The coefficient of static friction between a block of mass m and an incline is `mu_s=0.3` What can be the maximum angle `theta` of the incline with the horizontal so that the block does not slip on the plane ?

Explanation

The situation is shown in ure.
a. the forces on the block are
the weight mg downwasrd by the earth
the normal contact force N by the incline, and
the friction f paralel to the incline up the plane, by the incline

As the block is ast rest, these fores should add up to zero. Also, since `theta` is the maximum angle to prevent slipping, this is acase oflimiting equilibrium and so `f=mu_sN`.
Taking components perpendicular to the incline
`N-mg costheta=0`
or, `N=mg cos theta`..............i
Taking components parallel to the incline
`f-mg sin theta=0`
or `mu_sN=mg sin theta` ..........ii ltbr. Dividing ii by i `mu_2=tan theta`
or, `theta=tan^-1mu_s=tan^-1(0.3)`
b. If the angle of incline is reduced to `theta/2` the equilibrium is not limiting and hence the force of static friction f is less than`mu_sN`. To know the value of f we proceed as in part a. get the equations
`N=mgcos(theta/2)`
and `f=mgsin(theta/2)`
Thus, the force of friction is `mg sin (theta/2)`.