Question: An electric motor provides 6.6 KW to a belt driven pulley 125 mm in diameter. The speed of the pulley is 1050 rpm. Calculate the tight side and slack side tension of the belt if they are in the ratio 7:2?
Power, P = 6.6KW
Diameter, d = 125×10^-3m
Speed of the pulley, N = 1050 rpm
Let T1 be the tight side tension and T2 be the slack side tension. Also, given that, T1/T2 = 7/2 or T1 = T2 . 7/2
We have to find out T1 and T2.
Power transmitted by the pulley, P = (T1- T2) . π dN/60.
But, given that Power, P = 6.6 KW
(T1- T2) . π dN/60 = 6.6 x 10^3
Substituting the value of T1 and thus solving for T1 and T2, we get,
(7/2T2 – T2) . π . 125 . 10^-3 . 1050/60 = 6.6 . 10^3
i.e, T2 = 6.6 . 10^3 / 17.181 = 384.15 N
T1 = T2 . 7/2 = 384.15 . 7/2
= 1344.53 N
∴ Tight side tension, T1 = 1344.53 N
Slack side tension, T2 = 384.15 N