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  = e^µθ

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