Definitions

# Normal force

In physics, the normal force $F_n$ (or in some books N) is the component, perpendicular to the surface of contact, of the contact force exerted by, for example, the surface of a floor or wall, on an object, preventing the object from entering the floor or wall. In a static situation it is just enough to balance the force with which the object pushes, e.g. its weight on the floor, or a smaller force if somebody leans against a wall. If an object hits the surface with some speed, the normal force provides for a rapid negative acceleration, depending on how flexible the floor/wall is (and, of course, if it can provide enough force for stopping the object instead of breaking). Also, if the object is soft, only the outer part needs to decelerate rapidly, the inner part can do that more gradually, while the layer in between is compressed.

In general, the magnitude of the normal force is the projection of the surface traction, T, in the normal direction, n, and so the normal force vector can be found by scaling the normal direction by that force. The surface traction, in turn, is equal to the dot product of the unit normal with the stress tensor describing the stress state of the surface. That is,

$mathbf\left\{F\right\}=mathbf\left\{n\right\}, F = mathbf\left\{n\right\}, \left(mathbf\left\{T\right\}cdot mathbf\left\{n\right\}\right) = mathbf\left\{n\right\}, \left(mathbf\left\{n\right\}cdot mathbf\left\{tau\right\} cdot mathbf\left\{n\right\}\right).$
Or, in indicial notation,
$F_i = n_i F = n_i T_j n_j = n_i n_k tau_\left\{jk\right\} n_j.$

## Frictional force

The parallel shear component of the contact force is known as the frictional force ($F_fr$).

## Example

In a simple case such as a 40 kg object resting upon a table, the normal force on the object is equal but in opposite direction to the gravitational force applied on the object i.e. the weight of the object. In this case the normal force is given by, 40 kg · 9.81 m/s2=392.4 newtons where 9.81 m/s2 is equal to the acceleration due to gravity (near the Earth's surface).

In another case where the same object as mentioned above is on a 40 degree incline, we have to insert cos θ into the equation for normal force. Fnormal = mass · gravity · cos θ. So solving for the normal force, we get: FN = 40kg · 9.81m/s2 · cos 40° = 300.6 newtons

## Real-world applications

For a person standing in an elevator moving with constant velocity (including stopped), the normal force on the person's feet balances the person's weight. In an elevator that is accelerating upward, the normal force is greater than the weight and so the person's apparent weight increases (making the person feel heavier). In an elevator that is accelerating downward, the normal force is less than the weight and so a user's apparent weight decreases. If a user were to stand on a "weight scale", such as a conventional bathroom scale, onto the elevator, the scale would read either more or less than the person's actual weight when the elevator is accelerating up or down (respectively) because weight scales measure normal force (which varies as the lift accelerates), not gravitational force (which does not).

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