Connected Bodies (Pulleys) (Edexcel A Level Maths: Mechanics): Revision Note
Exam code: 9MA0
Connected bodies (pulleys)
What is a pulley?
A pulley is a wheel like device that rotates as a string passes over it allowing motion of any particles attached to the string
Pulleys allow a (inextensible) string to change its orientation
In A level mathematical models, pulleys will always be smooth and light, so there is no friction involved at the pulley and its mass is negligible
A peg is similar to a pulley but is a fixed point that a particle can be suspended from (like a nail in a wall)
How do I solve pulley questions?
In all pulley questions the particles are moving in different directions so are considered separately rather than the system being treated as one
If a particle is in motion in the direction being considered then Newton’s Laws of Motion apply so use “F = ma” (N2L)
For constant acceleration the ‘suvat’ equations could be involved
Step 1. Draw a series of diagrams
Label the forces and the positive direction of motion for each particle.
Colour coding forces acting on each particle may help
Step 2. Write equations of motion, using “F = ma”
Equations 1 and 2: Treating each particle separately
(↑) T - m1g = m1a
(↓) m2g - T = m2a
Step 3. Solve the relevant equation(s) and answer the question
Some trickier problems may lead to simultaneous equations
Worked Example
A block of mass 25 kg lies on a smooth horizontal table. A light, inextensible string passes over a smooth pulley attached to the edge of the table. The string connects the 25 kg block to another, of mass 60 kg, that hangs freely as shown in the diagram below.
The system is released from rest.
Find the tension, T N , in the string and the acceleration, a m s-2 ,of the system.
Answer:
Examiner Tips and Tricks
Sketch a diagram or add to a diagram given in a question.
All pulleys are smooth and light; in many questions the pulley itself can be ignored.
In pulley questions the particles will be moving in different directions - so each particle needs to be considered separately.
If one particle is on a horizontal surface (such as a desk or table) then the weight only need be considered if friction is involved (since F = μR and R is related to weight).
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Questions students ask about Connected Bodies (Pulleys)
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How do you find the equation of motion for a particle in a pulley system?
Treat each particle separately: take its own direction of motion as positive, then write F = ma for that particle alone.
- Draw a force diagram for each particle, labelling the tension T, the weight mg and the direction of the acceleration a
- Resolve in the direction that particle moves — forces that way are positive, forces against it negative
- Write F = ma using only that particle's mass
- Solve the two equations, usually simultaneously
For two particles hanging either side of a pulley (m₁ rising, m₂ falling):
- (↑) T − m₁g = m₁a
- (↓) m₂g − T = m₂a
T and a are the same in both, because the string is light and inextensible.
Page example: 25 kg block on a smooth table, 60 kg block hanging.
- (→) T = 25a
- (↓) 60g − T = 60a
- So 60g − 25a = 60a, giving 85a = 60g and a = 12g/17 = 6.92 m s⁻² (3 s.f.)
- T = 25 × 6.91764... = 173 N (3 s.f.)
The normal reaction R is vertical, so it never appears in the horizontal equation.
How do you find the force exerted on a pulley?
Find the tension T first, then add the two tension forces as vectors — the string pulls on the pulley with a force of T along each section.
- Both sections vertical: the two forces both act downwards, so the force on the pulley is T + T = 2T, vertically downwards
- One section horizontal, one vertical (the page's table set-up): the two forces are at right angles, so use Pythagoras — magnitude = √(T² + T²) = T√2, acting at 45° between the two strings
Using the page's example, where T = 172.941... N:
- Force on the pulley = 172.941... × √2 = 245 N (3 s.f.)
- It acts at 45° below the horizontal, into the corner of the table
Method:
- Write F = ma for each particle and solve for T
- Draw both tension arrows acting on the pulley, each pointing away along the string
- Add them as vectors to get the resultant
Always use the unrounded value of T.
How do you find the greatest height reached by a particle in a pulley system?
Split the motion into two stages: while the string is taut, and then after the other particle lands and the string goes slack.
Stage 1 — string taut
- Use F = ma on each particle to find the acceleration a
- Use suvat to find the speed v of the rising particle at the moment the other one hits the ground: v² = u² + 2as
Stage 2 — string slack
- The tension drops to zero, so only the weight acts and the particle decelerates at g
- At the highest point v = 0, so the extra height is s = v²/(2g)
Add the heights from both stages, plus any height the particle started at.
Example: 3 kg and 5 kg particles released from rest, with the 5 kg one 1.5 m above the ground and the 3 kg one starting on the ground.
- 5g − T = 5a and T − 3g = 3a, so 2g = 8a and a = 2.45 m s⁻²
- v² = 0 + 2 × 2.45 × 1.5 = 7.35
- Extra height = 7.35 ÷ (2 × 9.8) = 0.375 m
- The 3 kg particle rises 1.5 m while the string is taut, so its greatest height = 1.5 + 0.375 = 1.875 m ≈ 1.88 m above the ground
