# A horizontal drum of a belt drive carries the belt over a semicircle around it. It is rotated anticlockwise to transmit a torque of 300N-m.

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A horizontal drum of a belt drive carries the belt over a semicircle around it. It is rotated anticlockwise to transmit a torque of 300N-m. If the coefficient of friction between the belt and rope is 0.3, calculate the tension in the limbs 1 and 2 of the belt shown in figure, and the reaction on the bearing. The drum has a mass of 20Kg and the belt is assumed to be mass less.

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Given data:

Torque(t) = 300N-m

Coff. of friction(µ) = 0.3

Diameter of Drum (D) = 1m, R = 0.5m

Mass of drum(m) = 20Kg.

Since angle of contact = π rad

Torque = (T1 – T2).R

300 = (T1 – T2) X 0.5

T1– T2 = 600N ...(i)

And, T1/T2 = eµθ

T1/T2 = e(0.3)π

T1 = 2.566T2 ...(ii)

Solving (i) and (ii)

We get,

T1 = 983.14N

T2 = 383.14N

Now reaction on bearing is opposite to the mass of the body, and it is equal to

R = T1 + T2 + mg

R = 983.14 + 383.14 + 20 X 9.81

R = 1562.484N.