Sample Problem of Pulley – Belt - Friction
Sample Problem of Pulley β Belt - Friction The tension in the pulley is 110 N when stationary. Calculate the tension in each side and the power transmitted when the belt Is on the point of slipping on the smaller wheel. The wheel is 240 mm diameter and the coefficient of friction is 0.32. The angle of lap is 165º. The wheel speed is 1500 rev/min a) b) c) d) F1 = 140 N , F1 = 157 N , F1 = 130 N , F1 = 120 N , F2= 60 N , P= 1,500 W F2= 62 N , P= 1,786 W F2= 70 N , P= 1,632 W F2= 50 N , P= 1,720 W Solution Belt Velocity = π β π β π· = 3.14 β 1500 β 0.24 ÷ 60 = 18.85 π/π Lap angle = (165/180) β π = 2.88 π ππππππ (Initial Tension) Carlos Batista M. Sc. P. Eng. CCEA CBA Engineering Support Services www.cbaengineer.com F1 = F2 When Stationary F1 + F2 = 220 N Tension in belts F1/F2 = ππ π½ = ππ.ππ π.ππ = 2.513 N F1 = 2.513 × F2 = 2.513 x (220 β F1) = 552.9 β 2.513 x F1 F1 + 2.513 x F1 = 552.9 F1 = 157.4 N 3.513 x F1 = 552.9 When we obtained the first tension F1, in the FE exam is not necessary to continue solving F2 and Power (P) , because if we see the answers, we can identify rapidly the correct answer Carlos Batista M. Sc. P. Eng. CCEA CBA Engineering Support Services www.cbaengineer.com 2 However, Is wise idea to continue a little more to be sure that our Previous calculation is correct. F2 = 157 ÷ π. πππ = ππ. π π π π = F1 ÷ 2.513 Excellent, We are more Sure that answer b is the Correct answer P = V ( F1 β F2) = 18.85 m/s x ( 157.4 β 62.6) N = Be careful with units conversion Carlos Batista M. Sc. P. Eng. CCEA CBA Engineering Support Services www.cbaengineer.com 3 1,786 W
© Copyright 2024