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An open belt runs between two pulleys 400mm and 150mm diameter and their centers are 1000mm apart. If coefficient of friction for larger pulley is 0.3, then what should be the value of coefficient of friction for smaller pulley, so that the slipping is about to take place at both the pulley at the same time?

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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

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