A pulley rotating at 50 m/s transmits 40 kW. The safe pull in belt is 400 N/cm width of belt. The angle of lap is 170º. If coefficient of friction is 0.24, find required width of belt.

Data: Initial tension, To = 2000 N, coefficient of friction, µ = 0.3, Angle of lap, θ = 1500 = 1500 x П / 180 = 2.618 rad, Smaller pulley radius, R = 200 mm, hence, D = 400 mm, Speed of smaller pulley, N = 500 r.p.m. We know that the velocity of the belt, v = П = П = 10.47 m/sec (01 mark) Let T1 = Tension in the belt on the tight side, N Let T2 = Tension in the belt on the slack side, N We know that, T0 = Hence, 2000 = (T1 + T2) / 2 Thus, (T1 + T2) = 4000 N ....................... (1) We also know that, = therefore, = or = 2.2 ............. (2) From equations 1 and 2, T1 = 2750 N and T2 = 1250 N (02 marks) Power transmitted by belt, P = [T1 - T2] v = [2750 - 1250] 10.47 = 15700 watts = 15.7 kW