In the figure shown, the minimum force `F` to be applied perpendicular to the incline so that the block does not slide is

Options

A. `0`
B. `40N`
C. `120N`
D. `200N`

Answer

D

Explanation

In the …………
For block to be in equilibrium :
`F = N+mg cos 37^(@)`
`f = mg sin 37^(@)`
`FBD` :

For `F` to be minimum,
For no sliding `f le mu N`
`implies mg sin 37^(@) le (1)/(2) [F-mg cos 37^(@)]`
`implies 2mg sin 37^(@)+mg cos 37^(@) le F`
`implies F ge 2mg xx (3)/(5)+mgxx(4)/(5) implies F ge 2mg i.e. F ge 200 N`