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Calculate tension in limbs:
Q : A horizontal drum of belt drive carries the belt over semicircle around it. It is rotated counter clockwise to transmit a torque of 300N-m. If coefficient of friction between the belt and rope is 0.3, calculate tension in limbs 1 and 2 of belt shown in figure, and reaction on bearing. The drum is having mass of 20Kg and the belt is assumed to be mass less.
Sol: Given: 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 = p rad Torque = (T1 - T2).R
300 = (T1 - T2) X 0.5
T1- T2 = 600N ...(i)
And, T1/T2 = eµθ
T1/T2 = e(0.3)p
T1 = 2.566T2 ...(ii)
Solve (i) and (ii)
We get,
T1 = 983.14N .......ANS
T2 = 383.14N .......ANS
Now reaction on bearing is opposite to mass of body, and it is equal to
R = T1 + T2 + mg
R = 983.14 + 383.14 + 20 X 9.81
R = 1562.484N .......ANS
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