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in Physics by (58.2k points)

A belt drives a pulley of 200mm diameter such that the ratio of tensions in the tight side and slack side is 1.2. If the maximum tension in the belt is not to exceed 240KN. Find the safe power transmitted by the pulley at a speed of 60rpm.

1 Answer

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Best answer

Given that, 

D1 = Diameter of the driver = 200mm = 0.2m

T1/T2 = 1.2

Since between T1 and T2, T1 is always greater than T2,

Hence T1 = 240KN

N1 = Speed of the driver in R.P.M. = 60PRM

P = ?

We know that

T1/T2 = 1.2

T2 = T1/1.2 =240/1.2 = 200KN

 V = Velocity of the belt in m/sec

= πDN/60 m/sec, D is in meter and N is in RPM

= (3.14 X 0.2 X 60)/60 = 0.628 m/sec

P = (T1 - T2) X V

P = (240 - 200) X 0.628

P = 25.13KW

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