Weight vs Normal Force: What Is the Difference?

Weight is gravity pulling the object straight down, with magnitude mg. The normal force is a surface pushing perpendicular to itself, and its size is set by Newton's second law in that direction. They are equal only on level ground with no vertical acceleration and no other vertical force.

AP Physics: Unit 2 (topics 2.2 Forces and Free-Body Diagrams, 2.6 Gravitational Force, 2.7 Kinetic and Static Friction). This comparison sits across three AP Physics 1 Unit 2 topics. EK 2.6.A.3 defines weight as the gravitational force exerted by an astronomical body on a relatively small nearby object and gives the derived equation Weight = F_g = mg. EK 2.7.A.2.ii defines normal force as the perpendicular component of the force exerted on an object by the surface with which it is in contact, directed away from the surface. EK 2.6.C.1 identifies the magnitude of a system's apparent weight with the magnitude of the normal force, and EK 2.6.C.2 states that an accelerating system's apparent weight is not equal to the magnitude of the gravitational force exerted on it. The free-body-diagram rules quoted here come from the Topic 2.2 boundary statement. Unit 2 carries 18 to 23 percent of the multiple-choice section over about 22 to 27 class periods.

The distinction, stated once

Two forces, two different origins, and the AP Physics 1 CED defines them in two separate places.

Weight is gravitational. EK 2.6.A.3 calls the gravitational force exerted by an astronomical body on a relatively small nearby object its weight, and gives the derived equation $\text{Weight} = F_g = mg$ . No contact is required. Near Earth's surface it points straight down, toward the center of the Earth, and its magnitude depends on exactly two things: the object's mass and the local gravitational field strength.

The normal force is a contact force. EK 2.7.A.2.ii defines it as the perpendicular component of the force exerted on an object by the surface with which it is in contact, directed away from the surface. Read that definition for what is absent: no mass, no $g$ , no mention of gravity anywhere. The normal force points perpendicular to the surface, which is not a synonym for upward, and its magnitude is set by whatever Newton's second law perpendicular to that surface demands.

The two forces are not even answers to the same question. Weight answers how hard the Earth is pulling on this object. The normal force answers how hard this surface is having to push to keep the object out of it. They return the same number in the one arrangement every course opens with, and that coincidence is the whole reason the confusion survives to the exam.

This page is about telling them apart. The procedure for getting the number in each geometry belongs to how to find normal force, and the drawing rules belong to how to draw a free-body diagram.

Weight vs normal force, side by side

	Weight	Normal force
What it is	The gravitational force an astronomical body exerts on the object	The perpendicular component of the force a surface exerts on the object
Symbol	$F_g$ , sometimes $W$	$F_N$ , or $N$
Contact required	No	Yes, always
Direction	Straight down, toward the center of the Earth	Perpendicular to the surface, pointing away from it
Magnitude	$mg$ , in every situation on this page	Whatever $\sum F_\perp = ma_\perp$ returns
Depends on mass	Directly, and always	Only through the second law, and sometimes not at all
Changes when the object accelerates	No	Yes, whenever the acceleration has a component perpendicular to the surface
Can be zero	Not while the object is near a planet	Yes, the instant the object stops pressing on the surface
Newton's third law partner	The object's gravitational pull on the Earth	The object's push on the surface
On the AP Physics 1 equation sheet	No, $F_g = mg$ is a derived equation	No, it appears only inside $\lvert F_f \rvert \leq \lvert \mu F_N \rvert$
What a bathroom scale reads	Not this	This

Read the last row with the two rows above it and the rest of the page follows. Everything people call weight in everyday speech is the normal force, which is why the two words got welded together in the first place.

The case that separates them: tilt the surface

Put the same 12 kg crate on the same planet on five surfaces at different angles. The weight is $mg = (12\ \text{kg})(9.8\ \text{m/s}^2) = 117.6\ \text{N}$ in all five rows and never moves. Only the normal force responds.

Surface angle from horizontal	Normal force	Value	Fraction of the weight
$0^\circ$ , level ground	$mg$	117.6 N	1.000
$25^\circ$	$mg\cos 25^\circ$	106.6 N	0.906
$30^\circ$	$mg\cos 30^\circ$	101.8 N	0.866
$60^\circ$	$mg\cos 60^\circ$	58.8 N	0.500
$90^\circ$ , a vertical wall	0 from gravity alone	0 N	0.000

The cosine is not a decoration. Tilting the surface tilts the perpendicular direction away from vertical, so only part of the weight has to be balanced by the surface. The rest of the weight is still there, still 117.6 N in total, and the part the surface does not balance is what accelerates the crate down the slope.

That last row is the one worth staring at. Stand the surface up vertical and gravity produces no normal force at all, because gravity now has zero component perpendicular to the surface. Any normal force on a vertical wall has to come from something else pressing the object against it. The whole class of problems is worked in inclined plane problems.

They are not a Newton's third law pair

This is the error underneath the error. A block sits on a table, the weight points down, the normal force points up, they are equal, and it is enormously tempting to call them action and reaction. They are not, and the reason is one line: both of them act on the block.

A third law pair acts on two different objects. EK 2.2.A.1.i says a force exerted on an object is always due to the interaction of that object with another object, and the partner force is what the second object feels back. So:

Weight. The Earth pulls the block down with $mg$ . Partner: the block pulls the Earth up with $mg$ . That partner is a force on the Earth, and it never appears on the block's free-body diagram.
Normal force. The table pushes the block up with $F_N$ . Partner: the block pushes the table down with $F_N$ . That partner is a force on the table.

There is a clean test you can write in a justification. Third law pairs are equal in magnitude in every situation, with no exceptions. Weight and normal force are not equal in an elevator that is accelerating, so they cannot be a third law pair. One counterexample settles it.

Why this costs marks specifically: a free-response question that asks for the reaction to the normal force on the block wants "the downward force the block exerts on the table," and "the weight of the block" scores nothing. The two statements are not even close, because they name forces on different objects. Topic 2.3 is where the CED sets this up.

The elevator, and what the scale is actually reading

The CED gives this its own name. EK 2.6.C.1: the magnitude of the apparent weight of a system is the magnitude of the normal force exerted on the system. EK 2.6.C.2: if the system is accelerating, the apparent weight of the system is not equal to the magnitude of the gravitational force exerted on it.

So a bathroom scale never reads your weight. It reads the normal force it is exerting on you, which is your apparent weight, and those agree only when your vertical acceleration is zero. Take a 55 kg passenger, whose weight is $(55\ \text{kg})(9.8\ \text{m/s}^2) = 539\ \text{N}$ in every row of this table:

Elevator	Acceleration	Scale reads	Weight
At rest, or moving at constant velocity	0	539 N	539 N
Speeding up while going up, or slowing while going down	$2.4\ \text{m/s}^2$ upward	671 N	539 N
Slowing while going up, or speeding up while going down	$2.4\ \text{m/s}^2$ downward	407 N	539 N
Cable cut, in free fall	$9.8\ \text{m/s}^2$ downward	0 N	539 N

Rows two and three are where the marks go. The velocity never enters the calculation. An elevator moving downward and slowing has an upward acceleration, so the scale reads more, which feels backwards until you notice that only $a$ appears in $F_N - mg = ma$ .

The last row is EK 2.6.C.3, and it has two conditions, not one: a system appears weightless when there are no forces exerted on the system, or when the force of gravity is the only force exerted on it. Note the word appears. The 539 N of gravity is still acting; it is the normal force that has gone to zero. That is also the correct description of an astronaut in orbit.

When they do coincide, and why that lulls you

The normal force equals the weight when three conditions hold at the same time:

The surface is horizontal, so the perpendicular direction is vertical.
The acceleration has no vertical component.
No other force acting on the object has a vertical component.

A block on a level table satisfies all three, and so does most of the first chapter of any mechanics course. You can substitute $F_N = mg$ for weeks without being caught.

The first thing that catches you is friction, because $\lvert F_f \rvert \leq \lvert \mu F_N \rvert$ multiplies your error by the coefficient and carries it into the answer. After that, in rough order of how often they appear on the exam: inclines break condition 1, elevators and vertical circles break condition 2, and angled ropes, extra objects stacked on top, and people leaning on things break condition 3.

The reflex worth building is small. Never write $F_N = mg$ . Write the second law perpendicular to the surface and solve for $F_N$ . On level ground with $a_y = 0$ that produces $F_N = mg$ in a single line anyway, so the habit costs you nothing and saves you every time one of the three conditions quietly fails.

Where it costs a mark

Each of these is a real scoring event, not a philosophical point.

Writing $F_N = mg$ on an incline. The normal force is $mg\cos\theta$ . If the problem involves friction you now have two wrong numbers, because $F_f$ was built on $F_N$ .
Calling the normal force the reaction to the weight. Both act on the same object, so they are not a third law pair. Asked to name the partner of $F_N$ , say the force the object exerts on the surface.
Saying your weight changes in an elevator. Weight is $mg$ and does not move. What changes is the apparent weight, which EK 2.6.C.1 defines as the magnitude of the normal force.
Saying an astronaut in orbit has no weight. EK 2.6.C.3 says the system appears weightless, and it lists two ways that happens: no forces at all, or gravity as the only force. In orbit it is the second, which is a very different statement from gravity being absent.
Forgetting the normal force can hit zero. At the crest of a hill or the top of a loop, $F_N$ falls to zero at a specific speed, and past that speed the object leaves the surface. Worked below.
Drawing the normal force vertical on an incline. The Topic 2.2 boundary statement says individual forces on a free-body diagram must be drawn as individual straight arrows originating on the dot and pointing in the direction of the force, so a normal force drawn vertically is drawn wrong.
Drawing the weight and its two components as three arrows. The same boundary statement says AP Physics 1 only expects students to depict the forces exerted on objects, not the force components. Draw $F_g$ once, then resolve it in your algebra.
Using the normal force in a gravitational potential energy calculation. $\Delta U_g = mg\Delta y$ uses the weight. The normal force does no work at all when the object slides along the surface, because it stays perpendicular to the displacement.

What the equation sheet gives you, and what it does not

Neither force gets its own line on the AP Physics 1 equation sheet.

For the weight, the sheet prints the constants $g = 9.8\ \text{m/s}^2$ and $g = 9.8\ \text{N/kg}$ , and it prints the general gravitational force $\lvert \vec{F}_g \rvert = \frac{G m_1 m_2}{r^2}$ . The near-Earth form is a derived equation in the CED, listed under EK 2.6.A.3 as $\text{Weight} = F_g = mg$ , and it does not appear on the sheet. You write it yourself.

For the normal force, the sheet mentions $F_N$ once, inside the friction relation:

\lvert \vec{F}_f \rvert \leq \lvert \mu \vec{F}_N \rvert

which consumes the normal force and offers nothing about how to find it. There is no printed formula for $F_N$ anywhere on any of the four course sheets, because there could not be one: its value depends on the geometry, and geometry is what the sheet cannot know.

One more thing worth pinning down before you compute. A Topic 1.3 boundary statement says that for all situations in which a numerical quantity is required for $g$ , the value $g \approx 10\ \text{m/s}^2$ will be used, and then adds that students will not be penalized for correctly using the more precise commonly accepted values of $g = 9.81\ \text{m/s}^2$ or $g = 9.8\ \text{m/s}^2$ . The Table of Information prints 9.8. Every number on this page uses 9.8; is g 9.8 or 10 covers what that choice does to your arithmetic. The full sheet is here.

Same crate, tilted surface: the two forces come apart

A 12 kg crate rests on a frictionless ramp inclined at $25^\circ$ above the horizontal. Find the crate's weight and the normal force on it, state the difference between them, and find the crate's acceleration.

Set the convention. Take the axes along and perpendicular to the incline surface, with the positive direction down the slope. This is the choice EK 2.2.B.4 recommends, because the acceleration is along the slope.
Weight first, because it does not care about the ramp: $F_g = mg = (12\ \text{kg})(9.8\ \text{m/s}^2) = 117.6\ \text{N}$ , directed straight down. The ramp angle appears nowhere in that line.
Resolve the weight in the tilted axes. Perpendicular to the surface it contributes $mg\cos 25^\circ$ , and along the surface it contributes $mg\sin 25^\circ$ .
Apply the second law perpendicular to the surface. The crate does not accelerate into or off the ramp, so $a_\perp = 0$ and $F_N - mg\cos 25^\circ = 0$ . With $\cos 25^\circ = 0.9063$ , $F_N = (117.6\ \text{N})(0.9063) = 106.6\ \text{N}$ .
Compare. The weight is 117.6 N and the normal force is 106.6 N, a gap of 11.0 N. Nothing about the crate or the planet changed; the surface simply stopped being horizontal.
Now the along-slope direction, which is where the rest of the weight went. With $\sin 25^\circ = 0.4226$ , the unbalanced component is $mg\sin 25^\circ = (117.6\ \text{N})(0.4226) = 49.7\ \text{N}$ , so $a = 49.7\ \text{N} / 12\ \text{kg} = 4.14\ \text{m/s}^2$ down the slope. The mass cancels if you do it symbolically: $a = g\sin\theta = (9.8)(0.4226) = 4.14\ \text{m/s}^2$ .

Weight 117.6 N, normal force 106.6 N, and the crate accelerates down the slope at $4.14\ \text{m/s}^2$ . The 11.0 N gap is not weight that disappeared: the full 117.6 N is still acting. The surface only has to cancel the perpendicular part of it, and the 49.7 N left over along the slope is exactly what produces the acceleration.

A vertical wall, where the normal force has nothing to do with mg

A 3.0 kg book is held against a vertical wall by a horizontal push of 40 N. The book does not slide. Find the normal force from the wall on the book, the book's weight, the friction force holding it up, and the smallest coefficient of static friction that makes this possible.

Set the convention: positive $x$ is horizontally into the wall, positive $y$ is up. The wall's surface is vertical, so its perpendicular direction is horizontal and its normal force points horizontally, out of the wall.
Horizontal second law. The book does not accelerate into or away from the wall, so $F_N - 40\ \text{N} = 0$ and $F_N = 40\ \text{N}$ . The mass of the book never entered that line, and neither did $g$ .
Weight, unchanged by any of this: $F_g = (3.0\ \text{kg})(9.8\ \text{m/s}^2) = 29.4\ \text{N}$ , straight down. Note that it is perpendicular to the normal force here, not opposite it.
Vertical second law. Nothing else is vertical except friction, and the book is at rest, so $F_f - 29.4\ \text{N} = 0$ and static friction supplies 29.4 N upward. Friction is holding the book up, and friction is pointing up, which is the direction a static friction force takes whenever that is what prevents slipping.
Smallest coefficient. Static friction cannot exceed $\mu_s F_N$ , so $29.4\ \text{N} \leq \mu_s (40\ \text{N})$ , giving $\mu_s \geq 29.4/40 = 0.735$ .
Sanity check on the direction of the answer. Push harder and $F_N$ rises, so the friction ceiling rises while the 29.4 N requirement stays put. That matches the experience of pressing a book harder to stop it sliding down a wall.

Normal force 40 N horizontal, weight 29.4 N downward, friction 29.4 N upward, and $\mu_s \geq 0.735$ . The normal force equals the applied push, not $mg$ , and it is at right angles to the weight rather than opposite it. Any formula of the form $F_N = mg$ would have produced 29.4 N here, which is wrong in both magnitude and direction. The static and kinetic friction split is what decides which friction rule applies.

Over the crest of a hill, where the normal force runs out

A 1200 kg car drives over the crest of a hill that is the arc of a circle of radius 45 m. At the top the car is moving at 15 m/s. Find the car's weight and the normal force from the road, then find the speed at which the car would just lose contact with the road.

Set the convention: at the crest, take the positive direction as pointing toward the center of the circle, which is straight down. The car is moving in a circle, so its acceleration at the top is centripetal, $a_c = v^2/r$ , and points that way.
Weight: $F_g = (1200\ \text{kg})(9.8\ \text{m/s}^2) = 11760\ \text{N}$ , straight down. The hill's curvature does not change it.
Required net force: $ma_c = m v^2 / r = (1200\ \text{kg})(15\ \text{m/s})^2 / (45\ \text{m}) = (1200)(225)/45 = 6000\ \text{N}$ , directed downward.
Second law in the chosen direction. Down is positive, so $F_g - F_N = m v^2 / r$ , giving $F_N = 11760\ \text{N} - 6000\ \text{N} = 5760\ \text{N}$ .
Compare the two. The normal force is 5760 N against a weight of 11760 N, so the road is pushing with 49 percent of the weight and the passengers feel light. The weight did not change by a single newton.
Losing contact means $F_N = 0$ , which leaves gravity as the only force and forces it to supply the whole centripetal requirement: $mg = m v^2 / r$ , so $v = \sqrt{gr} = \sqrt{(9.8\ \text{m/s}^2)(45\ \text{m})} = \sqrt{441} = 21.0\ \text{m/s}$ .
Check the mass cancellation. The 1200 kg dropped out of that last line, so every vehicle leaves this crest at the same 21.0 m/s regardless of how heavy it is.

Weight 11760 N, normal force 5760 N, and contact is lost at 21.0 m/s. This is the cleanest demonstration that the two are separate quantities: one of them can be driven all the way to zero while the other sits unchanged at 11760 N. Push past 21.0 m/s and the road cannot help at all, because a surface can push but never pull. The circular-motion side of this is in centripetal force.

Frequently asked questions

Is the normal force always equal to the weight?

No. They are equal only when three conditions hold together: the surface is horizontal, the object has no vertical acceleration, and no other force on the object has a vertical component. Break any one and they part company. On an incline of angle theta the normal force is mg cos theta, which is smaller. In an elevator accelerating upward it is m(g + a), which is larger. On a vertical wall it has nothing to do with mg at all and equals whatever is pressing the object against the wall. The safe habit is to write Newton's second law perpendicular to the surface and solve for the normal force every time, rather than assuming it equals mg.

Are weight and normal force a Newton's third law action-reaction pair?

No, and this is one of the most common free-body-diagram errors. Both forces act on the same object, while a third law pair always acts on two different objects. The partner of the weight is the gravitational pull the object exerts on the Earth. The partner of the normal force is the push the object exerts on the surface. A quick proof: third law pairs are equal in magnitude in every situation without exception, but the normal force and the weight are unequal in an accelerating elevator, so they cannot be a pair.

Why is the normal force mg cos theta on an incline?

Because the normal force is perpendicular to the surface, and tilting the surface tilts that perpendicular direction away from vertical. Only the component of gravity perpendicular to the incline, mg cos theta, has to be cancelled by the surface, since the object has no acceleration into or off the ramp. The remaining component, mg sin theta, is left unbalanced along the slope, which is why the object accelerates down it at g sin theta. The weight itself is still the full mg; it is only the part the surface has to handle that shrinks with the cosine.

Does your weight change in an elevator?

No. Weight is the gravitational force mg, and neither your mass nor Earth's gravitational field changes when an elevator starts moving. What changes is the normal force from the floor or the scale, which the AP Physics 1 CED calls your apparent weight in EK 2.6.C.1. Accelerating upward at a gives a reading of m(g + a); accelerating downward at a gives m(g - a); in free fall the reading is zero. The elevator's direction of travel is irrelevant, because only the acceleration appears in the equation, so an elevator moving down and slowing makes the scale read more, not less.

Is the normal force always vertical, or always upward?

Neither. The normal force is perpendicular to the surface the object is touching and points away from that surface, so its direction is fixed by the surface rather than by gravity. On an incline it tilts with the ramp. On a vertical wall it is horizontal. On the inside of the top of a loop it points straight down, toward the center of the loop. The word normal in the name means perpendicular, not upward, and reading it as upward is what produces vertically drawn normal arrows on incline diagrams.

What is apparent weight, and how is it different from weight?

Apparent weight is what a scale reads, and the AP Physics 1 CED defines its magnitude as the magnitude of the normal force exerted on the system, in EK 2.6.C.1. Weight is the gravitational force mg. The two agree only when there is no vertical acceleration; EK 2.6.C.2 states that if the system is accelerating, its apparent weight is not equal to the magnitude of the gravitational force exerted on it. This is why everyday language and physics language collide here: what most people mean by their weight is technically their apparent weight, which is a normal force.

Can the normal force be zero while the weight is not?

Yes, and it is a standard exam setup. Any time an object stops pressing on a surface the normal force vanishes while gravity keeps acting: an elevator in free fall, a car at the crest of a hill above a critical speed, an object at the top of a vertical loop moving at the minimum speed, an astronaut in orbit. In every one of those cases the weight is unchanged and the normal force is zero. EK 2.6.C.3 describes this as appearing weightless, and the wording matters, because gravity is not absent. A surface can push but never pull, so the normal force can reach zero but cannot go negative.