Apply Newtons third law and separate interacting bodies
Identify equal and opposite interaction pairs acting on different bodies, and draw separate force balances for stacked or contacting particles.
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
- Choose the body for a free-body diagram.
- Use horizontal and vertical equilibrium equations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A particle pushes the ground down by 30 N. What is its contact partner?
What you will learn
- Identify Newton’s third-law pairs by their two interacting bodies.
- Use separate equilibrium diagrams to calculate internal contact and ground reaction.
A third-law pair acts on two different bodies
If body A exerts a force on body B, body B exerts an equal-magnitude opposite-direction force on A in the same interaction. These are simultaneous forces on different bodies, so they do not cancel in the free-body diagram of either one alone. For a 3 kg particle resting on horizontal ground with g = 10 m s⁻², ground pushes it up by 30 N and the particle pushes the ground down by 30 N. Those contact forces are a third-law pair.
The particle’s downward weight is Earth’s gravitational force on it; its third-law partner is the particle’s gravitational force on Earth, not the ground’s normal reaction. Weight and normal may have equal magnitudes in this particular equilibrium but act on the same particle and arise from different interactions. Equality in a force balance does not by itself identify a third-law pair. Contact partners can have different masses; the paired force magnitudes are still equal.
Draw each body separately to find contact forces
A particle of weight 20 N rests on another of weight 30 N, which rests on horizontal ground. The upper body’s equilibrium gives an upward contact force 20 N from the lower. By the third law the upper pushes the lower down with 20 N. The lower’s ground reaction must balance 30 + 20 = 50 N. Treating both as one combined system also gives ground reaction 50 N, but does not by itself determine their internal contact.
If a vertical string supports the upper particle with tension 8 N, its contact becomes 20 − 8 = 12 N, and the lower’s ground reaction becomes 30 + 12 = 42 N. Use the changed upper balance before transferring its equal opposite contact to the lower. For a light taut string, equal end tensions require equilibrium of the massless string together with the third-law interactions at its ends; forces exerted on two bodies by a connecting string are not automatically a direct third-law pair with each other.
Pause and explain
Which fact distinguishes a third-law pair from two forces balancing one particle?
Worked example
A 3 kg particle rests on horizontal ground, g = 10 m s⁻². Identify its normal-contact third-law pair and explain why weight and normal are not that pair.
Show the worked solution
- Particle equilibrium gives upward ground-on-particle reaction 30 N and downward weight 30 N.
- The contact partner is particle-on-ground force 30 N down, acting on a different body.
- Weight is Earth-on-particle gravity, with partner particle-on-Earth gravity; weight and normal act on the same particle and are different interactions.
Answer Ground-on-particle 30 N up pairs with particle-on-ground 30 N down; weight and normal are not a third-law pair.
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 particle is pushed horizontally into a vertical wall by 15 N. Its wall-normal direction is in equilibrium. Identify the two forces of the wall-contact third-law pair.
Give me a hint
Name the receiving body for each force and reverse its direction.
Compare my reasoning
- The wall exerts 15 N normal force on the particle, away from the wall.
- The particle exerts 15 N normal force on the wall, into the wall.
- These forces have equal magnitudes and opposite directions on different bodies; the particle’s weight is not their contact partner.
Wall on particle: 15 N away from wall; particle on wall: 15 N into wall.
Look for these in your work
- Name both interacting bodies and directions.
- Keep the particle’s weight separate from the wall-contact pair.
An upper particle of weight 20 N rests on a lower particle of weight 30 N, which rests on horizontal ground. Find their contact and the ground reaction using separate diagrams.
Give me a hint
Solve the upper particle first, then apply its equal opposite contact to the lower.
Compare my reasoning
- Upper equilibrium requires contact 20 N upwards from the lower.
- The upper therefore exerts 20 N downwards on the lower.
- Lower equilibrium gives ground reaction 30 + 20 = 50 N upwards.
Inter-particle contact 20 N; ground reaction 50 N upwards on the lower.
Look for these in your work
- Find the internal contact from the upper body’s balance.
- Use its third-law partner in the lower body’s separate balance.
In the same stacked arrangement, weights are 20 N upper and 30 N lower, but a vertical string pulls the upper up with tension 8 N. Find the inter-particle contact and ground reaction, and check contact remains possible.
Give me a hint
The string supports part of the upper weight, changing both contact balances.
Compare my reasoning
- Upper equilibrium gives C + 8 = 20, so C = 12 N upwards on the upper.
- Its contact partner is 12 N downwards on the lower.
- Lower balance gives ground reaction 30 + 12 = 42 N; both contact magnitudes are positive.
Inter-particle contact 12 N; ground reaction 42 N; both contacts remain possible.
Look for these in your work
- Include the new support before calculating internal contact.
- Transfer the opposite contact and check non-negative reactions.
Common mistakes
- Third-law partners act on different bodies.
- Do not use equal magnitude alone to identify a pair.
- Transfer a calculated contact to the other body with equal magnitude and opposite direction.
What can you explain now?
Why do a particle’s weight and ground reaction not form a Newton’s third-law pair?
Compare with the explanation
They both act on the same particle and belong to different interactions.
Their equality here comes from equilibrium, whereas a third-law pair acts on two interacting bodies.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Name the two bodies of an interaction pair, draw each body’s own forces and use equilibrium and the third law without mixing their diagrams.
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.