395. A plate in the form of a triangle, the base of which is $4 \mathrm{~cm}$, and the height is $3 \mathrm{~cm}$, is vertically submerged in water. Find the force of water pressure on this plate, if its vertex lies on the water surface (Fig. 200).
Solution. Again, we will use formula (5). Here $a=0$, $b=3 \mathrm{~cm}=0.03 \mathrm{~m}$, $|A C|=4 \mathrm{~cm}=0.04 \mathrm{~m}$, $\gamma=1000 \mathrm{kr} / \mathrm{m}^{3}$.

From the similarity of triangles $D B E$ and $A B C$ (Fig. 200), we have $\frac{|D F|}{|A C|}=$ $=\frac{|B K|}{|B E|}$, or $\frac{y}{0.04}=\frac{x}{0.03}$. Therefore, $y=\frac{0.04 x}{0.03}$, or $y=\frac{4}{3} x$.

Substituting all the data into the calculation formula, we get

$$
\begin{gathered}
P=9810 \int_{0}^{0.03} x \cdot \frac{4}{3} x d x=9810 \cdot \frac{4}{3} \int_{0}^{0.03} x^{2} d x=\left.\frac{9810 \cdot 4}{3} \cdot \frac{x^{3}}{3}\right|_{0} ^{0.03}= \\
=\frac{9810 \cdot 4}{9} \cdot 0.03^{3} \approx 0.117 \quad(\mathrm{N})
\end{gathered}
$$