A-level · Mechanics · Beta

Connected particles and pulleys

Compare the forces on two connected masses and derive their common acceleration.

Adjust both masses. Trace the motion from rest using the time slider.

a=(m2−m1)gm1+m2a=\frac{(m_2-m_1)g}{m_1+m_2}
Two masses joined by a light inextensible string over a smooth fixed pulleyFrom rest; positive direction: mass 2 down2 kgMass 13 kgMass 2Mass 1 forcesT = 23.52 N19.6 NMass 2 forcesT = 23.52 N29.4 NEqual tension; equal acceleration magnitudes.

Scroll the diagram sideways to see every label.

Mass 2 accelerates down; mass 1 accelerates up.

Signed acceleration (m/s²)
1.96
Tension (N)
23.52
Model time (s)
0
Mass 2 displacement down (m)
0
Mass 2 velocity down (m/s)
0
Build the equations

Take upwards positive for mass 1 and downwards positive for mass 2.

T−m1g=m1a,m2g−T=m2aT-m_1g=m_1a,\quad m_2g-T=m_2a

Add to eliminate tension: (m2−m1)g=(m1+m2)a(m_2-m_1)g=(m_1+m_2)a

Then T=m1(g+a)=m2(g−a)T=m_1(g+a)=m_2(g-a). A negative a reverses the assumed directions.

The common tension follows from the light string and smooth pulley. Do not set T equal to a weight unless that particle has zero acceleration.

Ideal particles, light inextensible taut string, smooth fixed pulley, no drag; g = 9.8 m/s². Released from rest. Trace ends at 1 m travel in either direction; no collision is simulated. Force diagrams share an arrow scale. Values rounded to 3 decimal places.

Predict, test and explain

Step 1 of 3

Predict

Is the tension equal to the weight of the heavier particle while the system accelerates?

InvestigateMake the two masses equal. What happens to tension and acceleration?

Explore and compare

Choose the investigation above. Predict what will change before you move a control.

Try it in the simulation. Change one thing at a time.

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Hints, self-check and connections
Try a self-check

Fixed values, separate from the diagram controls.

Is the tension equal to the weight of the heavier particle while the system accelerates?

a=(3−2)9.83+2a=\frac{(3-2)9.8}{3+2}

Hint

Draw forces on each particle and choose a consistent positive direction for each.

Reveal answer

For the descending heavier particle, mg − T = ma. Nonzero acceleration means tension is less than its weight.

a=1.96,T=23.52a=1.96,\quad T=23.52

Build an explanation

Hint 1

Draw forces on each particle and choose a consistent positive direction for each.

Hint 2

For the descending heavier particle, mg − T = ma. Nonzero acceleration means tension is less than its weight.

Hint 3

State the idealised assumptions before applying the equations.

Worked example

This example uses fixed values, separate from the diagram controls.

Two masses of 2 kg and 3 kg hang over a smooth pulley on a light inextensible string. Find acceleration and tension, with g = 9.8 m/s².

  1. For the 2 kg mass upwards: T − 19.6 = 2a. For the 3 kg mass downwards: 29.4 − T = 3a.
  2. Add: 9.8 = 5a, so a = 1.96 m/s².
  3. Substitute back: T = 19.6 + 2(1.96) = 23.52 N.

Answer: Acceleration 1.96 m/s², with the 3 kg mass descending; tension 23.52 N.

Connect this idea

Watch out: For the descending heavier particle, mg − T = ma. Nonzero acceleration means tension is less than its weight.

Specification and learning route

AQA R4 · Edexcel Mechanics 8.4

AS and A-level. This model illustrates selected content; references are not a claim of full coverage or exam-board endorsement.

Useful starting knowledge: Newton’s second law; Weight = mg; Simultaneous equations.

AQA specification · Edexcel specification