Lesson 4 of 5Forces

Unit 2 · Lesson 4 of 5

Friction and the Coefficient of Friction

Where friction comes from, why static friction beats kinetic, and how one number — the coefficient — links friction to the normal force.

3 min read10 practice questionsAsk AI about this →

Key things to remember

1Friction opposes relative sliding between surfaces and always acts along the surface.
2 $μ = \frac{F _{f r i c t i o n}}{F _{n or ma l}}$ , so $F_{f r i c t i o n} = μ F_{n or ma l}$ . It is a ratio, so it has no units.
3Static friction adjusts to match the applied force up to a maximum; kinetic friction is constant and lower.
4 $μ_{s t a t i c} > μ_{kin e t i c}$ , which is why an object lurches as it breaks free.
5Friction depends on the surface pair and the normal force — not on the apparent contact area.
6On level ground with no extra vertical force, $F_{N} = m g$ .
7The coefficient describes two surfaces together, never one material alone.

Friction is the force that opposes relative sliding between two surfaces in contact. It comes from the surfaces being rough at a microscopic scale and from molecular attraction where they actually touch, which is why the real contact area matters more than the apparent one. It always acts along the surface, and always against the relative sliding.

Static and kinetic friction

Fig 1Static friction grows to match the push, then kinetic friction takes over at a lower, steady value.

Push a heavy box gently and it does not move: static friction has risen to exactly cancel your push. Push harder and it still does not move — static friction has risen again. Static friction is not a fixed number; it adjusts itself up to a maximum. Once you exceed that maximum the box breaks free and kinetic friction takes over, at a value that is both constant and lower than the maximum static value.

Why the box lurches when it finally moves

Kinetic friction is smaller than maximum static friction. At the instant the box breaks free, the opposing force suddenly drops while your push stays the same — so the net force jumps and the box accelerates away. That lurch is the two kinds of friction handing over.

AStatic friction

Acts while the surfaces are not sliding
Adjusts itself to match the applied force
Has a maximum value it cannot exceed
The larger of the two coefficients
What lets you walk, and what grips a car's tyres when rolling

BKinetic friction

Acts while the surfaces are sliding
Essentially constant regardless of speed
Always less than maximum static friction
The smaller coefficient
What slows a sliding box, or a skidding car

The coefficient of friction

μ = \frac{F _{f r i c t i o n}}{F _{n or ma l}} ⟺ F_{f r i c t i o n} = μ F_{n or ma l}

The Greek letter mu. A ratio of two forces, so it has no units at all.

The coefficient of friction $μ$ is a single number describing a pair of surfaces — not one material on its own. Rubber on dry concrete has a high $μ$ ; steel on ice has a very low one. Because it is a ratio of two forces, the units cancel and $μ$ is a pure number, typically between about $0.05$ and $1.2$ .

Surfaces	Approximate $μ$ (kinetic)	Consequence
Rubber on dry concrete	0.7 – 0.9	Good braking and grip
Rubber on wet concrete	0.4 – 0.6	Stopping distance roughly doubles
Wood on wood	0.3	Drawers slide but do not run free
Steel on steel, oiled	0.05	Why bearings are lubricated
Steel on ice	0.02 – 0.04	Skating is possible
Teflon on steel	0.04	Non-stick cookware

Friction does not depend on the contact area

Turning a brick on its side changes the apparent contact area but not the friction, because the same weight is now spread over a different area at the same total pressure — the real microscopic contact is unchanged. This surprises people, and it is exactly what

F_{f} = μ F_{N}

says: only the normal force and the surface pair matter. Wide racing tyres are about heat and wear, not about this equation.

The normal force is what friction scales with, so anything that changes $F_{N}$ changes the friction. Loading a truck increases $F_{N}$ and therefore the friction. Pushing down on a box while sliding it does too. And on level ground with nothing else pushing vertically, $F_{N} = F_{g} = m g$ , which is the case most problems use.

Friction in a net-force problem

Find the normal force first. On level ground with no extra vertical push, $F_{N} = m g$ .
Calculate the friction: $F_{f} = μ F_{N}$ .
Put friction into the net force as a negative, since it opposes the motion.
Apply Newton's second law, $F_{n e t} = m a$ , as usual.

Friction is not the enemy

Without it you could not walk, drive, brake, hold a pencil or tie a knot. Engineering is about increasing it where it is wanted — tyre tread, brake pads, grip tape — and reducing it where it is not, with lubricants, bearings and polished surfaces. Roughly a fifth of an engine's fuel goes into overcoming its own internal friction.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

25 kg

crate slides across a level floor where

μ_{k} = 0.35

. Find the force of kinetic friction.

Example 2

40 kg

box is pushed with

180 N

across a floor with

μ_{k} = 0.30

. Find its acceleration.

Example 3

12 kg

object needs

47 N

of horizontal force to keep it sliding at constant velocity. Find the coefficient of kinetic friction.

Where marks get lost

Giving $μ$ a unit. It is a ratio of two forces, so the units cancel completely.
Assuming $F_{N} = m g$ when something else also pushes vertically, or when the surface is not level.
Thinking a larger contact area means more friction. Only the normal force and the surface pair matter.
Using $μ_{s}$ when the object is already sliding, or $μ_{k}$ when deciding whether it will start to move.
Forgetting that 'constant velocity' means the applied force exactly equals friction, which is usually the information the question is handing you.
Treating friction as a property of one material. The coefficient always describes a pair of surfaces.
Adding friction to the applied force instead of subtracting it. Friction opposes the relative sliding.

Flashcards 9

Friction

Coefficient of friction

Static friction

Kinetic friction

+ 5 more

Practice

30 kg

crate sits on a level floor with

μ_{k} = 0.40

. What is the kinetic friction force, in newtons, once it is sliding?

20 kg

box needs

59 N

to slide at constant velocity. What is the coefficient of kinetic friction?

A brick is turned from lying flat to standing on its narrow end. What happens to the friction force as it slides?

Which statement about static and kinetic friction is correct?

What are the units of the coefficient of friction?

50 kg

crate is pushed with

250 N

on a floor where

μ_{k} = 0.35

. What is its acceleration, in

m/s^{2}

m/s^2

Why does a car's stopping distance increase sharply on a wet road?

Match each surface pair to its approximate coefficient of friction.

Rubber on dry concrete

Wood on wood

Oiled steel on steel

Steel on ice

Order the steps for solving a net-force problem that includes friction.

1Calculate the friction force with $F_{f} = μ F_{N}$
2Apply $F_{n e t} = m a$ and solve
3Add the horizontal forces with signs, friction opposing the motion
4Find the normal force — on level ground with no extra vertical push, $F_{N} = m g$

A box is pushed with

60 N

and does not move. Maximum static friction is

75 N

. What is the friction force acting?

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