# What is the friction force?

Friction calculator: the friction force f = μN, static or kinetic, from a coefficient and the normal force (or a mass on level ground or a slope), or the coefficient μ = f ÷ N.

- Page: https://www.acalculator.org/physics/friction-calculator
- JSON spec: https://www.acalculator.org/physics/friction-calculator.json
- Version: 2ef8f021f803

## Default answer

Example with the default inputs (Find Friction force, Friction Kinetic (sliding), Surfaces Wood on wood, Normal force from Mass, level, Mass (m) 44.0924524369755 lb, Gravity (g, m/s²) 9.80665): f = 58.8399 N (13.2277 lbf), with a normal force of 196.133 N.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| find | Find | Work out the friction force, or the coefficient of friction from a measured force. |
| kind | Friction | Kinetic friction acts on a sliding object; static friction holds a still one, up to μ_s N. |
| surf | Surfaces | The two surfaces in contact, which set μ (OpenStax Table 6.1). |
| mu | Coefficient of friction (μ) | The coefficient for the two surfaces, a pure number. |
| f | Friction force (f) | The measured friction force on the sliding object, or the largest static force just before it slips. |
| by | Normal force from | Type the normal force, or work it out from a mass on level ground or on a slope. |
| N | Normal force (N) | The force pressing the surfaces together, square to them. |
| m | Mass (m) | The mass of the object resting on the surface. |
| th | Slope angle (θ, degrees) | The angle of the slope above level, from 0° to 90°. |
| g | Gravity (g, m/s²) | The acceleration of gravity: 9.80665 m/s² is standard on Earth. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| result | Answer | The friction force (N and lbf) or the coefficient, to 6 significant figures. |
| friction | Friction force (N) | f = μ × N, in newtons: the kinetic force, or the largest static force. |
| lbf | Friction force (lbf) | f in pounds-force. |
| mu | Coefficient of friction (μ) | μ = f ÷ N, a pure number. |
| normal | Normal force (N) | N, in newtons: m × g on level ground, m × g × cos θ on a slope. |
| angle | Slope where it starts to slide (°) | arctan μ: with static friction, the steepest slope that holds; with kinetic, the slope of a steady slide. |
| along | Weight along the slope (N) | m × g × sin θ: the pull of gravity down the slope. |
| slides | Does it slide? | On a slope with static friction: whether μ_s N can hold the weight along the slope. |
| accel | Acceleration down the slope (m/s²) | a = g × (sin θ − μ_k cos θ) for an object sliding down; negative means it slows. |

## Method

f = μ × N (kinetic; static friction up to μ_s × N); N = m × g, or m × g × cos θ on a slope; μ = f ÷ N; on a slope it holds while μ_s ≥ tan θ, and slides with a = g(sin θ − μ_k cos θ).

## Assumptions

- Friction is μ times the normal force and does not depend on the contact area or (for kinetic friction) the speed; the coefficients of Table 6.1 are approximate.
- On a slope, the object rests on the plane with no other forces pressing it in or pulling it off.

## Worked examples

1. find = force, kind = static, surf = custom, mu = 0.7, by = mass, m = 20, g = 9.8 gives N = 196, f = 137.2, result = f = 137.2 N (30.8438 lbf). Source: OpenStax, University Physics Volume 1, §6.2 Friction (Equations 6.1 and 6.2; Table 6.1; Example 6.10: a 20.0 kg crate with μ_s = 0.700 and μ_k = 0.600 has N = 196 N, 137 N of static and 118 N of kinetic friction; Example 6.11: 45.0 N on a 62 kg skier on a 25° slope is μ_k = 0.082; Example 6.14: a = g(sin θ − μ_k cos θ) = 0.29 m/s² at 13° and μ_k = 0.20), https://openstax.org/books/university-physics-volume-1/pages/6-2-friction (retrieved 2026-10-19).
2. find = force, kind = kinetic, surf = custom, mu = 0.6, by = force, N = 196 gives f = 117.6. Source: OpenStax, University Physics Volume 1, §6.2 Friction (Equations 6.1 and 6.2; Table 6.1; Example 6.10: a 20.0 kg crate with μ_s = 0.700 and μ_k = 0.600 has N = 196 N, 137 N of static and 118 N of kinetic friction; Example 6.11: 45.0 N on a 62 kg skier on a 25° slope is μ_k = 0.082; Example 6.14: a = g(sin θ − μ_k cos θ) = 0.29 m/s² at 13° and μ_k = 0.20), https://openstax.org/books/university-physics-volume-1/pages/6-2-friction (retrieved 2026-10-19).
3. find = coefficient, kind = kinetic, f = 45, by = incline, m = 62, th = 25, g = 9.8 gives mu = 0.081718, result = μ = 0.0817182. Source: OpenStax, University Physics Volume 1, §6.2 Friction (Equations 6.1 and 6.2; Table 6.1; Example 6.10: a 20.0 kg crate with μ_s = 0.700 and μ_k = 0.600 has N = 196 N, 137 N of static and 118 N of kinetic friction; Example 6.11: 45.0 N on a 62 kg skier on a 25° slope is μ_k = 0.082; Example 6.14: a = g(sin θ − μ_k cos θ) = 0.29 m/s² at 13° and μ_k = 0.20), https://openstax.org/books/university-physics-volume-1/pages/6-2-friction (retrieved 2026-10-19).
4. find = force, kind = kinetic, surf = custom, mu = 0.2, by = incline, m = 60, th = 13, g = 9.8 gives accel = 0.294755. Source: OpenStax, University Physics Volume 1, §6.2 Friction (Equations 6.1 and 6.2; Table 6.1; Example 6.10: a 20.0 kg crate with μ_s = 0.700 and μ_k = 0.600 has N = 196 N, 137 N of static and 118 N of kinetic friction; Example 6.11: 45.0 N on a 62 kg skier on a 25° slope is μ_k = 0.082; Example 6.14: a = g(sin θ − μ_k cos θ) = 0.29 m/s² at 13° and μ_k = 0.20), https://openstax.org/books/university-physics-volume-1/pages/6-2-friction (retrieved 2026-10-19).
5. find = force, kind = static, surf = woodWood, by = incline, m = 10, th = 30, g = 9.80665 gives th = 26.565051, slides = Yes: it slides, along = 49.03325. Source: OpenStax, University Physics Volume 1, §6.2 Friction (Equations 6.1 and 6.2; Table 6.1; Example 6.10: a 20.0 kg crate with μ_s = 0.700 and μ_k = 0.600 has N = 196 N, 137 N of static and 118 N of kinetic friction; Example 6.11: 45.0 N on a 62 kg skier on a 25° slope is μ_k = 0.082; Example 6.14: a = g(sin θ − μ_k cos θ) = 0.29 m/s² at 13° and μ_k = 0.20), https://openstax.org/books/university-physics-volume-1/pages/6-2-friction (retrieved 2026-10-19), Table 6.1 (wood on wood: μ_s = 0.5).

## FAQ

### How do you calculate the friction force?

Multiply the coefficient of friction by the normal force: f = μN. A 20.0 kg crate on a level floor presses down with N = 20 × 9.8 = 196 N; with μ_k = 0.600 the sliding friction is 0.6 × 196 = 117.6 N.

### What is the difference between static and kinetic friction?

Static friction holds a still object and grows with the push, up to μ_s N; past that the object slips. Kinetic friction acts on a sliding object and is μ_k N. μ_s is usually larger, so it takes more force to start a slide than to keep it going: the crate above needs up to 137.2 N to start (μ_s = 0.700).

### How do you find the coefficient of friction?

Divide the friction force by the normal force: μ = f ÷ N. A 62 kg skier sliding down a 25° slope with 45.0 N of friction has N = 62 × 9.8 × cos 25° = 550.7 N, so μ_k = 45 ÷ 550.7 = 0.082.

### How does a slope change friction?

On a slope the normal force is only m g cos θ, so friction is smaller, while gravity pulls down the slope with m g sin θ. An object stays put while μ_s ≥ tan θ; a wooden block (μ_s = 0.5) holds up to 26.57° and slides on a 30° wooden ramp.

### How fast does an object slide down a slope?

With kinetic friction, a = g (sin θ − μ_k cos θ). A snowboarder on a 13° slope with μ_k = 0.20 speeds up at 9.8 × (0.2250 − 0.1949) = 0.29 m/s². A negative result means a sliding object slows down.

### Does friction depend on the contact area?

In this model, no: f = μN uses only the coefficient and the normal force, which is why a brick slides the same on any face. The coefficients are approximate and depend on the surfaces, whether they are wet or oiled, and their condition.

## Sources

- OpenStax, University Physics Volume 1, §6.2 Friction: Equation 6.1 (f_s ≤ μ_s N), Equation 6.2 (f_k = μ_k N); Table 6.1 (approximate coefficients of static and kinetic friction); Example 6.10 (a 20.0 kg crate, N = 196 N, f_s up to 137 N, f_k = 118 N); Example 6.11 (45.0 N on a 62 kg skier on a 25° slope gives μ_k = 0.082); μ_k = tan θ at a steady slide; Example 6.14 (a_x = g(sin θ − μ_k cos θ) = 0.29 m/s²), retrieved 2026-10-19. https://openstax.org/books/university-physics-volume-1/pages/6-2-friction
- OpenStax, University Physics Volume 1, §5.6 Common Forces: the normal force N = mg on a level surface and N = mg cos θ on a slope; the weight along the slope mg sin θ, retrieved 2026-10-19. https://openstax.org/books/university-physics-volume-1/pages/5-6-common-forces
- NIST, CODATA 2022: standard acceleration of gravity g_n = 9.80665 m/s² (exact), retrieved 2026-10-19. https://physics.nist.gov/cgi-bin/cuu/Value?gn
- NIST Special Publication 811 (2008), Appendix B.8: 1 lb = 0.45359237 kg, 1 oz = 0.028349523125 kg, 1 lbf = 4.4482216152605 N, 1 kgf = 9.80665 N, retrieved 2026-10-19. https://www.nist.gov/pml/special-publication-811