Question

A 5.0 kg wooden block is placed at an adjustable inclined plane. what is the angle of incline above which the block will start to slide down the plane ?

Answer 1

The value of the angle of the incline $\theta$ at which the block starts to slide is the angle at which the component of the weight parallel to the incline becomes equal to the frictional force that keeps the block on the incline:
$mg \sin \theta = \mu N$
where the term on the left is the component of the weight parallel to the incline, and the term on the right is the frictional force, which is the product between the coefficient of friction $\mu$ and the normal reaction of the incline N.

The normal reaction of the incline, N, is equal to the component of the weight perpendicular to the incline:
$N=mg \cos \theta$
Therefore, the initial equation becomes
$mg \sin \theta = \mu mg \cos \theta$
From which we find
$\tan \theta = \mu$
$\theta = \arctan \mu$

For angles above this value, the block will start sliding down, otherwise the block will stay on the incline.