A block of mass m is placed on a rough inclined plane of inclination $θ$ and co – efficient of friction $μ .$ The block remains stationary due to friction. A force $\vec{F}$ is applied on the block to slide the block. Mark the correct option

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If $\vec{F} = 0,$ the direction of contact force applied by the inclined surface on the block is vertical

If $\vec{F} \neq 0$ and still block remains stationary, the magnitude of frictional force on the block may be greater than $μ m g \cos θ$

If block is accelerated up the incline plane due to action of force $\vec{F},$ the contact force by the inclined plane on the block may be directed upwards

If $\vec{F} = 0,$ the frictional force acting on the block is $μ m g \cos θ$

answer is A, C.

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Detailed Solution

If $\vec{F} = 0,$ the weight of the block should be balanced by the contact force (resultant of frictional force and the normal force). Thus, in this case the contact force should be vertically upwards. Under this situation, the frictional force may have any value between zero and limiting friction.
If $\vec{F} \neq 0,$ such that $\vec{F}$ acts to increase normal force by the inclined plane on the block, then the frictional force acting may be greater than $μ m g \cos θ$
If the block is accelerated up the incline plane, the frictional force will act on the block along down the incline plane. The resultant of normal and the frictional force cannot be in vertical direction in that case.