**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**