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Mechanics · Forces & Newton's laws

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Pulleys

Why a light inextensible string over a smooth pulley gives the two masses the same tension and the same size of acceleration in opposite directions, the two-equations-add method for finding a and T, and the classic twist — when one mass lands the string goes slack and the other flies on freely under gravity.

Mechanics · Forces & Newton's laws

Pulleys

Why a light inextensible string over a smooth pulley gives the two masses the same tension and the same size of acceleration in opposite directions, the two-equations-add method for finding a and T, and the classic twist — when one mass lands the string goes slack and the other flies on freely under gravity.

Why it works

A pulley just changes the direction of a string's pull. Over a smooth (frictionless) pulley, a light, inextensible string keeps both the properties from [[forces.connected-particles]], with one new wrinkle:
  • Same tension throughout the string — the pulley being smooth means it doesn't
grip the string, so TT is the same on both sides.
  • Same size of acceleration — inextensible string, so as one mass goes down the
other comes up by the same amount, at the same rate. But the accelerations point in opposite directions: if the heavier side accelerates down at aa, the lighter side accelerates up at aa.3 kg5 kgT3gT5g*Two masses over a smooth pulley (not accurately drawn). One string, so one tension TT; the 5 kg5\,\text{kg} mass descends and the 3 kg3\,\text{kg} mass rises, both with the same acceleration aa.*

The method: one equation per body, then add. Apply F=maF = ma to each mass along its own direction of motion, taking each body's acceleration as positive the way it actually moves. For masses M>mM > m hanging either side, with MM descending: Mg−T=Ma(heavier, going down)Mg - T = Ma \qquad\text{(heavier, going down)} T−mg=ma(lighter, going up).T - mg = ma \qquad\text{(lighter, going up)}. Add the two equations and TT cancels, leaving (M−m)g=(M+m)a(M - m)g = (M + m)a, so a=(M−m)gM+m.a = \frac{(M - m)g}{M + m}. Put that back into either equation for TT. Adding the equations is the pulley version of "whole system for the acceleration" — the tension is internal to the pair.

A mass on a table, a mass hanging. Same idea with a corner-turn. The hanging mass is pulled down by gravity and up by TT; the table mass (on a smooth table) is pulled horizontally by TT, while its weight is balanced by the normal reaction RR — so the weight of the table mass does not drive the motion.ABRMgTTmg*Mass AA on a smooth table, connected over a pulley to a hanging mass BB (not accurately drawn). BB's weight drives the system; AA's weight is cancelled by RR.*

The classic twist: the string goes slack. When the descending mass hits the floor, the string stops pulling — there's nothing to keep it taut. The other mass is still moving upward, so it now becomes a particle moving freely under gravity ([[kinematics.vertical-motion]]): it decelerates at gg, rises a little further, stops, and falls back. To find how much higher it climbs, take its speed at the instant of slackening as the launch speed and use v2=u2−2gsv^2 = u^2 - 2gs with v=0v = 0. Forgetting this hand-off — assuming the mass stops dead, or keeps its speed — is the most common way to drop the last few marks.