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## Physics library

### Course: Physics library > Unit 3

Lesson 2: Normal force and contact force# What is normal force?

When two objects touch, they exert a force on each other.

## What is normal force?

Ever turn too quickly and walk straight into a wall? I have. It hurts and makes me feel dumb. We can blame the

**normal force**for the pain we feel when running into solid objects. The normal force is the force that surfaces exert to prevent solid objects from passing through each other.Normal force is a contact force. If two surfaces are not in contact, they can't exert a normal force on each other. For example, the surfaces of a table and a box cannot exert normal forces on each other if they are not in contact.

However, when two surfaces are in contact (e.g. a box and a table) they exert a normal force on each other, perpendicular to the contacting surfaces. This normal force will be as large as necessary to prevent the surfaces from penetrating each other.

The word "normal" in

**normal force**is not referring to ordinary or commonplace. The "normal" here refers to**perpendicular.**This is because the normal force, usually represented with F, start subscript, n, end subscript or just N, is a force that is directed perpendicular to the two surfaces in contact. It makes sense that the force is perpendicular to the surface since the normal force is what prevents solid objects from passing through each other. Surfaces can also exert contact forces in the direction parallel to the surfaces, but we would typically call those forces frictional forces (since they work to prevent the surfaces from sliding across each other) instead of calling them normal forces.## How do inanimate surfaces "know" to exert a normal force?

It makes sense for most people that a person would have to exert an upward force with their hands when carrying a heavy bag of dog food as seen in Figure 3(a) below.

But some people find it hard to believe that an inanimate object like a table can exert an upward normal force on a bag of dog food as seen in Figure 3(b) seen below. Sometimes people believe that the table is not really exerting an upward force at all, but merely "getting in the way" of the dog food falling down. But that's not how Newton's laws work. If there was only a downward force of gravity on the dog food, the dog food would have to accelerate downward. The table must do more than "get in the way". The table must exert an upward force to prevent the dog food from falling through the table.

Strangely, if a heavier object is placed on a table, the table must exert more normal force to prevent the weight from passing through the table. How does the table know to exert just the right amount of force to prevent the object from passing through it?

Essentially, the table "knows" how much force to exert based on how much the surface/object is compressed or deformed. When solid objects deform they typically try to restore themselves and "spring back" to their natural shape. The heavier the weight, the greater the deformation, the greater the restoring force trying to bring the surface back to its natural shape. This deformation would be noticeable if the load were placed on a card table, but even rigid objects deform when a force is applied to them. Unless the object is deformed beyond its limit, it will exert a restoring force much like a deformed spring (or trampoline or diving board). So when the load is placed on the table, the table sags until the restoring force becomes as large as the weight of the load. At this point the net external force on the load is zero. That is the situation when the load is stationary on the table. The table sags quickly, and the sag is slight so we typically do not notice it.

*Figure 3: (a) The person holding the bag of dog food must supply an upward force F, start subscript, start text, h, a, n, d, end text, end subscript equal in magnitude and opposite in direction to the weight of the food W. (b) The card table sags when the dog food is placed on it, much like a stiff trampoline. Elastic restoring forces in the table grow as it sags until they supply a normal force N or F, start subscript, n, end subscript equal in magnitude and opposite in direction to the weight of the load. (Image Credit: Openstax College Physics)*

## How do you solve for normal force?

There isn't really a formula made specifically for finding the normal force. To find the normal force we typically use the fact that we know something about the acceleration perpendicular to the surfaces (since we assume the surfaces can't pass through each other). Therefore, we almost always use Newton's second law to solve for normal force by using this strategy.

- Draw a force diagram showing all forces acting on the object in question.
- Choose the direction for Newton's second law in the same direction as the normal force (i.e. perpendicular to the contacting surfaces)
- Plug in the acceleration, mass, and forces acting, into Newton's second law left parenthesis, a, equals, start fraction, \Sigma, F, divided by, m, end fraction, right parenthesis for that direction.
- Solve for the normal force F, start subscript, n, end subscript.

Essentially we are solving for normal force by assuming the normal force will be as large or small as it needs to be to prevent the surfaces from penetrating each other.

Let's apply this strategy to the following simple example. Consider the simple case of a box of mass m that's sitting on a table at rest, as seen below.

Following the procedure we get,

0, equals, start fraction, start color #e84d39, F, start subscript, n, end subscript, end color #e84d39, minus, start color #6495ed, F, start subscript, g, end subscript, end color #6495ed, divided by, m, end fraction, start text, left parenthesis, p, l, u, g, space, i, n, space, v, e, r, t, i, c, a, l, space, a, c, c, e, l, e, r, a, t, i, o, n, comma, space, a, n, d, space, v, e, r, t, i, c, a, l, space, f, o, r, c, e, s, right parenthesis, end text

In this simple case of an object sitting on a horizontal surface, the normal force will be equal to the force of gravity F, start subscript, n, end subscript, equals, m, g.

**The normal force will not always equal m, g**. If we consider a more complicated case where the contact surface is not horizontal, or there are extra vertical forces present, or there is vertical acceleration, the normal force will not necessarily equal m, g. However, even in a more complicated case, we would still solve for the normal force using the process shown above. We might plug in a different acceleration, or there might be more forces to include, but the overall

**problem solving strategy of finding normal force**using Newton's second law would still be the same.

## What do solved examples look like involving normal force?

### Example 1: Elevator normal force

A 4, point, 5, start text, space, k, g, end text package of kiwi flavored bubble gum is being delivered to the top floor of an office building. The box sits on the floor of an elevator which accelerates upward with an acceleration of magnitude a, equals, 3, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction. The delivery person is also resting one foot on the package exerting a downward force on the package of magnitude 5, start text, space, N, end text.

**What is the normal force on the package exerted by the floor of the elevator?**

First we draw a force diagram showing all forces on the package (we don't include acceleration in the diagram since acceleration is not a force. Also, we don't include an extra

*elevator force*since the normal force*is*the force exerted on the box by the elevator).Note, if we had just naively used F, start subscript, n, end subscript, equals, m, g, equals, 44, point, 1, start text, space, N, end text we would have found the wrong answer. The normal force here is different from m, g since there was a vertical acceleration and an additional vertical force.

### Example 2: Normal force with diagonal force

A person is pushing a 1, point, 0, start text, space, k, g, end text box of mint chocolate chip cookies across a frictionless table with a downward diagonal force F, start subscript, A, end subscript, equals, 10, start text, space, N, end text at an angle of theta, equals, 30, start superscript, o, end superscript as seen below.

**What is the normal force exerted on the box of cookies by the table?**

Even though this seems like a different type of problem, we attack it with the same strategy as before. First we draw a force diagram of all the forces acting on the box.

## Want to join the conversation?

- it is said that even rigid object would be deformed during contact, does that mean all objects are naturally elastic?(21 votes)
- What happens to a rigid object depends on the atomic and molecular forces. If these forces allow the object to undergo elastic deformation then they will deform and then go back to their original shape once the force is removed. If the forces between the atoms and molecules shift without breaking then the object undergoes plastic deformation and stays in the new shape. If the forces do not stretch or shift then the object cracks. Most objects that we conciser rigid will react to a force on them with a combination of all three responses.(35 votes)

- In Example 1 : why we didn't take the force that causes the elevator to accelerate in consideration?????(13 votes)
- Because we're just calculating for the things inside the elevator not the elevator itself, and since the objects we're calculating the normal for are in the elevator, they automatically get the acceleration and velocity of the elevator.(0 votes)

- Hi, when I solved for example one, I used a as -9.8 m/s2, because isn't acceleration from gravity always downward and in this case to the opposite of upward (considered positive) direction, thus should be negative ? I got my answer wrong because of this and don't understand why the plug a as a positive value. Please clarify. Thank you.(4 votes)
- There is NO acceleration because the table is pushing up on the object with as much as the object is pushing on the table. The object is not accelerating at all, so a is equal to zero.(12 votes)

- Isn't the normal force same as the reaction force in Newton's 3rd Law of Motion?(8 votes)
- in this context, 'normal' just means perpendicular to a line or surface(1 vote)

- Does the normal force exist due to the deformation of the object exerting the normal force, or the repulsion between the electrons in the two objects?(5 votes)
- i'm not sure but i think repulsion between electrons is the root cause of both

when we concern the shape of contacting objects, we tag them deformed by that very repulsion

when it comes to forces (and thus acceleration), we see them exerting normal forces each other by the same cause of repulsion

in other words, the two (deformation, normal force) might be 2 symptoms from 1 cause (electron repulsion)(2 votes)

- In the first example why is 13.5N not negative since it was subtracted to be placed on the other side of the equation.(4 votes)
- It is algebra.

Ok, since you are confused, I will explain step by step.

It says 13.5N=Fn-mg-5N

We will solve for Fn

-Fn=-mg-5N-13.5N Multiply by -1

Fn=mg+5N+13.5N

Guess what? it is positive 13.5N!(4 votes)

- So I've asked this question in the normal force video about the shoe on the floor, but the same type of problem came up in this article. In the first question, the box of bubble gum is inside an elevator that is accelerating upward. If we assumed the elevator wasn't moving, then the normal force exerted by the elevator floor onto the box would equal in magnitude the downward forces acting on the box of bubble gum. What I don't understand is, why would the normal force of the floor change at all if the elevator was accelerating? It's not like the elevator floor is exerting more force to move upwards; it's the cables attached to the elevator that are exerting more force to accelerate isn't it?(2 votes)
- The normal force must change because now the floor must not only oppose gravity but also provide enough extra force to accelerate the object upward.

To understand the motion of the gum we don't need to know anything about the cables, we just need to know that the gum is getting pushed on by the floor. How the floor is managing to do that is irrelevant (could be cables, could be hydraulic, could be giant magic hand reaching down from sky...)(5 votes)

- After going over the problem again, is it because we can ignore the horizontal force because the normal force in this question only applies to vertical forces, therefore making F sub A of x irrelevant?(2 votes)
- Normal force is perpendicular to the surface of contact, so yes.(5 votes)

- Another question I had was if the normal force was greater than that of the force of gravity then wouldn't the object be floating in the air?(3 votes)
- If it is floating in the air, there can't be a normal force on it.(3 votes)

- Could it be considered that when an object is in free fall it could be perpendicular to the air? I was wondering if that could be what causes air resistance at higher altitudes.(1 vote)
- You could think of it that way but it's not going to help you figure out the force.(5 votes)