Equilibrium of a Rigid Body
When a rigid body neither accelerates nor rotates, the forces and moments acting on it must cancel completely. This chapter turns that idea into a working method: isolate the body, account for every force and reaction, and solve for the unknowns.
Learning objectives
- Develop the equations of equilibrium for a rigid body in two and three dimensions.
- Introduce the free-body diagram as the starting point of every analysis.
- Solve rigid-body equilibrium problems for support reactions and member forces.
- Classify supports by the reactions they produce and judge whether a body is properly constrained.
This chapter is the backbone of the whole statics course. If you can reliably draw a correct free-body diagram and pick a smart moment point, almost every later topic (trusses, frames, internal loads) becomes routine. Don't rush sections 5.1–5.3 — slow mastery here pays off everywhere later.
Conditions for Rigid-Body Equilibrium
Every external force and couple moment acting on a body can be reduced to a single resultant force and a single resultant couple moment at any chosen point. A body is in equilibrium precisely when both of these resultants vanish. Internal forces between particles never appear here: by Newton's third law they occur in equal, opposite, collinear pairs and cancel.
Rigid-Body Equilibrium
A rigid body is in equilibrium when the resultant of all external forces is zero and the resultant couple moment about any point is also zero. These two conditions are both necessary and sufficient.
The second condition holds about any point, not just the chosen one. Summing moments about a different point \(A\) gives \(\sum \mathbf{M}_A = \mathbf{r}\times\mathbf{F}_R + (\mathbf{M}_R)_O\); if the resultant force and resultant couple are already zero, this is automatically satisfied. That freedom to pick the moment point is the single most useful tactic in the chapter.
The Rigidity Assumption
Real bodies deform under load, but engineering materials such as steel and concrete deform so little that treating them as rigid introduces negligible error. Keeping the geometry fixed means force directions and moment arms stay the same before and after loading.
Free-Body Diagrams
The free-body diagram (FBD) is the single most important step in any equilibrium problem. It is a sketch of the body isolated from its surroundings, showing every external force and couple moment that acts on it. Get the FBD right and the algebra follows; get it wrong and no amount of calculation recovers.
Free-Body Diagram
A diagram of a body drawn "free" of its supports, with all external forces and couple moments shown — applied loads, support reactions, and weight. Internal forces are omitted because they cancel in pairs.
Support reactions
Two simple rules generate every reaction. A support that prevents translation in a direction exerts a force in the opposite direction; a support that prevents rotation exerts a couple moment opposing that rotation. The more motion a support blocks, the more reaction components it produces.
Reaction Rule
Each motion a support blocks corresponds to one reaction. Block translation → get a force. Block rotation → get a couple moment. A roller blocks one direction (1 unknown); a pin blocks two (2 unknowns); a fixed support blocks translation and rotation (3 unknowns in 2D).
| Support type | Reaction produced | Unknowns |
|---|---|---|
| Cable | Tension directed along the cable, away from the body | 1 |
| Smooth surface / roller / rocker | Force perpendicular to the contact surface | 1 |
| Pin in a smooth slot | Force perpendicular to the slot | 1 |
| Smooth pin or hinge | Two force components (or magnitude & direction) | 2 |
| Member fixed to a collar on a smooth rod | One force ⟂ rod and a couple moment | 2 |
| Fixed support | Two force components and a couple moment | 3 |
Internal Forces, Weight & Idealized Models
Internal forces never appear on the FBD of a whole body — they cancel in collinear pairs. Weight is an external force acting at the center of gravity \(G\) (at the geometric centroid for uniform bodies). Before analysis, the real object is replaced by an idealized model: real connections become pins, rollers, or fixed supports, and the material is treated as rigid.
Never solve an equilibrium problem without first drawing the FBD. Removing a support that blocks translation introduces a force there; removing one that blocks rotation introduces a couple moment. Couple moments are free vectors and may be placed anywhere; forces are sliding vectors and may be placed anywhere along their line of action.
Equations of Equilibrium
For a coplanar (2D) force system, the vector conditions collapse to three independent scalar equations. These solve for at most three unknowns per body.
Alternative Equation Sets
Two moment equations may replace force equations. The set \(\sum F_x=0,\ \sum M_A=0,\ \sum M_B=0\) works provided line \(AB\) is not perpendicular to the \(x\) axis. The all-moment set \(\sum M_A=0,\ \sum M_B=0,\ \sum M_C=0\) works provided \(A\), \(B\), \(C\) are not collinear.
Draw the FBD with axes, all loads, and dimensions. Then sum moments about a point where the lines of action of two unknowns intersect — this isolates the third unknown directly. Orient the \(x\)–\(y\) axes to simplify force resolution. A negative result simply means the assumed sense was reversed.
Exam tip: choosing the right moment point is what separates a five-line solution from a page of simultaneous equations. Always look for the point where two unknown forces cross — sum moments there first, and you'll often get one unknown immediately with no system to solve.
Two- and Three-Force Members
Recognizing how many points a force acts at can shortcut an entire analysis. Two special cases recur constantly.
If a member carries forces at only two points (and no couple), the two forces must be equal in magnitude, opposite in direction, and collinear along the line joining the two points. Force equilibrium forces them equal and opposite; moment equilibrium forces them onto the same line of action. A pin-ended link with negligible weight is the classic example.
If a member carries forces at only three points, moment equilibrium requires the three forces to be either concurrent (lines of action meet at a common point) or parallel (the concurrency point is at infinity). Locating that meeting point often fixes an unknown force direction without any algebra.
Free-Body Diagrams (3D)
The procedure mirrors the 2D case: isolate the body, then show all forces and couple moments referenced to an \(x, y, z\) coordinate system. The catalog of supports simply grows richer because there are now three translations and three rotations a support might block.
3D Support Reactions
Each blocked translation produces a force component; each blocked rotation produces a couple-moment component. A ball-and-socket joint blocks all three translations (3 force components, no moments). Bearings, hinges, and pins block additional rotations and so add moment components.
| Connection | Reaction | Unknowns |
|---|---|---|
| Cable | One force along the cable | 1 |
| Smooth surface / roller | One force ⟂ surface | 1 |
| Ball-and-socket joint | Three force components | 3 |
| Single journal bearing / hinge / pin | Force and couple-moment components ⟂ the axis | 4–5 |
| Single thrust bearing | Three forces and two couple moments | 5 |
| Fixed support | Three forces and three couple moments | 6 |
Redundant Moment Reactions
The couple-moment reactions at a single bearing, pin, or hinge are usually not drawn when the body is also held by other properly aligned supports. Those supports together prevent rotation, so the moment components become redundant and the force reactions alone suffice.
The hardest part of 3D problems isn't the math — it's deciding which moment reactions to keep. My rule of thumb: if two or more aligned bearings/hinges share an axis, drop the couple moments at each and treat them as force-only supports. Re-read Example 9 (5.19) until this clicks; it's the cleanest illustration in the chapter.
Equations of Equilibrium (3D)
In three dimensions the two vector conditions expand into six independent scalar equations — three for force, three for moment — solving for up to six unknowns.
Scalar vs. Vector Analysis
When the force and moment components are easy to read off the geometry, apply the six scalar equations directly. When forces point at awkward angles, express each as a Cartesian vector and use \(\mathbf{M}_O = \sum (\mathbf{r}\times\mathbf{F})\). Choosing a moment axis that intersects many unknowns zeroes their moments and isolates the rest.
Constraints and Statical Determinacy
Satisfying the equilibrium equations is not enough — the supports must also hold the body in a stable, well-posed way. Three failure modes matter.
Statically Indeterminate (Redundant Constraints)
When a body has more support reactions than independent equilibrium equations, it is statically indeterminate: there are more unknowns than equations. The extra equations come from deformation conditions, studied in mechanics of materials.
Improper Constraints
Even with the right count of reactions, a body is improperly constrained — and unstable — if all reactive forces are concurrent (meet at a point), pass through a common axis (in 3D), or are all parallel. In each case an applied load produces a net moment or net force the supports cannot resist.
Partial Constraints
When a body has fewer reactions than necessary equilibrium equations, it is only partially constrained and cannot maintain equilibrium under general loading.
A body is statically determinate when supported by the minimum reactions needed for equilibrium, and indeterminate when it has more. For a stable, proper constraint the lines of action of the reactive forces must not all be parallel and must not all intersect a common point or axis.
Before you ever write an equation, count: number of reactions vs. number of equilibrium equations (3 in 2D, 6 in 3D). Fewer ⇒ partial (unstable). Equal ⇒ likely determinate, but still check the reactions aren't all parallel or concurrent. More ⇒ indeterminate, and statics alone won't solve it. This 30-second check saves enormous wasted effort.
Physical meaning
Equilibrium means no translational or angular acceleration in the stated reference frame. Zero net force alone can still permit rotation, and zero net moment alone can still permit translation.
Problem-solving strategy
- Isolate one purposeful body and replace every support with only the reactions it can transmit.
- Place weight, applied forces, and couples at their actual lines of action.
- Count unknowns against independent equilibrium equations and inspect support geometry for stability.
- Take a moment about a point or axis that eliminates several unknown reactions.
- Use remaining force equations, then audit the unused equilibrium relation.
Recognition strategy
This is a planar rigid body with three reaction components. Taking moments about the pin removes both pin-force unknowns immediately.
Given information
- Span \(AB=5.00\) m, load 4.00 kN downward at \(x=2.00\) m, clockwise couple 3.00 kN·m.
Model and interpretation
The pin supplies \(A_x,A_y\); the level roller supplies \(B_y\). Positive force is right/up and positive moment is counterclockwise.
Verification
Units: Each moment term is kN times m or a stated kN·m couple, so the moment equation is dimensionally consistent.
Signs and direction: Both solved vertical reactions are positive, agreeing with the assumed upward arrows; no horizontal external load requires \(A_x=0\).
Plausibility: Reactions total 4.00 kN and the clockwise couple makes the right support carry more than it would under the load alone.
Engineering interpretation
The couple shifts vertical reaction toward B without changing total vertical load. Anchor and bearing selection must consider this reaction plus credible installation and service load combinations.
Recognition strategy
The cable has one unknown magnitude but known direction. Isolating the boom and taking moments at A avoids solving pin components before tension.
Given information
- Cable force at B is \(\mathbf T=T(-3\mathbf i+4\mathbf j)/5\); all loads are in kN.
Model and interpretation
The boom is rigid and weightless for this exercise. The pin allows two force components but no reaction couple.
Verification
Units: Lever arms in metres multiply kN components to kN·m, while both force equations retain kN consistently.
Signs and direction: The cable pulls up-left, so the pin reaction is rightward and upward; both positive solved values match the assumed arrows.
Plausibility: The cable vertical component is 2.33 kN, less than the 3.00 kN load because A supplies the remaining 0.667 kN upward.
Engineering interpretation
Cable angle controls both support thrust and tension; a flatter cable would increase tension sharply. Actual boom design must also check cable capacity, pin bearing, buckling, and dynamic handling loads.
Recognition strategy
A fixed 3D support can transmit three force and three moment components. Reduce the external loading at O, then apply vector equilibrium.
Given information
- Axes are right-handed; reaction vectors are initially assumed positive in their named component directions.
Model and interpretation
The bracket is rigid and the tool load is a point force. The supplied couple is a free vector and therefore enters the moment equation unchanged.
Verification
Units: The cross product uses metres times newtons and therefore gives N·m, the same unit as the applied couple.
Signs and direction: The force at positive x acting downward creates a negative k moment; the fixed support reaction must be positive k to cancel the net external moment.
Plausibility: The 120 N·m applied couple partly offsets the 320 N·m force moment, leaving the smaller 200 N·m support moment.
Engineering interpretation
The support must resist a combined shear and torsional demand, not merely the tool force. A production bracket requires fastener-group, weld, fatigue, stiffness, and code-based safety review.
Chapter recap
- Equilibrium of a planar rigid body requires force balance in two directions and moment balance; 3D requires three of each.
- Before moving on, verify that you can draw a support-correct FBD, choose a strategic moment center, interpret reaction signs, and distinguish determinacy from stability.