Atwood Machine

The setup

Hang a mass m1 on one end of a light string, a mass m2 on the other, and drape the string over a pulley so the two sides hang vertically. That is the entire Atwood machine. If m1 > m2, the heavier mass falls, the lighter rises, and the whole system accelerates — but more gently than a free drop. The cleverness is in how much more gently: you control the acceleration just by choosing the masses.

In the idealized version we assume the string is massless and inextensible, the pulley is massless and turns without friction, and the string does not slip. Those assumptions are not just laziness — they let the algebra reveal the physics cleanly, and we'll relax them one at a time afterward.

Newton's second law, twice

The key constraint is the string: because it cannot stretch, both masses move with the same speed and the same magnitude of acceleration, a. When m1 goes down by some distance, m2 goes up by exactly the same distance. So we write Newton's second law for each mass separately, picking "downward positive for m1" and "upward positive for m2" so both accelerations are +a.

For the heavy mass m1 (moving down), the forces are its weight m1·g down and the tension T up:

m1·g − T = m1·a

For the light mass m2 (moving up), tension T is up and its weight m2·g is down:

T − m2·g = m2·a

Add the two equations — the tension cancels because an ideal massless pulley passes it through unchanged:

(m1 − m2)·g = (m1 + m2)·a
a = (m1 − m2)·g / (m1 + m2)

Substitute that back to find the tension:

T = 2·m1·m2·g / (m1 + m2)

Notice T is the harmonic mean of the two weights m1·g and m2·g — it always lands strictly between them. That has to be true: if T equaled the heavy weight, the heavy side would feel no net force and couldn't accelerate.

Reading the formula

The acceleration formula is worth staring at, because every limit tells a story:

Case	Acceleration a	What happens
m1 = m2	0	Perfectly balanced — the system stays at rest or drifts at constant speed.
m1 slightly > m2	small fraction of g	Slow, easily-timed descent (Atwood's whole point).
m1 = 2·m2	g/3 ≈ 3.27 m/s²	The heavy side falls at one-third of free fall.
m1 = 3·m2	g/2 ≈ 4.90 m/s²	Half of free fall.
m1 ≫ m2	→ g ≈ 9.81 m/s²	The tiny mass is dragged along; the heavy one is in near free fall.

The factor (m1 − m2)/(m1 + m2) is a dial between 0 and 1. Set it to 0.05 and an object that would normally hit 9.8 m/s² instead crawls at 0.49 m/s² — slow enough to time with a stopwatch and a meter stick, which is exactly how 18th-century students measured g.

A worked example

Take m1 = 0.60 kg and m2 = 0.50 kg, with g = 9.81 m/s².

a = (0.60 − 0.50)·9.81 / (0.60 + 0.50)
  = 0.10 · 9.81 / 1.10
  = 0.892 m/s²

T = 2 · 0.60 · 0.50 · 9.81 / 1.10
  = 5.886 / 1.10
  = 5.35 N

Sanity check the tension: m2's weight is 0.50·9.81 = 4.91 N and m1's weight is 0.60·9.81 = 5.89 N. The tension 5.35 N sits neatly between them. And the acceleration, 0.892 m/s², is under a tenth of g — a 1.10 kg system inching downward, easy to film and time.

When the pulley has mass

A real pulley resists being spun. Model it as a uniform disk of mass M and radius R, so its rotational inertia is I = (1/2)MR². To angularly accelerate it, the two string tensions must now differ — T1 on the heavy side, T2 on the light side — and that difference supplies the torque:

m1·g − T1 = m1·a            (heavy mass)
T2 − m2·g = m2·a            (light mass)
(T1 − T2)·R = I·α = I·(a/R)  (pulley rotation, no slip → α = a/R)

Solving the three together gives:

a = (m1 − m2)·g / (m1 + m2 + I/R²)
  = (m1 − m2)·g / (m1 + m2 + M/2)

The pulley behaves like an extra M/2 of mass bolted onto the moving system. A 0.20 kg pulley in our worked example adds 0.10 kg to the denominator, dropping the acceleration from 0.892 to 0.818 m/s² — a real, measurable slowdown, and a favorite exam twist.

Model	Acceleration formula	Tension behavior
Ideal pulley (massless, frictionless)	a = (m1 − m2)g / (m1 + m2)	Single T = 2·m1·m2·g/(m1+m2)
Massive pulley (disk M)	a = (m1 − m2)g / (m1 + m2 + M/2)	T1 > T2; difference spins the wheel
With axle friction torque τ_f	a = [(m1 − m2)g − τ_f/R] / (m1 + m2 + M/2)	Friction further reduces a until it can stall

The energy view

You can sidestep forces entirely with energy. After the system has moved a distance d, the heavy mass has dropped d and the light one risen d, so the net change in gravitational potential energy is −(m1 − m2)g·d. With an ideal pulley all of that becomes kinetic energy shared by both masses (which move at the same speed v):

(m1 − m2)·g·d = ½·(m1 + m2)·v²

Differentiate, or use v² = 2·a·d, and you recover a = (m1 − m2)g/(m1 + m2) — the same answer. With a massive pulley you simply add the wheel's rotational kinetic energy ½·I·ω² to the right side, which is why M/2 reappears in the denominator. The two routes — forces and energy — must agree, and checking that they do is a good way to catch an algebra slip.

JavaScript — Atwood calculations

const g = 9.81;

// Ideal Atwood machine
function atwood(m1, m2) {
  const a = (m1 - m2) * g / (m1 + m2);
  const T = 2 * m1 * m2 * g / (m1 + m2);
  return { a, T };
}

const { a, T } = atwood(0.60, 0.50);
console.log(`a = ${a.toFixed(3)} m/s²`);   // 0.892
console.log(`T = ${T.toFixed(2)} N`);      // 5.35

// Tension always sits between the two weights
console.log(0.50 * g, T, 0.60 * g);        // 4.91 < 5.35 < 5.89

// With a uniform-disk pulley of mass M (radius cancels)
function atwoodWithPulley(m1, m2, M) {
  return (m1 - m2) * g / (m1 + m2 + M / 2);
}
console.log(atwoodWithPulley(0.60, 0.50, 0.20).toFixed(3)); // 0.818

// Back out g from a measured acceleration (Atwood's original use)
function measureG(m1, m2, aMeasured) {
  return aMeasured * (m1 + m2) / (m1 - m2);
}
console.log(measureG(0.60, 0.50, 0.892).toFixed(2)); // 9.81

// Time to descend a height h from rest: h = ½at²
function descentTime(m1, m2, h) {
  const { a } = atwood(m1, m2);
  return Math.sqrt(2 * h / a);
}
console.log(descentTime(0.60, 0.50, 1.0).toFixed(2)); // 1.50 s over 1 m

Where the Atwood machine shows up

• Measuring g. Atwood's 1784 purpose — turn fast free fall into a slow, timeable descent and solve back for the acceleration due to gravity.
• Elevators. A counterweight nearly equal to the loaded car means the motor only fights the small mass difference, not the full weight — an Atwood machine with a motor on the axle.
• Funicular and cable railways. Two cars on a slope counterbalance each other; one descending helps haul the other up.
• Sash windows and garage doors. Hidden counterweights or springs balance the panel so it feels nearly weightless.
• Teaching Newton's second law. The cleanest two-body coupled system in the curriculum — forces, tension, constraints, and rotational inertia all in one apparatus.
• Cranes and theatrical fly systems. Counterweighted rigging lets a stagehand raise heavy scenery by hand.

Common mistakes

• Assuming the tension equals a weight. T is neither m1·g nor m2·g — it lies strictly between them. Setting T = m1·g forces a = 0, which is wrong whenever the masses differ.
• Forgetting the masses share one acceleration. The inextensible string ties their speeds and accelerations together; treating them independently breaks the whole derivation.
• Sign-convention slips. Pick "motion-forward positive" for each mass (down for the heavy one, up for the light one) so both accelerations are +a, then add the equations.
• Ignoring the pulley's inertia. A massive pulley adds M/2 to the denominator and makes the two tensions unequal. Treating it as ideal overestimates the acceleration.
• Mixing up potential-energy bookkeeping. Only the mass difference changes the net potential energy — the common mass that goes up cancels the part that goes down.
• Confusing acceleration with effective gravity. The masses do accelerate at a, but a is the effective gravity they feel; it's smaller than g, not equal to it.