Equilibrium of a rigid body
Balance coplanar forces and moments, locate centres of mass and test whether a body will slide or topple.
Section Paper 3 · Further Mechanics · Topics 3.1–3.6
Exam scope and assessment notes
Mapped to Cambridge International AS & A Level Further Mathematics 9231. These original explanations and self-checks are independent study support, not Cambridge questions, marks or grades. The syllabus assumes prior knowledge of the relevant 9709 Mathematics content; check each component's prerequisites with your teacher. Full-paper assessment and human subject-specialist review remain outstanding.
Before you begin
- Cambridge expects learners to know the whole Cambridge International A Level Mathematics (9709) content. Paper 1 further pure mathematics draws on 9709 Papers 1 and 3. Confirm your exam year and component route with your school.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What makes a Further Mathematics solution convincing?
What you will learn
- Calculate moments about a point and use centres of mass
- Apply zero resultant force and zero resultant moment to a rigid body
- Solve simple toppling and sliding equilibrium problems
Choose a pivot that removes unknown forces
The moment of a force about a point is force times perpendicular distance from the point to the line of action. Choose a pivot through unknown reactions where possible; their moments are then zero.
For a body in coplanar equilibrium, resolve forces horizontally and vertically and set the total moment about a convenient point to zero. A force's line of action matters more than the point where it is drawn on a diagram.
Centre of mass and loss of equilibrium
For particles with masses mᵢ at positions xᵢ along a line, the centre of mass is x̄=Σmᵢxᵢ/Σmᵢ. Use given centres for simple uniform shapes and symmetry for standard bodies.
A body can slide if friction is insufficient or topple if the vertical line through its centre of mass leaves the support base. At limiting equilibrium, friction is at its maximum and the reaction may act at the edge of the base.
Pause and explain
Which habit best checks your work on equilibrium of a rigid body?
Worked example
A uniform horizontal beam is 6 m long, hinged at its left end and held by a vertical cable at its right end. The beam weighs 120 N, and a 60 N load hangs 2 m from the hinge. Find the cable tension and the vertical hinge reaction.
Show the worked solution
- The beam's weight acts at its centre, 3 m from the hinge. Taking moments about the hinge removes the unknown hinge force.
- Clockwise moments are 120(3)+60(2)=480 N m. The cable gives an anticlockwise moment 6T, so T=80 N.
- Vertical equilibrium gives R+80−120−60=0, so the hinge reaction R=100 N upward.
Answer Cable tension 80 N; vertical hinge reaction 100 N upward.
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.
Two point masses, 2 kg at x=0 and 6 kg at x=3 m, are joined by a light rod. Find their centre of mass from x=0.
Give me a hint
Use the mass-weighted mean position.
Compare my reasoning
- Total mass is 2+6=8 kg.
- The weighted position sum is 2(0)+6(3)=18 kg m.
- Divide by total mass.
x̄=18/8=2.25 m.
Look for these in your work
- Masses weighted by position
- Total mass used
- Position measured from stated origin
A uniform horizontal beam is 5 m long, hinged at its left end and supported by a vertical cable at its right end. The beam weighs 150 N and a 50 N load hangs 1 m from the hinge. Find the cable tension and vertical hinge reaction.
Give me a hint
Use moments about the hinge, then vertical force balance.
Compare my reasoning
- The beam's weight acts 2.5 m from the hinge, so clockwise moment is 150(2.5)+50(1)=425 N m.
- If cable tension is T, moments give 5T=425, so T=85 N.
- Vertical balance gives R+85−150−50=0, so R=115 N upward.
T=85 N; hinge reaction=115 N upward.
Look for these in your work
- Beam weight at its centre
- Correct moment arms
- Force and moment balances both satisfied
A uniform block rests on a rough horizontal floor. Explain the geometric condition for toppling as a horizontal force increases.
Give me a hint
Track the vertical line through the centre of mass relative to the support base.
Compare my reasoning
- The weight continues to act vertically through the centre of mass.
- As the applied force increases, the contact reaction shifts toward an edge.
- At limiting toppling, the resultant support acts at the edge; beyond it, the line of action of weight lies outside the base.
Toppling begins when the vertical line through the centre of mass passes outside the support base; the limiting reaction acts at an edge.
Look for these in your work
- Centre of mass line identified
- Support region considered
- Limiting condition stated
Common mistakes
- Using a sloping distance instead of the perpendicular moment arm
- Treating a pin or smooth support as able to provide an unspecified couple
- Assuming a body is stable just because the force resultant is zero
What can you explain now?
What two conditions must hold for a coplanar rigid body to be in equilibrium?
Compare with the explanation
Zero resultant force and zero resultant moment
Both translation and rotation must be balanced.
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
How can the choice of pivot simplify an equilibrium calculation, and what distinguishes sliding from toppling?
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.