Distinguish towing tension from braking thrust
Find acceleration and the connecting force for a car and trailer, then decide whether a light rope or a rigid tow-bar can sustain the required force.
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.4. Use g = 10 m s⁻² throughout these problems. Each particle has positive constant mass. Forces and acceleration are constant within each stated stage; linear motion only. Use the resultant force in F = ma. A light taut inextensible string over a smooth pulley has one tension and equal acceleration magnitudes; those constraints cease if the string becomes slack. A rope can pull but cannot push; a light rigid tow-bar may exert tension or thrust. Resistance other than the stated friction is neglected unless explicitly supplied. In our rough-contact model, sliding friction has magnitude μR and opposes motion; at rest, friction adjusts with magnitude at most μR. Check contact and motion assumptions before continuing a solution. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Apply Newton’s laws to separate and combined systems.
- Distinguish tension, thrust and Newton’s third-law pairs.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which connecting model can transmit either tension or thrust?
What you will learn
- Use the combined system to find acceleration and the trailer to find the connection force.
- Interpret a negative signed pulling force as compression and distinguish rope from rod.
Internal connection forces cancel in the whole system
A 1000 kg car tows a 500 kg trailer on a horizontal road. With forward drive 3000 N, car resistance 500 N and trailer resistance 250 N, the combined resultant is 2250 N and total mass is 1500 kg. Thus a = 1.5 m s⁻² forward. On the trailer, T − 250 = 500(1.5), giving T = 1000 N. On the car, 3000 − 500 − 1000 = 1000(1.5), independently checking the result.
The connection pulls the trailer forward and the car backward with equal magnitudes. These third-law partners act on different bodies, so each separate diagram includes its own force. A light taut rope or a light rigid tow-bar can both sustain this tension state in the ideal model. Do not use the combined-system equation alone to infer the connection force; it has cancelled out there.
A rod may push, while a rope cannot
During a braking stage, suppose a 1000 kg car has 1750 N total backward external force and a 500 kg trailer has 250 N backward resistance. Treating a rigid connection as maintaining common acceleration gives a = −2000/1500 = −4/3 m s⁻² with forward positive. Let C be the signed forward connection force on the trailer. C − 250 = 500(−4/3), so C = −1250/3 N. The rod pushes the trailer backward with thrust 1250/3 N and exerts the opposite forward force on the car.
A rope cannot supply that negative pulling force. If no other constraint is stated, it becomes slack rather than maintaining the assumed shared acceleration. The car’s own backward acceleration is then −1750/1000 = −1.75, and the trailer’s is −250/500 = −0.5 m s⁻² during the stated forward motion. These separate-stage equations apply only until another contact, force change or rest event; do not describe negative tension as a working rope model.
Pause and explain
A trailer equation gives a signed forward connection force C < 0. For a rigid tow-bar, what does it mean?
Worked example
A 1000 kg car and 500 kg trailer move forwards on a horizontal road. Total backward external forces are 1750 N on the car and 250 N on the trailer. A light rigid tow-bar keeps them connected. Find a and the tow-bar force.
Show the worked solution
- For the combined system, −1750 − 250 = 1500a, so a = −4/3 m s⁻².
- For the trailer, C − 250 = 500a, giving C = −1250/3 N.
- This is thrust 1250/3 N backward on the trailer and forward on the car; a rope could not sustain it.
Answer a = −4/3 m s⁻²; tow-bar thrust 1250/3 N.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A 1000 kg car tows a 500 kg trailer with a light taut connection. Forward drive is 3000 N; resistances are 500 N on the car and 250 N on the trailer. Find a and tension.
Give me a hint
Use the whole system for acceleration and the trailer alone for tension.
Compare my reasoning
- Combined resultant is 3000 − 500 − 250 = 2250 N, so a = 2250/1500 = 1.5 m s⁻².
- Trailer balance is T − 250 = 500(1.5), so T = 1000 N.
- Car balance gives 3000 − 500 − 1000 = 1500 N = 1000a, confirming the solution.
a = +1.5 m s⁻²; T = 1000 N.
Look for these in your work
- Include both external resistances and both masses.
- Check the separate car and trailer equations.
A 1000 kg car and 500 kg trailer are joined by a light rigid tow-bar while moving forwards. Backward external forces are 1750 N on the car and 250 N on the trailer. Find common acceleration and the bar’s thrust.
Give me a hint
Define a signed forward connection force on the trailer and interpret its sign.
Compare my reasoning
- Combined balance gives a = −2000/1500 = −4/3 m s⁻².
- Trailer balance C − 250 = 500(−4/3) gives C = −1250/3 N.
- Thus the bar’s thrust is 1250/3 N backward on the trailer and forward on the car.
a = −4/3 m s⁻²; thrust 1250/3 N.
Look for these in your work
- Retain negative acceleration during the forward braking stage.
- Interpret compression as thrust, not as a negative physical magnitude.
Replace that braking tow-bar with a light flexible rope. The car still has 1750 N backward external force and the trailer 250 N, with masses 1000 kg and 500 kg. Test a taut-rope model and find each acceleration immediately after it slackens, before any new event.
Give me a hint
A required backward connection force on the trailer would make a rope push.
Compare my reasoning
- A common-acceleration calculation requires C = −1250/3 N, which is impossible for a taut rope.
- The rope therefore cannot maintain the shared-acceleration constraint in this model.
- With no rope force, car acceleration is −1750/1000 = −1.75 and trailer acceleration is −250/500 = −0.5 m s⁻² while they still move forwards.
Taut-rope model fails; after slackening, car a = −1.75 and trailer a = −0.5 m s⁻².
Look for these in your work
- Reject compression for a flexible rope.
- Recompute the two accelerations using separate force systems.
Common mistakes
- Both masses contribute to the common acceleration.
- The trailer needs its own equation to find the connecting force.
- A negative rope tension means the taut-rope assumption fails.
What can you explain now?
Can a light rope supply the compressive force required to keep a braking trailer with its car?
Compare with the explanation
No; a rope pulls but cannot push, so that taut-rope model fails.
A light rigid tow-bar can transmit thrust; a rope requires a new slack-connection stage.
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Make it stick
Add the bodies for common acceleration, resolve one for the connection force, then check whether the connection can pull or push as required.
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