Question

A belt is running over a pulley of 1.5m diameters at 250RPM. The angle of contact is 120º and the coefficient of friction is 0.30. If the maximum tension in the belt is 400N, find the power transmitted by the belt.

Answer 1

Given data 

Diameter of pulley(D) = 1.5m 

Speed of the driver(N) = 250RPM 

Angle of contact(?) = 1200 = 1200 X (π/180º) = 2.09 rad 

Coefficient of friction(µ) = 0.3 

Maximum tension(Tmax) = 400N = T₁ 

Power (P) = ? 

Since P = (T₁ – T₂) X V Watt ............(i) 

T₁ is given, and for finding the value of T₂, using the formula 

Ratio of belt tension = T₁/T₂ = e^µθ 

400/T₂ = e(0.3)(2.09) 

T₂ = 213.4N ...........(ii) 

Now We know that V = πDN/60 m/sec

 V = [3.14 X 1.5 X 250]/60 = 19.64 m/sec ...(iii)

Now putting all the value in equation (i) 

P = (400 – 213.4) X 19.64 watt 

P = 3663.88 Watt or 3.66K W.