Find the pivot support force on a balanced beam
Include beam weight in the force balance even when it has zero moment about the pivot.
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
- Balance one turning force each side.
- Add all forces on the selected beam.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What moment does beam weight acting through the pivot produce about it?
What you will learn
- Calculate the support force of a balanced beam.
- Distinguish a zero moment from an absent force.
The pivot reaction can vanish from one equation
A uniform beam has its weight acting at its centre of gravity. If the pivot is at that point, the beam's weight has zero moment about the pivot. The upward support reaction also passes through it and has zero moment there. Neither force has disappeared: both remain in the vertical force balance.
For a beam with weight 8 N acting at its central pivot, loads 20 N at 0.60 m left and 30 N at 0.40 m right give equal opposing 12 N m moments. All downward forces total 20 + 30 + 8 = 58 N. Complete resting equilibrium therefore requires an upward support force of 58 N.
State the body and the support assumption
Draw or describe forces on the beam, including the downward forces the hanging loads exert on it. The upward pivot reaction is a force on that beam. Its equal-and-opposite force on the support belongs to a different body and must not be added to the beam's own force list as another cancelling contribution.
The 58 N support is the value required by the ideal equilibrium equations, not a certified safe load for actual apparatus. A missing beam weight is not zero unless the question says it is negligible. If the beam's centre of gravity is away from the pivot, its weight also creates a turning moment; use the separately labelled Extended method for that additional turning force.
Pause and explain
Should beam weight through the pivot appear in the vertical force balance?
Worked example
A beam weighs 8 N through its central pivot. Downward loads 20 N at 0.60 m left and 30 N at 0.40 m right balance its moments. Find the upward pivot reaction.
Show the worked solution
- Opposing load moments are 12 N m each.
- Downward forces total 20 + 30 + 8 = 58 N.
- An upward reaction of 58 N gives zero resultant force.
Answer 58 N upward.
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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 support for balanced 20 N and 30 N downward loads plus an 8 N beam weight through the pivot.
Give me a hint
Sum every downward force on the beam before choosing an upward reaction.
Compare my reasoning
- Loads contribute 20 + 30 = 50 N down.
- The beam adds 8 N down, giving 58 N.
- Reaction must be 58 N up for zero resultant force.
58 N upward.
Look for these in your work
- Include beam weight.
- State force direction.
A beam weighs 6 N through its pivot. An 8 N load at 0.30 m left is balanced by a load at 0.20 m right. Find that load and the upward reaction.
Give me a hint
First find the unknown load from moments, then include all weights in the force sum.
Compare my reasoning
- Right load = (8 × 0.30)/0.20 = 12 N.
- Downward total = 8 + 12 + 6 = 26 N.
- Support reaction = 26 N upward.
12 N right-hand load; 26 N upward reaction.
Look for these in your work
- Use moments for the missing load.
- Use force balance for the support.
A student cancels the beam's weight from both moment and force equations because its line passes through the pivot. Explain the distinction.
Give me a hint
A lever arm belongs to the moment equation rather than the force sum.
Compare my reasoning
- The weight has zero moment because its perpendicular distance is zero.
- It remains a non-zero downward force acting on the beam.
- The support reaction must still include that weight in the force balance.
Remove its zero moment only; retain its weight in the vertical force balance.
Look for these in your work
- Explain the zero lever arm.
- Preserve the non-zero force on the selected body.
Common mistakes
- Zero moment does not mean zero force.
- Include beam weight in the resultant-force equation.
- Do not add forces on a different body to the beam's list.
What can you explain now?
Would omitting the 8 N beam weight give a valid pivot reaction here?
Compare with the explanation
No; it would underestimate the reaction by 8 N.
The weight still contributes to vertical force balance despite its zero lever arm.
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Make it stick
Write separate moment and force balances for a uniform beam with its central weight and pivot reaction.
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.