LearnYMA01WME01 · Mechanics 1Statics of a particleFriction and limiting equilibrium

Statics of a particle · Subtopic 2 of 2

Friction and Limiting Equilibrium

A particle on a rough surface stays at rest while the friction needed is at most $\mu R$ . When it is "on the point of moving" (limiting equilibrium), friction equals $\mu R$ and acts against the direction it would move. On a slope, the direction of that friction depends on whether it is about to slip up or down.

Edexcel YMA01 WME01≈ 40 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Decide whether a particle moves by comparing the needed friction with $\mu R$ .
Use $F=\mu R$ in limiting equilibrium.
Handle forces at an angle that change $R$ .
Solve "about to slip up" and "about to slip down" problems on slopes.

Pause and think

A block rests on a rough slope. You push gently up the slope, then harder and harder. Which way does friction act at first, and at the end?

Section 2 of 14

Core idea

Start herePushing a heavy box

Push a heavy box gently: nothing. Push harder: still nothing, friction is matching you. Push harder still: it lurches into motion. Just before it moves, friction is at its maximum. That edge is limiting equilibrium.

The big idea

Static friction adjusts up to $\mu R$ . On the point of moving: $F=\mu R$ . Angled forces change $R$ , and so change the maximum friction.

1. Friction rising to its limit

Friction on a 5 kg block as a horizontal push P increases (μ = 0.4)

\mu R=0.4\times5\times9.8=19.6

N. Pushes below this are fully balanced; above it the block accelerates.

0\le F\le\mu R

$F$: friction actually acting.
$\mu R$: maximum possible friction.

Worked example5 kg block, $\mu=0.4$

Horizontal pushes of 15 N and 25 N.

Goal: What happens.
Strategy: Compare with $F_{\max}$ .

$F_{\max}=0.4\times49=19.6$
$R=mg$ .
$15<19.6:\ F=15,\ \text{at rest}$
Friction only as large as needed.
$a=\frac{25-19.6}5=1.08$
Exceeds the limit: slides.

AnswerAt rest; then

1.08

m s

^{-2}

✓ CheckFriction is

15

N, not

19.6

N, in the first case ✓.

Worked examplePushed at $30^\circ$ below horizontal

Least $P$ to move the 5 kg block.

Goal: $P$ .
Strategy: The push increases $R$ .

$R=49+P\sin30^\circ=49+0.5P$
Vertical balance.
$P\cos30^\circ=0.4(49+0.5P)$
Limiting.
$P(0.866-0.2)=19.6\Rightarrow P=29.4$
Collect.

Answer

29.4

✓ CheckMore than the

19.6

N horizontal push needs ✓.

Worked exampleRough slope, $\mu=0.3$ , $20^\circ$

Will a block at rest slide?

Goal: Yes/no.
Strategy: Compare down-slope pull with $F_{\max}$ .

$mg\sin20^\circ=0.342mg$
Pull.
$0.3mg\cos20^\circ=0.282mg$
Maximum friction.

Answer

0.342>0.282

: it slides.

✓ Check

\tan20^\circ=0.364>0.3

✓.

Common misconceptionAlways $F=\mu R$ Only when limiting or sliding.

Pause and think

Why does pushing down at an angle need more force than pulling up?

Summary

Limiting: $F=\mu R$ .
Find $R$ including angled forces.

Limiting equilibrium: On the point of moving.

HistoryCoulomb's friction law. Charles-Augustin de Coulomb (better known for electric charge) tested friction carefully in 1781 for the French navy, which needed to haul ships. He confirmed

F\le\mu R

and distinguished friction at rest from friction while sliding.

Fun factStatic beats kinetic. It usually takes more force to start something sliding than to keep it sliding: the coefficient at rest is slightly bigger than when moving. That's why a heavy box suddenly "gives" and lurches forward when it starts to move.

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