Hong Life CalculatorsUnderstanding the pulley and Atwood machine calculator — concept and principle
A pulley redirects or multiplies force through a rope. This calculator covers two classic set-ups: the Atwood machine, where two masses hang on either side of one fixed pulley, and a block and tackle, where several rope segments share a load. It gives the Atwood acceleration and rope tension, plus the ideal effort to lift a load with a given number of supporting ropes. It is used for physics homework, lab experiments and rough rigging estimates.
How it is calculated
With g = 9.80665 m/s², masses in kg:
- Atwood acceleration: a = g (m₁ − m₂) ÷ (m₁ + m₂). A positive value means m₁ moves down.
- Rope tension: T = 2 m₁ m₂ g ÷ (m₁ + m₂), in newtons.
- Block and tackle effort: F = m₁ g ÷ n, where n is the number of rope segments supporting the moving block.
The principle
In the Atwood machine the rope is assumed massless and inextensible and the pulley frictionless and weightless, so the tension is the same on both sides and both masses share one acceleration. Newton's second law for each mass gives m₁g − T = m₁a and T − m₂g = m₂a. Adding them eliminates T: the driving force is the weight difference (m₁ − m₂)g, but it must accelerate the total mass m₁ + m₂. That is why the machine slows free fall down and made gravity easy to measure in the 18th century. For a block and tackle, n rope segments each carry the same tension, so each takes 1/n of the load; in exchange you pull n times as much rope, and the work done is unchanged.
Worked example
- Inputs: m₁ = 5 kg, m₂ = 3 kg, n = 2.
- Acceleration: 9.80665 × (5 − 3) ÷ 8 = 2.4517 m/s².
- Tension: 2 × 5 × 3 × 9.80665 ÷ 8 = 36.775 N. Check: 5 × (9.80665 − 2.4517) = 36.775 N.
- Block and tackle: 5 × 9.80665 ÷ 2 = 24.517 N.
| Mass 1 (kg) | Atwood acceleration (m/s²) |
|---|---|
| 1.25 | -4.038032353 |
| 2.5 | -0.8915136364 |
| 3.75 | 1.089627778 |
| 5 | 2.4516625 |
| 6.25 | 3.44557973 |
| 7.5 | 4.20285 |
| 10 | 5.280503846 |
How to read the result
The tension always lies between the two weights (here 29.4 N and 49.0 N), because the heavier mass is not in free fall and the lighter one is being lifted. If m₁ = m₂, the acceleration is zero; if m₂ = 0, it becomes g. For the block and tackle, the ideal effort is half the 49.0 N load with two ropes; real pulleys with bearing friction typically need noticeably more.
Common mistakes
- Counting the rope you pull as a supporting segment when it runs down from a fixed pulley.
- Expecting the tension to equal one of the weights.
- Ignoring pulley mass and friction in precise lab work.
Frequently asked questions
Does a single fixed pulley reduce the effort?
No. It only changes the direction of the pull; the mechanical advantage is 1.
How do I find g with an Atwood machine?
Measure the time t for a known drop d, compute a = 2d ÷ t², then g = a (m₁ + m₂) ÷ (m₁ − m₂).
How far must I pull the rope with four supporting ropes?
Four times the lifting height, ideally with a quarter of the force.
Sources: George Atwood, A Treatise on the Rectilinear Motion and Rotation of Bodies, Cambridge, 1784 · Halliday, Resnick and Walker, Fundamentals of Physics, Wiley