Apply the principle of moments to equilibrium
Balance clockwise and anticlockwise moments and check both force balance and turning balance for equilibrium.
Section 4 · Forces, density and pressure
AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 4. These original explanations and numerical practice support theory study; they do not demonstrate supervised practical competence. Full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use SI units, resolve perpendicular components, and distinguish force from pressure.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For rotational equilibrium, how do total clockwise and anticlockwise moments compare?
What you will learn
- State and apply the principle of moments
- Use balanced moments to solve equilibrium problems
- Recognise that equilibrium requires zero resultant force and zero resultant torque
Balance moments about a convenient point
For an object in rotational equilibrium, the sum of clockwise moments about a point equals the sum of anticlockwise moments about that point. This is the principle of moments. Choosing a pivot through an unknown force can remove that force's moment from the equation.
Calculate each moment with its perpendicular lever arm. Keep forces and distances consistent, and include the weight acting at the centre of gravity. The equation balances turning effects; it does not by itself prove that the resultant force is zero.
Full equilibrium needs force and torque balance
A system is in equilibrium when there is no resultant force and no resultant torque. It then has no linear or angular acceleration. A stationary system is in static equilibrium; an object can also move with constant velocity and remain in translational equilibrium.
After balancing moments, separately check horizontal and vertical force components. If either the net force or net torque is non-zero, the system is not in complete equilibrium.
Pause and explain
What two conditions define full mechanical equilibrium here?
Worked example
A light beam is pivoted at its centre. A 20 N load is 0.60 m to the left. What downward force at 0.40 m to the right balances the moments?
Show the worked solution
- The left moment magnitude is 20 × 0.60 = 12 N m.
- For rotational equilibrium, the right moment must also be 12 N m.
- F × 0.40 = 12, so F = 30 N.
Answer 30 N downward gives equal and opposite moments. Check support forces separately for complete equilibrium.
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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 15 N weight acts 0.20 m to the left of a pivot. What force acting downward 0.50 m to the right balances the moments?
Give me a hint
Set the clockwise and anticlockwise moment magnitudes equal.
Compare my reasoning
- Left moment = 15 × 0.20 = 3.0 N m.
- Let the right force be F: F × 0.50 = 3.0.
- F = 3.0/0.50.
6.0 N downward on the right balances the moments.
Look for these in your work
- Moment calculated
- Balance equation formed
- Unknown force solved
A light beam is supported at a pivot at x = 0 and a second support at x = 0.60 m. Loads of 40 N and 20 N act downward at x = 0.30 m and x = 0.90 m. Find the two upward support reactions.
Give me a hint
Take moments about the pivot first; then use vertical force balance.
Compare my reasoning
- Downward moment total about the pivot = 40(0.30) + 20(0.90) = 30 N m.
- For moment balance, the second support reaction R₂ satisfies R₂(0.60) = 30, so R₂ = 50 N upward.
- For zero resultant vertical force, R₁ + 50 = 40 + 20, so R₁ = 10 N upward.
The support at 0.60 m provides 50 N upward; the pivot provides 10 N upward. Their total balances the 60 N downward load, and the moments balance about the pivot.
Look for these in your work
- Both load moments included
- Second reaction found from moment balance
- Pivot reaction found from force balance
A hanging sign is stationary with zero total torque, but a horizontal resultant force of 5.0 N acts on it. Is it in equilibrium? Explain.
Give me a hint
Check both equilibrium conditions rather than torque alone.
Compare my reasoning
- The zero total torque means there is no resultant turning effect.
- The horizontal resultant force is not zero.
- Both zero resultant force and zero resultant torque are required.
No. Its torque is balanced, but the non-zero 5.0 N resultant force means it is not in full equilibrium.
Look for these in your work
- Torque condition checked
- Force condition checked
- Conclusion follows from both
Common mistakes
- Balancing moments but ignoring resultant force
- Using distances that are not perpendicular to the force
- Assuming every force through a pivot is zero
What can you explain now?
For rotational equilibrium, how do total clockwise and anticlockwise moments compare?
Compare with the explanation
They are equal
Their signed sum is zero.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Why must you check resultant force after applying the principle of moments?
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.