Virtual Work

A virtual displacement $\delta r$ is an imaginary, infinitesimal displacement of a point, consistent with the constraints of the system, taken at a fixed instant of time with no change in the applied forces. Virtual work $\delta U$ is the work done by the forces during a virtual displacement.

Principle for a particle

A particle is in equilibrium if and only if the total virtual work of all forces acting on it is zero for every virtual displacement:

$$\delta U = \sum \mathbf{F}_i \cdot \delta \mathbf{r} = 0$$

This is equivalent to $\Sigma F = 0$, because $\delta \mathbf{r}$ is arbitrary.

Principle for a rigid body

A rigid body is in equilibrium if the total virtual work of all external forces and couples is zero for every virtual displacement (translation and rotation) consistent with the constraints:

$$\delta U = \sum \mathbf{F}_i \cdot \delta \mathbf{r}_i + \sum M_j\, \delta\theta_j = 0$$

A couple $M$ does virtual work $M\,\delta\theta$ for a virtual rotation $\delta\theta$. For a system of connected rigid bodies the equation is written in terms of one independent coordinate $\theta$ (or $x$), so that $\delta U = Q\,\delta\theta = 0$.

Forces that do no virtual work are left out: reactions at smooth pins and smooth surfaces, the internal forces of a rigid body, and forces at fixed supports.

Uses

1. Finding unknown forces or couples in machines and linkages (jacks, toggles, scissors) without dismembering them, since the unknown reactions at smooth supports never appear.
2. Finding the equilibrium position of a system under given loads, for example the angle at which a ladder or a spring-loaded linkage rests.

Method: the floor friction force $F$ at A does work only if A is allowed to move, so give the ladder a virtual rotation $\delta\theta$ and equate the total virtual work to zero. The normal reaction at A and the wall reaction at B (smooth) do no work because they are perpendicular to the displacements of A and B (B moves only along the wall, A only along the floor).

        wall
         |\ B
         | \
         |  \  W at G
         |   \
         |  60\
   ------+------A------> floor
            F <-- (friction toward wall)

Let the ladder make angle $\theta$ with the floor and length $L = 5$ m. Take the horizontal distance of A from the wall as $x_A = L\cos\theta$ and the height of the centre of gravity G as $y_G = \tfrac{L}{2}\sin\theta$.

$$\delta x_A = -L\sin\theta\,\delta\theta \qquad \delta y_G = \tfrac{L}{2}\cos\theta\,\delta\theta$$

Friction at A acts toward the wall (opposite to the sense in which A tends to slip), i.e. in the $-x$ direction, so its virtual work is $-F\,\delta x_A$. The weight acts downward, so its virtual work is $-W\,\delta y_G$.

$$\begin{aligned} \delta U &= -F\,\delta x_A - W\,\delta y_G = 0 \\ F L \sin\theta\,\delta\theta &- W\tfrac{L}{2}\cos\theta\,\delta\theta = 0 \\ F &= \frac{W}{2\tan\theta} \end{aligned}$$

Substituting $W = 400$ N and $\theta = 60^\circ$:

$$F = \frac{400}{2\tan 60^\circ} = \frac{400}{3.464} = 115.5\ \text{N}$$

Check by statics: wall reaction $N_B = F$ (horizontal equilibrium). Moments about A: $N_B L\sin\theta = W\tfrac{L}{2}\cos\theta$, giving $N_B = 400/(2\tan 60^\circ) = 115.5$ N. This agrees.

Answer: Friction force at A = 115.5 N, directed toward the wall.