Cap 08 by Cristian Manrique

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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. ΣFy = 0:

N − WB cos 30° = 0,

N = WB cos 30°

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

F = µ s N = 0.35WB cos30° ΣFx = 0:

F − 25 lb + WB sin 30° = 0

( 0.35cos30° + sin 30° )WB = 25 lb WB min = 31.1 lb (b) For largest WB , slip impends down the incline, and F = − µ s N = − 0.35 WB cos30° ΣFx = 0:

Fs + WB sin 30° − 25 lb = 0

( sin 30° − 0.35cos30° )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 = µ s 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 = µk N = 0.25 ( 47.128 lb )

F = 11.78 lb

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 ΣF y = 0 :

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

N = 78.503 N Fmax = µ s N = ( 0.30 )( 78.503 N ) = 23.551 N For equilibrium:

ΣFx = 0:

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

Feq = 4.0355 N < Fmax , F = Feq

∴ 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 ΣFy = 0:

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

Fmax = µ s N = ( 0.30 )( 76.008 N ) = 22.802 N For equilibrium:

ΣFx = 0:

( 62.5 N ) cos15° − ( 98.1 N ) sin 20° − F = 0

Feq = 26.818 N > Fmax

so no equilibrium,

and block slides up the incline

Fslip = µ x N = ( 0.25 )( 76.008 N ) = 19.00 N F = 19.00 N

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

ΣFy = 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 = − µ s N = − 0.3 N and solving (1) and (2),

PU = 59.2 N

so, for equilibrium

7.56 N ≤ P ≤ 59.2 N

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Chapter 8, Solution 6. FBD Block:

(

)

W = ( 20 kg ) 9.81 m/s 2 = 196.2 N

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

ΣFx = 0:

(1) + ( 2 ):

N − ( 220 N ) sin θ − (196.2 N ) cos 35° = 0 F = µ s N = 0.3 ( 220 sin θ + 196.2 cos 35° ) N

(1)

( 220 N ) cosθ

(2)

− F − (196.2 N ) sin 35° = 0

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

220cosθ − 66sin θ = 160.751

Solving numerically:

θ = 28.9°

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Chapter 8, Solution 7. FBD Block: For Pmin 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 ) cos 59.29° = 81.71 N (a)

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

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)

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 = θ1,2 − φ s 10 lb 15 lb = sinφs sinβ1,2

From force triangle:

 15 lb  33.854° sin ( 21.801° )  =  10 lb  146.146°

β1,2 = sin −1 

55.655° So θ1,2 = β1,2 + φ s =  167.947° So

(a)

equilibrium for

0 ≤ θ ≤ 55.7°

(b)

equilibrium for

167.9° ≤ θ ≤ 180°

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

ΣFy = 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 ) sinφ s

2sin φ s = sin (θ − φ s )

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

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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 = µ s N1 = 0.4 (196.2 N )

F1 = 78.48 N FBD bottom block:

ΣFy = 0:

N 2 − 196.2 N − 294.3 N = 0

N 2 = 490.5 N F2 = µ s N 2 = 0.4 ( 490.5 N ) = 196.2 N ΣFx = 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 ΣFy = 0:

N − 490.5 N = 0,

Impending slip: ΣFx = 0:

N = 490.5 N

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

P = 196.2 N

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Chapter 8, Solution 12.

FBD top block:

Note that, since φ s = tan −1 µ s = 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 F1 = µ s N ΣFy′ = 0:

N1 − W1 cos15° = 0,

N1 = W1 cos15° = 189.515 N

F1 = µ s N1 = ( 0.40 )W1 cos15° = 0.38637 W1

F1 = 75.806 N FBD bottom block:

ΣFx′ = 0:

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

(

)

W2 = ( 30 kg ) 9.81 m/s 2 = 294.3 N ΣFy = 0 :

N 2 − (189.515 N ) cos (15° ) − 294.3 N + ( 75.806 N ) sin15° = 0

N 2 = 457.74 N F2 = µ s N 2 = ( 0.40 )( 457.74 N ) = 183.096 N

FBD block:

ΣFx = 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 = µ s N = ( 0.40 )( 490.5 N ) = 196.2 N ΣFx = 0:

− P + 196.2 N = 0

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 sin θ Impending slip:

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

F1 = 3.2 lb − ( 0.2 ) T sin θ

(1)

F1 − T cosθ = 0

ΣFx = 0:

(2)

FBD B: ΣFy = 0:

N 2 − N1 − 24 lb = 0,

N 2 = N1 + 24 lb = 30 lb − T sin θ

Impending slip:

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

ΣFx = 0:

10 lb − F1 − F2 = 0 10 lb = µ s ( N1 + N 2 ) = ( 0.2 )  N1 + ( N1 + 24 lb ) 

10 lb = 0.4 N1 + 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 cosθ = 2.6 lb

Dividing tan θ =

3 , 2.6

θ = tan −1

3 = 49.1° 2.6

θ = 49.1°

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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 = µ s N1 = ( 0.25 )( 20 lb ) = 5 lb

ΣFx = 0:

− T + 5 lb = 0,

T = 5 lb

ΣFy′ = 0:

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

B: Impending slip: ΣFx′ = 0:

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

Solving numerically,

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/s 2 = 392.4 N For impending slip φ = φs = 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)

α s = tan −1

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

α s = 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 α max will increase.

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

ΣFy = 0:

N + P sin 40° − W = 0,

N = W − P sin 40°

F = µ s N = 0.35 (W − P sin 40° ) ΣFx = 0:

F − P cos 40° = 0

0.35W = P ( cos 40° + 0.35sin 40° ) Ps = 0.35317 W

(1)

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

Pt = 0.4458W > 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

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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

FA = µ A N A = µ Aµ B N B

ΣFy = 0:

N B + FA − W = 0 1

NB =

and

FB = µ B N B =

1 + µ Aµ B

µB W 1 + µ Aµ B

FA = µ Aµ B N B =

µ Aµ B W 1 + µ Aµ B

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

µA = 0

N B (1 + µ Aµ B ) = W

and

M = Wr µ B

1 + µA 1 + µ Aµ B

µ B = 0.36 M = 0.360Wr

(b) For

µ A = 0.30

and

µ B = 0.36 M = 0.422Wr

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Chapter 8, Solution 18. FBD’s: FBD Drum:

(a)

ΣM D = 0:

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

F = 60 lb Impending slip: N =

µs

60 lb = 150 lb 0.40

FBD arm:

ΣM A = 0:

( 6 in.) C + ( 6 in.) F − (18 in.) N

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

C = 60 lb + 3 (150 lb ) = 510 lb Cccw = 510 lb

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Chapter 8, Solution 19. For slipping, F = µ k N = 0.30 N

FBD’s:

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

ΣM A = 0:

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

F =

F =0 0.30

600 lb 9

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

M cw = 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

F =0 0.30

F =

600 lb 11

 10  600  Moment about D =  ft  lb  = 45.45 lb ⋅ ft  12  11  M ccw = 45.5 lb ⋅ ft

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Chapter 8, Solution 20. FBD:

(a)

ΣM C = 0:

r ( F − T ) = 0,

T = F

Impending slip: F = µ s N or N =

ΣFx = 0:

µs

F + T cos ( 25° + θ ) − W sin 25° = 0

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

(1)

N − W cos 25° + T sin ( 25° + θ ) = 0

 1  + sin ( 25° + θ )  = W cos 25° T 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 = 0.252W

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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 =

AD =

1 L 2

1 L 4

For impending slip: FA = µ s N A ,

FC = µ s NC

 12  Also θ = tan −1   − 15° = 52.380°  5 FA − W sin15° + FC cosθ − NC sin θ = 0

ΣFx = 0:

FA = W sin15° − µ s

10 10 W cosθ + W sin θ 39 39

= ( 0.46192 − 0.15652µ s )W ΣFy = 0:

N A − W cos15° + FC sin θ + NC cosθ = 0 N A = W cos15° − µ s

10 10 W sin θ − W cosθ 39 39

= ( 0.80941 − 0.20310µ s )W But FA = µ N A :

0.46192 − 0.15652µ s = 0.80941µ s − 0.20310µ s2

µ s2 − 4.7559µ s + 2.2743 µ s = 0.539, 4.2166 µ s min = 0.539

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Chapter 8, Solution 22. FBD ladder: Slip impends at both A and B, FA = µ s N A , FB = µ s N B ΣFx = 0: ΣFy = 0:

FA − N B = 0,

N B = FA = µ s N A

N A − W + FB = 0,

N A + FB = W

N A + µs N B = W

(

)

N A 1 + µ s2 = W ΣM O = 0:

( 6 m ) N B + 

5  5  m W −  m  N A = 0 4   2 

6µ s N A +

µ s2 +

(

)

5 5 N A 1 + µ s2 − N A = 0 4 2

24 µs − 1 = 0 5

µ s = − 2.4 ± 2.6

µ s min = 0.200

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Chapter 8, Solution 23. FBD rod: (a) Geometry:

BE =

L cosθ 2

EF = L sin θ So or Also, or

L  DE =  cosθ  tan β 2 

DF =

L cosθ 2 tan φ s

1  L cosθ L  cosθ tan β + sin θ  = 2  2 tan φs

tan β + 2 tan θ =

1 1 1 = = = 2.5 tan φ s µ s 0.4

(1)

L sin θ + L sin β = L sin θ + sin β = 1

Solving Eqs. (1) and (2) numerically

θ1 = 4.62°

(2)

β1 = 66.85°

θ 2 = 48.20° β 2 = 14.75° θ = 4.62° and θ = 48.2°

Therefore, (b) Now and

φ s = tan −1 µ s = tan −1 0.4 = 21.801° T W = sin φs sin ( 90 + β − φ s ) T =W

sin φs sin ( 90 + β − φ s )

For

θ = 4.62°

T = 0.526W

θ = 48.2°

T = 0.374W

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Chapter 8, Solution 24. FBD: Assume the weight of the slender rod is negligible compared to P. First consider impending slip upward at B. The friction forces will be directed as shown and FB,C = µ s N B,C ΣM B = 0:

( L sinθ ) P − 

a  sin θ

NC = P ΣFx = 0:

  NC = 0 

L 2 sin θ a

NC sin θ + FC cosθ − N B = 0 NC ( sin θ + µ s cosθ ) = N B

so ΣFy = 0:

NB = P

L 2 sin θ ( sin θ + µ s cosθ ) a

− P + NC cosθ − FC sin θ − FB = 0

P = NC cosθ − µ s NC sin θ − µ s N B so P = P

L 2 L sin θ ( cosθ − µ s sin θ ) − µ s P sin 2 θ ( sin θ + µ s cosθ ) a a

Using θ = 35° and µ s = 0.20, solve for

(1)

L = 13.63. a

To consider impending slip downward at B, the friction forces will be reversed. This can be accomplished by substituting µ s = − 0.20 in L = 3.46. equation (1). Then solve for a Thus, equilibrium is maintained for

3.46 ≤

L ≤ 13.63 a

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Chapter 8, Solution 25. FBD ABC:

ΣM C = 0:

0.045 m + ( 0.30 m ) sin 30° ( 400 N ) sin 30° + 0.030 m + ( 0.30 m ) cos 30° ( 400 N ) cos 30°  12   5  − ( 0.03 m )  FBD  − ( 0.045 m )  FBD  = 0  13   13 

FBD = 3097.64 N

FBD Blade:

ΣFx = 0:

N −

25 ( 3097.6 N ) = 0 65

N = 1191.4

F = µ s N = 0.20 (1191.4 N ) = 238.3 N ΣFy = 0:

P+F−

60 ( 3097.6 N ) = 0 65

P = 2859.3 − 238.3 = 2621.0 N Force by blade

P = 2620 N

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Chapter 8, Solution 26. FBD CD: Note: The plate is a 3-force member, and for minimum µ s , slip impends at C and D, so the reactions there are at angle φ s from the normal. From the FBD,

OCG = 20° + φ s

and

ODG = 20° − φ s

Then OG = ( 0.5 in.) tan ( 20° + φs )

 1.2 in.  + 0.5 in.  tan ( 20° − φs ) and OG =   sin70°  Equating, tan ( 20° + φs ) = 3.5540 tan ( 20° − φs ) Solving numerically,

φ s = 10.5652°

µ s = tan φs = tan (10.5652° ) µ s = 0.1865

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Chapter 8, Solution 27. FBD pin A:

From FBD Whole the force at A = 750 lb

4 ( FAB′ − FAB ) = 0, FAB′ = FAB 5

ΣFx = 0:

3 750 lb − 2 FAB = 0, FAB = 625 lb 5

ΣFy = 0: FBD Casting:

ΣFx = 0:

N D′ − N D = 0, N D′ = N D = N

FD = FD′ = µ N D , or N D =

Impending slip

ΣFy = 0:

µs

2FD − 750 lb = 0, FD = 375 lb ND =

375 lb

µs

FBD ABCD:

ΣM C = 0:

(12 in.)

375 lb

µs

4 5

(12 in.) N − ( 6 in.) F − ( 42.75 in.) ( 625 lb ) = 0 = ( 6 in.)( 375 lb ) + ( 42.75 in.)

4 ( 625 lb ) = 0 5

µ s = 0.1900

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Chapter 8, Solution 28. From FBD Whole, and neglecting weight of clamp compared to 550 lb plate, P = − W Since AB is a two-force member, B is vertical and B = W .

FBD BCD: ΣM C = 0:

(1.85 in.)W − ( 2.3 in.) D cos 40° − ( 0.3 in.) D sin 40° = 0,

D = 0.94642W

FBD EG:

( 0.9 in.) NG − (1.3 in.) FG − (1.3 in.) N D cos 40° = 0

ΣM E = 0: Impending slip:

FG = µ s NG

Solving: ( 0.9 − 1.3µ s ) NG = 0.94250W

(1)

FBD Plate: By symmetry NG = NG′ , ΣFy = 0:

FG = FG′ = µ s NG

2 FG − W = 0,

Substitute in (1): ( 0.9 − 1.3µ s )

FG =

W , 2

NG =

W 2µ s

W = 0.94250W 2µ s

Solving, µ s = 0.283,

µsm = 0.283

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Chapter 8, Solution 29. FBD table + child:

( ) = 16 kg ( 9.81 m/s ) = 156.96 N

WC = 18 kg 9.81 m/s 2 = 176.58 N WT

(a) Impending tipping about E , N F = FF = 0, and

ΣM E = 0:

( 0.05 m )(176.58 N ) − ( 0.4 m )(156.96 N ) + ( 0.5 m ) P cosθ − ( 0.7 m ) P sin θ

33cosθ − 46.2sin θ = 53.955

Solving numerically

θ = −36.3°

and

θ = −72.6° −72.6° ≤ θ ≤ −36.3°

Therefore Impending tipping about F is not possible (b) For impending slip:

FE = µ s N E = 0.2 N E

ΣFx = 0: FE + FF − P cosθ = 0

FF = µ s N F = 0.2 N F

0.2 ( N E + N F ) = ( 66 N ) cos θ

ΣFy = 0: N E + N F − 176.58 N − 156.96 N − P sin θ = 0 N E + N F = ( 66sin θ + 333.54 ) N So Solving numerically,

330 cosθ = 66sin θ + 333.54

θ = −3.66°

and

θ = −18.96°

Therefore,

−18.96° ≤ θ ≤ −3.66°

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Chapter 8, Solution 30. Geometry of four-bar:

Considering the geometry when α = 0, 1/ 2

2 2 LCD = ( 60 mm − 52 mm ) + ( 36 mm + 22 mm )   

= 58.549 mm

In general, 52 mm − ( 36 mm ) sin α = 60 mm − ( 58.549 mm ) sin β  36sin α + 8  so β = sin −1    58.549  (a) FBD ACE:

α =0

β = 7.8533°, note that the links at E and K are prevented from pivoting downward by the small blocks FE FCD sin β − FE = 0, FCD = ΣFy = 0: sin 7.8533° ΣM A = 0:

( 60 mm ) 

FE   cos 7.8533° − ( 32 mm ) FE − ( 212 mm ) N E = 0  sin 7.8533° 

Impending slip on pad N E =

µs

, so

 212   435.00 − 32 −  FE = 0 µs  

µ s = 0.526

(b) α = 30°, β = 26.364° ΣFx = 0: ΣFy = 0:

−

3 FAB + FCD cos 26.364° − N E = 0 2

1 − FAB + FCD sin 26.364° − FE = 0 2

Eliminating FAB ,

FCD ( 0.89599 − 0.76916 ) − N E + FE = 0

Impending slip FE = µ s N E , so 0.126834 FE = (1 − µ s ) N E

( 60 mm ) FCD cos 26.364° − ( 212 mm ) N E − ( 32 mm ) µ s N E 53.759 FCD = ( 212 − 32µ s ) N E = 0

ΣM A = 0:

212 − 32µ s 53.759 = 1 − µs 0.12634

µ s = 0.277

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Chapter 8, Solution 31. FBD ABD:

(15 mm ) N A − (110 mm ) FA = 0

ΣΜ D = 0: Impending slip:

FA = µ SA N A

15 − 110µ SA = 0 µ SA = 0.136364

µ SA = 0.1364 FA − Dx = 0, Dx = FA = µ SA N A

ΣFx = 0:

FBD Pipe:

r = 60 mm ΣFy = 0:

NC − N A = 0,

NC = N A

FBD DF:

ΣM F = 0: Impending slip: So,

( 550 mm ) FC − (15 mm ) NC − ( 500 mm ) Dx = 0 FC = µ SC NC = µ SC N A

550µ SC N A − 15N A − 500µ SA N A = 0

550µ SC = 15 + 500 ( 0.136364 )

µ SC = 0.1512

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Chapter 8, Solution 32. FBD Plate:

Assume reactions as shown, at ends of sleeves, FA = µ s N A ,

For impending slip

FB = µ s N B

P sin θ − µ s N A − µ s N B = 0

ΣFx = 0:

N A + N B = 2.5 P sin θ ΣFy = 0:

N A − N B − P cosθ = 0,

Solving: N A = ΣM B = 0:

P ( 2.5sin θ + cosθ ) , 2

N A − N B = P cosθ NB =

P ( 2.5sin θ − cosθ ) (1) 2

( 23.5 in.) P sin θ − (16 in.) N A + (1 in.) FA = 0

( 23.5 in.) P sin θ

P − 16 in. − 0.4 (1 in.)  ( 2.5sin θ + cosθ ) = 0 2

4sin θ − 7.8cosθ = 0,

(2)

θ = 62.9°

For θ > 62.9°, the panel will be self locking, ∴ motion for θ ≤ 62.9°. As θ decreases, N B will reverse direction at 2.5sin θ − cosθ = 0, (see equ. 1) or at θ = 21.8°. So for θ ≤ 21.8°

ΣFx = 0 :

P sin θ − µ s ( N A + N B ) = 0 N A + N B = 2.5 P sin θ

ΣFy = 0:

N A + N B − P cosθ = 0,

∴ 2.5sin θ = cosθ ,

N A + N B = P cosθ

θ = 21.8°

So impending motion for 21.8° ≤ θ ≤ 62.9° W

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Chapter 8, Solution 33. FBD Plate:

Assuming reactions as shown, at ends of sleeves, For impending slip

FB = µ s FB

P sin θ − µ s ( N A + N B ) = 0

ΣFx = 0: ΣFy = 0:

FA = µ s N A ,

N A + N B = 2.5 P sin θ

(1)

N A − N B = P cosθ

(2)

N A − N B − P cosθ = 0, NA =

Solving:

P ( 2.5sin θ + cosθ ) , 2

NB =

P ( 2.5sin θ − cosθ ) 2

Note that, for θ < 21.8°, N B becomes negative, so we must change equ. 2 to N A + N B = P cosθ , ( 2′ ) but equ. (1) does not change. Solving (1) and ( 2′ ) gives P cosθ = 2.5P sin θ , or θ = 21.8°, so the lower limit for impending slip is θ = 21.8°. For θ ≥ 21.8°, the forces are as shown, and ΣM B = 0:

( 23.5 in.) P sin θ

+ xP cosθ + (1 in.) FA − (16 in.) N A = 0

P + x P cosθ + 0.4 (1 in.) − (16 in.)  ( 2.5sin θ + cosθ ) = 0 2 x 4sin θ − ( 7.8 in.) − x  cosθ = 0, tan θ = 1.950 − 4 in.

( 23.5 in.) P sin θ or

(a) For x = 4 in., tan θ = 1.950, θ = 43.5°. For θ > 43.5° self locking

∴ impending motion for 21.8° ≤ θ ≤ 43.5° W (b) As x increases from 4 in., the upper bound for θ decreases, becoming 21.8° ( tan θ = 0.4000 ) when x = ( 4 in.)(1.950 − 0.400 ) = 6.2 in. Thus xmax = 6.20 in. W at which θ must equal 21.8°.

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Chapter 8, Solution 34. FBD Collar:

Impending motion down:

a −a cosθ

Stretch of spring x = AB − a =

 a   1  − a  = (1.5 kN/m )( 0.5 m )  − 1 Fs = kx = k  cos cos θ θ      1  = ( 0.75 kN )  − 1  cos θ 

N − Fs cosθ = 0

ΣFx = 0:

N = Fs cosθ = ( 0.75 kN )(1 − cosθ )

Impending motion up: Impending slip:

F = µ s N = ( 0.4 )( 0.75 kN )(1 − cosθ ) = ( 0.3 kN )(1 − cosθ )

+ down, – up ΣFy = 0:

Fs sin θ ± F − W = 0

( 0.75 kN )( tan θ or

− sin θ ) ± ( 0.3 kN )(1 − cosθ ) − W = 0

W = ( 0.3 kN ) [ 2.5 ( tan θ − sin θ ) ± (1 − cosθ )]

with θ = 30°:

Wup = 0.01782 kN Wdown = 0.0982 kN

( OK ) ( OK )

Equilibrium if 17.82 N ≤ W ≤ 98.2 N W

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Chapter 8, Solution 35. Geometry: 1 m − ( 0.5 m ) cos α  tan θ = ( 0.5 m ) sin α

tan θ ( 2 − cos α ) = sin α

θ = 30° → α = 60° then LAB = (1 m ) cos 30° = FBD B:

3 m 2

1   kN   3 m − m Fs = k ( LAB − L0 ) = 1.5   m  2 2   Fs = 0.75

(

)

3 − 1 kN = 549.04 N

ΣFx = 0:

F + W sin 60° − 549.04 N = 0 F = 549.04 N −

ΣFy = 0:

W 2

N − W cos 60° = 0,

N =

3 W 2

For impending slip upward, F is as shown and F = µ s N , so 549.04 N −

W 3 W, = 0.40 2 2

Wmin = 648.61 N

For impending slip downward, F is reversed, or F = − µ s N , so 549.04 N − m=

( 9.81 m/s ) 2

W 3 W, = − 0.40 2 2 so

Wmax = 3575 N 66.1 kg ≤ m ≤ 364 kg W

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Chapter 8, Solution 36. FBD Collar:

Note: BC is a two-force member, and for M max , slip will impend to the right. ΣFy = 0: Impending slip: ΣFx = 0:

FBC cosθ − N = 0,

N = FBC cosθ

F = µ s N = µ s FBC cosθ FBC sin θ − F − P = 0 FBC ( sin θ − µ s cosθ ) = P

FBD AB:

ΣM A = 0:

M − ( 2l ) FAB cosθ = 0 M = 2l cosθ

P sin θ − µ s cosθ M max =

2 Pl W tan θ − µ s

For µ s = tan θ , M max = ∞   self locking W For µ s > tan θ , M max < 0 

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Chapter 8, Solution 37. Geometry:

FBD AB:

θ = cos −1

L L L + − 2 4 2 = 60° L 2

For min

a L

slip will impend to right and reactions will be at

φ s = tan −1 µ s = tan −1 ( 0.35 ) = 19.2900° from normal. Note: AB is a three-force member

CD = a tan ( 60 + φs ) = ( L − a ) tan ( 60° − φ s ) a tan ( 79.29° ) = ( L − a ) tan ( 40.71° )

6.1449 =

L −1 a

a = 0.13996 L min

a = 0.1400 W L

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Chapter 8, Solution 38.

FBD A:

Note: Rod is a two force member. For impending slip the reactions are at angle

φ s = tan −1 µ s = tan −1 ( 0.40 ) = 21.801°

FBD B:

Consider first impending slip to right 9 lb FAB = = 3.8572 lb tan 66.801 ΣFy = 0:

N B − ( 3.8522 lb ) sin 30° − ( 6 lb ) cos 30° = 0 N B = 7.1223 lb,

FB = µ s N B = 0.40 ( 7.1223 lb ) FB = 2.8489 lb

ΣFx = 0:

− 2.8489 lb + ( 3.8572 lb ) cos 30° − ( 6 lb ) sin 30° − P = 0 Pmin = − 2.508 lb

FBD A:

Next consider impending slip to left FAB = ( 9 lb ) tan 66.801° = 21.000 lb FBD B:

ΣFy = 0:

N B − ( 21 lb ) sin 30° − ( 6 lb ) cos 30° = 0,

N B = 15.6959 lb

FB = µ s N B = 0.4 (15.6959 lb ) = 6.2784 lb ΣFx = 0:

6.2784 lb + ( 21 lb ) cos 30° − ( 6 lb ) sin 30° − P = 0 Pmax = 21.465 lb equilibrium for − 2.51 lb ≤ P ≤ 21.5 lb W

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Chapter 8, Solution 39. FBD AB:

ΣM A = 0:

8 in 2 + 4 in 2 ( N ) − M A = 0

N =

(12 lb ⋅ ft )(12 in./ft ) 8.9443 in.

= 16.100 lb

F = µ s N = 0.3 (16.100 lb ) = 4.83 lb

Impending motion:

Note: For max MC, need F in direction shown; see FBD BC. FBD BC + collar:

ΣM C = 0: or

MC =

M C − (17 in.)

1 2 2 N − ( 8 in.) N − (13 in.) F =0 5 5 5

17 in. 16 in. 26 in. (16.100 lb ) + (16.100 lb ) + ( 4.830 lb ) = 293.77 lb ⋅ in. 5 5 5

( MC )max

= 24.5 lb ⋅ ft

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Chapter 8, Solution 40. FBD yoke:

ΣFx = 0:

P − N = 0,

N = P = 8 lb F = µ s N = 125 ( 8 lb )

For impending slip,

F = 2 lb For M max , F on yoke is down as shown FBD wheel and slider:

For M min , F on yoke is up.

(a) For M max the 2 lb force is up as shown. ΣM B = 0:

M B − ( 3 in.) sin 65° ( 8 lb ) − ( 3 in.) cos 65° ( 2 lb ) = 0

M B max = 24.3 lb ⋅ in. W (b) For M min the 2 lb force is reversed, and ΣM B = 0:

M B − ( 3in .) sin 65°  ( 8 lb ) + ( 3 in.) cos 65°  ( 2 lb ) = 0

M B min = 19.22 lb ⋅ in.

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Chapter 8, Solution 41. FBD Rod:

ΣM A = 0:

( 20 in.) N1 − (12.5 in.)(12 lb ) = 0 N1 = 7.5 lb.

FBD Cylinder:

ΣFy = 0:

N 2 − 7.5 lb − 36 lb = 0,

N 2 = 43.5 lb

since µ1 = µ2 and N1 < N 2 , slip will impend at top of cylinder first, so F1 = µ s N1 . F1 = 0.35 ( 7.5 lb ) = 2.625 lb ΣM D = 0:

( 4.25 in.) P − (12.5 in.)( 2.625 lb ) = 0,

P = 7.7206 lb Pmax = 7.72 lb W

To check slip analysis above, ΣFy = 0:

N 2 − 36 lb − 7.5 lb = 0 N 2 = 43.5 lb

F2max = µ s N 2 = 0.35 ( 43.5 lb ) = 15.225 lb ΣFx = 0:

P − F1 − F2 = 0,

7.72 lb − 2.625 lb − F2 = 0

F2 = 5.095 lb < Fmax ,

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Chapter 8, Solution 42. FBD pulley: Note that φSA = tan −1 µ SA = tan −1 ( 0.5 ) = 26.565° < 30°, Cable is needed to keep A from sliding downward.

ΣFy = 0:

2T − WB = 0,

T =

WB , 2

WB = 2T

(1)

FBD block A: (a) For minimum WB , there will be impending slip of block A downward, and FA = µ SA N A as shown.

ΣFy′ = 0:

N A − WA cos 30° = 0,

N A = WA cos 30°

= 23.544 N cos 30° = 20.390 N

(

)

WA = ( 2.4 kg ) 9.81 m/s 2 = 23.544 N

FA = ( 0.50 )( 20.390 N ) = 10.195 N

FBD block C:

ΣFx′ = 0:

T − WA sin 30° + FA = 0

T = ( 23.544 N ) sin 30° − 10.195 N = 1.577 N From (1)

WB = 2T = 3.154 N,

mB =

3.154 N

(

9.81 m/s 2

)

= 0.322 kg,

mB min = 322 g

ΣFy = 0:

NC − WC = 0,

NC = 58.86 N

FC max = µ SC NC = 0.30 ( 58.86 N ) = 17.658 N Since T = 1.577 N < FC max , block B doesn’t slip and above answer for

mB min is correct.

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(b) For mB max assume impending slip of block C to left, FC = Fmax

ΣFx = 0:

− T + FC = 0,

From (1) WB = 2T = 35.316 N,

T = FC = FC max = 17.658 N

mB =

WB 35.316 N = = 3.6 kg g 9.81 m/s 2

From FBD block A,

ΣFx = 0:

T − WA sin 30° + FA = 0,

FA = WA sin 30° − T

FA = ( 23.544 N ) sin 30° − 17.658 N = − 5.886, Since FA < FA max , A does not slip

FAmax = 10.195 N M B max = 3.6 kg

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Chapter 8, Solution 43. FBD A: For impending motion A must start up and C down the incline. Since the normal force between A and B is less than that between B and C, and the friction coefficients are the same, Fmax will be reached first between A and B, and B and C will stay together. ΣFy = 0:

Impending slip: ΣFx = 0:

N1 − ( 4 lb ) cos 30° = 0,

F1 = µ s N1 = 2 3µ s lb

T − ( 4 lb ) sin 30° − 2 3µ s lb = 0

(

FBD B and C:

N1 = 2 3 lb

)

T = 2 1 + 3µ s lb ΣFy = 0:

N 2 − 2 3 lb − ( 3 lb + 8 lb ) cos 30° = 0

N2 = Impending slip: ΣFx = 0:

(1)

15 3 lb 2

15 3µ s lb 2  15   3  µs  lb − ( 3 + 8) lbsin 30° = 0 T +  2 3 +   2 

F2 = µ s N 2 =

 11 19  3µ s  lb T = − 2 2 

FBD B:

Equating (1) and (2):

(

(2)

)

4 1 + 3µ s lb = 11 − 19 3µ s

23 3 µ s = 7,

µ s min = 0.1757 W

To check slip reasoning above: ΣFy = 0:

N3 − 2 3 lb − ( 3 lb ) cos 30° = 0, F3max = µ s N3 =

ΣFx = 0:

N3 =

7 3 lb 2

7 3µ s 2

− ( 3 lb ) sin 30° + 2 3µ s lb − F3 = 0

F3 = 2 3 ( 0.1757 ) lb −

3 lb = − 0.891 lb 2

F3 < F3max , OK

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Chapter 8, Solution 44. FBD rod: 3 in. N B − ( 4.5 in.) cosθ  W = 0 cosθ

ΣM A = 0: or

N B = (1.5cos 2 θ )W

Impending motion:

FB = µ s N B = (1.5µ s cos 2 θ )W

= ( 0.3cos 2 θ )W ΣFx = 0:

N A − N B sin θ + FB cosθ = 0 N A = (1.5cos 2 θ )W ( sin θ − 0.2 cosθ )

or Impending motion:

FA = µ s N A = ( 0.3cos 2 θ )W ( sin θ − 0.2 cosθ )

ΣFy = 0: FA + N B cosθ + FB sinθ − W = 0

(

FA = W 1 − 1.5cos3 θ − 0.3cos 2 θ sin θ

)

Equating FA’s 0.3cos 2 θ ( sin θ − 0.2cosθ ) = 1 − 1.5cos3 θ − 0.3cos 2 θ sinθ 0.6cos 2 θ sin θ + 1.44cos3 θ = 1 Solving numerically

θ = 35.8° W

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Chapter 8, Solution 45. FBD pin A:

ΣFx = 0: ΣFy = 0: Solving:

12 3 FAB − FAC = 0 13 5 5 4 FAB + FAC − P = 0 13 5 FAB =

13 P, 21

FAC =

20 P 21

FBD B:

ΣFx = 0:

NB −

12 13 ⋅ P = 0, 13 21

NB =

For Pmin slip of B impends down, so FB = µ s N B = ΣFy = 0: FBD C:

11 12 5 13 ⋅ P− ⋅ P − 18 lb = 0, 20 21 13 21

12 P 21

11 NB 20 Pmin = 236.25 lb

(For P < 236.25 lb, A will slip down)

ΣFy = 0:

NC − 80 lb −

4 20 16 ⋅ P = 0, NC = 80 lb + P 5 21 21

For Pmax slip of C impends to right, FC = µ s NC FC =

or ΣFx = 0:

11  16  44 P  = 44 lb + P  80 lb + 20  21  105

3 20 ⋅ P − FC = 0, 5 21

12 44 P = 44 lb + P 21 105 Pmax = 288.75 lb

∴ equilibrium 236 ≤ P ≤ 289 W

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Chapter 8, Solution 46. φ s = tan −1 µ s = tan −1 ( 0.4 ) = 21.801°, slip impends at wedge/block wedge/wedge and block/incline FBD Block:

R2 530 lb = sin 41.801° sin 46.398°

R2 = 487.84 lb

FBD Wedge:

P 487.84 lb = sin 51.602° sin 60.199° P = 441 lb

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Chapter 8, Solution 47. φ s = tan −1 µ s = tan −1 ( 0.40 ) = 21.801°, and slip impends at wedge/lower block, wedge/wedge, and upper block/incline interfaces. FBD Upper block and wedge: R2 530 lb = sin 41.801° sin 38.398° R2 = 568.76 lb

FBD Lower wedge:

P 568.76 lb = sin 51.602° sin 68.199° P = 480 lb

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Chapter 8, Solution 48.

(

)

WD = (18 kg ) 9.81 m/s 2 = 176.58 N Fs = kx = ( 3.5 kN/m )( 0.1 m ) = 0.35 kN = 350 N

φ s = tan −1 µ s = tan −1 ( 0.25 ) = 14.0362° FBD Lever:

ΣM C = 0:

( 0.3 m )( 350 N ) − ( 0.4 m )(176.58 N ) − ( 0.525 m ) RA cos 4.0362° + ( 0.05 m ) RA sin 4.0362° = 0 RA = 66.070 N

ΣFx = 0:

ΣFy = 0:

( 66.07 N ) sin 4.0362° + Cx = 0,

Cx = − 4.65 N

( 66.07 N ) cos 4.0362° − 350 N − 176.58 N = 0

FBD Wedge:

P 66.070 N = sin18.072° sin 75.964° P = 21.1 lb (a) (b)

P = 21.1 lb C x = 4.65 N C y = 461 N

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Chapter 8, Solution 49.

(

)

WD = (18 kg ) 9.81 m/s 2 = 176.58 N Fs = kx = ( 3.5 kN/m )( 0.1 m ) = 0.35 kN = 350 N

φ s = tan −1 µ s = tan −1 ( 0.25 ) = 14.0362° FBD Lever:

ΣM C = 0:

( 0.3 m )( 350 N ) − ( 0.4 m )(176.58 N ) − ( 0.525 m ) RA cos 24.036° − ( 0.05 m ) RA sin 24.036° = 0 RA = 68.758 N

ΣFx = 0: ΣFy = 0:

C x − ( 68.758 N ) sin 24.036° = 0,

Cx = 28.0 N

C y − 350 N − 176.58 N + ( 68.758 N ) cos 24.036° = 0 C y = 464 N

FBD Wedge: P 68.758 N = sin 38.072° sin 75.964°

(a) (b)

P = 43.7 N

C x = 28.0 N C y = 464 N

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Chapter 8, Solution 50. For steel/steel contact,

φ s1 = tan −1 µ s1 = tan −1 ( 0.3) = 16.6992°

For steel/concrete interface, φ s2 = tan −1 µ s2 = tan −1 ( 0.6 ) = 30.964°

FBD Plate CD: ΣFy = 0:

N − 90 kN = 0,

F = µ s1 N = 0.3 ( 90 kN ) = 27 kN

Impending slip: ΣFx = 0:

F = 90 kN

F − Q = 0,

Q = F = 27 kN

FBD Top wedge assuming impending slip between wedges: ΣFy = 0:

Rw = 100.74 kN

Rw cos 26.699° − 90 kN = 0,

ΣFx = 0:

P − 27 kN − (100.74 kN ) sin 26.699° = 0

P = 72.265 kN,

(a)

P = 72.3 kN

(b)

Q = 27.0 kN

To check above assumption; note that bottom wedge is a two-force member so the reaction of the floor on that wedge is Rw, at 26.699° from the vertical. This is less than φ s2 = 30.964°, so the bottom wedge doesn’t slip on the concrete.

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Chapter 8, Solution 51. For steel/steel contact, φ s1 = tan −1 µ s1 = tan −1 ( 0.30 ) = 16.6992° For steel/concrete contact, φ s2 = tan −1 µ s2 = tan −1 ( 0.60 ) = 30.964°

FBD Plate CD and top wedge: Q = 90 kN tan 26.6992° = 45.264 kN Rw =

90 kN = 100.741 kN cos 26.6992°

FBD Bottom wedge: slip impends at both surfaces

P 100.714 kN = sin 57.663° sin 59.036°

(a)

P = 99.3 kN

(b)

Q = 45.3 kN

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Chapter 8, Solution 52. FBD Wedge:

φ s = tan −1 µ s = tan −1 ( 0.4 ) = 21.801° By symmetry RB = RC ΣFy = 0:

2RC sin ( 29.801° ) − P = 0,

P = 0.9940 RC

FBD Block C: RC 175 lb , = sin 41.801° sin18.397 lb

P = 367.3 lb

(a)

P = 367 lb

b) Note: That increasing friction between B and the incline will mean that block B will not slip, but the above calculations will not change.

(b)

P = 367 lb

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Chapter 8, Solution 53. FBD Block C:

φ s = tan −1 µ s = tan −1 ( 0.4 ) = 21.8014° ΣFx = 0:

RACx − RCFx = 0

ΣFy = 0: so

RCFy − RACy − 175 lb = 0

RCFy RCFx

−

RACy RACx

175 lb RACx

cot ( 20° + φ ) − cot ( 32.2° ) > 0

φ < 12.2° < φ s = 21.8° so block C does not slip (or impend)

FBD Block B: (a) φ B = tan −1 µ B = tan −1 ( 0.4 ) = 21.8014° RB 175 lb , = sin 41.8014° sin 46.3972°

RB = 161.083 lb

(b) φ B = tan −1 µ B = tan −1 ( 0.6 ) = 30.9638° RB 175 lb , = sin 50.9638° sin 37.2330°

RB = 224.65 lb

FBD Wedge: P RB = , sin 59.6028° sin 52.1986°

(a)

RB = 161.083 lb,

(b)

RB = 224.65 lb,

P = 1.09163 RB P = 175.8 lb P = 245 lb

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Chapter 8, Solution 54. Since vertical forces are equal and µ s ground > µ s wood, assume no impending motion of board. Then there will be impending slip at all wood/wood contacts, φ s = tan −1 µ s = tan −1 ( 0.35 ) = 19.2900°

FBD Top wedge: R1 =

8 kN = 8.4758 kN cos19.29° R1 P = sin 52.710° cos 56.580° P = 8.892 kN

To check assumption, consider

FBD wedges + board: F1 = µ1 8 kN = 0.35 ( 8 kN ) = 2.8 kN

ΣFy = 0:

NG − 8 kN = 0,

NG = 8 kN

FG max = µG NG = ( 0.6 )( 8 kN ) = 4.8 kN

ΣFx = 0:

FG − F1 = 0,

FG = F1 = 2.8 kN FG < FG max ,

∴ P = 8.89 kN

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Chapter 8, Solution 55. Assume no impending motion of board on ground. Then there will be impending slip at all wood/wood interfaces.

FBD Top wedge: Wedge is a two-force member so R 2 = − R1 and θ = 2φs = 2 tan −1 µ s = 2 tan −1 ( 0.35 )

θ = 38.6°

To check assumption, consider

FBD wedges + board: F1 = µ1 8 kN = 0.35 ( 8 kN ) = 2.8 kN

ΣFy = 0:

NG − 8 kN = 0,

NG = 8 kN

FG max = µG NG = ( 0.6 )( 8 kN ) = 4.8 kN

ΣFx = 0:

FG − F1 = 0,

FG = F1 = 2.8 kN FG < FG max ,

∴ P = 8.89 kN

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Chapter 8, Solution 56. FBD Cylinder: Slip impends at B

φ SC = tan −1 ( 0.35 ) = 19.2900° ΣM A = 0:

r RC cos (12° + 19.29° ) −

4r W =0 3π

RC = 0.49665, W = 124.163 lb

FBD Wedge:

φ SF = tan −1 µ SF = tan −1 ( 0.50 ) = 26.565° P 124.163 lb = sin 58.855° sin 63.435° P = 117.5 lb

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Chapter 8, Solution 57. FBD tip of screwdriver:

φs = tan −1 µ s = tan −1 ( 0.12 ) = 6.8428° by symmetry R1 = R2

ΣFy = 0:

2R1 sin ( 6.8428° + 8° ) − 3.5 N = 0 R1 = R2 = 6.8315 N

If P is removed quickly, the vertical components of R1 and R2 vanish, leaving the horizontal components

H1 = H 2 = ( 6.8315 N ) cos14.8428° = 6.6035 N Side forces = 6.60 N This is only instantaneous, since 8° > φ s , so the screwdriver will be forced out.

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Chapter 8, Solution 58.

As the plates are moved, the angle θ will decrease. (a)

φ s = tan −1 µ s = tan −1 0.2 = 11.31°. As θ decreases, the minimum angle at the contact approaches 12.5° > φs = 11.31°, so the wedge will slide up and out from the slot.

(b)

φ s = tan −1 µ s = tan −1 0.3 = 16.70°. As θ decreases, the angle at one contact reaches 16.7°. (At this time the angle at the other contact is 25° − 16.7° = 8.3° < φ s ) The wedge binds in the slot.

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Chapter 8, Solution 59. FBD Wedge:

φ s = tan −1 µ s = tan −1 ( 0.35 ) = 19.2900° by symmetry R1 = R2 ΣFy = 0:

2 R1 sin 22.29° − 60 lb = 0 R2 = 79.094 lb

When P is removed, the vertical component of R1 and R2 will vanish, leaving the horizontal components H1 = H 2 = ( 79.094 lb ) cos 22.29°

= 73.184 lb Final forces H1 = H 2 = 73.2 lb Since these are at 3° ( < φs ) from the normal, the wedge is self-locking and will remain in place.

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Chapter 8, Solution 60. FBD Cylinder:

(

)

W = ( 80 kg ) 9.81 m/s 2 = 784.8 N

ΣM G = 0:

FA − FB = 0,

ΣM D = 0:

dN B − dN A + rW = 0,

N A > NB,

FA = FB

(1) N A = NB +

W 3

(2)

FA max > FB max

∴ slip impends first at B. FB = µ s N B = 0.25 N B ΣM A = 0: ( r cos 30° ) N B − ( r sin 30° )W − r (1 + sin 30° )( 0.25 N B ) = 0 r = note d = tan 30

N B = 1.01828W = 799.15 N

FB = 0.25 N B = 199.786 N

From (2) above, N A = 799.15 N +

784.8 N = 1252.25 N 3

From (1), FA = FB = 199.786 N

ΣFy = 0:

FBD Wedge:

NC − (1252.25 N ) cos10° + 199.786 N sin10° = 0 NC = 1198.53 N

Impending slip FC = µ s NC = 0.25 (1198.53 N ) = 299.63 N

ΣFx = 0:

P − 299.63 N − (199.786 N ) cos10°

− (1252.25 N ) sin10° = 0

P = 714 N

20.0°

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Chapter 8, Solution 61. FBD Cylinder:

(

)

W = ( 80 kg ) 9.81 m/s 2 = 784.8 N For impending slip at B, ΣM A = 0:

FB = µ sB N B = 0.30 N B

( r cos 30° ) N B − r (1 + sin 30° )( 0.30 N B ) − r sin 30°W = 0

N B = 1.20185W = 943.21 N FB = 0.30 N B = 0.36055W ΣM G = 0: ΣFx = 0:

r ( FA − FB ) = 0,

FA = FB = 0.36055W

N A sin 30° + FA cos30° − N B = 0 NA =

− ( 0.36055W ) cos30° + 1.20185W sin 30°

N A = 1.77920W For minimum µ A , slip impends at A, so

µ A min =

FA 0.36055W = = 0.2026 N A 1.77920W

µ A min = 0.203

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Chapter 8, Solution 62. FBD plank + wedge:

(8 ft ) N B − (1.5 ft )( 48 lb/ft )( 3 ft )

ΣM A = 0:

− ( 2 ft )

1 ( 48 lb/ft )( 3 ft ) 2

 5  1 −  3 +  ft  ( 96 lb/ft )( 5 ft ) = 0 3  2 

N B = 185 lb ΣFy = 0:

 48 + 96  NW + 185 lb −  lb/ft  ( 3ft ) 2   +

1 ( 96 lb/ft )( 5 ft ) = 0 2 NW = 271 lb

Since NW > N B , and all µ s are equal, assume slip impends at B and between wedge and floor, and not at A. Then FW = µ s NW = 0.45 ( 271 lb ) = 121.95 lb FB = µ s N B = 0.45 (185 lb ) = 83.25 lb ΣFx = 0:

P − 121.95 lb − 83.25 lb = 0,

P = 205.20 lb

Check Wedge for assumption

271 lb − RA cosθ = 0

ΣFy = 0: ΣFx = 0: so tan θ =

205.2 lb − 121.95 lb − RA sin θ = 0

83.25 = 0.3072 < µ s + tan 9° 271 so no slip here ∴ (a)

(b)

P = 205 lb

impending slip at B

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Chapter 8, Solution 63. FBD plank + wedge:

ΣM A = 0:

(8 ft ) N B − (1.5 ft )( 48 lb/ft )( 3 ft ) 1 ( 48 lb/ft )( 3 ft ) 2  5  −  3 +  ft  ( 96 lb/ft )( 5 ft ) = 0 3  

− ( 2 ft )

NW = 185 lb

ΣFy = 0:

 48 + 96  N A + 185 lb −  lb/ft  ( 3ft ) 2   1 − ( 96 lb/ft )( 5 ft ) = 0 2 N A = 271 lb

Since N A > NW , and all µ s are equal, assume impending slip at top and bottom of wedge and not at A. Then FW = µ s NW = 0.45 (185 N )

FW = 83.25 lb FBD Wedge:

φ s = tan −1 µ s = tan −1 ( 0.45 ) = 24.228° ΣFy = 0:

185 lb − RB cos ( 24.228° + 9° ) = 0 RB = 221.16 lb

ΣFx = 0:

( 221.16 lb ) sin 33.228° + 83.25 lb − P = 0 P = 204.44 lb

Check assumption using plank/wedge FBD ΣFx = 0: FA + FW − P = 0,

FA = 204.44 lb − 83.25 lb = 121.19 lb

FA max = µ s N A = 0.45 ( 271 lb ) = 121.95 lb

FA < FA max , OK

∴ (a) (b)

P = 204 lb

no impending slip at A

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Chapter 8, Solution 64.

(

)

(

)

WA = (10 kg ) 9.81 m/s 2 = 98.1 N, WB = ( 50 kg ) 9.81 m/s 2 = 490.5 N Slip must impend at all surfaces simultaneously, F = µ s N FBD I: A + B

ΣFy = 0:

N B − 150 N − 98.1 N − 490.5 N = 0,

impending slip:

FBD II: A

N B = 738.6 N

FB = µ s N B = ( 738.6 N ) µ s N A = ( 738.6 N ) µ s

ΣFx = 0:

N A − FB = 0,

ΣFy′ = 0:

FAB + ( 738.6µ s ) N  sin 20° − (150 N + 98.1 N ) cos 20° = 0 FAB =  233.14 − ( 252.62 ) µ s  N

ΣFx′ = 0:

( 738.6µ s ) N  cos 20° − (150 N + 98.1 N ) sin 20° − N AB = 0 N AB = 84.855 + ( 694.06 ) µ s  N

µs =

FAB 233.14 − 252.62µ s = N AB 84.855 + 694.06µ s

µ s2 = 0.48623µ s − 0.33591 = 0

µ s = − 0.24312 ± 0.62850 Positive root

µ s = 0.385

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Chapter 8, Solution 65.

(

)

(

)

WA = (10 kg ) 9.81 m/s 2 = 98.1 N, WB = ( 50 kg ) 9.81 m/s 2 = 490.5 N Slip impends at all surfaces simultaneously FBD I: A + B

ΣFx = 0: ΣFy = 0:

N A − FB = 0,

N A = FB = µ s N B

(1)

FA − (150 N + 98.1 N + 490.5 N ) + N B = 0

µ s N A + N B = 738.6 N Solving (1) and (2) N B = FBD II: B

ΣFx′ = 0:

FB =

738.6 µ s N 1 + µ s2

N AB + ( 490.5 N ) cos 70° − N B cos 70° − FB sin 70° = 0 N AB =

ΣFy′ = 0:

738.6 N , 1 + µ s2

(2)

738.6 N ( cos 70° + µ s sin 70° ) − ( 490.5 N ) cos 70° (1) 1 + µ s2

−FAB − ( 490.5 N ) sin 70° + N B sin 70° − FB cos 70° = 0 FAB =

738.6 N ( sin 70° − µs cos 70° ) − ( 490.5 N ) sin 70° = 0 1 + µ s2

Setting FAB = µ s N AB ,

µ s3 − 6.8847µ s2 − 2.0116µ s + 1.38970 = 0 Solving numerically, µ s = − 0.586, 0.332, 7.14 Physically meaningful solution:

µ s = 0.332

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Chapter 8, Solution 66. FBD jack handle:

See Section 8.6 ΣM C = 0:

aP − rQ = 0 or P =

r Q a

FBD block on incline:

(a)

Raising load

Q = W tan (θ + φ s )

r W tan (θ + φs ) a continued

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PROBLEM 8.66 CONTINUED (b)

Lowering load if screw is self-locking ( i.e.: if φs > θ )

Q = W tan (φs − θ ) P= (c)

r W tan (φs − θ ) a

Holding load is screw is not self-locking ( i.e: if φs < θ )

Q = W tan (θ − φs ) P=

r W tan (θ − φ s ) a

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Chapter 8, Solution 67. FBD large gear:

ΣM C = 0:

(12 in.)W

− 7.2 kip ⋅ in. = 0,

W = 0.600 kips = 600 lb

Block on incline:

θ = tan −1

0.375 in. = 2.2785° 2π (1.5 in.)

φ s = tan −1 µ s = tan −1 0.12 = 6.8428° Q = W tan (θ + φ s ) = ( 600 lb ) tan 9.1213° = 96.333 lb FBD worm gear:

r = 1.5 in. ΣM B = 0:

(1.5 in.)( 96.333 lb ) − M

M = 144.5 lb ⋅ in.

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Chapter 8, Solution 68. FBD large gear:

ΣM C = 0:

(12 in.)W

− 7.2 kip ⋅ in. = 0

W = 0.600 kips = 600 lb

Block on incline:

θ = tan −1

0.375 in. = 2.2785° 2π (1.5 in.)

φ s = tan −1 µ s = tan −1 0.12 = 6.8428° Q = W tan (φ s − θ ) = ( 600 lb ) tan 4.5643° = 47.898 lb FBD worm gear:

r = 1.5 in. ΣM B = 0:

M − (1.5 in.)( 47.898 lb ) = 0 M = 71.8 lb ⋅ in.

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Chapter 8, Solution 69. Block/incline analysis:

θ = tan −1

0.125 in. = 2.4238° 2.9531 in.

φ s = tan −1 ( 0.35 ) = 19.2900°

Q = 47250 tan ( 21.714° ) = 18.816 lb Couple =

d  0.94  in.  (18.516 lb ) = 8844 lb ⋅ in. Q= 2  2  Couple = 7.37 lb ⋅ ft

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Chapter 8, Solution 70. FBD joint D:

FAD = FCD

By symmetry:

ΣFy = 0: 2FAD sin 25° − 4 kN = 0

FAD = FCD = 4.7324 kN FBD joint A:

FAE = FAD

By symmetry:

ΣFx = 0: FAC − 2 ( 4.7324 kN ) cos 25° = 0

FAC = 8.5780 kN Block and incline A:

θ = tan −1

2 mm = 4.8518° π ( 7.5 mm )

φ s = tan −1 µ s = tan −1 0.15 = 8.5308°

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PROBLEM 8.70 CONTINUED

Q = ( 8.578 kN ) tan (13.3826° ) = 2.0408 kN Couple at A:

M A = rQ  7.5  = mm  ( 2.0408 kN ) 2   = 7.653 N ⋅ m

By symmetry: Couple at C:

M C = 7.653 N ⋅ m Total couple M = 2 ( 7.653 N ⋅ m )

M = 15.31 N ⋅ m

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Chapter 8, Solution 71. FBD joint D:

By symmetry:

FAD = FCD ΣFy = 0: 2FAD sin 25° − 4 kN = 0

FAD = FCD = 4.7324 kN FBD joint A:

By symmetry:

FAE = FAD ΣFx = 0: FAC − 2 ( 4.7324 kN ) cos 25° = 0

FAC = 8.5780 kN Block and incline at A:

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PROBLEM 8.71 CONTINUED θ = tan −1

2 mm = 4.8518° π ( 7.5 mm )

φ s = tan −1 µ s = tan −1 0.15

φ s = 8.5308°

φ s − θ = 3.679° Q = ( 8.5780 kN ) tan 3.679° Q = 0.55156 kN Couple at A: M A = Qr  7.5 mm  = ( 0.55156 kN )    2  = 2.0683 N ⋅ m By symmetry:

Couple at C : M C = 2.0683 N ⋅ m Total couple M = 2 ( 2.0683 N ⋅ m )

M = 4.14 N ⋅ m

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Chapter 8, Solution 72. FBD lower jaw:

By symmetry B = 540 N ΣFy = 0:

− 540 N + A − 540 N = 0,

A = 1080 N

(a) since A > B when finished, adjust A first when there will be no force Block/incline at B:

(b) θ = tan −1

4 mm = 6.0566° 12π mm

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

Q = ( 540 N ) tan 25.3466° = 255.80 N Couple = rQ = ( 6 mm )( 255.80 N ) = 1535 N ⋅ mm

M = 1.535 N ⋅ m

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Chapter 8, Solution 73. FBD lower jaw:

By symmetry B = 540 N ΣFy = 0:

− 540 N + A − 540 N = 0,

A = 1080 N

since A > B, A should be adjusted first when no force is required. If instead, B is adjusted first, Block/incline at A:

θ = tan −1

4 mm = 6.0566° 12π mm

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

Q = (1080 N ) tan 25.3466° = 511.59 N Couple = rQ = ( 6 mm )( 511.59 N ) = 3069.5 N ⋅ mm

M = 3.07 N ⋅ m Note that this is twice that required if A is adjusted first.

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Chapter 8, Solution 74. Block/incline:

θ = tan −1

0.25 in. = 2.4302° 1.875π in.

φ s = tan −1 µ s = tan −1 ( 0.10 ) = 5.7106°

Q = (1000 lb ) tan ( 8.1408° ) = 143.048 lb Couple = rQ = ( 0.9375 in.)(143.048 lb ) = 134.108 lb ⋅ in.

M = 134.1 lb ⋅ in.

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Chapter 8, Solution 75. FBD Bucket:

(

) 0.30 ) = 0.05172 m

rf = r sin φ s = r sin tan −1 µ s

(

= ( 0.18 m ) sin tan −1 ΣM A = 0:

(1.6 m + 0.05172 m ) T − ( 0.05172 m )W

T = 0.031314W  kN  = 0.031314 ( 50 Mg )  9.81  Mg  

= 15.360 kN

T = 15.36 kN !

NOTE FOR PROBLEMS 8.75–8.89

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)

Note to instructors: In this manual, the simplification sin tan −1 µ ≈ µ is NOT used in the solution of journal bearing and axle friction problems. While this approximation may be valid for very small values of µ , there is little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79, µ s = 0.50, and the error made by using the approximation is about 11.8%.

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Chapter 8, Solution 76. FBD Windlass:

(

rf = rb sin φs = rb sin tan −1 µ s

(

)

= (1.5 in.) sin tan −1 0.5 = 0.67082 in.

ΣM A = 0:

( 8 − 0.67082 ) in. P − ( 5 + 0.67082 ) in. 160 lb = 0

P = 123.797 lb P = 123.8 lb

NOTE FOR PROBLEMS 8.75–8.89

(

)

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Chapter 8, Solution 77. FBD Windlass:

(

rf = r sin φs = r sin tan −1 µ s

(

)

= (1.5 in.) sin tan −1 0.5 = 0.67082 in.

ΣM A = 0:

( 8 + 0.67082 ) in. P − ( 5 + 0.67082 ) in. (160 lb ) = 0 P = 104.6 lb

NOTE FOR PROBLEMS 8.75–8.89

(

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Chapter 8, Solution 78. FBD Windlass:

(

rf = r sin φs = r sin tan −1 µ s

(

)

= (1.5 in.) sin tan −1 0.50 = 0.67082 in.

ΣM A = 0:

( 8 + 0.67082 ) in. P − ( 5 − 0.67082 ) in. (160 lb ) = 0 P = 79.9 lb

NOTE FOR PROBLEMS 8.75–8.89

(

)

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Chapter 8, Solution 79. FBD Windlass:

(

rf = r sin φs = r sin tan −1 µ s

(

)

= (1.5 in.) sin tan −1 0.50 = 0.67082 in.

ΣM A = 0:

( 8 − 0.67082 ) in. P − ( 5 − 0.67082 ) in. (160 lb ) = 0 P = 94.5 lb

NOTE FOR PROBLEMS 8.75–8.89

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Chapter 8, Solution 80. (a) FBD lever (Impending CW rotation): ΣM C = 0:

( 0.2 m + rf ) ( 75 N ) − ( 0.12 m − rf ) (130 N ) = 0 rf = 0.0029268 m = 2.9268 mm sin φs =

rf rs

*  −1 rf   −1 2.9268 mm  µ s = tan φs = tan  sin  = tan  sin  18 mm  rs   

= 0.34389

µ s = 0.344 (b) FBD lever (Impending CCW rotation):

( 0.20 m − 0.0029268 m )( 75 N )

ΣM D = 0:

− ( 0.12 m + 0.0029268 m ) P = 0 P = 120.2 N

NOTE FOR PROBLEMS 8.75–8.89

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Chapter 8, Solution 81. Pulley FBD’s:

rp = 30 mm

Left:

(

rf = raxle sin φk = raxle sin tan −1 µ k

)

(

= ( 5 mm ) sin tan −1 0.2

)

= 0.98058 mm Left: ΣM C = 0:

Right:

( rp − rf ) ( 600 lb ) − 2rpTAB = 0 TAB =

30 mm − 0.98058 mm ( 600 N ) = 290.19 N 2 ( 30 mm ) TAB = 290 N

ΣFy = 0:

290.19 N − 600 N + TCD = 0 TCD = 309.81 N

TCD = 310 N

Right:

( rp + rf ) TCD − ( rp − rf ) TEF

ΣM G = 0: or

TEF =

30 mm + 0.98058 mm ( 309.81 N ) = 330.75 N 30 mm − 0.98058 mm TEF = 331 N

NOTE FOR PROBLEMS 8.75–8.89

(

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Chapter 8, Solution 82. Pulley FBDs: Left:

rp = 30 mm

(

rf = raxle sin φk = raxle sin tan −1 µ k

)

(

= ( 5 mm ) sin tan −1 0.2

)

= 0.98058 mm

( rp + rf ) ( 600 N ) − 2rpTAB = 0

ΣM C = 0:

30 mm + 0.98058 mm ( 600 N ) = 309.81 N 2 ( 30 mm )

TAB =

TAB = 310 N

ΣFy = 0:

Right:

TAB − 600 N + TCD = 0 TCD = 600 N − 309.81 N = 290.19 N

TCD = 290 N

( rp − rf ) TCD − ( rp + rf ) TEF

ΣM H = 0:

TEF =

30 mm − 0.98058 mm ( 290.19 N ) 30 mm + 0.98058 mm TEF = 272 N

NOTE FOR PROBLEMS 8.75–8.89

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Chapter 8, Solution 83. FBD link AB: Note: That AB is a two-force member. For impending motion, the pin forces are tangent to the friction circles. r θ = sin −1 f 25 in. where

(

rf = rp sin φs = rp sin tan −1 µ s

)

(

)

= (1.5 in.) sin tan −1 0.2 = 0.29417 in. Then

θ = sin −1

0.29417 in. = 1.3485° 12.5 in. (b)

Rvert = R cosθ

θ = 1.349°

Rhoriz = R sin θ

Rhoriz = Rvert tan θ = ( 50 kips ) tan1.3485° = 1.177 kips (a) Rhoriz = 1.177 kips

NOTE FOR PROBLEMS 8.75–8.89

(

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Chapter 8, Solution 84. FBD gate:

(

)

(

)

W1 = 66 kg 9.81 m/s 2 = 647.46 N W2 = 24 kg 9.81 m/s 2 = 235.44 N

(

rf = rs sin φs = rs sin tan −1 µ s

(

)

= ( 0.012 m ) sin tan −1 0.2 = 0.0023534 m ΣM C = 0:

( 0.6 m − rf )W1 + ( 0.15 m − rf ) P − (1.8 m + rf )W2 = 0 P=

(1.80235 m )( 235.44 N ) − ( 0.59765 m )( 647.46 N ) ( 0.14765 m )

= 253.2 N P = 253 N !

NOTE FOR PROBLEMS 8.75–8.89

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Chapter 8, Solution 85. It is convenient to replace the ( 66 kg ) g and ( 24 kg ) g weights with a single combined weight of

( 90 kg ) ( 9.81 m/s2 ) = 882.9 N,

located at a distance x =

right of B.

(

rf = rs sin φs = rs sin tan −1 µ s

)

(1.8 m )( 24 kg ) − ( 0.6 m )( 66 kg ) 90 kg

(

= ( 0.012 m ) sin tan −1 0.2

= 0.04 m to the

)

= 0.0023534 m FBD pulley + gate:

α = tan −1 β = sin −1

rf OB

= sin −1

0.04 m = 14.931° 0.15 m

OB =

0.0023534 m = 0.8686° 0.15524 m

0.15 = 0.15524 m cos α then

θ = α + β = 15.800°

P = W tan θ = 249.8 N P = 250 N !

NOTE FOR PROBLEMS 8.75–8.89

(

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Chapter 8, Solution 86. FBD gate:

( ) = 24 kg ( 9.81 m/s ) = 235.44 N

W1 = 66 kg 9.81 m/s 2 = 647.46 N W2

(

rf = rs sin φs = rs sin tan −1 µ s

(

)

= ( 0.012 m ) sin tan −1 0.2 = 0.0023534 m ΣM C = 0:

( 0.6 m + rf )W1 + ( 0.15 m + rf ) P − (1.8 m − rf )W2 = 0 P=

(1.79765 m )( 235.44 N ) − ( 0.60235 m )( 647.46 N ) 0.15235 m

= 218.19 N P = 218 N !

NOTE FOR PROBLEMS 8.75–8.89

(

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COSMOS: Complete Online Solutions Manual Organization System

Chapter 8, Solution 87. It is convenient to replace the ( 66 kg ) g and ( 24 kg ) g weights with a single weight of

( 90 kg )( 9.81 N/kg ) = 882.9 N,

(1.8 m )( 24 kg ) − ( 0.15 m )( 66 kg )

located at a distance x =

90 kg

= 0.04 m to the

right of B. FBD pulley + gate:

(

rf = rs sin φs = rs sin tan −1 µ s

)

(

= ( 0.012 m ) sin tan −1 0.2

)

rf = 0.0023534 m

α = tan −1 β = sin −1

rf OB

= sin −1

0.04 m = 14.931° 0.15 m

OB =

0.0023534 m = 0.8686° 0.15524 m

0.15 m = 0.15524 m cos α then

θ = α − β = 14.062°

P = W tan θ = 221.1 N P = 221 N !

NOTE FOR PROBLEMS 8.75–8.89

(

)

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Chapter 8, Solution 88. FBD Each wheel:

(

rf = raxle sin φ = raxle sin tan −1 µ

ΣFx = 0:

P − R sin θ = 0 4 R cosθ −

ΣFy = 0:

)

W =0 4

∴ tan θ = sin θ =

but

P W rf rw

P = W tanθ

or =

raxle sin tan −1 µ rw

(

)

(a) For impending motion, use µ s = 0.12 sin θ =

0.5 in. sin tan −1 0.12 5 in.

(

)

θ = 0.68267°

P = W tan θ = ( 500 lb ) tan ( 0.68267° ) P = 5.96 lb (b) For constant speed, use µ k = 0.08

sin θ =

1 sin tan − 1 0.08 10

(

)

θ = 0.45691°

P = ( 500 lb ) tan ( 0.45691° ) P = 3.99 lb

NOTE FOR PROBLEMS 8.75–8.89

(

)

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Chapter 8, Solution 89. FBD Each wheel:

For equilibrium (constant speed) the two forces R and

W must be equal 2

and opposite, tangent to the friction circle, so rf sin θ = where θ = tan −1 ( slope ) rw

(

)

sin tan −1 0.03 =

rw = (12.5 mm )

(

rB sin tan −1 µ k

( sin ( tan

)

) = 49.666 mm 0.03)

sin tan −1 0.12 −1

d w = 99.3 mm

NOTE FOR PROBLEMS 8.75–8.89

(

)

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Chapter 8, Solution 90. FBD

(8 in.) Q − M

ΣM O = 0:

= 0,

M 8 in.

but, from equ. 8.9,

M =

2 2  7 in.  µk WR = ( 0.60 )(10.1 lb )   3 3  2 

= 14.14 lb so,

14.14 , 8

Q = 1.768 lb

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Chapter 8, Solution 91. 2 R3 − R13 1 D23 − D13 µ s P 22 µ = P s 3 3 R2 − R12 D22 − D12

Eqn. 8.8 gives

M =

1 M = ( 0.15 )( 80 kg ) 9.81 m/s 2 3

(

( 0.030 m )3 − ( 0.024 m )3 ) ( 0.030 m )2 − ( 0.024 m )2

M = 1.596 N ⋅ m

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Chapter 8, Solution 92.

Let the normal force on ∆A be ∆N , and

∆N k = r ∆A

∆F = µ∆N , ∆M = r ∆F

As in the text The total normal force

2π  R k  P = lim Σ∆N = ∫ 0  ∫ 0 rdr  dθ ∆A → 0  r 

(

)

P = 2π ∫ 0 kdr = 2π kR The total couple

P 2π R

k 2π  R  M worn = lim Σ∆M = ∫ 0  ∫ 0 r µ rdr  dθ ∆A → 0 r   R

M worn = 2πµ k ∫ 0 rdr = 2πµ k

R2 P R2 = 2πµ 2 2π R 2

M worn =

1 µ PR 2

Now

M new =

2 µ PR 3

Thus

k =

M worn = M new

[Eq. (8.9)] 1 2 2 3

µ PR 3 = = 75% µ PR 4

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Chapter 8, Solution 93.

Let normal force on ∆A be ∆N , and

∆N k = ∆A r

∆F = µ∆N , ∆M = r ∆F

As in the text The total normal force P is

2π  R k  P = lim Σ∆N = ∫ 0  ∫ R 2 rdr  dθ 1 r ∆A → 0  

P = 2π ∫R 2 kdr = 2π k ( R2 − R1 ) R

k =

P 2π ( R2 − R1 )

k 2π  R  M worn = lim Σ∆M = ∫ 0  ∫R 2 r µ rdr  dθ ∆A → 0 r  1 

The total couple is

R2 R1

M worn = 2πµ k ∫

( rdr ) = πµ k (

R22

−

R12

2π ( R2 − R1 )

M worn =

(

πµ P R22 − R12

)

1 µ P ( R2 + R1 ) 2

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Chapter 8, Solution 94.

Let normal force on ∆A be ∆N , and

∆A = r ∆s∆φ

∆N = k ∆A

∆N = k, ∆A ∆s =

∆r sin θ

where φ is the azimuthal angle around the symmetry axis of rotation ∆Fy = ∆N sin θ = kr ∆r ∆φ P = lim Σ∆Fy

Total vertical force

∆A → 0

2π

P = ∫0

(

(∫

R2 krdr R1

P = π k R22 − R12

Friction force

)

) dφ = 2π k ∫

R2 rdr R1

k =

Total couple

M = 2π

(

− R12

)

∆F = µ∆N = µ k ∆A ∆M = r ∆F = r µ kr

Moment

P R22

∆r ∆φ sin θ

2π  R µ k  M = lim Σ∆M = ∫ 0  ∫R 2 r 2dr  dφ ∆A → 0  1 sin θ 

µ k R2 2 2 πµ P r dr = ∫ R1 2 sin θ 3 sin θ π R2 − R32

(

)

M =

3 2

− R33

)

2 µ P R23 − R13 3 sin θ R22 − R12

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Chapter 8, Solution 95. If normal force per unit area (pressure) of the center is PO , then as a function

r  of r, P = PO 1 −  R   r 2π R  ΣFN = W = ∫ PdA = ∫ 0 ∫ 0 PO 1 −  rdrdθ R  2 R3  R2 2π  R W = PO ∫ 0  d P − = θ 2 π  O 3R  6  2

so PO =

3W π R2

For slipping, dF = µk ( PdA)

r 2π R  Moment = ∫ rdF = µk PO ∫ 0 ∫ 0 r  1 −  rdrdθ R  3 R4  R3 2π  R d P = µ k PO ∫ 0  − = θ 2 πµ  k O 4 R  12  3

so M = 2πµ k ΣM O = 0:

3W R3 1 = µ k WR π R 2 12 2

(8 in.) Q − M

1  7 in.  0.6 )(10.1 lb )  (  M 2  2  Q= = 8 in. (8 in.) Q = 1.326 lb

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Chapter 8, Solution 96. FBD pipe:

θ = sin −1

0.025 in. + 0.0625 in. = 1.00257° 5 in.

P = W tan θ for each pipe, so also for total P = ( 2000 lb ) tan (1.00257° ) P = 35.0 lb

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Chapter 8, Solution 97. FBD disk:

tan θ = slope = 0.02 b = r tan θ = ( 60 mm )( 0.02 ) b = 1.200 mm

COSMOS: Complete Online Solutions Manual Organization System

Chapter 8, Solution 98. FBD wheel:

r = 230 mm b = 1 mm

θ = sin −1

b r

b  P = W tan θ = W tan  sin −1  for each wheel, so for total r 

(

)

1   P = (1000 kg ) 9.81 m/s 2 tan  sin −1  230   P = 42.7 N

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Chapter 8, Solution 99. FBD wheel:

(

)

rf = raxle sin φ = raxle sin tan −1 µ , rw =

rf sin θ

b tan θ

For small θ , sin θ

tan θ , so tan θ

R cosθ −

ΣFy = 0:

rf + b rw

W =0 4

− R sin θ +

ΣFx = 0:

P =0 4

P W

Solving: tan θ =

so P = W tan θ = W (a) For impending slip, use µ s ,

µs or µk

rf + b rw

 0.5 in.  −1 rf =   sin tan 0.12 = 0.029786 in.  2 

(

so P = ( 500 lb )

)

0.02986 in. + 0.25 in. = 55.96 lb 2.5 in. P = 56.0 lb

(b) For constant speed, use µ k ,

 0.5 in.  −1 rf =   sin tan 0.08 = 0.019936 in.  2 

so P = ( 500 lb )

(

( 0.019936 + 0.25) in. 2.5 in.

)

= 53.99 lb P = 54.0 lb

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Chapter 8, Solution 100. FBD wheel: For equilibrium (constant speed), R and

W are equal and opposite 2

and tangent to the friction circle as shown

(

)

(

rf = raxle sin tan −1 µk = (12.5 mm ) sin tan −1 0.12

)

rf = 1.48932 mm From diagram, rw = For small θ , sin θ

tan θ = slope

rw =

rf sin θ

b tan θ

tan θ , so rw

rf + b tan θ

1.48932 mm + 1.75 mm = 107.977 mm 0.03 d w = 216 mm

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Chapter 8, Solution 101. β = 4π rad

Two full turns of rope → (a)

µ s β = ln

µs =

T2 T1

µs =

T2 T1

1 20 000 N = 0.329066 ln 4π 320 N

µ s = 0.329 (b)

β = =

µs

T2 T1

1 80 000 N ln 0.329066 320 N = 16.799 rad

β = 2.67 turns

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Chapter 8, Solution 102. FBD A:

(

)

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

ΣFx = 0:

TA − WA sin 30° = 0,

TA =

WA 2

FBD B:

ΣFx′ = 0:

WB sin 30° − TB = 0,

TB =

WB 2

(a) Motion of B impends up incline and mB = 8 kg

TA = eµs β , TB

µs = =

TA 1 W = ln A TB β WB

mA 3  10 kg  = ln   π  8 kg  mB

From hint, β is not dependent on shape of support

µ s = 0.21309 µ s = 0.213 (b) For maximum mB , motion of B impend down incline

TB = eµs β , TA

TB = TAe

0.21309

π 3

= 1.250TA

∴ WB = 1.25WA and mB = 1.25 mA = 1.25 (10 kg )

mB max = 12.50 kg

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Chapter 8, Solution 103. FBD A:

ΣFx = 0:

TA − WA sin 30° = 0,

TA =

WA 2

FBD B:

ΣFx′ = 0:

WB sin 30° − TB = 0,

TB =

WB 2

For mB min , motion of B impends up incline π

And

But

0.50 TA = e 3 = 1.68809 TB

m A WA T = = A = 1.68809 mB WB TB so mB min = 5.9238 kg

From hint, β is not dependent on shape of C For mB max , motion of B impends down incline π

0.50 mB W T µ = B = B = e s β = e 3 = 1.68809 mA WA TA

so mB max = 16.881 kg For equilibrium

5.92 kg ≤ mB ≤ 16.88 kg

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Chapter 8, Solution 104.

β = 1.5 turns = 3π rad For impending motion of W up

P = We µs β = (1177.2 N ) e(

0.15 )3π

= 4839.7 N

For impending motion of W down − 0.15 3π P = We− µs β = (1177.2 N ) e ( )

= 286.3 N

For equilibrium

286 N ≤ P ≤ 4.84 kN

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Chapter 8, Solution 105. Horizontal pipe:

Contact angles β H =

Vertical pipe

Contact angle βV = π

µ sH = 0.25

µ sV = 0.2

For P to impend downward,

(

)

(

)

µ π µ π µ π µ π P =  e sH 2  Q =  e sH 2  e µsV π R =  e sH 2  e µsV π  e sH 2  (100 lb )         π µ +µ Pmax = e ( sH sV )  (100 lb ) = (100 lb ) e0.45π = 411.12 lb  

For 100 lb to impend downward, the ratios are reversed, so

100 lb = Pe0.45π ,

Pmin = 24.324 lb So, for equilibrium,

24.3 lb ≤ P ≤ 411 lb

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Chapter 8, Solution 106. Horizontal pipe

Vertical pipe

Contact angles β H =

Contact angle βV = π

µ sH = 0.30

µ sV = ?

For Pmin , the 100 lb force impends downward, and µ 100 lb =  e 

R =  e µ    

(

 eµ  

)Q =  e

µ s π π2

(

 eµ  

) e

µ sH π2

P  

π 0.30 + µ sV )  100 lb = e ( ( 20 lb ) , so eπ ( 0.30 + µsV ) = 5  

(a) For Pmax the force P impends downward, and the ratios are reversed,

Pmax = 5 (100 lb ) = 500 lb

(b) π ( 0.30 + µ sV ) = ln 5

µ sV =

ln 5 − 0.30 = 0.21230

µ sV = 0.212

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Chapter 8, Solution 107. FBD motor and mount: Impending belt slip: cw rotation

T2 = T1e µs β = T1e0.40π = 3.5136 T1 ΣM D = 0:

(12 in.)(175 lb ) − ( 7 in.) T2 − (13 in.) T1 = 0 2100 lb = ( 7 in.)( 3.5136 ) + 13 in. T1

T1 = 55.858 lb,

T2 = 3.5136 T1 = 196.263 lb

FBD drum at B:

ΣM B = 0:

M B − ( 3 in.)(196.263 lb − 55.858 lb ) = 0 M B = 421 lb ⋅ in.

r = 3 in.

(Compare to 857 lb ⋅ in. using V-belt, Problem 8.130)

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Chapter 8, Solution 108. FBD motor and mount: Impending belt slip: ccw rotation

T1 = T2e µs β = T2e0.40π = 3.5136 T2 ΣM D = 0:

(12 in.)(175 lb ) − (13 in.) T1 − ( 7 in.) T2

2100 lb = (13 in.)( 3.5136 ) + 7 in. T2 = 0 T2 = 39.866 lb,

T1 = 3.5136 T2 = 140.072 lb

FBD drum at B:

ΣM B = 0:

( 3 in.)(140.072 lb − 39.866 lb ) − M B

M B = 301 lb ⋅ in.

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Chapter 8, Solution 109. FBD lower portion of belt:

ΣFy = 0:

48 N − N D = 0,

N D = 48 N

Slip on both platen and wood

FD = µkD N D = 0.10 ( 48 N ) = 4.8 N FE = µkE N E = ( 48 N ) µ kE FBD Drum A (assume free to rotate)

ΣFx = 0:

TA − TB − 4.8 N − µ kE ( 48 N ) = 0 TB = TA + 4.8 N + µ kE ( 48 N )

(1)

ΣM A = 0:

rA (TA − TT ) = 0,

(2)

ΣM B = 0:

M B + r (TT − TB ) = 0

TT = TA

FBD Drive drum B

TB = TT +

2.4 N ⋅ m = TT + 96 N 0.025 N

Impending slip on drum, TB = TT e µs β = TT e0.35π so TT + 96 N = TT e0.35π , TT = 47.932 N

TB = 143.932 N From (2) above, TA = TT , so

(a)

Tmin lower = 47.9 N

From (1) above, 143.932 N = 47.932 N + 4.8 N + µkE ( 48 N ) So

(b)

µ kE = 1.900

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Chapter 8, Solution 110. FBD Flywheel:

( 0.225 m )(TB − TA ) − 12.60 N ⋅ m = 0

ΣM C = 0:

TB − TA = 56 N,

TB = TA + 56 N

Also, since the belt doesn’t change length, the additional stretch in spring B equals the decrease in stretch of spring A. Thus the increase in TB equals the decrease in TA. Thus TB + TA = ( 70 N + ∆T ) + ( 70 N − ∆T ) = 140 N

(TA + 56 N ) + TA

= 140 N,

TA = 42 N

TB = 42 N + 56 N = 98 N (a)

TA = 42.0 N

TB = 98.0 N For slip TB = TAe µk β , or µ k =

µk =

TB TA

98 = 0.2697 42 (b)

µ k = 0.230

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Chapter 8, Solution 111. FBD Flywheel: Slip of belt: TB = TA e µk β = TA e0.20π Also, since the belt doesn’t change length, the increase in stretch of spring B equals the decrease in stretch of spring A. Therefore the increase in TB equals the decrease in TA , and the sum is unchanged, so TA + TB = 80 N + 80 N = 160 N

(

)

∴ TA 1 + e0.20π = 160 N, so TA = 55.663 N TB = 104.337 N

(a)

TA = 55.7 N TB = 104.3 N

ΣM C = 0:

( 0.225 m )(TB

− TA ) − M C = 0

M C = ( 0.225 m )(104.337 N − 55.663 N ) (b) M C = 10.95 N ⋅ m

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Chapter 8, Solution 112. FBD Lever:

ΣM E = 0:

( 60 mm )( 240 N ) − ( 40 mm ) FBD cos 30° = 0

FBD = 415.69 N FBD Drum: Belt slip:

T2 = T1 e µk β

= ( 415.69 N ) e

0.25( 5.5851)

= 1679.44 N

ΣM C = 0:

r (T2 − T1 ) − M = 0

( 0.08 m )(1679.44 N − 415.69 N ) − M

M = 101.1 N ⋅ m

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Chapter 8, Solution 113. FBD Drum: (a) With M E = 125 lb ⋅ ft

ΣM E = 0:

( 7 in.) (TA − TC ) − (125 lb ⋅ ft ) = 0

TA − TC = 214.29 lb Belt slip: TA = TC e µk β = TC e

( )

0.30 76π

= 3.0028 TC

so 2.0028 TC = 214.9 lb,

TC = 106.995 lb

TA = 321.28 lb FBD Lever:

ΣM B = 0: P=

(15 in.) P + ( 2 in.) TC − ( 7.5 in.) TA = 0

(1)

( 7.5 in.)( 321.28 lb ) − ( 2 in.)(106.995 lb ) 17 in.

P = 129.2 lb

(b) With M E = 125 lb ⋅ ft , the drum analysis will be reversed, and will yield TA = 106.995 lb,

TC = 321.28 lb Eqn. (1) will remain the same, so

( 7.5 in.)(106.995 lb ) − ( 2 in.)( 321.28 lb ) 17 in.

P = 9.41 lb

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Chapter 8, Solution 114. FBD Lever: If brake is self-locking, no force P is required

( 2 in.) TC − ( 7.5 in.) TA = 0

ΣM B = 0:

TC = 3.75 TA For impending slip on drum: TC = TA e µs β

∴ e µs β = 3.75, or µ s = With β =

ln 3.75

7π , 6

µ s = 0.361

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Chapter 8, Solution 115. FBD Lever:

ΣM B = 0:

( 40 mm ) TC − (100 mm ) TA = 0,

TC = 2.5 TA

FBD Drum: (a) For impending slip ccw: TC = Tmax = 4.5 kN so

TA =

TC = 1.8 kN 2.5

M D + ( 0.16 m )(1.8 kN − 4.5 kN ) = 0

ΣM D = 0:

M D = 0.432 kN ⋅ m M D = 432 N ⋅ m (b) For impending slip ccw, TC = TA e µs β or µ s =

TC 3 ln 2.5 = 0.21875 = TA 4π

µ s = 0.219

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Chapter 8, Solution 116. (a) For minimum mC with blocks at rest, impending slip of A is down/left. Note: φs = tan −1 µ s = tan −1 0.30 = 16.7° < 30°, so mC min > 0 FBD A:

(

)

WA = ( 6 kg ) 9.81 m/s 2 = 58.86 N ΣFy = 0:

N A − WA cos 30° = 0,

N A = WA cos 30°

Impending slip: FA = µ s N A = 0.30WA cos 30°

ΣFx = 0:

TA + FA − WA sin 30° = 0, TA = WA ( sin 30° − 0.30 cos 30° ) = 14.1377 N

FBD Drum: If blocks don’t move, belt slips on drum, so 0.2 0.87266 ) 14.1377 N = TA = TC e µk β = TC e ( = 1.19069 TC

TC = 11.8735 N

FBD C:

ΣFy′ = 0:

NC − WC cos 20° = 0,

NC = WC cos 20°

Impending slip: FC = µ s NC = 0.30WC cos 20°

ΣFx′ = 0:

0.30WC cos 20° + WC sin 20° − 11.8735 N = 0

WC = 19.0302 N,

mC =

WC = 1.93988 kg 9.81 m/s 2 mC = 1.940 kg continued

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(b) For motion of A to impend up/right FBD A: As in part (a) N A = WA cos 30°, FA = 0.30WA cos 30°

ΣFx = 0:

TA − WA ( sin 30° + 0.30cos 30° ) = 0 TA = 44.722 N

Also, as in part (a) TA = TC e µk β = 1.19069 TC , so TC =

44.722 N 1.19069

TC = 37.560 N

FBD C: As in part (a) FC = 0.30WC cos 20°

ΣFx′ = 0:

WC ( sin 20° − 0.30cos 20° ) − 37.560 N = 0

WC = 624.83 N,

mC =

WC = 63.69 kg 9.81 m/s 2 mC = 63.7 kg

(c) For uniform motion of A up and B down, and minimum mC , there will be impending slip of the rope on the drum. FBD A is same as in (b) but FA = µk N A = 0.20WA cos30° and

ΣFx = 0:

TA − WA ( sin 30° + 0.20cos 30° ) = 0,

TA = 39.625 N

Drum analysis, with impending slip, TA = TC e µs β 39.625 N = TC e

0.30( 0.87266 )

= 1.29926 TC

or TC = 30.498 N FBD C is same as in (b), but FC = µ k NC = 0.20WC cos 20° and

ΣFx′ = 0:

WC ( sin 20° − 0.20 cos 20° ) − TC = 0 WC =

30.498 N = 197.933 N, 0.154082

mC =

197.934 N 9.81 m/s 2 mC = 20.2 kg

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Chapter 8, Solution 117. Geometry and force rotation: Let

EBC = α = cos −1

50 mm = 60° = 100 mm

s DBE , FAE , GAE

Then contact angles are β B = 360° − 120° = 240° = upper cylinder, and β A = 30° =

π 6

4π rad for cord on 3

rad for each cord

contact on lower cylinder. Let the force in section FC = TF Let the force in section DG = TG With A fixed and the cord moving,

TG = We µk β A = We

( )

0.25 π6

= 1.13985W

For maximum W, slip impends on drum B, so

TB = TF e µs β B or TF = TG e− µs β B TF = 1.13985We

( )

−0.30 43π

= 0.32441W

For slip at F

W1 = TF e µk β A = 0.32441We

( )

0.25 π6

= 0.36978W

so W = 2.7043W1 and m = 2.7043 in.

= 2.7043 ( 75 kg ) m = 203 kg

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Chapter 8, Solution 118. Geometry and force notation:

Note: θ = sin −1

βC = β D =

π 2

π r = 30° = rad, so contact angles are: 2r 6

π 6

2π , 3

βE = π

(a) For all pulleys locked, slip impends at all contacts If WA impends downward, T1 = (16 lb ) e µs β E , T2 = T1e µs β D , WA = T2e µ s βC 0.20( 73π ) µ β +β +β so WA = (16 lb ) e s ( C D E ) = (16 lb ) e = 69.315 lb

If WA impends upward all ratios are inverted, so WA = (16 lb ) e

( )

−0.20 73π

= 3.6933 lb For equilibrium,

3.69 lb ≤ WA ≤ 69.3 lb W

(b) If pulley D is free to rotate, T1 = T2 while the other ratios remain as in (a) 0.20( 53π ) µ β +β For WA impending down WA = (16 lb ) e s ( C E ) = (16 lb ) e

WA = 45.594 lb For WA impending upward, WA = (16 lb ) e

( )

−0.2 53π

= 5.6147 lb For equilibrium

5.61 lb ≤ WA ≤ 45.6 lb W

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Chapter 8, Solution 119. Geometry and force notation:

θ = sin −1

r π = 30° = , so contact angles are: 2r 6

βC = β D =

π 2

π 6

2π , 3

βE = π

(a) D and E fixed, so slip on these surfaces. For maximum N A , slip impends on pulley C

WA = T2e µs βC , and T1 = T2e µk β D ,

(16 lb ) = T1e µk β E

WA = (16 lb ) e

− µk ( β E + β D ) µ s βC

= (16 lb ) e

( )e0.20( 23π )

−0.15 53π

= 11.09 lb

(b) C and D fixed, so slip there. For maximum WA , slip impends on E so T1 = (16 lb ) e µs β E , T1 = T2e µk β D , T2 = WAe µk βC so WA = (16 lb ) e µs β E e

− µk ( βC + β D )

= (16 lb ) e0.20π e

( )

−0.15 43π

= 16 lb WA = 16.00 lb

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Chapter 8, Solution 120. Geometry and force notation:

θ = sin −1

5 in. π = 30° = rad, so contact angles are: 10 in. 6

βC = π −

π 6

5π π π 2π = , βD = + , 6 2 6 3

βE =

π 2

(a) All pulleys locked with impending slip at all. If WA impends upward, T1 = WAe µs βC ,

T2 = T1e µs β D , WA = (16 lb ) e

(16 lb ) = T2e µs β E , − µ s ( βC + β D + β E )

= (16 lb ) e

(

)

−0.20 56 + 46 + 63 π

WA = 4.5538 lb

If WA impends downward all ratios are inverted so WA = (16 lb ) e

+0.20( 2π )

= 56.217 lb 4.55 lb ≤ WA ≤ 56.2 lb

For equilibrium, (b) Pulley D is free to rotate so T1 = T2 , other ratios are the same If WA impends upward, WA = (16 lb ) e

− µ s ( βC + β E )

= (16 lb ) e

( )

−0.20 43π

WA = 6.9229 lb

If WA impends downward, ratios are inverted, WA = (16 lb ) e

( )

+0.20 43π

WA = 36.979 lb For equilibrium

6.92 lb ≤ WA ≤ 37.0 lb

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Chapter 8, Solution 121. Geometry and force notation:

θ = sin −1

βC = π −

5 in. π = 30° = rad, so contact angles are: 10 in. 6

π 6

5π π π 2π = , βD = + , 6 2 6 3

βE =

π 2

(a) D and E fixed, so slip at these surfaces, For maximum WA , slip impends on C.

WA = T1e µs βC , T2 = T1e µk β D , 16 lb = T2e µk β E so WA = (16 lb ) e

= (16 lb ) e

− µ k ( β D + β E ) µ s βC

( ) e0.20( 56π )

−0.15 76π

= 15.5866 lb WA max = 15.59 lb

(b) C and D fixed, so slip at these surfaces—impending slip on E

T1 = WAe µk βC , T2 = T1e µk β D , so WA = (16 lb ) e

− µk ( βC + β D ) µ s β E e

T2 = (16 lb ) e µs β E = (16 lb ) e

( ) e0.20( π2 )

−0.15 32π

WA = 10.8037 lb,

WA max = 10.80 lb

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Chapter 8, Solution 122. FBD drum B:

ΣM B = 0:

( 0.02 m )(TA − T ) − 0.30 N ⋅ m = 0 TA − T =

0.30 N ⋅ m = 15 N 0.02 m

Impending slip: TA = Te µs β = Te0.40π

(

)

Solving; T e0.40π − 1 = 15 N

T = 5.9676 N If C is free to rotate P = T Pmin = 5.97 N

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Chapter 8, Solution 123. FBD drum B:

ΣM B = 0:

( 0.02 m )(TA − T ) − 0.3 N ⋅ m = 0 TA − T = 15 N

Impending slip:

(

TA = Te µs β B = Te0.40π

)

Solving, T e0.40π − 1 = 15 N

T = 5.9676 N If C is frozen, tape must slip there, so

P = Te µk βC = ( 5.9676 N ) e

( )

0.30 π2

= 9.5599 N Pmin = 9.56 N

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Chapter 8, Solution 124. FBD pin B: T1 = T2

(a) By symmetry:  2  ΣFy = 0: B − 2  T =0  2 1   

Drum:

2T1 =

2T2

(1)

For impending rotation : T3 > T1 = T2 > T4 , so T3 = Tmax = 5.6 kN

(

−0.25 π4 + π6

Then

T1 = T3e− µs β L = ( 5.6 kN ) e

T1 = 4.03706 kN = T2

and

T4 = T2e− µs β R = ( 4.03706 kN ) e

T4 = 2.23998 kN

)

( )

−0.25 34π

ΣM F = 0: M 0 + r (T4 − T3 + T2 − T1 ) = 0 or

M 0 = ( 0.16 m )( 5.6 kN − 2.23998 kN ) = 0.5376 kN ⋅ m

Lever:

M 0 = 538 N ⋅ m (b) Using Equation (1) B=

2T1 =

2 ( 4.03706 kN )

= 5.70927 kN ΣM D = 0:

( 0.05 m )( 5.70927 kN ) − ( 0.25 m ) P = 0 P = 1.142 kN

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Chapter 8, Solution 125. FBD pin B: T1 = T2

(a) By symmetry:  2  ΣFy = 0: B − 2  T1  = 0  2 

FBD Drum:

(1)

2T1

For impending rotation : T4 > T2 = T1 > T3 , so T4 = Tmax = 5.6 kN

( )

−0.25 34π

Then

T2 = T4e− µ s β R = ( 5.6 kN ) e

T2 = 3.10719 kN = T1

and

T3 = T1e− µs β L = ( 3.10719 kN ) e

T3 = 2.23999 kN

(

−0.25 π4 + π6

)

ΣM F = 0: M 0 + r (T2 − T1 + T3 − T4 ) = 0 M 0 = (160 mm )( 5.6 kN − 2.23999 kN ) = 537.6 N ⋅ m

FBD Lever:

M 0 = 538 N ⋅ m (b) Using Equation (1) B=

2T1 =

2 ( 3.10719 kN )

B = 4.3942 kN ΣM D = 0:

( 0.05 m )( 4.3942 kN ) − ( 0.25 m ) P = 0 P = 879 N

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Chapter 8, Solution 126. FBD wrench:

Note: EC =

( 0.2 m ) , sin 65°

EA = EC − 0.03 m

θ = 65° so β = 295° = 5.1487 rad

ΣM E = 0:

 0.20 m   0.20 m  − 0.03 m  F −  cos 65° − 0.03 m  T = 0   sin 65°   sin 65°  T = 3.01408F

ΣFx = 0:

N sin 65° + F cos 65° − T = 0

Impending slip: N =

 sin 65°  , so F =  + cos 65°  = T µs  µs 

sin 65°

µs

+ cos 65° = 3.01408

µ s = 0.3497 Must still check slip of belt on pipe

FBD small portion of belt at A: ΣFn = 0:

N1 − N 2 = 0

Impending slip, both sides: F1 = µ s N1, so

F2 = µ s N 2

F1 = F2 = F

ΣFt = 0:

2 F − TA = 0, TA = 2F

Impending slip of belt on pipe: T = TAe µs β or µ s =

T 1 3.01408 = = 0.0797 ln 2F 5.1487 2

Above controls, so for self-locking, need

µ s = 0.350

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Chapter 8, Solution 127. FBD wrench Note: EC =

( 0.20 m ) , sin 75°

EA = EC − 0.03 m

θ = 75° so β = 285° = 4.9742 rad ΣM E = 0:

 0.20 m   0.20 m  − 0.03 m  F −  cos 75° − 0.03 m  T = 0   sin 75°   sin 75°  T = 7.5056 F

ΣFx = 0:

N sin 75° + F cos 75° − T = 0

Impending slip: N =

 sin 75°  , so F =  + cos 75°  = T = 7.5056 F µs  µs 

sin 75°

µs

+ cos 75° = 7.5056

µ s = 0.1333 Must still check impending slip of belt on pipe

FBD small portion of belt at A ΣFn = 0:

N1 − N 2 = 0

Impending slip F1 = µ s N1,

F2 = µ s N 2

F1 = F2 = F

so ΣFt = 0:

2 F − TA = 0, TA = 2 F

Impending slip of belt on pipe T = TAe µ s β or µ s =

T 1 7.5056 = = 0.2659 ln 2F 4.9742 2

This controls, so for self locking,

µ s min = 0.267

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Chapter 8, Solution 128.

ΣFn = 0:

∆θ ∆N − T + (T + ∆T )  sin =0 2

∆N = ( 2T + ∆T ) sin

ΣFt = 0:

∆θ (T + ∆T ) − T  cos − ∆F = 0 2

∆F = ∆T cos

Impending slipping:

∆F = µ s ∆N

In limit as

and

∆T cos

∆θ 2

∆θ ∆θ sin ∆θ = µ s 2T sin + µ s ∆T 2 2 2

∆θ → 0: dT = µ sTdθ ,

dT = µ s dθ T

T2 β ∫ T1 T = ∫ 0 µ s dθ ;

T2 = µs β T1 or T2 = T1e µs β

Note: Nothing above depends on the shape of the surface, except it is assumed to be a smooth curve.

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Chapter 8, Solution 129. Small belt section: Side view:

ΣFy = 0: ΣFx = 0:

∆N α ∆θ sin − T + (T + ∆T )  sin =0 2 2 2

∆θ (T + ∆T ) − T  cos − ∆F = 0 2 ∆F = µ s ∆N ⇒ ∆T cos

Impending slipping:

In limit as ∆θ → 0:

End view:

dT =

µ sTdθ α sin

dT µs = dθ α T sin 2

2 dT

∆θ 2T + ∆T ∆θ = µs sin α 2 2 sin 2

T2 β s ∫ T1 T = α ∫ 0 dθ sin

T2 µβ = s α T1 sin 2 T2 = T1e

µ s β /sin α2

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Chapter 8, Solution 130. FBD motor and mount: Impending belt slip, cw rotation µs β

T2 = T1e

sin α2

( 0.40π )

T2 = T1e sin18° = 58.356T1

ΣM D = 0:

(12 in.)(175 lb ) − (13 in.) T1 − ( 7 in.) T2

2100 lb = 13 in. + ( 7 in.)( 58.356 )  T1 T1 = 4.9823 lb,

T2 = 58.356T1 = 290.75 lb

FBD drum at B:

ΣM B = 0:

M B + ( 3 in.)( 4.9823 lb − 290.75 lb ) = 0 M B = 857 lb ⋅ in.

(Compare to 421 lb ⋅ in. using flat belt, Problem 8.107)

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Chapter 8, Solution 131. Geometry:

θ = sin −1

2 in. = 7.1808° = 0.12533 rad 16 in.

β A = π − 2θ = 2.8909 rad Since β B > β A , impending slip on A will control the maximum couple transmitted FBD A:

ΣM A = 0:

60 lb ⋅ in. + ( 2 in.)(T1 − T2 ) = 0 T2 − T1 = 30 lb µs β

Impending slip: T2 = T1e

sin α2

 ( 0.35)( 2.8909 )  so T1  e sin18° − 1 = 30 lb     T1 = 1.17995 lb T2 = 31.180 lb FBD B:

ΣFx = 0:

P − ( 31.180 lb + 1.17995 lb ) cos 7.1808° = 0 P = 32.1 lb

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Chapter 8, Solution 132. FBD block:

ΣFn = 0:

N − (1000 N ) cos 30° − ( 200 N ) sin 30° = 0 N = 966.03 N

Assume equilibrium:

ΣFt = 0:

F + ( 200 N ) cos 30° − (1000 N ) sin 30° = 0 F = 326.8 N = Feq.

But

Fmax = µ s N = ( 0.3) 966 N = 290 N Feq. > Fmax

and

impossible ⇒ Block moves

F = µk N = ( 0.2 )( 966.03 N )

Block slides down

F = 193.2 N

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Chapter 8, Solution 133. FBD block (impending motion to the right)

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

P W = sin φs sin (θ − φ s )

sin (θ − φs ) =

(a)

W sin φs P

W = mg

(

)

 ( 30 kg ) 9.81 m/s 2  m = 30 kg: θ − φ s = sin −1  sin14.036°    120 N

= 36.499°

∴ θ = 36.499° + 14.036° (b)

m = 40 kg: θ − φs = sin

or θ = 50.5°

(

 ( 40 kg ) 9.81 m/s 2

−1 



120 N

) sin14.036° 

= 52.474° ∴ θ = 52.474° + 14.036°

or θ = 66.5°

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Chapter 8, Solution 134. FBDs Top block:

(a) Note: With the cable, motion must impend at both contact surfaces. ΣFy = 0:

N1 − 40 lb = 0

F1 = µ s N1 = 0.4 ( 40 lb ) = 16 lb

Impending slip: ΣFx = 0:

Bottom block:

ΣFy = 0:

T − F1 = 0

T − 16 lb = 0

N 2 − 40 lb − 60 lb = 0

Impending slip:

T = 16 lb

N 2 = 100 lb

F2 = µ s N 2 = 0.4 (100 lb ) = 40 lb

ΣFx = 0:

FBD blocks:

N1 = 40 lb

− P + 16 lb + 16 lb + 40 lb = 0

P = 72.0 lb (b) Without the cable, both blocks will stay together and motion will impend only at the floor. ΣFy = 0: Impending slip: ΣFx = 0:

N − 40 lb − 60 lb = 0

N = 100 lb

F = µ s N = 0.4 (100 lb ) = 40 lb 40 lb − P = 0

P = 40.0 lb

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Chapter 8, Solution 135. FBD ladder:

Motion impends at both A and B, so FA = µ s N A ΣM A = 0:

NB =

a W =0 2

lN B −

ΣFx = 0:

FA +

ΣFy = 0:

2.5W 13

5 12 FB − NB = 0 13 13

NA −

l = 19.5 ft

a 7.5 ft W = W 2l 39 ft

2.5 W 13

µs N A +

a = 7.5 ft

NB =

FB = µ s N B = µ s

Then

a 5 = l 13

FB = µ s N B

and

12.5

(13)

µ sW −

(13)2

W =0

( 30 − 12.5µs )

(13)

NA − W +

µs

12 5 FB + NB = 0 13 13

 30 − 12.5µ s  W + 30µ s + 12.5  =W  2 µs   (13)

b 12 = l 13

µ s2 − 5.6333µ s + 1 = 0 µ s = 2.8167 ± 2.6332

µ s = 0.1835

and

µ s = 5.45

The larger value is very unlikely unless the surface is treated with some “non-skid” material. In any event, the smallest value for equilibrium is µ s = 0.1835

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Chapter 8, Solution 136. FBD window:

(

)

T = ( 2 kg ) 9.81 m/s 2 = 19.62 N =

ΣFx = 0:

N A − ND = 0

FA = µ s N A

Impending motion: ΣM D = 0:

(

W = ( 4 kg ) 9.81 m/s

NA =

) = 39.24 N ΣFy = 0:

N A = ND

FD = µ s N D

( 0.36 m )W − ( 0.54 m ) N A − ( 0.72 m ) FA W =

W 2

3 N A + 2µ s N A 2 2W 3 + 4µ s

FA − W + T + FD = 0 FA + FD = W − T =

Now Then

W 2

FA + FD = µ s ( N A + N D ) = 2µ s N A W 2W = 2µ s 2 3 + 4µ s

µ s = 0.750 W

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Chapter 8, Solution 137. FBD Collar:

Stretch of spring x = AB − a =

a −a cosθ

 a   1  − a  = (1.5 kN/m )( 0.5 m )  − 1 Fs = k  θ θ cos cos      1  = ( 0.75 kN )  − 1 = ( 750 N )( sec θ − 1) θ cos   Fs cosθ − W + N = 0

ΣFy = 0:

W = N + ( 750 N ) (1 − cosθ )

or Impending slip:

F = µ s N (F must be +, but N may be positive or negative)

Fs sin θ − F = 0

ΣFx = 0: or

F = Fs sin θ = ( 750 N )( tan θ − sin θ )

(a) θ = 20°:

F = ( 750 N )( tan 20° − sin 20° ) = 16.4626 N

Impending motion: (Note: for

N =

µs

16.4626 N = 41.156 N 0.4

N < 41.156 N, motion will occur, equilibrium for

N > 41.156)

W = N + ( 750 N )(1 − cos 20° ) = N + 45.231 N

But

So equilibrium for W ≤ 4.07 N and W ≥ 86.4 N W (b) θ = 30°:

F = ( 750 N )( tan 30° − sin 30° ) = 58.013 N

Impending motion:

N =

µs

58.013 = 145.032 N 0.4

W = N + ( 750 N )(1 − cos 30° ) = N ± 145.03 N = −44.55 N ( impossible ) , 245.51 N Equilibrium for W ≥ 246 N W

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Chapter 8, Solution 138. FBD pin C:

FAB = P sin10° = 0.173648P FBC = P cos10° = 0.98481P ΣFy = 0: FBD block A:

N A − W − FAB sin 30° = 0 N A = W + 0.173648P sin 30° = W + 0.086824P

or ΣFx = 0:

FA − FAB cos 30° = 0 FA = 0.173648P cos30° = 0.150384P

FA = µ s N A

For impending motion at A: NA =

Then

µs

: W + 0.086824 P =

0.150384 P 0.3

P = 2.413W

or ΣFy = 0:

N B − W − FBC cos 30° = 0 N B = W + 0.98481P cos30° = W + 0.85287 P

FBD block B:

ΣFx = 0:

FBC sin 30° − FB = 0 FB = 0.98481P sin 30° = 0.4924 P FB = µ s N B

For impending motion at B: Then or

NB =

µs

: W + 0.85287 P =

0.4924P 0.3

P = 1.268W

Thus, maximum P for equilibrium

Pmax = 1.268W W

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Chapter 8, Solution 139. φs = tan −1 µ s = tan −1 0.25 = 14.036° FBD block A:

R2 750 lb = sin104.036° sin16.928° R2 = 2499.0 lb FBD wedge B:

P 2499.0 = sin 73.072° sin 75.964°

P = 2464 lb P = 2.46 kips

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Chapter 8, Solution 140. Block on incline:

θ = tan −1

0.1 in. = 3.0368° 2π ( 0.3 in.)

φs = tan −1 µ s = tan −1 0.12 = 6.8428°

Q = ( 500 lb ) tan 9.8796° = 87.08 lb Couple on each side

M = rQ = ( 0.3 in.)( 87.08 lb ) = 26.12 lb ⋅ in. Couple to turn = 2M = 52.2 lb ⋅ in.

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Chapter 8, Solution 141. FBD pulley:

ΣFy = 0:

R − 103.005 N − 49.05 N − 98.1 N = 0 R = 250.155 N

ΣM O = 0:

( 0.12 m )(103.005 N − 98.1 N ) − rf ( 250.155 N ) = 0 rf = 0.0023529 m = 2.3529 mm

φ s = sin −1 

µ s = tan φs = tan  sin −1 

rf rs

rf   −1 2.3529 mm   = tan  sin  rs  30 mm  

µ s = 0.0787 W

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Chapter 8, Solution 142. FBD wheel:

ΣM E = 0:

M E = ( 7.5 in.)(T2 − T1 )

FBD lever:

ΣM C = 0:

Impending slipping:

( 4 in.)(T1 + T2 ) − (16 in.)( 25 lb ) = 0 T1 + T2 = 100 lb

− M E + ( 7.5 in.)(T2 − T1 ) = 0

T2 = T1e µs β T2 = T1e

( )

0.25 32π

= 3.2482T1

T1 (1 + 3.2482 ) = 100 lb T1 = 23.539 lb

and

M E = ( 7.5 in.)( 3.2482 − 1)( 23.539 lb ) = 396.9 lb ⋅ in. M E = 397 lb ⋅ in. W

Changing the direction of rotation will change the direction of M E and will switch the magnitudes of T1 and T2 . The magnitude of the couple applied will not change. W

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Chapter 8, Solution 143. FBD block:

ΣFn = 0:

NC − ( 200 lb ) cos 30° = 0; N = 100 3 lb

ΣFt = 0:

TC − ( 200 lb ) sin 30° ∓ FC = 0 TC = 100 lb ± FC

FBD Drum:

(1)

where the upper signs apply when FC acts (a) For impending motion of block , FC

, and

(

)

FC = µ s NC = 0.35 100 3 lb = 35 3 lb

(

)

TC = 100 − 35 3 lb

So, from Equation (1):

TC = WAe µk β

But belt slips on drum, so

−0.25( WA =  100 − 35 3 lb  e  

(

)

2π 3

) WA = 23.3 lb W

and FC = µ s NC = 35 3 lb

(b) For impending motion of block , FC From Equation (1): Belt still slips, so

(

)

TC = 100 + 35 3 lb −0.25( WA = TC e− µk β =  100 + 35 3 lb  e  

(

)

2π 3

)

WA = 95.1 lb W continued

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PROBLEM 8.143 CONTINUED (c) For steady motion of block , FC

, and FC = µk NC = 25 3 lb

(

)

T = 100 + 25 3 lb.

Then, from Equation (1):

Also, belt is not slipping on drum, so −0.35( WA = TC e− µ s β =  100 + 25 3 lb  e  

(

)

2π 3

)

WA = 68.8 lb W