Locate centre of gravity and calculate moments
Represent an object's weight at its centre of gravity and calculate the turning effect of a force using perpendicular distance.
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
Which distance is used to calculate the moment of a force?
What you will learn
- Describe centre of gravity as the point where an object's weight may be taken to act
- Define and calculate the moment of a force
- Use perpendicular distance from a pivot to the force's line of action
Model the weight at one point
An extended object contains mass throughout its shape, but its weight may be treated as acting at a single point: its centre of gravity. This representation makes turning-effect calculations manageable.
The centre of gravity depends on how the object's mass is distributed. For a uniform, symmetric object it is at the geometric centre, but it can lie away from that point when the shape or density is not uniform.
A moment measures turning effect about a pivot
The moment of a force about a point is the force multiplied by the perpendicular distance from that point to the force's line of action. Moment = Fd⊥. Its SI unit is N m.
Use the shortest perpendicular distance, not automatically the distance along the object. State whether the force tends to turn the object clockwise or anticlockwise when a sign convention is needed. A force whose line of action passes through the pivot has zero moment about that pivot.
Pause and explain
What is the moment of a force whose line of action passes through the pivot?
Worked example
A 12 N force acts perpendicular to a horizontal bar 0.25 m from its pivot. Find its moment magnitude.
Show the worked solution
- The force is perpendicular, so the perpendicular distance is the given 0.25 m.
- Moment = Fd⊥ = 12 × 0.25.
- Write the unit as N m.
Answer 3.0 N m about the pivot.
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 20 N force acts perpendicular to a spanner 0.15 m from its pivot. Calculate the moment.
Give me a hint
Use the stated perpendicular distance.
Compare my reasoning
- Moment = Fd⊥.
- Moment = 20 × 0.15.
- Include the moment unit.
3.0 N m.
Look for these in your work
- Force multiplied by perpendicular distance
- Arithmetic correct
- Unit N m
A 30 N force acts at a point 0.40 m from a pivot. The angle between the bar and force is 30°. Find the moment magnitude.
Give me a hint
The perpendicular lever arm is the distance times the sine of the angle between the bar and force.
Compare my reasoning
- d⊥ = 0.40 sin 30° = 0.20 m.
- Moment = 30 × 0.20.
- Report the magnitude with units.
6.0 N m.
Look for these in your work
- Perpendicular distance resolved
- Moment relationship used
- Magnitude and unit correct
A uniform 2.0 m beam is supported at its centre. Its weight is 40 N. State where the weight acts and its moment about the central support.
Give me a hint
For a uniform beam the centre of gravity is at its midpoint.
Compare my reasoning
- The centre of gravity is at the midpoint, 1.0 m from either end.
- The weight may be represented as acting at that midpoint.
- Its line of action passes through the support, so the perpendicular distance is zero.
The 40 N weight acts at the beam's centre of gravity, at the central support. Its moment about that support is 0 N m.
Look for these in your work
- Centre of gravity located
- Weight represented at that point
- Zero lever arm and moment explained
Common mistakes
- Using the distance along a sloping bar instead of perpendicular distance
- Forgetting the reference pivot
- Giving the moment unit as N or N m²
What can you explain now?
Which distance is used to calculate the moment of a force?
Compare with the explanation
The perpendicular distance from the pivot to the force's line of action
Moment uses the shortest perpendicular lever arm.
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Make it stick
Why is perpendicular distance essential when finding a force's moment?
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.