Physics I: Mechanics › Forces › full formula sheet

a = (m1−m2)g / (m1+m2)Say it: “the acceleration is the mass difference times g over the mass sum”

Two masses, one string, one pulley — the cleanest Newton's-second-law system you'll ever meet.

Notation on this page: m1 is the heavier mass, m2 the lighter, T is the string tension in N, a is the acceleration magnitude in m/s².

Before this lesson: Newton's second law

Where it comes from

In 1784 the Reverend George Atwood had a problem: free fall is fast. A dropped stone hits the ground in under a second — far too quick for 18th-century clocks to time accurately. His fix was brilliantly simple: hang two nearly-equal masses over a pulley. The heavier one still wins, but only barely — the whole system creeps downward at a small fraction of g, slow enough to measure.

Before reading on: m1 = m2 exactly. What does the system do? And if m2 shrinks to nearly zero, what should the acceleration approach?

Both predictions fall out of one formula. Equal masses balance — a = 0, the system hangs in equilibrium. And with m2 → 0, the lone mass m1 is essentially in free fall — a → g. The formula that satisfies both:

a = (m1−m2)g / (m1+m2)the mass difference drives it; the mass sum resists itSay it: “a equals m-one minus m-two, times g, over m-one plus m-two”

Read it as a story: (m1−m2)g is the unbalanced weight — the net external pull — and (m1+m2) is the total inertia being dragged along. F = ma for the whole system, in one line.

Derivation

Write F = ma for each mass separately, with one consistent sign convention. The idealizations: a massless, frictionless pulley (same tension T on both sides) and an inextensible string (both masses share the same speed and acceleration magnitude).

m1g − T
=
m1a
Step 1 — the heavy mass. Take downward as positive for m1: weight pulls down (+m1g), tension pulls up (−T), and it accelerates downward (+a).
T − m2g
=
m2a
Step 2 — the light mass. It moves up, so take upward as positive here: tension pulls up (+T), weight pulls down (−m2g), acceleration is upward (+a). Same magnitude a — the string doesn't stretch.
(m1−m2)g
=
(m1+m2)a
Step 3 — add the equations. The internal tension cancels (+T − T = 0): only the external weights survive. Divide by the mass sum:
a
=
(m1−m2)g/(m1+m2)
Step 4 — the acceleration. Then back-substitute into Step 2: T = m2(g+a) = 2m1m2g/(m1+m2). ∎

Check the tension formula against intuition: it always lands between the two weights (m2g < T < m1g). The heavy mass falls but is partially held up, so T < m1g; the light mass rises, so it must be pulled up harder than its weight, T > m2g.

How to use it

The procedure, every time:

  1. Label the heavier mass m1. If you mix them up, the sign of a flips — decide once, at the start.
  2. Write F = ma per mass with a consistent convention: positive in each mass's direction of motion (down for m1, up for m2).
  3. Add to kill T and solve for a = (m1−m2)g/(m1+m2).
  4. Find T from either equation: T = m2(g+a) = 2m1m2g/(m1+m2).
  5. Sanity-check the limits: m1 = m2 → a = 0; m2 → 0 → a → g, T → 0. If your numbers violate these, back up.
Common mistake: writing T = m1g (“the string holds the heavy mass”). It doesn't — the mass is accelerating downward, so the string holds less than the full weight: T < m1g, always.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: m1 = 3 kg, m2 = 2 kg

  1. Imbalance ratio. (m1−m2)/(m1+m2) = (3−2)/(3+2) = 1/5 = 0.2.
  2. Accelerate. a = 0.2 × 9.8 = 1.96 m/s² — the 3 kg mass descends, the 2 kg mass rises.
  3. Sense. One-fifth of g: a modest imbalance gives a gentle acceleration, exactly Atwood's point.
Common mistake: a = g (“things fall at g”). Only in free fall — here the lighter mass and the string fight back, leaving a fraction of g.
Your turn — m1 = 4 kg, m2 = 2 kg. Acceleration?

Answer: about 3.27 m/s². (4−2)/(4+2) = 2/6 = 1/3; a = 9.8/3 ≈ 3.267 ≈ 3.27 m/s².

Before reading on: for the 3 kg / 2 kg machine, is the tension closer to 19.6 N, 29.4 N, or right in the middle? Guess, then compute.

Example 2 — the tension: same 3 kg / 2 kg machine

  1. Formula. T = 2m1m2g/(m1+m2) = 2 × 3 × 2 × 9.8/5.
  2. Compute. T = 117.6/5 = 23.52 N.
  3. Check the bracket. m2g = 19.6 N < 23.52 < 29.4 = m1g ✓ — between the weights, as it must be.
Common mistake: T = m1g = 29.4 N. The heavy side accelerates downward — the string carries less than the full weight. T < m1g is a law, not a suggestion.
Your turn — m1 = 4 kg, m2 = 2 kg. Tension?

Answer: about 26.1 N. T = 2 × 4 × 2 × 9.8/6 = 156.8/6 ≈ 26.13 ≈ 26.1 N (between 19.6 N and 39.2 N ✓).

Example 3 — how far in 2 s: 3 kg / 2 kg, released from rest

  1. Acceleration. a = 1.96 m/s² (from Example 1) — constant, so kinematics applies.
  2. Distance. d = ½at² = 0.5 × 1.96 × 4 = 3.92 m.
  3. Atwood's payoff: in 2 seconds the mass drops nearly 4 m — easily timed with a stopwatch, where free fall would cover 19.6 m in the same time.
Common mistake: d = ½gt² = 19.6 m — using g instead of the system's actual acceleration. The machine's whole purpose is that a ≠ g.
Your turn — 4 kg / 2 kg machine (a ≈ 3.27 m/s²), released from rest. Distance in 2 s?

Answer: about 6.5 m. d = 0.5 × 3.267 × 4 ≈ 6.53 ≈ 6.5 m.

Example 4 — Atwood's original trick: 5.0 kg vs 4.9 kg

  1. Imbalance. (5.0−4.9)/(5.0+4.9) = 0.1/9.9 ≈ 0.0101.
  2. Accelerate. a = 0.0101 × 9.8 ≈ 0.099 m/s² — about 1% of g.
  3. The point: nearly-equal masses turn a sub-second blur into a stately, measurable crawl. That was the whole invention.
Common mistake: “the masses are basically equal, so a = 0.” Basically equal isn't equal — the tiny difference still drives a tiny, real acceleration.
Your turn — 2.0 kg vs 1.9 kg. Acceleration?

Answer: about 0.25 m/s². (2.0−1.9)/(2.0+1.9) = 0.1/3.9 ≈ 0.02564; a ≈ 0.02564 × 9.8 ≈ 0.251 ≈ 0.25 m/s².

Memorization tips

  • Say it aloud: “difference over sum, times g.” Six words, the whole acceleration formula.
  • The story version: unbalanced weight (difference) drives; total inertia (sum) resists. It is F = ma for the system wearing a costume.
  • Tension bracket: m2g < T < m1g, always. If your T escapes the bracket, the answer is wrong.
  • Limit checks: equal masses → a = 0; one mass vanishes → a = g. Two seconds, catches sign flips and upside-down fractions.
  • Heavier = m1: label it first. Every sign error in Atwood problems is a labelling error in disguise.
  • Same T, same a: ideal pulley → one tension; inextensible string → one acceleration magnitude. Those two sentences are the entire derivation's engine.

Final challenge

Five mixed questions — accelerations, tensions, limits, and the traps, all in one. Score 5/5 and the Atwood machine is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is an Atwood machine?

Two masses hanging on a string over a pulley. George Atwood invented it in 1784 to slow free fall down to measurable speeds, letting early physicists study acceleration with crude clocks.

What is the acceleration of an Atwood machine?

a = (m1−m2)g/(m1+m2), with m1 the heavier mass. The difference of the masses drives the motion; the sum resists it.

What is the tension in an Atwood machine?

T = 2m1m2g/(m1+m2). It always lies strictly between the two weights m2g and m1g — the heavier mass falls but is partially supported, so T < m1g.

Why is the tension the same on both sides?

For an ideal (massless, frictionless) pulley, the string pulls equally on both ends — the pulley merely redirects the force. A real pulley with mass would give different tensions on each side.

What happens when the two masses are equal?

The system balances: a = 0 and T = m1g = m2g. Each mass hangs in equilibrium, and the formula confirms it — (m1−m2) = 0 kills the acceleration.

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