Equilibrium equations

A beam bolted to a wall, a crane boom held by cables, a bracket carrying a shelf — none of these move, yet forces pull on them from every direction. Statics answers a simple question about all of them: what must the hidden reactions (pin forces, cable tensions, roller pushes) be so that nothing accelerates and nothing rotates? The answer is the equilibrium equations — the net force is zero and the net moment about any point is zero. In two dimensions this collapses to three scalar equations you can solve by hand; in three dimensions it becomes six. This article builds both forms from the free-body diagram you already know how to draw, shows how a smart choice of moment point turns algebra into arithmetic, and works a plane beam and a 3-D bracket problem end to end.

The core insight

A rigid body is in static equilibrium exactly when its net force and net moment about any point both vanish, giving three independent scalar equations in 2-D ( $\sum F_x=0$ , $\sum F_y=0$ , $\sum M_O=0$ ) and six in 3-D.

The core idea

Equilibrium is a vector statement, not a formula to memorise piece by piece: a rigid body at rest, or moving at constant velocity, has

\sum \vec F = 0 \qquad \sum \vec M_O = 0

where: $\vec F$ ranges over every applied load and support reaction on the free-body diagram, and $\vec M_O$ is the moment of each force about a chosen point $O$ . Both conditions must hold together — zero net force alone permits the body to spin, and zero net moment alone permits it to translate.

In 2-D, the vector statement splits into three independent scalar equations:

\sum F_x = 0, \qquad \sum F_y = 0, \qquad \sum M_O = 0

where $O$ can be any point in the plane, on the body or off it. Placing $O$ at an unknown reaction's line of action makes that reaction's moment arm zero, so it drops out of the moment equation entirely — this is the single biggest algebra-saving move in the whole topic. Because there are exactly three independent equations, a 2-D problem is statically determinate only if it has exactly three unknown reaction components: fewer leaves the body under-constrained (a mechanism, not a structure), more makes it statically indeterminate, which needs deformation equations beyond this course.

In 3-D, the same two vector conditions expand into six scalar equations:

\sum F_x=0,\ \ \sum F_y=0,\ \ \sum F_z=0, \qquad \sum M_x=0,\ \ \sum M_y=0,\ \ \sum M_z=0

with the moment sum usually built from $\sum(\vec r \times \vec F) = 0$ , summing $\vec r \times \vec F$ for every force about the same point $O$ and reading off its three components. Two shortcuts save enormous work once recognised. A two-force member — loaded at exactly two points — is in equilibrium only if those two forces are equal, opposite, and collinear along the line joining the points; it carries pure tension or compression, nothing else. A three-force member — loaded at exactly three points, no couples — is in equilibrium only if the three forces are either all concurrent at one point or all mutually parallel; knowing two lines of action pins down the third. Particle equilibrium is the special case of a body with no size: only $\sum \vec F = 0$ applies (no independent moment equation), since every force on a particle passes through the same point.

Figure

A horizontal beam pinned at the left end and resting on a roller at the right end, with two downward loads and the pin reaction shown as horizontal and vertical arrows and the roller reaction shown as a single vertical arrow

Derivation ASolving a beam's support reactions from the three 2-D equilibrium equations

The goal is to show how the abstract conditions $\sum F_x=0$ , $\sum F_y=0$ , $\sum M_O=0$ turn into an actual number for every unknown reaction on a simply supported beam, and why the order in which you apply them — moment first, then the two force sums — is a deliberate strategy, not an accident, because it isolates one unknown per equation instead of leaving two unknowns tangled together in a single line.

Step 1. Draw the free-body diagram and name the unknowns. A horizontal beam $AB$ , $5\ \text{m}$ long, is pinned at $A$ (which resists both horizontal and vertical motion, giving unknowns $A_x$ and $A_y$ ) and rests on a roller at $B$ (which resists only vertical motion, giving unknown $B_y$ ). Three unknowns, three equations — the problem is statically determinate.

Step 2. Apply $\sum F_x = 0$ first, since it is the simplest. No applied load has a horizontal component here, so $A_x = 0$ immediately — one unknown solved with almost no work.

Step 3. Apply $\sum M_A = 0$ next, choosing $A$ as the moment centre specifically because both $A_x$ and $A_y$ pass through it and vanish from this equation, leaving only $B_y$ . With the two downward loads of $800\ \text{N}$ at $3\ \text{m}$ from $A$ and $400\ \text{N}$ at $B$ (the full $5\ \text{m}$ from $A$ ), and taking counter-clockwise as positive,

B_y(5) - 800(3) - 400(5) = 0 \ \Rightarrow \ 5B_y = 4400 \ \Rightarrow \ B_y = 880\ \text{N}

Step 4. With $B_y$ known, apply $\sum F_y = 0$ to find the last unknown:

A_y + B_y - 800 - 400 = 0 \ \Rightarrow \ A_y = 1200 - 880 = 320\ \text{N}

Step 5. Check the full solution by re-summing vertical forces: $320 + 880 - 800 - 400 = 0$ ✓, confirming $A_x = 0$ , $A_y = 320\ \text{N}$ upward, $B_y = 880\ \text{N}$ upward is consistent with all three equilibrium equations at once.

Worked example

A rigid bracket in 3-D is held by a ball-and-socket joint at $A = (0,0,0)$ and two cables. Cable 1 exerts $\vec T_1 = -300\hat j + 400\hat k\ \text{N}$ at point $B$ ; cable 2 exerts $\vec T_2 = -500\hat j\ \text{N}$ at point $C$ ; a $200\ \text{N}$ weight acts as $-200\hat j$ at point $D$ . Find the reaction at $A$ .

Step 1. Identify what the ball-and-socket can resist. Unlike a pin, a ball-and-socket joint resists translation in all three directions but transmits no moment, so its reaction is a pure force $\vec R_A = A_x\hat i + A_y\hat j + A_z\hat k$ with no unknown moment component to solve for.

Step 2. Apply $\sum \vec F = 0$ , summing the reaction with every applied force:

\vec R_A + \vec T_1 + \vec T_2 - 200\hat j = 0

Step 3. Collect components one axis at a time. In $\hat i$ : $A_x = 0$ , since neither cable nor the weight has a horizontal component. In $\hat j$ : $A_y - 300 - 500 - 200 = 0 \Rightarrow A_y = 1000\ \text{N}$ . In $\hat k$ : $A_z + 400 = 0 \Rightarrow A_z = -400\ \text{N}$ .

Step 4. Assemble and size the reaction. $\vec R_A = 1000\hat j - 400\hat k\ \text{N}$ , with magnitude $\sqrt{1000^2 + 400^2} \approx 1077\ \text{N}$ . A full check would also verify $\sum \vec M_A = 0$ using the position vectors to $B$ , $C$ , $D$ , but since the ball-and-socket contributes no moment unknown, force balance alone was sufficient here to pin down $\vec R_A$ . Answer: $\vec R_A \approx 1077\ \text{N}$ , directed $1000\hat j - 400\hat k\ \text{N}$ .

Common traps

Choosing the moment-reference point carelessly — pick $O$ where the greatest number of unknown reactions act, so their moment arms vanish and only one unknown survives in that equation.
Writing more than three independent equations in 2-D (or six in 3-D) and expecting new information — any further equation is either redundant or, if it disagrees, signals an arithmetic slip.
Assuming a three-force member's forces are concurrent without checking the conditions — concurrency is a consequence of equilibrium for exactly three forces and no couples, and it fails the moment a couple or a fourth force appears.
Forgetting that a determinate 2-D problem needs exactly three unknown reaction components — fewer means an unstable mechanism that cannot be solved by statics at all; more means a statically indeterminate problem outside this course's scope.

Ask the tutor: If picking which equilibrium equation to write first, or where to put the moment point, feels like guesswork on a new reaction-finding problem, ask the AI tutor to walk through the strategy with you before diving into the algebra.

Why does the choice of moment point not affect the final answer, even though different points give different-looking equations?

Because $\sum \vec M_O = 0$ is true about every point simultaneously once a rigid body is in equilibrium — it is one physical condition, not a family of different conditions. Different choices of $O$ just repackage the same underlying balance into equations with different unknowns cancelled out, so any valid choice, followed correctly, must land on the same reaction values. The only thing that changes is how much algebra you do to get there, which is exactly why choosing $O$ at an unknown reaction is worth the moment of thought it takes.

If a 2-D problem has four unknown reactions instead of three, is it always unsolvable?

Not unsolvable in reality — the beam still sits there in equilibrium — but unsolvable by the three 2-D equilibrium equations alone. With four unknowns and only three independent equations, the system is statically indeterminate: you need additional equations from how the material deforms (beyond rigid-body statics) to pin down every reaction. What you can still extract from statics alone are combinations of the unknowns, or you can check whether the problem was actually meant to model something simpler, like a two-force member, where a geometric shortcut removes an unknown.

In the 3-D bracket example, why did the ball-and-socket reaction need no moment equation at all?

Because a ball-and-socket joint is a frictionless pivot: it can push or pull the bracket in any direction but it cannot twist it, so it contributes zero unknown moment components by its very nature. That leaves $\sum \vec F = 0$ , three scalar equations, to solve for the three unknown force components $A_x$ , $A_y$ , $A_z$ directly — no need to even set up $\sum \vec M_A = 0$ for this particular question. Had the joint been a rigid weld instead, it would resist moments too, adding three more unknowns that would require the moment equation to solve.

Equilibrium equations

The core idea

Worked example

Common traps

Before this

Where this leads