Statics · Module 11 of 11 · Advanced
Virtual Work
Equilibrium can also be checked through imagined motion and work balance.
Readiness check
From the whole course so far. Tick only what you can do closed-notes.
- Solve rigid-body equilibrium with forces and moments confidently.
- Compute work: W = F·d for a force along a displacement, W = M·θ for a moment.
- Relate arc length to angle: s = rθ for small rotations.
- Differentiate simple functions (for the potential-energy sections).
- Explain why reaction forces at a fixed pin do no work.
The core idea
Imagine a tiny allowed motion. In equilibrium, total work done is zero.
δU = 0One virtual displacement δθ moves the whole connected mechanism; only active forces (loads, springs, applied efforts) do work. Pins, rollers, and other constraint reactions drop out for free; that is the method's superpower.
The method
What single coordinate describes the mechanism's motion?
Write each force's point position as a function of it.
Mark the virtual displacement of every active force.
δU = 0; divide out δθ; solve for the unknown.
Worked example: a lever, solved without reactions
A rigid lever pivots at O. A 200 N load W hangs 0.2 m from the pivot; an effort P pushes down on the other side, 0.8 m from the pivot. Find P for equilibrium using virtual work.
- ProblemFind the effort P that balances the lever in Figure 1.
- Given / findW = 200 N at a = 0.2 m; effort arm b = 0.8 m. Find P.
- AssumptionsRigid lever, frictionless pivot, small virtual rotation, both forces stay vertical.
- ModelFigure 2: give the lever a virtual rotation δθ about O. The load end rises by a·δθ while the effort end descends by b·δθ. The pivot reaction does no work.
- EquationsδU = P(b·δθ) − W(a·δθ) = 0
- SolveDivide by δθ (it is arbitrary): P·b = W·a, so P = 200 × 0.2/0.8 = 50 N.
- CheckClassic moment balance about O gives P(0.8) = 200(0.2), so P = 50 N. Same answer, but virtual work never asked about the pivot reaction. Mechanical advantage 4:1 matches the arm ratio.
- ConclusionFor one lever, both methods cost the same. For a six-bar jack or a toggle press, δU = 0 gives the one force you want in a single line. That is why mechanism designers reach for it.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Virtual displacement violates a constraint | Reactions appear in δU; equation unsolvable or wrong | "Could the real mechanism actually move this way?" | The virtual motion must respect every pin, slot, and contact. Parameterize with the mechanism's own degree of freedom. |
| Sign errors in work terms | P comes out negative or the load "helps" itself | "Does this force point along or against its point's motion?" | Work is +F·δs when force and displacement align, −F·δs when opposed. Track each point's direction of travel. |
| Including forces that do no work | Cluttered equations with reaction terms | "Does this force's application point move? Perpendicular?" | Fixed pins: no motion, no work. Rollers: motion perpendicular to reaction, no work. Only active forces enter δU. |
| Mixing up the geometry derivative | Displacement relations like δy = δθ instead of δy = b·δθ | "What is each point's position as a function of my coordinate?" | Write positions x(θ), y(θ) first; virtual displacements are their derivatives times δθ. |
Practice ladder
A crowbar: effort arm 0.75 m, load arm 0.05 m. What effort lifts a 900 N load? Solve with δU = 0.
Show answer
P(0.75 δθ) = 900(0.05 δθ), so P = 60 N. Mechanical advantage 15:1.
For the worked-example lever, suppose the effort P is applied at 30° from vertical instead of straight down. What P is needed now?
Show answer
Only the component along the end's motion (vertical) works: P cos 30° × 0.8 δθ = 200 × 0.2 δθ, so P = 57.7 N. Angled effort is wasted effort.
A scissor jack raises a car: by geometry, when the handle screw advances δs, the platform rises δh = δs/8 at the current position. What handle force F holds a 4000 N corner load? Why is this almost independent of friction for holding but not for lifting?
Show answer
F·δs = 4000·δh = 4000·δs/8, so F = 500 N (ideal). Lifting must also overcome friction work (real F > 500 N); holding is helped by friction: screw jacks are self-locking, so the holding force can be near zero.
Pick a real mechanism (bottle jack, bolt cutter, hand brake, gym machine). Identify its single motion coordinate, write the displacement ratio between input and output points, and compute its ideal mechanical advantage with δU = 0. Compare with the manufacturer's claim if available.
What good work looks like
A kinematic sketch, the input-output displacement ratio measured or derived, the δU = 0 line, the resulting force ratio, and one paragraph on where friction makes reality worse than ideal.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Capstone for Statics: a complete analysis of one real mechanism (your Level 4 pick). Combine the whole course: FBD and reactions of one member (Module 5), the internal force in one link (Modules 6 and 7), and the input-output force ratio via virtual work. Two pages, professional format.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. State the principle of virtual work for a rigid body.
A body (or connected system) is in equilibrium if and only if the total virtual work of all active forces vanishes (δU = 0) for every virtual displacement compatible with the constraints.
2. Which forces never appear in δU, and why?
Constraint reactions at fixed points (no displacement) and forces perpendicular to their point's motion (rollers, normals): their work is identically zero.
3. When is virtual work clearly better than force and moment equilibrium?
Multi-member linked mechanisms where you want one input-output force relation; one δU = 0 equation replaces dismembering every link.
4. What is the potential-energy criterion for equilibrium?
dV/dq = 0 at equilibrium: the potential energy is stationary with respect to the position coordinate.
5. How does the second derivative of V classify stability?
d²V/dq² > 0: stable (energy valley). Less than 0: unstable (energy hill). Equal to 0: neutral; investigate higher terms.
Textbook mapping
| Item | Mapping |
|---|---|
| Main textbook | R.C. Hibbeler, Engineering Mechanics: Statics, Chapter 11, Virtual Work |
| Core sections | 11.1 Definition of Work · 11.2 Principle of Virtual Work · 11.3 Principle of Virtual Work for a System of Connected Rigid Bodies |
| Recommended problems | Fundamental Problems F11-1 onward (partial solutions in the back). Mechanism problems with one degree of freedom are the exam staple. |
| Skip on first pass | 11.4 to 11.7 (conservative forces, potential energy, stability); preview the stability criterion conceptually now; it returns in Dynamics and Machine Design. |
| Read next | You finished the book. Next course: Mechanics of Materials: your N, V, M values become stresses. |