A block of mass `m = 2kg` is resting on a inclined plane of inclination `30^(@)` as shown in figure 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?

Answer

8

Explanation

Since `mg sin 37^(@) gt mu mg cos37^(@)` the block has a tendency to slip downward
Let F be the minimum force applied on it , so that it does not slip Then
`N = T + mg cos37^(@)`
`mg sin 37^(@)= mu N = mu (F - mg cos 37^(@))`
`or F = (mg sin 37^(@))/(mu) - mg cos 37^(@)`
`= ((2)(10)(3//5))/(0.5) - (2)(10) ((4)/(5)) = 8 N`