Atwood Machine Calculator
Solve the two mass pulley for acceleration, tension, and fall time.
What it does
A rope over a pulley turns two weights into one clean equation. The heavier side wins, but not by much when the masses nearly match: the driving imbalance stays small while the total inertia stays large. That ratio is the whole machine, and it measures g in teaching labs to this day.
The tool above solves the rig. Enter both masses with a drop height in metric or imperial to get acceleration, tension, fall time, and which side descends. Defaults hang 5 kg against 3 kg over 2 m, so the page opens solved at 2.45 m/s2, 36.79 N, and 1.28 s.
Tension surprises first timers because it lands between the two weights, never above the heavy one. The descending mass must accelerate down, so its weight exceeds tension; the rising mass accelerates up, so tension exceeds its weight. At 2 vs 4 kg the pair gives 3.27 m/s2 with 26.16 N, matching the OpenStax worked example.
Both modes agree: imperial 11.02 vs 6.61 lb over 6.56 ft gives 8.05 ft/s2 and 8.26 lbf with the same 1.28 s fall. A steep 10 vs 2 kg rig runs at 6.54 m/s2 with 32.70 N and falls 3 m in 0.96 s. Equal masses report 0.00 with 39.24 N and no motion.
Use the table for classic pairings, then run your rig. The page assumes a light frictionless pulley and a non stretching string; heavy wheels, drag, and bearing friction all slow real falls. Within range it answers lab questions fast: acceleration, tension, direction, and time.
How to use
- Pick metric or imperial.
- Enter both hanging masses.
- Enter the drop height.
- Read acceleration and which side descends.
- Read tension and fall time.
- Add pulley inertia for heavy wheels.
Formula
- m1
- First hanging mass.
- m2
- Second hanging mass.
- g
- Gravity: 9.81 metric, 32.2 imperial.
- T
- String tension in N or lbf.
- h
- Drop height for the fall time.
Masses share one unit system per run (kg with meters, lb with feet) and g is 9.81 or 32.2 to match. The pulley is assumed light, massless, and frictionless with an inextensible string. Balanced masses give a = 0, reported as no motion with tension equal to each weight.
Worked example
Example: 5 kg against 3 kg
Masses of 5 kg and 3 kg hang over a light frictionless pulley with a 2 m drop. Find motion and tension.
- Net pull = |5 - 3| x 9.81 = 19.62 N over 8 kg total.
- a = 19.62 / 8 = 2.45 m/s2, with mass 1 descending.
- Tension = 2 x 5 x 3 x 9.81 / 8 = 36.79 N.
- Fall over 2 m: t = sqrt(2 x 2 / 2.45) = 1.28 s.
- Imperial mirror gives 8.05 ft/s2 and 8.26 lbf.
- Equal masses balance with zero acceleration.
Reference table
| Masses | a | Tension | Fall |
|---|---|---|---|
| 5 vs 3 kg | 2.45 m/s2 | 36.79 N | 1.28 s |
| 2 vs 4 kg | 3.27 m/s2 | 26.16 N | 1.11 s |
| 10 vs 2 kg | 6.54 m/s2 | 32.70 N | 0.78 s |
| 4 vs 4 kg | 0.00 m/s2 | 39.24 N | no motion |
Limits
- Light frictionless pulley assumed.
- Inextensible massless string.
- No air drag or bearing friction.
- Balanced case reports no fall time.
Common mistakes
- Setting tension equal to a weight.
- Mixing kg with lb in one run.
- Reading g as 9.81 in imperial mode.
- Expecting fall time when balanced.
- Ignoring heavy pulley inertia.
Try it yourself
Quick checks. Answers sit tucked under each question.
OpenStax check: 2 kg against 4 kg over 2 m?
a = 3.27 m/s2. T = 26.16 N. Fall 1.11 s.
Steep rig: 10 kg against 2 kg over 3 m?
a = 6.54 m/s2. T = 32.70 N. Fall 0.96 s.
Show the 10 vs 2 kg acceleration by hand.
Unbalanced force = (10-2) x 9.81 = 78.48 N over 12 kg gives 6.54 m/s2. T = 32.70 N.
Questions
What is an Atwood machine?
Two masses over a pulley: a = |m1 - m2|/(m1 + m2) x g, and tension T = 2 m1 m2 g/(m1 + m2). The heavy side descends.
Why is tension below the weight?
Tension sits below the hanging weight of the descending mass because that mass accelerates down. At 5 vs 3 kg it is 36.79 N, under 49.05 N.
What if the masses match?
Zero. Equal masses balance, tension equals each weight (39.24 N for 4 kg each), and the tool reports no motion.
Does it work in pounds and feet?
The same ratio with g = 32.2: 11.02 vs 6.61 lb gives 8.05 ft/s2 and 8.26 lbf over 6.56 ft.
What about a heavy pulley?
Add half m_pulley r squared style terms. This page assumes a light massless frictionless pulley.
Is it exact for real rigs?
No. String stretch, air drag, and bearing friction all trim real acceleration below theory.
Sources
- OpenStax University Physics: Newton laws problems
- OpenStax College Physics: Newton laws applications