Friction

Friction is the contact force a surface exerts parallel to the surface. It always opposes the sliding (or attempted sliding) between the two surfaces, and it comes in two kinds.

Static friction acts while the surfaces are not slipping. It is not a fixed number: it adjusts to balance other forces, up to a maximum.

f_s \le \mu_s N.

Kinetic friction acts once the surfaces are sliding, and has a single value:

f_k = \mu_k N.

The coefficients $\mu_s$ and $\mu_k$ describe the pair of surfaces, and typically $\mu_k < \mu_s$ . Both laws scale with the normal force $N$ , which quantifies how hard the surfaces press together. Note that the contact area of the surfaces does not appear in either case, and kinetic friction does not depend on how quickly the surfaces slide past each other.

Click and drag to pull Newt. While he sits still, friction exactly balances your applied force. When the static maximum is reached, Newt begins to slide, and the friction drops to its kinetic value. Check unlock angle to also push Newt down into the floor or lift Newt upward: the normal force changes, along with the corresponding friction.

roughness0.5weight20 Nunlock angle

These are empirical rules, found by measuring various surfaces sliding (or almost sliding) past each other. Really understanding how friction arises is a rather deep rabbit hole. If you’re interested in peeking down it, the remainder of this page uses a similar microscopic picture used for the normal force and tension. Consider the following: Why should friction depend on $N$ ? Why should it not depend on area? And what physical property of the two surfaces is a number like $\mu$ measuring?

Microscopic Modeling

Many apparently smooth surfaces are rough when magnified. A surface is a landscape of tiny peaks, called asperities, and when two solids meet, they touch only where peaks from one meet the other. For ordinary solids the true contact area is a tiny fraction (often well under 1%) of the apparent area of contact. The picture below is magnified, and exaggerates the touching fraction so that you can see it. For simplicity, we’ll leave the top surface as smooth.

Where the peaks touch, the atoms are pressed together like the springs in the normal-force picture, but these peaks carry a lot of load across a very small area. The pressure can be large enough to flatten them. A solid can only withstand so much pressure before it yields, and that limit is the hardness $H$ of the softer material. Each tip therefore flattens until it just supports its share of the load. The true contact area is proportional to the load and inversely proportional to the hardness:

A_{\text{true}} = \frac{N}{H}.

load N40hardness1.0block width (%)100

Note that the actual area of the block doesn’t matter. If it decreases in size, there are fewer asperities in contact, and the higher pressure flattens them more.

Along this effective area, the atoms in the two surfaces are close enough to temporarily bond. As with the normal force and tension, we’ll model this bond interaction with imaginary springs. As we begin to slide the surfaces, these temporary bonds will stretch until breaking.

Below, a block is held against the floor while the floor is dragged along beneath it. Each bond stretches as the floor moves, and eventually breaks at some limit. New bonds may form when a floor atom passes under a block atom, so a freshly formed bond starts out unstretched. The block feels the sum of all the bonds’ pulls.

The above trace illustrates the difference between static and sliding friction:

The static peak. A block that has been resting has all of its bonds formed and unstretched. They are stretched together until the weakest begin to give, and the total force reaches a maximum just before the bonds start letting go.
The sliding average. Once sliding, the bonds are in different stages: some freshly formed and barely stretched, some about to break. On average they carry less than they did when all were fresh and in place together. This is why kinetic friction is smaller than static friction.

Finally, let’s combine the ideas of true contact area and atomic adhesion. The shear strength (per unit of contact area) $\tau$ depends on how easily these bonds break. The maximum friction is given by:

f_{\max} = \tau A_{\text{true}} = \tau \frac{N}{H} = \mu_s N, \qquad \mu_s \approx \frac{\tau}{H}.

So the coefficient of friction depends on two properties of the interface: the strength of bonds between the two surfaces, and the hardness of the softer one. A clean, strongly bonding pair of metals has a large $\tau$ and a large $\mu$ . Ice, Teflon, or a layer of lubricant gives weak bonds and a small $\mu$ . A soft material flattens more under a given load, increasing the contact and so the friction.

A few cautions go with the model:

The rules $f_s \le \mu_s N$ and $f_k = \mu_k N$ are approximations. Rubber and other soft polymers, lubricated and very clean or very smooth surfaces, and high speeds all depart from them.
The model gives $\mu$ a deeper physical meaning, but $\mu$ is still an empirical value, measured for a given pair of surfaces in given conditions.
The static peak depends on how long the surfaces have been at rest, since bonds can strengthen with time.

Problem Solving

Starting and Keeping a Crate Moving

Problem

A $25\ \text{kg}$ crate rests on a level floor with $\mu_s = 0.40$ and $\mu_k = 0.30$ . What horizontal push is needed to start it moving? What push keeps it moving at constant speed? Use $g = 9.8\ \text{m/s}^2$ .

Use

The vertical forces balance, so $N = mg$ . To start the crate, the push must just exceed the maximum static friction. Once it slides at constant velocity, $\sum F_x = 0$ and the push equals the kinetic friction.

F_{\text{start}} = \mu_s N, \qquad F_{\text{steady}} = \mu_k N.

Solve

With $N = (25)(9.8) = 245\ \text{N}$ ,

F_{\text{start}} = (0.40)(245) = 98\ \text{N}, \qquad F_{\text{steady}} = (0.30)(245) = 73.5\ \text{N}.

Check

Starting takes more than keeping going, as it should since $\mu_s > \mu_k$ . If you keep pushing at $98\ \text{N}$ after it slips, the net force is $24.5\ \text{N}$ and the crate speeds up at $1.0\ \text{m/s}^2$ .

Pressing a Block Against a Wall

Problem

You hold a $2.0\ \text{kg}$ block against a vertical wall by pushing on it horizontally. The coefficient of static friction between the block and the wall is $0.50$ . What is the smallest push that keeps the block from sliding down?

Use

Draw the block as the system. Horizontally, the wall’s normal force balances your push, so $N = F$ . Vertically, static friction must hold up the weight:

f_s = mg, \qquad f_s \le \mu_s N = \mu_s F.

Solve

The least push has friction at its maximum, $\mu_s F = mg$ :

F = \frac{mg}{\mu_s} = \frac{(2.0)(9.8)}{0.50} = 39\ \text{N}.

Check

The weight is not what sets $N$ here. The push is, which is why a harder squeeze is what stops the block from slipping. At $39\ \text{N}$ the friction is $19.6\ \text{N}$ , equal to the weight, so the net force is zero.

Same Weight, Different Face

Problem

A $3.0\ \text{kg}$ brick sits on a table with $\mu_s = 0.40$ . How much horizontal force is needed to start it sliding when it lies flat? When it stands on its small end?

Use

In both positions the vertical forces balance, $N = mg$ , so the same relation applies to both:

F_{\text{start}} = \mu_s N = \mu_s mg.

Solve

F_{\text{start}} = (0.40)(3.0)(9.8) = 12\ \text{N}

for either face.

Check

Standing the brick on its end puts the same load on a smaller area, so the asperities are squeezed harder and flatten more. The true contact area, and so the friction, is unchanged.

Friction Checkpoint1 of 5

A heavy box sits on the floor while you push on it with $20\ \text{N}$, and it does not move. What is the friction force on it?