Maths › Mechanics › Friction and inclined planes
Friction and inclined planes
Friction is lazy: it matches whatever is asked of it until it cannot, and its ceiling is a fraction of how hard the surfaces press together. Add a slope, resolve the weight, and the whole of statics and sliding opens up.
Builds on Forces and Newton's laws.
IN THIS TOPIC
- Use F ≤ μR, with equality only in limiting equilibrium or sliding.
- Resolve forces along and perpendicular to an inclined plane.
- Decide whether an object on a rough slope slips, and find its acceleration when it does.
WHAT YOU PROBABLY THINK
Friction on a stationary object always equals μR.
How friction behaves
Push gently and friction pushes back exactly as hard: no motion. Push harder and it keeps matching, until the demand reaches μR, the coefficient of friction times the normal reaction. At that ceiling the object is in limiting equilibrium; beyond it, the object slides and friction locks at F = μR, opposing the sliding. The inequality is the law; the equality is earned only at the limit or in motion.
On a slope
Axes along and perpendicular to the slope pay for themselves instantly. The weight resolves into mg sin θ pulling down the slope and mg cos θ pressing into it, so on a plain slope R = mg cos θ, and friction's ceiling is μmg cos θ.
WORKED EXAMPLE
Does it slide, and how fast?
A 2 kg block sits on a rough 25° slope, μ = 0.3. Show it slides, and find its acceleration.
Down-slope pull: mg sin 25° = 8.28 N. Friction's maximum: 0.3 × mg cos 25° = 5.33 N.
8.28 > 5.33, so equilibrium is impossible: it slides.
Sliding: ma = 8.28 − 5.33 = 2.95 N, so a = 1.48 m s⁻² down the slope.
The check: tan 25° = 0.466 > 0.3 = μ says the same thing, since a block slips exactly when tan θ beats μ.
THE EXAM BIT
- Write F ≤ μR and say when equality holds; using F = μR for a resting object mid-question is the classic error.
- On slopes, resolve the weight into sin and cos parts and leave the axes tilted.
- “On the point of slipping” is code for limiting equilibrium: F = μR exactly.
- Friction opposes the sliding that happens or threatens: down-slope for a block pushed up, up-slope for one slipping down.
CHECK YOURSELF
A 4 kg box rests on a rough horizontal floor, μ = 0.5. A horizontal force of 15 N is applied. Show the box does not move, and state the friction force acting.
Show a hint
Compare the demand with the ceiling.
Show the answer
Ceiling: μR = 0.5 × 4 × 9.8 = 19.6 N, and the demand is 15 N < 19.6 N, so the box stays put.
Friction matches the push exactly: 15 N, not 19.6 N.
Friction matches the demand up to μR, and only sliding or the limit makes that an equality.
On a slope: mg sin θ along, mg cos θ into, and compare tan θ with μ to see if it holds.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Use F ≤ μR, with equality only in limiting equilibrium or sliding.
- Resolve forces along and perpendicular to an inclined plane.
- Decide whether an object on a rough slope slips, and find its acceleration when it does.
No animated video for this topic yet; these notes stand alone.