An Atwood machine is two masses connected by an inextensible string over a pulley; in AP Physics C Mechanics it's the classic Newton's second law setup, where the heavier mass falls and the system accelerates at a = (m₂ − m₁)g / (m₁ + m₂), with the same tension throughout an ideal string.
An Atwood machine is the simplest two-body dynamics problem in physics. Hang two masses on either end of an inextensible (non-stretching) string draped over a pulley. If the masses are unequal, the heavier one falls and the lighter one rises, and because the string can't stretch, both masses move with the same magnitude of acceleration.
The whole point of the device is that it dilutes gravity. Instead of free fall at g, the system accelerates at a = (m₂ − m₁)g / (m₁ + m₂), which can be made as slow as you want by choosing nearly equal masses. That's actually why George Atwood built it in 1784, to measure g with the slow-motion timing tools of his era. On the AP exam, you analyze it two ways. You can draw a free-body diagram for each mass and write Newton's second law twice (the tension T appears in both equations), or you can treat both masses as a single system where the net force is (m₂ − m₁)g acting on total mass (m₁ + m₂). The system approach gets you acceleration fast; the individual approach is what you need when the question asks for tension.
The Atwood machine lives in Topic 2.3, Newton's Laws of Motion, and it's the test bed for almost every Unit 2 skill at once. It forces you to draw correct free-body diagrams, apply ΣF = ma to multiple objects, use the constraint that connected objects share an acceleration, and recognize that an ideal string transmits the same tension to both masses. It's also a Newton's third law showcase, since the string pulls up on each mass and each mass pulls down on the string with equal-magnitude paired forces.
It matters beyond Unit 2, too. The same setup returns in energy conservation problems (find the speed after the heavy mass falls a height h) and then again in rotational dynamics, where the pulley finally gets mass and rotational inertia and the two tensions are no longer equal. If you can handle an Atwood machine at every level, you've basically traced the spine of AP Physics C Mechanics.
Keep studying AP Physics C: Mechanics Unit 2
Tension (Unit 2)
Tension is the variable that makes Atwood problems interesting. In an ideal massless string, T is the same on both sides, and it always falls between the two weights, less than m₂g (so the heavy mass can fall) but greater than m₁g (so the light mass can rise). If a problem hands you T outside that range, you made a sign error.
Pulley System (Unit 2 and Unit 5)
In Unit 2 the pulley is an idealized direction-changer with no mass. In Unit 5, the pulley gets rotational inertia, and suddenly the tensions on the two sides must differ so a net torque can spin it. Same picture, upgraded physics. Recognizing which version you're in is half the battle.
Equilibrium (Unit 2)
An Atwood machine with equal masses is in equilibrium, with a = 0 and T = mg on each side. That special case is a great sanity check. Plug m₁ = m₂ into your acceleration formula and you should get zero. If you don't, your algebra broke somewhere.
Frictional Force (Unit 2)
Swap one hanging mass for a block on a table and you get the modified (half) Atwood machine, where friction on the block joins the force equation. The method is identical, write ΣF = ma for each object and link them through tension and shared acceleration, but now friction shrinks the net force.
Atwood machines show up constantly in multiple-choice and as building blocks of dynamics FRQs. Typical asks include deriving the acceleration and tension symbolically, ranking tensions or accelerations across different mass combinations, predicting what the heavy mass's acceleration approaches as m₂ ≫ m₁ (it approaches g, never exceeds it), and analyzing a modified Atwood with a block on a rough table. In Physics C, expect calculus-flavored twists like a pulley with rotational inertia or graphing velocity versus time from the constant acceleration. The grading currency is your free-body diagrams and your two Newton's second law equations with consistent sign conventions. Pick a positive direction for the system (heavy side down, light side up) and stick with it, then show the algebra that eliminates T. Even when the FRQ doesn't say "Atwood machine," connected masses over a pulley are an Atwood machine in disguise.
A true Atwood machine has both masses hanging vertically, so gravity drives both sides and a = (m₂ − m₁)g / (m₁ + m₂). A modified or half-Atwood has one mass on a horizontal table and one hanging, so only the hanging weight drives the system and (on a frictionless table) a = m_hang·g / (m_hang + m_table). Students often memorize one formula and apply it to the wrong setup. Don't memorize either one. Derive from free-body diagrams every time and the geometry takes care of itself.
An Atwood machine is two masses connected by an inextensible string over a pulley, and it is the standard AP setup for applying Newton's second law to connected objects.
Because the string can't stretch, both masses have the same magnitude of acceleration, given by a = (m₂ − m₁)g / (m₁ + m₂).
In an ideal massless, frictionless pulley setup, tension is the same throughout the string and its value lands between the two weights, m₁g < T < m₂g.
You can find acceleration quickly by treating both masses as one system, but you must analyze a single mass with its own free-body diagram to solve for tension.
Equal masses give equilibrium (a = 0, T = mg), and a very large m₂ gives an acceleration that approaches g but never exceeds it; both limits are quick checks for your answer.
When the pulley has rotational inertia (Unit 5), the two tensions are no longer equal because a net torque is needed to angularly accelerate the pulley.
It's a device with two masses connected by an inextensible string over a pulley, used to study Newton's laws. The heavier mass falls, both masses share the same acceleration magnitude, and a = (m₂ − m₁)g / (m₁ + m₂).
Yes, but only if the string is massless and the pulley is massless and frictionless, which is the standard Unit 2 assumption. Once the pulley has rotational inertia (Unit 5), the tensions must differ to provide the net torque that spins the pulley.
No. Even if one mass is enormous, the formula a = (m₂ − m₁)g / (m₁ + m₂) caps acceleration below g, because the falling mass always has to drag the other mass along. An answer bigger than g on the exam means something went wrong.
In a regular Atwood machine both masses hang vertically, so both weights enter the equation. In a modified (half) Atwood, one mass sits on a table, so only the hanging weight drives the motion and the table mass may add friction. Same method, different free-body diagrams.
Because the masses are accelerating. The tension must be less than m₂g so the heavy mass has a net downward force, and greater than m₁g so the light mass has a net upward force. T equals the weight only in the equal-mass equilibrium case.