A block of mass `m = 4 kg` is placed over a rough inclined plane as shown in fig . The coefficient of friction between the block and the plane `mu = 0.5 ` A force `F = 10N` is applied on the block at an angle of `30^(@)`. Find the contact force between the block and the plane.
`k`

Answer

` = 13.5 sqrt(5) N`

Explanation

Drawing free - body diagram of block , we get
`N + F sin 30^(@) = mg cos 37^(@)`
`N = mg cos 37^(@) - F sin 30^(@)`
`= (4)(10) ((4)/(5)) - (10) ((1)/(2)) = 27 N `…(i)
` f_(max) = mu N = 0.5 xx 27 = 13.5 N`
` mg sin 37^(@) = (4)(10)((3)/(5)) = 24 N`
`and F cos 30^(@) = (10)(sqrt(3)/(5)) = 5 sqrt(3) N`
Now since `mg sin37^(@) gt f_(max) = F cos 30^(@)`
Therefore , block will slide down and friction will be maximum .Therefore net contact force is

`F_(c) = sqrt(N^(2) + (f_(max))^(2))`
`= sqrt((27)^(2) + (13.5)^(2)) = 13.5 sqrt(5) N`