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Two objects joined by a string, or pushing against each other, move as one — but the force between them is usually what the question wants. The technique is to treat the system as a single body to find the acceleration, then look at one object alone to find the internal force.
The big picture
This two-step method works because of the modelling assumptions, and it is worth seeing why. An string forces both objects to have the same acceleration, which is what allows one value of to describe the whole system. A string has the same tension throughout, so one value of suffices. And when the system is treated as a whole, the tension appears twice with opposite signs and cancels — which is exactly why the system equation contains no unknown internal force. Take away either assumption and the method collapses.
What you'll be able to do
Treating the system as one body means the internal forces — the tension pulling each object towards the other — appear as an action-reaction pair and cancel. Only forces remain, so the equation contains no unknown tension.
That gives the acceleration immediately, using the total mass. With known, isolating either object gives an equation containing the tension as the only unknown.
Choose the simpler object for the second step. If one has fewer forces acting on it, the equation is shorter and there is less to go wrong.
Tip — Analysing the 2 kg block gives in one line; analysing the 3 kg block gives and the same answer with more work. Pick the object with fewer forces.
For two masses hanging over a smooth pulley, the heavier one descends and the lighter rises, with the same magnitude of acceleration. The string being light and the pulley smooth mean the tension is the same on both sides.
The system approach works if you take "positive" as the direction of motion around the pulley — clockwise or anticlockwise. Then the heavier weight drives the motion and the lighter weight opposes it, so the net external force is the of the weights.
Isolating either mass then gives the tension, which always lies between the two weights: greater than the lighter weight (which is accelerating upwards) and less than the heavier (which is accelerating downwards).
Check the tension against both weights: N and N, and lies between them. If your tension falls outside that range, a sign has gone wrong.
A common arrangement has one mass on a horizontal table connected over a pulley at the edge to a mass hanging freely. The hanging weight drives the system; the table mass contributes inertia but no driving force if the table is smooth.
For the whole system, the external driving force is the hanging weight alone. The normal reaction and the table mass’s weight are vertical and perpendicular to that mass’s motion, so they do not contribute.
If the table is rough, friction opposes the motion and is subtracted from the driving force — which is the subject of the next lesson.
Tip — The table block’s weight never enters the system equation on a smooth table — it acts vertically while the block moves horizontally. Including it is a frequent error.
Two blocks pushed along in contact behave identically to a connected pair, with the between them playing the role of the tension.
The system approach gives the acceleration, and isolating the rear block gives the contact force. The key difference from a string is direction: a contact force pushes, whereas a tension pulls.
That means the order matters. Pushing a 5 kg block against a 3 kg block gives a different contact force from pushing the 3 kg against the 5 kg, even with the same applied force.
If the force were applied to the 4 kg block instead, the acceleration would still be but the contact force would be N — the contact force must accelerate whatever mass lies ahead of it.
Think like an examiner
Common misconceptions
Connected systems
Stretch yourself
A 5 kg block on a smooth horizontal table is connected by a light inextensible string over a smooth pulley at the table edge to a 3 kg mass hanging freely. The system is released from rest. Find the acceleration, the tension, and the speed of the hanging mass after it has descended 0.5 m. Take .
Hint — Two steps for the forces, then a suvat equation for the last part — the acceleration is constant.
Questions students ask
Key takeaways
How this fits the course
Test yourself
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