Solution Manual - Vector Mechanics Engineers Dynamics 8th Beer Chapter 16 PDF

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COSMOS: Complete Online Solutions Manual Organization System Chapter 16, Solution 1. M A N B (1 m ) sin ( 294 N )( 0 m )( cos ) ( ) ( 30 kg ) 4 2 ( sin )( 0 m ) N B 144 N ( Fx N B F 30 kg 4 2 ) F 24 N (a) RA N A2 F 2 295 N or R A 295 N tan 1 ! 294 24 and B 145 N (b) ! F 24 0 N A 294 or 0 ! Vector Me...

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Chapter 16, Solution 1.

∑ M A = N B (1.5 m ) sin 60° − ( 294.3 N )( 0.75 m )( cos60° )

(

)

= ( 30 kg ) 4 m/s2 (sin 60° )(0.75 m ) N B = 144.96 N

(

∑ Fx = NB − F = 30 kg 4 m/s2

)

F = 24.96 N

(a)

RA =

N A2 + F 2 = 295.36 N

or R A = 295 N

α = tan −1

85.2° !

294.3 = 85.2° 24.96

and B = 145.0 N (b)

µ=

!

24.96 F = = 0.08481 N A 294.3

or µ = 0.0848 !

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 16, Solution 2.

Alternative ∑ M A = ∑ M A eff :

mg

L L cos60 ° = ma sin 60 ° 2 2 g So a = tan 60o

∑ MA = ( 294.3 N )( 0.75 m )( cos 60° ) = ( 30 kg )( a)( 0.75 m )( sin 60 °) a = 5.664 m/s2

or a = 5.66 m/s 2

(

)

∑ Fx = F = ( 30 kg ) 5.664 m/s2 = 169.9 N

(a) (b)

!

µ=

F 169.9 N = = 0.5773 N A 294.3 N

or µ = 0.577 !

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 16, Solution 3.

∑ Fx = ma

(a)

∑ Fx = mg ( cos 60° ) = ma

a=

g 2 a = 16.10 ft/s2 !

∑ M C = ( ∑ M C )eff

(b)

∑ M C = − ( 1.8 lb)( 2 in.) + R A ( cos30° ) ( 8 in.) − R A ( sin 30° )( 8 in .) ⎛ −1.8 ⎞ ⎛ 1.8 ⎞ =⎜ ⎟ ( a )( cos30° )( 6 in.) − ⎜ ⎟ ( a )( sin 30°)( 2 in.) ⎝ g ⎠ ⎝ g ⎠ −3.6 + RA (2.9282 ) = −(11.1531)

a g

⎛ 1⎞ 3.6 − 11.1531 ⎜ ⎟ ⎝ 2⎠ RA = = − 0.675 lb 2.9282

or R A = 0.675 lb

60° !

∑ Fy = RA + RC − (1.8 lb) cos30° = 0 RA + RC = 1.5588 R C = 1.5588 − (− 0.675) = 2.234 lb

or R C = 2.23 lb

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.

60° !

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Chapter 16, Solution 4.

See the free body diagram for problem 16.3 ∑ Fx = mg sin θ = ma a = g sin θ ∑ M G = 0 = RC cosθ (2 in. ) − RC sin θ ( 6 in. )

(a) (b)

tanθ =

1 3

(

or θ = 18.43° !

)

a = 32.2 ft/s2 ( sin18.43° ) = 10.180 ft/s2

or a = 10.18 ft/s2

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.

18.43° !

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Chapter 16, Solution 5.

(a) If rear-wheel brakes fail to operate:

ΣM A= Σ NB =

(M A )eff :

N B (12 ft − ) W ( 5 ft=)

ma ( 4 ft )

5 1W W + a 12 3g

ΣFx = Σ (Fx )eff :

FB = ma, µk N B =

W a g

 5 1W  W 0.699 W + a = a g 3 g   12 5 0.699   32.2 ft/s2  12  a = a = 12.227 ft/s2 1 − 0.233 Uniformly accelerated motion

(

v2 = v02 + 2 ax

)

(

) (

)

0 = 30 ft/s2 − 2 12.227 ft/s2 x x = 36.8 ft !

(b) If front-wheel brakes fail to operate:

ΣM B = Σ ( M B )eff : W ( 7 ft ) − N A (12 ft ) = ma ( 4 ft ) NA =

7 1W W − a 12 3 g continued

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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ΣFx= Σ

(Fx )eff :

F=A

ma , µk N=A

W a g

 7 1W  W 0.699  W − a = a 12 3 g  g   7 0.699   32.2 ft/s2  12  a= 1 + 0.233

(

)

a = 10.648 ft/s2

Uniformly accelerated motion v2 = v02 + 2 ax

(

) (

)

0 = 30 ft/s2 − 2 10.648 ft/s2 x

x = 42.3 ft !

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 16, Solution 6.

(a) Four-wheel drive:

ΣFy= Thus:

0: N A+ N B− W=

N A+ N B= W= mg

0

F A + FB = µ k N A + µ k N B = µ k ( N A + N B ) = µ kW = 0.80mg

ΣFx = Σ (Fx )eff : FA + FB = ma 0.80mg = ma

(

)

a = 0.80 g = 0.80 9.81 m/s 2 = 7.848 m/s2 a = 7.85 m/s2 ! (b) Rear-wheel drive:

ΣM B = Σ ( M B ) eff :

( 1 m)W − ( 1.5 m) N A

= − ( 0.5 m) ma

N A = 0.4W + 0.2ma Thus:

F A = µk N B = 0.80 ( 0.4W + 0.2ma ) = 0.32mg + 0.16ma ΣF x = Σ (F x ) eff : F A = ma 0.32mg + 0.16ma = ma 0.32 g = 0.84a

a =

0.32 9.81 m/s 2 = 3.7371 m/s 2 0.84

(

)

or a = 3.74 m/s 2 ! continued

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

(c) Front-wheel drive:

Σ M A= Σ

=− ( M A )eff : (2.5 m ) N−B (1.5 m )W

(0.5 m ) ma

NB = 0.6W − 0.2ma Thus:

F B = µkN B = 0.80 ( 0.6W − 0.2ma ) = 0.48mg − 0.16ma ΣFx= Σ

( Fx )eff :

F=B

ma

0.48mg − 0.16ma = ma 0.48 g = 1.16a a =

0.48 9.81 m/s2 = 4.0593 m/s2 1.16

(

)

or a = 4.06 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.

!

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Chapter 16, Solution 7.

(a) Sliding impends:

ΣF y= Σ

( F x )eff :

=F

ma cos30 °

ΣFy= Σ

( Fy )eff :

− N

mg =

ma sin °30

N = m ( g + as in 30 ° )

µs =

F ; N

0.25 =

ma cos 30° ; m ( g + asin 30 ° )

g + a sin 30 ° = 3.3333 a cos30 °

g = 2.3867a

a = 0.419g

30° !

(b) Tipping impends:

⎛ h⎞ ⎛ d⎞ ΣM G = Σ( M G )eff : F ⎜ ⎟ − N ⎜ ⎟ = 0 ⎝ 2⎠ ⎝ 2⎠ F d = N h

µ=

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.

F ; N

0.3 =

d ; h

h = 3.33 ! d

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Chapter 16, Solution 8.

(a) Sliding impends:

ΣFx= Σ

( Fx )eff :

=F

ma cos30 °

ΣFy = Σ

( Fy ) eff :

−N

mg =−

ma sin °30

N = m ( g − a sin 30°)

µs =

F ; N

0.3 =

ma cos 30° m ( g − asin 30 °)

g − a sin 30° = 3.3333a cos30° 1 a = = 0.29527 g 3.3333cos30° + sin 30 °

a = 0.295

!

(b) Tipping impends:

ΣM G= Σ

( M G)eff

⎛ h⎞ ⎛d⎞ F⎜ ⎟ = W⎜ ⎟; ⎝ 2⎠ ⎝ 2⎠

F d = N h

µ=

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.

F ; N

0.30 =

d ; h

h = 3.33 ! d

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Chapter 16, Solution 9.

(a) Acceleration ΣF x= Σ

(F x )eff :

25 lb = ma 25 lb =

50 lb a 32 ft/s2 a = 16.10 ft/s 2 !

(b) For tipping to impend

;

A=0 ΣM B= Σ

(M B )eff :

(25 lb )h − (50 lb )(12 in. ) = ma (36 in. ) For tipping to impend

;

B=0 ΣMA= Σ

( 25 lb) h + ( 50 lb)(12 in.)

( MA ) eff :

= ma ( 36 )

or

h = 12 in.

cabinet will not tip for 12 in. ≤ h ≤ 60 in.!

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 16, Solution 10.

(a) Acceleration

ΣFy= 0

N A+ NB− W= 0

N A + N B = 50 lb

But F = µ N , Thus FA + FB = µ s ( 50 lb) , ΣFx= Σ

µ s = 0.25

(Fx )eff :

25 lb − ( FA + FB ) = ma ⎛ 50 lb ⎞ 25 lb − ⎣⎡( 0.25)( 50 lb) ⎦⎤ = ⎜ ⎟a ⎝ 32.2 ft/s2 ⎠ a = 8.05 ft/s2

(b) Tipping

For tipping to impend

: NA = 0 ΣM B= Σ

( M B)eff : 50 lb

( 25 lb ) h − (50 lb )(12 in. ) =

(8.05 ft/s ) (36 in. ) 2

32.2 ft/s2

h = 42 in.

For tipping to impend

:

NB = 0 ΣM A= Σ

( 25 lb ) + (50 lb )(12 in. ) =

(M A )eff : 50 lb

h = −6 in.

( 8.05 ft/s ) ( 36 in.) 2

32.2 ft/s 2

impossible

cabinet will not tip if h ≤ 42 in. !

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 16, Solution 11.

ΣM G= 0⇒ FA= 0

Σ F⊥BG = ( mg)(sin 30° ) = ma a=

g 2

or

a = 16.1 ft/s 2

∑ F& BG = FB − (16 lb )(cos30° ) = 0 FB = 13.856 lb

(a)

a = 16.10 ft/s2

(b)

F A = 0, FB = 13.86 lb compression W

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.

30°°W

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(c) Front-wheel drive:

ΣM A= Σ

(M A )eff : (2.5 m ) N−B (1.5 m )W= − (0.5 m )ma

N B = 0.6W − 0.2ma

Thus:

F B = µ k N B = 0.80 (0.6W − 0.2 ma ) = 0.48 mg − 0.16 ma Σ Fx= Σ

( Fx )eff :

F=B

ma

0.48mg − 0.16ma = ma 0.48 g = 1.16 a a =

0.48 9.81 m/s2 = 4.0593 m/s2 1.16

(

)

or a = 4.06 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.

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Chapter 16, Solution 12.

Analysis of linkage Since members ACE and DCB are of negligible mass, their effective forces may also be neglected and the methods of statics may be applied to their analysis. Free body: Entire linkage:

ΣM D= 0:

(B (B

) − E ) cos 30° − B

y−

y

E (30 sin 30° ) − B x (30 sin 30° ) = 0 x

sin 30° = 0

(1)

Free body: member ACE

ΣM C = 0: A( 15 cos30°) − E ( 15 cos30°) = 0 ; E = A

carrying into eq. (1):

(B

y

)

− A cos 30 °− Bx sin 30 °= 0

Equations of motion for member AB

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.

(2)

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(a)

+

60° ΣF−t ∑

(B

y

(F t )eff :

)

− A cos30°− Bx sin 30° + W cos30°= ma

Recalling equation (2), we have, W cos30° = ma

(

a =

W cos30° = g cos 30° m

)

a = 27.9 ft/s 2

a = 32.2 ft/s2 cos30° (b)

60° !

ΣM B = ∑ ( M B )eff : W (7.5 in. ) − A (30 in. ) cos30 ° = ( ma sin 60 °)(7.5 in.) − ( ma cos 60 °)( 2.5 in.) But ma = W cos30° As found above.

Thus:

7.5 W − 30 A cos30 ° = W cos30° (7.5 sin 60 ° − 2.5 cos60° )   1 1 1 A =W  cos 60 ° = 0.11384 W − sin 60 ° + 12  4 cos30 ° 4 

Recalling that W = 20 lb A = 0.11384 (20 lb ) = 2.2768 lb A = 2.28 lb !

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 16, Solution 13.

Bar AC ΣM C= R1 =

0⇒

6N ⋅ m− R1 (0.45 m= ) 0

6 N⋅ m 40 = N 0.45 m 3

Bar AB ΣF x=

(a)

40 N− 3

(4 )(9.81 ) sin 30° = −

4a

a = 1.5717 m/s 2

or a = 1.572 m/s 2 (b)

30° !

∑ M A = ( ∑ M A )eff : ∑ M A = ( 39.24 N )( 0.3 m ) − TEB ( cos30° )( 0.6 m ) = ma ( sin 30 ° )(0.3 m ) TAB =

− ( 4 )(1.5717 )(0.15 ) + 11.772 = 2 0.841 N ( 0.866 )( 0.6 )

or TAB = 20.8 N !

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 16, Solution 14.

Bar AC ΣMC= 0 0 = M − R1 ( 0.45) , ΣFx=

so R1 =

M 0.45

M − ( 3 9.24 N)( sin 30 ° )= − 0.45 m

( 4 kg) ( 4 m/s2 ) M = 1.629 N ⋅ m !

(a) (b)

ΣM A=

( 39.24 N)( 0.3 m)−

T EB ( cos30 ° )( 0.6 m)= ma ( sin 30 ° )( 0.3 m)

TEB = 18.036 N

or TEB = 18.04 N !

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 16, Solution 15.

We will need the acceleration of the center of gravity of the bar and since it is translating let’s find the acceleration of point B.

ω = 180 rpm = 18.85 rad/s Since ω is constant aB = aBn = rω 2 = ( 0.2 m )(18.85 rad/s)

2

= 71.061 m/s2

Since AB is translating

a G = a B = 71.061 m/s 2 60°

Apply Newton’s second law to the system = Bar BC

ΣMB = ( ΣMB )eff ⇒ Cy ( 0.75) − mg ( 0.375) = − maG sin 60 °( 0.375) C y = 0.5 mg − maG sin 60 °( 0.5 ) = 0 .5( 7.5)( 9.81) − 7.5( 71.061) sin 60° ( 0.5) = −193.99 N Cy = 194.0 N !

(

ΣFy = ΣFy

)eff ⇒ By − mg + Cy = − maG sin 60 °

B y = mg − C y − maG sin 60 ° = 7.5 ( 9.81) + 193.99 − 7.5 ( 71.061) sin 60 ° = −193.99 N B y = 194.0 N !

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 16, Solution 16.

∑ MG = 0 = 200 (0.6 )( 0.866 ) − 300 ( 0.6)( 0.866) + P( 0.6)( 0.5) P = 173.2 N

(a)

!

∑ F y = 300 − 5 (9.81) + 200 = 5 (0.4 )ω 2 ω = 15.02 rad/s !

(b)

∑ Fx = P = 5 ( 0.4 )α = 173.2 (c)

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.

α = 86.6 rad/s 2

!

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Chapter 16, Solution 17.

Bar AB Free body diagram

Kinetic diagram

an = 0; Released from rest

∑ Fx = ( ∑ Fx )eff  8  u   −V =     32.2  1.5 

g cos30° 2 

−V = 2.3094 u − V = 2.3094 u

At u = 1.5 ft

V = − 3.464 lb;

u  +∑ Mp = M = mat  2  

 8   u  g  u  M =   cos30°   2   32.2   1.5  2   M = 1.1547 u 2 At

u = 1.5 ft; Shear

M = 2.598 ft ⋅ lb Moment

"

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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Bar BE

 8  u mg =   32.2 1.5

  u  g = 8 1.5    

 8   u  g  ma x =     cos30° = 2.3094 u  32.2   1.5  2 

∑ Fx = ( ∑ Fx ) eff  u  cos30 − V + 8 ° = 2.3094 u   1.5 

V = 2.3094 u +

      ∑M p = M + 8  cos30°  = 2.3094 u    1.5   2  2

u

u

u

M = −1.1547 u2

Shear

Moment

"

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 16, Solution 18.

From Problem 16.15

aG = 71.061 m/s2 60° a y = 61.541 m/s 2

Distributed mass per unit length =

7...