How do you resolve weight on a slope?

Split mg into two perpendicular parts: mg sinθ acting down the slope and mg cosθ acting into the slope, where θ is the angle of the slope to the horizontal. At θ = 0 the whole weight presses into the ground; at 90° it all acts along the surface.

Why does mass not change the acceleration on a slope?

Every force along the slope is proportional to the mass: mg sinθ and the friction μk mg cosθ. Dividing the net force by m cancels it, leaving a = g(sinθ − μk cosθ).

What is the angle of repose?

The steepest angle at which an object can rest on a slope without sliding. At that angle mg sinθ equals the maximum static friction μs mg cosθ, so tanθ = μs. For μs = 0.40 it is 21.8°.

What is the difference between static and kinetic friction?

Static friction acts on a still object and is only as big as it needs to be, up to μs N. Kinetic friction acts on a sliding object and has the fixed size μk N. Usually μk is a little smaller than μs, which is why it is harder to start something sliding than to keep it going.

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Forces on a slope

Tilt the slope and the block's weight splits into a part pulling it down the slope and a part pressing it into the surface. Friction decides whether it moves.

Velocity down the slope vs time

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What's happening

Weight mg acts straight down, but on a slope it is easier to split it into two components: mg sinθ along the slope and mg cosθ into the surface. The surface pushes back with a normal force N = mg cosθ, and friction acts along the surface, against the motion. While the block is still, static friction matches whatever is needed to hold it, up to a limit of μs N. So the block stays put as long as mg sinθ ≤ μs mg cosθ, which means tanθ ≤ μs: the steepest angle that holds is the angle of repose. Once it slides, kinetic friction μk N takes over and the net force down the slope is mg sinθ − μk mg cosθ. The mass cancels, so the acceleration is a = g(sinθ − μk cosθ) for any block, and the distance from rest grows as ½at².

N = mg cosθa = g(sinθ − μk cosθ)tanθ = μs (on the point of slipping)s = ½at²

GCSE Physics (AQA, Edexcel, OCR): forces, resultant force and Newton's second law. A-Level Physics and Maths Mechanics: resolving forces on an inclined plane, friction F ≤ μR.

Work through the numbers with SUVAT Solver and Physics Formulas.

Challenge

Predict first: with μs = 0.40, what is the steepest angle at which the block stays still? Then, at 30° with μk = 0.30, what is its acceleration once it slides? Type it in, check, then press Start.

FAQ

How do you resolve weight on a slope?: Split mg into two perpendicular parts: mg sinθ acting down the slope and mg cosθ acting into the slope, where θ is the angle of the slope to the horizontal. At θ = 0 the whole weight presses into the ground; at 90° it all acts along the surface.
Why does mass not change the acceleration on a slope?: Every force along the slope is proportional to the mass: mg sinθ and the friction μk mg cosθ. Dividing the net force by m cancels it, leaving a = g(sinθ − μk cosθ).
What is the angle of repose?: The steepest angle at which an object can rest on a slope without sliding. At that angle mg sinθ equals the maximum static friction μs mg cosθ, so tanθ = μs. For μs = 0.40 it is 21.8°.
What is the difference between static and kinetic friction?: Static friction acts on a still object and is only as big as it needs to be, up to μs N. Kinetic friction acts on a sliding object and has the fixed size μk N. Usually μk is a little smaller than μs, which is why it is harder to start something sliding than to keep it going.