On one side of a mountain lake, they built a slipway, the lower edge of which is 5.4 m below the lake surface, and its width is $1.5 \mathrm{~m}$. What is the magnitude of the force exerted on the slipway? Where is the point of application of this force?
$A B D C$ the lateral pressure exerted on a vertical dam is equal to the weight of the water enclosed by the $A E B C F D$ parallelogram, if we imagine it rotated 90 degrees to lie horizontally $\longrightarrow$.

![](https://cdn.mathpix.com/cropped/2024_05_02_5aceb0d9e33eebafd6aag-1.jpg?height=612&width=718&top_left_y=236&top_left_x=693)

$P=\frac{1}{2} 54^{2} \cdot 15 \mathrm{~kg}$ weight $=21870 \mathrm{~kg}$ weight. The pressure force $P$ passes through the center of gravity $S$ and reaches the dam at point $T$. The point $S$ lies at the midpoint of the line connecting the centers of gravity of the two triangles $a$ and $a'$. The centers of gravity of the triangles are found by drawing the medians. Since these medians intersect at $\frac{2}{3}$ of the distance from the vertex, the distance $B R$ in the same coordinate system is also $\frac{2}{3}$ of $B A$. Therefore, the point of attack $T$ is $\frac{1.5}{2}=0.7 \mathrm{~m}$ from the edges of the dam and $\frac{2}{3} \cdot 5.4 \mathrm{~m}=3.6 \mathrm{~m}$ below the water surface.

(P̈̈zel Tibor, Budapest.)

The problem was also solved by: Bánó L., Freund E., Heimlich P., Jánosy Gy., Koritsánszky I., Kürth R., Pető L., Ruvald S., Stagl A., Székely J., Szilas O.