Friction

Understanding resistive forces, the friction model F ≤ μR, and solving problems on rough surfaces

SPEC 8.6 • FORCES AND NEWTON'S LAWS

What is Friction?

Definition

Friction is a contact force that opposes the relative motion (or potential motion) between two surfaces in contact.

Friction arises from the microscopic interactions between surfaces. In mechanics, we use a simplified model.

Critical Rule

Friction ALWAYS acts in the OPPOSITE direction to motion (or potential motion).

If an object moves RIGHT → Friction acts LEFT

If an object tends to slide DOWN a slope → Friction acts UP the slope

If a force pushes an object UP a slope → Friction acts DOWN the slope

Key Properties

Direction: ALWAYS opposes motion or the tendency to move

Magnitude: Adjusts up to a maximum value

Contact force: Only exists between surfaces in contact

Terminology

Rough surface: Friction is present (μ > 0)

Smooth surface: Friction is negligible (μ = 0)

The Friction Model

The friction model relates the frictional force to the normal reaction force:

Friction Inequality

F \leq μ R

where equality holds when the object is moving or on the point of moving

SymbolMeaningUnits

F

Frictional forceNewtons (N)

μ

Coefficient of friction (pronounced "mew")Dimensionless (no units)

R

Normal reaction forceNewtons (N)

About μ

The coefficient of friction μ is always between 0 and 1 for A-Level problems. A value of 0 means a smooth (frictionless) surface; values closer to 1 indicate rougher surfaces.

Static and Kinetic Friction

Static Friction

Object at rest

Friction adjusts to balance applied forces:

F < μ R

The object remains in equilibrium.

Limiting Equilibrium

On the point of moving

Friction reaches its maximum value:

F = μ R

Any increase in applied force causes motion.

Kinetic Friction

When an object is moving, friction takes its maximum value:

F = μ R

In A-Level, we typically assume static and kinetic coefficients are equal.

Friction on Horizontal Surfaces

For an object on a horizontal rough surface with no vertical acceleration:

Key Results

R(↑):

R - m g = 0

→

R = m g

Maximum friction:

F_{m a x} = μ R = μ m g

Decision Process

Compare applied force P to maximum friction μmg:

If P ≤ μmg: Object stays at rest (equilibrium)

R(→):

P - F = 0

→

F = P

If P > μmg: Object accelerates

R(→), F = ma:

P - μ m g = ma

01Worked Example

A 5 kg box rests on a rough horizontal floor. The coefficient of friction between the box and the floor is 0.4. A horizontal force of 15 N is applied to the box. Determine whether the box moves, and find the frictional force acting on it.

1

Resolving vertically:

•

R (↑) : R - m g = 0

•

R = m g = 5 \times 9.8 = 49

N

2

Maximum friction:

F_{m a x} = μ R = 0.4 \times 49 = 19.6

N

3

Compare: Applied force (15 N) < Maximum friction (19.6 N)

4

Since applied force < maximum friction, block does not move.

5

Resolving horizontally (equilibrium):

•

R (\to) : 15 - F = 0

•

F = 15

N

AnswerThe box does not move. Frictional force = 15 N

02Worked Example

A 4 kg block is pushed across a rough horizontal surface with a constant horizontal force of 30 N. The coefficient of friction is 0.3. Find the acceleration of the block.

1

Resolving vertically:

•

R (↑) : R - m g = 0

•

R = m g = 4 \times 9.8 = 39.2

N

2

Maximum friction:

F = μ R = 0.3 \times 39.2 = 11.76

N

3

Since 30 N > 11.76 N, the block moves with friction at maximum (11.76 N)

4

Resolving horizontally:

•

R (\to), F = ma :

•

30 - 11.76 = 4 a

•

18.24 = 4 a

•

a = 4.56

m/s²

AnswerAcceleration = 4.56 m/s² (or 4.6 m/s² to 2 s.f.)

03Worked Example

A 2 kg block rests on a rough horizontal table. A horizontal force of 8 N is just sufficient to move the block. Find the coefficient of friction between the block and the table.

1

"Just sufficient to move" means at limiting equilibrium:

F = μ R

2

Resolving vertically:

•

R (↑) : R - m g = 0

•

R = m g = 2 \times 9.8 = 19.6

N

3

Resolving horizontally (equilibrium):

•

R (\to) : 8 - F = 0

•

F = 8

N

4

At limiting equilibrium:

F = μ R

•

8 = μ \times 19.6

•

μ = \frac{8}{19.6} = 0.408

Answerμ = 0.41 (to 2 s.f.)

Forces Applied at an Angle

When a force P is applied at an angle to the horizontal, it has both horizontal and vertical components. This changes the normal reaction R.

Critical Point

When P has a vertical component, R ≠ mg!

P angled upward: P partly lifts the object → R is reduced → Less friction available

P angled downward: P pushes object into surface → R is increased → More friction available

Resolving Vertically

P above horizontal:

R(↑):

R + P sin θ - m g = 0

→

R = m g - P sin θ

P below horizontal:

R(↑):

R - P sin θ - m g = 0

→

R = m g + P sin θ

08Worked Example

A mass of 8 kg rests on a rough horizontal plane with μ = 0.5. A force P acts at 60° above the horizontal. Find the maximum value of P for which the mass remains in equilibrium.

1

At limiting equilibrium (max P before motion):

F = μ R

2

Resolving vertically:

•

R (↑) : R + P sin 60° - m g = 0

•

R = m g - P sin 60° = 78.4 - 0.866 P

3

Maximum friction:

F = μ R = 0.5 (78.4 - 0.866 P) = 39.2 - 0.433 P

4

Resolving horizontally (equilibrium):

•

R (\to) : P cos 60° - F = 0

•

0.5 P = 39.2 - 0.433 P

5

Solve:

•

0.5 P + 0.433 P = 39.2

•

0.933 P = 39.2

•

P = 42.0

N

Key: R is reduced because P has an upward component!

AnswerP = 42 N (to 2 s.f.)

Friction on Inclined Planes

When resolving forces on an inclined plane, work perpendicular and parallel to the slope:

Key Equations

R(↖):

R - m g cos θ = 0

→

R = m g cos θ

Maximum friction:

F_{m a x} = μ R = μ m g cos θ

Component down slope:

m g sin θ

Direction of Friction

Object tending to slide DOWN: Friction acts UP the slope

Object being pushed UP: Friction acts DOWN the slope

Always opposite to the direction of (potential) motion!

Limiting Equilibrium on Inclines

When an object is on the point of sliding down a rough inclined plane:

At Limiting Equilibrium

Friction acts up the slope at its maximum value. Resolving parallel to slope:

R(↗):

F - m g sin θ = 0

With

F = μ R = μ m g cos θ

:

μ m g cos θ = m g sin θ

Important Result

Dividing both sides by

m g cos θ

:

μ = tan θ

where θ is the angle of the slope at limiting equilibrium.

Application

This result allows us to find μ experimentally by gradually increasing the angle of a slope until an object just begins to slide.

Remember

This formula only applies when the object is on the point of sliding down (no applied force).

04Worked Example

A block is placed on a rough plane inclined at angle θ to the horizontal. The coefficient of friction is 0.5. Find the maximum angle θ at which the block can remain in equilibrium.

1

At limiting equilibrium, the block is on the point of sliding down.

2

Resolving perpendicular to slope:

•

R (↖) : R - m g cos θ = 0

•

R = m g cos θ

3

Resolving parallel to slope (equilibrium):

•

R (↗) : F - m g sin θ = 0

•

F = m g sin θ

4

At limiting equilibrium:

F = μ R

•

m g sin θ = μ \times m g cos θ

•

tan θ = μ = 0.5

•

θ = arctan (0.5) = 26.57°

Answerθ = 26.6° (to 3 s.f.) or θ = 27° (to 2 s.f.)

05Worked Example

A 3 kg block rests on a rough plane inclined at 30° to the horizontal. A force of 10 N is applied to the block parallel to and up the slope. Given μ = 0.4, determine whether the block moves up, down, or remains at rest.

1

Resolving perpendicular to slope:

•

R (↖) : R - m g cos 30° = 0

•

R = 3 \times 9.8 \times cos 30° = 25.46

N

2

Maximum friction:

F_{m a x} = μ R = 0.4 \times 25.46 = 10.18

N

3

Net force parallel to slope (ignoring friction):

•

Force up = 10 N, Weight component down =

m g sin 30° = 14.7

N

•

Net = 10 - 14.7 = -4.7 N (tendency to slide down)

4

Since |4.7| < 10.18, friction can balance this. Block stays at rest.

5

Resolving parallel to slope (equilibrium):

•

R (↗) : 10 + F - m g sin 30° = 0

•

F = 14.7 - 10 = 4.7

N (up slope)

AnswerThe block remains at rest. Friction = 4.7 N acting up the slope.

07Worked Example

A block of mass 2 kg rests on a rough plane inclined at 25° to the horizontal. The coefficient of friction is μ = 0.4. A light string applies a tension T to the block, parallel to and up the slope. Find the range of values of T for which the block remains in equilibrium.

1

Resolving perpendicular to slope:

•

R = m g cos 25° = 2 \times 9.8 \times cos 25° = 17.77

N

2

Maximum friction:

F_{ma x} = μ R = 0.4 \times 17.77 = 7.11

N

3

Weight component down slope:

m g sin 25° = 2 \times 9.8 \times sin 25° = 8.28

N

4

Case 1: Block on point of sliding DOWN (friction acts UP slope)

•

T + F_{ma x} = m g sin 25°

•

T_{min} = 8.28 - 7.11 = 1.17

N

5

Case 2: Block on point of sliding UP (friction acts DOWN slope)

•

T = m g sin 25° + F_{ma x}

•

T_{ma x} = 8.28 + 7.11 = 15.39

N

Note: If T < 1.17 N, the block slides down. If T > 15.4 N, the block slides up.

Answer

1.17 N \leq T \leq 15.4 N

(or 1.2 N ≤ T ≤ 15 N to 2 s.f.)

08Worked Example

A 5 kg block rests on a rough plane inclined at 20° to the horizontal. A force of 40 N acts on the block at an angle of 25° above the slope (i.e., 45° to the horizontal). Given μ = 0.3, find the acceleration of the block up the slope.

1

The 40 N force acts at 25° to the slope. Resolve into components:

•

Component parallel to slope (up):

40 cos 25° = 36.25

N

•

Component perpendicular to slope (into):

40 sin 25° = 16.90

N

2

Resolving perpendicular to slope:

•

R (↖) : R + 40 sin 25° = m g cos 20°

•

R = 5 \times 9.8 \times cos 20° - 40 sin 25°

•

R = 46.05 - 16.90 = 29.15

N

3

Friction acts DOWN the slope (opposes motion UP):

•

F = μ R = 0.3 \times 29.15 = 8.75

N

4

Resolving parallel to slope (UP positive):

•

R (↗), F = ma :

•

40 cos 25° - F - m g sin 20° = ma

•

36.25 - 8.75 - 16.76 = 5 a

•

10.74 = 5 a

•

a = 2.15

m/s²

Note: The angled force reduces R (and hence friction), giving greater acceleration than a parallel force.

AnswerAcceleration = 2.1 m/s² (to 2 s.f.) up the slope

Practice

Question 1: A 6 kg crate sits on a rough horizontal floor with μ = 0.35. A horizontal force P is applied. Find the minimum value of P required to move the crate.

1

Resolving vertically:

•

R (↑) : R - m g = 0

•

R = m g = 6 \times 9.8 = 58.8

N

2

Maximum friction:

F_{m a x} = μ R = 0.35 \times 58.8 = 20.58

N

3

At limiting equilibrium (point of moving):

•

R (\to) : P - F = 0

•

P = F_{m a x} = 20.58

N

AnswerP > 20.6 N (minimum force to cause motion)

Practice

Question 2: A particle of mass 2 kg rests in equilibrium on a rough plane inclined at 35° to the horizontal. Find the magnitude of the frictional force.

1

Resolving perpendicular to slope:

•

R (↖) : R - m g cos 35° = 0

•

R = 2 \times 9.8 \times cos 35° = 16.06

N

2

Resolving parallel to slope (equilibrium):

•

R (↗) : F - m g sin 35° = 0

•

F = m g sin 35° = 2 \times 9.8 \times sin 35°

•

F = 11.24

N

AnswerFriction = 11.2 N (acting up the slope)

Practice

Question 3: A 4 kg block slides down a rough plane inclined at 40° to the horizontal. Given μ = 0.25, find the acceleration of the block.

1

Resolving perpendicular to slope:

•

R (↖) : R - m g cos 40° = 0

•

R = 4 \times 9.8 \times cos 40° = 30.02

N

2

Friction (acts up slope, opposes motion down):

•

F = μ R = 0.25 \times 30.02 = 7.51

N

3

Resolving parallel to slope (taking DOWN as positive):

•

R (↘), F = ma :

•

m g sin 40° - F = ma

•

25.19 - 7.51 = 4 a

•

17.68 = 4 a

•

a = 4.42

m/s²

AnswerAcceleration = 4.4 m/s² down the slope

Practice

Question 4: A block of mass 4 kg rests on a rough horizontal table with μ = 0.3. A force of 6 N acts on the block at an angle of 30° to the horizontal in a downward direction. The block is on the point of slipping. Find the value of μ.

1

Resolving vertically:

•

R (↑) : R - m g - P sin 30° = 0

•

R = m g + P sin 30° = 39.2 + 3 = 42.2

N

2

Resolving horizontally (equilibrium at limiting friction):

•

R (\to) : P cos 30° - F = 0

•

F = P cos 30° = 6 \times cos 30° = 5.196

N

3

At limiting equilibrium:

F = μ R

•

5.196 = μ \times 42.2

•

μ = \frac{5.196}{42.2} = 0.123

Note: R is increased because P pushes down into the surface.

Answerμ = 0.12 (to 2 s.f.)

Summary

Key Concepts

ConceptKey Result

Friction Model

F \leq μ R

Limiting Equilibrium

F = μ R

Horizontal Surface

R = m g

, so

F_{m a x} = μ m g

Inclined Plane

R = m g cos θ

, so

F_{m a x} = μ m g cos θ

Limiting Angle

μ = tan θ

Angled Force (up)

R = m g - P sin θ

(R reduced)

Angled Force (down)

R = m g + P sin θ

(R increased)

Problem-Solving Strategy

Draw a clear force diagram

Resolve perpendicular to surface to find R

Calculate maximum friction = μR

Compare forces to determine if motion occurs

Apply Newton's Second Law if accelerating

Key Exam Points

Friction ALWAYS opposes motion. If moving/about to move UP → Friction acts DOWN. If moving/about to move DOWN → Friction acts UP.