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CHAPTER 8 - PROBLEM 8.1 Determine whether the block shown...

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PROBLEM 8.1 Determine whether the block shown is in equilibrium, and find the magnitude and direction of the friction force when o 30 θ = and 200 P = N. SOLUTION FBD block: ( ) ( ) 0: 1000 N cos30 200 N sin30 0 n F N Σ = ° − ° = 966.03 N N = Assume equilibrium: ( ) ( ) 0: 200 N cos30 1000 N sin30 0 t F F Σ = + ° − ° = eq. 326.8 N F F = = But ( ) max 0.3 966 N 290 N s F N µ = = = eq. max impossible F F > Block moves W and k F N µ = ( )( ) 0.2 966.03 N = Block slides down 193.2 N = F W
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PROBLEM 8.2 Determine whether the block shown is in equilibrium, and find the magnitude and direction of the friction force when o 35 θ = and 400 P = N. SOLUTION FBD block: ( ) ( ) 0: 1000 N cos35 400 N sin35 0 n F N Σ = ° − ° = 1048.6 N N = Assume equilibrium: ( ) ( ) 0: 1000 N sin35 400 N cos35 0 t F F Σ = ° + ° = eq. 246 N F F = = ( )( ) max 0.3 1048.6 N 314 N s F N µ = = = eq. max OK equilibrium F F < W 246 N = F W
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PROBLEM 8.3 Determine whether the 20-lb block shown is in equilibrium, and find the magnitude and direction of the friction force when 8 lb P = and 20 . θ = ° SOLUTION FBD block: ( ) ( ) 0: 20 lb cos20 8 lb sin 20 0 n F N Σ = ° + ° = 16.0577 lb N = ( )( ) max 0.3 16.0577 lb 4.817 lb s F N µ = = = Assume equilibrium: ( ) ( ) 0: 8 lb cos20 20 lb sin 20 0 t F F Σ = ° − ° − = eq. 0.6771lb F F = = eq. max OK equilibrium F F < W and 0.677 lb = F W
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PROBLEM 8.4 Determine whether the 20-lb block shown is in equilibrium, and find the magnitude and direction of the friction force when 12.5 lb P = and 15 . θ = ° SOLUTION FBD block: ( ) ( ) 0: 20 lb cos20 12.5 lb sin15 0 n F N Σ = ° + ° = 15.559 lb N = ( )( ) max 0.3 15.559 lb 4.668 lb s F N µ = = = Assume equilibrium: ( ) ( ) 0: 12.5 lb cos15 20 lb sin 20 0 t F F Σ = ° − ° − = eq. 5.23 lb F F = = eq. max but impossible, so block slides up F F > W and ( )( ) 0.25 15.559 lb k F N µ = = 3.89 lb = F W
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PROBLEM 8.5 Knowing that 25 , θ = ° determine the range of values of P for which equilibrium is maintained. SOLUTION FBD block: Block is in equilibrium: ( ) 0: 20 lb cos20 sin 25 0 n F N P Σ = ° + ° = 18.794 lb sin 25 N P = ° ( ) 0: 20 lb sin 20 cos25 0 t F F P Σ = ° + ° = or 6.840 lb cos25 F P = ° Impending motion up: ; Impending motion down: s s F N F N µ µ = = − Therefore, ( )( ) 6.840 lb cos25 0.3 18.794 lb sin 25 P P ° = ± ° up down 12.08 lb 1.542 lb P P = = eq. 1.542 lb 12.08 lb P W
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PROBLEM 8.6 Knowing that the coefficient of friction between the 60-lb block and the incline is 0.25, s µ = determine ( a ) the smallest value of P for which motion of the block up the incline is impending, ( b ) the corresponding value of β . SOLUTION FBD block (impending motion up) ( ) 1 1 tan tan 0.25 14.04 s s φ µ = = = ° ( a ) Note: For minimum P , so s β φ = P R Then ( ) sin 30 s P W φ = ° + ( ) 60 lb sin 44.04 41.71lb = ° = min 41.7 lb P = W ( b ) Have s β φ = 14.04 β = ° W
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PROBLEM 8.7 Considering only values of θ less than 90 ° , determine the smallest value of θ for which motion of the block to the right is impending when ( a ) 30 kg, m = ( b ) 40 kg. m = SOLUTION FBD block (impending motion to the right) ( ) 1 1 tan tan 0.25 14.036 s s φ µ = = = ° ( ) sin sin s s P W φ θ φ = ( ) sin sin s s W W mg P θ φ φ = = ( a ) ( ) ( ) 2 1 30 kg 9.81m/s 30 kg: sin sin14.036 120 N s m θ φ = = ° 36.499 = ° 36.499 14.036 θ = ° + ° or 50.5 θ = ° W ( b ) ( ) ( ) 2 1 40 kg 9.81m/s 40 kg: sin sin14.036 120 N s m θ φ = = ° 52.474 = ° 52.474 14.036 θ = ° + ° or 66.5 θ = ° W
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PROBLEM 8.8 Knowing that the coefficient of friction between the 30-lb block and the
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