Three-dimensional equilibrium problems

A shaft held by two bearings, a plate slung from three cables, a bracket pulled from several directions — none of these loads lie neatly in one plane, so the flat, three-equation equilibrium you learned for 2-D problems is not enough. Three-dimensional equilibrium keeps exactly the same physics — a rigid body at rest has zero net force and zero net moment — but now both statements are vector equations in space, and each vector equation hides three independent scalar equations. You need the tools of vector algebra: writing a force as a magnitude times a unit vector built from geometry, and computing a moment as a cross product $\vec r \times \vec F$ rather than a single "force times perpendicular distance." This article builds those tools from the 2-D case you already know, derives the general moment expression, and solves a fully three-dimensional cable-and-plate problem end to end.

The core insight

A rigid body in 3-D equilibrium demands six independent scalar equations — $\sum F_x=\sum F_y=\sum F_z=0$ and $\sum M_x=\sum M_y=\sum M_z=0$ — obtained by resolving every force into Cartesian components and every moment via $\vec M_O=\vec r\times\vec F$ .

The core idea

Two vector equations, six scalar equations. For a rigid body in equilibrium, the vector sum of all forces and the vector sum of all moments about any point must both vanish:

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

where: $\vec F$ is any applied or reaction force, $\vec M_O$ is the moment of a force about a chosen point $O$ . Because each of these is a vector equation in three-dimensional space, it supplies three independent scalar conditions, for six in total:

\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

Writing a force from geometry. A force rarely arrives pre-resolved; usually you know only its magnitude and the two points its line of action passes through. If a force of magnitude $F$ acts from point $A$ toward point $B$ ,

\vec F = F\,\hat u_{AB} = F\,\frac{\vec r_{AB}}{|\vec r_{AB}|}

where: $\vec r_{AB}$ is the position vector from $A$ to $B$ and $\hat u_{AB}$ is the unit vector along it — dividing by the length $|\vec r_{AB}|$ is what makes $\hat u_{AB}$ a pure direction with magnitude 1.

Reading the right reactions. Each support removes some of a body's freedom to move and supplies a reaction only along the directions it restrains: a ball-and-socket joint supplies three force components and no moment; a journal bearing supplies force perpendicular to the shaft axis but no axial force and no moment; a thrust bearing adds the axial force a journal bearing lacks; a cable supplies one tension along its own length. Counting reaction unknowns against six available equations tells you whether a support system is statically determinate (unknowns = 6), indeterminate (unknowns > 6), or unstable (unknowns < 6).

Figure

A rigid plate in space held by three vertical cables at three corners with coordinate axes x y and z shown, illustrating that a 3-D equilibrium problem needs all three axes tracked at once

Derivation AExpanding the moment vector into three scalar equations

The goal is to show precisely how the single vector statement $\sum \vec M_O = \vec 0$ turns into three separate scalar equations you can actually solve, by expanding the cross product $\vec r \times \vec F$ into its Cartesian components rather than treating "moment" as a single perpendicular-distance number the way 2-D problems allow.

Step 1. Write the moment of one force $\vec F = F_x\hat i + F_y\hat j + F_z\hat k$ about point $O$ , where its line of action passes through a point located at $\vec r = r_x\hat i + r_y\hat j + r_z\hat k$ relative to $O$ . By definition $\vec M_O = \vec r \times \vec F$ , and the cross product of two vectors is most reliably computed as a determinant:

\vec M_O = \vec r \times \vec F = \begin{vmatrix} \hat i & \hat j & \hat k \\ r_x & r_y & r_z \\ F_x & F_y & F_z \end{vmatrix}

Step 2. Expand the determinant along its top row. Each cofactor multiplies one unit vector by a two-by-two determinant of the remaining components:

\vec M_O = (r_y F_z - r_z F_y)\hat i - (r_x F_z - r_z F_x)\hat j + (r_x F_y - r_y F_x)\hat k

This single vector expansion is exactly why a 3-D moment carries three numbers instead of one: a force offset from $O$ in more than one direction produces turning tendency about more than one axis simultaneously.

Step 3. Read off the three scalar moment components directly from the expansion: $M_x = r_yF_z - r_zF_y$ , $M_y = r_zF_x - r_xF_z$ , and $M_z = r_xF_y - r_yF_x$ . Setting $\sum \vec M_O = \vec 0$ for a rigid body means each of these three sums, taken over every force acting on the body, must independently equal zero — giving the three moment equations that complete the six-equation set alongside $\sum F_x=\sum F_y=\sum F_z=0$ .

Step 4. When a support can only resist rotation about one particular axis (a hinge with a free axis, for instance), you do not need the full vector $\vec M_O$ — only its component along that axis, found with the scalar triple product $M_{\text{axis}} = \hat u\cdot(\vec r\times\vec F)$ , where $\hat u$ is the unit vector along the axis of interest. This is the same $\vec M_O$ from Step 2, simply projected onto one direction. Check: for a force purely along $\hat k$ (i.e. $F_x=F_y=0$ ) acting at $\vec r = r_x\hat i$ , Step 2 gives $\vec M_O = r_xF_z\hat j$ — a moment only about the $y$ -axis, matching the intuitive picture of a vertical force at the end of a horizontal arm twisting about the perpendicular horizontal axis.

Worked example

A uniform rectangular plate of weight $W=600$ N lies in a horizontal plane, held by three vertical cables at corners $A(0,0,0)$ , $B(4,0,0)$ and $C(0,3,0)$ (in meters), with its center of gravity at $G(2,1.5,0)$ . Find the tension in each cable.

Step 1. Let $T_A,\,T_B,\,T_C$ be the (unknown, vertical) cable tensions, and let $W=600$ N act downward through $G$ . All forces act in the $z$ -direction only, so the force equation is one-dimensional even though the geometry is not.

Step 2. Sum forces in $z$ : $\sum F_z=0$ gives $T_A+T_B+T_C-600=0$ , equation (1). This alone has three unknowns, so two moment equations are needed to pin them down.

Step 3. Sum moments about the $x$ -axis through $A$ (the edge $AB$ ), which balances $T_C$ (acting at $y=3$ ) against $W$ (acting at $y=1.5$ ): $\sum M_x=0$ gives $T_C(3)-600(1.5)=0$ , so $T_C=300$ N.

Step 4. Sum moments about the $y$ -axis through $A$ (the edge $AC$ ), which balances $T_B$ (at $x=4$ ) against $W$ (at $x=2$ ): $\sum M_y=0$ gives $-T_B(4)+600(2)=0$ , so $T_B=300$ N. Substituting both into equation (1): $T_A=600-300-300=0$ N.

Final answer: $T_A=0$ N, $T_B=300$ N, $T_C=300$ N. Geometrically this makes sense — $G(2,1.5,0)$ sits exactly on the diagonal from $A$ to the opposite corner, so cable $A$ carries no load at all.

Competitive stretch

The same six-equation machinery handles supports that are not simple cables. Take a horizontal shaft along the $x$ -axis with a journal bearing at $A$ (resists $y,z$ forces only) and a thrust bearing at $B$ , $1.2$ m away (resists $x,y,z$ ). A pulley at the midpoint applies $200$ N in $-y$ and $150$ N in $-z$ . Because no torque acts about the shaft's own axis, $\sum M_x=0$ is automatically satisfied, leaving five unknowns and five usable equations. Moments about $A$ give $B_y=100$ N and $B_z=75$ N (each pulley load splits according to the $0.6\ \text{m}:1.2\ \text{m}$ lever ratio), and back-substitution into the force equations gives $A_y=100$ N, $A_z=75$ N, $B_x=0$ . The load splits evenly between the two bearings because the pulley sits at the shaft's exact midpoint — the same "balance moments first, then back-substitute forces" strategy as the plate problem, just with different reaction directions.

Common traps

Using only three equations out of 2-D habit. In 3-D always check all six; skipping one because "it looks like zero" without actually verifying is the most common source of a missed unknown.
Forgetting a bearing does not resist its own axis. A journal bearing gives no axial force and no moment reaction; only a thrust bearing adds the axial component.
Computing $\vec r\times\vec F$ from the wrong point. The position vector must run from the moment center to any point on the force's line of action — recheck $\vec r$ every time you change reference point.
Forgetting to normalize $\vec r_{AB}$ . Skipping the division by $|\vec r_{AB}|$ when building $\hat u_{AB}$ silently scales every component of the force you write down.

Ask the tutor: if setting up the determinant for a 3-D moment feels error-prone, ask the AI tutor to walk through $\vec r\times\vec F$ with your own numbers, or to check your reaction-count for a support combination you're unsure about.

Why do I need a cross product instead of just multiplying force by perpendicular distance like in 2-D

In 2-D, every moment is about the single axis perpendicular to the page, so "force times perpendicular distance" is a complete answer with a sign for direction. In 3-D the moment of one force can have components about all three axes simultaneously, because the position vector and the force can point in different, non-coplanar directions. The cross product $\vec r\times\vec F$ automatically produces all three components correctly, including the sign of each, which is why the determinant method replaces the simple 2-D shortcut once you leave the plane.

How do I know which reaction components to include for a given support

Match the support to what it physically restrains. A ball-and-socket joint permits rotation about any axis but no translation, so it contributes three unknown force components and zero moment. A journal bearing permits sliding along and rotation about the shaft axis, so it contributes force perpendicular to that axis only. A thrust bearing additionally blocks axial sliding, adding one more force component. A smooth surface can only push, contributing one force component along its normal. Always ask "which motions does this support prevent" — each prevented translation gives one unknown force component, and each prevented rotation gives one unknown moment component.

Do I always need all six equations, or can some problems use fewer

You need as many independent equations as there are unknowns, up to six. Some problems genuinely have fewer effective equations — for instance, a shaft with a journal bearing at one end and no applied torque about its own axis automatically satisfies $\sum M_x=0$ , leaving five real unknowns and five usable equations, as in the bearing example above. The rule is to count unknown reaction components first, then confirm you have that many independent equations before solving; do not assume all six are needed, but do not skip any without checking.

Three-dimensional equilibrium problems

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads