A cubical block of side a and density ρ slides down an inclined plane of angle θ with constant speed v. A uniform thin Newtonian liquid film of thickness d and viscosity η separates the block from the plane. Assuming a linear velocity gradient in the film and neglecting all other resistive forces, what is d?
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Question

A cubical block of side a and density ρ slides down an inclined plane of angle θ with constant speed v. A uniform thin Newtonian liquid film of thickness d and viscosity η separates the block from the plane. Assuming a linear velocity gradient in the film and neglecting all other resistive forces, what is d?

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G1 Physics · Accepted Answer

The block mass is m = ρa³. The component of its weight down the plane is F(down) = mg sinθ = ρa³g sinθ. The contact area is A = a². For a liquid film of thickness d, the viscous resistance is F(viscous) = ηAv/d = ηa²v/d. The speed is constant, so the net force along the plane is zero: ρa³g sinθ = ηa²v/d. Cancel a²: ρag sinθ = ηv/d. Therefore, d = ηv/[ρag sinθ]. Correct option: A.