Radius of each pulley = 50mm,
R1 = R2 = 50mm
R1 = Radius of the driver = 50mm
R2 = Radius of the driven = 50mm
θ = Angle of contact(In radian) = 1800 = p,
µ = Coefficient of friction = 0.3
T1 = Allowable tension = 3KN
T1 always greater than T2
Using the relation T1/T2 = eµθ
Putting all the value,
3/T2 = e(0.3)(π)
On solving T2 = 1.169KN
Since Radius of both pulley is same;
So, Torque exerted on both pulley is same an
= (T1–T2).R1 = (T1–T2).R2
Putting all the value we get,
(3 – 1.169) X 50 = 91.55 KN-mm