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InClassLecture25

# InClassLecture25 - 9-86 A rope attached to a ZZO—kg block...

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Unformatted text preview: 9-86* A rope attached to a ZZO—kg block passes over a fixed drum as shown in Fig. P9—86. If the coefficient of friction between.the rope and the drum is 0.30, determine (a) The minimum force ﬁ that must be used to keep the block from falling. (b) The minimum force § that must be used to begin to raise the block. Fig.P9-86 SOLUTION From a free-body diagram for the block: ‘Y’ +T2Fy=T—mg=0 T = mg = 220(9.807) = 2158 N W3 From Eq. 9—16b: _ MB _ o.3o(n/2) _ T2 — Tie — Tie — 1.6020'1‘1 (a) During lowering of the block: T2 = 2158 N T2 2158 Pmin = T,1 = m = m = 1347.1 N g 1347 N Ans. (b) During raising of the block: T1 = 2158 N P , = T = 1.6020T1 = 1.6020(2158) = 3457 N ¥ 3460 N Ans. min 2 9—89 The electric motor shown in Fig. P9-89 weighs 30 lb and delivers 50 in.-lb of torque to pulley A of a furnace blower by means of a V—belt. The effective diameters of the 36 pulleys are 5 in. The coefficient of friction is 0.30. Determine the minimum distance a to prevent slipping of the belt if the rotation of the motor is clockwise. SOLUTION . _ ___Ji___., __Jlﬁ¥L__ _ For a pulley' “enh ' sin («/2) ‘ sin (36/2) “ 0'9708 u 5 = 0.9708(n) = 3.050 enh From Eq. 9—17: T2 = T1e3'050 = 21.11T1 From a free-body diagram for the blower pulley: + D 2MB = Tarp — TirP — cA 21.11T1(2.5) — T1(2.5) - 50 = 0 T = 0.9945 lb T = 21.11T1 21.1l(0.9945) = 20.99 lb From a free-body diagram for the motor: + C EMA = T2(8 + rP) + T1(8 - pp) — WM(a) = 20.99(10.5) + 0.9945(5.5) — 30a = 0 7.529 in. E 7.53 in. Ans. 9: u 9-90 The band brake of Fig P9-90 is used to control the rotation of a drum. The coefficient of friction between the belt‘and the drum is 0.35 and the weight of the handle is 15 N. If a force of 200 N is applied to the end of the handle, determine the maximum torque for which no motion occurs if the torque is applied (a) Clockwise. (b) Counterclockwise. SOLUTION 180 + 30 360 3.665 rad. (211') B: 0.35(3.665) = 1.2828 #3 = From Eq. 9-16b: . 8 T =Te =Te1282 = 2 1 1 3.6067T1 (a) From a free-body diagram of the handle when the rotation of the drum is 200N clockwise: + C EMA = 200(675) — 15(225) — T2(75) = 0 T2 = 1755.0 N T l _ 2 _ 1755.0 _ 1 T1 ' 3.607 ‘ 3.606 ' 486‘6 N Tmax = (T2 - T1)R = (1755.0 - 486.6)(0.150) = 190.3 N'm Ans T, 200); (b) From a free—body diagram of the handle when the rotation of the drum is counterclockwise: WSantl T2 = 1755.0 N n 5 \s N T2 = 3.6067T1 = 3.6067(1755.0) = 6329.8 N T = (T2 - T1)R = (6329.8 — 1755.0)(0.150) = 686 N'm Ans. 9-96* A 100—N crate is sitting on a ZOO—N crate as shown in Fig. P9-96. The rope that joins the crates passes around a frictionless pulley and a fixed drum. The coefficient of friction is 0.30 at all surfaces. Determine the minimum force F that must be applied to the ZOO-N crate to start motion. SOLUTION L2“ ,1“ From a free—body diagram .T -r of the 100—N crate: g I . . . . m "I For impending sllding: No )ooN 3 )ooN + T XFy = An — 100 = 0 __ n B A = 100 N R «3’ Bn " "n A =uA =0.3(100> =30N u TVqu f n For impending tipping: + C 2MB = T(2.5) - 100(0.6) = 0 T = 24 N < 30 N (crate tips) 'yi From Eq. 9—16b: T2 : T1eu6 = 24eo.3o(2N/3) = 45 N From a free-body diagram of the ZOO—N crate: 5" cn=300N + —e ZFX = P — Af - Cf - T2 = 0 P , = Ar + Cf + T2 = 24 + 0.3(300) + 45 = 159.0 N Ans. Min ...
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