A train pulls many carriages; a tow rope drags a stranded car; two boxes pushed together slide as one. When objects are linked, they share a single acceleration, but the rope or contact between them carries a force you can solve for. Master one trick: treat the whole system to find the acceleration, then look at one body alone to find the tension.
A $2\ \text{kg}$ block and a $3\ \text{kg}$ block sit on a smooth table joined by a light string. You pull the $3\ \text{kg}$ block with a steady force.
Before reading: do the two blocks have the same acceleration or different accelerations? Write your prediction and one sentence explaining why.
Linked bodies move together, so they share a single acceleration
Because a taut rope or a solid contact forces two objects to move together, they must speed up at exactly the same rate, they share one common acceleration.
The key insight: a rope that does not stretch keeps two bodies moving together, so they have the same velocity and the same acceleration at every instant. To find that acceleration, treat the connected bodies as a single system. Only forces from outside the system matter; the internal rope tension is one of the action–reaction pair and cancels within the system.
$$a = \frac{F_{net,\ \text{external}}}{m_{total}}$$
Connected bodies share one common acceleration. Find it by treating them as a single system: $a = \dfrac{F_{net,\ \text{external}}}{m_{total}}$, where $m_{total}$ is the sum of all the connected masses and internal tensions are ignored.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? True or false: two blocks joined by an inextensible string and pulled along a table have the same acceleration.
Isolate one body and apply Newton's second law to it alone
We just saw how to find the shared acceleration from the whole system. That raises a question: what force does the connecting rope actually carry? This card answers it → isolate one body, draw its free-body diagram, and apply $F = ma$ to that body alone.
To find the rope's tension, zoom in on just one of the bodies: the only horizontal force on it is the tension, and it must supply that body's share of the acceleration.
Once you know the common acceleration, focus on one body. Draw its free-body diagram and apply Newton's second law to that body using its mass and the forces acting on it. The tension is whatever force is needed to give that single body the shared acceleration.
A $2.0\ \text{kg}$ block (front) and a $3.0\ \text{kg}$ block (back) sit on a frictionless table joined by a light string. A horizontal force of $20\ \text{N}$ pulls the $2.0\ \text{kg}$ block forward, dragging the $3.0\ \text{kg}$ block behind it. Find the acceleration and the tension in the string.
Two-step method: (1) whole system gives $a = \dfrac{F}{m_{total}}$; (2) isolate one body and apply $F = ma$ to it alone to get the tension. Here $a = 20/5 = 4.0\ \text{m s}^{-2}$ and $T = m_2 a = 3.0 \times 4.0 = 12\ \text{N}$.
Add the highlighted two-step method to your notes before the check below.
Quick check: Two blocks, $1.0\ \text{kg}$ and $4.0\ \text{kg}$, joined by a string on a frictionless table, accelerate at $2.0\ \text{m s}^{-2}$ when the $1.0\ \text{kg}$ block is pulled. What is the string tension dragging the $4.0\ \text{kg}$ block?
Pattern A, Calculate then reason
A car of mass $1000\ \text{kg}$ tows a trailer of mass $500\ \text{kg}$ on a level road. The driving force on the car is $4500\ \text{N}$; ignore friction. In your book: (1) find the acceleration of the car–trailer system; (2) draw a free-body diagram of the trailer alone and find the tension in the tow bar; (3) explain in one sentence why the tow-bar tension is less than the $4500\ \text{N}$ driving force.
Two truths, one lie: which statement about connected bodies is the lie?
Dot point: Analyse the forces acting on connected bodies in physical contact and linked by ropes using vector diagrams and Newton's laws of motion
When boxes are pushed together, a contact force replaces the tension
We just learned to find a rope's tension. That raises a question: what if there is no rope, just two boxes pushed together so they slide as one? This card answers it → exactly the same method, with a contact force instead of a tension.
Pushing two touching boxes works the same as pulling roped ones: the surface where they touch passes on a contact force, found by isolating one box.
If you push two boxes that are touching, the same logic applies. The whole system accelerates at $a = F/m_{total}$. To find the force between them, isolate the box that is not directly pushed: the contact force from its neighbour is the only horizontal force on it, so contact force $= m \times a$ for that box. By Newton's third law, that box pushes back on the first box with an equal and opposite contact force.
This is why the front coupling of a train carries the most force: it must drag all the carriages behind it, whereas the rear coupling drags only the last carriage.
Bodies in direct contact share one acceleration just like roped bodies. Isolate the box that is not directly pushed; the contact force on it equals (its mass) × (the common acceleration). A coupling or contact carries more force the more mass it has to accelerate.
Pause, copy the highlighted reasoning into your book before the check below.
Last check: Decide if the statement is true or false.
In a train, the front coupling carries more force than the rear coupling because it must accelerate more mass.
Fill the gaps. Connected bodies share the same magnitude of . To find it, apply $F = ma$ to the using the total mass. To find the tension, then apply $F = ma$ to alone.
Dot point: Analyse the forces acting on connected bodies in physical contact and linked by ropes using vector diagrams and Newton's laws of motion
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 5(4 marks) 1. A $5.0\ \text{kg}$ block and a $1.0\ \text{kg}$ block, joined by a light string on a frictionless surface, are pulled by a $24\ \text{N}$ horizontal force applied to the $1.0\ \text{kg}$ block. Calculate the acceleration of the system and the tension in the string dragging the $5.0\ \text{kg}$ block.
1 mark: uses total mass for acceleration · 1 mark: $a = 24/6.0 = 4.0\ \text{m s}^{-2}$ · 1 mark: isolates the $5.0\ \text{kg}$ block, $T = m a$ · 1 mark: $T = 5.0 \times 4.0 = 20\ \text{N}$
ApplyBand 6(4 marks) 2. Two trolleys, $3.0\ \text{kg}$ and $2.0\ \text{kg}$, are joined by a light rope on a frictionless surface. A force of $20\ \text{N}$ pulls the $3.0\ \text{kg}$ trolley. Calculate the acceleration of the system and the tension in the rope.
1 mark: treats the system as one body of $5.0\ \text{kg}$ · 1 mark: $a = 4.0\ \text{m s}^{-2}$ · 1 mark: applies $F = ma$ to the $2.0\ \text{kg}$ trolley alone · 1 mark: $T = 8.0\ \text{N}$
Dot point: Analyse the forces acting on connected bodies in physical contact and linked by ropes using vector diagrams and Newton's laws of motion
UnderstandBand 5(3 marks) 3. Explain why the tension in the connecting rope does not appear when the two connected bodies are treated as a single system, and why it is smaller than the applied force.
1 mark: tension acts as an equal and opposite internal third-law pair · 1 mark: internal forces cancel in the system equation · 1 mark: the tension only has to accelerate the rear body, which is part of the total mass
Dot point: Analyse the forces acting on connected bodies in physical contact and linked by ropes using vector diagrams and Newton's laws of motion
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): Treat the two blocks as one system with total mass $6.0\ \text{kg}$. The common acceleration is $a = \dfrac{F}{m_{total}} = \dfrac{24}{6.0} = 4.0\ \text{m s}^{-2}$. To find the tension, isolate the $5.0\ \text{kg}$ block: the only horizontal force on it is the string tension $T$, which must give it the same $4.0\ \text{m s}^{-2}$. Applying Newton's second law to that block, $T = m a = 5.0 \times 4.0 = 20\ \text{N}$. The tension is less than the $24\ \text{N}$ applied force because it only has to accelerate the $5.0\ \text{kg}$ block.
Q2 (4 marks): Because the rope is inextensible, both trolleys have the same magnitude of acceleration, so the pair can first be treated as a single body of mass $3.0 + 2.0 = 5.0\ \text{kg}$. The rope tension is internal to that system and cancels, so $$a = \frac{F_{net}}{m_{total}} = \frac{20}{5.0} = 4.0\ \text{m s}^{-2}$$ To find the tension, apply Newton’s second law to one trolley alone. The $2.0\ \text{kg}$ trolley is the simpler choice because the only horizontal force on it is the tension: $$T = m a = 2.0 \times 4.0 = 8.0\ \text{N}$$ Checking with the other trolley gives $20 - T = 3.0 \times 4.0 = 12$, so $T = 8.0\ \text{N}$ as well, which confirms the answer.
Q3 (3 marks): The rope pulls forward on the rear trolley and backward on the front trolley with equal magnitude: it is a third-law pair, and both of its members are internal to the chosen system. When the two bodies are treated as one, those two forces are summed together and cancel exactly, so only the external $20\ \text{N}$ survives in the system equation. The tension is smaller than the applied force because it does a smaller job: the applied force must accelerate the whole $5.0\ \text{kg}$, while the tension only has to accelerate the $2.0\ \text{kg}$ rear trolley. The ratio of tension to applied force is therefore the ratio of the rear mass to the total mass.
You can now find the shared acceleration of connected bodies and the tension or contact force between them. Next lesson adds friction to the picture, the force that resists sliding.
↩ Back to focus-area hubAt the start you predicted whether the two joined blocks share an acceleration. The answer is they share one acceleration, because an inextensible string keeps them moving together. Find it from the total mass, then isolate one body to get the tension. If you thought each block accelerated differently, revisit Card 1.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
Tell us what you would like help with. Register interest; support details will be confirmed before enrolment.
Register interest →