Maths › Mechanics › Friction and inclined planes
Friction and inclined planes
Static friction adjusts to oppose impending motion up to its limiting value, F ≤ μR, where R is the normal reaction. At limiting equilibrium, and while sliding, F = μR. Add a slope, resolve the weight along and perpendicular to it, and the same model covers both statics and sliding.
Builds on Forces and Newton's laws.
IN THIS TOPIC
- Use F ≤ μR, with equality only in limiting equilibrium or during sliding.
- Work out R properly when the applied force has a vertical component.
- 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.
COMMON MISCONCEPTION
Friction on a stationary object always equals μR.
How friction behaves
Push gently and friction acts back with equal magnitude, so nothing moves. Push harder and it continues to match, until the applied force reaches μR, the coefficient of friction times the normal reaction. At that value the object is in limiting equilibrium. Beyond it, the object slides and friction takes the value F = μR, opposing the sliding. The inequality holds in general. Equality holds only at the limiting value or while sliding.
R is the quantity most often got wrong, because candidates write R = mg out of habit. R must be found from the equation perpendicular to the surface. A rope pulling at an angle above the horizontal reduces it; a push angled downwards increases it, so a heavy suitcase is easier to pull than to push.
WORKED EXAMPLE
An angled pull on rough ground
A 20 kg crate on rough horizontal ground, μ = 0.2, is pulled by a rope at 25° above the horizontal with tension 80 N. Find the acceleration.
Perpendicular: R + 80 sin 25° = 20g, so R = 196 − 33.8 = 162 N, not 196 N.
Friction: F = μR = 0.2 × 162.2 = 32.4 N.
Along the ground: 80 cos 25° − 32.4 = 72.5 − 32.4 = 40.1 N, so a = 40.1/20 = 2.00 m s⁻².
Using R = 196 N instead would have given F = 39.2 N and a = 1.67, and every later part of the question would inherit the error.
On a slope
Axes along and perpendicular to the slope simplify the working immediately. 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 the limiting friction is μmg cos θ. Sin goes down the slope. That is the one to memorise, because the sin and cos get swapped more often than anything else in the paper.
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 component: mg sin 25° = 8.28 N. Limiting friction: 0.3 × mg cos 25° = 5.33 N.
8.28 > 5.33, so equilibrium is impossible and it slides.
Sliding: ma = 8.28 − 5.33 = 2.95 N, so a = 1.48 m s⁻² down the slope.
Faster check: tan 25° = 0.466 > 0.3 = μ gives the same conclusion, because a block on a plain slope slips exactly when tan θ exceeds μ.
Which way does friction point?
Friction opposes the motion that is happening, or the motion that would happen. A block being pushed up a slope has friction acting down the slope. The same block, left alone and slipping, has friction acting up it. Deciding this before writing the equation is a ten-second job that saves a whole page.
Watch out for the two-part question that pushes a block up a slope and then removes the force. Friction reverses at the moment the block stops, so the deceleration going up and the acceleration coming back down are different numbers, and a question that gets the same answer twice has gone wrong.
ASSESSMENT FOCUS
- Write F ≤ μR and say when equality holds. Using F = μR for a resting object mid-question is the error examiners look for first.
- Never assume R = mg. Resolve perpendicular to the surface and see what comes out.
- 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, so F = μR exactly.
- Say which way friction acts before you write the equation, and justify it in one clause.
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 applied force with the limiting friction.
Show the answer
Limiting friction: μR = 0.5 × 4 × 9.8 = 19.6 N, and the applied force is 15 N < 19.6 N, so the box does not move.
Friction equals the applied force: 15 N, not 19.6 N.
Friction matches the applied force up to μR, and only sliding or the limiting case makes that an equality.
On a slope, mg sin θ along and mg cos θ into it, and compare tan θ with μ to see whether 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.
Or read them with their worked answers on the friction and inclined planes questions page.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.
- Use F ≤ μR, with equality only in limiting equilibrium or during sliding.
- Work out R properly when the applied force has a vertical component.
- 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.
Open the full revision checklist to see every objective in the course in one place.