Subtopic: Measurement of force, stress, torque, speed, displacement and acceleration on machine elements – shafts, couplings, keys, bearings and knuckle joints
Question Statement:
A V-belt drive transmits 10 kW at belt velocity 8 m/s. Angle of contact on smaller pulley = 170°, groove angle = 38°, coefficient of friction = 0.28, maximum permissible stress = 4 MPa (neglect centrifugal effect). Find the minimum cross-sectional area of the belt (mm²). (Rounded to 1 decimal place)
Official Answer Key:
337.0 to 340.0
Step-by-Step Solution & Explanation
T₁−T₂ = Power/velocity = 10000/8 = 1250 N. For a V-belt: T₁/T₂ = e^(μθ/sin(β/2)), θ=170°=2.967 rad, β/2=19° (sin19°=0.3256). Exponent = 0.28×2.967/0.3256 = 2.551 ⇒ T₁/T₂ = e^2.551 = 12.83. Solving with T₁−T₂=1250: T₂=105.7 N, T₁=1355.7 N. Minimum area = T₁/σ_max = 1355.7/4 ≈ 338.9 mm².
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.