Friction — AS Mechanics

Section 1Why Friction Matters

"Why do objects stop moving?" — a more useful question than "why do they move?"

When we first study motion we naturally ask what causes objects to move? But this way of thinking made it hard for scientists to understand motion for centuries. A far more powerful question is:

Why do objects stop moving?

When you slide a book across a carpet it quickly stops. On an icy surface the same push sends it far further. On a perfectly smooth surface — in theory — it would never stop at all.

This leads to Newton's First Law: an object moving with constant velocity will continue to do so unless a force acts on it. The force that slows most everyday objects is friction.

Make sure you are familiar with Newton's Laws of Motion before proceeding.

🧊 Smooth surface

Little friction → a moving object travels a long way (before stopping).

🪵 Rough surface

Large friction → a moving object slows over a short distance. Friction always acts opposite to the direction of motion.

To determine the friction, think of the forces in the order: weight ($mg$), contact force ($R$), pulling external force ($P$), friction ($F$).

Forces on a block on a horizontal surface — watch each force appear in turn

Two Key Situations. Their effect on size of friction force.

Object in Motion

Friction acts in the opposite direction to motion and equals its maximum value: $F = \mu R$

Object at Rest

Friction for a stationary object subject to an external force is always equal to the external force (up to its maximum). It acts opposite to the direction the object would move if the friction were not present.

Forces on a block on a horizontal surface: normal contact force $R$, weight $mg$, applied force $P$ and friction $F$

Key principle: Friction always acts in the direction that opposes the tendency to move. If the block moves (or would move) to the right, friction acts to the left.

Section 2The Friction Law

The Coefficient of Friction

The roughness between two surfaces is measured by the coefficient of friction, denoted by $\mu$ (pronounced "myoo").

Surface condition	Typical $\mu$
Perfectly smooth (theoretical)	0
Ice on ice	≈ 0.03
Wood on wood	≈ 0.3 – 0.5
Rubber on concrete	≈ 0.7 – 0.8

The rougher the surfaces, the larger $\mu$. For smooth surfaces in mechanics problems, $\mu = 0$.

Maximum Friction Force

The limiting (maximum) frictional force depends on two things:

The normal contact force $R$ between the surfaces
The combined roughness resulting from the material of the surfaces in contact (i.e. $\mu$)

The Friction Law

$$F_{\max} = \mu R$$

where $\mu$ is the coefficient of friction and $R$ is the normal contact force between the surfaces.

In general: $F \leq \mu R$. Equality holds when the object is moving or on the point of moving.

Three States of a Block

Stationary, no push

No tendency to move. Friction force $F = 0$.

Stationary, pushed

Friction increases to match push, provided $F < \mu R$.

Moving / limiting

$F = \mu R$. Block accelerates if push exceeds $\mu R$.

Watch out: "Limiting equilibrium" means the body is on the point of moving. It may be at rest or moving with constant velocity — in both cases $F = \mu R$.

Normal Contact Force on an Inclined Plane

On a slope inclined at angle $\alpha$ to the horizontal, with no additional vertical forces:

$$R = mg\cos\alpha$$

so the maximum friction becomes:

$$F_{\max} = \mu mg\cos\alpha$$

Block on a rough inclined plane. Friction $F$ acts up the slope if the block tends to slide down.

Section 3Horizontal Plane Problems

When a block rests on a horizontal surface with no vertical component of applied force, the normal contact force simply equals the weight:

$$R = mg$$

so $F_{\max} = \mu mg$.

Example 1 Block on rough horizontal ground — three applied forces ▾

A block of mass 5 kg lies on rough horizontal ground. The coefficient of friction between the block and the ground is 0.4. A horizontal force $P$ is applied to the block. Find the frictional force and the acceleration of the block when $P$ equals: (a) 10 N, (b) 19.6 N, (c) 30 N.

Force diagram: 5 kg block on rough horizontal surface

1

Find the maximum (limiting) friction.

Since the surface is horizontal with no vertical applied force: $R = mg = 5 \times 9.8 = 49\text{ N}$

F_max = μR = 0.4 × 49 = 19.6 N

Part (a) — P = 10 N

a

$P = 10\text{ N} < F_{\max} = 19.6\text{ N}$, so friction is enough to prevent motion.

Friction adjusts to balance $P$ exactly.

Friction = 10 N | Acceleration = 0

Part (b) — P = 19.6 N

b

$P = 19.6\text{ N} = F_{\max}$. The block is in limiting equilibrium — on the point of moving.

Friction = 19.6 N | Acceleration = 0

Part (c) — P = 30 N

c

$P = 30\text{ N} > F_{\max}$. The block moves; friction is at its limiting value 19.6 N.

Apply Newton's Second Law ($\rightarrow$):

30 − 19.6 = 5a

a = 10.4 ÷ 5 = 2.08 m s⁻²

Friction = 19.6 N | a = 2.1 m s⁻² (2 s.f.)

Example 2 Sled pulled at 45° to the horizontal ▾

A sled of mass 10 kg is being pulled along a rough horizontal plane by a force $P$ that acts at an angle of 45° to the horizontal. The coefficient of friction is 0.1. Given that the sled accelerates at 0.3 m s⁻², find the value of $P$.

Force diagram: sled on rough horizontal surface, force P at 45° above horizontal

Key idea: When a external force applied to an object has an upward component it reduces the normal contact force, which in turn reduces friction.

1

Resolve vertically (↑ positive):

$$R + P\sin 45° = 10g$$ $$R = 98 - P\sin 45° = 98 - \frac{P}{\sqrt{2}} \quad \cdots (1)$$

2

Friction at limiting value (sled is accelerating):

$$F = \mu R = 0.1\left(98 - \frac{P}{\sqrt{2}}\right)$$

3

Resolve horizontally (→, $F = ma$):

$$P\cos 45° - \mu R = 10 \times 0.3 = 3$$ $$\frac{P}{\sqrt{2}} - 0.1\!\left(98 - \frac{P}{\sqrt{2}}\right) = 3$$ $$\frac{P}{\sqrt{2}} + \frac{0.1P}{\sqrt{2}} - 9.8 = 3$$ $$P\!\left(\frac{1}{\sqrt{2}} + \frac{0.1}{\sqrt{2}}\right) = 12.8$$ $$P \cdot \frac{1.1}{\sqrt{2}} = 12.8$$

4

P = 12.8 × √2 ÷ 1.1 = 16.45 N

P = 16.4 N (3 s.f.)

Section 4Rough Inclined Planes

Setting Up the Equations

For a particle on a slope inclined at $\alpha$ to the horizontal, resolve forces parallel and perpendicular to the slope:

Standard Results

Perpendicular to slope: $\quad R = mg\cos\alpha$

Parallel to slope (sliding down): $\quad mg\sin\alpha - F = ma$

Friction (moving): $\quad F = \mu R = \mu mg\cos\alpha$

Example 3 Particle sliding down a rough slope — find μ ▾

A particle of mass 2 kg is sliding down a rough slope inclined at 30° to the horizontal. The acceleration of the particle is 1 m s⁻². Find the coefficient of friction $\mu$ between the particle and the slope.

Force diagram: particle sliding down rough 30° slope

1

Resolve perpendicular to slope (no acceleration in this direction):

$$R = 2g\cos 30° = 2 \times 9.8 \times \frac{\sqrt{3}}{2} = 16.97\ldots\text{ N}$$

2

Particle is moving, so $F = \mu R$ (friction acts up the slope, opposing downward motion).

3

Resolve parallel to slope (↙ positive, direction of motion):

$$2g\sin 30° - \mu R = 2 \times 1$$ $$9.8 - \mu(16.97) = 2$$ $$\mu = \frac{9.8 - 2}{16.97} = \frac{7.8}{16.97}$$

4

μ = 7.8 ÷ 16.97 ≈ 0.46

μ = 0.46 (2 s.f.)

Example 4 Particle released on a rough slope — find acceleration and distance ▾

A particle is held at rest on a rough plane inclined at angle $\alpha$ to the horizontal, where $\tan\alpha = 0.75$. The coefficient of friction is 0.5. The particle is released and slides down the plane. Find: (a) the acceleration, (b) the distance travelled in the first 2 seconds.

Force diagram: particle on rough slope (tan α = 0.75), released from rest

★

Find trig values from tan α = 0.75 = 3/4. Use a 3-4-5 triangle:

sin α = 3/5 = 0.6 | cos α = 4/5 = 0.8

1

Perpendicular to slope:

$$R = mg\cos\alpha = 0.8mg$$

2

Particle is sliding so $F = \mu R = 0.5 \times 0.8mg = 0.4mg$

3

Parallel to slope (down the slope positive):

$$mg\sin\alpha - F = ma$$ $$0.6mg - 0.4mg = ma$$ $$0.2g = a$$

4

a = 0.2 × 9.8 = 1.96 m s⁻²

a = 2.0 m s⁻² (2 s.f.)

Part (b) — Distance in 2 seconds

5

Use $s = ut + \frac{1}{2}at^2$ with $u = 0$, $a = 0.2g$, $t = 2$:

s = 0 + ½ × 0.2 × 9.8 × 4 = 3.92 m

s = 3.9 m (2 s.f.)

Example 5 Particle projected up a rough slope — deceleration and return speed ▾

A particle $P$ is projected up a rough plane inclined at angle $\alpha$ to the horizontal, where $\tan\alpha = \tfrac{3}{4}$. The coefficient of friction is $\tfrac{1}{3}$. The particle is projected from point $A$ with speed 20 m s⁻¹ and comes to instantaneous rest at point $B$. (a) Show the deceleration while moving up is $\tfrac{13g}{15}$. (b) Find $AB$. (c) Find the time from $A$ to $B$. (d) Find the speed when the particle returns to $A$.

Force diagram: particle projected up rough slope — friction acts down the slope

Note: $\tan\alpha = \tfrac{3}{4}$, so $\sin\alpha = \tfrac{3}{5}$, $\cos\alpha = \tfrac{4}{5}$.

Part (a) — Deceleration going up

1

Perpendicular: $R = mg\cos\alpha = \tfrac{4}{5}mg$

Moving, so $F = \mu R = \tfrac{1}{3} \cdot \tfrac{4}{5}mg = \tfrac{4}{15}mg$

Going up: both gravity component and friction act down the slope.

$$-\frac{3mg}{5} - \frac{4mg}{15} = ma$$ $$-\frac{9g}{15} - \frac{4g}{15} = a$$

a = −13g/15 (deceleration of 13g/15) ✓

Part (b) — Distance AB

2

$u = 20$, $v = 0$, $a = -\tfrac{13g}{15}$. Use $v^2 = u^2 + 2as$:

0 = 400 + 2(−13×9.8/15)s

s = 400 ÷ (26×9.8/15) = 400 ÷ 16.987 ≈ 23.5 m

AB ≈ 24 m (2 s.f.)

Part (c) — Time A to B

3

Use $v = u + at$:

0 = 20 − (13×9.8/15)t

t = 20 ÷ 8.493 ≈ 2.35 s

t ≈ 2.4 s (2 s.f.)

Part (d) — Speed returning to A

4

Coming down: gravity acts down the slope, friction now acts up the slope (opposing downward motion).

$$\frac{3mg}{5} - \frac{4mg}{15} = ma_{\text{down}}$$ $$\frac{9g}{15} - \frac{4g}{15} = \frac{5g}{15} = \frac{g}{3}$$

5

From $B$ with $u=0$, $a = g/3$, $s = AB \approx 23.5$ m. Use $v^2 = 2as$:

v² = 2 × (9.8/3) × 23.5 = 153.3

v = √153.3 ≈ 12.4 m s⁻¹

Speed at A ≈ 12 m s⁻¹ (2 s.f.)

Section 5Forces at Angles & Static Particles

When a force is applied at an angle (not parallel to the surface), its vertical component affects the normal contact force $R$, which in turn changes the maximum friction.

Remember: Always use the normal contact force $R$, not the weight, when calculating $F_{\max} = \mu R$.

Example 6 Maximum force before slipping — horizontal then angled ▾

A mass of 8 kg rests on a rough horizontal plane with $\mu = 0.5$. Find the maximum force $P$ that can be applied without causing motion if: (a) $P$ is horizontal, (b) $P$ acts at 60° above the horizontal.

Force diagram: 8 kg mass on rough horizontal plane

Part (a) — Horizontal force

1

$R = 8g = 78.4\text{ N}$ (no vertical component from $P$)

Limiting: $F = \mu R = 0.5 \times 78.4 = 39.2\text{ N}$

Equilibrium: $P = F$

P_max = 39 N (2 s.f.)

Part (b) — Force at 60° above horizontal

2

Vertical: $R + P\sin 60° = 8g \;\Rightarrow\; R = 8g - P\sin 60°$

Limiting friction: $F = 0.5(8g - P\sin 60°)$

3

Horizontal equilibrium: $P\cos 60° = F$

$$P\cos 60° = 0.5(8g - P\sin 60°)$$ $$P\cos 60° + 0.5P\sin 60° = 0.5 \times 8g = 4g$$ $$P(\cos 60° + 0.5\sin 60°) = 4g$$

4

P = 4g ÷ (cos60° + 0.5sin60°)

P = 39.2 ÷ (0.5 + 0.433) = 39.2 ÷ 0.933

P = 42 N (2 s.f.)

The upward component of $P$ reduces $R$, so friction also reduces — meaning a larger $P$ is needed to reach the limit of friction. Hence the answer in (b) is larger than in (a).

Example 7 Particle on inclined plane — force up line of greatest slope ▾

A particle of mass 1.5 kg rests in equilibrium on a rough plane inclined at 25° to the horizontal, under the action of a force $X$ N acting up the line of greatest slope. The particle is on the point of moving up the plane. The coefficient of friction is 0.25. Find: (a) the normal contact force, (b) the value of $X$.

Force diagram: particle in limiting equilibrium on 25° slope, force X up slope, friction F down slope

1

Perpendicular to slope:

$$R = 1.5g\cos 25°$$

R = 1.5 × 9.8 × cos25° = 13.3 N

R = 13.3 N

2

On the point of moving up, so friction acts down the slope (opposing upward tendency).

$F = \mu R = 0.25 \times 13.3 = 3.325\text{ N}$

3

Parallel to slope (up positive, equilibrium):

$$X - 1.5g\sin 25° - \mu R = 0$$ $$X = 1.5g\sin 25° + 0.25(13.3)$$

X = 1.5×9.8×sin25° + 3.325 = 6.21 + 3.325

X = 9.54 N (3 s.f.)

Example 8 Particle sliding down slope pulled back by force P ▾

A particle of mass 2 kg is sliding down a rough slope inclined at 20° to the horizontal. A force $P$ acts parallel to the slope and attempts to pull the particle up the slope. The acceleration is 0.2 m s⁻² down the slope. The coefficient of friction is 0.3. Find $P$.

Force diagram: particle sliding down rough 20° slope, friction F and applied force P both acting up slope

1

Perpendicular: $R = 2g\cos 20° = 18.42\text{ N}$

Particle moving down, so friction acts up: $F = 0.3 \times 18.42 = 5.53\text{ N}$

2

Resolve down the slope ($a = 0.2$ m s⁻² down):

$$2g\sin 20° - F - P = 2 \times 0.2$$ $$6.70 - 5.53 - P = 0.4$$

P = 6.70 − 5.53 − 0.40 = 0.77 N

P = 0.77 N

Section 6Practice Questions

Attempt each question before revealing the answer. Use $g = 9.8\text{ m s}^{-2}$.

Question 1 — 3 marks

A block of mass 4 kg rests on a rough horizontal table. A force of 6 N acts on the block at an angle of 30° below the horizontal. The block is on the point of slipping. Find the value of the coefficient of friction $\mu$.

Q1: 4 kg block on rough horizontal table — 6 N force at 30° below horizontal

Resolve vertically: $R = 4g + 6\sin 30° = 39.2 + 3 = 42.2\text{ N}$

Limiting friction: $F = \mu R$ and $F = 6\cos 30° = 5.196\text{ N}$

$\mu = \dfrac{5.196}{42.2}$

μ = 0.12 (2 s.f.)

Question 2 — 4 marks

A book of mass 2 kg slides down a rough plane inclined at 20° to the horizontal. The acceleration of the book is 1.5 m s⁻². Find the coefficient of friction between the book and the plane.

Q2: Book sliding down rough 20° slope — find coefficient of friction

Perpendicular: $R = 2g\cos 20° = 18.42\text{ N}$

Parallel (down slope): $2g\sin 20° - \mu R = 2 \times 1.5$

$6.704 - 18.42\mu = 3$

$18.42\mu = 3.704$

μ = 0.20 (2 s.f.)

Question 3 — 5 marks

A force of 30 N pulls a particle of mass 6 kg up a rough slope inclined at 15° to the horizontal. The force acts along the slope. Given that the particle moves at constant speed, find the magnitude of the frictional force and the coefficient of friction.

Q3: 6 kg particle pulled up rough 15° slope at constant speed by force P

Constant speed ⟹ zero acceleration ⟹ resultant force = 0.

Perpendicular: $R = 6g\cos 15° = 56.77\text{ N}$

Parallel: $30 - 6g\sin 15° - F_r = 0$

$F_r = 30 - 6 \times 9.8 \times \sin 15° = 30 - 15.22 = 14.78\text{ N}$

$\mu = F_r / R = 14.78 / 56.77$

Friction = 14.8 N | μ = 0.26 (2 s.f.)

Question 4 — 5 marks

A boy of mass 40 kg slides from rest down a straight slide of length 5 m. The slide is inclined at 20° to the horizontal. The coefficient of friction between the boy and the slide is 0.1. Find: (a) the acceleration of the boy, (b) the speed of the boy at the bottom of the slide.

Q4: Boy sliding down rough 20° slide of length 5 m

Perpendicular: $R = 40g\cos 20° = 368.2\text{ N}$

Friction: $F = 0.1 \times 368.2 = 36.82\text{ N}$

Parallel (down): $40g\sin 20° - 36.82 = 40a$

$134.0 - 36.82 = 40a \;\Rightarrow\; a = 2.43\text{ m s}^{-2}$

Speed: $v^2 = 2 \times 2.43 \times 5 = 24.3 \;\Rightarrow\; v = 4.93\text{ m s}^{-1}$

a = 2.4 m s⁻² | v = 4.9 m s⁻¹ (2 s.f.)

Question 5 — 7 marks

A particle of mass 0.5 kg is being pulled up a rough slope inclined at 30° to the horizontal by a force of 5 N. The force acts at 30° above the slope. The coefficient of friction is 0.1. Calculate the acceleration of the particle.

Q5: 0.5 kg particle pulled up rough 30° slope — force of 5 N at 30° above the slope

Perpendicular to slope:

$R + 5\sin 30° = 0.5g\cos 30°$

$R = 0.5 \times 9.8 \times \frac{\sqrt{3}}{2} - 2.5 = 4.243 - 2.5 = 1.743\text{ N}$

Friction (accelerating): $F = 0.1 \times 1.743 = 0.174\text{ N}$

Parallel to slope (up positive):

$5\cos 30° - 0.5g\sin 30° - 0.174 = 0.5a$

$4.330 - 2.450 - 0.174 = 0.5a$

$1.706 = 0.5a$

a = 3.41 m s⁻² up the slope

Question 6 — 6 marks

A 5 kg body lying on a rough horizontal surface has a coefficient of friction of $\tfrac{1}{7}$ with the surface. A vertical downward force of 14 N acts on the body in addition to its weight. A horizontal force also acts. Determine whether the body moves if the horizontal force is (a) 6 N, (b) 9 N, (c) 12 N, and find the acceleration in each case where movement occurs.

Q6: 5 kg body on rough horizontal surface — extra 14 N downward force, horizontal force P

$R = 5g + 14 = 49 + 14 = 63\text{ N}$

$F_{\max} = \tfrac{1}{7} \times 63 = 9\text{ N}$

(a) 6 N: $6 < 9$ → body remains at rest. Friction = 6 N.

(b) 9 N: $9 = F_{\max}$ → body in limiting equilibrium (no acceleration). Friction = 9 N.

(c) 12 N: $12 > 9$ → body moves. $F = ma$: $12 - 9 = 5a$

(a) rest | (b) limiting equilibrium | (c) a = 0.6 m s⁻²