Normal force

In mechanics, the normal force ${\displaystyle F_{n}\ }$ is the component of a contact force that is perpendicular to the surface that an object contacts.[1] For example, the surface of a floor or table that prevents an object from falling. In this instance normal is used in the geometric sense and means perpendicular, as opposed the common language use of normal meaning common or expected. For example, a person standing still on flat ground is supported by a ground reaction force that consists of only a normal force. If the person stands on a slope and does not slide down it, then the total ground reaction force can be divided into two components: a normal force perpendicular to the ground and a frictional force parallel to the ground. In another common situation, if an object hits a surface with some speed, and the surface can withstand it, the normal force provides for a rapid deceleration, which will depend on the flexibility of the surface and the object.

F_N represents the normal force

Equations

Weight (W), the frictional force (F_r), and the normal force (F_n) acting on a block. Weight is the product of mass (m) and the acceleration of gravity (g).

In a simple case such as an 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 (or the weight of the object), that is, ${\displaystyle N=mg}$, where m is mass, and g is the gravitational field strength (about 9.81 m/s² on Earth). The normal force here represents the force applied by the table against the object that prevents it from sinking through the table and requires that the table is sturdy enough to deliver this normal force without breaking. However, it is easy to assume that the normal force and weight are action-reaction force pairs (a common mistake). In this case, the normal force and weight need to be equal in magnitude to explain why there is no upward acceleration of the object. For example, a ball that bounces upwards accelerates upwards because the normal force acting on the ball is larger in magnitude than the weight of the ball.

Where an object rests on an incline, the normal force is perpendicular to the plane the object rests on. Still, the normal force will be as large as necessary to prevent sinking through the surface, presuming the surface is sturdy enough. The strength of the force can be calculated as:

${\displaystyle N=mg\cos(\theta )}$

where N is the normal force, m is the mass of the object, g is the gravitational field strength, and θ is the angle of the inclined surface measured from the horizontal.

The normal force is one of the several forces which act on the object. In the simple situations so far considered, the most important other forces acting on it are friction and the force of gravity.

Using vectors

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

${\displaystyle \mathbf {N} =\mathbf {n} \,N=\mathbf {n} \,(\mathbf {T} \cdot \mathbf {n} )=\mathbf {n} \,(\mathbf {n} \cdot \mathbf {\tau } \cdot \mathbf {n} ).}$

or, in indicial notation,

${\displaystyle \ N_{i}=n_{i}N=n_{i}T_{j}n_{j}=n_{i}n_{k}\tau _{jk}n_{j}.}$

The parallel shear component of the contact force is known as the frictional force (${\displaystyle F_{f}r\ }$).

The static coefficient of friction for an object on an inclined plane can be calculated as follows:[2]

${\displaystyle \mu _{s}=\tan(\theta )}$

for an object on the point of sliding where ${\displaystyle \theta }$ is the angle between the slope and the horizontal.

Physical origin

As explained before, the normal force is a contact force, that is, the resultant of the forces between the constituents of the actual constraint, and the constituents of the object. When the object and the constraint (a wall, a table, another object, or any surface) are pressed against each other, the van der Waals force, a force that grows very large very quickly, opposes this and for this reason, the two objects cannot penetrate each other. If the two objects are then separated, even by a very small distance, the normal force disappears. This is because the intermolecular force quickly vanishes past the point of equilibrium.[3]

Applications in real life

In an elevator either stationary or moving at constant velocity, 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 person's ground weight and so the person's perceived weight increases (making the person feel heavier). In an elevator that is accelerating downward, the normal force is less than the person's ground weight and so a passenger's perceived weight decreases. If a passenger were to stand on a weighing scale, such as a conventional bathroom scale, while riding the elevator, the scale will be reading the normal force it delivers to the passenger's feet, and will be different than the person's ground weight if the elevator cab is accelerating up or down. The weighing scale measures normal force (which varies as the elevator cab accelerates), not gravitational force (which does not vary as the cab accelerates).

When we define upward to be the positive direction, constructing Newton's second law and solving for the normal force on a passenger yields the following equation:

${\displaystyle N=m(g+a)}$

In a gravitron ride, the static friction caused by and perpendicular to the normal force acting on the passengers against the walls results in suspension of the passengers above the floor as the ride rotates. In such a scenario, the walls of the ride apply normal force to the passengers in the direction of the center, which is a result of the centripetal force applied to the passengers as the ride rotates. As a result of the normal force experienced by the passengers, the static friction between the passengers and the walls of the ride counteracts the pull of gravity on the passengers, resulting in suspension above ground of the passengers throughout the duration of the ride.

When we define the center of the ride to be the positive direction, solving for the normal force on a passenger that is suspended above ground yields the following equation:

${\displaystyle N={\frac {mv^{2}}{r}}}$

where ${\displaystyle N}$ is the normal force on the passenger, ${\displaystyle m}$ is the mass of the passenger, ${\displaystyle v}$ is the tangential velocity of the passenger and ${\displaystyle r}$ is the distance of the passenger from the center of the ride.

With the normal force known, we can solve for the static coefficient of friction needed to maintain a net force of zero in the vertical direction:

${\displaystyle \mu ={\frac {mg}{N}}}$

where ${\displaystyle \mu }$ is the static coefficient of friction, and ${\displaystyle g}$ is the gravitational field strength.

References

1. Physics, Pearson IIT Foundation Series (2018). Pearson IIT Foundation Series: Physics. India: Pearson. pp. 3.1–3.37. ISBN 9789353430917.
2. Nichols, Edward Leamington; Franklin, William Suddards (1898). The Elements of Physics. 1. Macmillan. p. 101.
3. Bettini, Alessandro. A Course in Classical Physics 1 - Mechanics. Springer. p. 110. ISBN 978-3-319-29256-4.