A block of mass m placed on a rough inclined plane of inclination `theta = 30^@` can just be prevented from sliding down by applying a force Fi up the plane and it can be just made to slide up the plane by applying a force `F_2` up the plane. If the coefficient of friction between the block and the inclined plane is `(1)/( 2 sqrt(2sqrt(3))` the relation between `F_1`and `F_2` is

Options

A. `F_2 =4F_1`
B. `f_2 - 3F_1`
C. `F_2-2F_1`
D. `F_2 -F_1`

Answer

B

Explanation

The force `F_1` requied to prevent to block from sliding down is
` F_1 = mg sin theta - mu mg cos theta `
The force ` F_2` requied to make the block move up the plane is
` F_2 = mg sin theta + mu mg cos theta`
adding (i) and (ii) , we get
` F_2 -F_1 = 2 mu mg cos theta`
` therefore (F_2+F_1)/(F_2 -F_1) = ( 2 mg sin theta)/( 2 mu mg cos theta) = ( tan theta)/( mu) = ( tan 30^@)/(1 // 2 sqrt(3)) =2`
` or F_2 + F_1 = 2F_2 - 2F_1 `
` or 3F_1 = F_2 or F_2 =3F_1`