Pulleys

Analysing systems of particles connected by strings passing over pulleys

SPEC 8.4 • MECHANICS

Introduction to Pulley Systems

A pulley is a wheel with a groove that allows a rope or string to pass over it. In A-Level mechanics, we use idealised pulleys to simplify calculations.

Key Assumptions

Light pulley: The pulley has negligible mass

Smooth pulley: The pulley is frictionless, so tension is the same throughout the string

Inextensible string: The string does not stretch, so connected particles have the same magnitude of acceleration

String remains taut: There is always tension in the string

Critical: Resolving Direction

Always resolve in the direction of motion, taking this as positive. This means the acceleration

a

will be positive. For connected particles, one moves in one direction while the other moves in the opposite direction - resolve each particle in its own direction of motion.

Exam Tip

Exams often ask: "State how you have used the fact that the string is inextensible." The answer is: "The acceleration of both particles has the same magnitude."

Horizontal Surface & Hanging Particle

The simplest pulley system has one particle on a horizontal surface and another hanging vertically.

For Particle A

R (\to)

: Resolve horizontally:

T = m_{A} a

For Particle B

R (↓)

: Resolve vertically:

m_{B} g - T = m_{B} a

Key Formulas: Smooth Horizontal Surface

Substitute to eliminate T and find acceleration:

Derivation

From A,

R (\to)

:

T = m_{A} a

... (1)

From B,

R (↓)

:

m_{B} g - T = m_{B} a

... (2)

Substitute (1) into (2):

m_{B} g - m_{A} a = m_{B} a

m_{B} g = m_{A} a + m_{B} a = (m_{A} + m_{B}) a

Acceleration

a = \frac{m _{B} g}{m _{A} + m _{B}}

Tension

T = \frac{m _{A} m _{B} g}{m _{A} + m _{B}}

Exam Tip

Substitute the acceleration back into equation (1) to find T. Notice that T is always less than

m_{B} g

- if it were equal, there would be no acceleration!

01Worked Example

Particle A of mass 3 kg rests on a smooth horizontal table. It is connected by a light inextensible string, which passes over a smooth pulley at the edge of the table, to particle B of mass 2 kg hanging freely. The system is released from rest. Find:

(a) the acceleration of the system

(b) the tension in the string

(c) the speed of the particles after 1.5 seconds

1

Resolve and apply F = ma:

For A,

R (\to)

:

T = 3 a

... (1)

For B,

R (↓)

:

2 g - T = 2 a

... (2)

2

(a) Substitute (1) into (2):

2 g - 3 a = 2 a

2 g = 5 a

a = \frac{2 \times 9.8}{5} = 3.92 m s^{- 2}

3

(b) Substitute a back into (1):

T = 3 \times 3.92 = 11.76 N

4

(c) Find speed using SUVAT:

v = u + a t = 0 + 3.92 \times 1.5 = 5.88 m s^{- 1}

Answer(a) 3.92 m s⁻², (b) 11.76 N, (c) 5.88 m s⁻¹

Force Exerted on Pulley by String

The string exerts a force on the pulley. For a smooth pulley, tension is the same on both sides.

Perpendicular Strings

R = T 2

At 45° to horizontal

Parallel Strings

R = 2 T

Acting vertically downward

General Formula

The force on the pulley is the vector sum of the two tensions:

R = 2 T cos θ

where θ is the angle each string makes with the line of symmetry.

Rough Horizontal Surface

When the horizontal surface is rough, friction opposes the motion of particle A.

For Particle A

R (\to)

: With friction

F = μ R

:

T - μ m_{A} g = m_{A} a

For Particle B

R (↓)

: Resolve vertically:

m_{B} g - T = m_{B} a

Substitution Method

From (1):

T = m_{A} a + μ m_{A} g

. Substitute into (2) to find:

a = \frac{( m _{B} - μ m _{A} ) g}{m _{A} + m _{B}}

Worked Example02

Particle A of mass 4 kg rests on a rough horizontal table with coefficient of friction μ = 0.3. It is connected by a light inextensible string over a smooth pulley to particle B of mass 5 kg. Find the acceleration, tension, and the force exerted on the pulley.

1

Calculate friction:

R (↑)

for A:

R = 4 g = 39.2 N

F = μ R = 0.3 \times 39.2 = 11.76 N

2

Resolve and apply F = ma:

For A,

R (\to)

:

T - 11.76 = 4 a

... (1)

For B,

R (↓)

:

49 - T = 5 a

... (2)

3

From (1): $T = 4 a + 11.76$ . Substitute into (2):

49 - (4 a + 11.76) = 5 a

37.24 = 9 a

, so

a = 4.14 m s^{- 2}

4

Substitute a back into (1):

T = 4 \times 4.14 + 11.76 = 28.3 N

5

Force on pulley: Strings are perpendicular, so:

P = T 2 = 28.3 \times 2 = 40.0 N

(at 45° below horizontal)

Answera = 4.14 m s⁻², T = 28.3 N, P = 40.0 N

Two Hanging Particles

When two particles hang vertically on either side of a pulley, the heavier particle descends while the lighter one ascends.

For A (descending)

R (↓)

:

m_{A} g - T = m_{A} a

For B (ascending)

R (↑)

:

T - m_{B} g = m_{B} a

Deriving the Formulas

Substitute to eliminate T:

Derivation

From A,

R (↓)

:

m_{A} g - T = m_{A} a

... (1)

From B,

R (↑)

:

T - m_{B} g = m_{B} a

... (2)

From (2):

T = m_{B} a + m_{B} g

. Substitute into (1):

m_{A} g - (m_{B} a + m_{B} g) = m_{A} a

(m_{A} - m_{B}) g = (m_{A} + m_{B}) a

Acceleration

a = \frac{( m _{A} - m _{B} ) g}{m _{A} + m _{B}}

Tension

T = \frac{2 m _{A} m _{B} g}{m _{A} + m _{B}}

Important

Substitute a back into (2) to find T. Note that

m_{B} g < T < m_{A} g

- the tension is between the two weights.

Worked Example03

Two particles A and B of masses 7 kg and 3 kg are connected by a light inextensible string passing over a smooth fixed pulley. The system is released from rest. Find (a) the acceleration, (b) the tension, (c) the force on the pulley.

1

Resolve and apply F = ma:

For A (7 kg),

R (↓)

:

7 g - T = 7 a

... (1)

For B (3 kg),

R (↑)

:

T - 3 g = 3 a

... (2)

2

(a) From (2): $T = 3 a + 3 g$ . Substitute into (1):

7 g - (3 a + 3 g) = 7 a

4 g = 10 a

, so

a = 0.4 g = 3.92 m s^{- 2}

3

(b) Substitute a back into (2):

T = 3 (3.92) + 3 (9.8) = 11.76 + 29.4 = 41.2 N

4

(c) Force on pulley: Strings are vertical, so:

P = 2 T = 2 \times 41.2 = 82.4 N

(downward)

Answer(a) a = 3.92 m s⁻², (b) T = 41.2 N, (c) P = 82.4 N

04Worked Example

Two particles P and Q have masses 5 kg and 3 kg respectively. They are connected by a light inextensible string over a smooth pulley. Initially P is 2 m above the ground. The system is released from rest. Find:

(a) the time for P to reach the ground

(b) the speed of P when it hits the ground

1

Resolve and apply F = ma:

For P,

R (↓)

:

5 g - T = 5 a

... (1)

For Q,

R (↑)

:

T - 3 g = 3 a

... (2)

From (2):

T = 3 a + 3 g

. Sub into (1):

5 g - 3 a - 3 g = 5 a

2 g = 8 a

, so

a = \frac{g}{4} = 2.45 m s^{- 2}

2

(a) Find time using $s = u t + \frac{1}{2} a t^{2}$ :

2 = 0 + \frac{1}{2} \times 2.45 \times t^{2}

t^{2} = \frac{4}{2.45} = 1.633

, so

t = 1.28 s

3

(b) Find speed using $v^{2} = u^{2} + 2 a s$ :

v^{2} = 0 + 2 \times 2.45 \times 2 = 9.8

v = 3.13 m s^{- 1}

Answer(a) t = 1.28 s, (b) v = 3.13 m s⁻¹

Pulley on Inclined Plane

When one particle is on an inclined plane, we must resolve forces parallel and perpendicular to the plane.

For A (smooth plane)

R (↗)

: Up the plane:

T - m_{A} g sin θ = m_{A} a

For B (hanging)

R (↓)

: Vertically down:

m_{B} g - T = m_{B} a

Normal Reaction

R (↖)

: Perpendicular to plane:

R = m_{A} g cos θ

. This is needed if the plane is rough.

Worked Example05

Particle A of mass 4 kg is on a smooth plane inclined at 30° to the horizontal. It is connected by a string over a smooth pulley to particle B of mass 3 kg hanging freely. Find the acceleration and tension.

1

Weight components:

4 g sin 30° = 19.6

N,

3 g = 29.4

N

Since 29.4 > 19.6, B descends, A moves up plane.

2

Resolve and apply F = ma:

For A,

R (↗)

:

T - 19.6 = 4 a

... (1)

For B,

R (↓)

:

29.4 - T = 3 a

... (2)

3

From (1): $T = 4 a + 19.6$ . Substitute into (2):

29.4 - (4 a + 19.6) = 3 a

9.8 = 7 a

, so

a = 1.4 m s^{- 2}

4

Substitute a back into (1):

T = 4 (1.4) + 19.6 = 25.2 N

Answera = 1.4 m s⁻², T = 25.2 N

Worked Example06

Particle A (5 kg) is on a rough plane inclined at 40° with μ = 0.2. Particle B (4 kg) hangs freely. Find the acceleration and tension.

1

Calculate forces:

Down plane:

5 g sin 40° = 31.5

N. Weight of B:

4 g = 39.2

N

Since 39.2 > 31.5, B descends, A moves up.

2

Friction (opposes motion up plane):

R (↖)

:

R = 5 g cos 40° = 37.5

N

F = μ R = 0.2 \times 37.5 = 7.5

N (down plane)

3

Resolve and apply F = ma:

For A,

R (↗)

:

T - 31.5 - 7.5 = 5 a

→

T - 39 = 5 a

... (1)

For B,

R (↓)

:

39.2 - T = 4 a

... (2)

4

From (1): $T = 5 a + 39$ . Substitute into (2):

39.2 - (5 a + 39) = 4 a

0.2 = 9 a

, so

a = 0.022 m s^{- 2}

5

Substitute back:

T = 5 (0.022) + 39 = 39.1 N

Answera = 0.02 m s⁻² (very slow!), T = 39.1 N

Subsequent Motion

Exam questions often ask about what happens after an event changes the system - such as the string breaking or a particle hitting the ground.

String Breaks

Each particle moves independently:

Hanging particle: falls freely under gravity (a = g)

Particle on surface: continues with current velocity, decelerates due to friction (or no force if smooth)

Particle Hits Ground

For the other particle:

String goes slack (T = 0)

Continues with velocity at that instant

Use SUVAT with new acceleration (often a = g or a = -g sin θ)

Problem-Solving Strategy

Find acceleration and velocity during connected phase

Identify the "event" that changes the system

New initial conditions: velocity at that moment

Apply Newton's laws with new forces (no tension)

Use SUVAT for subsequent motion

Worked Example07

Particle P (4 kg) rests on a rough horizontal table with μ = 0.3. It is connected by a light string over a smooth pulley to particle Q (3 kg) hanging freely 1.2 m above the ground. The system is released from rest. Find the total distance P travels before coming to rest.

1

Phase 1 - Both particles moving:

Normal reaction:

R = 4 g = 39.2

N

Friction:

F = μ R = 0.3 \times 39.2 = 11.76

N

For P:

T - 11.76 = 4 a

... (1)

For Q:

3 g - T = 3 a

→

29.4 - T = 3 a

... (2)

2

Adding (1) and (2):

29.4 - 11.76 = 7 a

a = 2.52 m s^{- 2}

3

Speed when Q hits ground (s = 1.2 m):

v^{2} = u^{2} + 2 a s = 0 + 2 \times 2.52 \times 1.2 = 6.048

v = 2.46 m s^{- 1}

4

Phase 2 - P alone on rough surface:

String slack, only friction acts on P.

Deceleration:

a = μg = 0.3 \times 9.8 = 2.94 m s^{- 2}

v^{2} = u^{2} + 2 a s

with

v = 0, u = 2.46

:

0 = 6.048 - 2 \times 2.94 \times s

s = 1.03

m

5

Total distance:

Phase 1: P moves 1.2 m (same as Q falls)

Phase 2: P moves 1.03 m further

Total = 1.2 + 1.03 = 2.23 m

AnswerP travels 2.23 m before coming to rest

Worked Example08

Two particles A (3 kg) and B (5 kg) are connected by a light inextensible string passing over a smooth pulley. Initially A is 2 m above the ground and B is 0.8 m above the ground. The system is released from rest. Find the maximum height above the ground that A reaches.

1

Phase 1 - Both particles moving (B descends, A rises):

For B:

5 g - T = 5 a

... (1)

For A:

T - 3 g = 3 a

... (2)

Adding:

5 g - 3 g = 8 a

, so

a = \frac{g}{4} = 2.45 m s^{- 2}

2

Speed when B hits ground (s = 0.8 m):

v^{2} = u^{2} + 2 a s = 0 + 2 \times 2.45 \times 0.8 = 3.92

v = 1.98 m s^{- 1}

3

Position of A when B hits ground:

A has risen 0.8 m (same distance B fell)

A is now at 2 + 0.8 = 2.8 m above ground

4

Phase 2 - A continues upward under gravity:

String slack, A decelerates at g = 9.8 m s⁻²

v^{2} = u^{2} + 2 a s

with

v = 0, u = 1.98, a = - 9.8

:

0 = 3.92 - 2 \times 9.8 \times s

s = 0.2

m

5

Maximum height of A:

2.8 + 0.2 = 3.0 m above ground

AnswerA reaches a maximum height of 3.0 m above the ground

Practice

Question 1: A particle A of mass 6 kg is on a rough horizontal table (μ = 0.25) connected by a string over a smooth pulley to particle B of mass 4 kg hanging freely. Find the acceleration and tension in the string.

1

Friction:

F = μ R = 0.25 \times 6 g = 14.7 N

2

Resolve and apply F = ma:

For A,

R (\to)

:

T - 14.7 = 6 a

... (1)

For B,

R (↓)

:

39.2 - T = 4 a

... (2)

3

From (1):

T = 6 a + 14.7

. Substitute into (2):

39.2 - (6 a + 14.7) = 4 a

→

24.5 = 10 a

→

a = 2.45 m s^{- 2}

4

Substitute back:

T = 6 (2.45) + 14.7 = 29.4 N

Answera = 2.45 m s⁻², T = 29.4 N

Practice

Question 2: Two particles of masses 5 kg and 2 kg are connected by a string over a smooth pulley. The 5 kg particle is 3 m above the ground. Find: (a) the acceleration, (b) the time for the 5 kg particle to reach the ground, (c) the maximum height reached by the 2 kg particle above its starting position.

1

Resolve and apply F = ma:

For 5 kg,

R (↓)

:

5 g - T = 5 a

... (1)

For 2 kg,

R (↑)

:

T - 2 g = 2 a

... (2)

From (2):

T = 2 a + 2 g

. Sub into (1):

5 g - 2 a - 2 g = 5 a

3 g = 7 a

, so

a = \frac{3 g}{7} = 4.2 m s^{- 2}

2

(b) Time:

s = \frac{1}{2} a t^{2}

→

3 = \frac{1}{2} (4.2) t^{2}

→

t = 1.20 s

3

Speed at impact:

v = a t = 4.2 \times 1.2 = 5.04 m s^{- 1}

4

(c) 2 kg rises 3 m while connected, then continues with v = 5.04 m s⁻¹.

Extra height:

s = \frac{v ^{2}}{2 g} = \frac{25.4}{19.6} = 1.30 m

Total height = 3 + 1.30 = 4.30 m

Answer(a) 4.2 m s⁻², (b) 1.20 s, (c) 4.30 m

Practice

Question 3: Particle A (3 kg) is on a smooth plane inclined at 25° to the horizontal. It is connected by a string over a smooth pulley at the top of the plane to particle B (2 kg) hanging vertically. Find the acceleration and state which direction the system moves.

1

A's weight down plane:

3 g sin 25° = 12.4

N. Weight of B:

2 g = 19.6

N

Since 19.6 > 12.4, B descends and A moves up the plane.

2

Resolve and apply F = ma:

For A,

R (↗)

:

T - 12.4 = 3 a

... (1)

For B,

R (↓)

:

19.6 - T = 2 a

... (2)

3

From (1):

T = 3 a + 12.4

. Substitute into (2):

19.6 - (3 a + 12.4) = 2 a

→

7.2 = 5 a

→

a = 1.44 m s^{- 2}

Answera = 1.44 m s⁻², B descends, A moves up the plane

Summary

Key Pulley Formulas

ConfigurationAccelerationNotes

Horizontal (smooth)

a = \frac{m _{B} g}{m _{A} + m _{B}}

B hangs, A on surface

Horizontal (rough)

a = \frac{( m _{B} - μ m _{A} ) g}{m _{A} + m _{B}}

Include friction on A

Two hanging particles

a = \frac{( m _{A} - m _{B} ) g}{m _{A} + m _{B}}

m_A > m_B

Inclined plane

a = \frac{m _{B} g - m _{A} g s i n θ}{m _{A} + m _{B}}

A on plane, B hangs

Force on Pulley

R = 2 T

(vertical strings)

R = 2 T cos θ

(angled strings)

Exam Strategy

Draw force diagrams with cut string

Resolve:

R (\to)

,

R (↓)

,

R (↗)

Apply F = ma for each particle

Substitute to find a, then T