A homogeneous mass-distributed, cube-shaped body is attached to the wall of a container along one of its edges, and it can rotate around this edge, which serves as a horizontal axis. The container is filled with a fluid up to the height of the axis. What will be the angle $\varphi$ between the fluid surface and the bottom face of the cube if the specific gravity of the cube is one-third that of the fluid?

![](https://cdn.mathpix.com/cropped/2024_05_02_545060835aec60ec969fg-1.jpg?height=344&width=336&top_left_y=283&top_left_x=881)
A cube is subjected to three forces:

1. The gravitational force acting vertically downward at the center of the cube:

$$
G=a^{3} \cdot \gamma \quad (\gamma \text { is the specific weight of the cube })
$$

2. The buoyant force acting vertically upward at the center of gravity of the submerged part of the cube:

$$
F=a^{3} \frac{\operatorname{tg} \varphi}{2} \cdot 3 \gamma \quad (3 \gamma \text { is the specific weight of the fluid })
$$

3. The force $K$ acting at the axis, which forces the cube into an inclined position.

![](https://cdn.mathpix.com/cropped/2024_05_02_e9d8a781e51b00c1cac4g-1.jpg?height=510&width=662&top_left_y=550&top_left_x=726)

The equilibrium condition is that the resultant of all forces acting on the cube and the sum of the torques must be zero. If we write the torques with respect to the axis, then to determine the angle, it is sufficient to consider only the latter equation, because the torque of the unknown force $K$ is zero.

The lever arm of the gravitational force:

$$
k=a \cos 45^{\circ} \cos \left(45^{\circ}-\varphi\right) .
$$

The lever arm of the buoyant force is equal to the $x$-coordinate of the center of gravity of the submerged part, if we take the $x$-axis parallel to the surface of the fluid:

$$
s_{x}=\frac{a \cos \varphi+a / \cos \varphi}{3}
$$

where we used the fact that the position vector of the centroid of the triangle $ABC$ is given by:

$$
\vec{s}=\frac{\vec{a}+\vec{b}+\vec{c}}{3}
$$

($\vec{a}, \vec{b}$, and $\vec{c}$ are the position vectors of the vertices).

Let's write the equation for the equality of the torques:

$$
\begin{gathered}
a^{3} \gamma \cdot a \cos 45^{\circ} \cos \left(45^{\circ}-\varphi\right)=a^{3} \frac{\operatorname{tg} \varphi}{2} 3 \gamma \cdot a \frac{\cos \varphi+1 / \cos \varphi}{3} \\
\cos \left(45^{\circ}-\varphi\right)=\frac{1}{\sqrt{2}}(\cos \varphi+\sin \varphi), \operatorname{tg} \varphi=\sin \varphi / \cos \varphi
\end{gathered}
$$

Using these, and simplifying the equation:

$$
\cos \varphi+\sin \varphi=\sin \varphi+\frac{\sin \varphi}{\cos ^{2} \varphi}
$$

From this, $\cos ^{3} \varphi=\sin \varphi$ (or $\cos ^{2} \varphi=\operatorname{tg} \varphi$ ).

Squaring and rearranging:

$$
\left(\cos ^{2} \varphi\right)^{3}+\cos ^{2} \varphi-1=0 \quad\left(\text { or } \operatorname{tg}^{3} \varphi+\operatorname{tg} \varphi-1=0\right)
$$

Graphing the equation or solving it by trial and error or using the Cardano formula, we get a single real root:

$$
\operatorname{tg} \varphi=\cos ^{2} \varphi \approx 0.6823
$$

Thus, the face of the cube makes an angle of $\varphi=34^{\circ} 18.5^{\prime}$ with the horizontal.

János Pipek (Budapest, I. István Gymnasium, 3rd year)

Note. Most solvers overlooked the force at the axis and calculated the angle based on the assumed "force equilibrium." Although this solution deviates from the correct value by only half a degree, such a significant conceptual error could not be accepted. The emphasis here was not on the numerical solution of the equation.