An Atwood machine is a classic physics setup with two hanging masses connected by a light string over a pulley. It is useful because it turns Newton's second law into a clear, measurable motion problem. When the masses are unequal, the heavier mass accelerates downward while the lighter mass accelerates upward.
The model helps students connect force diagrams, acceleration, and string tension in one system.
To solve an Atwood machine, draw a free-body diagram for each mass and choose the positive direction along each mass's motion. Gravity pulls each mass downward, while the string tension pulls upward on both masses. If the string is massless and the pulley is frictionless, both masses have the same magnitude of acceleration and the same tension acts throughout the string.
Combining the two Newton's second law equations gives formulas for the acceleration and tension.
Understanding Physics: The Atwood Machine
A useful way to understand this device is to choose the whole pair of masses as one system first. The pull in the string acts between parts of that system, so it cannot be the cause of the pair's overall change in motion. Those pulls cancel when the two masses are considered together.
What remains is the difference between the two weights. This explains an important result. A large pair of nearly equal masses can move slowly, even though each weight is large.
The driving force is small compared with the total mass that must be accelerated. Students often assume that tension must equal the weight of one mass.
That is only true when that mass has no acceleration. During motion, tension sits between the two weights.
The direction choices in the two separate force equations need careful attention. It is simplest to call downward positive for the mass that moves down, then call upward positive for the mass that moves up. In each equation, positive means the direction that particular mass is moving, not one universal direction on the page.
This makes the acceleration value positive in both equations. If one direction convention is used for both masses, one acceleration must carry a negative sign. Either approach works when it is used consistently.
A negative answer is not automatically wrong. It can mean that the assumed direction was opposite to the actual motion. Checking extreme cases helps.
If the masses become equal, the acceleration should approach zero. If one mass becomes much larger than the other, the acceleration should approach the gravitational acceleration without quite reaching it.
In a school experiment, the motion can be measured with a timer, a motion sensor, or video frames. The moving mass should travel far enough for a reliable time measurement, but not so far that it hits the floor before data are collected. A graph of position against time has a curved shape for steady acceleration.
A graph of velocity against time is straighter, and its slope gives the acceleration. Repeating trials with different mass differences shows that changing the imbalance changes the acceleration.
Keeping the total mass fixed is especially helpful because it isolates the effect of the weight difference. Measurements can be used to estimate gravitational acceleration, though the result is usually less accurate than the accepted value because real equipment is not ideal.
A real pulley has mass, so some of the driving force must turn it. Its rotation has inertia, which reduces the acceleration. Friction in the axle resists motion and may make a small mass difference unable to start the system.
A real string can stretch slightly or have noticeable mass. With a massive pulley, the tension on one side is not exactly the same as the tension on the other side because a tension difference is needed to rotate the pulley. These effects matter when comparing theory with data.
Students should state the model assumptions before calculating, label every force in a diagram, include units, and keep enough digits until the final answer. The most common mistakes are reversing a force direction, using the total weight instead of the weight difference, or mixing acceleration signs.
Key Facts
- For masses m1 and m2 with m2 > m1, the acceleration magnitude is a = (m2 - m1)g / (m1 + m2).
- The tension can be found from the lighter mass: T - m1g = m1a, so T = m1(g + a).
- The tension can be found from the heavier mass: m2g - T = m2a, so T = m2(g - a).
- Both masses have the same acceleration magnitude because they are connected by the same taut, inextensible string.
- The net external driving force on the two-mass system is (m2 - m1)g when pulley friction and string mass are ignored.
- If m1 = m2, then a = 0 and T = mg for each mass, so the system is in equilibrium if released from rest.
Vocabulary
- Atwood machine
- A system of two masses connected by a string over a pulley, used to study Newton's second law and tension.
- Tension
- The pulling force transmitted through a stretched string, rope, or cable.
- Free-body diagram
- A diagram that shows all external forces acting on one object.
- Acceleration
- The rate at which velocity changes, including changes in speed or direction.
- Net force
- The vector sum of all forces acting on an object or system.
Common Mistakes to Avoid
- Using different acceleration magnitudes for the two masses is wrong because a taut, massless string makes both masses move together with the same magnitude of acceleration.
- Writing the same force equation for both masses is wrong because one mass accelerates upward and the other downward, so the signs must match the chosen positive directions.
- Setting tension equal to weight for a moving mass is wrong because T = mg only when that mass has zero acceleration.
- Forgetting that tension is an internal force for the two-mass system is wrong because it cancels when the two masses are treated as one combined system.
Practice Questions
- 1 An Atwood machine has m1 = 2.0 kg and m2 = 5.0 kg. Using g = 9.8 m/s^2, find the acceleration magnitude and the string tension.
- 2 Two masses, 3.0 kg and 4.0 kg, are connected over a frictionless pulley. Find the acceleration of the system and the tension in the string using g = 9.8 m/s^2.
- 3 If the pulley has friction or the string has noticeable mass, explain why the simple formulas a = (m2 - m1)g / (m1 + m2) and T = constant may no longer be valid.