Example

Question:

A mass of \(4\,\mathrm{kg}\) rests on a horizontal plane. The plane is gradually inclined until at an angle \(\theta = 15^\circ\) with the horizontal, the mass just begins to slide.
What is the coefficient of static friction between the block and the surface?

Solution:

The forces acting on a block of mass \(m\) at rest on an inclined plane are:
(i) the weight \(mg\) acting vertically downwards,
(ii) the normal force \(N\) of the plane on the block,
(iii) the static frictional force \(f_s\) opposing impending motion.

Resolving the weight along the shown directions: \[ mg \sin\theta = f_s,\qquad mg \cos\theta = N \] As \(\theta\) increases, \(f_s\) increases until at \(\theta = \theta_{\text{max}}\), \(f_s\) achieves its maximum value, \(f_{s,\text{max}} = \mu_s N\).
Therefore, \[ \tan \theta_{\text{max}} = \mu_s \qquad \text{or} \qquad \theta_{\text{max}} = \tan^{-1} \mu_s \] For \(\theta_{\text{max}} = 15^\circ\): \[ \mu_s = \tan 15^\circ = 0.27 \] This value is independent of the mass of the block.