Four bars of length $l$, with negligible weight, form a frame with hinges at the corners. We form a soap film within the frame, then suspend it by holding one corner and attaching a weight $G$ to the opposite corner. What shape will the frame take if the surface tension of the liquid is $\sigma$?
I. Solution. The shape of the frame can be clearly characterized by the angle $\alpha$ that the legs make with the horizontal. The forces acting on the frame are shown in Figure 1.

![](https://cdn.mathpix.com/cropped/2024_05_02_20743ec66873f4034457g-1.jpg?height=477&width=445&top_left_y=260&top_left_x=821)

Figure 1

Since the liquid film has two free surfaces, $F=2 l \sigma$. If we write the equilibrium of external forces and torques for the entire frame, we obtain an identity. The relationship between $F, G$, and $\alpha$ can only be determined if we use the fact that each part of the frame is also in equilibrium individually. Let's write the equilibrium condition for the segment $AB$! The hinges at the ends of the rod press the rod with the forces shown in Figure 2.

![](https://cdn.mathpix.com/cropped/2024_05_02_20743ec66873f4034457g-1.jpg?height=412&width=418&top_left_y=1041&top_left_x=840)

Figure 2

Due to symmetry, only a vertical force acts at point $B$. Therefore, we have

$$
\begin{gathered}
F \cos \alpha + F_{3} - F_{1} = 0 \\
F_{2} - F \sin \alpha = 0
\end{gathered}
$$

Let's write the torque equation about point $B$!

$$
F(l / 2) - F_{1} l \cos \alpha - F_{2} l \sin \alpha = 0
$$

In these three equations, three unknown forces and the angle $\alpha$ appear. The missing equation is provided by the equilibrium of forces at the hinge at point $A$.

![](https://cdn.mathpix.com/cropped/2024_05_02_20743ec66873f4034457g-1.jpg?height=353&width=185&top_left_y=2008&top_left_x=951)

Due to the symmetric arrangement, the vertical components of the reaction forces in the rods $AB$ and $AD$ are of the same magnitude (Figure 3):

$$
2 F_{1} - G = 0
$$

From the system of equations (1)-(4), we obtain the following equation for $\cos \alpha$:

$$
2 F \cos^2 \alpha - G \cos \alpha - F = 0
$$

from which

$$
\cos \alpha = \frac{G}{4 F} \pm \sqrt{\left(\frac{G}{4 F}\right)^2 + \frac{1}{2}}
$$

Since $0 < \cos \alpha < 1$, we select the positive root:

$$
\cos \alpha = \frac{G}{4 F} + \sqrt{\left(\frac{G}{4 F}\right)^2 + \frac{1}{2}}
$$

This equation is valid for $G \leq 2 l \sigma$. For $G > 2 l \sigma$, this condition is never met, while for $G \leq 2 l \sigma$, it is met at some angle $\alpha$. This angle can be determined using differential calculus, as at $\alpha_{0}$:

$$
E' = 4 l^2 \sigma \cos 2 \alpha_{0} - 2 l G \cos \alpha_{0} = 0
$$

From this, with simple transformations, we can obtain equation (5).

In the limiting case, if $G = 2 l \sigma$, then $\alpha_{0} = 0^{\circ}$.

Joseph Klebniezki (Szeged, Ságvári E. Gymnasium, 3rd grade)