--- title: "Connected Particles and Pulleys" description: "For two hanging masses joined by a taut, light, inextensible string over a fixed smooth pulley, acceleration magnitudes match and tension is uniform. Write F = ma for each mass, then solve the equatio" canonical: https://lightmysky.com/learn/mathematics/connected-particles-and-pulleys-mt_pZ6zV0RFdC source: https://lightmysky.com/learn/mathematics/connected-particles-and-pulleys-mt_pZ6zV0RFdC.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Connected Particles and Pulleys For two hanging masses joined by a taut, light, inextensible string over a fixed smooth pulley, acceleration magnitudes match and tension is uniform. Write F = ma for each mass, then solve the equations for acceleration and tension. Subject: Mathematics · Area: Mechanics · Ages 17 to 18 Page: https://lightmysky.com/learn/mathematics/connected-particles-and-pulleys-mt_pZ6zV0RFdC ## Ready when they can - Write the equation of motion for each of two connected masses - Solve the pair for the acceleration magnitude and the tension - Explain why tension has the same magnitude throughout a light string over a smooth pulley ## Lesson: One string, one shared pull As with one body, choose a positive direction before writing F = ma. Here two hanging masses use a taut, light, inextensible string over a fixed smooth pulley. Taut means no slack; inextensible means no stretch. Their acceleration magnitudes match, but their directions are opposite. Light means negligible string mass. A smooth pulley supplies no friction along the string, so tension has one magnitude throughout. Ignore air resistance. This is our ideal pulley model. For each hanging mass, weight acts down and tension acts up. Let M be the heavier mass and m the lighter mass. Take down as positive for M and up for m. Write Mg - T = Ma and T - mg = ma, where a is the acceleration magnitude. Add the equations: -T and +T cancel, giving (M - m)g = (M + m)a. Divide by M + m to find a. Then use T = mg + ma to find T. **Example.** Hang 9 kg and 1 kg in the ideal pulley model, with g = 10 N/kg. Their weights are 90 N and 10 N. Write 90 - T = 9a and T - 10 = a. Add the left sides and the right sides: 80 = 10a, so a = 8 m/s squared. Substitute into T - 10 = a: T = 18 N. Check the heavier mass: 90 - 18 = 9 times 8. *(drawing: The 9 kg and 1 kg example gives a = 8 m/s squared and T = 18 N. The sequence shows the algebra, not a pulley diagram.)* Equal hanging masses give a = 0. They may rest or move at constant speed. Equal tension does not balance unequal weights: T lies between their weights. With both axes positive down, the heavier mass has acceleration +a and the lighter mass -a: they accelerate down and up respectively. These signs describe acceleration, regardless of velocity. Keep each axis fixed. A mass p on a smooth horizontal table can connect to a hanging mass q over a fixed smooth pulley at the table edge. Use a taut, light, inextensible string with its table section horizontal. Ignore air resistance. Take positive toward the pulley for p and down for q. Write T = pa and qg - T = qa. Adding gives qg = (p + q)a. Find a, then use T = pa. The same equations can find a missing mass or g. In the ideal hanging-mass model, if the lighter mass m, g and acceleration magnitude a are known, first find T = m(g + a). Then M(g - a) = T, so M = T/(g - a). If both masses are known but g is missing, start with (M - m)g = (M + m)a. Divide by M - m to find g. Check the result in both original equations. **Recap.** Write F = ma for each mass with clear positive directions. Use equal acceleration magnitudes and one tension in the ideal model, then solve and check both equations. *Lesson, practice questions and flashcards written for this dataset.* ## Practice 14 questions on this page, each with its working shown. ## Needs first - [Simultaneous equations](https://lightmysky.com/learn/mathematics/simultaneous-equations-mt_mqgu72aCMz) - [Newton's Second Law in Two Dimensions](https://lightmysky.com/learn/mathematics/newtons-second-law-in-two-dimensions-mt_yNUFF3lxlj)