Physics · Laws Of Motion · NEET
It just starts to slide when the incline angle θ equals the angle of repose (call it α). At this angle the down-the-slope pull mg sinθ exactly equals the maximum static friction μs mg cosθ. Setting them equal: mg sinθ = μs mg cosθ, so tanθ = μs. Therefore the just-slides angle is θ = tan⁻¹(μs). Below this angle friction holds it still; at this angle it is on the verge of sliding.
On a rough incline the weight mg splits into two parts: mg sinθ along the slope (tries to slide the body down) and mg cosθ into the slope. The normal reaction is N = mg cosθ, so maximum static friction = μs N = μs mg cosθ. 'Just slides' is the exact moment these balance: mg sinθ = μs mg cosθ. The mg cancels, leaving tanθ = μs. That is why the tangent of the just-slides angle equals the friction coefficient.
They are numerically equal but defined differently. Angle of repose (α) is the maximum incline angle at which a body stays at rest, given by tan α = μs. Angle of friction (λ) is the angle between the normal reaction and the resultant of normal reaction plus limiting friction, also given by tan λ = μs. Since both equal tan⁻¹(μs), angle of repose = angle of friction. For NEET, just remember both give tanθ = μs.
No. In tanθ = μs there is no mass term, because mg cancels from both sides of mg sinθ = μs mg cosθ. A heavy block and a light block of the same material slide at the same critical angle. Mass only affects the actual forces (heavier means more sliding force AND more friction, in the same ratio), not the angle. This is a common NEET trap answer.
If θ is larger than tan⁻¹(μs), then mg sinθ becomes greater than the maximum friction μs mg cosθ, so the body cannot stay at rest and accelerates down. Once moving, kinetic friction acts, and the acceleration is a = g(sinθ − μk cosθ), using the kinetic coefficient μk. If θ is smaller than the angle of repose the body stays put with static friction adjusting to exactly balance mg sinθ.
Two inclined surfaces of equal length L have the same inclination 45° with the horizontal; one is rough and the other perfectly smooth. A body takes 2 times as much time to slide down the rough surface as the smooth one. The coefficient of kinetic friction (μk) between the body and the rough surface is close to:
A block of mass 2 kg is placed on a rough inclined surface AC of coefficient of friction μ. It remains at rest. Taking g = 10 m/s², the net force on the block is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
tanθ = μs, so θ = tan⁻¹(μs), where μs is the coefficient of static friction. This angle is called the angle of repose.
Yes, numerically. Both are given by tan⁻¹(μs), so angle of repose equals angle of friction for the same surfaces.
Once it moves, a = g(sinθ − μk cosθ), where μk is the coefficient of kinetic friction. If the incline is smooth, a = g sinθ.
Both the sliding force (mg sinθ) and the maximum friction (μs mg cosθ) contain mg. Dividing one by the other removes mg, leaving tanθ = μs, so mass does not matter.
NEET often gives an incline with friction and asks for the coefficient from a time ratio (2025), or whether the net force is zero when a block rests on an incline (2023). Knowing tanθ = μs and a = g(sinθ − μk cosθ) solves most of them.
