Practice portal › Laws of Motion › Friction: Blocks, Belts and Inclines
Asked in CBSE PMT 1993 · Friction on an incline
Given: the block slides down at constant speed, so it is in equilibrium.
Idea: constant speed means zero acceleration, so friction up the slope exactly balances the weight component down it.
Down the slope: mg sin θ.
Up the slope: f=μₖN=μₖmg cos θ.
Setting them equal: mg sin θ=μₖmg cos θ.
The mg cancels: μₖ=(sin θ)/(cos θ).
μₖ=tan θ. This angle is the angle of repose.
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