How to Find Normal Force: Flat Ground, Inclines, Elevators

Q: Is the normal force always equal to mg?

No. The normal force equals mg only for an object on a horizontal surface with zero vertical acceleration and no other vertical forces. On an incline it is mg cos(theta), in an elevator accelerating upward it is m(g + a), and a downward push on the object increases it. Always get the normal force from a free-body diagram and Newton's second law, not from a memorized formula.

Q: What is the normal force in an elevator moving at constant velocity?

Exactly mg, whether the elevator moves up or down. Constant velocity means zero acceleration, so the net force is zero and the normal force must balance gravity. Only acceleration changes what a scale reads, never the direction of motion by itself.

Q: Why is the normal force mg cos(theta) on an incline and not mg sin(theta)?

The normal force balances the component of gravity perpendicular to the surface, which is mg cos(theta). The mg sin(theta) component points along the slope and is handled by friction, tension, or acceleration instead. Sanity check: at theta = 0 the surface is flat, cos(0) = 1, and you get back mg, which is correct.

Q: Are the normal force and gravity a Newton's third law pair?

No, and this is a favorite AP trap. Both forces act on the same object, while third law pairs always act on two different objects. The true partner of the floor pushing up on a box is the box pushing down on the floor. Normal force and gravity merely happen to balance in the special case of a flat surface with no vertical acceleration.

Q: Can the normal force be zero or negative?

It can be zero: in free fall, at the instant a car leaves the crest of a hill, or in an elevator falling freely. It can never be negative, because a surface can only push, not pull. If your algebra produces a negative normal force, the object has actually left the surface and your free-body diagram needs revising.

Draw a free-body diagram and apply Newton's second law perpendicular to the surface. On flat ground with no other vertical forces, N = mg. On an incline, N = mg cos(theta). In an elevator accelerating up, N = m(g + a); accelerating down, N = m(g - a). Normal force is not always mg.

AP Physics: Unit 2 (topics 2.2 Forces and Free-Body Diagrams, 2.4 Newton's First Law, 2.5 Newton's Second Law). Normal force problems sit in AP Physics 1 Unit 2 (Force and Translational Dynamics), especially topics 2.4 and 2.5, where you apply Newton's laws with free-body diagrams. The same skills carry directly into Unit 2 of AP Physics C: Mechanics.

Get your number first

Enter the mass (and an incline angle or extra vertical push) to read the normal force with every step shown. Open the full calculator.

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

The Method That Always Works

To find the normal force, draw a free-body diagram, point one axis perpendicular to the surface, and apply Newton's second law along that axis. The normal force $F_N$ is whatever value balances that equation. That is the whole method, and it works in every situation the AP exam can throw at you.

The normal force is a contact force. A surface pushes on any object touching it, and the push is always perpendicular (normal) to the surface. The surface supplies exactly as much force as the situation demands: more if something presses the object down, less if something lifts up on it, zero if contact is lost.

Memorizing $F_N = mg$ as a universal formula will cost you points, because it only holds in one special case. If your diagrams are shaky, start with how to draw a free-body diagram, then come back here.

Case 1: Flat Ground, No Extra Vertical Forces

A box of mass $m$ sits on a horizontal floor. Two forces act on it: gravity $F_g = mg$ pulling down and the normal force $F_N$ pushing up. The box does not accelerate vertically, so Newton's second law in the vertical direction gives

F_N - mg = 0 \quad \Rightarrow \quad F_N = mg

For a 12 kg box, $F_N = 12 \times 9.8 = 117.6$ N, about 118 N. This is the only case where $F_N = mg$ is guaranteed: horizontal surface, zero vertical acceleration, and no forces besides gravity and the normal force with vertical components. Change any one of those conditions and you must go back to the free-body diagram. This setup is a direct application of topic 2.4 in AP Physics 1: the box is in equilibrium, so the forces on it sum to zero.

Case 2: Flat Ground with an Applied Vertical Force

Push down on the box and the floor pushes back harder. Pull up on it and the floor eases off. The perpendicular equation picks this up automatically:

Downward push of magnitude $F$ : $F_N = mg + F$
Upward pull of magnitude $F$ : $F_N = mg - F$
Rope pulling at angle $\theta$ above horizontal: $F_N = mg - F\sin\theta$

Suppose a 12 kg box is pressed down with 40 N. Then $F_N = 12 \times 9.8 + 40 = 117.6 + 40 = 157.6$ N, about 158 N. This matters for friction problems, because the maximum friction force scales with the normal force through $|F_f| \leq |\mu F_N|$ . Pressing down on a box makes it harder to slide, and lifting on it makes it easier. See how to find the coefficient of friction for that next step.

Case 3: On an Incline, N = mg cos theta

On an incline at angle $\theta$ , tilt your axes so one runs along the surface and the other is perpendicular to it. Gravity then splits into two components: $mg\sin\theta$ pointing down the slope and $mg\cos\theta$ pressing into the slope. The object never accelerates into or out of the surface, so the perpendicular equation reads

F_N - mg\cos\theta = 0 \quad \Rightarrow \quad F_N = mg\cos\theta

A quick sanity check: at $\theta = 0$ this reduces to $F_N = mg$ (flat ground), and at $\theta = 90^\circ$ it gives $F_N = 0$ (a vertical wall cannot support weight). If you ever mix up sine and cosine, this check catches it instantly. For the full treatment, including friction and acceleration down the slope, work through inclined plane problems, and build intuition by adjusting the angle in the inclined plane simulator.

Case 4: In an Elevator

Stand on a scale in an elevator and the scale reads the normal force, not your weight. Take up as positive and apply Newton's second law vertically: $F_N - mg = ma$ , so

F_N = m(g + a)

where $a$ is positive for upward acceleration and negative for downward acceleration.

Accelerating upward: $F_N > mg$ , you feel heavy
Constant velocity (up or down): $a = 0$ , so $F_N = mg$
Accelerating downward: $F_N < mg$ , you feel light
Free fall ( $a = -g$ ): $F_N = 0$ , apparent weightlessness

Watch the wording on exam questions. An elevator moving downward at constant speed has zero acceleration, so the scale reads plain $mg$ . Only acceleration changes the normal force, which is exactly what Newton's second law (topic 2.5) says. The direction of the velocity is irrelevant.

Quick Reference and Common Traps

Situation	Normal force
Flat ground, nothing else	$F_N = mg$
Flat ground, downward push $F$	$F_N = mg + F$
Flat ground, upward pull $F$	$F_N = mg - F$
Incline at angle $\theta$	$F_N = mg\cos\theta$
Elevator, acceleration $a$ (up positive)	$F_N = m(g + a)$

Three traps to avoid. First, writing $F_N = mg$ on an incline. Second, assuming a downward-moving elevator has a reduced normal force (only downward acceleration does that). Third, calling the normal force and gravity a Newton's third law pair; they act on the same object, so they cannot be one. You can check any of these setups with the normal force calculator, which covers all four cases, or combine $F_N$ with other forces using how to find net force. Normal force sits in Unit 2, one of the heaviest-weighted units, so making it automatic is part of managing how hard AP Physics 1 actually is.

Box on a 30 Degree Incline

A 5.0 kg box rests on a ramp inclined at $30^\circ$ . Find the normal force on the box and compare it to the box's weight.

Draw the free-body diagram: gravity $mg$ straight down and the normal force $F_N$ perpendicular to the ramp surface. Tilt the axes to match the ramp.
Resolve gravity into components: $mg\cos\theta$ into the ramp and $mg\sin\theta$ along the ramp. Only the perpendicular component determines $F_N$ .
Compute the weight: $mg = 5.0 \times 9.8 = 49.0$ N.
Apply Newton's second law perpendicular to the ramp, where the acceleration is zero: $F_N = mg\cos 30^\circ = 49.0 \times 0.866 = 42.4$ N.

$F_N = 42.4$ N, noticeably less than the 49.0 N weight. On any incline the normal force is smaller than $mg$ , because the surface only supports the perpendicular component of gravity.

Person in an Accelerating Elevator

A 60.0 kg student stands on a bathroom scale in an elevator. What does the scale read when the elevator (a) accelerates upward at $2.0 \ \mathrm{m/s^2}$ , (b) moves upward at constant velocity, and (c) accelerates downward at $2.0 \ \mathrm{m/s^2}$ ?

The scale reads the normal force. Free-body diagram: $F_N$ up, $mg$ down. Newton's second law with up positive: $F_N - mg = ma$ , so $F_N = m(g + a)$ .
Weight first: $mg = 60.0 \times 9.8 = 588$ N.
(a) Upward acceleration, $a = +2.0 \ \mathrm{m/s^2}$ : $F_N = 60.0 \times (9.8 + 2.0) = 60.0 \times 11.8 = 708$ N.
(b) Constant velocity, $a = 0$ : $F_N = 60.0 \times 9.8 = 588$ N. Moving does not matter; only accelerating does.
(c) Downward acceleration, $a = -2.0 \ \mathrm{m/s^2}$ : $F_N = 60.0 \times (9.8 - 2.0) = 60.0 \times 7.8 = 468$ N.

(a) 708 N, (b) 588 N, (c) 468 N. The scale reads more than the true weight of 588 N while accelerating up, exactly the weight at constant velocity, and less while accelerating down.

Crate Pulled by an Angled Rope

A 10.0 kg crate sits on a horizontal floor. A rope pulls on it with a force of 25.0 N at $40^\circ$ above the horizontal. Find the normal force on the crate.

Free-body diagram: $mg$ down, $F_N$ up, and tension $T = 25.0$ N at $40^\circ$ above horizontal. Split the tension into components: $T\cos 40^\circ$ horizontal and $T\sin 40^\circ$ vertical.
Vertical component of the rope force: $T\sin 40^\circ = 25.0 \times 0.643 = 16.1$ N, pointing up.
Weight: $mg = 10.0 \times 9.8 = 98.0$ N.
The crate has no vertical acceleration, so $F_N + T\sin 40^\circ - mg = 0$ , which gives $F_N = 98.0 - 16.1 = 81.9$ N.

$F_N = 81.9$ N, about 82 N. The upward pull of the rope takes over part of the job of supporting the crate, so the floor pushes with less than $mg$ . If the rope pushed down at that angle instead, you would add the 16.1 N rather than subtract it.

Frequently asked questions

Is the normal force always equal to mg?

No. The normal force equals mg only for an object on a horizontal surface with zero vertical acceleration and no other vertical forces. On an incline it is mg cos(theta), in an elevator accelerating upward it is m(g + a), and a downward push on the object increases it. Always get the normal force from a free-body diagram and Newton's second law, not from a memorized formula.

What is the normal force in an elevator moving at constant velocity?

Exactly mg, whether the elevator moves up or down. Constant velocity means zero acceleration, so the net force is zero and the normal force must balance gravity. Only acceleration changes what a scale reads, never the direction of motion by itself.

Why is the normal force mg cos(theta) on an incline and not mg sin(theta)?

The normal force balances the component of gravity perpendicular to the surface, which is mg cos(theta). The mg sin(theta) component points along the slope and is handled by friction, tension, or acceleration instead. Sanity check: at theta = 0 the surface is flat, cos(0) = 1, and you get back mg, which is correct.

Are the normal force and gravity a Newton's third law pair?

No, and this is a favorite AP trap. Both forces act on the same object, while third law pairs always act on two different objects. The true partner of the floor pushing up on a box is the box pushing down on the floor. Normal force and gravity merely happen to balance in the special case of a flat surface with no vertical acceleration.

Can the normal force be zero or negative?

It can be zero: in free fall, at the instant a car leaves the crest of a hill, or in an elevator falling freely. It can never be negative, because a surface can only push, not pull. If your algebra produces a negative normal force, the object has actually left the surface and your free-body diagram needs revising.