Calculate moments using perpendicular distance
Use the shortest distance from the pivot to the force line, with consistent units.
Exam scope and assessment notes
Core and Extended · Physics 0625, 2026–2028, section 1.5.2. Describe and calculate turning effects using perpendicular distance, balance a beam with one turning force each side, and require both zero resultant force and zero resultant moment for equilibrium. Multi-force moment balances and the equilibrium experiment are Supplement work in separately labelled Extended lessons. Written practice is self-checked; supervised practical experience, cumulative assessment and human subject-teacher review remain pending.
Before you begin
- Identify the pivot and force line.
- Convert centimetres to metres.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is 25 cm in metres?
What you will learn
- Recall moment = force × perpendicular distance.
- Distinguish a lever's length from the required perpendicular distance.
Measure to the line of action
The moment of a force is its magnitude multiplied by the perpendicular distance from the pivot to its line of action. For a vertical force, this distance is horizontal; for a horizontal force, it is vertical. The distance to the force-application point is only usable directly when the force is perpendicular to the pivot-to-point line.
In the original diagram, the application point is 0.30 m horizontally and 0.40 m vertically from the pivot. Its distance from the pivot is 0.50 m, but a vertical 12 N force uses the horizontal separation 0.30 m. Its moment is 12 × 0.30 = 3.6 N m, clockwise for the downward force shown. Using 0.50 m would incorrectly give 6.0 N m.
Rearrange without changing the distance definition
A perpendicular 8 N force at 25 cm gives 8 × 0.25 = 2.0 N m. To produce moment 6.0 N m with a 15 N force, the required perpendicular distance is 6.0/15 = 0.40 m. To produce the same moment at distance 0.20 m, the force is 6.0/0.20 = 30 N.
Write the moment unit as N m, and state clockwise or anticlockwise when the situation supplies a direction. Although joules also have the dimensions N m, a moment describes a turning effect rather than an energy transfer; use N m for this answer. If the geometry supplies no perpendicular distance or equivalent information, do not invent one.
Pause and explain
For a vertical force at an offset point, which distance is needed from the pivot?
Worked example
A downward 12 N force acts 0.30 m right and 0.40 m above a pivot. Its application point is 0.50 m from the pivot. Calculate moment.
Show the worked solution
- The force line is vertical.
- The perpendicular separation is the horizontal 0.30 m.
- Moment = 12 × 0.30 = 3.6 N m clockwise.
Answer 3.6 N m clockwise.
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.
Find moment of the downward 12 N force whose vertical line is 0.30 m right of the pivot.
Give me a hint
Choose the separation perpendicular to a vertical line.
Compare my reasoning
- Perpendicular distance = 0.30 m.
- Moment = 12 × 0.30 = 3.6 N m.
- A downward force right of the pivot gives clockwise turning.
3.6 N m clockwise.
Look for these in your work
- Choose the correct distance.
- State turning direction.
Find the moment of a perpendicular 8 N force at 25 cm; then the perpendicular distance needed for 15 N to produce 6.0 N m.
Give me a hint
Convert the first distance and rearrange the second moment equation.
Compare my reasoning
- 25 cm = 0.25 m, so moment = 8 × 0.25 = 2.0 N m.
- Distance = moment/force = 6.0/15.
- Required perpendicular distance = 0.40 m.
2.0 N m; 0.40 m.
Look for these in your work
- Convert the distance unit.
- Rearrange with a length as the answer.
A student multiplies 12 N by the sloping 0.50 m arm in the diagram, obtaining 6.0 N m. Explain and correct the error.
Give me a hint
Test whether the arm is perpendicular to the force line.
Compare my reasoning
- The force is vertical but the sloping arm is not perpendicular to it.
- The correct horizontal separation is 0.30 m.
- Moment = 12 × 0.30 = 3.6 N m, clockwise for the downward force.
Use perpendicular distance 0.30 m; moment 3.6 N m clockwise.
Look for these in your work
- Explain why 0.50 m is invalid here.
- Supply the corrected value and direction.
Common mistakes
- Use distance to the force line.
- Convert cm before reporting N m.
- Do not substitute the sloping arm length when it is not perpendicular to force.
What can you explain now?
What force gives a 6.0 N m moment at perpendicular distance 0.20 m?
Compare with the explanation
30 N.
Rearrange force = moment / perpendicular distance.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Draw a force line and the shortest perpendicular segment from a pivot; then rearrange the moment equation.
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.