Solution Manual - Vector Mechanics Engineers Dynamics 8th Beer Chapter 08 PDF

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COSMOS: Complete Online Solutions Manual Organization System Chapter 8, Solution 1. FBD Block B: Tension in cord is equal to W A 25 lb from of block A and pulley. 0: N WB cos 0, N WB cos (a) For smallest WB , slip impends up the incline, and F s N 0 0: F 25 lb WB sin 0 ( sin )WB 25 lb WB min 31 lb (...

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COSMOS: Complete Online Solutions Manual Organization System

Chapter 8, Solution 1. FBD Block B: Tension in cord is equal to W A = 25 lb from FBD’s of block A and pulley. ΣF y = 0:

N − W B cos30° = 0,

N = WB cos 30°

(a) For smallest WB , slip impends up the incline, and

F = µ s N = 0.35WB cos 30° ΣFx = 0:

F − 25 lb + WB sin 30° = 0

(0.35cos 30 ° + sin 30 °) WB

= 25 lb WB min = 31.1 lb 

(b) For largest WB , slip impends down the incline, and F = − µs N = − 0.35 WB cos30° ΣFx = 0:

Fs + WB sin 30° − 25 lb = 0

( sin 30° − 0.35 cos 30°) WB = 25 lb W B max = 127.0 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 8, Solution 2. FBD Block B:

Tension in cord is equal to WA = 40 lb from FBD’s of block A and pulley. (a)

ΣFy = 0:

N − ( 52 lb) cos 25° = 0,

N = 47.128 lb

Fmax = µs N = 0.35( 47.128 lb) = 16.495 lb ΣFx = 0:

Feq − 40 lb + (52 lb ) sin 25° = 0

So, for equilibrium, Feq = 18.024 lb Since Feq > Fmax , the block must slip (up since F > 0)

∴ There is no equilibrium  (b) With slip,

F = µ k N = 0.25 ( 47.128 lb )

F = 11.78 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.

35° 

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Chapter 8, Solution 3. FBD Block: Tension in cord is equal to P = 40 N, from FBD of pulley.

(

)

W = (10 kg ) 9.81 m/s 2 = 98.1 N Σ Fy = 0 :

N − (98.1 N) cos 20° + (40 N) sin 20° = 0

N = 78.503 N Fmax = µs N = (0.30)( 78.503 N ) = 23.551 N For equilibrium:

ΣFx = 0:

(40 N) cos 20° − (98.1 N ) sin 20° − F

Feq = 4.0355 N < Fmax, F = Feq

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.

=0

∴ Equilibrium exists  F = 4.04 N

20° 

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Chapter 8, Solution 4.

Tension in cord is equal to P = 62.5 N, from FBD of pulley.

(

)

W = (10 kg ) 9.81 m/s 2 = 98.1 N ΣF y = 0:

N − (98.1 N )cos 20 ° + (62.5 N ) sin15 ° = 0 N = 76.008 N

Fmax = µs N = (0.30 )( 76.008 N ) = 22.802 N For equilibrium:

ΣFx = 0:

( 62.5 N ) cos15 ° − (98.1 N ) sin 20 ° − F

Feq = 26.818 N > Fmax

=0

so no equilibrium,

and block slides up the incline 

Fslip = µ xN = (0.25 )( 76.008 N ) = 19.00 N F = 19.00 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.

20° 

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Chapter 8, Solution 5.

Tension in cord is equal to P from FBD of pulley.

(

)

W = (10 kg ) 9.81 m/s 2 = 98.1 N

ΣF y = 0:

N − ( 98.1 N ) cos 20° + P sin 25° = 0

(1)

ΣFx = 0:

P cos 25° − (98.1 N )sin 20° + F = 0

(2)

For impending slip down the incline, F = µ s N = 0.3 N and solving (1) and (2),

PD = 7.56 N

For impending slip up the incline, F = − µs N = − 0.3 N and solving PU = 59.2 N (1) and (2), so, for equilibrium

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.

7.56 N ≤ P ≤ 59.2 N 

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Chapter 8, Solution 6. FBD Block:

(

)

W = ( 20 kg) 9.81 m/s2 = 196.2 N

For θ min motion will impend up the incline, so F is downward and F = µs N ΣF y = 0:

ΣFx = 0:

(1) + ( 2) :

N − ( 220 N ) sinθ − (196.2 N) cos35° = 0 F = µsN = 0.3 ( 220 sin θ + 196.2 cos35 ° ) N

(1)

( 220 N ) cosθ

(2)

− F − (196.2 N ) sin 35° = 0

0.3( 220 sin θ + 196.2 cosθ ) N = ( 220 cosθ ) N − ( 196.2sin 35°) N

or

220 cosθ − 66sin θ = 160.751

Solving numerically:

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.

θ = 28.9° 

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Chapter 8, Solution 7. FBD Block: For P min motion will impend down the incline, and the reaction force R will make the angle

φ s = tan−1 µ s = tan−1 (0.35) = 19.2900° with the normal, as shown.

Note, for minimum P, P must be ⊥ to R, i.e. β = φs (angle between P and x equals angle between R and normal).

β = 19.29° 

(b) then P = (160 N) cos ( β + 40°) = ( 160 N ) cos59.29° = 81.71 N (a)

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.

Pmin = 81.7 N 

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Chapter 8, Solution 8.

FBD block (impending motion downward)

φ s = tan −1 µ s = tan −1 ( 0.25 ) = 14.036°

(a) Note: For minimum P,

P⊥ R

So

β = α = 90° − (30° + 14.036° ) = 45.964°

and

P = ( 30 lb )sin α = ( 30 lb) sin ( 45.964 °) = 21.567 lb P = 21.6 lb 

(b)

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.

β = 46.0° 

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Chapter 8, Solution 9. FBD Block:

For impending motion. φ s = tan −1 µ s = tan −1 ( 0.40)

φ s = 21.801°

Note β1,2 = θ 1,2 − φ s 10 lb 15 lb = sin φs sinβ1,2

From force triangle:

15 lb  sin ( 21.801°)  = 10 lb 

β1,2 = sin −1 

33.854 °  146.146°

55.655° So θ 1,2 = β1,2 + φs =  167.947° So

(a)

equilibrium for

0 ≤ θ ≤ 55.7° 

(b)

equilibrium for

167.9° ≤ θ ≤ 180° 

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 8, Solution 10. FBD A with pulley: Tension in cord is T throughout from pulley FBD’s

ΣF y = 0:

2T − 20 lb = 0,

T = 10 lb

FBD E with pulley: For θ max , motion impends to right, and

φ s = tan −1 µ s = tan −1 ( 0.35) = 19.2900°

From force triangle, 20 lb 10 lb = , sin (θ − φs ) sinφs

2 sinφ s = sin (θ − φs )

θ = sin− 1( 2sin19.2900° ) + 19.2900° − 60.64° θ max = 60.6° 

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 8, Solution 11. FBD top block: ΣFy = 0:

N1 − 196.2 N = 0

N1 = 196.2 N (a) With cable in place, impending motion of bottom block requires impending slip between blocks, so F1 = µs N1 = 0.4 (196.2 N )

F1 = 78.48 N FBD bottom block:

ΣFy = 0:

N2 − 196.2 N − 294.3 N = 0

N2 = 490.5 N F2 = µ s N2 = 0.4( 490.5 N) = 196.2 N ΣF x = 0:

− P + 78.48 N + 196.2 N = 0

P = 275 N



FBD block: (b) Without cable AB, top and bottom blocks will move together ΣF y = 0:

N − 490.5 N = 0,

Impending slip: ΣFx = 0:

N = 490.5 N

F = µ sN = 0.40 (490.5 N ) = 196.2 N − P + 196.2 N = 0

P = 196.2 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 8, Solution 12.

FBD top block:

Note that, since φs = tan− 1 µs = tan− 1 (0.40) = 21.8 ° > 15 °, no motion will impend if P = 0, with or without cable AB. (a) With cable, impending motion of bottom block requires impending slip between blocks, so F 1 = µs N ΣFy ′ = 0:

N1 − W1 cos15° = 0,

N1 = W1 cos15° = 189.515 N

F1 = µs N1 = ( 0.40 )W1 cos15° = 0.38637W1

F1 = 75.806 N FBD bottom block:

ΣF x′ = 0:

T − F1 − W1 sin15 ° = 0 T = 75.806 N + 50.780 N = 126.586 N

(

)

W2 = ( 30 kg) 9.81 m/s2 = 294.3 N ΣFy = 0 :

N2 − (189.515 N )cos ( 15 °) − 294.3 N + ( 75.806 N) sin15° = 0

N2 = 457.74 N F2 = µs N2 = ( 0.40) ( 457.74 N ) = 183.096 N

FBD block:

ΣF x = 0:

− P + ( 189.515 N) + (75.806 N ) cos15° + 126.586 N + 183.096 N = 0

P = 361 N



(b) Without cable, blocks remain together ΣFy = 0:

N − W1 − W2 = 0

N = 196.2 N + 294.3 N = 490.5 N

F = µs N = ( 0.40)( 490.5 N) = 196.2 N ΣF x = 0:

− P + 196.2 N = 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.

P = 196.2 N



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Chapter 8, Solution 13. FBD A:

Note that slip must impend at both surfaces simultaneously. N1 + T sin θ − 16 lb = 0

ΣFy = 0:

N1 = 16 lb − T sinθ Impending slip:

F1 = µs N1 = ( 0.20) (16 lb − T sin θ )

F1 = 3.2 lb − (0.2 ) T sin θ

(1)

F1 − T cosθ = 0

(2)

ΣF x = 0:

FBD B: ΣFy = 0:

N2 − N1 − 24 lb = 0,

N2 = N1 + 24 lb = 30 lb − T sinθ

Impending slip:

F2 = µs N 2 = ( 0.20 ) (30 lb − T sin θ ) = 6 lb − 0.2 T sin θ

ΣF x = 0:

10 lb − F1 − F2 = 0 10 lb = µ s ( N1 + N2 ) = ( 0.2)  N1 + ( N1 + 24 lb)

10 lb = 0.4 N 1 + 4.8 lb, Then Then

N1 = 13 lb

F1 = µ s N1 = ( 0.2 )(13 lb ) = 2.6 lb

(1) :

T sin θ = 3.0 lb

( 2) :

T cos θ = 2.6 lb

Dividing tan θ =

3 , 2.6

θ = tan −1

3 = 49.1° 2.6

θ = 49.1° 

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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 8, Solution 14. FBD’s:

Note: Slip must impend at both surfaces simultaneously.

A: ΣFy = 0:

N1 − 20 lb = 0,

Impending slip:

N1 = 20 lb

F1 = µs N1 = ( 0.25) ( 20 lb) = 5 lb

ΣF x = 0:

− T + 5 lb = 0,

T = 5 lb

ΣF y ′ = 0:

N 2 − ( 20 lb + 40 lb) cosθ − ( 5 lb) sinθ = 0 N2 = ( 60 lb) cos θ − ( 5 lb ) sin θ

B: Impending slip: Σ Fx′ = 0:

F2 = µ s N 2 = ( 0.25 )( 60cosθ − 5sin θ )lb − F2 − 5 lb − (5 lb ) cos θ + ( 20 lb + 40 lb ) sin θ = 0 − 20cosθ + 58.75sinθ − 5 = 0

Solving numerically,

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.

θ = 23.4° 

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Chapter 8, Solution 15. FBD: For impending tip the floor reaction is at C.

(

)

W = ( 40 kg) 9.81 m/s2 = 392.4 N For impending slip φ = φ s = tan− 1 µ s = tan− 1( 0.35 )

φ = 19.2900 ° tan φ =

0.8 m EG

,

EG =

0.4 m = 1.14286 m 0.35

EF = EG − 0.5 m = 0.64286 m (a )

αs = tan− 1

EF 0.64286 m = tan− 1 = 58.109° 0.4 m 0.4 m

αs = 58.1°  (b)

P W = sin19.29° sin128.820 P = (392.4 N )( 0.424 ) = 166.379 N P = 166.4 N  Once slipping begins, φ will reduce to φk = tan− 1 µk . Then αmax will increase.

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 8, Solution 16. First assume slip impends without tipping, so F = µ sN FBD

ΣFy = 0:

N + P sin 40° − W = 0,

N = W − P sin 40°

F = µs N = 0.35 (W − Psin 40° ) ΣFx = 0:

F − P cos 40° = 0

0.35W = P ( cos 40 ° + 0.35sin 40 °) (1)

Ps = 0.35317 W Next assume tip impends without slipping, R acts at C.

ΣM A = 0:

( 0.8 m ) P sin 40° + (0.5 m ) P cos 40° − (0.4 m )W

=0

Pt = 0.4458 W > Ps from (1)

(

∴ Pmax = Ps = 0.35317 (40 kg ) 9.81 m/s 2

)

= 138.584 N (a)

Pmax = 138.6 N ⊳

(b) Slip is impending ⊳

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 8, Solution 17. FBD Cylinder: For maximum M, motion impends at both A and B FA = µ A N A ; ΣFx = 0:

FB = µ B N B N A = FB = µ B N B

N A − FB = 0

F A = µ AN A = µ Aµ BN B

ΣFy = 0:

NB (1 + µ A µB ) = W

N B + FA − W = 0 1

or

NB =

and

FB = µ B NB =

1 + µ Aµ B

W

µB

1 + µ Aµ B

µA µB W 1 + µA µB

FA = µA µB N B =

ΣM C = 0: M − r ( FA + FB ) = 0 (a) For

µA = 0

and

W

M = Wr µ B

1+ µ A 1 + µA µ B

µ B = 0.36 M = 0.360Wr 

(b) For

µ A = 0.30

and

µ B = 0.36 M = 0.422Wr 

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 8, Solution 18. FBD’s: (a)

FBD Drum:

ΣM D = 0:

 10  F  ft  − 50 lb ⋅ ft = 0  12 

F = 60 lb Impending slip: N =

F

µs

=

60 lb = 150 lb 0.40

FBD arm:

ΣM A = 0:

( 6 in.) C + ( 6 in.) F − ( 18 in.) N

= 0

C = −60 lb + 3 (150 lb ) = 390 lb Ccw = 390 lb ⊳ (b) Reversing the 50 lb ⋅ ft couple reverses the direction of F , but the magnitudes of F and N are not changed. Then, using the FBD arm:

ΣM A = 0:

( 6 in.) C − ( 6 in.) F − ( 18 in.) N

= 0

C = 60 lb + 3(150 lb ) = 510 lb Cccw = 510 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 8, Solution 19. For slipping, F = µ kN = 0.30 N

FBD’s:

(a) For cw rotation of drum, the friction force F is as shown. From FBD arm:

ΣM

A

(6 in. )(600 lb ) + (6 in. )F − (18 in. )N

= 0:

600 lb + F − 3

=0

F =0 0.30

600 lb 9

F =

Moment about D = (10 in. ) F = 666.67 lb ⋅in.

Mcw = 55.6 lb ⋅ft

⊳

(b) For ccw rotation of drum, the friction force F is reversed

ΣM A = 0:

(6 in.)( 600 lb ) − (6 in. ) F − (18 in.) N 600 lb − F − 3

=0

F =0 0.30

F =

600 lb 11

 10  600  Moment about D =  ft  lb = 45.45 lb ⋅ft  12  11  Mccw = 45.5 lb ⋅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 8, Solution 20. FBD:

(a)

ΣM C = 0:

r (F − T ) = 0,

T =F

Impending slip: F = µs N or N =

Σ F x = 0:

F

µs

=

T

µs

F + T cos (25° + θ ) − W sin 25° = 0

T 1 + cos ( 25 ° + θ )  = W sin 25 ° Σ F y = 0:

(1)

N − W cos 25° + T sin( 25° + θ ) = 0

 1  T + sin( 25° + θ )  = W cos 25° 0.35   Dividing (1) by (2):

(2)

1 + cos ( 25 ° + θ ) = tan 25° 1 + sin( 25° + θ) 0.35

Solving numerically, 25° + θ = 42.53°

θ = 17.53° ⊳ (b) From (1)

T (1 + cos 42.53° ) = W sin 25 ° T = 0.252 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.

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Chapter 8, Solution 21. FBD ladder: Note: slope of ladder =

L = 6.5 m, so AC =

4.5 m 12 13 = , so AC = ( 4.5 m ) = 4.875 1.875 m 5 12

4.875 m 3 = L, 6.5 m 4

and DC = BD =