A block of mass `m` is at rest on a rough wedge as shown in figure. What is the force exerted by the wedge on the block?

Options

A. mg
B. mg cos theta
C. mg sin theta
D. mg/2

Answer

A

Explanation

Since, the block is permanently at rest, it is in equilibrium . Net force on it should be zero. In this case only two forces are acting on the block.
(i) Weight=mg( downward)
(ii) Contact force (resultant of normal reaction and friction force) applied by the wedge on the block.
For the block to be in equilibrium these two forces should be equal and opposite. Therefore, force exerted by the wedge on the bock is mg (upward).
Note From Newton's third law of motion- force exerted by the block on the wedge is also mg but downwards. The result can also be obtained in a different manner. The normal force on the block is `N=mg cos theta` and the friction force on the block is `f= mg sin theta`( not `mu mg cos theta`) because it is not the case of limiting friction.
These two forces are mutually perpendicular.
`therefore` Net contact force would be `sqrt(N^(2)+f^(2))" or " sqrt((mg cos theta)^(2)+(mg sin theta)^(2))` which is equal to mg.