An object of mass 20kg slides down an inclined plane at an angle of 30o to the horizontal. The coefficient of static friction is? [g = 10$ms^{-2}$]

Analyze the question

A 20 kg object slides down a plane inclined at 30° to the horizontal. We are asked for the coefficient of static friction. The question is incomplete as worded, because it doesn't say that the object is just about to slide (limiting equilibrium) or that it moves at constant speed. The standard JAMB interpretation is that the object is on the verge of sliding, so the friction is at its limit.

Concept

At the point where sliding just begins:

• Component of weight down the plane = mg sin θ
• Limiting friction up the plane = μR = μ mg cos θ

Setting them equal:

μ mg cos θ = mg sin θ μ = tan θ

Solution

μ = tan 30° = 1/√3 ≈ 0.577

The mass (20 kg) and g cancel out, so they aren't needed.

Matching to the options

0.577 is closest to 0.6, which is option D. Option C (0.5) would correspond to an angle of about 26.6°, so it isn't right for 30°.

Tip: For any "just about to slide" or "constant velocity down a slope" question, use μ = tan θ immediately. Remember the common values: tan 30° ≈ 0.58, tan 45° = 1, tan 60° ≈ 1.73. Mass cancels out, so ignore it as a distractor.