Atwood Machine Calculator
Calculate the acceleration and rope tension in an Atwood machine. Enter the two masses (kg or g) and gravitational acceleration to instantly solve this classic two-mass pulley system.
Find Acceleration and Tension
Atwood Machine Formula
Where:
- \(a\) — acceleration of the system (m/s²)
- \(T\) — tension in the rope (N)
- \(m_1\), \(m_2\) — the two hanging masses (kg)
- \(g\) — gravitational acceleration (m/s²)
The first equation comes from applying Newton's second law to the two-mass system as a whole: the net driving force is the weight difference \((m_1 - m_2)g\), and it accelerates the combined inertia \(m_1 + m_2\). The second equation gives the rope tension, which is the same everywhere in an ideal (massless, inextensible) rope. By convention, \(a\) is positive when \(m_1\) moves downward and \(m_2\) moves upward.
Atwood Machine – Calculation Example
Example: an Atwood machine carries a mass of \(m_1 = 5\ \text{kg}\) on one side and \(m_2 = 3\ \text{kg}\) on the other, with \(g = 9.81\ \text{m/s}^2\).
\[ a = \frac{(5 - 3) \times 9.81}{5 + 3} = \frac{19.62}{8} = 2.4525\ \text{m/s}^2 \]
\[ T = \frac{2 \times 5 \times 3 \times 9.81}{5 + 3} = \frac{294.3}{8} = 36.7875\ \text{N} \]
Result: the system accelerates at 2.4525 m/s² (the heavier 5 kg mass moves down), and the rope tension is 36.7875 N.
How to Use the Atwood Machine Calculator
- Enter the first mass in the Mass 1 (m₁) field and choose its unit —
kgorg— from the dropdown. This is the mass expected to move downward. - Enter the second mass in the Mass 2 (m₂) field and select its unit the same way.
- Set the Gravitational Acceleration (g) in m/s². Leave the default 9.81 for Earth, or use 1.62 for the Moon, 3.71 for Mars, and so on.
- Read the results: Acceleration is shown in m/s² and Tension in Rope in newtons. If both masses are equal, the calculator reports that the system is in equilibrium and shows the tension \(T = m \cdot g\).
What Is an Atwood Machine?
An Atwood machine, invented by English mathematician George Atwood in 1784, is one of the simplest and most instructive setups in classical mechanics: two masses connected by a light rope that runs over a pulley. Because the masses are unequal, gravity produces a net force on the system and both masses accelerate — the heavier one downward, the lighter one upward — with the same magnitude of acceleration.
The machine is a favorite teaching tool because it "dilutes" gravity. The system accelerates more slowly than a freely falling object (\(a < g\) always), which made it possible to measure acceleration with the timing instruments of the 18th century. It also isolates the key ideas of Newton's second law: net force, total inertial mass, and the constraint that a rope transmits a single tension to both sides.
Interpreting the results is straightforward. The acceleration is zero when the masses are equal, and it approaches \(g\) when one mass is much larger than the other (the small mass is then dragged up almost like a free-falling object). The tension always lies between the two weights: \(m_2 g < T m_2\). Note that the tension is not simply the average of the weights — it depends on both masses and on the acceleration they share.
The Atwood machine is also the building block of more complex systems: half-pulleys in elevators and cable systems, multi-pulley arrangements, and laboratory experiments that measure \(g\) or verify Newton's second law. If you want to compare it with related setups, see the Pulley Advantage Calculator for systems where pulleys trade speed for force, or the Free Fall Calculator for the limiting case \(m_2 \to 0\).
Assumptions, Limitations and Practical Tips
- Frictionless, massless pulley: a real pulley has mass and friction; part of the driving force goes into spinning it, so measured acceleration is slightly lower than predicted. With a pulley of moment of inertia \(I\) and radius \(r\), replace \(m_1 + m_2\) with \(m_1 + m_2 + I/r^2\).
- Massless, inextensible rope: the rope's own weight and stretch are neglected, so both masses share the same acceleration and the tension is uniform along the rope.
- Equal masses: if \(m_1 = m_2\), the net force is zero, the system stays at rest (or moves at constant speed), and the tension equals the weight of either mass, \(T = mg\).
- Sign convention: the calculator reports the magnitude of the acceleration; the heavier mass always moves downward. Enter the larger mass as \(m_1\) if you want the positive result to match the direction of motion.
- Other planets: because \(a\) and \(T\) both scale linearly with \(g\), you can model the Moon or Mars simply by changing the gravitational acceleration field — or check values with our Weight on Other Planets Calculator.
- Related physics: to explore the forces involved further, try the Gravitational Force Calculator for Newton's law of gravitation, or the Hooke's Law Calculator if the connection is a spring instead of a rope.
Frequently Asked Questions
How do I solve an Atwood machine problem by hand without the formula?
Write Newton's second law for each mass separately. For the heavier mass: \(m_1 g - T = m_1 a\). For the lighter mass: \(T - m_2 g = m_2 a\). Add the two equations to eliminate \(T\), giving \(a = (m_1 - m_2)g/(m_1 + m_2)\), then substitute back to find \(T\). This method works for any pulley problem, including ones with friction or a massive pulley.
What happens if one of the masses is zero?
If \(m_2 = 0\), the formula gives \(a = g\) and \(T = 0\). The remaining mass is in free fall — with nothing on the other end, the rope exerts no tension and the pulley does nothing. This is a useful sanity check: the Atwood machine smoothly approaches the free-fall limit as one mass shrinks toward zero.
What happens if I enter the smaller mass as m₁?
Nothing breaks — the calculator reports the magnitude of the acceleration, and the heavier mass always moves down regardless of which field it is in. If you compute by hand, entering the smaller mass as \(m_1\) simply gives a negative value for \(a\), which indicates the motion is opposite to the assumed direction. The tension \(T\) is unaffected because it is symmetric in \(m_1\) and \(m_2\).
Why isn't the tension equal to the weight of one of the masses?
The tension equals a weight only when the system is in equilibrium (equal masses). When the masses accelerate, the heavier mass is losing some support from the rope, so \(T\) is less than \(m_1 g\); meanwhile the lighter mass is being pulled upward, so \(T\) is greater than \(m_2 g\). That is why \(T\) always falls strictly between the two weights when the masses differ.
How do I convert masses given in grams?
Divide grams by 1000 to get kilograms: \(500\ \text{g} = 0.5\ \text{kg}\). Alternatively, the calculator's unit dropdown converts automatically, so you can enter 500 and select 'g'. Keep everything in kilograms and newtons when solving by hand — mixing grams with \(g\) in m/s² is one of the most common mistakes and produces results off by a factor of 1000.
How do I find the velocity or distance traveled after some time?
Once you have the acceleration \(a\), use the constant-acceleration kinematics equations: velocity after time \(t\) is \(v = a t\) (starting from rest), and the distance either mass travels is \(s = \tfrac{1}{2} a t^2\). The heavier mass descends and the lighter one rises by the same distance, since the rope keeps their motions linked.
How does a real pulley with mass change the results?
A pulley with moment of inertia \(I\) and radius \(r\) resists being spun up, which reduces the acceleration. In practice, replace the denominator \(m_1 + m_2\) with \(m_1 + m_2 + I/r^2\) in both formulas. For a solid disk pulley of mass \(M\), \(I = \tfrac{1}{2}Mr^2\), so the pulley effectively adds \(M/2\) to the system's inertia. Friction in the axle lowers the measured acceleration further.