Two force balances, one acceleration

Define a as positive downward for the right mass. The left-up and right-down equations are T − m₁g = m₁a and m₂g − T = m₂a. Adding them eliminates the internal tension:

The two accelerations have equal magnitudes and opposite vertical directions. With release from rest and constant gravity, the right mass has velocity v = at and displacement s = at²/2. Its hanging length is 20 + s metres; the left length is 20 − s. These two hanging segments always sum to 40 metres.

Predict the forces before changing the time

Set Left mass m1 to 4 kg, Right mass m2 to 6 kg and Gravity g to 10 m/s². Both masses start from rest. At Elapsed time t = 2 s:

1. The system has a 60 − 40 = 20 N driving imbalance and 10 kg total mass, so a = 20/10 = 2 m/s² downward on the right.
2. The left mass requires T = 40 + 4 × 2 = 48 N. The right independently gives T = 60 − 6 × 2 = 48 N.
3. The right velocity is 2 × 2 = 4 m/s and its displacement is 2 × 2²/2 = 4 m.
4. The left and right hanging lengths become 16 m and 24 m. Their sum stays 40 m.

As a boundary check, choose two 3 kg masses at the same gravity. Their weights are both 30 N, a = 0 and T = 30 N. Zero net force does not mean that each individual force vanishes.

Read the arrows, then test the prediction

Before changing a control, predict which mass will accelerate downward and which individual forces remain nonzero. Compare those predictions with the arrow lengths and numeric results; changing time alone should not change either force balance.

1. Open the Atwood Machine interactive module. Select the masses and gravity from the worked example.
2. Keep Show force vectors enabled. Green T arrows point upward; the blue and rose mg arrows point downward. The right weight exceeds tension, while tension exceeds the left weight.
3. Change Elapsed time t from 1 to 2 seconds. Displacement grows from 1 to 4 m; acceleration and tension remain 2 m/s² and 48 N.
4. Make both masses 3 kg. Compare the equal upward and downward arrows and the 30 N tension readout. Set Gravity g to 0 to distinguish balanced nonzero forces from zero forces.
5. Toggle the force vectors off and back on; this changes the annotations, not the calculation. Drag the drawing or use the arrow keys to pan. Reset view and Home restore only the view, retaining the selected physical values.

What the picture does and does not measure

The string is massless and inextensible, the pulley is fixed, massless and frictionless, gravity is constant, and air resistance is ignored. A real pulley can have rotational inertia or friction, so its acceleration and two tensions need not match this ideal model.

The dropdowns select independent snapshots from rest, not an instantaneous physical mass swap during motion. The model permits masses from 1 to 10 kg and times from 0 to 2 s. At t = 0, velocity and displacement are zero even if acceleration is nonzero. At g = 0, all displayed forces and motion from rest are zero.

Mass icons, pulley size and clearances are schematic and resize with the viewport. Read physical lengths from Results. Within one configuration, all force arrows use the same magnitude scale, fitted to the panel; compare their relative lengths rather than pixels between configurations. Zero force has no arrow. Arrow labels round to 0.1 N; numeric results show up to 12 significant digits.

References and sign conventions

These sources support the ideal force model and its limiting cases. Their coordinate conventions can differ, so compare physical directions and explicitly chosen signs rather than copying an acceleration sign without its definition.

• OpenStax, University Physics Volume 1, section 6.1, example 6.5: the paired free-body diagrams, opposite accelerations and ideal-pulley force equations. The numerical experiment above is calculated independently for Riano.
• MIT OpenCourseWare, 6.055J/2.038J Homework 6, problem 8: symmetry, limiting cases and tension in an ideal Atwood machine. This module explicitly uses right-down as positive; reversing that convention reverses the acceleration sign.