A block of mass m = 4 kg is placed over a rough inclined plane as shown in figure. The coefficient of friction between the block and the plane is u=0.5. A force F= 10 N is applied on the block at an angle of 30°. The friction force between the block and wedge is

Options

A. static in nature in the direction up the plane and have the value 30.2 N
B. static in nature in the direction down the plane and have the value 30.2 N
C. kinetic in nature in the direction up the plane and have the value 13.5 N
D. None of these

Answer

c

Explanation

Drawing free body diagram of block, `sumF_(y)=0`

`impliesN+F sin30^(@)=mgcos37^(@)`
`or,N=mgcos37^(@)-Fsin30^(@)`
`(4)(10)(4/5)-(10)(1/2)`
`or, N=27 N`
`f_(max)=muN=0.5 xx 27 =13.5 N`
`mg sin 3e7^(@)=(4)(10)(4/5)=32 N`
`and F cos 30^(@)=(10)((sqrt3)/2)=8.66 N`
Now since mg `sin 37^(@)>f_(max)+Fcos30_(@)`
Therefore block will slide down and friction will be kinetic.