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 $F_n$F, start subscript, n, end subscript or just $N$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.

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.

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.

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 $m$m that's sitting on a table at rest, as seen below.

Following the procedure we get,

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

The normal force will not always equal $mg$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 $mg$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.

A $4.5\text{ kg}$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 $a=3.0\dfrac{\text m}{\text{s}^2}$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\text{ N}$5, start text, space, N, end text.

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 $F_n=mg=44.1 \text{ N}$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 $mg$m, g since there was a vertical acceleration and an additional vertical force.

A person is pushing a $1.0 \text{ kg}$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 $F_A=10\text{ N}$F, start subscript, A, end subscript, equals, 10, start text, space, N, end text at an angle of $\theta=30^o$theta, equals, 30, start superscript, o, end superscript as seen below.

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.