The coefficient of static friction between a block of mass m and an incline is `mu_s=0.3`, a. What can be the maximum angle `theta` of the incline with the horizontal so that the block does not slip on the plane? b. If the incline makes an angle `theta/2` with the horizontal, find the frictional force on the block.

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)`.