Free-body diagrams

Before you can write a single equilibrium equation, you must isolate the body of interest and draw every external force and moment acting on it. That sketch is the free-body diagram (FBD) — arguably the single most important skill in statics, because almost every wrong answer in an equilibrium problem traces back to a missing, extra, or mis-typed reaction on the FBD, not to a mistake in the algebra that follows. This node follows force-system reduction in the "Force systems and equilibrium foundations" chapter and bridges directly to writing equilibrium equations. The habit it builds — isolate the body, remove each support, replace it with exactly the reaction it can physically exert — is a prerequisite for every equilibrium problem that follows, especially structures and friction later in the course. Get the FBD right and the rest is arithmetic; get it wrong and no amount of careful algebra saves you.

The core insight

A free-body diagram isolates a body and replaces every support or contact it touches with exactly the reaction force or moment that support type is physically capable of exerting — nothing more, nothing less.

The core idea

A free-body diagram is a sketch of a body, completely separated ("freed") from everything else it touches, with every external force and moment drawn explicitly: applied loads, the body's own weight where relevant, and the reaction(s) from every support or contact that was removed. The discipline is mechanical once known: identify the support type, and let the support type — not intuition — tell you what it can and cannot resist.

Reaction rules by support type.

Pin/hinge (2-D): resists translation in any direction but not rotation — two unknown force components ( $A_x$ , $A_y$ ), no moment.
Roller (2-D): resists translation only perpendicular to the rolling surface — one unknown force along the surface normal, no moment, and never a component along the surface.
Fixed/built-in (2-D): resists translation and rotation — two force components and a reaction moment $M_A$ .
Smooth (frictionless) surface contact: a single force, normal to the surface, with no friction component.
Cable or rope: can only pull — a single tension force, directed along the cable, pulling on the body (never pushing).
3-D analogues: a ball-and-socket resists translation in all three directions (three force components, no moment); a 3-D pin/hinge resists all translations but not rotation about its own axis; a fully fixed 3-D support resists everything (three force and three moment components).

Sign convention. When a reaction's true sense is unknown, assume a direction, solve the equilibrium equations, and accept a negative result as meaning the opposite sense. Never guess a direction and then silently flip it mid-calculation — let the algebra tell you.

What belongs on the diagram. Every external force and moment acting on the isolated body: applied loads, weight (if not stated negligible), and support reactions. Internal forces between parts of the same body never appear — unless you deliberately cut through the body, as in method-of-sections problems for trusses and frames, where the cut exposes an internal force as external to the isolated piece. For the body as a whole, equilibrium requires both the force and moment sums to vanish:

\sum \vec{F} = 0, \qquad \sum \vec{M}_P = 0. \qquad (1)

where: $\sum \vec{F}$ is the vector sum of every force on the FBD (applied loads, weight, reactions), and $\sum \vec{M}_P$ is the sum of moments of every force (plus any reaction moments) about any chosen point $P$ . Equation (1) is why the FBD exists: it is the complete list of forces and moments you feed into these two conditions, and a correct FBD is precisely what makes the equations solvable — for 2-D problems, three independent scalar equations ( $\sum F_x = 0$ , $\sum F_y = 0$ , $\sum M = 0$ ) for at most three unknowns.

Figure

A horizontal beam pinned at A and resting on a roller at B with a downward load at midpoint C and the reaction arrows Ax Ay and By drawn at the supports

Two-force and three-force members

A two-force member is loaded at exactly two points, with no load in between. Its FBD carries only two forces, one at each point, and equilibrium forces both to be equal in magnitude, opposite in sense, and collinear along the line joining the two points — the only arrangement for which $\sum \vec{F} = 0$ and $\sum \vec{M} = 0$ hold simultaneously with no other loads present. A three-force member, by contrast, carries three non-parallel forces, and equilibrium requires them to be either concurrent (meeting at one point) or all mutually parallel.

Derivation AWhy a two-force member's forces must be collinear

The goal is to show, starting from nothing but the two equilibrium conditions in equation (1), that a rigid body loaded at exactly two points A and B with no other force or moment can only be in equilibrium if the two forces are equal in magnitude, opposite in direction, and act along the single line joining A and B — the fact every truss-analysis shortcut in the next chapter relies on.

Step 1. Let $\vec{F}_A$ be the resultant external force at point A and $\vec{F}_B$ the resultant at point B; by definition of a two-force member these are the only two forces on the body — no third force, no applied moment anywhere.

Step 2. Force equilibrium, $\sum \vec{F} = 0$ , gives $\vec{F}_A + \vec{F}_B = 0$ , so $\vec{F}_B = -\vec{F}_A$ : the two forces are already equal in magnitude and opposite in direction, though their common direction relative to the body is not yet fixed.

Step 3. Moment equilibrium about A: $\vec{F}_A$ itself contributes zero moment (zero lever arm at A), so $\sum \vec{M}_A = 0$ reduces to $\vec{r}_{AB} \times \vec{F}_B = 0$ , where $\vec{r}_{AB}$ is the position vector from A to B.

Step 4. A cross product vanishes only if the vectors are parallel, so $\vec{F}_B$ must lie along $\vec{r}_{AB}$ — the line joining A and B. With Step 2, $\vec{F}_A$ lies along that same line, equal and opposite to $\vec{F}_B$ : the collinear pair stated above, true for any two-force member shape since the argument never used the member's geometry, only the two load points.

Worked example

A horizontal beam $AB$ , $4\ \text{m}$ long, is pinned at $A$ and rests on a roller at $B$ , both on a horizontal surface. A $600\ \text{N}$ downward force acts at midpoint $C$ , $2\ \text{m}$ from $A$ . Construct the FBD.

Step 1. Isolate the beam by mentally removing the pin at $A$ and the roller at $B$ , leaving only the beam itself as the free body.

Step 2. Reaction at $A$ . A pin resists translation in any direction but not rotation, so it contributes two unknowns: $A_x$ (assumed $+x$ ) and $A_y$ (assumed $+y$ ), no moment.

Step 3. Reaction at $B$ . A roller on a horizontal surface resists translation only perpendicular to that surface, so it contributes one unknown, vertical: $B_y$ (assumed $+y$ ), with no horizontal component — a roller cannot resist sliding along its own surface.

Step 4. Applied load. The $600\ \text{N}$ force acts downward at $C$ , $2\ \text{m}$ from $A$ ; it is drawn exactly as given, with no unknowns.

Step 5. Assemble the complete FBD: $A_x$ and $A_y$ at $A$ , $B_y$ at $B$ , and $600\ \text{N}$ downward at $C$ . $A_x$ will turn out to equal zero once solved (no applied horizontal load exists), but it must still be drawn as an unknown — assuming a reaction is zero before solving is not valid FBD practice. This four-force FBD feeds directly into $\sum F_x = 0$ , $\sum F_y = 0$ , $\sum M = 0$ — three equations, three unknowns, fully solvable.

Competitive stretch

Now suppose the same beam instead has a fixed (built-in) support at $A$ and a cable at $B$ pulling at $30°$ above horizontal with tension $250\ \text{N}$ , and no roller. The FBD changes character: $A$ now contributes $A_x$ , $A_y$ , and a reaction moment $M_A$ (three unknowns, since a fixed support resists rotation too), while $B$ contributes a single $250\ \text{N}$ tension along the cable, pulling away from the beam (never pushing). Total unknowns: $A_x$ , $A_y$ , $M_A$ — exactly three, matching the three 2-D equilibrium equations, so the problem stays solvable. The support type alone fixes both the number of unknowns and whether a moment reaction appears.

Common traps

Giving a roller or smooth surface a force component along the surface — rollers and smooth contacts cannot resist tangential (friction) forces unless friction is explicitly stated in the problem.
Giving a pin or hinge a reaction moment — a pin cannot resist rotation; only a fixed/built-in support carries a reaction moment.
Drawing a cable's reaction pushing on the body, or not aligned along the cable — cables can only pull, along their own length, never push or angle off-axis.
Omitting the body's own weight when it is not stated negligible, or including internal forces on a fully isolated body's FBD — internal forces appear only when a cut is deliberately made through the body, as in method-of-sections problems.

Ask the tutor: if you're unsure which support type applies to a picture in your textbook, or whether a reaction needs a moment component, ask the AI tutor to check your FBD before you move on to the equilibrium equations.

Why doesn't a roller get a horizontal reaction if the surface is horizontal?

A roller's whole purpose is to allow free sliding along the surface it sits on while resisting motion only perpendicular to that surface. On a horizontal surface it can slide horizontally without resistance, so it supplies just a vertical reaction. On an inclined surface the same roller's single reaction would be perpendicular to that incline instead. The direction of a roller's one unknown force always follows the surface it touches, never a fixed compass direction.

If I assume a reaction direction and my answer comes out negative, did I do something wrong?

No — a negative result is a normal, expected outcome. You must assume some direction for an unknown reaction before writing equilibrium equations, since you often cannot tell the true sense by inspection. Solve honestly with that assumption; a negative number simply means the reaction points opposite to what you assumed. Report the correct final direction in your answer, but never go back and flip signs mid-calculation — that is how sign errors creep in.

How do I know when to include the weight of the body on the FBD?

Read the problem statement: if it says weight is negligible, or gives no mass data for that member, leave weight off. If a mass or weight is given, or the problem is clearly about a heavy object, include it as a single downward force at the body's centroid, in addition to every support reaction and applied load. Phrases like "weight of the bracket is negligible" are an instruction, not a detail you're free to ignore either way.