Moment of a force about a point

The turning idea. A force can make a body rotate about a point. How strongly it turns depends on two things: how big the force is, and how far its line of action passes from the turning point. The product of these is the moment.

$\text{moment} = F \times d$

where:

• $F$ is the magnitude of the force (in newtons, N),
• $d$ is the perpendicular distance from the point to the line of action of the force (in metres, m).

The units are therefore newton metres (N m).

Perpendicular distance is the whole game. The distance $d$ is not the distance from the point to where the force is applied — it is the shortest (perpendicular) distance from the point to the straight line along which the force acts. Picture a door: pushing far from the hinge (large $d$) opens it easily; pushing right next to the hinge (small $d$) barely moves it; pushing along the line straight at the hinge ($d = 0$) does nothing at all.

Cambridge tip. Whenever you write a moment, check that the distance you multiplied by is genuinely perpendicular to the force. Mixing up the application point with the perpendicular distance is the single most common error in this topic.

A moment tends to rotate the body either clockwise or anticlockwise. These are opposite turning effects, so a moment is a signed quantity.

Convention. Choose one sense as positive — the usual choice in 9231 is anticlockwise positive (matching the positive direction of angle measurement). Then:

• a force tending to turn the body anticlockwise contributes a positive moment,
• a force tending to turn the body clockwise contributes a negative moment.

You are free to choose clockwise-positive instead; just state your convention and stick to it for the whole question.

Reading the sense. To see which way a force turns the body about a point $O$, imagine the body pinned at $O$ and ask: would this force swing it clockwise or anticlockwise? A horizontal force acting to the right, applied above $O$, swings the top to the right — that is clockwise about $O$.

Cambridge tip. Examiners do not mind which convention you pick, but they penalise inconsistency. Write "(taking anticlockwise as positive)" once at the top and every sign falls into place.

Finding the perpendicular distance for a slanted force can be fiddly. There is a cleaner, almost foolproof route: resolve the force into two perpendicular components and take the moment of each separately. The total moment is the sum.

This works because of the principle of moments for components: the moment of a force about a point equals the sum of the moments of its components about that point.

The standard setup. Suppose a force $F$ acts at a point $P$, and $P$ is at horizontal distance $x$ and vertical distance $y$ from $O$. Resolve $F$ into a horizontal component $F_x$ and a vertical component $F_y$. Then, taking anticlockwise as positive:

$\text{moment about } O = F_y \, x - F_x \, y$

The vertical component $F_y$, acting at horizontal distance $x$, gives an anticlockwise moment $F_y x$; the horizontal component $F_x$, acting at vertical distance $y$, gives a clockwise moment $-F_x y$ (signs depend on directions — always check by inspection).

Why this is safer. Horizontal and vertical distances are usually given directly in the diagram, whereas the perpendicular distance to a slanted line needs trigonometry. Resolving converts a trig problem into simple multiplication and addition.

Cambridge tip. If a force is given as magnitude-and-angle (e.g. "$20$ N at $30°$ above the horizontal"), resolve it immediately into components $20\cos 30°$ and $20\sin 30°$. Then take moments component by component — you will rarely need to hunt for a perpendicular distance.

When several coplanar forces act on a body, each has its own moment about a chosen point $O$. The resultant (total) moment about $O$ is the algebraic sum of the individual moments — anticlockwise ones positive, clockwise ones negative (with your chosen convention).

$M_O = \sum (\pm F_i d_i)$

Step-by-step.

1. Fix a sense convention (anticlockwise positive).
2. For each force, find its moment $F_i d_i$ about $O$ (resolving into components if slanted).
3. Attach the correct sign (anticlockwise $+$, clockwise $-$).
4. Add them all up.

A positive total means the system tends to turn the body anticlockwise about $O$; a negative total, clockwise; zero means there is no net turning effect about $O$ (the body is in rotational equilibrium about that point — a key idea in subtopic 9.4).

The same force, different points. A force has a different moment about different points, because the perpendicular distance changes. So always state which point you are taking moments about: "taking moments about $A$ …".

Cambridge tip. Lay your work out as a clear sum, e.g. $M_O = (+F_1 d_1) + (-F_2 d_2) + \ldots$, so the marker can see each term and its sign. A tidy signed sum protects your method marks even if one number is wrong.