Learn9MA0Paper 3 · Statistics and MechanicsForces and Newton's lawsFriction

Forces and Newton's laws · Subtopic 6 of 8

Friction

Friction is lazy: it only pushes back as hard as it needs to, up to a maximum of $\mu R$ . Reach that maximum and the object is on the point of moving.

Cambridge 9709 Paper 4Edexcel 9MA0 Paper 3AQA 7357 Paper 2≈ 35 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Know that friction opposes motion (or the tendency to move) along the surface.
Use $F\le\mu R$ , with $F=\mu R$ when friction is limiting.
Decide whether an object stays at rest by comparing the force needed with $\mu R$ .
Solve limiting-friction problems where the applied force is at an angle.

Pause and think

A heavy box needs $30$ N to start it sliding. You push it with $12$ N and it does not move. How big is the friction force?

Section 2 of 14

Core idea

Start hereWhy heavy boxes are hard to start

Push a heavy box gently: nothing happens. Push harder: still nothing. Then suddenly it slides. Friction grew to match your push, up to a limit, then gave way. That limit depends on how hard the surfaces press together.

The big idea

Friction $F$ opposes motion (or the tendency to move) along the surface. $F\le\mu R$ ; at the point of slipping (limiting) or when sliding, $F=\mu R$ .

1. The model

Friction against applied force for a box with μR = 30 N

Friction equals the push until the limit, then stays at

\mu R

once the box slides.

F\le\mu R\ \ (\text{static}),\qquad F=\mu R\ \ (\text{limiting or sliding})

$R$: normal reaction: the surface pushing perpendicular to itself.
$\mu$: coefficient of friction (depends on the surfaces; "smooth" means 0).

Forces on a box pulled along rough ground

Friction acts along the surface, against the motion.

R

is perpendicular to the surface.

Worked example10 kg box, $\mu=0.3$

What happens with pushes of 20 N and 40 N horizontally?

Goal: Motion and friction.
Strategy: Find $F_{\max}$ , compare.

$R=100\text{ N},\ F_{\max}=30\text{ N}$
Weight $mg$ .
$20<30:\ F=20,\ \text{at rest}$
Friction only as big as needed.
$40>30:\ F=30,\ a=\tfrac{10}{10}=1\text{ m s}^{-2}$
Slides.

AnswerStays still at 20 N; accelerates at

1

m s

^{-2}

at 40 N.

✓ CheckFriction never exceeds the push when at rest ✓.

Worked examplePulling at an angle

Same box pulled by 40 N at $30^\circ$ above horizontal.

Goal: Acceleration.
Strategy: The pull reduces $R$ .

$R+40\sin30^\circ=100\Rightarrow R=80$
Vertical.
$F_{\max}=24;\ 40\cos30^\circ=34.6>24$
It moves.
$a=\tfrac{34.6-24}{10}=1.06$
$F=ma$ .

Answer

1.06

m s

^{-2}

✓ CheckA horizontal 40 N pull gave

a=1

. Angling it loses

5.4

N of forward pull but cuts friction by

6

N, so

a

is slightly larger ✓.

Common misconceptionAlways $R=mg$ ? Only if nothing else pushes or pulls vertically. Angled forces and slopes change

R

Pause and think

A 6 kg block just moves under 15 N horizontal. $\mu$ ?

Summary

$F\le\mu R$ .
Find $R$ first.

Limiting equilibrium: About to slip.

HistoryLeonardo's notebooks. Around 1493 Leonardo da Vinci sketched blocks being dragged across tables and noted that friction does not depend on the area of contact. His notes went unpublished, and the laws were rediscovered by Guillaume Amontons in 1699 and refined by Charles-Augustin de Coulomb in 1785, which is why

F\le\mu R

is often called Coulomb friction.

Fun factArea does not matter (mostly). A brick needs the same force to slide whether it lies on its large face or stands on its end. The larger face presses more gently over a bigger area, and the two effects cancel. That surprising fact is built into

F=\mu R

, which has no area in it.

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