9 = 18 rad/s 0.15 0.1 0.45 = 9 rad/s 0.15 0.1

At =

1 rev = radians, 2

d = d = 38.675 rad/s 2

9 d 18

= 0 d

92 182 = ( )( ) 2 2

= 38.7 rad/s 2a A = rA = 0.15 ( 38.675 ) = 5.8012 m/s 2 a B = rB = 0.1( 38.675 ) = 3.8675 m/s 2 or or 5.8012 m/s 2 3.8675 m/s 2

(a )

a A/B = a A a B = 1.9337 m/s 2

or

1.9337 m/s 2a A/B = 1.934 m/s 2

(b)

x A/B = v A/B

(

)0 t + 1 (a A/B ) t 2 21 ( 1.9337 )( 0.3)2 = 0.1830 m 2

= ( 0.9 )( 0.3) +

x A/B = 183.0 mm

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Chapter 15, Solution 29.

(a) Motion of pulley.

( v E )0 = ( v A )0 = 8 in./sFixed axis rotation.

( a E )t

= a A = 10 in./s 2

( vE )0 = rA0( aE )tSince is constant,

0 = =

( vE )0rA

= =

8 in./s = 2 rad/s 4 in.

= rA

( a E )trA

10 in./s 2 = 2.5 rad/s 2 4 in.

= 0 + t = 2 + 2.5t = 0 + 0t + t 2 = 0 + 2t + 1.25t 2For t = 3s,

1 2

= 2 + ( 2.5 )( 3) = 9.5 rad/s = 0 + ( 2 )( 3) + (1.25 )( 3) = 17.25 rad2

In revolutions, (b) Motion of load B.

=

17.25 2

= 2.75 rev

t = 3s

vB = rB = ( 6 )( 9.5 )yB = rB = ( 6 )(17.25 )(c) Acceleration of point D. t=0

v B = 57.0 in./s

yB = 103.5 in. = 2.5 rad/s 2

= 0 = 2 rad/s

( a D )t

= rD = ( 6 )( 2.5 ) = 15 in./s 2

( a D )n = rD 2 = ( 6 )( 2 )2 = 24 in./s2a D = 28.3 in./s 232.0

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Chapter 15, Solution 30.

= 2.4 rad/s 2Use equations for constant angular acceleration.

0 = 0

= 0 + t = 2.4t = 0 + 0t + t 2 = 1.2t 2At t = 4s,

1 2

= ( 2.4 )( 4 ) = 9.6 rad/s = 1.2 ( 4 ) = 19.2 rad(a) Load A. at t = 4s,2

rA = 4 in.

v A = rA = ( 4 in.)( 9.6 rad/s ) y A = rA = ( 4 in.)(19.2 rad )(b) Load B. at t = 4s,rB = 6 in.

v A = 38.4 in./s

y A = 76.8 in.

vB = rB = ( 6 in.)( 9.6 rad/s ) yB = rB = ( 6 in.)(19.2 rad )

v B = 57.6 in./s y B = 115.2 in.

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Chapter 15, Solution 31.

When contact is made,

A = 240 rpm = 8 rad/s

Let C be the contact point between the two gears.

vC = rA A = ( 0.15 )( 8 ) = 1.2 m/s 2

B =

vC 1.2 = = 6 rad/s 0.2 rB

A = 8 = t rad/s

B = 6 = ( t 2 ) rad/sSubtracting,

2 = ( )( 2 )

= rad/s 2 = 3.14 rad/s 2

(a ) (b)t = 8

=

8

= 8s

t = 8.00 s

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Chapter 15, Solution 32.

( A )0

= 240 rpm = 8 rad/s3

( A )1 = 8

A t1

3 1 1r 1 0.15 2 B = 4 = B t12 = A A t12 = A t1 2 2 rB 2 0.2

A t12 = ( 8 )

0.2 = 59.574 radians 0.15

3

( B )1 = B t1 = Let vC be the velocity at the contact point.

0.15 A t1 = 0.421875 A t1 0.2

3

vC = rA A = ( 0.15 )( 8 A t1 ) = 1.2 0.15 A t1and

vC = rB B = ( 0.2 )( 0.421875 A t1 ) = 0.084375 A t1

Equating the two expressions for vC ,

1.2 0.15 A t1 = 0.084375 A t1Then,

or

A t1 = 16.0850 rad/s

t1 =

A t12 59.574 = = 3.7037 s A t1 16.0850 A = 4.34 rad/s 2 B = 1.832 rad/s 2t1 = 3.70 s

(a )

A =

16.0850 = 4.3429 rad/s 2 3.70373

0.15 2 B = ( 4.3429 ) = 1.83218 rad/s 0.2

(b) From above,

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Chapter 15, Solution 33.

Motion of disk B.

( B )0 = 500 rpm = 52.360 rad/s B = ( B )0 + B t

Assume that the angular acceleration of disk B is constant.

At

t = 60 s,

B = 0B = B ( B ) 0t = 0 52.360 = 0.87266 rad/s 2 60

vB = rB B = ( 3)( 52.360 0.87266 t ) = 157.08 2.618 t in./sMotion of disk A.

( A )0 = 0,

A = 3 rad/s 2

A = ( A )0 + At = 3 tv A = rA A = ( 2.5 )( 3 t ) = 7.5 t in./sIf disks are not to slip,

v A = vB

7.5 t = 157.08 2.618 t(a) (b)

t = 15.52 s

A = ( 3)(15.52 ) = 46.6 rad/s B = 52.360 ( 0.87266 )(15.52 ) = 38.8 rad/s A = 445 rpm,

B = 371 rpm

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Chapter 15, Solution 34.

Wheel B. At Angular acceleration.

( B )0 = 300 rpm = 31.416 rad/st = 12s, B = 75 rpm = 7.854 rad/s=

7.854 31.416 = 1.9635 rad/s 12 t Velocity at contact point with disk A at t = 12 s:

B =

B ( B ) 0

vB = rB B = ( 3)( 7.854 ) = 23.562 in./s

( A )0 = 300 rpm = 31.416 rad/s Wheel A. Assume that slipping ends when t = 12 s.Then,

v A = 23.562 in./s

v A 23.562 = = 9.4248 rad/s rA 2.5 A = 9.4248 rad/s = 9.4248 rad/s

A =

A =(a)

A ( A ) 0t

=

9.4248 31.416 = 3.4034 rad/s 2 12 A = 3.40 rad/s 2 B = 1.963 rad/s 2

(b)

Time when A is zero.

A = ( A ) 0 + A t = 0

t=

A ( A )0 0 31.416 = A 3.4034

t = 9.23 s

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Chapter 15, Solution 35.

Let one layer of tape be wound and let v be the tape speed.

vt = 2 r

and

r = b

r bv b = = t 2 r 2For the reel:

d d v 1 dv d 1 = = +v dt dt r r dt dt r =

a v dr a v b 2 = 2 r r dt r r 2b 2 1 a =0 r 2

=

a=

2 b 0 2

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Chapter 15, Solution 36.

Let one layer of paper be unrolled.

vt = 2 r

and

r = b

dr r bv = = t 2 r dt

=

d d v 1 dv d 1 v dr = = +v =0 2 dt dt r r dt dt r r dt =bv 2 2 r 3

bv 2 v bv = 2 = 3 r 2 r 2 r

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Chapter 15, Solution 37.

Velocity analysis.v B = 150 mm/s vA = vA

15

v B/ A = 500

50

Plane motion = Translation with B + Rotation about B. v A = v B + v A/BDraw velocity vector diagram.

= 180 50 75 = 55Law of sines.

v A/B sin 75

=

vA vB = sin sin 50

(a )

v A/B =

vB sin 75 150 sin 75 = = 189.14 mm/s sin 50 sin 50v A/B l AB

=

=

189.14 = 0.378 rad/s 500

= 0.378 rad/s

(b)

vA =

vB sin 150 sin 55 = = 160.4 mm/s sin 50 sin 50v A = 160.4 mm/s

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Chapter 15, Solution 38.

Velocity analysis.v A = 225 mm/s

v B = vB

15 30

v B/ A = vB/ A

Plane motion = Translation with A + Rotation about A. v B = v A + v B/ ADraw velocity vector diagram.

= 180 60 75 = 45Law of sines.

vB/ A sin 75

=

vB v = A sin 60 sin

(a )

vB/ A =

v A sin 75 225sin 75 = = 307.36 mm/s sin sin 45vB/ A l AB

=

=

307.36 = 0.615 rad/s 500

= 0.615 rad/s

(b)

vB =

v A sin 60 225sin 60 = = 276 mm/s sin sin 45v B = 276 mm/s 15

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Chapter 15, Solution 39.

Geometry.

sin =tan =Velocity analysis.

12 , 2012 , 5

= 36.87 = 67.38

v A = 4.2 ft/s

v B/ A = r AB =v B = vB

20 AB 12

Plane motion = Translation with A + Rotation about A.

v B = v A + v B/ ADraw velocity vector diagram.

= 180 ( 90 ) = 59.49Law of sines.vB/ A sin = vB v = A sin ( 90 ) sin

vB/ A =

v A sin 4.2sin 67.38 = = 4.5 ft/s sin sin 59.49

(a )

AB =

4.5 = 2.7 rad/s 20 /12

AB = 2.70 rad/s(b)vB = v A cos 4.2cos 36.87 = = 3.90 ft/s sin sin 59.49v B = 3.90 ft/s 67.4

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Chapter 15, Solution 40.

Geometry.

sin =tan =Velocity analysis.

12 , 2012 , 5

= 36.87 = 67.38

AB = 4.2 rad/s 20 v B/ A = rB/ A AB = ( 4.2 ) = 7.0 ft/s 12 v B = vB v A = vAPlane motion = Translation with A + Rotation about A.

v B = v A + v B/ ADraw velocity vector diagram.

= 180 ( 90 ) = 59.49Law of sines.vB/ A vA vB = = sin sin ( 90 ) sin

(a )

vA =

vB/ A sin sin

=

7sin 59.49 = 6.53 ft/s sin 67.38

v A = 6.53 ft/s

(b)

vB =

vB/ A cos sin

=

7 cos 36.87 = 6.07 ft/s sin 67.38v B = 6.07 ft/s 67.4

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Chapter 15, Solution 41.

In units of m/s,

v B/ A = k rB/ A = k ( 0.6i + 0.6 j) = 0.6 i + 0.6 j

vC/ A = k rC/ A = k 1.2i = 1.2 j v B = v A + v B/ A7.4i + ( vB ) y j = ( v A ) x i 7 j 0.6 i + 0.6 j Components.

i : 7.4 = ( v A ) x 0.6j:

(1) (2)

( vB ) y

= 7 + 0.6

v C = v A + v C/ A1.4i + ( vC ) y j = ( v A ) x i 7 j + 1.2 j

Components.

i : 1.4 = ( v A ) xj:

(3) (4)

( vC ) y( vA )x

= 7 + 1.2 = 1.4 m/s = 10 rad/s, = 10.00 rad/s

From (3), (a) From (1), From (2), (b)

=

7.4 ( 1.4 ) 0.6

( vB ) y

= 7 + ( 0.6 )(10 ) = 1 m/s

v B = ( 7.40 m/s ) i (1.000 m/s ) j

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Chapter 15, Solution 42.

In units of m/s,

v B/ A = k rB/ A = k ( 0.6i + 0.6 j) = 0.6 i + 0.6 j

vC/ A = k rC/ A = k 1.2i = 1.2 j v B = v A + v B/ A7.4i + ( vB ) y j = ( v A ) x i 7 j 0.6 i + 0.6 j Components.

i : 7.4 = ( v A ) x 0.6j:

(1) (2)

( vB ) y

= 7 + 0.6

v C = v A + v C/ A1.4i + ( vC ) y = ( v A ) x i 7 j + 1.2 j Components.

i : 1.4 = ( v A ) xj:

(3) (4)

( vC ) y

= 7 + 1.2

From (3), From (1),(a)

( vA )x =

= 1.4 m/s,

v A = 1.4i 7 j = (10.00 rad/s ) k

7.4 ( 1.4 ) 0.6

= 10 rad/s,

vO = v A + vO/ A = v A + rO/ A = v A + 10k ( 0.6i )= 1.4i 7 j + 6 j = 1.4i 1j

vO = (1.400 m/s ) i (1.000 m/s ) j(b)

0 = vO + ( xi + yj) 0 = 1.4i 1j + 10k ( xi + yj) = 1.4i 1j + 10 xj 10 yj

Components.

i : 0 = 1.4 10 y, j : 0 = 1 + 10 x,

y = 0.14 m x = 0.1 m

y = 140.0 mm x = 100.0 mm

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Chapter 15, Solution 43.

In units of mm/s,

v B/ A = k rB/ A = k (125i + 75j) = 75 i + 125 j vC/ A = k rC/ A = k ( 50i + 150 j) = 150 i + 50 j

v B = v A + v B/ A

( vB ) x i 75j = 100i + ( vA ) y j 75 i + 125 jComponents.

i:

( vB ) x

= 100 75

(1) (2)

j : 75 = ( v A ) y + 125

v C = v A + v C/ A400i + ( vC ) y j = 100i + ( v A ) y j 150 i + 50 j Components. j: (a) (b)

i : 400 = 100 150

(3) (4) = ( 2 rad/s ) k

( vC ) y

= ( v A ) y + 125

From (3), From (2),

= 2 rad/s

( vA ) y

= 75 125 = 75 125 ( 2 ) = 175 mm/s

v A = (100.0 mm/s ) i + (175.0 mm/s ) j

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Chapter 15, Solution 44.

In units of mm/s,

v B/ A = k rB/ A = k (125i + 75j) = 75 i + 125 j

vC/ A = k rC/ A = k ( 50i + 150 j) = 150 i + 50 j

v B = v A + v B/ A

( vB ) x i 75j = 100i + ( vA ) y j 75 i + 125 jComponents.

i:

( vB ) x

= 100 75

(1) (2)

j : 75 = ( v A ) y + 125

v C = v A + v C/ A400i + ( vC ) y j = 100i + ( v A ) y j 150 i + 50 j Components.

i : 400 = 100 150j:

(3) (4)

( vC ) y

= ( v A ) y + 125

From (3), From (2),

= ( 2 rad/s ) k

( vA ) y

= 75 125 = 75 125 ( 2 ) = 175 mm/s

v A = (100 mm/s ) i + (175 mm/s ) jFind the point with zero velocity. Call it D.

vD = 0

v D = v A + v D/ A

or

0 = 100i + 175 j + ( 2k ) ( xi + yj)

0 = 100i + 175j + 2 xj 2 yi = 0Components.

i : 0 = 100 2 y, y = 50 mm j : 0 = 175 + 2 x, x = 87 mmv 200 = 100 mm 2

Radius of locus.

r =

=

Circle of 100.0 mm radius centered at x = 87.5 mm, y = 50.0 mm

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Chapter 15, Solution 45.

Slope angle of rod. AC =

tan =

7 = 0.7, 10

= 35

10 = 12.2066 in. cos Velocity analysis.v A = 25 in./s ,

CB = 20 AC = 7.7934 in.

vC = vC

vC/ A = AC AB

v C = v A + v C/ ADraw corresponding vector diagram.

vC/ A = v A sin = 25sin 35 = 14.34 in./s (a )

AB =

vC/ A AC

=

14.34 = 1.175 rad/s 12.2066

AB = 1.175 rad/svC = v A cos = 25cos = 20.479 in./s

vB/C = CB AB = ( 7.7934 )(1.175 ) = 9.1551 in./s

v B/C has same direction as vC/ A. v B = vC + v B/CDraw corresponding vector diagram.tan = vB/C vC = 9.1551 , 20.479

= 24.09

(b)

vB =

vC 20.479 = = 22.4 in./s = 1.869 ft/s cos cos 24.09

+ = 59.1v B = 1.869 ft/s 59.1

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

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Chapter 15, Solution 46.

Instantaneous geometry. Law of sines:

sin sin120 = 10 15 sin = 10 sin120 = 0.57735 15

= 35.264Velocity analysis.

v A = 1.2 ft/s

= 14.4 in./s60

v B/ A = 10 ABv B = vB

vB = v B/ A + v AUse the triangle construction to perform the vector addition.

= 60 = 24.736 = 90 + = 125.264vB/ A vB vA Law of sines. sin = sin 30 = sin

vB/ A =

v A sin 14.4 sin125.264 = = 28.10 in./s sin sin 24.736

(a) AB = (b) vB =

28.10 10

AB = 2.81 rad/s

v A sin 30 14.4 sin 30 = = 17.21 in./s sin sin 24.736v B = 1.434 ft/s35.3

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

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Chapter 15, Solution 47.

Label the contact point between gears A and B as 1, the center of gear B as 2, and the contact point between gears B and C as 3. Gear A: Arm AB: Gear B:

v1 = 80 A v2 = 120 AB v1 = v2 40 B v3 = v2 + 80 B

(1) (2) (3) (4) (5)

Gear C: Data: From (1), From (5), From (3), From (4),

v3 = 200 C

A = 0, C = 5 rad/sv1 = 0,

v3 = ( 200 )( 5 ) = 1000 mm/sv2 40 B = 0v2 + 80 B = 1000

(6) (7)

Solving (6) and (7) simultaneously, (a)

B =

1000 = 8.333 rad/s 120

B = 8.33 rad/s

v2 = (40)(8.333) = 333.33 mm/s(b) From (2),

AB =

333.33 = 2.78 rad/s 120

AB = 2.78 rad/s

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Chapter 15, Solution 48.

Label the contact point between gears A and B as 1, the center of gear B as 2, and the contact point between gears B and C as 3. Gear A: Arm AB: Gear B:

v1 = 80 A v2 = 120 AB v1 = v2 40 Bv3 = v2 + 80 B

(1) (2) (3) (4) (5)

Gear C: Data: From (5), From (4), From (3), From (1), (a) From (2), (b)

v3 = 200 C

B = 20 rad/s, C = 0v3 = 0.

v2 = 80 B = ( 80 )( 20 ) = 1600 mm/s v1 = 1600 ( 40 )( 20 ) = 2400 = 2400 mm/s

A = v1 / 80 = 30 rad/s A = 30.0 rad/s1600 = 120 AB

AB =

1600 = 13.33 rad/s 120 AB = 13.33 rad/s

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

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Chapter 15, Solution 49.

Data: A = 3600 rpm = 376.99 rad/s,

B = 0

rA =

1 d A = 1.25 in. 2

d = diameter of ball = 0.5 in.Velocity of point on inner race in contact with a ball.v A = rA A = (1.25)(376.99) = 471.24 in./s

Consider a ball with its center at point C.

v A = vB + v A/Bv A = 0 + C d

C =

v A 471.24 = d 0.5

= 942.48 rad/s

vC = vB + vC/B=0+(a) (b) Angular velocity of ball.

1 d = (0.25)(942.48) = 235.62 in./s 2vC = 236 in./s

C = 942.48 rad/s(c) Distance traveled by center of ball in 1 minute.

C = 9000 rpm

lC = vC t = (235.62)(60) = 14137.2 in.Circumference of circle: 2 r = 2 (1.25 + 0.25)

= 9.4248 in.Number of circles completed in 1 minute:

n=

l 2 r

=

14137.2 9.4248

n = 1500

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Chapter 15, Solution 50.

Contact point 1 between gears A and B. Contact point 2 between gears B and C. Gear B:

B = 6 rad/sv1 = (6 rad/s)(10 in.) = 60 in./s

(1) (2)

v2 = (6 rad/s)(5 in.) = 30 in./sArm ABC: ABC = ABC v A = 15 ABC v C = 15 ABC

Gear A:

A = 3 rad/sv1 = v A + (5)(3) = 15 ABC + 15(3)

Matching expressions (1) and (3) for v1, (a) Gear C: :60 = 15 ABC + 15 ABC = 3.00 rad/s

C = Cv2 = vC + 10 C = 15 + 10 C

(4)

Matching expressions (2) and (4) for v2 , (b) : 30 = 15 + 10 CC = 1.500 rad/s

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Chapter 15, Solution 51.

Let a be the radius of the central gear A, and let b be the radius of the planetary gears B, C, and D. The radius of the outer gear E is a + 2b. Label the contact point between gears A and B as 1, the center of gear B as 2, and the contact point between gears B and E as 3.Gear A: v1 = a A

(1) (2) (3) (4) (5) (6) (7)

Spider:Gear B:

v2 = ( a + b ) Sv2 = v1 + b B v3 = v2 + b B

Gear E: From (4) and (5),From (1) and (3), Solving for v2 and B ,

v3 = ( a + 2b ) E v2 + b B = ( a + 2b ) Ev2 b B = v1 = a A ( a + 2b ) E + a A v2 = 2 ( a + 2b ) E a A B = 2b

From (2),

S =

v2 a+b

( a + 2b ) E + a A S = 2(a + b)

Data: a = 60 mm, b = 60 mm, a + 2b = 180 mm, a + b = 120 mm (a )

B =

180 E 60 A = 1.5 E 0.5 A ( 2 )( 60 )

= (1.5 )(120 ) ( 0.5 )(150 ) = 105 rpm B = 105.0 rpm

(b)

S =

180 E + 60 A = 0.75 E + 0.25 A ( 2 )(120 )

= ( 0.75 )(120 ) + ( 0.25 )(150 ) = 127.5 rpm S = 127.5 rpm

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

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Chapter 15, Solution 52.

Let a be the radius of the central gear A, and let b be the radius of the planetary gears B, C, and D. The radius of the outer gear E is a + 2b. Label the contact point between gears A and B as 1, the center of gear B as 2, and the contact point between gears B and E as 3.

Gear A:

v1 = a A

(1) (2) (3) (4) (5) (6) (7)

Spider:Gear B:

v2 = ( a + b ) Sv2 = v1 + b B

v3 = v2 + b B

Gear E: From (4) and (5),From (1) and (3), Solving for v2 and B ,

v3 = ( a + 2b ) E v2 + b B = ( a + 2b ) Ev2 b B = v1 = a A ( a + 2b ) E + a A v2 = 2 ( a + 2b ) E a A B = 2b

From (2),

S =

v2 a+b

( a + 2b ) E + a A S = 2(a + b)

Data: E = 0, S =

1 A 5

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(a )

S =

( a + 2b )( 0 ) + a A 1 A = 5 2 ( a + b)

a 1 = , 2 ( a + b) 5

1 2 1+

(

b a

)

=

1 5

b 2 1 + = 5 a

b = 1.500 a

(b)

( a + 2b ) E a A = 0 Aa = B = 2b 2b

( 2 )(1.5)

A

=

A3

B =

1 A 3

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

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Chapter 15, Solution 53.

Label the contact point between gears A and B as 1 and that between gears B and C as 2. Rod ABC:

ABC = 75 rpm = 2.5 rad/svA = 0vB = (12)(2.5 ) = 30 rad/s vC = (12 + 7)(2.5 ) = 47.5 rad/s

Gear A: Gear B:

A = 0,

v A = 0,

v1 = 0

v1 = vB 4 B = 0

30 4 B = 0

B = 7.5 rad/sv2 = vB + 4 B

= 30 + 30 = 60 in./sGear C:

v2 = vC 3C60 = 47.5 3C

C = 4.1667 rad/sSummary:B = 225 rpm

C = 125.0 rpm

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

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Chapter 15, Solution 54.

Label the contact point between gears A and B as 1 and that between gears B and C as 2. Rod ABC:vC = 0, ABC = 80 rpm = 8 /3 rad/s vB = 7 ABC = (7)(8 /3) = 56 /3 in./s v A = (7 + 12) ABC = (19)(8 /3) = 152 /3 in./s

Gear C: Gear B:

C = 0,

vC = 0,

v2 = 0

v2 = vB 4 B = 56 /3 4 B = 0

B = 14 /3 rad/sv1 = vB + 4 B = 56 /3 + 56 /3

= 112 /3 in./sGear A:v1 = v A 8 A

112 /3 = 152 /3 8 A

A = 5 /3 rad/sSummary:

A = 50.0 rpmB = 140.0 rpm

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

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Chapter 15, Solution 55.

Geometry.

( OA) sin sin =Shaft and eccentric disk. (Rotation about O),

= ( AB ) sin 10 sin30 , 160

( OA) sin AB

=

= 1.79

OA = 900 rpm = 30 rad/s

v A = ( OA) OA = (10)( 30 ) = 300 mm/sRod AB. (Plane motion = Translation with A + Rotation about A.)

v B = v A + v B/ A

[ vB

] = [ vA

60] + v A/B

]

Draw velocity vector diagram.

90 = 88.21

= 180 60 88.21 = 31.79Law of sines.

vB vA = sin sin ( 90 )vB =

( 300 ) sin 31.79 v A sin = sin ( 90 ) sin 88.21v B = 497 mm/s

= 497 mm/s

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

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Chapter 15, Solution 56.

Geometry.

( OA) sin (180 ) = ( AB ) sin sin =Shaft and eccentric disk. (Rotation about O)

( OA) sin (180 )AB

=

10 sin 60 , 160

= 3.10

OA = 900 rpm = 30 rad/s

v A = ( OA) OA = (10)( 30 ) = 300 mm/sRod AB. (Plane motion = Translation with A + Rotation about A.)

v B = v A + v B/ A

[ vB

] = [300

30] + vB/ A

]

Draw velocity vector diagram.

90 + = 93.10

= 180 30 93.10 = 56.90Law of sines.

vB vA = sin sin ( 90 + )vB =

( 300 ) sin 56.90 v A sin = sin ( 90 + ) sin 93.10v B = 791 mm/s

= 791 mm/s

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

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Chapter 15, Solution 57.

Disk A

A = 15 rad/s ,Rotation about a fixed axis.

AB = 2.8 in.

vB = ( AB ) A = ( 2.8 )(15 ) = 42 in./s (a) = 0.v B = 42 in./s

sin =

2.8 , 10

= 16.260

v D = v B + v D/BBar BD :

vD = [ 42

] + vD/B

vD/B =

42 = 43.75 in./s cos 43.75 , 10v B = 42 in./s

DB =

vD/B DB

=

DB = 4.38 rad/sv D = 12.25 in./s

vD = vB tan , (b) = 90.sin =Bar BD :

5.6 , = 34.06 10

v D = v B + v D/BvD

= [ 42

] + vD/B

Components:

: vD/B = 0 : vD = v B

DB = 0v D = 42.0 in./s

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

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(c) = 180.sin = 2.8 , 10

vB = 42 in./s

= 16.26

Bar BD :vD = [ 42

v D = v B + v D/B ] + vD/B42 = 43.75 cos

vD/B =

DB =

vD/B DB

=

43.75 10

DB = 4.38 rad/svD = vB tan v D = 12.25 in./s

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

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Chapter 15, Solution 58.

rA = 2.8 in., From geometry,

lBD = 10 in.(1)

lBD sin = rA + rA sin

v B is tangent to the circular path of B,thusFor rod BD

v B = rA A

v B/D = lBD BD

v B = v D + v B/D = 0 + v B/D v B/D = v B (a) For matching direction = For = , sin = sin

orso that

= 180 + lBD sin = rA + rA sin

sin =

rA 2.8 , = lBD rA 10 2.8

= 22.9,

= 22.9

For = 180 + ,

sin = sin ,lBD sin = rA rA sin

sin =

rA 2.8 , = lBD + rA 10 + 2.8

= 12.6

= 192.6

(b) For matching magnitudes vD/B = vBlBD BD = rA A ,For = 22.9, For = 192.6,

BD =

( 2.8)( 20 ) = 5.6 rad/s rA A = lBD 10BD = 5.60 rad/s BD = 5.60 rad/s

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Chapter 15, Solution 59.

Rod BE:

vE = 0

v B = rB/E BE = (192)(4) = 768 mm/sRod ABD:v D = vD

v D = v B + v D/BvD = 768

+ 360 AD

60

Draw diagram for vector addition. (a) 768 = 360 AD sin 30

AD = 4.2667 rad/s AD = 4.27 rad/s(b) 768 = vD tan 30vD = 1330 mm/s v D = 1.330 m/s

(c) v A/B = 240 AD

60 = 1024 mm/s

60

v A = v B + v A/B = 768

+ 1024

60 = 1557 mm/s

34.7

v A = 1.557 m/s

34.7

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Chapter 15, Solution 60.

Rod BE: Rod ABD:

vE = 0

v B = vB

v D = 1.6 m/s

AD = AD

v B = v D + v B/DvB

= 1.6 + 0.360 AD

60

Draw diagram for vector addition. (a) 0.360 AD =

1.6 sin 60 AD = 5.13 rad/s

(b) vB = 1.6 tan 30 = 0.92376v B = 0.924 m/s

(c) v A/B = 0.240 AD

60 = 1.2317 m/s

6060

v A = v B + v B/ A = 0.92376

+ 1.2317

= [1.5396

] + [1.0667 ] = 1.873

34.7

v A = 1.873 m/s

34.7

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Chapter 15, Solution 61.

AB = 1000 rpm

=

(1000)( 2)60

= 104.72 rad/s

(a) = 0. Crank AB. (Rotation about A) rB/ A = 3 in.v B = vB/ A AB = ( 3)(104.72) = 314.16 in./sRod BD.

(Plane motion = Translation with B + Rotation about B) v D = v B + vD/BvD = [ 314.16

] + vD/B

]vP = 0

vD = 0,

vD/B = 314.16 in./s

vP = vD

BD =

vB 314.16 = l 8

BD = 39.3 rad/s

(b) = 90. Crank AB. (Rotation about A)

rB/ A = 3 in.

v B = rB/ A AB = ( 3)(104.72) = 314.16 in./sRod BD.

(Plane motion = Translation with B + Rotation about B.) v D = v B + v D/BvD = 314.16 + vD/B

]

vD/B = 0,

vD = 314.16 in./svD/B l ,BD = 0

BD =

v P = v D = 314.16 in./s

v P = 26.2 ft/s

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

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Chapter 15, Solution 62.

AB = 1000 rpm =

(1000)( 2)60

= 104.72 rad/s30 60

= 60, Crank AB. (Rotation about A) rB/ A = 3 in.v B = rB/ A AB = ( 3)(104.72) = 314.16 in./sRod BD. Geometry.

(Plane motion = Translation with B + Rotation about B.) l sin = r sin sin = r 3 sin = sin 60 l 8 = 18.95

v D = v B + v D/B

[ vD ] = [314.16Law of sines.

60] + vD/B

] Draw velocity vector diagram.

= 180 30 ( 90 ) = 78.95vD/B vD vB = = sin sin 30 sin ( 90 )vD = vB sin 314.16 sin 78.95 = = 326 in./s cos cos18.95

vP = vD

v P = 27.2 ft/s

vD/B =

vB sin 30 314.16sin 30 = = 166.08 in./s cos cos18.95 vD/B l

BD =

=

166.08 8

BD = 20.8 rad/s

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Chapter 15, Solution 63.

Bar AB. (Rotation about A)

v B = AB rB/ A = ( 4k ) ( 0.25j) = (1.00 m/s ) iBar ED. (Rotation about E)

v D = DE k rD/E = DE k ( 0.075i 0.15j) = 0.15 DE i 0.075 DE jBar BD. (Translation with B + Rotation about B.)

v D/B = BDk rD/B = BDk 0.2i = 0.2 BD j v D = v B + v D/B0.15 DE i 0.075 DE j = 1.00i + 0.2 BD j

Components:i : 0.15 DE = 1.00, j: 0.075 DE = 0.2 BD

DE = 6.6667 rad/s

DE = 6.67 rad/s

BD =

( 0.075 )( 6.6667 )0.2

BD = 2.50 rad/s

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Chapter 15, Solution 64.

Bar AB. Bar DE.

(Rotation about A.) (Rotation about E.)

v B = ( AB ) AB = 0.18 AB v D = ( DE ) DE = 0.18 DE

30

Bar BGD. (Plane motion = Translation with B + Rotation about B. )

v D = v B + v D/B

[ v D ] = [ vB

30] + vD/B

30 ]

Draw the velocity vector diagram.

Equilateral triangle.

vD/B = vB = 0.18 AB

BD =

vD/B lBD

=

0.18 AB = AB 0.18

vG/B = lGB BD =

1 vD/B = 0.09 AB 2

vG = v B + vG/B Draw vector diagram.

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Law of cosines2 vG = ( 0.18 AB ) + ( 0.09 AB ) 2 ( 0.18 AB )( 0.09 AB ) cos 60 2 2

2 vG = 0.0243AB2

AB = 6.415 vG = ( 6.416 )( 2.5 ) AB = 16.04 rad/s BD = 16.04 rad/svD = vB = 0.18 AB

DE =

vD 0.18 AB = = AB lDE 0.18

DE = 16.04 rad/s

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Chapter 15, Solution 65.

Bar AB.

(Rotation about A.)

AB = 25 rad/sv B = rB/ A AB = ( 8 )( 25 ) = 200 in./s

Bar ED.

(Rotation about E.)v D = vD vD = 8 DE

Plate BDHF.

(Translation with B + Rotation about B. )

v D = v B + v D/B

[ vD

] = [vB ] + vD/B

30 ]

Draw velocity vector diagram.vD/B = vB 200 = = 230.94 in./s cos 30 cos 30vB/D lBD

BDHF =(a)

=

230.94 = 14.4338 rad/s 16BDHF = 14.43 rad/s

vF /B = BF BDHF = ( 8 )(14.4338 ) = 115.47 in./sv F /B = 115.47 m/sv F = v B + v F /B = [ 200 in./s

30

] + [115.47 in./s

30]54.9

(b) v F = [142.265 in./s

] + [100 in./s

]

v F = 173.9 in./s

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

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Chapter 15, Solution 66.

Rod DE. (Rotation about E.)

DE = 35 rad/sv D = rD/E DE = ( 8 )( 35 ) = 280 in./s

Rod AB. (Rotation about A.)v B = vB vB = 8 AB

Plate BDHF.

(Translation with D + Rotation about D. )

v B = v D + v B/D

[ v B ] = [ vDvD/B =

] + vB/D

30]

Draw velocity vector diagram.vD 280 = = 560 in./s sin 30 sin 30 vD/B 560 = = = 35 rad/s lDB 16 BDHF = 35.0 rad/sPoint of zero velocity lies above point D.yC/D = vD 280 = 8 in. 35

BDHF(a)

BDHF

=

(b)

8 in. above point D.

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Chapter 15, Solution 67.

AB = (15 rad/s) k

BD = BD k

DE = DE k

rB/A = ( 0.2 m ) j

rD/B = ( 0.6 m ) i ( 0.25 m ) j

rD/E = ( 0.2 m ) i

v B = v A + AB rB/A = 0 + ( 15 k ) ( 0.25 j) = ( 3 m/s ) i v D/B = BD rD/B = ( BD k ) ( 6 i 0.2 j) = 0.25 BD i 0.6 BD jv D = v B + v D/B = 3 i + 0.25 BD i 0.6 BD j

(1) (2)

v D = v E + v D/E = 0 + DE rD/E = ( DE k ) ( 0.2 i ) = 0.2 DE jEquate the expressions (1) and (2) for vD and resolve into components.i : 3 + 0.25 BD = 0 j: 0.6 BD = 0.2 DE

BD = 12 rad/s DE = 36 rad/sBD = 12 rad/s

(a) Angular velocity of rod BD.(b) Velocity of the midpoint M of rod BD.

rM/B =

1 rD/B = ( 0.3 m ) i ( 0.125 m ) j 2

v M = v B + v M/B = v B + BD rM/B = 3 i + 12 k ( 0.3 i 0.125 j)= (1.5 m/s ) i ( 3.6 m/s ) j

v M = 3.90 m/s

67.4

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Chapter 15, Solution 68.

Bar AB.

rB/ A = ( 0.300 m ) i + ( 0.125 m ) j,vA = 0

AB = ( 3 rad/s ) k

v B = v A + v B/ A = 0 + AB rB/ A= 3k ( 0.3i + 0.125 j) = 0.375i + 0.9 jBar BD.

rD/B = ( 0.325 m ) j

BD = BDk

v D = v B + v D/B = 0.375i + 0.9 j + BDk ( 0.325 ) j= 0.375i + 0.9 j + 0.325 BD iBar DE.

rE/D = ( 0.150 m ) i + ( 0.200 m ) j,

DE = DE k

v E = v D + v E/D = v D + DE rE/D= v D + DE k ( 0.150i + 0.200 j) = 0 0.375i + 0.9 j + 0.325 BDi 0.15 DE j 0.2 DE i = 0Components: j: i:0.9 0.15 DE = 0

DE = 6 rad/s

0.375 + 0.325 BD 0.2 DE = 0

0.325 BD = 0.375 + ( 0.2 )( 6 )

BD = 4.85 rad/s

BD = 4.85 rad/s DE = 6.00 rad/s

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Chapter 15, Solution 69.

v A = 80 km/h = 22.222 m/s

vC = 0

d = 560 mm

r =

d = 280 mm = 0.28 m 2

=

vA 22.222 = = 79.364 rad/s r 0.28

vB/ A = vD/ A = vE/ A = r= ( 0.28 )( 79.364 ) = 22.222 m/sv B = v A + v B/ A = [ 22.222 m/s

] + [ 22.222 m/s ] + [ 22.222 m/s

]30 ]15.0

v B = 44.4 m/s

v D = v A + v D/ A = [ 22.222 m/s

v D = 42.9 m/s

v E = v A + v E/ A = [ 22.222 m/s

] + [ 22.222 m/s ]v E = 31.4 m/s45.0

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Chapter 15, Solution 70.

(a) = 0

Wheel AD.

vC = 0,

v D = 45 in./s

AD =

vD 45 = = 11.25 rad/s CD 4

CA = ( CD ) ( DA) = 4 2.5 = 1.5 in. v A = ( CA) AD = (1.5 )(11.25 ) = 16.875 in./sRod AB.

v B = v A + v B/ A

[ vB

] = [16.875

] + vB/ A

]

v B = 16.88 in./s

vB/ A = 0(b) = 90Wheel AD.v C = 0,

AB = 0 AD = 11.25 rad/s

tan =

DA 2.5 = , DC 4

= 32.005

CA =

DC = 4.7170 in. cos

v A = ( CA ) AD = ( 4.7170 )(11.25 ) = 53.066 in./s v A = [53.066 in./sRod AB.

32.005]

v B = vB

sin =

4 , = 18.663 12.5

Plane motion = Translation with A + Rotation about A.

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v B = v A + v B/ A

[ vB

] = [ vA

r ] + [ vB/A

]

Draw velocity vector diagram.

= 180 ( 90 + )= 90 32.005 18.663 = 39.332

Law of sines.

vB/ A vB vA = = sin sin sin ( 90 + ) vB =

( 53.066 ) sin 39.332 v A sin = sin ( 90 + ) sin 108.663v B = 35.5 in./s

= 35.5 in./s

vB/ A =

v A sin (53.066)sin 32.005 = sin ( 90 + ) sin108.663

= 29.686 in./s

AB =

vB/ A AB

=

29.686 = 2.37 rad/s 12.5

AB = 2.37 rad/s

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Chapter 15, Solution 71.

v 0 = 180 km/h = 50 m/s (180 )( 2 ) = 18.85 rad/s = 180 rpm = 60 Top View v0 = z

z=

v0

=

50 = 2.65 m 18.85x=0

Instantaneous axis is parallel to the y axis and passes through the point

z = 2.65 m

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Chapter 15, Solution 72.

=

vE vD 1.5 1.0 1 = = rad/s lED 1.5 3lCE = vD

=

1.01 3

= 3m

(a) (b)

l AC = 1.5 + 2 3 = 0.5 m

C lies 0.500 m to the right of Av A = 0.1667 m/s

1 v A = l AC = ( 0.5 ) = 0.1667 m/s 3

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Chapter 15, Solution 73.

Contact points: 1 2 Arm ABC: between gears A and B. between gears B and C.

ABC = 4 rad/s

v A = (15 )( 4 ) = 60 in./s vC = (15 )( 4 ) = 60 in./sGear B:

B = 8 rad/sv1 = (10 )( 8 ) = 80 in./s v2 = ( 5 )( 8 ) = 40 in./s

Gear A:

A =

v1 v A 80 60 = 5 5

A = 4 rad/slA =

A

vA

=

60 = 15 in. 4

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Gear C:

C =

vC v2 60 40 = 10 10

C = 2 rad/slC =

C

vC

=

60 = 30 in. 2

(a) Instantaneous centers. Gear A: 15 in. left of A Gear C: 30 in. left of C (b) Angular velocities.

A = 4.00 rad/s C = 2.00 rad/s

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Chapter 15, Solution 74.

Since the drum rolls without sliding, the point of contact C with the fixed surface is the instantaneous center. Let point A be the center of the cylinder and point B the point where the cord breaks contact with the cylinder.

vC = 0

vB = vD = 6 in./s

(a) Angular velocity of cylinder

vB = rB/C

=

vB 6 = = 6 rad/s rB/C 1

= 6.00 rad/s

(b) Velocity of point A.

v A = rA/C

= ( 5 )( 6 ) = 30v A = 30.0 in./s(c) Rate of winding of cord. Since v A > vB , the cord is wound up at rate of

v A vB = 30 6 = 24 in./s.Winding rate = 24.0 in./s.

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Chapter 15, Solution 75.

=

vO v A 12 1.5 = = 35 rad/s l AO 0.3vA

(a)

lCA =

=

1.5 = 42.86 103 m 35

= 42.86 mmC lies 42.9 mm below A.

(b)

lCB = 0.6 + 0.04286 = 0.64286 m

vB = lCB = ( 0.64286)( 35) = 22.5 m/sv B = 22.5 m/s

(c)

lOD = 0.3 m,

lCO = 0.3 + 0.04286 = 0.34286 m

lCD =

( 0.3)2 + ( 0.34286)2

= 0.45558 m

vD = lCD = ( 0.45558 )( 35 ) = 15.95 m/s

tan =

lOD 0.3 = , lCO 0.34286

= 41.2v D = 15.95 m/s41.2

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Chapter 15, Solution 76.

=l AC =

vO v A 12 + 1.5 = = 45 rad/s l AO 0.3vA

(a)

=

1.5 = 33.33 103 m 45= 33.33 mm

C lies 33.3 mm above A.

(b)

lCB = 0.3 + ( 0.3 0.0333) = 0.56667 m vB = lCB = ( 0.56667 )( 45) = 25.5 m/sv B = 25.5 m/s

(c)

lOE = 0.3 m,

lOC = 0.3 0.03333 = 0.26667 m

lCE =

( 0.3)2 + ( 0.26667)2

= 0.4014 m

vE = lCE = ( 0.4014)( 45) = 18.06 m/stan = lOE 0.3 , = lOC 0.26667

= 48.4v E = 18.06 m/s48.4

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Chapter 15, Solution 77.

(a) Location of instantaneous axis.

v E = v D = 8 in./sv A = 3 in./s

v E = v A + rE/A vE = v A + rE/A8 = 3 + 3

= 1.6667 rad/svC = v A + rC/A

0 = 3 1.6667 y(b) Velocity of point B.

y = 1.800 in. above point A.

v B = v A + rB/A

vB = 3 3 = 3 ( 3)(1.6667 ) = 2v B = 2.00 in./s

(c) Since vD = vE > v A , the paper unwinds. Rate of unwinding:

vE v A = 8 3 = 5 in./s

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

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Chapter 15, Solution 78.

v D = 10 in./s ,

v B = 8 in./s

=CD =

vD + vB 10 + 8 = = 4 rad/s BD 4.5 vD

=

10 = 2.5 in. 4

CA = 3.0 2.5 = 0.5 in.

(a) (b)

C lies 0.500 in. to the right of A.

v A = 0.5 = ( 0.5 )( 4 ) = 2 in./sv A = 2.00 in./s

(c)

v D v A = 12 in./sCord DE is unwrapped at 12.00 in./s.

v B v A = 6 in./sCord BF is unwrapped at 6.00 in./s.

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Chapter 15, Solution 79.

Rod AD.v B = rB/E BE = ( 0.192)( 4) = 0.768 m/s

(a) Instantaneous center C is located by noting that CD is perpendicular to vD and CB is perpendicular to vBrB/C = 0.360 sin 30 = 0.180 m

AD =

vB 0.768 = = 4.2667 0.180 rB/C AD = 4.27 rad/s

"

(b) Velocity of D.

rD / C = 0.360 cos30 = 0.31177 m vD = rD/C = ( 0.31177 )( 4.2667 ) v D = 1.330 m/s "

(c) Velocity of A.l AE = 0.240cos 30 = 0.20785 m lCE = 0.600sin 30 = 0.300 m

tan =

0.20785 0.300

= 34.7= 0.36497 m

lCA =

( 0.20785)2 + ( 0.300)2

v A = lCA AD = ( 0.36497 )( 4.2667 ) = 1.557 m/sv A = 1.557 m/s

34.7 "

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Chapter 15, Solution 80.

AB = 15 rad/svB = ( AB ) AB = ( 0.200 )(15 ) = 3 m/sv B = vB v D = vD

Locate the instantaneous center (point C) of bar BD by noting that velocity directions at points B and D are known. Draw BC perendicular to v B and DC perpendicular to v D .

(a )

BD =

vB 3 = = 12 rad/s BC 0.25

BD = 12.00 rad/s

(b) Locate point M, the midpoint of rod BD. Draw CM.

BD =

( 0.6 )2 + ( 0.25)2 = 22.62

= 0.65 m

tan =

0.25 0.6

90 = 67.38

CM = DM = MB =

1 ( BD ) = 0.325 m 2v M = 3.90 m/s

vM = ( CM ) = ( 0.325 )(12 ) = 3.9 m/s

67.4

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Chapter 15, Solution 81.

Bar DC. (rotation about D)

vC = CD (CD) = (18)(10)= 180 in./s

vC = 180 in./sBar AB. (rotation about A)

30

v B = vB

30

Locate the instantaneous center (point I) of bar BC by noting that velocity directions at two points are known. Extend lines AB and CD to intersect at I. For the given configuration, point I coincides with D.

IC = 10 in., IB = 10 3 in.

BC =

vC 180 = = 18 rad/s IC 10

vB = ( IB ) BC = 10 3 (18 ) = 311.77 in./s(a) (b) (c) Locate point M, the midpoint of bar BC. Triangle ICM is an equilateral triangle.IM = 10 in.

(

)

AB =

vB 311.77 = = 31.177 rad/s AB 10

AB = 31.2 rad/s

BC = 18.00 rad/s

vM = ( IM ) BC = (10 )(18 ) = 180 in./s

v M = 15.00 ft/s

30

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Chapter 15, Solution 82.

Bar AB. (rotation about A)v B = vB

60

Bar CD. (rotation about D)

vC = vCBar BC. Locate its instantaneous center (point I) by noting that velocity directions at two points are known. Extend lines AB and CD to intersect at I. For the given configuration, point I coincides with D. Locate point M, the midpoint of bar BC. From geometry, triangle ICM is an equilateral triangle.

IM = AB = CD = 10 in.,

IB = 10 3 in.

BC =(a)

vM ( 7.8)(12) = 9.36 rad/s = 10 IM

vB = ( IB ) BC = 10 3 ( 9.36 ) = 162.12 in./s

(

)

AB =(b) (c)

vB 162.12 = = 16.21 rad/s AB 10

AB = 16.21 rad/s BC = 9.36 rad/s

vC = ( IC ) BC = (10 )( 9.36 ) = 93.6 in./s

CD =

vC 93.6 = = 9.36 rad/s DC 10

CD = 9.36 rad/s

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Chapter 15, Solution 83.

v B = 800 mm/s

v A = vA

Locate the instantaneous center of rod ABD by noting that velocity directions at points A and B are known. Draw AC perpendicular to v A and BC perpendicular to v B . (a)

ABD =

vB 800 = = 3.0792 rad/s 300 cos 30 BC

ABD = 3.08 rad/slCD =

( 600 cos 30 )2 + ( 300sin 30 )2300sin 30 600 cos 30

= 540.83 mm90 = 73.9

tan =(b)

= 16.10

vD = lCD ABD = ( 540.83)( 3.0792 ) = 1.665 103mm/sv D = 1.665 m/s73.9

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Chapter 15, Solution 84.

= 40,

v B = 0.6 m/s ,

v A = vA

Locate the instantaneous center (point C) by noting that velocity directions at points A and B are known. Draw AC perpendicular to v A and BC perpendicular to vB .

BC = ( AB ) sin = 2 sin 40 = 1.28557 m(a)

ABD =

vB 0.6 = = 0.46672 rad/s BC 1.28557 ABD = 0.467 rad/s

rD/C = rB/C + rD/B= [1.28557 m= 2.9930 m

] + [2 m30.79

40]

= 30.79(b)vD = rD/C ABD = ( 2.9930 )( 0.46672 )

= 1.397 m/sv D = 1.397

90 = 59.2v D = 1.397 m/s59.2

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Chapter 15, Solution 85.

DE =

3202 + 2402 = 400 mm

tan =

240 = 0.75 320

= 36.87

vD = ( DE )( DE ) = ( 400 )(15 ) = 6000 mm/s = 6 m/sv D = 6 m/s

v B = vB

Locate point C, the instantaneous of bar DBF, by drawing BC perpendicular to vB and DC perpendicular to vD.

From the figure:

AC =

540 = 720 mm tan

BC = AC AB = 720 ( 320 + 100 ) = 300 mmSince triangles FCB and BDK are similar,

b DK 300 = = BC BK 100

b=

( 300 )( 300 )100

= 900 mm.b = 0.900 m

(a) Distance b.CD = 3002 + 4002 = 500 mm = 0.5 mvD 6 = = 1.2 rad/s CD 5

BDF =

vF = b BDF = ( 0.900 )(1.2 ) = 10.8 m/s (b) Velocity of point F.v F = 10.80 m/s

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Chapter 15, Solution 86.

Locate the instantaneous center I of rotation of bar ABD as the intersection of line AI perpendicular to v A and line BI perpendicular to v B Triangle IAB is equilateral.

lIA = lIB = 300 mm(a) ABD =

vA 900 mm/s = = 3 rad/s lIA 300 mm

ABD = 3.00 rad/s(b) By the law of cosines, lID = ( 600cos 30) mm

vD = lID ABD = ( 600cos 30 )( 3) = 1559 mm/sv D = 1.559 m/s

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Chapter 15, Solution 87.

v A = vA

45,

v B = 7.5 ft/s

Locate the instantaneous center (point C) of rod AB by noting that velocity directions at points A and B are known. Draw AC perpendicular to v A and BC perpendicular to v B . Letl = AB = 24 in. = 2 ft

Law of sines for triangle ABC.

b a l = = = 2.8284 ft sin 75 sin 60 sin 45a = 2.4495 ft, b = 2.73205 ft

=(a)

vB 7.5 = = 2.7452 rad/s b 2.73205

v A = a = ( 2.4495)( 2.7452) = 6.724 ft/sv A = 6.72 ft/s45.0

(b) (c) Let M be the midpoint of AB. Law of cosines for triangle CMB.2

= 2.75 rad/s

l l m 2 = b 2 + 2b cos 60 2 2 = ( 2.73205) + (1) ( 2)( 2.73205)(1) cos 602 2

m = 2.3942 ftLaw of sines.

sin l 2

=

sin 60 , m

sin =

1sin 60 , 2.3942

= 21.2

vM = m = ( 2.3942)( 2.7452) = 6.573 ft/s,v M = 6.57 ft/s21.2

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Chapter 15, Solution 88.

Bar DE.

vD = e ED = ( 24 )( 8 ) = 192 in./sv D = 192 in./s

Bar AB.

vB = a AB = 8 ABv B = 8 AB30

Locate the instantaneous center (point C) of bar BD by noting that velocity directions at points B and D are known. Draw BC perpendicular to v B and DC perpendicular to v D . Letl = BD = 24 in.

Law of sines for triangle CBD.

b d l 24 = = = = 48 in. sin120 sin 30 sin 30 sin 30b = 41.569 in., d = 24 in.

(a) (b)

BD =

vD 192 = = 8 rad/s d 24

BD = 8.00 rad/s

vB = b BD = ( 41.569 )( 8 ) = 332.55 in./s

AB =

vB 332.55 = = 41.6 rad/s a 82

AB = 41.6 rad/s

(c) Law of cosines for triangle CMD.

l l m 2 = d 2 + 2d cos120 2 2 = 242 + (12) ( 2)( 24)(12) cos1202

m = 31.749 in.Law of sines.

sin l 2

=

sin120 , m

sin =

(12) sin120 ,31.749

= 19.1

Velocity of M.

vM = m BD = ( 31.749 )( 8 ) = 253.99 in./sv M = 21.2 ft/s19.1

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Chapter 15, Solution 89.

vB = ( AB ) ABF ,v D = 200 mm/s

v B = vB

75

Locate the instantaneous center (point C) of bar DBE by noting that the velocity directions at points B and D are known. Draw BC perpendicular to v B and DC perpendicular to v D .

Law of sines for triangle BCD.

CD BC BD = = sin150 sin15 sin15 CD = 180sin150 = 347.73 mm sin15

BC = BD = 180 mm

DBE =

vD 200 = = 0.57515 rad/s CD 347.73

vB = ( BC ) DBE = (180 )( 0.57515 ) = 103.528 mm/s

ABF =

vB 103.528 = = 0.57515 rad/s AB 180

vF = ( AF ) ABF = ( 300 )( 0.57515 ) = 172.546 mm/sLaw of cosines for triangle DCE.

( CE )2

= ( CD ) + ( DE ) 2 ( CD )( DE ) cos15

2

2

( CE )2

= 347.732 + 3002 ( 2 )( 347.73)( 300 ) cos15,

CE = 96.889 mm

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EH = DE sin15 = 300 sin15

cos =(a)

EH 300 sin15 = CE 96.889

= 36.7

vE = ( CE ) BCD = ( 96.889 )( 0.57515 ) = 55.7 mm/s,v E = 55.7 mm/s

36.7 75.0

(b)

v F = 172.5 mm/s

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Chapter 15, Solution 90.

DE = 3 rad/svD = ( DE ) DE = (160 )( 3) = 480 mm/sv D is perpendicular to DE.v B = vB

Locate the instantaneous center (point C) of bar ABD by noting that velocity directions at points B and D are known. Draw BC perpendicular to v B and DC perpendicular to v D .

BD = 120 mm, DK = ( BD ) cos 30 = 120cos 30cos = DK 120 cos30 = , = 49.495, ED 160

= 180 30 = 100.505

Law of sines for triangle BCD.

CD BC BD = = sin 30 sin sin

CD = BC =

( BD ) sin 30sin

=

120sin 30 = 78.911 mm sin

( BD ) sin sin

=

120sin = 155.177 mm sin

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Law of cosines for triangle ABC.

( AC )2

= ( BC ) + ( AB ) 2 ( AB )( BC ) cos150

2

2

( AC )2Law of sines.

= 155.177 2 + 2002 ( 2 )(155.177 )( 200 ) cos150,sin sin150 = AB AC sin =

AC = 343.27 mm

200 sin150 , 343.27

= 16.9

(a)

ABD =

vD 480 = = 6.0828 rad/s CD 78.911

ABD = 6.08 rad/s(b)

v A = ( AC ) ABD = ( 343.27 )( 6.0828 ) = 2088 mm/sv A = 2.09 m/s

73.1

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

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