# What is the normal force?

Normal force calculator: N = mg on level ground, N = mg cos θ on a slope, with a push or pull at an angle (N = mg − F sin φ), or in an accelerating lift (N = m(g + a)).

- Page: https://www.acalculator.org/physics/normal-force-calculator
- JSON spec: https://www.acalculator.org/physics/normal-force-calculator.json
- Version: 34d7d9de5426

## Default answer

Example with the default inputs (Situation On a slope, Mass (m) 132.277357310927 lb, Gravity (g, m/s²) 9.80665, Slope angle (θ, degrees) 25): The normal force is 533.271 N (119.884 lbf), 0.906308 times the weight.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| on | Situation | Where the object rests and what else acts on it. |
| m | Mass (m) | The mass of the object. |
| g | Gravity (g, m/s²) | The acceleration of gravity: 9.80665 m/s² is standard on Earth. |
| th | Slope angle (θ, degrees) | The angle of the slope above level, from 0° to 90°. |
| F | Applied force (F) | A push or pull on the object, besides its weight. |
| phi | Force angle (φ, degrees) | The angle of the force above level: positive pulls up, negative pushes down into the ground. |
| a | Acceleration (a, m/s²) | The acceleration of the lift: positive up (starting up or stopping on the way down), negative down. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| result | Answer | The normal force to 6 significant figures, in newtons and pounds-force. |
| normal | Normal force (N) | N, the push of the surface square to it, in newtons. |
| lbf | Normal force (lbf) | N in pounds-force. |
| weight | Weight (N) | w = m × g, in newtons. |
| share | Normal force ÷ weight | N ÷ (m × g): 1 on level ground with nothing else acting. |
| along | Weight along the slope (N) | m × g × sin θ: the part of the weight down the slope. |
| lift | Force, up part (N) | F × sin φ: the part of the applied force that pulls up (negative pushes down). |
| push | Force, level part (N) | F × cos φ: the part of the applied force along the ground. |

## Method

N = m × g on level ground; N = m × g × cos θ on a slope; N = m × g − F × sin φ with a force F at φ above level; N = m × (g + a) in a lift accelerating at a (up positive).

## Assumptions

- The surface is rigid and the object stays on it; no other forces act square to the surface.
- On a slope the object rests on (or slides along) the plane; a push or pull acts on level ground.

## Worked examples

1. on = level, m = 20, g = 9.8 gives normal = 196, share = 1, result = N = 196 N (44.0626 lbf). Source: OpenStax, University Physics Volume 1, §5.6 Common Forces (w = mg; the normal force N = mg on a level surface and N = mg cos θ on an incline; Example 5.12: a 60.0 kg skier on a 25° slope), https://openstax.org/books/university-physics-volume-1/pages/5-6-common-forces (retrieved 2026-10-19).
2. on = incline, m = 60, g = 9.8, th = 25 gives normal = 532.908979, along = 248.499538. Source: OpenStax, University Physics Volume 1, §5.6 Common Forces (w = mg; the normal force N = mg on a level surface and N = mg cos θ on an incline; Example 5.12: a 60.0 kg skier on a 25° slope), https://openstax.org/books/university-physics-volume-1/pages/5-6-common-forces (retrieved 2026-10-19).
3. on = lift, m = 75, g = 9.8, a = 1.2 gives normal = 825, weight = 735. Source: OpenStax, University Physics Volume 1, §6.1 Solving Problems with Newton’s Laws (Example 6.3: a 75.0 kg person in an elevator accelerating up at 1.20 m/s² reads 825 N; 735 N at constant velocity), https://openstax.org/books/university-physics-volume-1/pages/6-1-solving-problems-with-newtons-laws (retrieved 2026-10-19).
4. on = force, m = 20, g = 9.8, F = 100, phi = 30 gives normal = 146, lift = 50, result = N = 146 N (32.8221 lbf). Source: OpenStax, University Physics Volume 1, §5.6 Common Forces (w = mg; the normal force N = mg on a level surface and N = mg cos θ on an incline; Example 5.12: a 60.0 kg skier on a 25° slope), https://openstax.org/books/university-physics-volume-1/pages/5-6-common-forces (retrieved 2026-10-19).
5. on = force, m = 20, g = 9.8, F = 100, phi = -30 gives normal = 246, lift = -50. Source: OpenStax, University Physics Volume 1, §5.6 Common Forces (w = mg; the normal force N = mg on a level surface and N = mg cos θ on an incline; Example 5.12: a 60.0 kg skier on a 25° slope), https://openstax.org/books/university-physics-volume-1/pages/5-6-common-forces (retrieved 2026-10-19).

## FAQ

### How do you calculate the normal force?

On level ground with nothing else pushing or pulling, the normal force equals the weight: N = mg. A 20.0 kg crate has N = 20 × 9.8 = 196 N.

### What is the normal force on an incline?

Only the part of the weight square to the slope: N = mg cos θ. A 60.0 kg skier on a 25° slope has N = 60 × 9.8 × cos 25° = 532.9 N, while 248.5 N (mg sin θ) pulls along the slope.

### How does pulling up at an angle change the normal force?

The up part of the pull, F sin φ, carries some of the weight: N = mg − F sin φ. A 100 N pull at 30° on the 196 N crate leaves N = 196 − 50 = 146 N. Pushing down at 30° adds instead: 246 N.

### What is the normal force in an elevator?

N = m(g + a), with a the elevator’s acceleration, up positive. A bathroom scale reads N: a 75.0 kg person reads 825 N while the lift speeds up going up at 1.20 m/s², and 735 N at a steady speed.

### Is the normal force always equal to the weight?

No. It is equal only on level ground with no other vertical forces and no vertical acceleration. It is smaller on a slope or with an upward pull, and larger with a downward push or an upward acceleration.

### When is the normal force zero?

When nothing presses the object onto the surface: a pull whose up part matches the weight, a lift falling at g, or a 90° wall. A bigger pull would lift the object off, and then there is no contact and no normal force.

## Sources

- OpenStax, University Physics Volume 1, §5.6 Common Forces: weight w = mg; the normal force N = mg on a level surface and N = mg cos θ on an incline, with the weight along the slope mg sin θ; Example 5.12 (a 60.0 kg skier on a 25° slope), retrieved 2026-10-19. https://openstax.org/books/university-physics-volume-1/pages/5-6-common-forces
- OpenStax, University Physics Volume 1, §5.3 Newton’s Second Law: Equations 5.3 to 5.5, ΣF = ma, applied to each direction (here the vertical one), retrieved 2026-10-19. https://openstax.org/books/university-physics-volume-1/pages/5-3-newtons-second-law
- OpenStax, University Physics Volume 1, §6.1 Solving Problems with Newton’s Laws: Example 6.3 (a 75.0 kg person in an elevator: 825 N accelerating up at 1.20 m/s², 735 N at constant velocity), retrieved 2026-10-19. https://openstax.org/books/university-physics-volume-1/pages/6-1-solving-problems-with-newtons-laws
- OpenStax, University Physics Volume 1, §6.2 Friction: Example 6.10 (a 20.0 kg crate on a level floor: N = w = 196 N), retrieved 2026-10-19. https://openstax.org/books/university-physics-volume-1/pages/6-2-friction
- 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