## Solution

![](https://cdn.mathpix.com/cropped/2024_05_22_31c1f87d4364af4121d5g-19.jpg?height=505&width=625&top_left_y=621&top_left_x=1298)

$$
c=b-2 F B ; \triangle F B G \text { is similar to }
$$

$$
\Delta E B C \Rightarrow c=b-x \frac{b-a}{h}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_31c1f87d4364af4121d5g-19.jpg?height=637&width=897&top_left_y=1258&top_left_x=1002)

$$
\begin{aligned}
& P=\rho g x ; P=\frac{F}{S} ; \Delta S=c \cdot \Delta x ; F_{x}=\rho g x \cdot c \cdot \Delta x \Rightarrow \\
& \Rightarrow F=\int_{0}^{h} \rho g x \cdot\left(b-x \frac{b-a}{h}\right) d x=\left.\left(\rho g b \frac{x^{2}}{2}-\rho g \frac{b-a}{h} \cdot \frac{x^{3}}{3}\right)\right|_{0} ^{h}= \\
& =\rho g\left(\frac{1}{2} \cdot b h^{2}-(b-a) \cdot \frac{h^{2}}{3}\right)= \\
& =\rho g h^{2}\left(\frac{b}{2}-\frac{b-a}{3}\right)= \\
& =1000 \cdot 10 \cdot 5.0^{2} \cdot\left(\frac{12.0}{2}-\frac{12.0-7.2}{3}\right)=1100000 \mathrm{N}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_

\%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+22-10$ »

Categories: Kuznetsov's Problem Book Integrals Problem 22 | Integrals

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## Problem Kuznetsov Integrals 22-11

## Material from PlusPi
## Problem Statement

Calculate the force with which water presses on a dam, the cross-section of which has the shape of an isosceles trapezoid (see figure). The density of water \(\rho = 1000 \mathrm{kg} / \mathrm{m}^{3}\), and the acceleration due to gravity \(g\) is taken to be \(10 \mathrm{m} / \mathrm{s}^{2}\).

Hint: The pressure at depth \(x\) is \(\rho g x\).

\[
a = 7.2 \mathrm{m}, b = 12.0 \mathrm{m}, h = 5.0 \mathrm{m}
\]