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.
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Velocity down the slope vs time
Ölçüm göstergeleri
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 cosθ
- a = g(sinθ − μk cosθ)
- tanθ = μs (on the point of slipping)
- s = ½at²
Müfredatta
- GCSE Physics AQA, Edexcel, OCRforces, resultant force and Newton's second law
- A-Level Physics and Maths Mechanicsresolving forces on an inclined plane, friction F ≤ μR
Görev
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.
It holds up to tan⁻¹(0.40) = 21.8°. At 30°, a = 9.81 × (sin 30° − 0.30 × cos 30°) = 9.81 × (0.500 − 0.260) = 2.36 m/s², whatever the mass.
Bunu doğru dürüst öğren
Bu konuda GCSE Combined and Separate Science kursundan dersler: adım adım çözümler, ardından her seferinde yeni sayılarla sınav tarzı sorular. Dersler İngilizcedir.
- Forces, mass and weightForces, work, elasticity, moments and pressureÜcretsiz
- Resultant forces and free body diagramsForces, work, elasticity, moments and pressureÜcretsiz
- Work done and energy transferForces, work, elasticity, moments and pressurePlus
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SSS
How do I find the coefficient of friction from the acceleration?
Rearrange a = g(sin θ − μk cos θ) to μk = (g sin θ − a) / (g cos θ). A block sliding down a 30° slope at 2.5 m/s² gives μk = (9.81 × 0.5 − 2.5) / (9.81 × 0.866) = 0.28. Pick μk from a under Solve for to see each step.
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.