A cubical block of side 0.20 m and density 2500 kg/m³ slides down a 30° incline with constant speed 0.40 m/s. A uniform oil film of viscosity 0.50 Pa·s separates the block from the plane. Take g = 10 m/s². If the film thickness is N × 10⁻² mm, find N.
Physics · Properties of solid and fluid · viscosity and terminal velocity
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Question

A cubical block of side 0.20 m and density 2500 kg/m³ slides down a 30° incline with constant speed 0.40 m/s. A uniform oil film of viscosity 0.50 Pa·s separates the block from the plane. Take g = 10 m/s². If the film thickness is N × 10⁻² mm, find N.

Physics · Properties of solid and fluid · viscosity and terminal velocity

G1 Physics · Accepted Answer

Block volume: V = (0.20)³ = 0.008 m³. Mass: m = ρV = 2500×0.008 = 20 kg. Component of weight down the incline: F(down) = mg sin30° = 20×10×1/2 = 100 N. Contact area: A = (0.20)² = 0.04 m². Viscous resistance: F(viscous) = ηAv/d. For constant speed, F(down) = F(viscous). Therefore, 100 = 0.50×0.04×0.40/d. The numerator on the right is 0.008. Hence, d = 0.008/100 = 8×10⁻⁵ m. Convert to millimetres: d = 0.08 mm = 8×10⁻² mm. Hence, N = 8.