Convert hanging masses into balancing weights
Use gravitational force in the moment equation and keep the local field assumption clear.
Section 1.5.2 · Turning effect of forces
Exam scope and assessment notes
Core and Extended · Physics 0972, 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. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Use weight = mass × gravitational field strength.
- Convert grams to kilograms and centimetres to metres.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In g = 10 N/kg, what is the weight of a 200 g load?
What you will learn
- Convert a hanging mass to its weight for a moment calculation.
- Explain when a common field strength cancels in a mass balance.
A mass produces a turning effect through its weight
A hanging 200 g mass is 0.200 kg. In the stated field g = 10 N/kg, its weight is 2.0 N. At perpendicular distance 30 cm or 0.30 m, it produces moment 0.60 N m. Mass in kilograms is not itself the force in newtons, so substituting 200 directly into a moment in N m is incorrect.
To balance that load using a 300 g mass on the other side in the same field, its weight is 3.0 N and the distance must be 0.60/3.0 = 0.20 m. Include a hanger's mass if it belongs to the suspended load; labels that exclude the hanger are not permission to ignore its actual weight.
Use cancellation only under the stated common conditions
For two hanging masses in the same approximately uniform gravitational field, mleft g dleft = mright g dright simplifies to mleft dleft = mright dright. You may then use both masses in the same unit, such as grams, to find their relative balancing distances. That cancellation predicts balance; it does not turn a numerical gram-distance product into N m.
For instance, 150 g at 40 cm balances 300 g at 20 cm under that common-field model. If a moment value in N m is requested, convert mass to kg and distance to m and use the stated g. Keep the beam's own turning effect absent or through the pivot in these one-load-each-side cases.
Pause and explain
Why can g cancel when comparing two hanging masses for balance?
Worked example
A 200 g hanging load acts 30 cm left of a pivot in a field of 10 N/kg. Where should a 300 g load hang on the right to balance it?
Show the worked solution
- Left weight = 0.200 × 10 = 2.0 N; left moment = 2.0 × 0.30 = 0.60 N m.
- Right weight = 0.300 × 10 = 3.0 N.
- Right distance = 0.60/3.0 = 0.20 m or 20 cm.
Answer 20 cm right of 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.
Balance 200 g at 30 cm left with 300 g on the right, using g = 10 N/kg and no other moments.
Give me a hint
Find the first moment in N m, then divide by the other weight.
Compare my reasoning
- Left moment = (0.200 × 10) × 0.30 = 0.60 N m.
- Right weight = 0.300 × 10 = 3.0 N.
- Distance = 0.60/3.0 = 0.20 m = 20 cm.
20 cm right.
Look for these in your work
- Convert both units.
- Use weight rather than mass as force.
A total 150 g load hangs 40 cm left. Find its moment for g = 10 N/kg and the balancing distance for a total 300 g load right.
Give me a hint
The stated loads already include their hangers; convert before finding the moment.
Compare my reasoning
- Left weight = 0.150 × 10 = 1.5 N.
- Left moment = 1.5 × 0.40 = 0.60 N m.
- Right weight is 3.0 N, so distance = 0.60/3.0 = 0.20 m.
0.60 N m; 20 cm right.
Look for these in your work
- Retain the force-distance unit.
- Find a matching opposite moment.
A student writes 200 × 30 = 6000 N m from grams and centimetres. Explain why their balance comparison might work but their moment unit is wrong.
Give me a hint
Distinguish common-unit relative products from a force-distance value in SI units.
Compare my reasoning
- Using the same mass and distance units on both sides can compare balance when g is common.
- But grams are not newtons and centimetres are not metres.
- For g = 10 N/kg the actual moment is 0.200 × 10 × 0.30 = 0.60 N m.
The relative balance product can be compared, but the actual moment is 0.60 N m, not 6000 N m.
Look for these in your work
- State the common-field condition.
- Correct the value and unit.
Common mistakes
- Mass is not force in the moment formula.
- Include any load-bearing hanger mass.
- Common-field cancellation determines balance but not a moment in N m.
What can you explain now?
In g = 10 N/kg, what moment does 150 g produce at 40 cm?
Compare with the explanation
0.60 N m.
Its weight is 1.5 N and its distance is 0.40 m.
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
Convert a suspended mass to weight and show why cancelling a common g does not supply a moment in N m.
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.