Maths › Mechanics › Connected particles and pulleys
Connected particles and pulleys
Two objects tied together are one system when you want the acceleration and two separate objects when you want the tension. Knowing which view to take, and when, decides the question.
Builds on Forces and Newton's laws.
IN THIS TOPIC
- Model connected objects sharing one acceleration through a light inextensible string.
- Choose between whole-system and single-particle equations deliberately.
- Analyse smooth-pulley systems, including a mass on a table, finding acceleration and tension.
COMMON MISCONCEPTION
The tension on the heavy side of a pulley is larger than on the light side.
System or separate
A light inextensible string forces two accelerations to be equal and gives one tension along its length. Internal forces cancel inside a system, so treating both objects as one gets the acceleration cheaply. The tension is invisible from there, and isolating one object is what brings it back.
Two blocks of 4 kg and 2 kg joined by a string and pulled by 18 N give a system of mass 6 kg, so a = 3 m s⁻². The rear block is pulled only by the string, so T = 2 × 3 = 6 N. Two viewpoints, two equations, everything found.
The system view still works over a pulley, provided you take the direction the string moves as positive all the way round. For 5 kg and 3 kg hanging, the driving force is 5g − 3g and the moving mass is 8 kg. What is illegal is adding both weights as though they pulled the same way, which is the version that produces a = g and no marks.
Over a smooth pulley
A smooth pulley redirects a string without changing its tension. With 5 kg and 3 kg hanging, take each mass's own direction of travel as positive and write Newton's second law twice.
WORKED EXAMPLE
Acceleration and tension
For the 5 kg mass: 5g − T = 5a. For the 3 kg mass: T − 3g = 3a.
Adding eliminates T: 2g = 8a, so a = g/4 = 2.45 m s⁻².
Then T = 3(g + a) = 3 × 12.25 = 36.75 N, about 37 N.
Check: T sits between 3g ≈ 29 N and 5g = 49 N, which is where the middleman between a falling heavy side and a rising light side has to sit.
If T had come out above 49 N or below 29 N, something is wrong, and that check takes four seconds.
A pulley at the table edge
The other standard arrangement puts one mass on a horizontal table with the string running over a pulley at the edge to a second mass hanging. The hanging weight drives the system, the table mass contributes inertia but no driving force, and any friction on the table resists.
WORKED EXAMPLE
Smooth table, then rough
A 3 kg block on a smooth table is joined over a smooth pulley to a 2 kg mass hanging freely. Find the acceleration and the tension.
Around the string: 2g = 5a, so a = 19.6/5 = 3.92 m s⁻².
Isolating the block: T = 3a = 11.76 N, and checking on the hanging mass, 2(g − a) = 2 × 5.88 = 11.76 N, which agrees.
If the table is rough with μ = 0.2, friction is 0.2 × 3g = 5.88 N, so 2g − 5.88 = 5a and a = 2.74 m s⁻². Note that R = 3g here, since nothing pulls the block upwards.
Watch for the second act. When the hanging mass lands, the string goes slack and the block carries on alone, decelerating under friction or continuing at constant speed on a smooth table. That is a new model and a new suvat list.
ASSESSMENT FOCUS
- State the modelling as you use it. Light string, so one tension. Inextensible, so one acceleration. Smooth pulley, so the tension is unchanged around it.
- Write one equation per particle, each in that particle's own positive direction, then add them to eliminate T.
- Multi-stage questions change the model partway. Re-read what happens when the string breaks or a mass lands.
- When both masses hang freely from a smooth pulley, check the tension lies between the two weights. In that ideal arrangement it must, and the check has caught many a sign error; with a table, a slope or any other force in play, the check does not apply.
CHECK YOURSELF
Masses of 7 kg and 3 kg hang over a smooth pulley. Write the two equations of motion and find the acceleration in terms of g.
Show a hint
Each mass gets its own direction of travel as positive.
Show the answer
7g − T = 7a and T − 3g = 3a.
Adding: 4g = 10a, so a = 0.4g ≈ 3.92 m s⁻².
System for the acceleration, single particle for the tension.
One equation per mass, own direction positive, then add to eliminate T.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the connected particles and pulleys questions page.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.
- Model connected objects sharing one acceleration through a light inextensible string.
- Choose between whole-system and single-particle equations deliberately.
- Analyse smooth-pulley systems, including a mass on a table, finding acceleration and tension.
Open the full revision checklist to see every objective in the course in one place.