Identify forces and draw a particle free body diagram
Choose the body being modelled, include each force acting on it and distinguish weight, normal contact, friction and string tension.
Section 4.1 · Forces and equilibrium
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.1. Paper 1 algebra and trigonometry are assumed. Bodies are modelled as particles: forces act at one point and rotational effects are not considered. Resolve numerically; a scale drawing is not a substitute for calculation. Force diagrams are qualitative unless stated otherwise. Where mass is supplied in this sequence, use the stated g = 10 m s⁻². Check non-negative tension and contact reaction and use F ≤ μR except at limiting friction. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Use signed directions and basic trigonometry.
- Distinguish mass in kilograms from force in newtons.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
With g = 10 m s⁻², what is the weight of a 3 kg particle?
What you will learn
- Identify external forces acting on a chosen particle.
- Draw labelled directions using the supplied contact and string assumptions.
Choose one body before drawing its forces
A free-body diagram shows forces acting on the chosen body, not forces it exerts on its surroundings. Model the body as a particle with all force arrows meeting at one point. Weight mg acts vertically downwards; a taut string pulls along its length away from the body; normal contact acts perpendicular to a surface and pushes rather than pulls. Friction, when present, acts along the surface and opposes actual or impending relative sliding. It can be zero when no tangential force is needed.
For a 5 kg particle at rest on a horizontal rough surface, pulled horizontally right by 12 N, use the stated g = 10 m s⁻². Its weight is 50 N down, normal reaction is 50 N up and equilibrium requires friction 12 N left. These magnitudes follow from balance; the model also needs enough available friction to support that rest state. Do not add an extra force called ‘motion’ or put the force exerted on the ground onto the particle’s own diagram.
Model assumptions determine which forces are present
A smooth contact has no friction, while a rough contact permits a tangential friction component. A light flexible string carries tension when taut and cannot push a particle; a slack string supplies no tension. Weight remains vertical on an inclined plane, whereas the normal contact tilts perpendicular to it. On a smooth plane inclined at 30°, a 3 kg particle held by a string parallel to the plane has weight 30 N, tension 15 N up the plane and reaction 15√3 N normal to it.
An extended object is treated as a particle in this Mechanics component. This removes rotational effects from the model; it does not prove a real object cannot tip. State the assumptions before using equilibrium equations. A horizontal surface does not automatically make R = mg: an additional upward pull reduces the reaction, and a downward push increases it. A missing contact or slack string changes the force diagram and invalidates equations that retain that force.
Pause and explain
Which force is absent at a smooth contact?
Worked example
A 5 kg particle rests on a rough horizontal surface under a horizontal pull of 12 N to the right. With g = 10 m s⁻², list the forces required for equilibrium.
Show the worked solution
- Weight is mg = 50 N vertically downwards.
- Vertical balance requires a 50 N upward normal reaction.
- Horizontal balance requires friction 12 N left, opposing the tendency to slide right; the available friction must be sufficient.
Answer Weight 50 N down; reaction 50 N up; pull 12 N right; friction 12 N left.
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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 3 kg lamp is at rest, suspended by one vertical taut light string. Use g = 10 m s⁻². Draw its force directions and find the tension.
Give me a hint
Choose the lamp alone and balance its two vertical forces.
Compare my reasoning
- Its weight is 3 × 10 = 30 N downwards.
- The string pulls the lamp vertically upwards with tension T.
- Equilibrium gives T − 30 = 0, so T = 30 N.
Weight 30 N down; tension 30 N up.
Look for these in your work
- Include the two forces on the lamp, not forces on the support.
- Use weight in newtons and check vertical balance.
A 4 kg particle rests on a smooth horizontal surface. A string pulls at 20 N, 30° above the rightward horizontal; a separate horizontal force keeps it at rest. Use g = 10 m s⁻². Find the reaction and balancing horizontal force.
Give me a hint
The pull has both horizontal and upward components; smooth contact supplies no friction.
Compare my reasoning
- Weight is 40 N, and the string’s upward component is 20sin30° = 10 N.
- Vertical balance gives R = 40 − 10 = 30 N.
- The rightward string component is 20cos30° = 10√3 N, so the balancing force is 10√3 N left.
Reaction 30 N up; balancing horizontal force 10√3 N left; no friction.
Look for these in your work
- Include the string’s vertical component in the reaction.
- Use the smooth assumption and balance the horizontal forces separately.
A 3 kg particle is held at rest on a smooth 30° incline by a taut string parallel to the plane. Use g = 10 m s⁻². List the three forces and find their magnitudes.
Give me a hint
Resolve the vertical weight parallel and perpendicular to the plane.
Compare my reasoning
- Weight is 30 N vertically downwards.
- The downslope component is 30sin30° = 15 N, so tension is 15 N upslope.
- The perpendicular component is 30cos30° = 15√3 N, so the normal reaction has that magnitude; friction is absent.
Weight 30 N vertical; tension 15 N upslope; reaction 15√3 N perpendicular to the plane.
Look for these in your work
- Draw distinct vertical and plane-normal directions.
- Use the smooth model without adding friction.
Common mistakes
- Include only forces acting on the chosen body.
- Keep weight vertical and normal reaction perpendicular to the surface.
- A string pulls and a smooth surface supplies no friction.
What can you explain now?
Does a particle resting on an inclined plane have weight directed down the plane?
Compare with the explanation
No: weight is vertically downwards.
Its parallel and normal components are obtained by resolving the vertical weight.
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Make it stick
Choose a body, state its assumptions and draw every external force with the correct direction before resolving.
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.