Normal Force Calculator

normal force = mass × gravity × cos(incline angle)

m

The mass resting on the surface, in kilograms, with grams, tonnes and pounds in the same field. It is a mass and not a weight: the weight is m × g and the page reports it as its own output row. The field rejects zero, because an object with no mass is not a small object, it is no object — and it would make every row on the page read zero

θ

The incline angle, measured between the surface and the horizontal, in degrees, radians or gradians. Flat ground is 0 and a vertical wall is 90, and the field accepts anything in between. At 30 degrees the normal force is 87 percent of the weight; at 90 it is nothing at all, because the surface is no longer underneath the object

g

Standard gravity, fixed at 9.80665 m/s² — the defined value, not a local measurement. It is deliberately not an input here, because it is not a quantity the reader is choosing; if you want to know what the same object weighs and stores on the Moon or on Mars, that is the potential energy page, where gravity is a field with a table of six bodies beside it

N

The normal force itself: perpendicular to the surface, pointing away from it, in newtons and in pounds-force. The word normal is the geometric one, meaning perpendicular — this force is defined by its direction, not by any property of the object. It is a response rather than a cause, and it is what the surface does rather than what the object does

F∥

The downhill component, mg sin θ, parallel to the surface and pointing down the slope. It is not a third force acting on the object — it is the rest of the weight, the part the surface is not holding up. It is the number the friction pages need, because whether an object slides comes down to comparing it with the friction the surface can supply

Use this page when a mass is sitting on a ramp, a roof, a loading slope or any other inclined surface and you need the force that surface is actually carrying: sizing a chock, checking whether a parked trailer holds, working out what a sled's runners press with, or finding the weight component along a slope before you go on to ask whether it slides. It is also the page to read when the answer surprises you — the normal force on an incline is not the weight, it is the weight scaled by the cosine of the incline angle, so a 100 kg crate on a 30 degree ramp presses with 849 N rather than 981 N — the 981 N has not gone anywhere, it has just been split, and 490 N of it is now tugging the crate downhill. That split is the whole subject, and seeing the two components next to the weight is the point of showing three rows. The two component rows are also the handoff to the friction calculators, which take the normal force on an incline as an input rather than computing it.

1. The defaults: 10 kg on a 30 degree slope

   1. Mass 10 kg, incline angle 30 degrees
   2. Weight: W = m × g = 10 × 9.80665 = 98.0665 N, which prints as 98.07 N
   3. Normal force: N = W × cos 30 = 98.0665 × 0.8660254 = 84.9281, which prints as 84.93 N
   4. Downhill component: F∥ = W × sin 30 = 98.0665 × 0.5 = 49.0333, which prints as 49.03 N
   5. Check: 84.93² + 49.03² = 7213.1 + 2403.9 = 9617, and 98.07² = 9617.7 — the two components reconstruct the weight
   6. In pounds-force: 84.9281 ÷ 4.4482216 = 19.0926, which prints as 19.09 lbf

   The check in the fifth step is the reason this page prints three force rows instead of one, and it is worth doing by hand once: the normal force and the downhill component are perpendicular, so their squares add to the square of the weight. On a 30 degree slope the surface carries 87 percent of the weight — 84.93 of 98.07 N — while the downhill pull is 49.03 N, which is half the weight and not the 13 percent the cosine leaves over. The two are perpendicular components rather than two halves of a sum, which is why 84.93² + 49.03² comes back to 98.07². That 49 N is the number to carry to the friction calculator, because whether the object stays put is decided by comparing it with the friction the surface can offer, which is the coefficient of friction times this 84.93 N and not times the weight.

2. Flat ground: 70 kg standing on the floor

   1. Mass 70 kg, incline angle 0 degrees — a person standing on a level floor
   2. Weight: W = 70 × 9.80665 = 686.4655 N
   3. Normal force: N = W × cos 0 = 686.4655 × 1 = 686.4655, so 686.47 N
   4. Downhill component: F∥ = W × sin 0 = 686.4655 × 0 = 0 N, exactly zero
   5. In pounds-force: 686.4655 ÷ 4.4482216 = 154.3236, which prints as 154.32 lbf

   At zero degrees the two top rows are the same number and the bottom row is zero, which is why the page looks redundant on flat ground and is not: this is the case where the normal force really does equal the weight, and it is the only case where that is true. The two rows separate as soon as the surface tilts — by 30 degrees they are 84.93 and 98.07 — and the point of printing both is that a reader who has only ever met the flat version tends to assume N = mg everywhere. The floor you are standing on is pushing up with 686 N, which is 154 pounds, and it has been doing that all day.

3. The limit: 1000 kg against a vertical wall

   1. Mass 1000 kg, incline angle 90 degrees — the surface has become a wall
   2. Weight: W = 1000 × 9.80665 = 9806.65 N
   3. Normal force: N = W × cos 90 = 9806.65 × 0 = 0 N
   4. Downhill component: F∥ = W × sin 90 = 9806.65 × 1 = 9806.65 N — all of it
   5. So the surface is carrying none of the weight and the whole 9806.65 N is pulling straight down

   This is the row that makes the formula make sense. A vertical surface carries no weight at all, because there is nothing underneath the object for it to push back with: cos 90 is zero, so the normal force is zero, and the whole weight has moved into the downhill component, which at 90 degrees points straight down. The page still prints the weight, because it is a property of the object rather than of the surface, but if you were sizing a wall to hold this crate up you would get no help from these rows — a wall that is not pushing on the crate provides no friction either, and holding it up would need a force this page does not model.

A straight plane is assumed, and the object is a point on it: a curved hill, or a crate long enough to touch the slope at two points, will not match. Only gravity is counted, so a rope pulling upward, a hand pressing down or a second object on top all change the normal force — a response, not a quantity this page can know in advance. Nothing here includes friction; the page computes the two components the friction calculators take as input. Gravity is fixed at the defined standard 9.80665 m/s².

What is the normal force formula?

N = m × g × cos θ, where θ is the incline angle measured from the horizontal. On level ground cos 0 is 1, so the normal force equals the weight; on a 30 degree slope cos 30 is 0.866, so a 10 kg object's normal force is 84.93 N against a weight of 98.07 N. The formula is the weight multiplied by the fraction of it that presses into the surface, and the rest of the weight is the component pulling the object down the slope, m × g × sin θ.

Why is the normal force less than the weight on a slope?

Because part of the weight is not pressing into the surface at all. Weight is a single downward force, and on a slope it is useful to split it into one part perpendicular to the surface and one part parallel to it; only the perpendicular part is the weight component that the surface has to hold up. At 30 degrees that is 87 percent of the weight, at 45 degrees 71 percent, and at 60 degrees 50 percent. The part that is left over, mg sin θ, is what pulls the object down the incline.

How do I find the normal force for a mass on a slope?

Enter the mass and the incline angle; the page needs nothing else, because gravity is fixed at 9.80665 m/s². For a 1500 kg car on a 10 degree ramp the normal force is 1500 × 9.80665 × cos 10 = 14486.5 N, and the downhill component is 2554.4 N. If you have the weight already rather than the mass, divide it by 9.80665 first, or just multiply it by cos θ yourself — the two routes give the same answer to the digits shown.

What is the normal force on flat ground?

It is exactly the weight, m × g. At an incline angle of 0 degrees the cosine is 1 and the sine is 0, so a 70 kg person standing on a level floor is held up by 686.47 N while the downhill component is zero. This is the only angle where the two are equal, which is why the flat-ground case is worth trying first: it tells you what the page is doing before the slope makes the two numbers different.

Which way does the normal force point?

Perpendicular to the surface, away from it — that is what the word normal means in geometry. On a 30 degree slope it points 30 degrees off vertical, not straight up, and it is not in the same direction as the weight. Because it is perpendicular to the surface it also does no work on an object sliding along that surface: the force and the motion are at right angles, so their product is zero. It is a response rather than a cause, and it takes whatever value is needed to stop the two surfaces from passing through each other.

• Normal, Tension, and Other Examples of Forces (College Physics 2e, §4.5) — the normal force as a perpendicular response, and why it equals mg only on level ground — OpenStax
• Normal force — the definition as the perpendicular component of the contact force, and the worked inclined-plane case — Wikipedia
• Inclined plane — resolving weight into components parallel and perpendicular to the surface, and the friction conditions that follow — Wikipedia
• NIST Guide for the Use of the International System of Units (SI), Appendix B.8 — Factors for Units Listed Alphabetically: the pound-force row (1 lbf = 4.448 222 N) that this page's second row is converted with — NIST