Virtual work & equilibrium

1. Use geometry to avoid solving every reaction

Consider a bead at rest on a smooth curved wire. Newton's equilibrium equation contains gravity, any applied force, and the unknown wire reaction. The reaction points normally to the wire. If we project the equation along the allowed tangent displacement, its contribution vanishes. We can find equilibrium without first calculating that reaction.

Virtual work generalizes this idea to many particles and arbitrary independent coordinates. It does not say the reactions are absent. It chooses a projection in which ideal reactions contribute zero.

Prerequisites: dot products and work from P1–P2, and virtual displacement from Chapter 4.

2. Virtual work and generalized forces

Definition. For forces $\mathbf F_a$ acting at particle positions $\mathbf r_a$, their virtual work in an admissible variation is $\delta W=\sum_a\mathbf F_a\cdot\delta\mathbf r_a$.

Virtual work is an algebraic first-order quantity with dimensions of energy. It need not be actual energy transferred, because the displacement is a fixed-time comparison rather than a physical motion. The symbol $\delta W$ is also not a claim that work is a state function.

For smooth holonomic transformations, the constraints are already incorporated in the independent coordinates. Virtual work can then be expressed entirely in terms of their variations.

Insert the displacement formula from Chapter 4 into the force sum:

The sums are finite, so we may interchange their order. The same $\delta q_i$ can move several particles. Grouping all their contributions gives

Define the generalized force conjugate to $q_i$ by

$Q_i$ measures how strongly the applied forces favour increasing $q_i$, with the geometry of all moving particles included. Since $Q_i\delta q_i$ is energy, $[Q_i]=\text{energy}/[q_i]$. A length coordinate has a force-like $Q_i$; an angle coordinate has a torque-like $Q_i$. A generalized force is not always measured in newtons.

3. Ideal constraints

An ideal constraint is one whose reactions have zero total virtual work for all admissible virtual displacements:

For a single smooth surface, the reaction is normal and the virtual displacement tangent, so each dot product vanishes. For connected systems, individual reaction works need not vanish; their sum may cancel. A massless inextensible string with an ideal pulley provides a useful example below.

The definition concerns total virtual work, not necessarily actual work and not necessarily the work of each reaction separately. A moving support can supply actual energy, as Chapter 4 demonstrated.

Smoothness is a physical assumption: no tangential friction from the surface. When sliding friction is present, include it among the forces contributing to virtual work. Do not call it ideal and then discard it. Ideal no-slip rolling on a stationary rigid surface can also have zero reaction virtual work when the complete body's admissible displacements are treated consistently; this is different from a sliding-friction model.

4. Principle of virtual work

An equilibrium condition that eliminates ideal reactions follows directly from force balance. Consider static equilibrium in an inertial frame, ideal bilateral constraints, and independent regular generalized coordinates. Applied forces include any non-ideal forces that cannot be eliminated. A moving constraint generally requires a dynamical treatment; here the ordinary equilibrium interpretation assumes stationary geometry.

At equilibrium each particle has zero acceleration:

Dot each force balance with its admissible virtual displacement and add:

The second sum is zero. Hence the applied-force virtual work obeys

This is the principle of virtual work. The word every matters. Finding one displacement perpendicular to a nonzero force proves nothing about equilibrium in other allowed directions.

Since $\delta W=\sum_iQ_i\delta q_i$, choose one variation nonzero and all the others zero. The only way the sum vanishes for every independent choice is

This states that there is no applied-force component along any allowed configuration direction. Under the stated regular bilateral ideal-constraint model, remaining normal components can be balanced by reactions. Thus it characterizes equilibrium provided those reactions are physically available. With unilateral contact, signs and contact conditions must also be checked; equality conditions alone are not sufficient.

Virtual work does not require every applied force to be zero. It requires their total tendency along every permissible direction to cancel. For an unconstrained particle all displacement directions are admissible; the condition then reduces to ordinary zero resultant force.

5. Conservative forces and potential energy

If applied forces are conservative, let $V(q)$ be the potential energy after substitution of the coordinate transformations. At fixed time, a first-order change is

Conservative-force virtual work is $\delta W=-\delta V$. Comparing coefficients with $\delta W=\sum_iQ_i\delta q_i$ gives

Consequently, equilibrium requires $\partial V/\partial q_i=0$. This is a stationary-potential condition, not automatically a minimum. A pendulum balanced upright is also an equilibrium, but a small perturbation moves it away. For a single coordinate, a strict local minimum usually gives stable equilibrium under the ordinary positive-kinetic-energy model; a maximum gives instability. If the second derivative is zero, further terms must be examined. Detailed stability analysis returns in Unit 4.

6. Applications

Example 1 · Equilibrium of a bead on an incline

A bead of mass $m$ rests on a smooth plane inclined at angle $\alpha$ to the horizontal. An applied force $P$ acts up the slope. Choose a coordinate $q$ increasing up the slope to determine the force required for equilibrium.

The virtual displacement is along the slope. The normal reaction does zero virtual work. The gravitational component along increasing $q$ is $-mg\sin\alpha$. Therefore $\delta W=(P-mg\sin\alpha)\delta q$. Since $\delta q$ can have either sign, equilibrium requires $P=mg\sin\alpha$.

For $m=2.0\,\mathrm{kg}$, $g=9.8\,\mathrm{m\,s^{-2}}$, and $\alpha=30^\circ$, $P=9.8\,\mathrm N$. At $\alpha=0$ the necessary holding force is zero, as expected for a smooth horizontal surface.

Example 2 · Virtual work of forces on a lever

A massless rigid lever is hinged at a fixed point and is horizontal in the configuration considered. Downward forces $F_1$ and $F_2$ act at distances $a$ to the left and $b$ to the right. A small rotation about the hinge relates the displacements of their points of application.

Choose a small positive counterclockwise rotation $\delta\theta$. The left endpoint has vertical displacement $-a\delta\theta$ and the right endpoint $b\delta\theta$. Thus the downward forces do virtual work

The fixed hinge does no virtual work because its point of application does not move. Equilibrium requires $F_1a=F_2b$. For $F_1=40\,\mathrm N$, $a=0.30\,\mathrm m$, and $b=0.60\,\mathrm m$, $F_2=20\,\mathrm N$. The generalized force is a torque, and the virtual-work condition reproduces the moment balance.

Example 3 · Pendulum in a horizontal force field

A pendulum of length $a$ and mass $m$ is acted on by a constant horizontal force $P$ to the right. Measure $\theta$ from the downward vertical and consider equilibrium on the lower branch.

The applied force is $(P,-mg)$ and $\partial\mathbf r/\partial\theta=(a\cos\theta,a\sin\theta)$. Their dot product gives

Set $Q_\theta=0$. On the lower branch $-\pi/2<\theta<\pi/2$, this yields $\tan\theta=P/(mg)$. If $P=mg$, $\theta=45^\circ$ to the right. The tension need not be solved to obtain the angle. Afterward it can be recovered from the force balance as $\mathcal T=\sqrt{P^2+m^2g^2}$ for the taut string.

For a rigid rod there is also an opposite stationary orientation, generally unstable. It must not be interpreted as a valid taut-string equilibrium requiring compression.

Example 4 · Cancellation of tension in an inextensible string

For an ideal Atwood machine, write the string constraint as $x_1+x_2=b$, with both coordinates measured downward. The tension has the same magnitude $\mathcal T$ in both segments. Its total virtual work can be found without first calculating that magnitude.

Admissible variations satisfy $\delta x_2=-\delta x_1$. The tension's total virtual work is $-\mathcal T\delta x_1-\mathcal T\delta x_2=0$. Each term need not vanish, but the sum does. Gravity contributes

Equilibrium therefore requires $m_1=m_2$. For unequal masses, the nonzero generalized force indicates a tendency to move; Chapter 6 will determine the acceleration.

7. Contact, friction, and equilibrium

If friction acts along a wire, its virtual work usually does not vanish. Keep it in $Q_i$. If a contact can only push, solve the equilibrium condition and then check the reaction's sign. If a system is already moving, zero generalized applied force does not imply it is instantaneously at rest or that it will remain at the same configuration. Equilibrium includes the appropriate zero-velocity state in stationary geometry.

Do not prove equilibrium by selecting $\delta q=0$. That makes virtual work zero for every system and says nothing. You need the identity to hold for all admissible variations.

8. Practice problems

See Chapter 5 hints and solutions.

1. 5.1 Is virtual work necessarily an actual transfer of energy?

2. 5.2 Must each ideal reaction do zero virtual work individually?

3. 5.3 What are the units of a generalized force conjugate to a length? To an angle?

4. 5.4 Does $\delta W=0$ for one chosen variation prove equilibrium?

5. 5.5 Define ideal constraints and explain why smooth fixed surfaces are examples.

6. 5.6 Why can generalized force include contributions from several particles?

7. 5.7 Why is a stationary point of potential not necessarily a stable equilibrium?

8. 5.8 Derive $Q_i=\sum_a\mathbf F_a\cdot\partial\mathbf r_a/\partial q_i$.

9. 5.9 Prove the principle of virtual work and obtain $Q_i=0$, stating all assumptions.

10. 5.10 A $3\,\mathrm{kg}$ bead is held on a smooth $30^\circ$ incline by an uphill force. Find that force for $g=9.8\,\mathrm{m\,s^{-2}}$.

11. 5.11 A lever supports $60\,\mathrm N$ at $0.20\,\mathrm m$ to the left. What downward force at $0.50\,\mathrm m$ to the right balances it?

12. 5.12 For a pendulum under constant horizontal force $P=mg/\sqrt3$, find the lower equilibrium angle.

13. 5.13 Develop virtual work and generalized forces, explain ideal constraints, and apply the method to a pendulum under a horizontal force.

14. 5.14 Compare Newtonian equilibrium and virtual-work equilibrium, illustrating cancellation of internal string tensions.

15. 5.15 A vertical spring of stiffness $k$ supports a mass $m$. Let $q$ be downward extension. Use virtual work to find equilibrium and its stability.

16. 5.16 A particle on a rough horizontal plane is at rest under a horizontal force $P$. Why is setting $P\delta x=0$ an incorrect equilibrium calculation?

9. Equilibrium and admissible directions

The equilibrium condition concerns the applied force along every admissible direction. A normal component need not vanish: it may be balanced by a constraint reaction. Virtual work removes that reaction from the calculation when its total contribution is zero for all allowed variations.

For independent coordinates, each generalized force must consequently vanish. This conclusion still requires a physically admissible reaction. A smooth floor cannot pull a particle downward, and a flexible string cannot supply compression. In connected systems, such as the Atwood machine, ideality may arise through cancellation of reaction work over the whole system rather than through a zero contribution from every particle separately.