What is the width of belt to transmit 15 HP to a pulley 35 cm in diameter. Pulley makes 1600 rev/min, coefficient of friction between belt is 0.22 and angle of contact is 20°. Maximum tension in belt not to exceed 8 kg per cm width.
a) 7.79cm
b) 9.79cm
c) 3.49cm
d) 8.79cm

Answer: d
Explanation: \(\frac{T1}{T2}\) = e^0.22*3.66=e^0.8065
T₁ = T₂ * 2.24
15*746 = \(\frac{[π*35*1600(T1-T2)]}{100*60} \)
2.24 T₂-T₁ = 381.63
T₂ = \(\frac{381.63}{1.24}\) = 307.76N
T₁ = 307.76*2.24=689.38N
Width of the belt = \(\frac{689.38}{8*9.8}\) = 8.79cm.