Subtopic: Design and selection of machine elements – gears, pulleys, chains and sprockets and belts
Question Statement:
A flat belt of negligible mass and thickness drives an output pulley of 0.6 m diameter, constantly rotating at 500 rpm without slip. The arc of contact is 190° for this pulley, while it is 170° for the smaller input pulley. The coefficient of friction between the belt and pulleys is 0.30. The measured slack-side tension is 250 N. Neglecting other losses, the power transmitted by the belt drive, in kW, is nearest to (Take π = 3.14)
Options:
(A) 4.8
(B) 5.6
(C) 6.5
(D) 7.3
Official Answer Key:
B
Step-by-Step Solution & Explanation
Step 1 — identify the governing pulley. Slip always begins on the pulley with the smaller arc of contact, because it has the least frictional grip. Hence use θ = 170°.
θ = 170 × 3.14/180 = 2.967 rad
Step 2 — ratio of tensions (belt friction / Eytelwein equation):
T₁/T₂ = e^(μθ) = e^(0.30 × 2.967) = e^0.8901 = 2.434
T₁ = 250 × 2.434 = 608.6 N
Step 3 — belt velocity (at the 0.6 m output pulley running at 500 rpm):
v = πDN/60 = 3.14 × 0.6 × 500/60 = 15.70 m/s
Step 4 — power transmitted:
P = (T₁ − T₂) v = (608.6 − 250) × 15.70 = 358.6 × 15.70 = 5630 W ≈ 5.6 kW
Common error: using the 190° arc gives ≈ 6.5 kW — option (C) is planted as the trap.
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