Friction

When two surfaces slide against each other, a dissipative force called friction acts to resist their motion. Although the degree of friction can vary widely between two different pairs of surfaces, all surfaces exhibit some degree of friction.

For instance, when a book is slid across a table, it immediately starts to slow down until it comes to a rest. Molecular forces between the surface of the book and the table resist sliding and exert a force opposite the direction of the object's motion.

Similarly, fluids can also exert friction: when a skydiver jumps out of an airplane, the magnitude of her acceleration immediately begins to decrease. As the skydiver falls through the air, she must transfer energy to air molecules to push them out of the way. Fluid-based friction is generally called drag or fluid resistance (or air resistance in the case of air) and is discussed separately. This article focuses on the friction forces between solid surfaces.

Although friction fundamentally results from complex interactions at the molecular level, for many everyday macroscopic materials, a simple intuitive model for friction suffices. Consider the static situation of a heavy box placed on a floor. In an attempt to move the box, suppose that a gradually increasing horizontal force of magnitude $F_\text{push}$ is applied to the box. As long as the box remains stationary, the net force on the box must be zero, and the friction force $f_\text{static}$ is exactly equal and opposite to that of the force applied to the box. In this case, the friction force is said to be static because the object is not moving: $F_\text{push} = f_\text{static}$.

In other words, as long as $F_\text{push}$ is below some "critical force" $F_\text{crit}$ required to dislodge the box, the static friction force simply acts against the applied force.

Empirically, one generally observes that the critical force is directly proportional to the magnitude of the normal force $N$ of the box against the floor. The constant of proportionality is called the coefficient of static friction as is generally symbolized by the Greek letter mu $\mu_s$ (with a subscript to indicate that the friction is static):

$$F_\text{crit} = \mu_s N.$$

Coefficients of friction are defined between pairs of surfaces. The coefficient of friction for metal on glass is about $0.1$, while the coefficient of friction for rubber on asphalt is about $0.7$.

Intuitively, objects with more normal force ought to have a higher critical force, as they should be more difficult to dislodge. But why is there no dependence on surface area? It might seem as if objects with a greater surface area might have more contact area with which to form friction-causing interactions. In general, it turns out that the effect of surface area is usually negligible for the simple reason that increasing the surface area decreases the normal force per unit area; in the end, what matters is the total normal force per unit area summed over the entire area. No matter what "juggling" of the normal force distribution is made, the total normal force is the same.

It is worth noting that the normal force exerted on the box need not simply be equal to its weight; additional forces can be applied to the box, which can increase or decrease the size of the critical force.

As the applied force increases, at some point $F_\text{push}$ eventually exceeds $F_\text{crit}$ and the object is dislodged. Here, the nature of the interaction between the two surfaces changes. Once the object starts moving, the nature of the molecular interactions between the two surfaces is no longer the same as it was during the static configuration. Instead, the friction is said to be kinetic.

The empirical observation is that the magnitude of the kinetic friction force $f_\text{kinetic}$ is constant and directly proportional to the magnitude of the normal force $N$:

$$f_\text{kinetic} = \mu_k N.$$

The constant of proportionality is the coefficient of kinetic friction $\mu_k$. Note that $f_\text{kinetic} \leq F_\text{critical}$, or else it would be impossible to get the box moving. As a result, $\mu_k \leq \mu_s$. In other words, it requires less force to push an object that is already moving than to push a static object.

The kinetic friction force continues to act in a direction opposite the direction of the object's motion so long as it continues to move, even if the applied force is removed. Kinetic friction is responsible for bringing moving objects to a halt.

Note that in both the static and kinetic cases, friction never moves an object on its own. Static friction simply provides the minimum force required to cancel the applied force, while kinetic friction provides a constant force to an already moving object.

As an object slides across a rough surface and slows down, friction does work against the object. To investigate where the dissipated energy goes, consider that at the molecular level atoms collide and weak bonds are broken, which leads to increased molecular motion. Ultimately, this results in heating of the two surfaces and the surrounding air.

In the case of constant kinetic friction, the magnitude of the total energy loss due for an object that slides a distance of $d$ is simply given by

$$W_\text{fric} = \mu_k N d.$$

A box moving at an initial speed of $v$ slides over a floor. The coefficient of kinetic friction between the box and the floor is $\mu_k$. Over what distance does the box slide before it comes to rest?

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The simplest solution is to note that all of the initial kinetic energy of the box is ultimately dissipated by friction into heat, so it must be the case that the stopping distance $d$ is given by

$$\frac{1}{2} mv^2 = \mu_k m g d,$$

so

$$d = \frac{v^2}{2 \mu_k g}.$$