Friction: static and kinetic

Why does a sliding box stop?

Push a box across a floor, then let go. It stops. A force acts along the floor, against the motion. This is friction. It comes from the tiny bumps of the two surfaces, which catch on each other, and from the pull between their atoms.

Friction is a nuisance in a machine, where it wastes energy and wears parts out. It is also why you can walk, hold a pen, or stop a bicycle.

Static friction

Push a heavy box gently. It does not move. So the floor must push back on it along the floor, just as hard as you push. Push harder: still no motion, so the friction has grown too. This is static friction. It adjusts itself to match your push, but only up to a limit, called limiting friction:

The number $\mu_s$ is the coefficient of static friction. It has no unit. Push harder than $\mu_s N$ and the box starts to slide.

Kinetic friction

Once the box slides, the friction is kinetic friction, nearly steady whatever the speed:

For the same two surfaces $\mu_k$ is a little smaller than $\mu_s$. So a box is harder to start than to keep going.

Try it, and watch the friction jump down the moment the box breaks free.

You should find that the friction follows your push up a straight line, stops at $\mu_s mg$, and then drops to $\mu_k mg$ and stays there.

Watch out. Friction is proportional to $N$, and $N$ is not always $mg$. Press down on the box, or pull it up at an angle, and $N$ changes, so friction changes too. And friction does not depend on the area of contact: a brick on its side or on its face needs the same push.

A worked example

A $24\ \text{kg}$ trunk is on a floor with $\mu_s = 0.38$ and $\mu_k = 0.29$.

1. The limit. $f_{s,\max} = 0.38 \times 24\ \text{kg} \times 9.8\ \text{m/s}^2 = 89.4\ \text{N}$.
2. A push of $70\ \text{N}$. That is below the limit, so the trunk stays put, and the friction is exactly $70\ \text{N}$, not $89.4\ \text{N}$.
3. A push of $100\ \text{N}$. Above the limit, so it slides. $f_k = 0.29 \times 24\ \text{kg} \times 9.8\ \text{m/s}^2 = 68.2\ \text{N}$, and $a = \dfrac{100\ \text{N} - 68.2\ \text{N}}{24\ \text{kg}} = 1.33\ \text{m/s}^2$.

Think first. Static friction is only as big as it needs to be. The formula $\mu_s N$ gives its largest value, not its value every time.

Practise with new numbers

Try it in Python

The program pushes a crate harder and harder, and reports the friction and the acceleration each time.

Remember these

• Static friction matches the push, up to $\mu_s N$.
• Kinetic friction is $\mu_k N$, nearly steady, and $\mu_k < \mu_s$.
• Friction depends on $N$ and on the two surfaces, not on the area of contact.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A $6.0\ \text{kg}$ box rests on a floor with $\mu_s = 0.55$. You push it with $18\ \text{N}$ and it does not move. What is the friction? ($g = 9.8\ \text{m/s}^2$.)

2. Why is it harder to start a heavy cupboard moving than to keep it moving?

3. A brick is pulled along a table first on its large face, then on its narrow side. Which needs the bigger force at a steady speed?

4. How does friction help you walk?