Question

Chapter 6, Problem 6/160 The motor at B supplies a constant torque M which is applied to a 395-mm-diameter internal drum around which is wound the cable shown. This cable then wraps around a pulley which has a mass of 86 kg and is attached to a cart which has a mass of 120 kg carrying 515 kg of rock. The motor is able to bring the loaded cart to a cruising speed of 2.7 m/s in 5 seconds. What torque M is the motor able to supply, and what is the average value of the tension in each side of the cable which is wrapped around the pulley at o during the speed-up period? The cable does not slip on the pulley and the centroidal radius of gyration of the pulley is 505 mm. What is the power output of the motor when the cart reaches its cruising speed? 790 mm 11° Answers: M = Nom Ttop =

Answer 1

Solutions Given that mass of pully (mp) = 86 kg mass of Cart (Mc) = 120 kg mass of rock (mg) = 515 kg crursing speed CV) - 2.7 m/s time (t) = 5 seconds radius of gyration pully. (Ko) = 505 mm = 6,505m diameter (d). - 790 mm) = 0.790 m Inclination angle (0) = 10 As shown below figure. 7 Zao mm The impulse momentum Equation given as ( Itop). t + (Toot) t -(mpt my+me)g.Snø.t =(mpt My tm() Trop. 5+ Tbot.5-(86+515+120).9.81* Sin six 5 - (86+515+120) 2.7

5 Top +5 Thot-1721x9.81X0.19080 5) = 1946.7. 5 Top + 5 Thot - 6747.6515 = 1946.7 5 Ttop + 5 Tbot = 1946:7 +67 47.6515 5 ( Trop tToot ) = 8694.3515 Teee Toot = f1 38. 87D3N0 Mass moment of inertia af pully To = mp kom : 86 x 10.505)? = 21,9321 kg-m? Now The Instid angular velocity of pully is Wp = 2v. 24.74 – 6-835 Yad/sec The angular moment Equation ig, Thot (/2) t-Trop (02) t = -To af 0.79 Thot (4:39)5 - Ttop (079)5=-(21-9321x 6.835) (Tbot - Ttop) 1.975= -149.9059 Tbot-Stop =-75.901710

O&o Now solve &q we get Txop TTbot = 1938,8703 Toot - Ty6p = -75.9017 2 Tbot = 1662.9686 | Toot = 831.48 N | from car? O +831.481738.8703 Trop = 90713903 N7 The given motor diameter is 395 mm dmetor 20.395 m Angular velocity of motor is given W moter = 4V - 4X 2,7 = 27.34 rad's a d meter 0:395 Torque M of the motor is given bas M= Ttop { $motor} M = 907.3903 ( 0-395 ) 1 M = 179.2095 Nom

power out put of motor is given as P = MW motor po 179. 2095 x 27.34 P = 4899.596 P: 4.899 kw Answers M = 179.2095 Nm Trop = 907.3903 N Tunat - 831.48N p= 4.899 kw

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