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. 40 N
C. 120 N
D. 200 N

Answer

D

Explanation

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

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