When one surface slides over another, friction acts against the sliding. For many pairs of dry surfaces, its size follows a simple rule that depends on the The Normal Force and Apparent Weight.
Key idea
Kinetic friction: $f_k = \mu_k F_N$, directed opposite to the sliding. The coefficient $\mu_k$ (no units) depends on the two surfaces; typical values run from about $0.03$ (oiled steel on steel) to about $0.7$ (rubber on dry concrete).
Why it works
Surfaces touch only at tiny bumps. Pressing them together harder flattens the bumps into more contact, and the extra contact resists sliding in proportion. So friction grows with how hard the surfaces press, which is the normal force, not the weight. For a box on level ground those are equal; on a ramp, with an extra push or with a slanted pull, they are not.
| Surfaces | μ_s (about) | μ_k (about) |
|---|---|---|
| rubber on dry concrete | 1.0 | 0.7 |
| wood on wood | 0.5 | 0.3 |
| steel on steel, dry | 0.6 | 0.3 |
| steel on steel, oiled | 0.05 | 0.03 |
| shoes on ice | 0.1 | 0.05 |
Worked examples
Worked example: Friction on a level floor
A $25\ \text{kg}$ box slides on a floor with $\mu_k = 0.3$. Find the friction force.
- $F_N = mg = 245\ \text{N}$.
- $f_k = 0.3 \times 245 = 73.5\ \text{N}$, against the sliding.
Answer: $73.5\ \text{N}$
Worked example: How far it slides
A box sliding at $6\ \text{m/s}$ on a floor with $\mu_k = 0.25$ coasts to a stop. How far does it go?
- Friction is the net force: $a = \dfrac{\mu_k mg}{m} = \mu_k g = 2.45\ \text{m/s}^2$ (slowing).
- $0 = 6^2 - 2(2.45)d$, so $d = \dfrac{36}{4.9} \approx 7.35\ \text{m}$.
Answer: about $7.35\ \text{m}$
Worked example: A push from above
You press down on a sliding $25\ \text{kg}$ box with $50\ \text{N}$ ($\mu_k = 0.3$). Find the friction.
- $F_N = 245 + 50 = 295\ \text{N}$.
- $f_k = 0.3 \times 295 = 88.5\ \text{N}$.
Answer: $88.5\ \text{N}$
Common mistake
Using $mg$ instead of the normal force when they differ. Check: find $F_N$ from the vertical (or perpendicular) forces first, every time; then multiply by $\mu_k$.