Learning outcomes
- Calculate moments using perpendicular distance.
- Apply the principle of moments in equilibrium.
- Describe an experiment verifying moments.
- Locate centre of gravity and explain stability.
9.1 Turning effect of a force
A moment is the turning effect of a force about a pivot. Moment = force × perpendicular distance from the pivot to the force’s line of action. The unit is N m. A door handle is far from the hinges so a modest force produces a large turning effect.
The distance must be perpendicular. If a force acts at an angle, do not simply use the length of the lever unless it is perpendicular to the force. Extend the force’s line of action if necessary and measure the shortest distance from the pivot.
9.2 Principle of moments
For an object in rotational equilibrium, total clockwise moment about a pivot equals total anticlockwise moment. The object must also have zero resultant force for complete equilibrium, although many O Level calculations focus on turning balance.
Choose a convenient pivot, often at an unknown support force, because a force through the pivot has zero moment and drops out of the equation. Include the weight of a non-negligible beam acting at its centre of gravity.
9.3 Verifying moments experimentally
Balance a metre rule on a pivot. Hang known weights at measured distances on both sides. Adjust one distance until the rule is horizontal. Calculate clockwise and anticlockwise moments and compare. Repeat with different combinations.
Measure distances from the pivot to the vertical line of action of each weight. If the metre rule’s own weight affects the balance, first locate its centre of gravity or arrange the pivot at the centre so its weight has no moment.
9.4 Centre of gravity
The centre of gravity is the point through which the entire weight of an object may be considered to act. For a uniform symmetrical object, it lies at the geometric centre. For an irregular plane lamina, suspend it freely and use a plumb line; the centre of gravity lies on the vertical line below the suspension point. Repeating from another point gives an intersection.
An object is more stable when it has a lower centre of gravity and a wider base. It topples when the vertical line through its centre of gravity falls outside the base. Leaning into a turn, spreading the feet and loading heavy objects low all improve stability.
Worked examples
Balanced beamA 6 N force acts 2.0 m from the pivot. A 4 N force must act 3.0 m on the other side because 6×2.0 = 4×3.0 = 12 N m.
Unknown forceA 20 N load acts 0.30 m left of a pivot. A force F acts 0.50 m right. F×0.50 = 20×0.30, so F = 12 N.
TopplingA tall narrow object begins to topple when the line of action of its weight passes beyond the edge of its base.
Practical focus
InvestigationVerify the principle of moments with a metre rule, pivot and slotted masses. Include a labelled apparatus diagram, a table of force and perpendicular distance, moment calculations and a comparison using percentage difference.
Examination guidance
- Use perpendicular distance, not distance along a sloping beam.
- State clockwise and anticlockwise directions.
- In a stability explanation, mention both the centre-of-gravity line and the base.
Check your understanding
- Define moment.
- Where does the weight of a uniform metre rule act?
- How can a vehicle be made more stable?
Answers
- Force multiplied by perpendicular distance from the pivot to the line of action.
- At its centre of gravity, approximately the 50 cm mark.
- Lower its centre of gravity and/or widen its base.
