A block of mass `m = 2kg` is resting on a inclined plane of inclination `30^(@)` The coefficient of friction between the block and the plane is `mu = 0.5` what minimum force F (in newton) should be applied perpendicular to the plane on the block so that the does not slip on the plane?

Explanation

Since mg `sin 37^(@) gt mu g cos 37^(@)`, the block has a tendency to slip downwards.
Let F be the minimum force applied on it, so that it does not slip. Then,
`N = F + mg cos 37^(@)` `:. Mg sin 37^(@) = mu N = mu (F + mg cos 37^(@))`
or `F = (mg sin 37^(@))/(mu) - m g cos 37^(@) = ((2) (10) (3//5))/(0.5) - (2) (10) ((4)/(5)) = 8N`