In the odjoining figure, angle of plane `theta` is increased from `0^(@)`to `90^(@)`. Plot force of friction `f` veraus `theta` graph

Answer

A

Explanation

Normal reation `N =mg cos theta`. Limiting value of static friction.
`f_(L) =mu_(s)N =mu_(s) mg cos theta`
Constant value of kinetic friction
`f_(K) =mu_(k)N =mu_(k) mg cos theta`
Driving force down the plane
`F =mg sin theta`
Now block remain stationary and `f=F` until `F` becomes equal to `f_(L)`
or `mg sin theta = mu_(s) mg cos theta`
or `tan theta =mu_(s)` or `theta =tan^(-1)(mu_(s))=theta`, (say)
After this block starts moving and constant value of friction will act. Thus, For `thetaletan^(-1) (mu_(s)) or theta_(r)`
`f prop sin theta`
At `theta =0^(@), f =0` and at theta =tan^(-1)(mu_(s)) or theta_(r)`
`f = mg sin theta` or `mu_(s) mg cos theta`
For `thetagttan^(-1) (mu_(s))` or `theta_(r)`
`f =f_(k) =mu_(k)mg cos theta` or `f prop cos theta`
At `theta=tan^(-1) (mu_(s))` or `theta_(r)`
`f =mu_(k)mg cos theta` and at `theta =90^(@)`
`f =0` Corresponding `f` versus `theta` graph is as shown in figure
In the figure `OP` is sine graph and `MN` is `cos` graph.
`f_(1) = mg sin theta_(r) =mu_(s) mg cos theta_(r)`
`f_(2) =mu_(k) mg cos theta_(r)`