Normal Force Calculator (Flat & Inclined Surfaces)

Normal force is the perpendicular push from a surface. On flat ground with no other vertical forces it equals mg. On an incline it drops to mg cos(theta), and an extra vertical push or pull changes it to mg plus or minus F. The calculator above handles all three cases using g = 9.8 m/s^2.

Mass m (kg)

normal force (N)

98 N

Steps

1.On flat ground with nothing else pushing vertically, N balances the weight: N = m g
2.N = m g = 10 kg x 9.8 m/s^2 = 98 N

AP Physics: Unit 2 (topics 2.5 Newton's Second Law, 2.2 Forces and Free-Body Diagrams). Normal force calculations come from Topic 2.5 (Newton's Second Law) in AP Physics 1 Unit 2, Force and Translational Dynamics, weighted at 18 to 23% of the multiple-choice section. The same analysis appears in AP Physics C: Mechanics, where Unit 2 carries 20 to 25%.

What the calculator above computes

The calculator above finds the normal force $F_N$ , the contact force a surface exerts on an object, directed perpendicular to the surface. It handles three setups: an object resting on flat ground ( $F_N = mg$ ), an object on an incline at angle $\theta$ ( $F_N = mg\cos\theta$ ), and an object on flat ground with an extra vertical push or pull ( $F_N = mg \pm F$ ). It uses $g = 9.8\ \text{m/s}^2$ , the value printed on the AP Physics 1 equation sheet.

Type in a mass and, where the setup needs it, an angle or an applied force, and you get $F_N$ in newtons. The sections below explain where each formula comes from, so you can reproduce the result by hand on a test, where the setup earns the points. For the full step-by-step method, see how to find normal force.

Why normal force is not always mg

Normal force is a constraint force: the surface pushes back with exactly the strength needed to keep the object from sinking into it, nothing more and nothing less. There is no standalone formula for $F_N$ the way there is for gravity. You always get it from Newton's second law applied perpendicular to the surface.

On flat ground with no other vertical forces, the perpendicular direction is vertical and the vertical acceleration is zero, so $F_N - mg = 0$ and $F_N = mg$ . Change anything about that picture and $F_N$ changes too:

Tilt the surface, and only part of gravity presses into it: $F_N = mg\cos\theta$ .
Push down on the object, and the surface must support the push as well: $F_N = mg + F$ .
Pull up on it, and the surface supports less: $F_N = mg - F$ .
Accelerate vertically, as in an elevator, and $F_N$ must supply that acceleration: $F_N = m(g + a)$ when the acceleration points upward.

Writing $F_N = mg$ automatically, without checking the situation, is one of the most common errors in Unit 2.

The three formulas at a glance

Every output of the calculator comes from one of these rows:

Setup	Normal force	Key condition
Flat surface, no extra vertical force	$F_N = mg$	No vertical acceleration
Incline at angle $\theta$ from horizontal	$F_N = mg\cos\theta$	No acceleration perpendicular to the ramp
Flat surface, extra vertical force $F$	$F_N = mg \pm F$	+ for a downward push, - for an upward pull

All three rows come from the same procedure: draw the free-body diagram, point one axis perpendicular to the surface, and set the net force along that axis to zero, because the object neither sinks into the surface nor jumps off it. If free-body diagrams still feel shaky, work through how to draw a free-body diagram first; every force problem in Unit 2 starts with one.

Normal force on an incline: why cosine

On a ramp at angle $\theta$ , gravity still points straight down, but the surface is tilted, so you split $mg$ into components along tilted axes: $mg\sin\theta$ parallel to the surface, pulling the object down the slope, and $mg\cos\theta$ perpendicular to it, pressing the object into the ramp. The normal force only has to balance the perpendicular part, so $F_N = mg\cos\theta$ .

A quick sanity check catches the classic sine and cosine mix-up. At $\theta = 0$ the ramp is flat, and $\cos 0 = 1$ gives $F_N = mg$ , which is correct. Sine would give $F_N = 0$ for a flat floor, which is clearly wrong. As the angle grows toward $90^\circ$ , $F_N$ shrinks toward zero, because an object barely presses against a near-vertical surface.

You can watch this happen live in the inclined plane simulator: drag the angle up and the normal force vector shrinks in real time. For complete ramp problems with friction and acceleration, see inclined plane problems.

Normal force feeds straight into friction

The main reason to get $F_N$ exactly right is that friction depends on it. The AP equation sheet gives friction as $|F_f| \leq |\mu F_N|$ , a single inequality that covers both static and kinetic cases. Any error in $F_N$ passes straight through to the friction force and from there into the acceleration.

This is where the flat-with-applied-force case matters most. Press down on a box and you increase $F_N$ , which raises the maximum friction force, which is why leaning on something makes it harder to slide. Pull upward on it and you decrease $F_N$ , which is part of why dragging a sled with an angled rope beats pushing down on it.

Once you have $F_N$ from the calculator above, send it into the friction calculator to get the friction force, or work backward from observed motion with how to find the coefficient of friction.

Where this sits in AP Physics 1

Normal force lives in Unit 2, Force and Translational Dynamics, which carries 18 to 23% of the multiple-choice section, tied with Unit 3 for the largest weight in AP Physics 1. The formulas on this page are Topic 2.5, Newton's Second Law, applied perpendicular to a surface, built on the free-body diagram skills of Topic 2.2.

On the exam, $F_N$ is almost never the final answer. It is the middle step: find the normal force, feed it into friction, then combine everything into net force and acceleration. A four-function, scientific, or graphing calculator is allowed on both sections, so the arithmetic is not what costs points; writing the wrong perpendicular equation is.

If you are taking the calculus-based course instead, nothing changes: AP Physics C: Mechanics covers the same normal force analysis in its own Unit 2, weighted at 20 to 25% of the multiple-choice section.

Backpack on a flat floor

A 12.0 kg backpack sits at rest on a horizontal floor with no other vertical forces acting on it. Find the normal force from the floor.

Draw the free-body diagram. Only two forces act: gravity $mg$ pointing down and the normal force $F_N$ pointing up.
Apply Newton's second law vertically. The backpack has zero vertical acceleration, so $F_N - mg = 0$ , which gives $F_N = mg$ .
Substitute the numbers: $F_N = (12.0\ \text{kg})(9.8\ \text{m/s}^2) = 117.6\ \text{N}$ .

$F_N = 118\ \text{N}$ (3 sig figs). This is the one clean case where normal force really does equal $mg$ : flat surface, no other vertical forces, no vertical acceleration.

Box on a 30 degree ramp

A 5.00 kg box rests on a frictionless ramp inclined at $30.0^\circ$ above the horizontal. Find the normal force from the ramp.

Tilt the axes so x runs along the ramp and y is perpendicular to it. Gravity splits into $mg\sin\theta$ down the slope and $mg\cos\theta$ pressing into the surface.
Perpendicular to the ramp the box does not accelerate, so $F_N - mg\cos\theta = 0$ , which gives $F_N = mg\cos\theta$ .
Compute the weight first: $mg = (5.00\ \text{kg})(9.8\ \text{m/s}^2) = 49.0\ \text{N}$ .
Multiply by the cosine: $F_N = 49.0\ \text{N} \times \cos 30.0^\circ = 49.0\ \text{N} \times 0.866 = 42.4\ \text{N}$ .

$F_N = 42.4\ \text{N}$ , noticeably less than the 49.0 N weight. The other component, $mg\sin 30.0^\circ = 24.5\ \text{N}$ , points down the slope and would accelerate the box if nothing held it in place.

Pressing down versus pulling up on a crate

An 8.00 kg crate sits on a horizontal floor. First a person presses straight down on it with a 25.0 N force. Then the person instead pulls straight up on it with 25.0 N. Find the normal force in each case.

Pressing down: three vertical forces act ( $F_N$ up, $mg$ down, applied $F$ down). With zero vertical acceleration, $F_N - mg - F = 0$ , so $F_N = mg + F$ .
Substitute: $F_N = (8.00\ \text{kg})(9.8\ \text{m/s}^2) + 25.0\ \text{N} = 78.4\ \text{N} + 25.0\ \text{N} = 103.4\ \text{N}$ .
Pulling up: the applied force now helps support the crate, so $F_N + F - mg = 0$ , which gives $F_N = mg - F$ .
Substitute: $F_N = 78.4\ \text{N} - 25.0\ \text{N} = 53.4\ \text{N}$ .

Pressing down: $F_N = 103\ \text{N}$ . Pulling up: $F_N = 53.4\ \text{N}$ . Same crate, same floor, and the normal forces differ by almost a factor of two, which is exactly why you check the free-body diagram before writing $F_N = mg$ .

Frequently asked questions

Is normal force always equal to mg?

No. F_N = mg only when the surface is horizontal, no other force has a vertical component, and the object is not accelerating vertically. On an incline it drops to mg cos(theta), an extra downward push raises it, an upward pull lowers it, and vertical acceleration (an elevator, for example) changes it too. Always get the normal force from Newton's second law perpendicular to the surface, never from memory.

Why does the incline formula use cosine instead of sine?

The normal force balances the component of gravity perpendicular to the surface, which is mg cos(theta) when the angle is measured from the horizontal. The mg sin(theta) component points along the slope and has nothing to do with the normal force. Quick check: at theta = 0 (flat ground) cosine gives F_N = mg, which is correct, while sine would give zero, which is not.

Can normal force be zero or negative?

It can be zero. If an upward pull on an object on flat ground reaches F = mg, the surface no longer presses on the object at all and it is on the verge of lifting off. It can never be negative: a surface can push but cannot pull. If your algebra returns a negative normal force, the object has actually left the surface and you need a new free-body diagram for the new situation.

What is the normal force in an accelerating elevator?

The vertical acceleration is no longer zero, so the normal force does not equal mg. Newton's second law gives F_N - mg = ma, so F_N = m(g + a) when the acceleration points upward (speeding up while going up, or slowing down while going down) and F_N = m(g - a) when it points downward. That changing normal force is exactly what a bathroom scale would read in the elevator.

What value of g does the calculator use?

It uses g = 9.8 m/s^2, the value printed on the AP Physics equation sheets, with mass in kilograms, forces in newtons, and angles in degrees. Use 9.8 rather than 9.81 in your own AP work so your numbers stay consistent with the equation sheet.