The mass of the belt is 0.9 kg per metre length and the maximum tension is not to exceed 2000 N. The coefficient of friction is 0.3. The 1.2 m pulley, which is the driver, runs at 200 rpm. Due to belt slip on one of the pulleys, the velocity of the driven shaft is only 450 rpm. Calculate the torque on each of the two shafts, the power transmitted, and power lost in friction. What is the efficiency of the drive?

Subject: Kinematics of Machinery

Topic: Belts, Chains and Brakes

Difficulty: High

Given : $d_{1}$ = 1.2 m or $r_{1}$ =0.6 m ; $d_{2}$= 0.5 m or $r_{2}$ =0.25 m; x =4 m; m =0.9 kg/m;

T =2000N; $\mu$ =0.3; $N_{1}$ =200 rpm. ; $N_{2}$ =450 rpm.

We know that velocity of the belt,

$v=\frac{\pi d_{1}N_{1}}{60}=\frac{\pi\times1.2\times 200}{60}$= 12.57 m/s.

and centrifugal tension, …