Equilibrium of a rigid body

Everything needed for this has now been assembled. The free-body diagram says which forces exist. Moments say how a force off-centre turns things. And the reduction says that whatever the loading, it comes down to one resultant force and one resultant couple. Put those together and the condition for a rigid body to stay put writes itself.

The two conditions

Any system of forces on a body is equivalent to a resultant force and a resultant couple at a chosen point. For the body not to accelerate, neither of those may survive:

The first stops it translating; the second stops it rotating. Neither implies the other, which is the whole reason this topic exists. A couple has no resultant force at all, so a body can satisfy the first condition perfectly and still be spun by a pure couple. That is why particle equilibrium could get away with force balance alone: when every force runs through one point, no couple can arise.

Now the question raised by the reduction: the moment condition names a point \(O\), so does the choice matter? It does not, and the reason is worth seeing. Moving the reference point from \(O\) to \(O'\) changes the resultant moment by exactly one term,

and in equilibrium both pieces on the right are already zero — the first by assumption, the second because \(\mathbf{F}_R = \mathbf{0}\). So if the moments balance about any one point, they balance about every point in space.

That is a licence, not a triviality. It means you may take moments about whichever point makes the arithmetic easiest — including points off the body entirely, points in mid-air, points chosen purely because three unknowns happen to pass through them. Most of the skill in this subject is spending that licence well.

Three equations in a plane

When all the forces lie in the \(x\)–\(y\) plane, the six scalar conditions collapse to three. Two of the force equations are automatically satisfied, and all moments point along \(z\), so what remains is

Three equations means at most three unknowns, which is why the support counting in free-body diagrams mattered so much.

That is the familiar set, but it is not the only valid one. Because moments balance about every point, you may trade force equations for moment equations. Two alternatives are standard, and both come with a condition that is easy to forget.

You may use one force equation and two moment equations — \(\sum F_x = 0\), \(\sum M_A = 0\), \(\sum M_B = 0\) — provided the line \(AB\) is not perpendicular to the \(x\) axis. Or you may use three moment equations, \(\sum M_A = \sum M_B = \sum M_C = 0\), provided \(A\), \(B\) and \(C\) do not lie on a straight line.

The restrictions are not fussiness. Three equations are only useful if they are independent, and in the forbidden arrangements the third is an algebraic consequence of the first two. It will be satisfied automatically, tell you nothing, and leave you an equation short with no warning that anything has gone wrong. If a system of equations comes out singular, this is usually why.

Choosing the moment centre

Three equations in three unknowns can always be ground through as simultaneous equations. It is almost never necessary, and the reason is the licence from the first section.

A force has no moment about any point on its own line of action. So if you take moments about a point where two unknowns intersect, both vanish from that equation and it contains a single unknown, solvable on sight. A pin support is the obvious candidate: both of its components pass through the pin, so moments about the pin involve neither.

For an ordinary beam on a pin at \(A\) and a roller at \(B\), the sequence is: take moments about \(A\) to get \(B_y\) alone, then \(\sum F_y = 0\) for \(A_y\), then \(\sum F_x = 0\) for \(A_x\). Three equations, each with one unknown, in that order. Nothing is ever solved simultaneously.

The habit generalises. Before writing anything, look at the diagram and ask where the unknowns cross; take moments there. If two unknown reactions are parallel rather than crossing, their intersection is at infinity and the equivalent move is to sum forces perpendicular to them. And take moments about a point even when you could avoid it — an equation you did not need makes an excellent check on one you did.

One warning about signs. Fix a positive sense, conventionally counter-clockwise, and apply it to every term including the ones you are confident about. A negative answer means the reaction acts opposite to the arrow you drew, which is information rather than an error — except for a cable or a one-way support, where it means the body has left the support and the model itself is wrong.

Solving by triangle

When a body carries exactly three forces, there is a route that avoids the equations entirely, and for hand work it is often faster than any of them.

Three non-parallel forces in equilibrium must be concurrent. The argument is short: take moments about where two of them cross. Those two contribute nothing, so the third must have no moment about that point either, which means its line passes through it as well.

That single fact is often enough to finish a problem. If you know the line of action of two forces — the weight acts vertically through the centre of gravity, a cable acts along its own length — then their crossing point is fixed by the drawing, and the direction of the third force, usually an unknown pin reaction, follows from geometry before any algebra is attempted. An unknown that would otherwise have been two components is now one magnitude in a known direction.

With all three directions known and one magnitude given, the forces must also close head-to-tail into a triangle, because they sum to zero. Solve that triangle with the law of sines and both unknown magnitudes drop out at once:

It is the same closed polygon that governed particle equilibrium, now earning its keep on a body with extent. Worth recognising on sight: a bracket held by one cable, a lever with a single load and a pin, a ladder against a smooth wall.

In three dimensions

In space the two vector conditions become six scalar ones — three against sliding, three against turning:

Six equations, so six unknown reaction components, and the counting rules carry over unchanged.

There are two ways to produce them. The scalar approach takes moments about each coordinate axis directly, using \(M_a = \mathbf{u}_a \cdot (\mathbf{r} \times \mathbf{F})\) or, more often, by inspection — noticing that a force parallel to an axis has no moment about it, and neither does one that intersects it. That is quick when the geometry is simple and error-prone when it is not.

The vector approach writes every force in components, forms \(\sum \mathbf{M}_O = \sum (\mathbf{r} \times \mathbf{F})\) with determinants, and reads off three equations from the \(\mathbf{i}\), \(\mathbf{j}\) and \(\mathbf{k}\) components. It is longer but nearly mechanical, and it does not require you to see anything. For a first course the honest advice is to use vectors whenever the geometry is not obvious, and to check the result by inspection afterwards.

The axis-selection trick survives into three dimensions and is even more valuable there. Taking moments about an axis that passes through two or three unknown reactions removes all of them at once, and a well-chosen axis can reduce a six-unknown problem to a single equation in one unknown. Look for the line joining two supports; it is almost always the right axis to try first.