## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.2 Fluid Pressure and Force - Exercises - Page 476: 25

#### Answer

$$\frac{\sqrt{3}}{9} w H^{3},\ \ \ \frac{\sqrt{3}}{3} \ell w H^{2}$$ where ($w=\rho g$).

#### Work Step by Step

Place the origin at the lower vertex of the trough and orient the positive $y$ -axis pointing upward. First, consider the faces at the front and back ends of the trough. A horizontal strip at height $y$ has a length of $\frac{2 y}{\sqrt{3}}$ and is at a depth of $H-y .$ Thus, \begin{align*} F&=w \int_{0}^{H}(H-y) \frac{2 y}{\sqrt{3}} d y\\ &=\left.w\left(\frac{H}{\sqrt{3}} y^{2}-\frac{2}{3 \sqrt{3}} y^{3}\right)\right|_{0} ^{H}\\ &=\frac{\sqrt{3}}{9} w H^{3} \end{align*} where ($w=\rho g$). For the slanted sides, we note that each side makes an angle of $60^{\circ}$ with the horizontal. If we let $\ell$ denote the length of the trough, then \begin{align*} F&=\frac{2 w \ell}{\sqrt{3}} \int_{0}^{H}(H-y) d y\\ &=\frac{\sqrt{3}}{3} \ell w H^{2} \end{align*}

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