Connect a rough horizontal particle to a hanging mass
Combine a surface friction calculation with the hanging particle’s gravity to find acceleration and tension, and test whether release starts motion.
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
- Calculate static and sliding friction on a horizontal surface.
- Solve ideal pulley equations with a common tension.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For a table particle whose string pulls horizontally, which normal reaction applies?
What you will learn
- Resolve the surface particle before joining its equation to the hanging mass.
- Distinguish a released resting system from a specified sliding state.
Friction on one particle resists the whole connected motion
A 4 kg particle on a rough horizontal table is connected over a smooth pulley to a hanging 2 kg mass by a light taut inextensible string. With μ = 1/4, the table reaction is 40 N and sliding friction is 10 N. If the 2 kg mass descends and the table particle slides towards the pulley, write T − 10 = 4a and 20 − T = 2a. Adding gives 10 = 6a, so a = 5/3 m s⁻² and T = 50/3 N.
The table particle’s weight and reaction balance vertically; they are not a horizontal driving force. The hanging weight supplies the gravitational drive along the string’s path. The common acceleration magnitude follows the taut inextensible connection; each particle’s diagram still needs its own external forces. After solving, 50/3 − 10 = 4(5/3) and 20 − 50/3 = 2(5/3) independently check the result.
At release, test static friction first
If the system starts at rest and the hanging weight is no more than the table’s static-friction limit, it may stay at rest. With 4 kg on the table, μ = 0.6 and 2 kg hanging, the maximum friction is 24 N. Rest needs T = 20 N and table friction 20 N, both consistent with the model. Setting friction to 24 N and predicting a reverse acceleration would choose a sliding branch that has not begun.
When hanging weight exceeds the static limit under our equal-coefficient model, the indicated descent starts and the sliding equations apply. If the table particle is already moving towards the pulley but gravity drive is weaker than sliding friction, acceleration can oppose the motion until it stops. Recheck the static state at that event. Equations must also change if the hanging particle hits the floor or the string slackens.
Pause and explain
A 4 kg table particle has μ = 0.6 and a 2 kg hanging partner. Released from rest, what is admissible?
Worked example
A 4 kg table particle is connected to a 2 kg hanging mass by an ideal pulley string. It moves towards the pulley as the hanging mass descends. With sliding μ = 1/4, find a and T before contact.
Show the worked solution
- Reaction is 40 N and sliding friction is 10 N.
- Equations T − 10 = 4a and 20 − T = 2a give a = 10/6 = 5/3 m s⁻².
- Substitution gives T = 10 + 4(5/3) = 50/3 N.
Answer a = 5/3 m s⁻² towards the pulley and down; T = 50/3 N.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
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 3 kg particle slides towards a pulley on a smooth horizontal table, connected to a descending 2 kg hanging mass by an ideal string. Find a and T.
Give me a hint
Smooth removes table friction, but both masses contribute to inertia.
Compare my reasoning
- For the table particle T = 3a; for the hanging mass 20 − T = 2a.
- Adding gives 20 = 5a, so a = 4 m s⁻².
- Then T = 3 × 4 = 12 N, satisfying both equations.
a = 4 m s⁻²; T = 12 N.
Look for these in your work
- Use both particles in the summed inertia.
- Retain tension in each individual balance.
A 4 kg particle slides towards a smooth pulley on a rough horizontal table with μ = 1/4, connected to a descending 2 kg mass by a light taut string. Find acceleration and tension.
Give me a hint
Calculate the table reaction and friction before adding the along-motion balances.
Compare my reasoning
- Reaction is 40 N and friction is 10 N.
- T − 10 = 4a and 20 − T = 2a give a = 5/3 m s⁻².
- T = 10 + 4(5/3) = 50/3 N.
a = 5/3 m s⁻²; T = 50/3 N.
Look for these in your work
- Use friction opposite the stated table motion.
- Check both the table and hanging equations after substitution.
A 4 kg table particle is connected to a 2 kg hanging mass, initially at rest. The static-friction limit is μR with μ = 0.6. Decide whether rest is possible and find its tension and friction.
Give me a hint
Test the friction required by equilibrium against the available maximum.
Compare my reasoning
- The table reaction is 40 N and static-friction limit is 24 N.
- The hanging mass can rest with T = 20 N.
- The table then requires friction 20 N opposing tension, which is below 24 N; a = 0 is admissible.
Rest is possible: T = 20 N; friction 20 N; a = 0.
Look for these in your work
- Avoid using the sliding maximum for an unstarted system.
- Check the equilibrium of both particles.
Common mistakes
- The table weight is not an along-table drive.
- Use a static test for a system released from rest.
- Check the equations only while the connection is taut and before contact.
What can you explain now?
If the table’s static limit exceeds the hanging weight, must a system released from rest move?
Compare with the explanation
No; suitable static friction can hold the system at rest.
Tension equals the hanging weight in that rest state, and the table needs only that friction value.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Find contact friction on the table, combine both balances and check whether the assumed motion can start or persist.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.