Question

Question #31590

Answer 1

Forces acting on the body are shown in the diagram above

Sum of the forces along the inclined plane:
`F_"net" = mg sin theta -f`
where `mg` is the weight of the body and `f` is force of friction.

There is no acceleration along the inclined plane, `F_"net" = 0`, therefore
`mg sin theta =f` ......(1)

Similarly in a direction perpendicular to the inclined plane:
`F_"net" = N -mg cos theta`
where `N` is normal reaction.

As per given condition above there is no acceleration in this direction either, `F_"net" = 0`, and therefore
`N =mg cos theta` ......(2)

If `mu_k` is coefficient of kinetic friction, then
`f = µ_kN`

Using (1) and (2) we get
`mg sin theta=mu_kxxmg cos theta`

Solving for `mu_k`
`mu_k=tan theta`
`mu_k=tan 30^@=1/sqrt3`