Atwood Machine and Connected Bodies

Newton’s second law for a system of bodies. If bodies are connected by an inextensible string, their accelerations have the same magnitude a. Write Newton’s second law for each body separately: ma = ΣF. Mass m₁ is acted on by its weight m₁g downwards and the string tension T₁ upwards; mass m₂ by m₂g downwards and T₂ upwards. If m₂ > m₁, the right mass goes down and the left one goes up: m₁a = T₁ − m₁g, m₂a = m₂g − T₂. For a massless pulley T₁ = T₂ = T. Add the equations and the tension cancels out: (m₁ + m₂)a = (m₂ − m₁)g.

Why a is less than g. Only the difference between the weights, (m₂ − m₁)g, drives the system, while both masses, m₁ + m₂, resist being accelerated. Hence a = (m₂ − m₁)g / (m₁ + m₂). When m₂ = m₁ the acceleration is zero: push the masses and they move uniformly, even though forces act on each of them. That is Newton’s first law: the forces balance and the velocity does not change. If m₂ is much heavier than m₁, the acceleration approaches g but never reaches it. The third graph, a(m₂), shows this dependence.

String tension. The string cannot pull both masses with a force equal to the weight of one of them. Substituting a gives T = 2m₁m₂g / (m₁ + m₂). This value always lies between m₁g and m₂g: the descending mass is pulled by the string less than its weight (T < m₂g), which is why it accelerates downwards, and the rising mass is pulled more than its weight (T > m₁g), which is why it accelerates upwards. In the simulator the force arrows are drawn to the same scale — compare the blue T arrows with the red mg ones.

Block on a table. In the second mode a block m₁ lies on a horizontal table and a mass m₂ hangs from a string passing over a pulley at the table edge. The block’s weight is balanced by the normal force N = m₁g; the weight m₂g drives the system and the sliding friction μN = μm₁g opposes it: a = (m₂ − μm₁)g / (m₁ + m₂). If m₂g ≤ μm₁g, the block does not move at all. Then the static friction is exactly m₂g, not μm₁g: static friction adjusts to the applied force and reaches μN only in the limiting case.

Pulley mass. A real pulley also has to be spun up. A uniform disc of mass M behaves as if a mass M/2 were added to the system, so a = (m₂ − m₁)g / (m₁ + m₂ + M/2). The tensions on the two sides of the pulley become different: T₂ − T₁ = (M/2)·a. It is this difference in tensions that creates the torque spinning the pulley. Give the pulley a mass and see T₁ and T₂ split in the “Measurements” tab.

The strobe and Galileo’s rule. The mass starts from rest and moves with constant acceleration, so its path grows as s = at²/2. In equal successive time intervals the body covers distances in the ratio of successive odd numbers: 1 : 3 : 5 : 7 … Galileo discovered this rule. It does not depend on the masses, on g or even on the planet, but it breaks down if the masses get an initial push v₀. Try it yourself.

Things to try. 1) Make the masses equal and set v₀ — uniform motion. 2) Push the heavier mass upwards: the velocity points up while the acceleration points down, so the mass stops and comes back. 3) Move the experiment to the Moon: the acceleration drops by the same factor as g, while the ratio a/g stays the same. 4) On the table, find the μ at which the block just starts to move: m₂ = μm₁.