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Atwood Machine Calculator

Enter the two masses to find the system's acceleration and string tension.

How it is calculated

Enter the two hanging masses in kilograms. The tool models an ideal setup: a string that does not stretch and a pulley with no mass and no friction. The acceleration is the net force divided by the total mass being moved, a = |m1 − m2| · g ÷ (m1 + m2), and the string tension is T = 2 · m1 · m2 · g ÷ (m1 + m2), with g = 9.80665 m/s². Acceleration is a positive magnitude in m/s², tension is in newtons, and a note tells which mass goes down. Both masses must be above zero; zero or negative values give no result. Equal masses give acceleration 0 and a tension equal to one weight. Results are rounded to three decimals. Real pulleys have mass and friction, so measured values come out a little lower.

Worked example

Mass 1 is 5 kg and mass 2 is 3 kg. The difference is 2 kg and the total is 8 kg. Acceleration: 2 × 9.80665 ÷ 8 = 2.452 m/s², with the 5 kg mass moving down. Tension: 2 × 5 × 3 × 9.80665 ÷ 8 = 36.775 N. Check: the lighter mass weighs 3 × 9.80665 = 29.42 N and the heavier 49.03 N, and the tension of 36.775 N lies between the two, as it must.

Try this example →

Common mistakes

  • Entering weights in newtons, or masses in grams. The fields take kilograms, so a 500 g mass is 0.5.
  • Expecting a negative acceleration when mass 2 is heavier. The tool reports only the size; the note shows which side descends.
  • Assuming the tension equals the weight of the heavier mass. When the masses differ it is lower than the heavier weight and higher than the lighter one.
  • Expecting a real lab result to match exactly. A pulley with mass or friction slows the motion, and the tool ignores both.
  • Entering 0 for a missing mass. Zero or negative masses give no result; a single hanging body is simply in free fall at g.

About this calculator

An Atwood machine is the classic physics setup of two masses, m1 and m2, connected by an inextensible string over a frictionless, massless pulley. Applying Newton's second law to both masses gives a constant acceleration of a = (m1 − m2)g / (m1 + m2), directed so the heavier mass falls.

The string tension follows from the same equations of motion: T = 2·m1·m2·g / (m1 + m2), a value less than the weight of either mass individually, which is exactly why the system accelerates instead of staying in static equilibrium.

Since Newton's time, the Atwood machine has been a standard teaching tool for demonstrating Newton's second law with two coupled bodies, because the low acceleration (compared to free fall) makes it easy to measure directly in an introductory physics lab.

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Comments (1)

  • Editorial team

    Acceleration is always the fraction (m1 minus m2) over (m1 plus m2) of g, so for classroom timing pick masses that give a slow, measurable fall. With 5 and 3 kg the acceleration is 2.452 m/s², one quarter of g, and the tension is 36.775 N. Measured values come out lower when the pulley has mass and friction. How far did your measured acceleration fall below the ideal value?

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