5. The younger sister asked the math student Gavrila to fix the swings in the yard. After the repair, they became a flat board (seat) rigidly attached to two parallel rods, the distance between which is $d$. The rods are mounted on a horizontal axis in a cylindrical hinge, meaning they can rotate relative to this axis. One of the rods turned out to be shorter than the other by $a$. At the same time, the normal to the seat lies in the same plane as the rods. The sister sat her favorite doll on the swings, which cannot hold onto the rods but is kept on the seat only by the force of dry friction with a coefficient $\mu$. At what angle can the swings be deflected from the vertical so that the doll does not slide off the seat?
5. Continue the line connecting the bases of the rods until it intersects with the axis of the swings. Between them, there is a constant angle $\alpha$, independent of the angle of deflection of the swings. The normal to the seat at any angle of deflection forms an angle $\pi / 2 + \alpha$ with the axis of the swings (see the left figure).

At the limiting angle of deflection, the angle between the normal to the seat and the vertical is the angle of friction, determined by the relation $\operatorname{tg} \gamma = \mu$, from which $\cos \gamma = 1 / \sqrt{1 + \mu^{2}}, \sin \gamma = \mu / \sqrt{1 + \mu^{2}}$.

Consider the trihedral angle formed by the normal to the seat, the ascending vertical, and the axis of the swings (see the right figure).

In it, all three dihedral angles are known: between the normal to the seat and the axis of the swings ($\pi / 2 + \alpha$), between the normal and the vertical ($\gamma$), and between the axis and the vertical ($\pi / 2$). We need to find the dihedral angle $\beta$ between the vertical plane and the plane of the swings (normal to the seat and axis). Opposite it lies the angle $\gamma$. By the cosine theorem for trihedral angles,

$$
\cos \gamma = \cos \alpha' \cos \pi / 2 + \sin \alpha' \sin \pi / 2 \cos \beta
$$

From this, $\cos \beta = \cos \gamma / \sin \alpha' = \cos \gamma / \cos \alpha$,

$$
\beta = \arccos \left(\frac{1}{\sqrt{1 + \mu^{2}} \cos \alpha}\right) = \arccos \left(\frac{\sqrt{d^{2} + a^{2}}}{\sqrt{1 + \mu^{2}} d}\right)
$$

Note that the answer is defined if $\cos \gamma \leq \cos \alpha$, i.e., $\alpha \leq \gamma$, and this condition is satisfied, given that by the problem's condition, in the lower position of the swings, the doll was held on the seat by dry friction forces.

![](https://cdn.mathpix.com/cropped/2024_05_06_26dc8c5bf5aedf0c0e50g-3.jpg?height=251&width=325&top_left_y=323&top_left_x=1668)