Friction – A Level Maths Year 2

1 The Friction Model

So far surfaces have been smooth. A rough surface exerts a frictional force along the surface, opposing motion or the tendency to move. DfE content area R · Forces and Newton's laws uses the simple model below.

The friction model

$F \leqslant \mu R$
$R$ = normal reaction, $\mu$ = coefficient of friction ($\mu \geqslant 0$)

Situation	Friction
At rest, not about to move	$F < \mu R$ — just big enough to stop motion. Find it by resolving.
At rest, on the point of moving (limiting equilibrium)	$F = \mu R$
Moving	$F = \mu R$, opposing the direction of motion

Friction is not always $\mu R$. $\mu R$ is the maximum friction available. A book resting on a level table with nothing pushing it has zero friction, however rough the table.

Worked Example 1 — Does it move?

A box of mass $10$ kg rests on rough horizontal ground, $\mu = 0.3$. A horizontal force of $20$ N is applied. Does the box move? Find the friction force.

①$R = 10g = 98$ N, so $F_{\max} = 0.3 \times 98 = 29.4$ N.

②$20 < 29.4$, so the box stays at rest.

③Resolving horizontally, the friction is exactly $20$ N — not $29.4$ N.

Worked Example 2 — Accelerating on a rough floor

The same box is now pulled with a horizontal force of $50$ N. Find its acceleration.

①$50 > 29.4$, so it moves and $F = 29.4$ N.

②$50 - 29.4 = 10a$, so $a = 2.06$ m s$^{-2}$.

Worked Example 3 — Pulling at an angle

A crate of mass $8$ kg is pulled along rough horizontal ground ($\mu = 0.4$) by a rope at $30^\circ$ above the horizontal with tension $40$ N. Find the acceleration.

①Vertical: $R + 40\sin 30^\circ = 8g$, so $R = 78.4 - 20 = 58.4$ N.

②Moving, so $F = 0.4 \times 58.4 = 23.36$ N.

③Horizontal: $40\cos 30^\circ - 23.36 = 8a \Rightarrow 34.64 - 23.36 = 8a$

④$a = 1.41$ m s$^{-2}$.

The upward pull reduces $R$, so it reduces friction. Using $R = mg$ here is the most common error in friction questions.

2 Rough Inclined Planes

Resolve parallel and perpendicular to the plane as in Resolving Forces, then add friction along the slope, opposite to the motion or the way the particle would move.

Particle on a rough plane inclined at $\alpha$, no other forces

$R = mg\cos\alpha$ $F_{\max} = \mu mg\cos\alpha$
on the point of slipping: $mg\sin\alpha = \mu mg\cos\alpha \;\Rightarrow\; \mu = \tan\alpha$
sliding down: $a = g(\sin\alpha - \mu\cos\alpha)$ sliding up: deceleration $g(\sin\alpha + \mu\cos\alpha)$

Worked Example 4 — About to slip

A particle rests on a rough plane inclined at $25^\circ$ and is on the point of slipping. Find $\mu$.

①Parallel: $F = mg\sin 25^\circ$. Perpendicular: $R = mg\cos 25^\circ$.

②Limiting, so $F = \mu R$: $\mu = \tan 25^\circ = 0.466$.

Worked Example 5 — Sliding down

A $3$ kg particle slides down a rough plane inclined at $30^\circ$ with $\mu = 0.2$. Find its acceleration.

①$R = 3g\cos 30^\circ$; friction $0.2 \times 3g\cos 30^\circ$ acts up the slope.

②$3g\sin 30^\circ - 0.2(3g\cos 30^\circ) = 3a$

③$a = 9.8(0.5 - 0.1732) = 3.20$ m s$^{-2}$.

Worked Example 6 — Projected up a rough slope

A $2$ kg particle is projected up a rough plane inclined at $20^\circ$ ($\mu = 0.25$) with speed $8$ m s$^{-1}$. How far up does it travel? Does it then slide back?

①Moving up, so friction acts down the slope, together with $mg\sin 20^\circ$.

②Deceleration $= g(\sin 20^\circ + 0.25\cos 20^\circ) = 5.654$ m s$^{-2}$.

③$0 = 8^2 - 2(5.654)s \Rightarrow s = 5.66$ m.

④At rest, it slips back if $mg\sin 20^\circ > \mu mg\cos 20^\circ$, i.e. if $\tan 20^\circ > \mu$. Since $0.364 > 0.25$, it does slide back down.

On the way back down friction changes direction, so the acceleration changes too. Never reuse the "up" acceleration for the "down" journey.

Worked Example 7 — A range of forces for equilibrium

A $5$ kg particle rests on a rough plane inclined at $30^\circ$ ($\mu = 0.3$). A force $P$ acts up the line of greatest slope. Find the range of $P$ for which the particle stays at rest.

①$R = 5g\cos 30^\circ = 42.44$ N, so $F_{\max} = 12.73$ N. Component of weight down the slope $= 5g\sin 30^\circ = 24.5$ N.

②About to slide down (friction up): $P + 12.73 = 24.5 \Rightarrow P = 11.8$ N.

③About to slide up (friction down): $P = 24.5 + 12.73 = 37.2$ N.

④Equilibrium for $11.8 \leqslant P \leqslant 37.2$ N.

3 Connected Particles on Rough Surfaces

Decide the direction of motion first, by comparing the driving forces with the maximum friction available. Friction then acts against that direction on every rough contact.

Worked Example 8 — Rough table and a hanging mass

Particle $A$ ($4$ kg) is on a rough horizontal table ($\mu = 0.25$), connected by a light inextensible string over a smooth pulley at the edge to particle $B$ ($2$ kg) hanging freely. Find the acceleration and tension.

①$R_A = 4g$, so $F_{\max} = 0.25 \times 4g = 9.8$ N. $B$'s weight $2g = 19.6$ N $> 9.8$ N, so the system moves.

②$B$: $2g - T = 2a$. $A$: $T - 9.8 = 4a$.

③Add: $19.6 - 9.8 = 6a \Rightarrow a = 1.63$ m s$^{-2}$.

④$T = 2(g - a) = 16.3$ N.

Worked Example 9 — Rough slope and a hanging mass

Particle $A$ ($5$ kg) is on a rough plane inclined at $30^\circ$ ($\mu = 0.2$), connected over a smooth pulley at the top to particle $B$ ($4$ kg) hanging freely. Find the acceleration and tension.

①$R_A = 5g\cos 30^\circ = 42.44$ N, so $F = 0.2 \times 42.44 = 8.49$ N.

②Pulling up the slope: $4g = 39.2$ N. Resisting: $5g\sin 30^\circ + F_{\max} = 24.5 + 8.49 = 32.99$ N. So $A$ moves up and friction acts down the slope.

③$B$: $4g - T = 4a$. $A$: $T - 24.5 - 8.49 = 5a$.

④Add: $39.2 - 32.99 = 9a \Rightarrow a = 0.690$ m s$^{-2}$.

⑤$T = 4(g - a) = 36.4$ N.

4 Ladders: Friction with Moments

A ladder leaning against a wall is the classic problem combining friction with moments. The standard model: a uniform rod, resting on rough horizontal ground against a smooth vertical wall.

Resolve vertically: $R = $ total weight (the smooth wall has no vertical force).
Resolve horizontally: $F = S$.
Take moments about the foot $A$: this removes both $R$ and $F$, leaving $S$ and the weights.
Use $F \leqslant \mu R$ (with $=$ if limiting).

Worked Example 10 — The least coefficient of friction

A uniform ladder of length $5$ m and mass $20$ kg rests against a smooth vertical wall, at $60^\circ$ to rough horizontal ground. Find the least value of $\mu$ for which the ladder can rest in equilibrium.

①Vertical: $R = 20g = 196$ N. Horizontal: $F = S$.

②Moments about $A$: $S$ has perpendicular distance $5\sin 60^\circ$; the weight has $2.5\cos 60^\circ$.

③$S \times 5\sin 60^\circ = 20g \times 2.5\cos 60^\circ \Rightarrow S = \dfrac{245}{4.330} = 56.6$ N.

④$F = 56.6$ N and $F \leqslant \mu R$: $\mu \geqslant \dfrac{56.6}{196} = 0.289$.

In general, for a uniform ladder at angle $\theta$ to the ground with a smooth wall, $\mu \geqslant \dfrac{1}{2\tan\theta}$. Check: $\dfrac{1}{2\tan 60^\circ} = 0.289$ ✓. The length and mass cancel — but in an exam, derive it rather than quoting it.

Worked Example 11 — How far up can she climb?

The same ladder now stands on ground with $\mu = 0.35$. A woman of mass $70$ kg climbs it. How far up the ladder can she go before it slips?

①Vertical: $R = 90g = 882$ N. Limiting: $F = 0.35 \times 882 = 308.7$ N $= S$.

②Let her be $d$ m from $A$. Moments about $A$:

③$S \times 5\sin 60^\circ = 20g \times 2.5\cos 60^\circ + 70g \times d\cos 60^\circ$

④$1336.7 = 245 + 343d \Rightarrow d = 3.18$ m.

⑤She can climb $3.18$ m along the ladder (out of $5$ m) before it slips.

Friction direction at the foot. The foot tends to slide away from the wall, so friction at the foot acts towards the wall. If the wall is also rough, friction at the top acts upwards and the vertical equation gains an extra term.

5 Quick Reference

Model

$F \leqslant \mu R$.

Limiting or moving

$F = \mu R$.

Direction

Opposes motion, or the tendency to move.

Find $R$ first

Resolve perpendicular, including any angled pulls.

About to slip on slope

$\mu = \tan\alpha$.

Down a rough slope

$a = g(\sin\alpha - \mu\cos\alpha)$.

Up a rough slope

Deceleration $g(\sin\alpha + \mu\cos\alpha)$.

Range of $P$

Two limiting cases: friction up, then down.

Ladder

Resolve twice, moments about the foot.

6 Practice Questions

Use $g = 9.8$ m s$^{-2}$ and give answers to 3 significant figures where they are not exact.

Question 1

A $15$ kg box rests on rough horizontal ground ($\mu = 0.4$) and is pulled by a horizontal force of $50$ N. Does it move? State the friction force.

▶ Show solution

$F_{\max} = 0.4 \times 15g = 58.8$ N. Since $50 < 58.8$ it does not move.

Friction $= 50$ N (equal to the pull).

Question 2

A sledge of mass $12$ kg is pulled across rough horizontal snow ($\mu = 0.3$) by a horizontal force of $70$ N. Find its acceleration.

▶ Show solution

$F = 0.3 \times 12g = 35.28$ N.

$70 - 35.28 = 12a \Rightarrow a = 2.89$ m s$^{-2}$.

Question 3

A particle is on the point of slipping down a rough plane inclined at $18^\circ$. Find $\mu$.

▶ Show solution

$mg\sin 18^\circ = \mu mg\cos 18^\circ$, so $\mu = \tan 18^\circ = 0.325$.

Question 4

A particle is released from rest on a rough plane inclined at $40^\circ$, with $\mu = 0.5$. Find its speed after $2$ s.

▶ Show solution

$\tan 40^\circ = 0.839 > 0.5$, so it slides.

$a = g(\sin 40^\circ - 0.5\cos 40^\circ) = 9.8(0.6428 - 0.3830) = 2.546$ m s$^{-2}$.

$v = 0 + 2.546 \times 2 = 5.09$ m s$^{-1}$.

Question 5

A puck slides across rough horizontal ice with initial speed $12$ m s$^{-1}$ and stops after $20$ m. Find $\mu$.

▶ Show solution

$0 = 12^2 + 2a(20) \Rightarrow a = -3.6$ m s$^{-2}$.

Friction is the only horizontal force: $\mu mg = 3.6m \Rightarrow \mu = \dfrac{3.6}{9.8} = 0.367$.

Question 6

A box of mass $6$ kg is pulled at constant speed across rough horizontal ground ($\mu = 0.35$) by a rope at $25^\circ$ above the horizontal. Find the tension.

▶ Show solution

Constant speed, so forces balance. Vertical: $R = 6g - T\sin 25^\circ$.

Horizontal: $T\cos 25^\circ = 0.35(6g - T\sin 25^\circ)$

$T(\cos 25^\circ + 0.35\sin 25^\circ) = 20.58 \Rightarrow T = \dfrac{20.58}{1.0542} = 19.5$ N.

Question 7

A $4$ kg particle is held at rest on a rough plane inclined at $30^\circ$ ($\mu = 0.25$) by a horizontal force $P$. Find the range of possible values of $P$.

▶ Show solution

Perpendicular: $R = 4g\cos 30^\circ + P\sin 30^\circ$. Up the slope, $P$ contributes $P\cos 30^\circ$.

About to slide down (friction up): $P\cos 30^\circ + 0.25(4g\cos 30^\circ + P\sin 30^\circ) = 4g\sin 30^\circ$

$P(\cos 30^\circ + 0.25\sin 30^\circ) = 4g(\sin 30^\circ - 0.25\cos 30^\circ) \Rightarrow P = \dfrac{11.11}{0.9910} = 11.2$ N.

About to slide up (friction down): $P\cos 30^\circ = 4g\sin 30^\circ + 0.25(4g\cos 30^\circ + P\sin 30^\circ)$

$P(\cos 30^\circ - 0.25\sin 30^\circ) = 4g(\sin 30^\circ + 0.25\cos 30^\circ) \Rightarrow P = \dfrac{28.09}{0.7410} = 37.9$ N.

So $11.2 \leqslant P \leqslant 37.9$ N.

Question 8

Particle $A$ ($3$ kg) is on a rough horizontal table, joined by a light inextensible string over a smooth pulley at the edge to particle $B$ ($2$ kg) hanging freely. The system accelerates at $2$ m s$^{-2}$. Find the tension and $\mu$.

▶ Show solution

$B$: $2g - T = 2 \times 2 \Rightarrow T = 19.6 - 4 = 15.6$ N.

$A$: $T - \mu(3g) = 3 \times 2 \Rightarrow 15.6 - 29.4\mu = 6$.

$\mu = \dfrac{9.6}{29.4} = 0.327$.

Question 9

A uniform ladder rests against a smooth vertical wall with its foot on rough horizontal ground, $\mu = 0.4$. Find the smallest angle the ladder can make with the ground.

▶ Show solution

Let the ladder have length $2a$, weight $W$, angle $\theta$. $R = W$, $F = S$.

Moments about the foot: $S \times 2a\sin\theta = W \times a\cos\theta \Rightarrow S = \dfrac{W}{2\tan\theta}$.

$F \leqslant \mu R$: $\dfrac{W}{2\tan\theta} \leqslant 0.4W \Rightarrow \tan\theta \geqslant 1.25$.

Smallest angle $\theta = \tan^{-1}1.25 = 51.3^\circ$.

Question 10

A uniform ladder $AB$ of length $6$ m and mass $15$ kg has its foot $A$ on rough horizontal ground and its top $B$ against a smooth vertical wall. The ladder makes angle $\theta$ with the ground, where $\tan\theta = \tfrac43$. A man of mass $75$ kg stands on the ladder at $C$, where $AC = 4.5$ m, and the ladder is on the point of slipping.

(a) Find the normal reaction at the ground. (b) Find the reaction at the wall. (c) Find $\mu$. (d) State two modelling assumptions. (e) Explain why a second person standing on the bottom rung makes the ladder safer.

▶ Show solution

$\tan\theta = \tfrac43 \Rightarrow \sin\theta = 0.8$, $\cos\theta = 0.6$.

(a) $R = 15g + 75g = 90g = 882$ N.

(b) Moments about $A$: $S \times 6\sin\theta = 15g \times 3\cos\theta + 75g \times 4.5\cos\theta$

$4.8S = 15g(1.8) + 75g(2.7) = 229.5g \Rightarrow S = 47.8125g = 469$ N.

(c) Limiting: $F = \mu R$ and $F = S$, so $\mu = \dfrac{47.8125g}{90g} = 0.531$.

(d) The ladder is a uniform rigid rod; the man is a particle; the wall is smooth; the ladder is in a vertical plane perpendicular to the wall.

(e) A person at $A$ increases $R$ (and so the maximum friction $\mu R$) but their weight acts through $A$, so it adds nothing to the moment that $S$ must balance. More friction is available for the same required $F$.

Friction · A Level Maths Year 2 · Created with MathJax