What is the normal force?
Pick the situation, then type the mass and the slope angle, the push or pull, or the lift's acceleration. The normal force calculator gives the push of the surface in newtons and pounds-force, with the weight and its parts.
- Answer
- N = 533.271 N (119.884 lbf)
The normal force is 533.271 N (119.884 lbf), 0.906308 times the weight.
- Normal force (N)
- 533.271
- Normal force (lbf)
- 119.884
- Weight (N)
- 588.399
- Normal force ÷ weight
- 0.906308
- Weight along the slope (N)
- 248.668
Answer: N = 533.271 N (119.884 lbf). The normal force is 533.271 N (119.884 lbf), 0.906308 times the weight.
How to calculate
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)).
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.
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).
- 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
Each example is checked against the calculator on every build.
- Situation Level ground, Mass (m) 44.09 lb, Gravity (g, m/s²) 9.8 gives Normal force (N) 196, Normal force ÷ weight 1, Answer 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)
- Situation On a slope, Mass (m) 132.3 lb, Gravity (g, m/s²) 9.8, Slope angle (θ, degrees) 25 gives Normal force (N) 532.908979, Weight along the slope (N) 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)
- Situation In a lift, Mass (m) 165.3 lb, Gravity (g, m/s²) 9.8, Acceleration (a, m/s²) 1.2 gives Normal force (N) 825, Weight (N) 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)
- Situation Push or pull, Mass (m) 44.09 lb, Gravity (g, m/s²) 9.8, Applied force (F) 22.48 lbf, Force angle (φ, degrees) 30 gives Normal force (N) 146, Force, up part (N) 50, Answer 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)
- Situation Push or pull, Mass (m) 44.09 lb, Gravity (g, m/s²) 9.8, Applied force (F) 22.48 lbf, Force angle (φ, degrees) -30 gives Normal force (N) 246, Force, up part (N) -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)
How it works
The normal force N is the push of a surface on an object, square to the surface. With the weight w = m × g:
- Level ground: N = m × g.
- On a slope at θ above level: N = m × g × cos θ. The page also shows the weight along the slope, m × g × sin θ.
- Push or pull on level ground: a force F at an angle φ above level (negative φ pushes down). The vertical forces balance, N + F sin φ = m × g, so N = m × g − F × sin φ. The page also shows the up part F × sin φ and the level part F × cos φ.
- In a lift accelerating at a (up positive): N − m × g = m × a, so N = m × (g + a). A scale in the lift reads N.
The page also shows the weight m × g and the share N ÷ (m × g).
Units (exact): 1 g = 0.001 kg; 1 t = 1,000 kg; 1 oz = 0.028349523125 kg; 1 lb = 0.45359237 kg; 1 short ton = 907.18474 kg; 1 kN = 1,000 N; 1 lbf = 4.4482216152605 N; 1 kgf = 9.80665 N. Angles are in degrees; g and a are in m/s² (9.80665 m/s² is standard gravity).
Rules
- Mass from 10⁻⁶ kg to 10¹² kg; g from 0.001 to 1,000 m/s²; slope from 0° to 90°; applied force from 0 to 10¹⁵ N; force angle from −90° to 90°; lift acceleration from −1,000 to 1,000 m/s². Limits are in SI units and include their ends; a value typed in another unit is converted first.
- A pull whose up part is more than the weight (m × g − F sin φ below 0), or a lift accelerating down faster than g (g + a below 0), lifts the object off the surface: there is no answer. Exactly at the limit, N is 0.
- A result that is exactly 0 shows as 0. Any other result too large or too small to show as a number gives no answer.
Output format
The answer line reads N = a N (b lbf). It and every number show to 6 significant figures, rounded half up, with trailing zeros dropped and thousands commas, and below 10⁻⁶ or from 10¹⁵ up as a × 10ⁿ; negative numbers show with a minus sign (−). Typed values are read back as the exact decimals you typed, and every unit size is exact. sin and cos are exact where they are rational (0, ±½ or ±1, at 0°, ±30°, 60° and ±90°), so those results are exact fractions rounded once and the lift-off test compares exact values; other angles use 64-bit floats.
Worked examples by hand
Level ground. OpenStax Example 6.10: a 20.0 kg crate with g = 9.80 m/s²: N = 20 × 9.8 = 196 N (44.0626 lbf), exactly the weight.
A slope. OpenStax Example 5.12: a 60.0 kg skier on a 25° slope, g = 9.80 m/s²: N = 60 × 9.8 × cos 25° = 588 × 0.906308 = 532.909 N, and the weight along the slope is 588 × sin 25° = 248.500 N.
A lift. OpenStax Example 6.3: 75.0 kg, accelerating up at 1.20 m/s², g = 9.80 m/s²: N = 75 × (9.8 + 1.2) = 825 N, against a weight of 735 N.
A pull at an angle. The 20 kg crate (196 N) pulled with 100 N at 30° above level: the up part is 100 × sin 30° = 50 N, so N = 196 − 50 = 146 N (32.8221 lbf). Pushing down at −30° instead: N = 196 + 50 = 246 N.
Other questions people ask
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.