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Question

A cubical block rests on a plane of μ=√3. The angle through which the plane be inclined to the horizontal so that the block just slides down will be

A
30
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B
45
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C
60
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D
75
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Solution

The correct option is C 60∘I− Method The block will have force mgsinθ along the incline plane and the frictional force will have fs=μN where N=mgcosθ fs=mgsinθ For equilibrium μ mgcosθ=mgsinθ tanθ=μ=√3 θ=60∘ II− Method This is the case of angle of repose. Angle of repose is defined as the minimum angle of the inclined plane and we have the formulae as tanθ=μ by putting the value of μ=√3 we get θ=60∘

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