Balance a beam with one force on each side
Equate opposite turning effects rather than assuming the forces themselves must be equal.
Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-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.
Before you begin
- Calculate force times perpendicular distance.
- Identify clockwise and anticlockwise turning tendencies.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which way does a downward right-hand load turn a horizontal beam?
What you will learn
- Apply the principle of moments to one turning force each side.
- Find an unknown force or distance for a balanced beam.
Balance the moments about one pivot
For rotational balance, the total clockwise moment equals the total anticlockwise moment. With one downward force on each side of a horizontal beam, use Fleft × dleft = Fright × dright, where each distance is perpendicular to the corresponding force line. A 20 N load 0.60 m left gives 12 N m anticlockwise; 30 N at 0.40 m right gives 12 N m clockwise.
Those unequal forces balance the turning effect because the smaller force has the longer lever arm. A pivot's support force acts through the pivot and contributes zero moment about it. In this example the beam's own weight is negligible or acts through the pivot; an off-centre beam weight would be an additional turning force and needs the Extended multi-force balance.
Solve and check the missing quantity
An 8 N load 0.30 m left has moment 2.4 N m. For a downward load at 0.20 m right, F × 0.20 = 2.4, so F = 12 N. Alternatively, a 10 N right-hand load must be 0.24 m from the pivot. Substitute the found value into the opposite side to check that the two moments really match.
Both moment directions must be opposite. Equal moments pointing the same way add instead of balance. A moment balance alone checks the turning condition; an appropriate support also has to balance all forces if the beam is to be in complete equilibrium. Keep that separate force check visible rather than silently assuming it follows from equal moments.
Sources for this lesson
Pause and explain
Can 20 N at 0.60 m left balance 30 N at 0.40 m right?
Worked example
A 20 N load acts 0.60 m left of a horizontal pivot. What downward force at 0.40 m right balances it, with no other turning forces?
Show the worked solution
- Anticlockwise moment = 20 × 0.60 = 12 N m.
- Required clockwise moment = F × 0.40 = 12.
- F = 30 N; substituting gives 30 × 0.40 = 12 N m.
Answer 30 N downward on the right.
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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.
Find the downward right-hand force at 0.40 m that balances 20 N at 0.60 m left, with no other turning forces.
Give me a hint
Equate clockwise and anticlockwise force-distance products.
Compare my reasoning
- Left moment = 20 × 0.60 = 12 N m.
- F × 0.40 = 12 N m.
- F = 30 N downward on the right.
30 N downward on the right.
Look for these in your work
- Set equal opposing moments.
- Check the calculated force by substitution.
An 8 N force acts 0.30 m left. Find the balancing downward force at 0.20 m right, then the right-hand distance if that force is instead 10 N.
Give me a hint
Both cases require the same clockwise moment, but have different unknowns.
Compare my reasoning
- Required moment = 8 × 0.30 = 2.4 N m.
- First force = 2.4/0.20 = 12 N.
- Second distance = 2.4/10 = 0.24 m.
12 N; 0.24 m for the 10 N load.
Look for these in your work
- Keep the two cases separate.
- Use the same target moment with the appropriate rearrangement.
Two equal 10 N downward loads act at 0.20 m left and 0.50 m right. Predict the turning tendency before release from rest, with no other moments.
Give me a hint
Calculate the products rather than cancelling equal force magnitudes.
Compare my reasoning
- Anticlockwise moment = 10 × 0.20 = 2 N m.
- Clockwise moment = 10 × 0.50 = 5 N m.
- Net moment is 3 N m clockwise.
3 N m clockwise; equal loads do not balance at unequal distances.
Look for these in your work
- Compare both moments.
- State the net turning direction.
Common mistakes
- Balance opposing moments rather than force magnitudes alone.
- State where the beam's own weight acts.
- Check the support-force condition separately.
What can you explain now?
At what right-hand distance does 10 N balance 8 N at 0.30 m left?
Compare with the explanation
0.24 m.
Divide the required moment 2.4 N m by the 10 N force.
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Make it stick
Write a one-force-each-side moment balance, solve a missing distance, and explain the beam-weight assumption.
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.