Atwood Machine Calculator (Acceleration & Tension)
Solve an Atwood machine: acceleration of two hanging masses over a pulley, string tensions, travel time and speed, with an optional solid-disc pulley mass for the rotational inertia.
How it works
a = (m₁ − m₂)·g / (m₁ + m₂ + M/2) · T₁ = m₁(g − a) · T₂ = m₂(g + a) · with a massless pulley T = 2m₁m₂g/(m₁ + m₂) · t = √(2d/a) · v = √(2ad)
An Atwood machine hangs two masses from a light string over a pulley. Newton’s second law applied to each mass (and to the pulley’s rotation, I = ½MR² for a solid disc) gives one equation for the shared acceleration: the net driving force (m₁ − m₂)g divided by all the inertia being accelerated. With a massless pulley both tensions are equal; a heavy pulley needs a larger tension on the descending side to spin it up. The motion starts from rest with constant acceleration, so the travel time and speed follow from the kinematic equations. This is the standard derivation in Halliday, Resnick & Walker and OpenStax University Physics.
Worked example
With m₁ = 5 kg, m₂ = 3 kg, a massless pulley and g = 9.81 m/s²: a = (5 − 3) × 9.81 / (5 + 3) = 2.453 m/s², mass 1 moves down. The tension is T = 5 × (9.81 − 2.4525) = 36.79 N on both sides (2 × 5 × 3 × 9.81 / 8). Starting from rest, the masses move 1 m in t = √(2 × 1 / 2.4525) = 0.903 s and reach v = √(2 × 2.4525 × 1) = 2.215 m/s. The pulley axle carries 2 × 36.79 = 73.58 N.
Sources: OpenStax University Physics Vol. 1 — Newton’s Laws: Applications (Atwood machine) · HyperPhysics — Atwood machine
Frequently asked questions
What is an Atwood machine?
Two masses connected by a string over a pulley. Invented by George Atwood in 1784 to slow free fall down so it could be measured, it is now a standard exercise for Newton’s second law and tension.
Why is the tension not equal to the weight of either mass?
The heavier mass accelerates downward, so the string pulls it with less than its weight; the lighter one accelerates upward, so the string pulls it with more than its weight. The tension always lies between m₂g and m₁g.
How does the pulley’s mass change the answer?
A pulley with mass must be spun up, which adds its rotational inertia I/R² to the denominator. For a solid disc I/R² = M/2, for a thin ring it is M. The acceleration falls and the two tensions become different.
What happens if the masses are equal?
The net force is zero, so the system stays at rest or keeps moving at constant speed, and the tension equals the weight of each mass, m·g.
Can I measure g with an Atwood machine?
Yes. Measure the time to fall a known distance, find a = 2d/t², and rearrange: g = a(m₁ + m₂ + M/2)/(m₁ − m₂). Friction in the bearing makes the estimate slightly low.
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Results assume the idealised textbook model stated above; real systems have losses and tolerances, so verify safety-critical values with measurements or a qualified engineer.