Given data
D1 = 400mm, R1 = 200mm
D2 = 150mm, R2 = 775mm
X = 1000mm
µ1 = 0.3 µ2 = ?
Sinα = (r1 – r2)/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 = T1/T2 = eµθ
It is equal for both the pulley, i.e., (T1/T2)larger pulley = (T1/T2)smaller pulley
or, eµ1θ1 = eµ2θ2 , or µ1θ1 = µ2θ2
putting all the value, we get,
(0.3)(Π + 2α) = (µ2)(Π – 2α)
(0.3)(Π + 2 × 0.1256) = (µ2)(Π – 2 X 0.1256)
on solving, µ2 = 0.352