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Moments and equilibrium

A force can turn as well as push, and the turning effect depends on where the force acts as much as how big it is. The principle of moments settles every balancing problem in the course, provided the distance you use is the perpendicular one.

Year 12AQA 3.4.1.2 CIE 4.1.1, 4.1.2, 4.1.3, 4.1.4, 4.2.1, 4.2.2 OCR A 3.2.3(a), 3.2.3(b), 3.2.3(c), 3.2.3(d), 3.2.3(e)EDEXCEL 23, 24

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Builds on Scalars and vectors.

The maths behind it: Moments on InkMaths.

IN THIS TOPIC

Calculate the moment of a force as force times perpendicular distance from the point to the line of action.
Recognise a couple, and give its torque as one force times the separation of the lines of action.
Apply the principle of moments, choosing the pivot that deletes an unknown force.
Place an object's weight at its centre of mass, which for a uniform regular solid is its geometric centre.
Separate centre of mass from centre of gravity, and for OCR, find the centre of gravity of an irregular lamina with a plumb line.

COMMON MISCONCEPTION

If the forces are equal, it balances.

The moment of a force

The turning effect of a force about a point is its moment.

\text{moment} = F d

ON THE AQA DATA SHEET

Here d is the perpendicular distance from the point to the line of action of the force, and the unit is the newton metre, N m. Every word of that definition earns marks. “Perpendicular” is the one that separates definitions which score from definitions which do not.

FIG. 1A force applied at an angle to a beam. The distance in the moment is not the distance along the beam but the perpendicular distance from the pivot to the force's line of action.

When a force is applied at an angle, extend its line of action as a construction line and drop a perpendicular from the pivot onto it. That perpendicular length, equal to L sin θ for a force at θ to the beam, is the d in Fd. Pushing along a line that passes through the pivot gives a moment of zero, however hard you push.

Couples

A couple is a pair of equal and opposite coplanar forces whose lines of action do not coincide, the grip of two hands on a steering wheel being the picture to hold. The forces cancel, so a couple produces no resultant force and no acceleration of the centre of mass. Turning is all it produces.

FIG. 2Two equal and opposite forces separated by a distance s. No resultant force, pure turning.

Its moment is force times the perpendicular distance between the two lines of action, F × s, taking one of the forces and not both. That moment is also called the torque of the couple. Doubling the answer by counting each force separately is the standard slip.

The principle of moments

An object in equilibrium obeys the principle of moments. About any point, the sum of the clockwise moments equals the sum of the anticlockwise moments. Equal forces are neither necessary nor sufficient. What balances is the products Fd.

FIG. 3A 6.0 N force at 0.20 m balances a 3.0 N force at 0.40 m: the moments match even though the forces do not.

The freedom to choose the pivot is the key to the technique. Moments may be taken about any point, so take them about the point where an unknown force acts. That force then has zero distance from the pivot, therefore zero moment, and it drops out of the equation entirely. One unknown instead of two, for no extra work.

WORKED EXAMPLE

Balancing a seesaw

A child weighing 400 N sits 1.5 m from a seesaw's pivot. Where must an adult weighing 600 N sit to balance it?

Take moments about the pivot, so the support force there vanishes from the start.

Balancing gives 400 × 1.5 = 600 × d, so d = 600/600 = 1.0 m on the other side.

Does it sense-check? The heavier person sits closer in, in inverse proportion to the weights. It is the product Fd that balances, never the force on its own.

GUIDED PRACTICE

Lifting a wheelbarrow

A loaded wheelbarrow carries 300 N acting 0.50 m from the wheel's axle; the handles are 1.2 m from the axle. Choose the pivot that removes the unknown ground force, then find the lift needed at the handles.

Show the working

Put the pivot at the axle, where the wheel's contact force acts and therefore vanishes.

Moments then give F × 1.2 = 300 × 0.50, so F = 150/1.2 = 125 N. Less than half the load, because the long handle arm multiplies whatever you supply.

INDEPENDENT PRACTICE

A trapdoor held open

A uniform trapdoor of weight 60 N is hinged along one edge and held horizontal by a vertical force applied three quarters of the way from the hinge to the far edge. Find the holding force.

Show the working

Uniform means the weight acts at the centre, half the length from the hinge. Take moments about the hinge to eliminate its unknown force.

Call the length L. Then F × 0.75L = 60 × 0.5L, so F = 30/0.75 = 40 N. The L cancels, so no length was ever needed. Only the ratio of the two distances carries any physics.

Centre of mass and centre of gravity

An object's weight is spread through it, but for moments it behaves as if all its weight acts at one point, the centre of mass. For a uniform regular solid, the centre of mass is at its geometric centre, so a uniform 4.0 m beam carries its weight at the 2.0 m mark. Placing the weight anywhere else is the commonest way a beam calculation goes wrong before it starts.

The centre of gravity is the same idea told through weight rather than mass, the single point at which the whole weight of the body may be taken to act. Where the gravitational field is uniform across the object the two points sit on top of one another, and everything in this course is small enough for that to hold, so in practice the two names label one point. They part company only for something tall enough that g measurably differs from its top to its bottom.

One board asks for a method as well. OCR wants the experimental determination of the centre of gravity, so the paragraph below is for that specification; the other three examine the two definitions only.

For an irregular lamina the apparatus is a pin and a plumb line. Hang the lamina from a small hole near its edge and let it settle. It can only come to rest with its centre of gravity vertically below the pin, because in any other position the weight has a moment about the pin and turns it further. Hang a plumb line from that same pin, mark two points along the thread, and rule the line joining them across the lamina. The centre of gravity lies somewhere on that line, though this one hanging cannot say where.

FIG. 4Hung from one hole, the lamina settles with its centre of gravity directly below the pin, and the plumb line marks a line it must lie on. Hung from a second hole, it settles differently and marks a second line. The two cross at G, and a third hole would draw a line through the same point.

Repeat from a second hole well away from the first and rule a second line. The centre of gravity is the point where the two lines cross, since it has to lie on both. A third hole is the check worth doing, because its line should pass through that same crossing, and a line that misses it says a hole was binding on the pin instead of hanging freely.

ASSESSMENT FOCUS

Define the moment with all the furniture. Force multiplied by the perpendicular distance from the point to the line of action. Definitions missing either phrase drop the mark, every series.
Take moments about the point where an unknown force acts, and that unknown disappears from the equation. It is the single most useful trick in the topic.
A uniform beam's weight acts at its centre. Forgetting the beam's own weight, or placing it at an end, wrecks otherwise sound working.
A couple's moment is one force times the full separation, not both forces. And a couple cannot be balanced by a single force, only by another couple.
AQA calls that moment the torque of the couple, so use their word when the paper does.
Check both conditions when asked whether a body is in equilibrium: resultant force zero and resultant moment zero. Each can hold without the other.
OCR asks for the plumb-line determination, and the reason carries the answer, not the recipe. A freely suspended body hangs with its centre of gravity below the point of suspension, because anywhere else the weight has a moment about that point. Two hangings give two lines, and the crossing is the answer.

CHECK YOURSELF

A uniform 4.0 m beam of weight 200 N rests on supports at each end. A 600 N load sits 1.0 m from the left support. Find the force from the left support.

Show a hint

Take moments about the right support, so its unknown force vanishes.

Show the answer

Take moments about the right support. Anticlockwise there is R_left × 4.0. Clockwise there are two, the beam's own weight of 200 N acting at its centre 2.0 m away, giving 400 N m, and the load of 600 N sitting 3.0 m from the right, giving 1800 N m.

So 4.0 R_left = 400 + 1800 = 2200, giving R_left = 550 N.

The pivot choice did all the work. The right support's force never once appeared, and a single equation held a single unknown.

Moments balance, not forces.

Use the perpendicular distance to the line of action.

Choose the pivot that eliminates the unknown.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

19 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the moments and equilibrium questions page.

11 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

Trigonometry and resolving vectors is the maths this lesson leans on, worked through from GCSE.

CHECK YOUR PROGRESS

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Calculate the moment of a force as force times perpendicular distance from the point to the line of action.
Recognise a couple, and give its torque as one force times the separation of the lines of action.
Apply the principle of moments, choosing the pivot that deletes an unknown force.
Place an object's weight at its centre of mass, which for a uniform regular solid is its geometric centre.
Separate centre of mass from centre of gravity, and for OCR, find the centre of gravity of an irregular lamina with a plumb line.

Open the full revision checklist to track your progress across the whole unit.