Solve two hanging particles joined over a smooth pulley
Use separate force diagrams and one taut-string constraint to solve a connected system’s acceleration and tension.
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.4. Use g = 10 m s⁻² throughout these problems. Each particle has positive constant mass. Forces and acceleration are constant within each stated stage; linear motion only. Use the resultant force in F = ma. A light taut inextensible string over a smooth pulley has one tension and equal acceleration magnitudes; those constraints cease if the string becomes slack. A rope can pull but cannot push; a light rigid tow-bar may exert tension or thrust. Resistance other than the stated friction is neglected unless explicitly supplied. In our rough-contact model, sliding friction has magnitude μR and opposes motion; at rest, friction adjusts with magnitude at most μR. Check contact and motion assumptions before continuing a solution. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Use signed vertical Newton equations.
- Eliminate an unknown from two simultaneous linear equations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For two masses on a taut light inextensible string over a smooth pulley, what is shared?
What you will learn
- Set equal acceleration magnitudes and a common tension under the stated ideal-string model.
- Solve the coupled equations and verify both particles separately.
The constraint links motions, not individual forces
Two positive masses hang on opposite sides of a light inextensible string over a smooth pulley. While the string stays taut, one rises as the other falls with equal acceleration magnitudes. The light string and smooth pulley give a common tension T in this model. Choose a positive direction along the expected motion separately for each particle; their vertical accelerations have opposite signs on a single upward axis.
For 3 kg and 5 kg released from rest with the 5 kg mass descending, the equations are 50 − T = 5a and T − 30 = 3a. Adding these signed along-motion equations eliminates the common tension and gives 20 = 8a, hence a = 2.5 m s⁻². Then T = 30 + 3(2.5) = 37.5 N. Tension is not equal to either weight because both particles accelerate.
Check both equations and the allowed interval
Substituting into the heavier equation gives 50 − 37.5 = 5(2.5), and into the lighter gives 37.5 − 30 = 3(2.5). This double check catches a sign error that a combined-system equation can hide. For positive unequal masses released under gravity, the heavier descends and acceleration magnitude is less than g; equal masses give zero acceleration, allowing rest or a specified constant speed.
These equations hold before a mass hits a floor, the string becomes slack or another force is introduced. The ideal common-tension assumption would not automatically apply to a heavy string or a non-smooth pulley. A negative computed tension is incompatible with a taut flexible string. State the assumptions instead of treating equal acceleration and equal tension as facts about every connection.
Pause and explain
For 50 − T = 5a and T − 30 = 3a, what follows on adding?
Worked example
Masses 3 kg and 5 kg hang on a light inextensible string over a smooth pulley and are released from rest. Find acceleration and tension before any contact.
Show the worked solution
- Choose 5 kg down and 3 kg up: 50 − T = 5a and T − 30 = 3a.
- Add to get 20 = 8a, so a = 2.5 m s⁻².
- Substitute into the 3 kg equation: T = 30 + 7.5 = 37.5 N, satisfying both balances.
Answer Acceleration 2.5 m s⁻², 5 kg down and 3 kg up; T = 37.5 N.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Masses 2 kg and 3 kg hang on a light inextensible string over a smooth pulley, released from rest. Find acceleration and tension before contact.
Give me a hint
Write one equation for the heavier descending mass and one for the lighter rising mass.
Compare my reasoning
- Equations are 30 − T = 3a and T − 20 = 2a.
- Adding gives 10 = 5a, so a = 2 m s⁻².
- Then T = 20 + 2 × 2 = 24 N, which also satisfies the heavier balance.
a = 2 m s⁻², 3 kg down; T = 24 N.
Look for these in your work
- Use opposite vertical motion directions consistently.
- Check tension in both equations.
Masses 2 kg and 6 kg are released on the same ideal taut-string pulley model. Find acceleration, tension and the distance the heavier mass travels in 1 s from rest, before any boundary.
Give me a hint
Total inertia is the sum of the two masses.
Compare my reasoning
- Adding 60 − T = 6a and T − 20 = 2a gives a = 40/8 = 5 m s⁻².
- T = 20 + 2 × 5 = 30 N.
- Distance from rest in 1 s is ½ × 5 × 1² = 2.5 m downwards.
a = 5 m s⁻², 6 kg down; T = 30 N; distance 2.5 m.
Look for these in your work
- Use the correct common acceleration magnitude.
- Apply the motion calculation only before contact changes the constraints.
A 2 kg mass rises with acceleration 2 m s⁻² while a mass m descends on a light taut inextensible string over a smooth pulley. Find T and m.
Give me a hint
Use the known rising mass to find tension before solving the other equation.
Compare my reasoning
- For the 2 kg mass, T − 20 = 2 × 2, so T = 24 N.
- For the descending mass, 10m − 24 = 2m.
- Thus 8m = 24 and m = 3 kg, larger than 2 kg as expected.
T = 24 N; m = 3 kg.
Look for these in your work
- Use the measured acceleration in each particle’s own equation.
- Check positive mass and the stated descending direction.
Common mistakes
- The two particles move in opposite vertical directions.
- Common tension requires the stated light-string and smooth-pulley model.
- Tension is not generally equal to a weight in an accelerating system.
What can you explain now?
Why can common tension disappear when the two along-motion equations are added?
Compare with the explanation
It has equal magnitude and opposite algebraic signs on the two chosen along-motion axes.
The two actual tension forces both point upward; this is equation elimination, not their vector cancellation on one body.
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Make it stick
Draw both particles, apply the taut-string constraints, add for acceleration and substitute for tension.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.