Ladder equilibrium by virtual work

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.