An Atwood machine is two masses hanging from a string that runs over a pulley, used in AP Physics 1 (Topic 2.6) to apply Newton's second law: the heavier mass falls, the lighter one rises, and both share the same acceleration magnitude and the same string tension.
An Atwood machine is the simplest two-object system on the AP exam. Take two masses, tie them to opposite ends of a string, and drape the string over a pulley. If the masses are unequal, the heavier side accelerates downward, the lighter side accelerates upward, and gravity does all the driving.
The standard AP assumptions make the math clean. The string is massless and doesn't stretch, so both blocks have the same acceleration magnitude. The pulley is massless and frictionless, so the tension is the same everywhere in the string. With those assumptions, you can treat the whole thing as one system: the net force is the difference in the weights, (m₂ − m₁)g, and the total mass being accelerated is the sum, (m₁ + m₂). That gives a = (m₂ − m₁)g / (m₁ + m₂). To find tension, zoom in on one block alone and apply Newton's second law to just that block.
The Atwood machine lives in Topic 2.6, Newton's Second Law, and it's the test of whether you actually understand F_net = ma or just memorized it. It forces you to do the two moves AP Physics 1 rewards over and over. First, draw a separate free-body diagram for each object. Second, choose your system wisely, because analyzing both blocks together gets you acceleration fast, while analyzing one block alone gets you tension. It also exposes the single most common dynamics misconception: tension is NOT equal to the weight of either block when the system accelerates. If you can explain why tension sits between m₁g and m₂g, you understand Unit 2.
Newton's Second Law (Unit 2)
The Atwood machine is basically Newton's second law with a built-in plot twist. The driving force is the difference between the weights, but the mass that resists acceleration is the total of both blocks. That's why a is always less than g.
Tension (Unit 2)
Tension is the connective tissue here. The string pulls up on both blocks with the same magnitude, and its value lands between the two weights, larger than the lighter block's weight, smaller than the heavier one's. Solving for T means isolating one block, not the whole system.
Equilibrium (Unit 2)
Set the two masses equal and the Atwood machine becomes an equilibrium problem. Net force is zero, acceleration is zero, and tension equals each block's weight exactly. It's a quick way to check your acceleration formula, since plugging in m₁ = m₂ should give a = 0.
Coefficient of Kinetic Friction (Unit 2)
The 'modified Atwood machine' puts one block on a horizontal table instead of hanging it. Add friction on the table and you get a classic exam variation where the hanging weight has to beat both the table block's inertia and the friction force.
No released FRQ has to name the device for it to show up; pulley-and-two-blocks setups are a staple of Unit 2 questions in both MCQ and FRQ form. Multiple-choice stems love conceptual traps, like asking whether tension equals the hanging weight (no, when accelerating) or what happens to acceleration if you add mass to both sides equally (it decreases, since net force stays the same while total mass grows). FRQs typically ask you to draw free-body diagrams for each block, derive the acceleration symbolically, then solve for tension. Always derive in symbols first, then plug in numbers. And remember the limiting-case check graders love: if m₁ = m₂, your acceleration expression must give zero, and if one mass is huge compared to the other, a should approach g.
A true Atwood machine has both masses hanging vertically, so gravity acts on both and the net force is the difference in weights. A modified Atwood machine has one block sliding on a horizontal surface while the other hangs, so only the hanging block's weight drives the system, giving a = m_hanging·g / (m₁ + m₂) when the surface is frictionless. Same free-body-diagram strategy, different driving force. Mixing up which weight (or weight difference) goes in the numerator is the classic error.
An Atwood machine is two masses connected by a string over a pulley, and the heavier mass falls while the lighter one rises with the same acceleration magnitude.
The acceleration is a = (m₂ − m₁)g / (m₁ + m₂), because the net force is the difference in weights but the inertia is the total mass of the system.
Tension is the same throughout the string (assuming a massless string and frictionless, massless pulley) and its value falls between the two weights when the system accelerates.
To find acceleration, treat both blocks as one system; to find tension, apply Newton's second law to one block by itself.
If the two masses are equal, the system is in equilibrium, acceleration is zero, and tension equals each block's weight.
Check your answer with limiting cases: equal masses should give a = 0, and a very lopsided pair should give acceleration approaching g.
It's two masses connected by a string that runs over a pulley, the standard Topic 2.6 setup for practicing Newton's second law. The unequal weights create a net force, and the formula a = (m₂ − m₁)g / (m₁ + m₂) gives the acceleration.
No, not when the system is accelerating. Tension lands between the two weights, less than the heavier block's weight (so it can fall) and more than the lighter block's weight (so it can rise). Tension only equals the weights when the masses are equal and the system sits in equilibrium.
In a true Atwood machine, both masses hang vertically, so the net force is the difference in their weights. In a modified (half) Atwood machine, one block slides on a table while the other hangs, so only the hanging block's weight drives the motion.
Because the string is assumed to be inextensible, meaning it doesn't stretch. If one block moves down 1 meter, the other must move up 1 meter in the same time, so their speeds and acceleration magnitudes are always equal even though their directions differ.
Only in the limit where one mass is enormous compared to the other, so it's basically in free fall dragging a negligible partner. For any real pair of masses, a is always less than g because the lighter block's weight pulls back against the motion.