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A centrifugal clutch is to transmit 35 kW at 1500 r.p.m. There are four shoes. The speed at which the engagement begins is 0.7 times the running speed. The inside radius of the pulley rim is 120 mm and the centre of gravity of the shoe lies at 80 mm from the centre of the spider. Coefficient of friction may be taken as 0.35. Determine intensity of pressure exerted on the shoes if the angle subtended by the shoes at the centre of the spider is 60° and the width of the shoes is 105.52 mm.

Correct Answer: 105.52 mm
Given : P = 35 kW = 35 × 103 W ; N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157.08 rad/s ; n = 4 ; R = 120 mm = 0.12 m ; r = 80 mm = 0.08 m ; µ = 0.35 ; ω1 = 0.7 ω ; θ = 60° = π/3 rad ; b = 105.52 mm. ω1 = 0.7 ω = 0.75 x 157.08 = 109.95 rad/s Power transmitted = Tω = T x 104.72 T = 222.816 N-m Pc = mω2r = m x 157.082 x 0.08 = 1973.93 m N Ps = mω12r = m x 109.952 x 0.08 = 967.12 m N Frictional force acting tangentially on each shoe, F = µ(Pc– Ps) = 0.35 (1973.93 m – 967.12 m) = 352.38 m N Torque transmitted (T), 190.98 = n.F.R = 4 × 352.38 m × 0.12 = 169.144 m m = 1.317 kg l = θ x R = π/3 x 120 = 125.66 mm l.b.p = Pc – Ps = 1006.81 m 125.66 x 105.52 x p = 1006.81 x 1.317 p = 0.1 N/mm2.

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