Balance several forces and an off-centre beam weight
Sum all clockwise and anticlockwise moments, then check the support reaction.
Exam scope and assessment notes
Mapped to the named 0973 section using teaching from the corresponding single-science course. A 0973-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade. This qualification reports two grades on the 9–1 scale; confirm the entry route for your zone and exam series.
Before you begin
- Calculate a signed moment for each force.
- Find support using a separate force balance.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the clockwise moment of 10 N downward acting 0.20 m right?
What you will learn
- Extend the principle of moments to several turning forces.
- Include an off-centre beam weight at its stated centre of gravity.
Every force needs its own lever arm
For equilibrium, add all clockwise moments and set their sum equal to all anticlockwise moments about the same pivot. A 15 N downward load 0.40 m left gives 6.0 N m anticlockwise. A uniform beam weight 10 N acting 0.20 m right gives 2.0 N m clockwise. A further downward load at 0.50 m right must supply the remaining 4.0 N m, so its force is 8.0 N.
The upward pivot reaction has zero moment about the pivot but must balance all the downward forces. For that example it is 15 + 10 + 8 = 33 N upward. The beam's centre is not at this pivot, so omitting its weight's moment would incorrectly predict a 12 N balancing load instead of 8 N.
Turn direction depends on force as well as side
With two left-hand downward loads 10 N at 0.30 m and 4 N at 0.60 m, the anticlockwise sum is 3.0 + 2.4 = 5.4 N m. A right-hand 12 N downward load at 0.20 m supplies 2.4 N m clockwise; another right-hand downward load at 0.50 m must be 6.0 N. These illustrative loads exclude beam turning effects unless separately supplied.
An upward force right of the pivot produces an anticlockwise moment, unlike a downward force there. For a weight-negligible beam, 8 N upward at 0.30 m right and 4 N downward at 0.40 m left together provide 4.0 N m anticlockwise, balancing 20 N downward at 0.20 m right. Their force balance requires a further 16 N upward reaction through the pivot. Classify each force's actual turning tendency instead of treating every right-hand force as clockwise.
Sources for this lesson
Pause and explain
Which moment must be included when the beam's centre of gravity is right of the pivot?
Worked example
A downward 15 N load is 0.40 m left, and a beam's 10 N weight acts 0.20 m right. Find the downward load at 0.50 m right for balance and the pivot reaction.
Show the worked solution
- Anticlockwise moment = 15 × 0.40 = 6.0 N m.
- Clockwise balance: 10 × 0.20 + F × 0.50 = 6.0, so F = 8.0 N.
- Upward reaction = 15 + 10 + 8 = 33 N.
Answer 8.0 N downward on the right; reaction 33 N upward.
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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.
Balance 15 N at 0.40 m left against beam weight 10 N at 0.20 m right and unknown load at 0.50 m right. Find load and reaction.
Give me a hint
Subtract the beam weight's clockwise moment from the required total before solving the load.
Compare my reasoning
- Required clockwise total = 15 × 0.40 = 6.0 N m.
- F = (6.0 − 10 × 0.20)/0.50 = 8.0 N.
- Reaction = 15 + 10 + 8 = 33 N upward.
8.0 N load; 33 N upward reaction.
Look for these in your work
- Include the beam weight moment.
- Check the independent force sum.
For a weight-negligible beam, left downward loads are 10 N at 0.30 m and 4 N at 0.60 m. Right downward loads are 12 N at 0.20 m and F at 0.50 m. Find F.
Give me a hint
Add both left moments and retain the known right moment.
Compare my reasoning
- Anticlockwise sum = 10 × 0.30 + 4 × 0.60 = 5.4 N m.
- Clockwise sum = 12 × 0.20 + F × 0.50.
- F = (5.4 − 2.4)/0.50 = 6.0 N.
6.0 N downward on the right.
Look for these in your work
- Add the two forces' separate moments.
- Subtract the known opposing contribution.
A weight-negligible beam has 8 N up at 0.30 m right, 20 N down at 0.20 m right, and F down at 0.40 m left. Find F and the upward pivot reaction.
Give me a hint
The upward force right and downward force left both turn anticlockwise.
Compare my reasoning
- Moment balance: 8 × 0.30 + F × 0.40 = 20 × 0.20.
- F = (4.0 − 2.4)/0.40 = 4.0 N downward.
- Force balance: reaction + 8 = 20 + 4, giving reaction 16 N upward.
4.0 N left-hand downward force; 16 N upward reaction.
Look for these in your work
- Classify upward and downward force directions correctly.
- Keep the reaction equation separate.
Common mistakes
- Include all turning forces, including off-centre beam weight.
- Use the same pivot for every moment.
- An upward right-hand force is anticlockwise.
What can you explain now?
Why does choosing the pivot remove its reaction from the moment equation?
Compare with the explanation
Its line of action passes through that pivot.
It still contributes to the separate resultant-force equation.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Draw a beam with an off-centre weight, classify each moment and check both equilibrium equations.
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.