Solve particle equilibrium in two perpendicular directions
Set both component sums to zero, solve for unknown tensions or forces and check that the solution is physically compatible with the model.
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
- Resolve oblique forces with sine and cosine.
- Solve simultaneous linear equations with trigonometric coefficients.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What component equation applies to equilibrium on a vertical axis?
What you will learn
- Set up independent equilibrium equations by resolving in two directions.
- Solve and verify tensions and applied forces using the supplied geometry.
Zero resultant requires balance in both directions
A particle in equilibrium has a zero sum of force components in every direction. Resolve along two perpendicular axes to obtain independent equations, such as rightward sum = 0 and upward sum = 0. Balancing the vertical components alone does not prove equilibrium if a horizontal resultant remains. A force triangle can express equivalent balance, but resolving is the method used here and no knowledge of Lami’s theorem is assumed.
A 60 N weight suspended by two equal-angle strings, each 30° above the horizontal, has equal tensions by horizontal balance. Vertical balance gives 2Tsin30° = 60, so each tension is 60 N. The tension magnitudes do not simply sum to the weight: only their vertical components do. The strings must be taut and capable of exerting those positive tensions.
Use asymmetric geometry and physical checks
For a 40 N weight with a left string 30° above the horizontal and right string 60° above it, horizontal balance gives Tₗcos30° = Tᵣcos60°. Hence Tᵣ = √3Tₗ. Vertical balance gives Tₗsin30° + Tᵣsin60° = 40, yielding Tₗ = 20 N and Tᵣ = 20√3 N. Substitute into both original equations, rather than checking only one solved relation.
A negative calculated tension contradicts a flexible-string model: a string cannot push. Similarly a negative normal reaction contradicts maintained non-adhesive contact. Such a result means the proposed force arrangement cannot support that equilibrium under the stated assumptions. Zero resultant is consistent with rest or constant straight-line velocity; here the supplied situations are at rest. Do not infer rest from force balance alone without an initial-state statement.
Pause and explain
A flexible string solution gives T = −5 N. What follows?
Worked example
A 60 N particle is suspended by two strings each 30° above the horizontal. Find both tensions using equilibrium.
Show the worked solution
- Horizontal balance gives equal tensions T because the angles are equal.
- Vertical balance gives 2Tsin30° = 60.
- Thus T = 60 N in each string; their horizontal components cancel and upward components sum to 60 N.
Answer Both tensions are 60 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 50 N particle is suspended by two strings, each 45° above the horizontal. Calculate the tensions.
Give me a hint
Symmetric horizontal components cancel when the tensions match.
Compare my reasoning
- Horizontal balance gives equal tensions T.
- Vertical balance is 2Tsin45° = 50.
- Thus T = 25√2 N in each string; each vertical component is 25 N.
Each tension is 25√2 N ≈ 35.4 N.
Look for these in your work
- Use both horizontal symmetry and vertical balance.
- Check the two upward components add to the weight.
A 40 N particle hangs from a left string at 30° and a right string at 60° above the horizontal. Find each tension.
Give me a hint
Use horizontal balance to express one tension in terms of the other.
Compare my reasoning
- Tₗcos30° = Tᵣcos60° gives Tᵣ = √3Tₗ.
- Vertical balance gives Tₗ/2 + (√3Tₗ)(√3/2) = 40, so 2Tₗ = 40.
- Hence Tₗ = 20 N and Tᵣ = 20√3 N; both original component sums vanish.
Left tension 20 N; right tension 20√3 N ≈ 34.6 N.
Look for these in your work
- Use each string’s own angle and direction.
- Verify the result in both equilibrium equations.
A particle of weight 30 N is at rest under a force P at 45° above the rightward horizontal and a separate horizontal force H leftwards. Find P and H.
Give me a hint
The oblique force supplies all the upward support and its rightward component must be balanced.
Compare my reasoning
- Vertical balance gives Psin45° = 30.
- Thus P = 30√2 N.
- Horizontal balance gives H = Pcos45° = 30 N leftwards.
P = 30√2 N; H = 30 N leftwards.
Look for these in your work
- Use both axes to determine the two unknown forces.
- Keep the specified leftward direction of H.
Common mistakes
- Set both component sums to zero.
- Resolve tensions before adding their upward contributions.
- Reject negative tension or reaction when the model only permits pulling or pushing contact.
What can you explain now?
Does a zero vertical resultant alone establish equilibrium?
Compare with the explanation
No: the horizontal resultant must also be zero.
Equilibrium requires the full force resultant to vanish in every direction.
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Make it stick
Draw the force directions, solve two component equilibrium equations and verify the resulting forces against the model’s sign constraints.
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