Atwood Machine

The Atwood machine connects two masses m₁ and m₂ by an inextensible string passing over a pulley. Its acceleration, a = g (m₁ − m₂)/(m₁ + m₂), dilutes gravity: it produces a slow, easily measurable motion that makes it possible to verify Newton’s second law.

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How to measure it in class

The setup requires only a pulley, a string and two masses. FizziQ comes in to measure the motion, either by video analysis or by acoustic recording of the passage of the masses.

Steps:

Mount the pulley at a height, run the string over it, attach two close masses (for example 100 g and 110 g) and measure m₁ and m₂ precisely with a balance
Place a graduated vertical ruler in the frame and film the descent with the smartphone fixed on a stand
Track the position of the descending mass frame by frame in FizziQ’s video analysis module
Plot y as a function of time: the curve should be a parabola; plot v as a function of t to obtain a straight line whose slope is a
Repeat with several pairs of masses and plot a as a function of (m₁ − m₂)/(m₁ + m₂)
Check that the graph is a straight line whose slope gives g. Choose a drop height giving at least one second of motion, so as to have about thirty points at 30 frames per second

Scientific activities on this topic

Atwood machine: verify Newton’s 2nd law - measure the acceleration of the system for several pairs of masses and deduce g
Free fall: measuring g
Galileo’s inclined plane: d ∝ t²

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Where the formula comes from

The reasoning is a model of its kind for a first-year physics class. First isolate each mass. For the descending mass m₁: m₁a = m₁g − T. For the rising mass m₂: m₂a = T − m₂g. Since the string is inextensible, the two masses have the same acceleration in absolute value; since the pulley is ideal (massless and frictionless), the tension T is the same on both sides.

Adding the two equations, T disappears:

(m₁ + m₂) a = (m₁ − m₂) g

hence a = g (m₁ − m₂)/(m₁ + m₂). One can then go back to T = 2m₁m₂g/(m₁ + m₂), which is always between m₂g and m₁g: the string “pulls” neither as much as the heavier weight, nor as little as the lighter one.

The two limiting cases that validate the model

If m₁ = m₂, then a = 0: the system remains in equilibrium in any position, or continues at constant speed if it has been set in motion. If m₂ = 0, then a = g: we recover free fall. Checking these two limits is the first reflex to have when facing any formula.

What the real experiment adds to the model

Three deviations, to be aware of in order to interpret the results honestly.

The pulley is neither massless nor frictionless. If it has a moment of inertia J and a radius r, it must also be set into rotation: the denominator becomes (m₁ + m₂ + J/r²). The term J/r² plays the role of an additional mass to accelerate, which reduces a. For a solid homogeneous pulley of mass M, J/r² = M/2. Hence the practical rule: the lighter and smaller the pulley, the more valid the simple model.

Friction in the axle. It opposes a resistive torque and lowers a further. Characteristic symptom: if the system cannot be made to start with a small mass difference, static friction is dominating.

The string. It must be inextensible - otherwise the two masses do not have the same acceleration - and of negligible mass, otherwise the mass of string on each side changes during the motion and the acceleration is no longer constant.

An equivalent energy balance

The same result can be obtained without writing any forces. When m₁ descends by a height h, m₂ rises by h: the potential energy of the system changes by −(m₁ − m₂)gh, entirely converted into kinetic energy ½(m₁ + m₂)v². From this we get v² = 2gh(m₁ − m₂)/(m₁ + m₂), that is v² = 2ah: the motion is indeed uniformly accelerated with the same value of a. Comparing the two methods is a good final-year exercise.

The historical context

The machine was designed by the English mathematician George Atwood (1745-1807) and described in 1784 in his treatise on rectilinear and rotational motion; its simplicity, and the clarity with which it isolates the mechanical principles at play, made it a basic instrument of physics teaching. In 1784, timing a free fall was out of reach: a ball dropped from 2 m hits the ground in 0.64 s, and no clock was fine enough. Atwood got around the obstacle by slowing down the fall instead of speeding up the measurement. His machine, equipped with a pendulum clock and removable trays, allowed him to verify that the distance traveled grows as t², that the speed grows as t, and to obtain a consistent value of g. It is a typical example of experimental progress obtained by a change of apparatus rather than by a better instrument.

Formula

Acceleration of the system, with ideal pulley and string:

a = g (m₁ − m₂) / (m₁ + m₂)

Tension of the string:

T = 2 m₁ m₂ g / (m₁ + m₂)

Acceleration taking the pulley into account (moment of inertia J, radius r):

a = g (m₁ − m₂) / (m₁ + m₂ + J/r²)

Equations of motion (starting from rest):

v = a t and h = ½ a t²

Speed after a descent of height h, from the energy balance:

v = √(2 a h)

where:

a: acceleration common to both masses (m·s⁻²)
m₁: descending mass, the heavier one (kg)
m₂: rising mass (kg)
T: tension of the string (N)
J: moment of inertia of the pulley (kg·m²)
r: radius of the pulley (m)
h: height traveled (m)
g: gravitational field strength, about 9.81 m·s⁻²

Application examples

With m₁ = 110 g and m₂ = 100 g: a = 9.81 × 10/210 ≈ 0.47 m·s⁻², twenty times less than g. The descent of one meter then lasts t = √(2h/a) ≈ 2.1 s, perfectly timeable.
With m₁ = 200 g and m₂ = 100 g: a = 9.81 × 100/300 = 3.27 m·s⁻², or g/3. The tension is T = 2 × 0.2 × 0.1 × 9.81 / 0.3 ≈ 1.31 N, well between 0.98 N (weight of m₂) and 1.96 N (weight of m₁).
A counterweighted elevator is a large-scale Atwood machine: the counterweight balances the car at half load, so the motor only has to supply the difference, which divides the required power several times over.
A funicular with two cars connected by a cable works on the same principle: the descending car helps the ascending one.
On a solid pulley of 50 g and radius 2 cm, J/r² = 25 g: with m₁ = 110 g and m₂ = 100 g, the acceleration drops from 0.47 to 0.42 m·s⁻², an 11% difference. Neglecting the pulley is therefore not harmless for such close masses.
The same setup can be used to measure a coefficient of friction: by making m₂ slide on a horizontal table instead of suspending it, the acceleration becomes a = g(m₁ − μm₂)/(m₁ + m₂), from which μ is deduced.

FAQ

Q: Why not measure g directly with a free fall? A: It can be done, but the fall is very fast: two meters are covered in 0.64 s, and an error of 0.05 s in the timing skews g by more than 15%. The Atwood machine slows the motion by a chosen factor, without changing the physics: the same timing error becomes negligible.

Q: Do the two masses really have the same acceleration? A: Yes, in absolute value, provided the string is inextensible and does not slip on the pulley. It is this geometric constraint that couples the two equations. The acceleration vectors are opposite: one goes down, the other goes up.

Q: Why do we always find a g that is too small? A: Because all the imperfections act in the same direction. The mass of the pulley adds inertia, its friction dissipates energy, the mass of the string does too. None of these effects can accelerate the system. A measured g greater than 9.81 m·s⁻² therefore signals a measurement or calculation error, not a particularly good setup.

Q: Is the tension of the string equal to the weight of the heavier mass? A: No, and this is a frequent error. If it were, m₁ would be in equilibrium and would not accelerate. The tension is 2m₁m₂g/(m₁ + m₂), always strictly between the two weights.

Q: What happens if the two masses are equal? A: The acceleration is zero. The system remains motionless wherever it is placed, and if given an impulse, it continues at constant speed - up to friction. This is a good way to evaluate experimentally the importance of this friction.

Uniformly Accelerated Linear Motion - Friction Force - Conservation of Energy - Free Fall