An object of mass 20kg slides down an inclined plane at an angle of 30° to the horizontal. The coefficient of static friction is ....[g = 10ms⁻²]

The problem describes an object of mass $m = 20 \text{ kg}$ on an inclined plane at an angle $\theta = 30^\circ$ to the horizontal. It asks for the coefficient of static friction, $\mu_s$. The phrase "slides down" combined with asking for "static friction" implies that the object is at the point where it is just about to slide or just begins to slide. In this critical condition, the static friction force reaches its maximum value. We are given $g = 10 \text{ m/s}^2$.

We resolve the gravitational force $mg$ into components parallel and perpendicular to the inclined plane.

• The component of gravity perpendicular to the plane is $mg \cos\theta$. This component is balanced by the normal force $N$.

• The component of gravity parallel to the plane is $mg \sin\theta$. This component tends to pull the object down the incline.

When the object is just about to slide, the maximum static friction force $f_{s,max}$ is equal to the component of gravity pulling the object down the incline.

The maximum static friction force is also defined as:

where $\mu_s$ is the coefficient of static friction.

Substitute $N = mg \cos\theta$ into the equation for $f_{s,max}$:

Divide both sides by $mg \cos\theta$:

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$.

The calculated value $\mu_s \approx 0.57735$ is closest to 0.6.