A machine built to slow down free fall
The Atwood machine, devised by the Reverend George Atwood in 1784, is about as simple as classical mechanics gets: two masses, m1 and m2, hang from opposite ends of a light string that passes over a pulley. If m1 and m2 differ, the heavier side falls and the lighter side rises, both moving with the same magnitude of acceleration because the inextensible string ties their motions together. Its enduring appeal in physics classrooms is that this one setup demands, and rewards, a careful application of Newton's second law to two connected bodies at once.
Two equations, one shared unknown
Write Newton's second law separately for each mass, taking the direction each mass actually moves as positive. String tension T is the same on both sides for an ideal massless, frictionless pulley and an inextensible, massless string:
m1 g - T = m1 a (heavier mass, m1 > m2, accelerating downward)
T - m2 g = m2 a (lighter mass, accelerating upward, same |a|)
adding the two equations eliminates T:
a = g (m1 - m2) / (m1 + m2)
substituting back:
T = 2 m1 m2 g / (m1 + m2)
Two sanity checks make the result trustworthy. Set m1 = m2 and a collapses to zero, with T = m1g = m2g - the masses balance and nothing moves, exactly as intuition demands. Send m2 → 0 (a single mass falling with an essentially free string over the pulley) and a → g, T → 0 - ordinary free fall, since there is effectively nothing on the other end to resist it.
Why the tension sits strictly between the two weights
A frequent intuition error is assuming the string tension equals one of the hanging weights. It cannot: if T equaled m1g exactly, the net force on mass 1 would be zero and it would not accelerate - but it does, at rate a, together with mass 2. The correct value, T = 2m1m2g/(m1+m2), can be shown algebraically to always lie strictly between m2g and m1g (for m1 ≠ m2): just enough larger than m2's weight to accelerate it upward, and just enough smaller than m1's weight to let m1 accelerate downward, at exactly the same shared rate a that keeps the inextensible string taut on both sides.
When the pulley itself has mass
A real pulley has mass and therefore its own moment of inertia I, and it must be spun up by the string moving over it - which means the string tension on the two sides is no longer exactly equal, since the small net torque (T1 - T2)·r is what accelerates the pulley's rotation. Working through both masses and the pulley's rotational equation together (with the string's linear acceleration a related to the pulley's angular acceleration by a = αr for a string that does not slip) gives a modified result:
a = g (m1 - m2) / (m1 + m2 + I/r²) I/r² acts exactly like EXTRA mass shared by both sides — a heavier or larger-radius pulley always makes the system accelerate more slowly for the same m1 and m2
The extra term I/r² in the denominator behaves exactly like additional inertial mass added symmetrically to the system: energy that would otherwise go entirely into linear kinetic energy of the two masses is instead partly diverted into the pulley's rotational kinetic energy, so for any given imbalance m1 - m2, a real pulley always yields a smaller acceleration than the idealized massless-pulley formula predicts - a nice, concrete illustration of how rotational inertia "steals" acceleration from a linear system whenever the two are coupled.
Why 18th-century physicists loved it
Free fall near Earth's surface happens at roughly 9.8 m/s² - far too fast for a person with a hand-held clock or pendulum timer to measure precisely over a short drop in Atwood's era. The genius of his machine is that a = g(m1-m2)/(m1+m2) can be made arbitrarily small simply by choosing m1 and m2 close to each other: with m1 only slightly heavier than m2, the acceleration might be a tiny fraction of g, giving a slow, easily timed fall over a measurable distance. Atwood used exactly this trick to produce some of the more precise experimental measurements of g available at the time, and the device remained a standard laboratory demonstration and measurement tool for well over a century afterward.
Frequently asked questions
Why isn't the string tension equal to either weight?
If tension equaled either mass's weight exactly, that mass would be in equilibrium and wouldn't accelerate - but both masses do accelerate together. The correct tension, T = 2m1m2g/(m1+m2), sits strictly between m1g and m2g, exactly enough net force to accelerate both masses at the shared rate a.
Why does the pulley's moment of inertia slow the system down?
Some of the gravitational potential energy released as the heavier mass falls has to go into spinning up the pulley itself, not just accelerating the two masses. A pulley with more rotational inertia (a solid disk versus a light ring, or simply a heavier pulley) soaks up more of that energy per unit acceleration, which shows up as a smaller a for the same masses.
How accurate was the Atwood machine for measuring g historically?
Very, for its era. By choosing m1 close to m2, the resulting acceleration a = g(m1-m2)/(m1+m2) could be made arbitrarily small and therefore easy to time accurately with 18th-century clocks, letting experimenters back-calculate g to good precision without needing to time a fast free-fall directly.
Try it live
Everything above runs in your browser - open Atwood Machine and change the parameters while it is running. Nothing is installed, nothing is uploaded, the whole model lives in one tab.
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