1.5.2 Turning effect of forces
- Syllabus
- 0625–2026–2027
- Topic
- 1.5.2
- Level
- —
Learning objectives
1.5.2.1Moment of a force as a measure of its• Describe the moment of a force as a measure of its turning effect and give everyday examples1.5.2.2Moment of a force as moment = force ו Define the moment of a force as moment = force × perpendicular distance from the pivot; recall/use this equation1.5.2.3Apply the principle of moments to• Apply the principle of moments to situations with one force each side of the pivot, including balancing of a beam1.5.2.4That, when there is no resultant force• State that, when there is no resultant force and no resultant moment, an object is in equilibrium1.5.2.5Apply the principle of moments to• Apply the principle of moments to other situations, including those with more than one force each side of the pivot1.5.2.6An experiment to demonstrate that• Describe an experiment to demonstrate that there is no resultant moment on an object in equilibrium
Recognise the turning effect of a force
The moment of a force about a point is a measure of the force's turning effect about that point or pivot.
| Example | Pivot | Turning action |
|---|---|---|
| door | hinge | push at the handle rotates the door |
| spanner | centre of nut | force on handle turns the nut |
| scissors | central joint | forces on handles rotate the blades |
| wheelbarrow | wheel axle | lifting handles turns the barrow about the wheel |
A larger force or a force acting farther from the pivot usually produces a greater turning effect.
A force through the pivot has no turning effect about that pivot. Do not confuse a moment with the force itself.
Calculate the moment of a force
The moment equals the force multiplied by the perpendicular distance from the pivot to the force's line of action.
M=Fd⊥
| Symbol | Meaning | SI unit |
|---|---|---|
| M | moment | N m |
| F | force | N |
| d⊥ | perpendicular distance from pivot to line of action | m |
Extend the force's line of action if necessary, draw the shortest perpendicular from the pivot to that line, convert the distance to metres, then multiply and state clockwise or anticlockwise.
A 20 N force with a perpendicular distance of 0.30 m produces a moment of 20×0.30=6.0 N m.
Do not use the sloping distance from pivot to the point of application unless it is perpendicular to the force.
Balance one force on each side of a pivot
For a balanced beam, the total clockwise moment about the pivot equals the total anticlockwise moment about the pivot.
F1d1=F2d2
Choose the pivot, label each force's turning direction, calculate each force × perpendicular distance, equate the two moments and solve for the unknown.
Include the beam's own weight if it is not negligible; for a uniform beam it acts at the beam's centre.
A smaller force can balance a larger force if its perpendicular distance from the pivot is proportionally larger.
Equal forces do not guarantee balance unless their moments are equal. Compare force–distance products, not distances alone.
State both conditions for equilibrium
An object is in equilibrium when it has no resultant force and no resultant moment.
| Condition | Consequence |
|---|---|
| resultant force = 0 | no linear acceleration |
| resultant moment = 0 | no angular acceleration |
Resolve or compare all forces so upward equals downward and left equals right; then check clockwise moments equal anticlockwise moments about any point.
Equilibrium can be static or dynamic: an object may remain at rest, or continue with constant velocity and constant rotational motion state.
Zero resultant force alone is incomplete: a pair of equal opposite forces can still produce a non-zero turning effect.
Apply moments with several forces
With several forces, equilibrium requires the sum of all clockwise moments to equal the sum of all anticlockwise moments about the chosen pivot.
∑Mclockwise=∑Manticlockwise
| Step | Action |
|---|---|
| 1 | choose a convenient pivot, often where an unknown support acts |
| 2 | include every force, including weights and support forces |
| 3 | find each perpendicular distance and turning direction |
| 4 | sum moments on each side and solve |
| 5 | use resultant force = 0 for any remaining support force |
For a bridge or beam with two supports, moments about one support can find the other reaction; vertical force balance then finds the first.
Do not omit the weight of a uniform beam: it acts at its midpoint. A force at the pivot contributes zero moment but may still matter to force balance.
Demonstrate zero resultant moment experimentally
Use a metre rule, pivot and known masses to test whether clockwise and anticlockwise moments are equal when the rule is in equilibrium.
| Step | Procedure |
|---|---|
| prepare | balance the unloaded metre rule on a pivot and record the pivot position |
| load | hang known masses on both sides at measured positions |
| adjust | move a mass until the rule is horizontal and stationary |
| measure | find each perpendicular distance from pivot to the weight's line of action |
| compare | convert masses to weights and calculate every Wd |
Add clockwise moments and anticlockwise moments separately. Within measurement uncertainty, the two totals should agree, so the resultant moment is zero.
Use a sharp low-friction pivot, read positions at eye level, keep strings vertical, repeat with other masses and distances, and include the rule's weight if it was not initially balanced at its centre of gravity.
A horizontal rule alone is not sufficient evidence: record forces and perpendicular distances and compare the calculated moment totals.