Pulleys and the Atwood Machine handled ropes over fixed pulleys, where everything moves together at one speed. Add a pulley that moves, and the parts move at different rates. The key is one geometric fact, the rope's fixed length, turned into a statement about rates with Leibniz Notation and Units.
Why it works
Write the rope's length as a sum of its straight pieces, each in terms of the positions of the objects. The total never changes, so its derivative is zero. For a load hanging from a movable pulley at height $y$, with one end tied to the ceiling (height $H$) and the other end pulled down past a fixed pulley to a hand at depth $s$ below it: $$L = 2(H - y) + s + \text{constant}.$$ Then $\dfrac{dL}{dt} = 0$ gives $\dfrac{ds}{dt} = 2\dfrac{dy}{dt}$, and once more, $a_{\text{hand}} = 2a_{\text{load}}$.
Key idea
To relate accelerations in a pulley system: (1) write the rope length in terms of positions, (2) differentiate twice. A movable pulley held by two strands moves half as far, half as fast and with half the acceleration of the free end of the rope. Then write Newton's second law for each object, using the constraint to cut the unknowns.
Worked example: A table and a movable pulley
A $2\ \text{kg}$ block on a frictionless table is tied to a rope that runs over a pulley at the edge, down around a movable pulley holding a $4\ \text{kg}$ block, and up to a hook under the table. Find both accelerations and the tension.
- Constraint: when the hanging block drops $d$, each of its two strands lengthens by $d$, so the table block slides $2d$: $a_1 = 2a_2$.
- Table block: $T = m_1a_1 = 4a_2$. Hanging block: $m_2g - 2T = m_2a_2$, so $39.2 - 8a_2 = 4a_2$.
- $a_2 = 39.2/12 = 3.27\ \text{m/s}^2$, $a_1 = 6.53\ \text{m/s}^2$, $T = 2(6.53) = 13.1\ \text{N}$.
Answer: $a_2 = 3.27\ \text{m/s}^2$, $a_1 = 6.53\ \text{m/s}^2$, $T = 13.1\ \text{N}$.
Worked example: Lifting with a counterweight
A $4\ \text{kg}$ load hangs from a movable pulley; the rope runs from the ceiling around it, over a fixed pulley, down to a $3\ \text{kg}$ counterweight. Find the load's acceleration.
- Load up $a$, counterweight down $2a$. Counterweight: $3g - T = 3(2a)$. Load: $2T - 4g = 4a$.
- From the first, $T = 3g - 6a$; then $6g - 12a - 4g = 4a$, so $a = \dfrac{2g}{16} = 1.23\ \text{m/s}^2$ upward.
- The lighter counterweight wins because the load is held by two strands.
Answer: $1.23\ \text{m/s}^2$ upward.
Common mistake
Giving every object the same acceleration, as with a single fixed pulley. With a movable pulley, that breaks the rope: if the load and the free end both moved 1 m, two strands would have to shorten by 1 m each while only 1 m of rope came out. Check your constraint by moving one object a small distance and asking how much rope each piece gains or loses.