A square block weighing 1.1 kN and 250 mm on an edge slides down an incline on a film of oil 6.0 µm thich (see Fig.). Assuming a linear velocity profile in the oil, what is the terminal speed of the block? The viscosity of the oil is 7 mPa • s.
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
Sol . τ = µ (dv/dy) = (7 × 10 –3 ) [v T / (6.0 × 10 –6 )] = 1167 v T
F f = τ A = (1167v T ) (0.250) 2 = 72.9v T
At the terminal condition, equilibrium occurs. Hence, 1100 sin 20º = 72.9v T , v T = 5.16 m/s.
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