"Kinetic friction formula" is searched by students in their first mechanics course who have been given f = μN and want to know which μ, which N, and what to do on a slope. The formula is short; the marks are lost on the difference between static and kinetic friction, on the normal force when the surface is tilted or something is pressing down, and on the two things friction does not depend on. This page has the formula, the two kinds of friction side by side, the coefficient table for common surfaces, the slope case, and eight worked examples of the exam kind. The printable has the table, the formulas and practice on two pages.
Download the friction formulas and coefficient table (PDF) (the formulas for static and kinetic friction, the normal force on flat ground, on a slope and with an extra load, the coefficient table for twelve surface pairs, the eight worked examples, and ten practice questions with answers; two pages, free).
The formula
Kinetic friction: f = μₖ N
- f: the friction force, in newtons, acting along the surface against the direction of sliding
- μₖ: the coefficient of kinetic friction, a number with no units, fixed for a pair of surfaces
- N: the normal force, the push of the surface on the object at right angles to the surface, in newtons
Kinetic means moving: this is the friction on an object that is sliding. On a flat surface with nothing else pressing, the normal force equals the weight, N = mg, and f = μₖmg. A 20 kg crate sliding on concrete with μₖ = 0.5: f = 0.5 × 20 × 9.8 = 98 N.
Static against kinetic
| Static friction | Kinetic friction | |
|---|---|---|
| When | the object is not sliding | the object is sliding |
| Size | whatever is needed to prevent sliding, up to a maximum: fₛ ≤ μₛN | fixed: f = μₖN |
| The coefficient | μₛ, nearly always larger than μₖ | μₖ |
| Direction | opposite to the way the object would slide | opposite to the sliding |
| Example | a parked car on a hill; a box you are pushing that has not moved yet | a sledge on snow; a skidding tyre; a box sliding across a floor |
Static friction is a response: push a box with 50 N and static friction pushes back with 50 N, push with 100 N and it pushes back with 100 N, until the push reaches the maximum μₛN; then the box slides and the friction drops to the kinetic value μₖN, which is why the box lurches forward when it starts. The sentence to write in an exam: "static friction is less than or equal to μₛN; it equals μₛN only on the point of sliding."
Two consequences: a box is harder to start than to keep moving, and a car stops in a shorter distance when its tyres roll (static friction with the road, up to μₛN) than when they skid (kinetic, μₖN), which is the whole point of anti-lock brakes.
The coefficient of friction table
Approximate textbook values; real surfaces vary with cleanliness, wetness and wear.
| Surfaces | μₛ (static) | μₖ (kinetic) |
|---|---|---|
| Rubber on dry concrete | 1.0 | 0.8 |
| Rubber on wet concrete | 0.7 | 0.5 |
| Glass on glass | 0.94 | 0.4 |
| Steel on steel (dry) | 0.74 | 0.57 |
| Aluminium on steel | 0.61 | 0.47 |
| Copper on steel | 0.53 | 0.36 |
| Wood on wood | 0.25 to 0.5 | 0.2 |
| Wood on concrete | 0.6 | 0.5 |
| Steel on steel (lubricated) | 0.15 | 0.06 |
| Waxed wood (ski) on wet snow | 0.14 | 0.1 |
| Waxed wood on dry snow | 0.04 | |
| Ice on ice | 0.1 | 0.03 |
| Teflon on Teflon | 0.04 | 0.04 |
| Human joint (synovial fluid) | 0.01 | 0.003 |
What the coefficient does and does not depend on:
- It depends on both surfaces together (rubber on concrete is a pair; "the coefficient of rubber" means nothing).
- It does not depend on the area of contact: a brick on its end and on its face feel the same friction, because the smaller area carries more pressure.
- Kinetic friction does not depend much on speed (in the school model it is constant).
- It is measured, by pulling with a force meter or tilting until sliding starts; it is not calculated from the materials.
The normal force: three cases
The normal force is the surface's push on the object, perpendicular to the surface. It is the weight only in the simplest case.
| Situation | Normal force N | Kinetic friction |
|---|---|---|
| Flat surface, nothing else vertical | mg | μₖmg |
| Flat surface, an extra downward push P (someone leaning on the box) | mg + P | μₖ(mg + P) |
| Flat surface, an upward pull or lift L (a rope at an angle) | mg − L | μₖ(mg − L) |
| A slope of angle θ | mg cos θ | μₖmg cos θ |
On a slope, gravity splits into mg sin θ down the slope and mg cos θ into it; the normal force balances the second, so friction uses mg cos θ. The down-slope pull mg sin θ grows with the angle and the maximum static friction μₛmg cos θ shrinks, and sliding begins where they are equal: tan θ = μₛ. That is a standard way to measure μₛ: tilt the surface until the object slips and take the tangent of the angle. For μₛ = 0.5, θ = 26.6°; for rubber on concrete at 1.0, 45°.
Worked examples
| The question | The working | The answer |
|---|---|---|
| A 10 kg box slides on a floor, μₖ = 0.3. Friction? | f = 0.3 × 10 × 9.8 | 29.4 N |
| Push needed to keep the same box sliding at constant speed | equal to the friction | 29.4 N |
| Push needed to start it moving, μₛ = 0.4 | μₛmg = 0.4 × 98 | 39.2 N |
| The box is pushed with 60 N while sliding: acceleration? | net = 60 − 29.4 = 30.6 N; a = F/m | 3.06 m/s² |
| A 1,200 kg car skids on dry road, μₖ = 0.8: deceleration and stopping distance from 20 m/s | f = 0.8 × 1,200 × 9.8 = 9,408 N; a = 7.84 m/s²; d = v²/2a | 25.5 m |
| The same car skids on a wet road, μₖ = 0.5 | a = 4.9 m/s²; d = 400/9.8 | 40.8 m |
| A 5 kg block on a 30° slope, μₖ = 0.2: friction, and does it accelerate? | N = 5 × 9.8 × cos 30 = 42.4 N; f = 8.5 N; down-slope pull = 5 × 9.8 × sin 30 = 24.5 N | net 16 N down the slope; a = 3.2 m/s² |
| At what angle does a book start to slide off a desk if μₛ = 0.35? | tan θ = 0.35 | 19.3° |
The pattern for every friction problem: draw the forces (weight, normal, friction, any push or pull), find N from the forces perpendicular to the surface, then friction from μN, then the net force along the surface, then Newton's second law.
What friction is, physically
The rough picture: two surfaces touch only at the tips of their bumps, those tips weld and tear as the surfaces slide, and the force needed to keep tearing them is friction. Pressing harder makes more tips touch (which is why friction is proportional to the normal force), and a bigger area spreads the same load over more tips, each touching less (which is why area cancels out). Lubricants work by keeping the tips apart with a film of liquid; polished surfaces can show more friction, not less, because flat faces have more atoms close enough to bond.
Where it sits in school
Friction as a force appears in grade 6 to 8 science; the formula f = μN and the static-kinetic distinction are grade 11 and 12 physics in the United States (Physics 1, AP Physics 1 unit 2), GCSE (qualitatively) and A level (Mechanics) in England, Year 11 and 12 Physics in Australia, and CBSE Class 11 Physics (Laws of Motion, including the slope and tan θ = μₛ). The common errors: using mg for N on a slope, using μₛ for an object that is already sliding, and forgetting that static friction is "up to" μₛN rather than always equal to it.
Practice: ten questions
- A 15 kg box slides on a surface with μₖ = 0.25. What is the friction?
- What horizontal push keeps it moving at constant speed?
- If μₛ = 0.35, what push starts it moving?
- The box is pushed with 80 N while sliding. Find its acceleration.
- A 900 kg car skids with μₖ = 0.7 from 25 m/s. Stopping distance?
- A sledge of 30 kg (with rider) slides on snow with μₖ = 0.05. Friction, and the pull needed for constant speed?
- A 4 kg block rests on a slope of 20°. If μₛ = 0.5, does it slide? Show the two forces.
- The same block on a 35° slope with μₖ = 0.3 is sliding. Find its acceleration.
- A person leans on a 10 kg box with an extra 50 N downward while it slides, μₖ = 0.3. Friction?
- A book slides off a tilted desk at 25°. What is μₛ for book on desk?
Answers: 1) 0.25 × 15 × 9.8 = 36.75 N. 2) 36.75 N. 3) 0.35 × 147 = 51.45 N. 4) (80 − 36.75) ÷ 15 = 2.88 m/s². 5) a = 0.7 × 9.8 = 6.86 m/s²; d = 625 ÷ 13.72 = 45.6 m. 6) 0.05 × 30 × 9.8 = 14.7 N; pull 14.7 N. 7) Down-slope pull 4 × 9.8 × sin 20 = 13.4 N; max static friction 0.5 × 4 × 9.8 × cos 20 = 18.4 N; no, it stays. 8) Pull 39.2 × sin 35 = 22.5 N; friction 0.3 × 39.2 × cos 35 = 9.6 N; net 12.9 N; a = 3.2 m/s². 9) N = 98 + 50 = 148 N; f = 44.4 N. 10) μₛ = tan 25° = 0.47.
Download the friction formulas and coefficient table (PDF)
Where to get help
A 1-on-1 online physics class with a tutor who holds a master's degree in science, for grade 11 and 12 physics, AP Physics 1, GCSE and A level, the Australian years and CBSE Class 11: force diagrams drawn on the shared screen for every problem, the normal force found first, then friction, then Newton's second law, until the method is automatic. The recording and notes after every class, the same tutor every week, US$10 to $13 an hour. The physics page has the courses; the free 30-minute first class is a real lesson, usually on the homework the student brings.
Questions parents ask
1What is the formula for kinetic friction?
2What is the difference between static and kinetic friction?
3What is the coefficient of friction?
4Does friction depend on the area of contact?
5How do you find friction on a slope?
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