Explain moments and everyday turning effects
Choose the pivot and explain how a force can turn a door, spanner or balanced beam.
Section P1.5.2 · Turning effect of forces
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
- Identify a force and its direction.
- Recognise a pivot about which an object can turn.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which part is the pivot when a door turns?
What you will learn
- Describe moment as a measure of turning effect.
- Compare everyday turning effects about a chosen pivot.
A turning effect belongs to a chosen pivot
A moment measures a force's turning effect about a pivot. For a door, the hinge provides the turning axis; for a spanner, the nut provides the pivot in the plane of the turn. Say which pivot you are using before describing clockwise or anticlockwise turning. A force is not automatically a moment of the same numerical size.
A 20 N force acting perpendicular to a straight lever at 0.20 m from its pivot gives moment 4.0 N m. The same perpendicular force at 0.60 m gives 12 N m. The longer perpendicular distance makes the turning effect greater, even though the force itself is unchanged. This compares the stated force direction and pivot, rather than any arbitrary distance from the object.
Position and direction both matter
On a horizontal beam, a downward force to the left of the pivot tends to turn it anticlockwise; a downward force to the right tends to turn it clockwise. Reversing a force's direction reverses its turning effect. A force whose line of action passes through the pivot has zero moment about that pivot, even if its magnitude is large.
Pushing on a door near the hinge is less effective than pushing at the handle with the same suitable direction. Pushing directly towards the hinge along a line through it produces no opening moment in the ideal example. Describe the perpendicular separation between pivot and force line, not simply the length of the handle or door.
Sources for this lesson
Pause and explain
A force's line of action passes through the pivot. What is its moment about that pivot?
Worked example
Compare a 20 N perpendicular force at 0.20 m and at 0.60 m from the same pivot.
Show the worked solution
- Use moment = force × perpendicular distance.
- The two moments are 20 × 0.20 = 4.0 N m and 20 × 0.60 = 12 N m.
- The second turning effect is three times the first.
Answer 4.0 N m and 12 N m; the second is three times greater.
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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.
Compare 20 N acting perpendicularly at 0.20 m and 0.60 m from one pivot.
Give me a hint
Keep the force unchanged and compare its two perpendicular distances.
Compare my reasoning
- First moment = 20 × 0.20 = 4.0 N m.
- Second moment = 20 × 0.60 = 12 N m.
- The second is three times as large.
4.0 N m and 12 N m; second is three times greater.
Look for these in your work
- Multiply force by distance.
- Compare moments rather than just forces.
State the turn caused by a downward force right of the pivot, then an upward force at the same position.
Give me a hint
Imagine which way the right end initially tends to move.
Compare my reasoning
- A downward right-hand force lowers the right end.
- That tendency is clockwise.
- Reversing the force gives an anticlockwise tendency.
Clockwise, then anticlockwise.
Look for these in your work
- Use one viewing plane and pivot.
- Reverse the moment when the force reverses.
A door is pushed towards its hinge along a line through that hinge. Explain why a large force still gives no opening moment in the ideal model.
Give me a hint
Find the distance to the force line rather than to the hand.
Compare my reasoning
- The pivot lies on the force's line of action.
- Its perpendicular distance to that line is zero.
- Force multiplied by zero gives zero opening moment.
Zero opening moment because the force line passes through the hinge.
Look for these in your work
- Identify the force line.
- Separate force magnitude from turning effect.
Common mistakes
- Name the pivot before describing a moment.
- Equal forces need not have equal turning effects.
- A force through the pivot gives zero moment about it.
What can you explain now?
Which way does a downward force on the left of a horizontal pivot turn the beam?
Compare with the explanation
Anticlockwise.
The left end tends to move down while the right end tends to rise.
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Make it stick
Name two everyday pivots and explain how force position and direction alter their turning effects.
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.