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9-3, 9-16, 9-20

# 9-3, 9-16, 9-20 - 9-3 workers are pulling a 400-1b crate up...

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Unformatted text preview: 9-3 workers are pulling a 400-1b crate up an incline as shown in Fig. P9-3. The coefficient of friction between the crate and the surface is 0.20, and the rope on which the workers are pulling is horizontal. {a} Determine the force F that the workers must exert to start sliding the crate up the incline. If one of the workers lets go of the rope for a moment, determine the minimum force the other workers must exert to keep the crate from sliding back down the incline. SOLUTION A free-body diagram for the block is shown at the right. (a) For impending motion of the block up the incline: + z” EFx p cos 15° - 400 sin 15° - 0.2An o n + “\ 2F? A - P sin 15° - 400 cos 15° = 0 Solving yields: A = 437.6 lb P = 19?.?8 lb 8 19?.8 lb TI For impending motion of the block down the incline: + /’ EFx 9 cos 15° - 400 sin 15° + 0.2An = o + “\ EFY A — P sin 15° - 400 cos 15° = 0 :1 Solving yields: A = 393.0 lb P = 25.30 lb The masses of blocks A and B of Fig. P9-16 are mA = 50 kg and '3 = 75 kg. If the coefficient of friction is 0.25 for both surfaces, determine the force F required to cause impending notion of block B. SOLUTION A free-body diagram for block A is shown at the right. HA mhg + 50(9.807l = 490.35 N For impending motion: + -» 2F 2 A - T cos 45° 3 f 0.25.4n - T cos 45° = o T = 0.3536An + T 2F = A + T sin 45°- H y n A = An + 0.3536 An sin 450 - 490.35 = 0 = 392.27 N A free-body diagram for block B is shown at the right. ”B = nag + 75(9.807} = 735.53 N For impending motion: + T SE = B - W - A y B 1'! h B - 735.53 - 392.27 = 0 1127.30 N P - AF — BF P - 0.25{392.27I - 0.25(1127.BO} = 0 380.02 N 8 380 N 9-20 A 30-kg box is sitting on an inclined surface as shown in Fig. PQ-ZO. If the coefficient of friction between the box and the surface is 0550 and the angle a = 60 . determine the maximum force T for which the box will be in equilibrium. Is impending motion by tipping or by slipping? SOLUTION W mg = 30(9.807} = 294.2 N From a free-body diagram for the rectangular block when motion is impending: For sliding: A = Aftmax) = “A“ = 0.50An f + /” 2Fx 0.5mn - T cos 60° ~ 294.2 sin 20° = 0 + “\ EFY A + T sin 60° - 294.2 cos 20° = 0 Solving yields: 241.55 N 40.31 N 8 40.3 N (for sliding} For tipping: + C Sun = T sin 60° (0.91 + T cos 60° {0.6} + 294.2 sin 20° {0.31 - 294.2 cos 20° (0.45) = 0 T = 87.29 N 3 87.3 N [for tipping) Therefore: T = 40.3 N max Impending motion is by slipping ...
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