JAMB Physics 2012 Paper

Concept:When an object is just about to slide down an inclined plane, the maximum static friction balances the component of its weight acting down the plane.This gives the condition $\mu = \tan\theta$.Explanation:The weight $mg$ acts vertically downward.Resolve it parallel and perpendicular to the plane.Component down the plane: $mg\sin\theta$.Component normal to the plane: $mg\cos\theta$.Therefore, the normal reaction is $N = mg\cos\theta$.At the limiting point of static friction, the friction force equals the downward component:$$\mu N = mg\sin\theta$$Substitute $N = mg\cos\theta$:$$\mu (mg\cos\theta) = mg\sin\theta$$Cancel $mg$ from both sides:$$\mu = \frac{\sin\theta}{\cos\theta} = \tan\theta$$The mass $20\text{ kg}$ and $g = 10\text{ ms}^{-2}$ cancel out, so they are not needed.Given $\theta = 30^\circ$:$$\mu = \tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577$$The nearest option to $0.577$ is $0.6$.Answer:D. 0.6