Given data

D_{1 }= 400mm, R_{1} = 200mm

D_{2} = 150mm, R_{2} = 775mm

X = 1000mm

µ_{1} = 0.3 µ_{2} = ?

Sinα = (r_{1} – r_{2})/X = (200 – 75)/1000

α = 7.18° = 7.18° × Π/180° α = 0.1256 rad ...(i)

We know that

For Open belt drive:

Angle of contact (θ) for larger pulley = Π + 2α

Angle of contact (θ) for smaller pulley = Π – 2α

Since, Ratio of belt tension = T_{1}/T_{2} = eµθ

It is equal for both the pulley, i.e., (T_{1}/T_{2})larger pulley = (T_{1}/T_{2})smaller pulley

or, e^{µ}1^{θ}1 = e^{µ}2^{θ}2 , or µ_{1}θ_{1 }= µ_{2}θ_{2}

putting all the value, we get,

(0.3)(Π + 2α) = (µ_{2})(Π – 2α)

(0.3)(Π + 2 × 0.1256) = (µ_{2})(Π – 2 X 0.1256)

on solving, µ_{2} = 0.352