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 = (T_{1} - T_{2}).R
300 = (T_{1} - T_{2}) X 0.5
T_{1}- T_{2} = 600N ...(i)
And, T_{1}/T_{2} = e^{µθ}
T_{1}/T_{2} = e(0.3)p
T_{1 }= 2.566T2 ...(ii)
Solve (i) and (ii)
We get,
T_{1 } = 983.14N .......ANS
T_{2} = 383.14N .......ANS
Now reaction on bearing is opposite to mass of body, and it is equal to
R = T_{1} + T_{2} + mg
R = 983.14 + 383.14 + 20 X 9.81
R = 1562.484N .......ANS