Free-body diagrams
A free-body diagram is not an illustration of the problem. It is the problem: the moment you decide which forces exist. Everything after it is arithmetic, and arithmetic cannot rescue a diagram that left something out.
Drawing forces on a point, as in equilibrium of a particle, is the easy case: everything meets at one place and every connection is a cable that pulls along its own length. Real bodies have extent. Forces arrive at different points. Supports do things a cable cannot: they push sideways, they resist turning. The diagram has to carry all of it. This page is about the drawing itself. Solving what you have drawn needs moments, which come next; but a diagram drawn wrongly cannot be solved at all.
Drawing the boundary
Every free-body diagram begins with a decision that is never written down: where the body ends. Draw a closed curve around whatever you intend to analyse. That curve is the entire content of the method, because it is what sorts forces into two piles.
Forces that act between two parts both inside the curve are internal. They occur in equal and opposite pairs by Newton’s third law, so they cancel in any sum you take over the body and they never appear on the diagram. Forces crossing the curve are external, and every one of them must be drawn: applied loads, reactions at supports, and the body’s own weight, acting through the centre of gravity.
The word internal is doing something subtle there. It is not a property of a force. It is a property of a force and a chosen boundary. Slide the curve so that it cuts through the middle of a beam and the forces the two halves were exchanging are suddenly external to the half you kept, drawn as arrows on the cut face and available as unknowns you can solve for. Nothing about the beam changed. Only the question did.
That is the whole method of sections, and it is why one structure yields dozens of legitimate free-body diagrams. Choosing well is a real skill: draw the boundary so it crosses the forces you want to find and avoids as many as possible of the ones you do not.
A beam, its load, and the things holding it up.
Supports and reactions
Once the boundary crosses a support, you have to say what that support does. There is one rule, and it is worth stating in italics because every entry in every table of supports is a consequence of it: if a support prevents motion in some direction, it exerts a force in that direction; if it prevents rotation, it exerts a couple moment.
So you never have to memorise the table. Look at the connection, ask what it stops, and count. A roller stops motion perpendicular to its surface and nothing else: one unknown force, normal to the surface. A smooth contacting surface is the same thing wearing different clothes. A pin or hinge stops translation in both directions in the plane but lets the body turn freely on the pin: two unknown force components, no couple. A fixed or built-in end stops everything: two force components and a couple moment, three unknowns.
Two habits are worth forming early. Draw pin reactions as \(A_x\) and \(A_y\) rather than as a magnitude at an unknown angle: components keep the equilibrium equations linear, and you can always recover the magnitude and direction afterwards. And when you do not know which way a reaction points, assume a sense, draw it, and hold to it. A negative answer is not an error; it is the algebra telling you the arrow points the other way. The exception, as always, is the cable, which cannot push. A negative cable tension means the cable has gone slack and the configuration you assumed does not exist.
A roller stops one direction, so it pushes one way.
Two- and three-force members
Some members announce their own answer, and spotting them saves more work than any other habit in statics.
A two-force member is one with forces applied at exactly two points and nowhere else: no load in between, and light enough that its weight is neglected. Force balance immediately makes the two forces equal and opposite. Moment balance then does something stronger: taking moments about one of the two points, the force there contributes nothing, so the force at the other point must have no moment about it either. It has to act along the line joining the two points.
The consequence is that its direction is known from the drawing before you write a single equation. A pinned link that would otherwise carry two unknown components carries one unknown scalar instead, and you know its line of action from two coordinates. This is exactly why trusses are tractable, and why a link, a strut, or a hydraulic ram pinned at both ends is always worth identifying first.
A three-force member gives you less, but still something. If only three non-parallel forces act on a body in equilibrium, their lines of action must be concurrent: all three pass through one point. The argument is the same: take moments about where two of them cross, and those two contribute nothing, so the third must pass through that point as well. When two of the three directions are known, the geometry hands you the third for free. If instead the three are parallel, they are concurrent at infinity, which is the degenerate case of the same statement.
A member loaded at two points and nowhere else.
Counting the unknowns
Before solving anything, count. A rigid body in a plane gives three independent equilibrium equations: two force sums and one moment sum:
Compare that with the number of unknown reaction components \(r\) on your diagram. Fewer unknowns than equations and the body is partially constrained: it is not held, and it moves. Equal numbers and the problem is statically determinate, solvable from statics alone. More unknowns than equations and it is statically indeterminate. The extra supports are real and carry real load, but finding how the load divides between them requires knowing how much each one stretches, which is a question about the material rather than about equilibrium.
The counting rule is necessary but not sufficient, and this is where exams are won. Three unknowns and three equations can still fail, in exactly two ways. If all the reactions are parallel, nothing resists a load applied across them, and the body slides. If all the reactions are concurrent (their lines of action meet at a common point), then every one of them has zero moment about that point, the moment equation about it reduces to \(0 = 0\), and the body is free to rotate. Supports arranged either way are called improper constraints. The arithmetic will not warn you; the geometry will, if you look.
Three unknowns, three equations. Solvable.
In three dimensions
Nothing changes but the count. A rigid body in space has six degrees of freedom (three translations and three rotations), so equilibrium is six scalar equations:
The same rule generates the supports. A cable still gives one unknown, along its own length. A smooth surface or a roller gives one, along its normal. A ball-and-socket blocks all three translations while allowing any rotation, so it gives three force components and no couples. A journal bearing or a smooth pin blocks the two translations perpendicular to its axis, giving two. A fixed support gives all six.
One judgement call recurs. A journal bearing that is long enough also resists being tilted, adding two couple moments to its tally. Whether to include them depends on the rest of the structure: if other supports already prevent the tilt, including the bearing couples makes the problem indeterminate for no physical reason, and the usual idealisation drops them. Say which assumption you are making. A bearing modelled two ways gives two different answers, and only one of them matches how the machine was designed.
Draw everything in components along \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\) rather than as resultants at angles, and label the unknowns before you start. Then count once more. Six against six is the shape of a problem that will come out.