Question

a block of mass M lies on a rough surface of coefficient of friction mew a force is applied on it at an angle theta to the horizontal on one of the top edge the block remains at rest the frictional force acting on the block will be

Answer 1

answer : $\textbf{frictional force}=\frac{\mu Mgsin\theta}{cos\theta-\mu sin\theta}$

explanation : see figure, A block of mass M lies on a rough surface of coefficient of friction μ, a force F is applied on it at an angle θ to the horizontal on one of the top edge the block remains at rest.

at equilibrium,

friction force = horizontal component of force

or, μ N = Fcosθ .......(1)

where N is normal reaction acting on block.

at equilibrium,

vertical upward force = vertical downward force

or, N = Mg + Fsinθ, put it in equation (1),

μ ( Mg + Fsinθ ) = Fcosθ

or, μMg + μFsinθ = Fcosθ

or, μMg = Fcosθ - μFsinθ

or, F = μMg/(cosθ - μsinθ)

so, frictional force = Fsinθ

= μMg/(cosθ - μsinθ) × sinθ

= μMgsinθ/(cosθ - μsinθ)