Moments and couples

Everything so far has been about forces adding up to nothing. That was enough while every force passed through a single point, because concurrent forces cannot turn a body about the point they share. Give the body some size and that stops being true. A bracket can have its forces balance perfectly and still shear off the wall, because the forces balance at different places. This is the missing half of equilibrium, and it starts with a single quantity.

The moment of a force

The moment of a force about a point \(O\) measures how hard that force tends to rotate the body about \(O\). In a plane it is a product of two numbers:

and the whole subject turns on what \(d\) is. It is the perpendicular distance from \(O\) to the line of action of the force — not the distance to where the force happens to be applied. Those two coincide only when the force acts square to the lever, which is exactly the case textbooks draw first and exams then avoid.

Two consequences follow immediately, and both are useful. Because \(d\) is measured to the line, sliding the force anywhere along that line changes nothing — the same transmissibility you met in force vectors. And if the line of action passes through \(O\), then \(d = 0\) and the moment is zero however large the force is. Pushing a door straight at its hinge will never close it.

Units are newton-metres, the same combination as energy, and the coincidence is worth noticing so you can dismiss it: a moment is a force times a perpendicular distance, energy is a force times a distance along the force. Different products, different quantities.

A moment has a sense as well as a size — clockwise or counter-clockwise — and in planar work you fix a convention, conventionally counter-clockwise positive, and stay with it. Finding \(d\) by geometry gets unpleasant fast, so the usual dodge is to resolve the force into components and add their moments separately, which is legitimate and is called Varignon’s theorem:

The moment of a force equals the sum of the moments of its components, because the cross product distributes over addition. Horizontal and vertical components have obvious moment arms, so this usually turns an awkward trigonometry problem into two easy multiplications.

The cross product

In three dimensions, “clockwise” stops meaning anything, and hunting for perpendicular distances by eye becomes hopeless. The vector formulation does both jobs at once:

Here \(\mathbf{r}\) is a position vector from \(O\) to any point on the line of action of \(\mathbf{F}\). That freedom is not a footnote — it is the practical heart of the method. Pick whichever point makes the arithmetic cleanest, usually one whose coordinates you already know.

The result is a vector, which is what makes it work in space. Its magnitude is

so it agrees with the scalar method: \(r\sin\theta\) is precisely the perpendicular distance. Its direction is normal to the plane containing \(\mathbf{r}\) and \(\mathbf{F}\), with the sense given by the right-hand rule, and it encodes the axis about which the turning happens.

In practice you evaluate it as a determinant:

Order matters. The cross product is not commutative but anti-commutative, \(\mathbf{F} \times \mathbf{r} = -\mathbf{r} \times \mathbf{F}\), so writing the rows the wrong way round reverses every moment in your problem. It is the single most common three-dimensional slip, and the fastest check is to confirm the result points the way your hand says it should.

Moment about an axis

Sometimes the question is narrower than “how much does this turn about a point”. A shaft in bearings can only rotate about its own axis. A door can only swing about its hinge line. What you want is the part of the moment that acts about that particular axis, and the rest is somebody else’s problem.

Take the moment about any convenient point \(O\) on the axis, then project it onto the axis direction with a dot product:

That combination is the scalar triple product, and it can be written as a single determinant with the axis direction in the first row. Because the projection kills whatever is perpendicular to \(\mathbf{u}_a\), it does not matter which point on the axis you chose — every choice gives the same \(M_a\).

The discarded part is not fictional. It is the moment that tries to bend or tilt the shaft rather than spin it, and it is carried by the bearings. Recognising which part of a load turns a mechanism and which part merely loads its supports is most of what this calculation is for.

Couples

A couple is two forces of equal magnitude, opposite direction, and separated by some distance — parallel, but not sharing a line. Its defining property is a strange one: it has no resultant force at all. The two cancel exactly, so a couple cannot push a body anywhere. All it can do is turn it.

Its moment is the product of one of the forces with the perpendicular distance between the two lines of action:

Note carefully that \(F\) is one of the forces, not the two added together, and \(d\) is the gap between the lines, not the distance from anywhere in particular. That is because a couple has a property no single force has: the moment it produces is the same about every point in space. Take moments about a point between the lines and you get \(Fa + F(d-a) = Fd\). Take them about a point outside and the two contributions subtract instead of adding, leaving \(Fd\) again. The reference point cancels out of the algebra entirely.

A couple moment is therefore a free vector. It has magnitude and direction but no line of action and no point of application, and it may be moved anywhere on the body without changing its effect. Couples add as ordinary vectors, which in a plane means adding signed numbers. This freedom is what makes them so convenient, and it is worth holding on to: a force must be told where it acts, a couple never does.

Moving a force

Transmissibility lets you slide a force along its own line. It says nothing about moving it off that line, which of course changes the moment. But there is a way to do it honestly, and it is the most useful single manoeuvre in rigid-body statics.

Suppose \(\mathbf{F}\) acts at \(A\) and you would rather have it at \(B\). At \(B\), add two forces: \(\mathbf{F}\) and \(-\mathbf{F}\). They cancel, so nothing whatsoever has been changed about the loading. Now read the three arrows differently: the new \(\mathbf{F}\) at \(B\) is the force you wanted, and the original at \(A\) together with the \(-\mathbf{F}\) at \(B\) is a couple.

So a force can be moved to any point you like, provided you add the couple you owe. The pair is called a force-couple system, and the two together are equivalent to the original in every external respect.

This is why a bolt group can be checked at its centroid, why a wind load on a wall can be quoted at the base as a shear plus an overturning moment, and why an offset load on a column is reported as an axial force plus a bending moment. It is also the engine of the next topic: apply it to every force in a system, gather all the forces at one point, and add all the couples, and any collection of forces whatever reduces to one force and one couple.