Force system resultants
A machine may be loaded by fifty forces in fifty places, and not one of them is convenient. All fifty can be replaced by a single force and a single couple, and often, if you put the force in the right place, by the force alone.
The previous topic ended with a small piece of sleight of hand: a force can be moved to any point you like, provided you add the couple you owe for moving it. That trick was worth learning on its own, but its real purpose is what follows. Apply it to every force in a system at once and something remarkable falls out: no matter how many forces there are, or how awkwardly they are placed, the entire collection is equivalent to exactly two things.
The general reduction
Choose any point \(O\). Drag every force in the system to it, and for each one write down the couple that the move costs. Then gather up what you have: a pile of forces all acting at \(O\), and a pile of couples. Forces add as vectors. Couples, being free vectors, add as vectors too. So the whole system becomes
The first sum is over every force. The second has two parts, and forgetting the trailing one is the classic error: you must add both the moments generated by shifting each force to \(O\), and any couples that were already applied to the body to begin with. Couples that were there from the start are simply added in, unchanged, because they are the same about every point.
That is the whole result. Fifty forces become one force and one couple. The pair is not the same as the original system in every conceivable way (the internal stresses differ, just as with transmissibility), but it is identical in every external effect, which is to say it produces the same reactions at the same supports and the same tendency to translate and rotate. For a rigid body that is all anyone asks.
Forces scattered over a body, each at its own place.
What the point changes
The reduction started with “choose any point”, which should raise a suspicion: what happens if somebody else chooses a different one? The two halves of the answer behave quite differently.
The resultant force does not care. It is a plain sum of vectors with no reference to position anywhere in it, so \(\mathbf{F}_R\) is the same whichever point you pick. It is an invariant of the system.
The resultant couple very much does care. Move from \(O\) to \(O'\) and every force in the system gets a slightly different moment arm. Summing the changes gives a single correction term:
which is the force-couple move again, now applied to the resultant itself. So the couple is not a property of the system alone; it is a property of the system and the point you asked about. Quoting a resultant moment without saying which point it is about is meaningless, and it is worth being pedantic about the subscript.
The useful reading of that equation is backwards. If \(\mathbf{M}_R\) depends on the point, perhaps there is a point where it becomes zero. If there is, the system is equivalent to a single force acting there and nothing else. That is the question the next two sections answer.
The system, read from O.
Down to a single force
For the correction term to cancel the couple, it has to be able to point the right way. Since \(\mathbf{r} \times \mathbf{F}_R\) is always perpendicular to \(\mathbf{F}_R\), no choice of point can ever remove a couple that lies along the force. The condition for a system to reduce to a single force is therefore exactly that the two be perpendicular:
Three common families of systems satisfy this automatically, which is why they are singled out in every textbook.
A concurrent system has all its lines of action through one point. Reduce about that point and every moment arm is zero, so the couple is zero to begin with: this is the case handled in equilibrium of a particle. A coplanar system has all its forces in one plane, so every moment is perpendicular to that plane and hence perpendicular to the resultant, which lies in it. A parallel system points all its forces one way, and again the moments come out square to them.
There is one exception hiding in all three, and it is worth naming because it looks like an error when it appears. If the forces happen to cancel (\(\mathbf{F}_R = \mathbf{0}\)) but the moment does not, there is no force to relocate and the system reduces to a pure couple. It cannot be simplified further, and the couple is then the same about every point. A body loaded this way will not go anywhere, but it will certainly turn.
Concurrent: every line already runs through one point.
Locating the line of action
Knowing a system reduces to one force is only half of it. The other half is saying where that force acts, and the arithmetic is short.
Having reduced the system to \(\mathbf{F}_R\) and \(M_{R_O}\) at some point, slide the force onto a parallel line a perpendicular distance \(d\) away. That move generates a moment of \(F_R d\) about \(O\). Choose \(d\) so it is exactly the couple you were trying to get rid of:
and pick the side so the senses agree. The couple is paid off and a single force remains. In vector form the same statement is \(\mathbf{r} \times \mathbf{F}_R = \mathbf{M}_{R_O}\), solved for the position \(\mathbf{r}\).
Notice that this only ever locates a line, never a point. The answer is a line of action, and the force may sit anywhere along it: the same transmissibility as ever. Asking “where does the resultant act” and expecting a unique point is asking the wrong question; asking for the line, or for where the line crosses a particular axis, is asking the right one.
This is the calculation behind a great many practical results. Where a crane’s combined load lands relative to its tipping edge, where the total pressure on a wall effectively acts, where a bolt pattern must be centred so it carries no moment: all of them are this one division.
A force at O, and a couple you would rather be rid of.
The wrench
That leaves the general three-dimensional system, where the resultant force and the resultant couple point in unrelated directions and the dot product is not zero. Something irreducible remains, and it has a name.
Split the couple into a part along the force and a part across it:
The perpendicular part can be dealt with exactly as before: shift the line of action of \(\mathbf{F}_R\) and it cancels. The parallel part cannot, for the reason already given: moving the force only ever produces couples square to it, so it can never touch a couple that lies along it. What survives is a force and a couple sharing a single line, and that combination is called a wrench, or a screw.
It is a good name. It describes exactly what a screwdriver does: push along an axis while twisting about the same axis. The general result is that any system of forces in space, however complicated, is equivalent to some wrench on some line. Simpler cases are simply wrenches whose twist happens to be zero.
Two quantities survive every change of reference point: the resultant force \(\mathbf{F}_R\), and the scalar product \(\mathbf{F}_R \cdot \mathbf{M}_R\). The second is what decides whether a system can be reduced to a single force at all, and it is why the perpendicularity test earlier was not an arbitrary condition but the only one available.