Question

The crane shown in the figure is lifting a 463-kg crate upward with an acceleration of 2.60 m/s2. The cable from the crate passes over a solid cylindrical pulley at the top of the boom. The pulley has a mass of 111 kg. The cable is then wound onto a hollow cylindrical drum that is mounted on the deck of the crane. The mass of the drum is 124 kg, and its radius (the same as that of the pulley) is 0.520 m. The engine applies a counterclockwise torque to the drum in order to wind up the cable. What is the magnitude of this torque? Ignore the mass of the cable. Pulley- Drum Boom LEEE Number Number Units

Answer 1

The solution is

mass of coate, m'= 463 kg Private acceleration, as 2.6 mls?. mass of pulley, m 111 kg mass of drum, ma = 124kg. radius of drouma radius y palley.rs 0.52m. V Torque to be applied by the engine, ī - torque to pull the cable + torque to rotate - the pulley & porque to rotate the drum. T = Tot To + To Now the formula of torque is, T = fxr Torque to pull the cable, Te - Tension in the rope x radius of drum - mg +ma) x ♡ smcg+a) xo = 463% [ 9.8+2.6) *0.52 = 2985.424 Nm

n Torque to notate the pulley, - Tp = Ipdp. where I moment of inertia .. Mis j tudi 7 angulon acceleration i Ip = my (for disc) in = 0.52). 1 = 15 kg? txa T do z 0:52 : -5 reid (s?. - Tpe Ip«р . Tp = 75 Nm

Torque to notate drum, Id = Id ad Id = marz sa 2 = 124 x 10.5072 - - 2 Id = 16.7848 A legm² Xd = a 2.6 = 0.52 s road is. so il Id 16.7848x55 83.824 Nm torque applied by engine, T = Te + Tp + Ta = 22985. 424 + 75783.824 T = 3 144 • 248 Nm Cams)

Thank you.