An Atwood's machine is a system of two masses connected by a string over a pulley; in AP Physics 1 you analyze it with Newton's second law to find that the system accelerates at a = (m₂ − m₁)g / (m₁ + m₂), with the same tension and acceleration magnitude shared by both masses.
An Atwood's machine is two blocks tied together by a string that runs over a pulley. The heavier block falls, the lighter block rises, and because the string doesn't stretch, both blocks move with the same magnitude of acceleration. In AP Physics 1, the string and pulley are usually treated as ideal (massless string, frictionless pulley), which means the tension is the same everywhere in the string.
Think of it as gravity in slow motion. Instead of dropping a ball and watching it accelerate at the full 9.8 m/s², the two masses partially balance each other out, so the system accelerates at some fraction of g. That's exactly why Atwood built it in 1784, to measure g with equipment that couldn't time a free fall. On the AP exam, it's the go-to setup for testing whether you can draw free-body diagrams for two connected objects, write Newton's second law for each one, and solve the equations together (or treat both masses as one system).
Atwood's machines live in the dynamics unit of AP Physics 1, where you apply Newton's laws to systems of connected objects. They're the cleanest possible test of three skills the CED cares about. First, drawing a correct free-body diagram for each mass (only two forces act on each block, tension up and weight down). Second, applying ΣF = ma consistently, including picking a consistent positive direction so the signs work out. Third, choosing your system wisely. You can analyze each block separately, where tension is an external force, or treat both blocks as one system, where tension becomes an internal force and drops out entirely. That system-vs-object choice shows up constantly across AP Physics 1, and the Atwood's machine is where most people learn it.
Keep studying AP Physics 1 Unit 2
Tension (Unit 2)
The tension in an Atwood's machine is the force that ties the two equations together, and it's always between the two weights. It must be bigger than m₁g (to accelerate the light mass up) and smaller than m₂g (to let the heavy mass accelerate down). If you ever calculate a tension equal to either weight in an accelerating Atwood's machine, something went wrong.
Internal vs. External Forces (Unit 2)
Treat both blocks plus the string as one system and the tension becomes internal, so it cancels out. The only external forces left are the two weights, which gives you a = (m₂ − m₁)g / (m₁ + m₂) in one line. This system trick is the same logic you'll use later for momentum and energy problems.
Acceleration due to gravity (Units 1-2)
The whole point of the original device was to 'dilute' g into something measurable. The acceleration is g scaled by the fraction (m₂ − m₁)/(m₁ + m₂), so nearly equal masses move slowly and very unequal masses approach free fall. Checking those limiting cases is a classic AP reasoning move.
Equilibrium (Unit 2)
Set m₁ = m₂ and the Atwood's machine becomes an equilibrium problem. The net force is zero, acceleration is zero, and the tension equals each block's weight. It's a built-in special case for checking whether your general answer makes sense.
Atwood's machines (and their modified cousins) are a staple of force MCQs and FRQs. Multiple-choice stems ask you to rank tensions, predict what happens to acceleration when a mass changes, or pick the correct free-body diagram. FRQs go further. Expect to draw free-body diagrams with correctly labeled forces, derive the acceleration and tension symbolically (answers in terms of m₁, m₂, and g, not numbers), and justify your reasoning in words. A favorite trap is asking whether the tension equals the heavy block's weight (it doesn't, if the system accelerates). Another favorite is a limiting-case check, like explaining what the acceleration approaches if m₂ becomes much larger than m₁. Practice writing ΣF = ma for each block with a consistent sign convention, because that's where most lost points come from.
A classic Atwood's machine has both masses hanging vertically, so gravity pulls on both and they fight each other. A modified Atwood's machine puts one mass on a horizontal table (or incline) with the string bending over a pulley at the edge. Now only the hanging mass's weight drives the motion, and the table mass adds normal force and possibly friction to its free-body diagram. The solving strategy is identical, two free-body diagrams and two ΣF = ma equations, but the forces in each diagram are different. Mixing up which weight actually drives the system is a common error.
An Atwood's machine is two masses connected by a string over a pulley, and it's the standard AP Physics 1 setup for Newton's second law with connected objects.
Because the string doesn't stretch, both masses share the same magnitude of acceleration, given by a = (m₂ − m₁)g / (m₁ + m₂) for an ideal pulley and string.
The tension is the same throughout an ideal string, and its value sits between the two weights, larger than the lighter mass's weight and smaller than the heavier mass's weight.
You can solve it two ways, by writing ΣF = ma separately for each block or by treating both blocks as one system so the tension cancels as an internal force.
Always check limiting cases. Equal masses give zero acceleration (equilibrium), and one mass much larger than the other gives acceleration approaching g.
On FRQs, derive answers symbolically and keep a consistent sign convention (for example, the heavy mass's downward direction as positive for the whole system).
It's two masses connected by a string over a pulley. The heavier mass falls and the lighter one rises, both with the same acceleration, a = (m₂ − m₁)g / (m₁ + m₂). It's the classic AP setup for applying Newton's second law to connected objects.
No, not if the system is accelerating. The tension is T = 2m₁m₂g / (m₁ + m₂), which falls between the two weights. Tension only equals the weights when the masses are equal and the system sits in equilibrium.
In a classic Atwood's machine both masses hang vertically, so both weights pull on the system. In a modified version, one mass sits on a horizontal table while the other hangs, so only the hanging weight drives the motion and the table mass picks up a normal force (and maybe friction).
The string connecting them doesn't stretch, so when one mass moves down 1 meter, the other moves up exactly 1 meter in the same time. Their speeds and acceleration magnitudes are locked together, even though they point in opposite directions.
Yes. Atwood-style systems show up regularly in force multiple-choice questions and free-response problems. You'll be asked to draw free-body diagrams for each mass, derive the acceleration and tension symbolically, and justify your reasoning with Newton's second law.