Physics · Laws Of Motion · NEET
They are defined differently but have the same value. The angle of friction λ is defined from the contact force: tan λ = (limiting friction)/(normal reaction) = μs N / N = μs. The angle of repose θ is defined from the incline: a body just about to slide down needs tan θ = μs. Since both give tan(angle) = μs, we get λ = θ = tan⁻¹(μs). So numerically they are equal, but one comes from the force triangle and the other from the incline geometry.
Put a body on a rough incline of angle θ and increase θ until it is just about to slide. Take axes along and perpendicular to the incline. Perpendicular: N = mg cos θ. Along the incline (just about to slide, so friction is limiting and acts up): mg sin θ = μs N = μs mg cos θ. Cancel mg: sin θ = μs cos θ, which gives tan θ = μs. So the angle of repose θ = tan⁻¹(μs).
When a body is on the verge of sliding on a horizontal surface, the normal reaction N and the limiting friction (μs N) add to give one resultant contact force. The angle λ this resultant makes with N obeys tan λ = μs N / N = μs. So the angle of friction λ = tan⁻¹(μs). This is a pure ratio of two forces, so it has no units.
No. In the derivation mg appears on both sides and cancels: tan θ = μs. The angle of repose depends only on the coefficient of static friction between the two surfaces, not on the mass, weight, or size of the body. A heavy box and a light box of the same material start to slide at the same tilt angle.
Both angles describe the moment the body is 'just about to move' but has not yet started moving. At that instant the static friction has reached its maximum (limiting) value μs N. That is why both formulas use μs, the coefficient of static friction. Once the body is actually sliding, kinetic friction μk takes over, but that is a different situation.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
tan λ = μs, so the angle of friction λ = tan⁻¹(μs). A larger coefficient of friction means a larger angle of friction.
Yes. If μs is greater than 1, then tan θ = μs > 1, so θ > 45°. For most everyday surfaces μs is less than 1, so the angle of repose is usually below 45°.
The body stays at rest. The needed friction mg sin θ is less than the maximum available μs mg cos θ, so static friction holds it. The body slides only when the incline angle exceeds the angle of repose.
They are defined separately. The static angle of friction uses μs (tan⁻¹μs) and the kinetic angle of friction uses μk (tan⁻¹μk). Since μk is less than μs, the kinetic angle of friction is smaller. For the angle of repose, use μs.
