Atwood Machine
Solve Atwood a = g|m₂−m₁|/(m₁+m₂) and T = 2m₁m₂g/(m₁+m₂), or find an unknown mass. Free pulley calculator with step-by-step working.
Key formula
a = g|m2-m1|/(m1+m2), T = 2 m1 m2 g/(m1+m2)
How do you solve an Atwood machine?
For an ideal Atwood machine, a = g|m₂ − m₁|/(m₁ + m₂) and T = 2 m₁ m₂ g /(m₁ + m₂). The heavier mass descends. Tension is the same throughout the string and lies between the two weights. Use g from the question (often 9.8 m/s²).
About this tool
An Atwood machine connects two hanging masses with a string over a pulley. This page keeps a = g|m₂−m₁|/(m₁+m₂) and T = 2m₁m₂g/(m₁+m₂) visible so pulley homework stays consistent on free-body signs and the ideal-pulley assumptions.
How to use this calculator
- Choose the variable you want to solve for and confirm which values are already known.
- Enter inputs in consistent units before calculating so the formula is applied correctly.
- Check whether the result is physically reasonable by comparing it with the size of the inputs.
- Use the step-by-step breakdown to see how the equation was rearranged and evaluated.
Worked examples
Example 1: 1.0 kg and 3.0 kg
With g = 9.8 m/s², a = 9.8|3−1|/(1+3) = 4.9 m/s². T = 2(1)(3)(9.8)/4 = 14.7 N. The 3.0 kg mass descends. Check: T = 1(9.8+4.9) = 14.7 N on the lighter mass.
Example 2: equal masses
m₁ = m₂ = 2.0 kg, g = 9.8. Then a = 0 and T = mg = 19.6 N. The system is in equilibrium; each mass hangs at rest or moves at constant velocity if already moving.
Example 3: find m₂ from a = 2.45 m/s²
m₁ = 2.0 kg, a = 2.45 m/s², g = 9.8, with m₂ heavier. m₂ = m₁(g+a)/(g−a) = 2.0(9.8+2.45)/(9.8−2.45) = 24.5/7.35 ≈ 3.333 kg. Check: a = 9.8|3.333−2.0|/(5.333) ≈ 2.45 m/s².
Frequently asked questions
What is the Atwood machine acceleration formula?
a = g|m₂ − m₁|/(m₁ + m₂) for an ideal pulley. The heavier mass descends. Acceleration is always less than g unless one mass is zero.
What is the tension in an Atwood machine?
T = 2 m₁ m₂ g /(m₁ + m₂). Equivalently T = m_light(g + a) = m_heavy(g − a). Tension is the same throughout an ideal string.
Why is tension not equal to mg?
Each mass is accelerating, so net force is not zero. For the descending mass, mg − T = ma; for the rising mass, T − mg = ma. T sits between the two weights.
What if the masses are equal?
Then a = 0 and T = mg. The pulley does not start from rest into accelerated motion; any existing motion continues at constant speed.
Does this include pulley mass or friction?
No. This calculator is the ideal intro-physics Atwood: massless frictionless pulley and a light inextensible string. Real lab pulleys need extra rotational inertia terms.