Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 15 - Fluids and Elasticity - Exercises and Problems - Page 437: 45

Answer

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Work Step by Step

a) To find the force exerted by the liquid on the bottom of the aquarium, we need to find the pressure exerted by this liquid there since we know that $P=F/A$. Hence, $$F=PA$$ where $A$ is the area of the bottom and is given by $LW$, as we see in the figure below. Thus, $$\boxed{F=D\rho g LW}$$ __________________________________________________________ b) By the same approach, the liquid force on the front window of the aquarium is given by $$dF=P(dA)=y\rho g (Ldy)$$ Hence, $$dF=L\rho gydy$$ integrating both sides; $$\int_0^FdF=\int_0^DL\rho gydy=L\rho g\int_0^Dydy$$ $$ F= L\rho g\left[\dfrac{y^2}{2}\right]_0^D$$ $$\boxed{ F= \frac{1}{2}L\rho gD^2} $$ __________________________________________________________ c) Plugging the given into the boxed formula above. $\bullet$ The liquid force on the bottom of the aquarium is $$F_{bottom}=D\rho g LW=(0.4) (1000)(9.8) (1.0)(0.35)$$ $$F_{bottom}=\color{red}{\bf 1372}\;\rm N$$ $\bullet$ The liquid force on the front window of the aquarium is $$F_{front}=\frac{1}{2}L\rho gD^2=\frac{1}{2} (1.0) (1000)(9.8) (0.4)^2$$ $$F_{front}=\color{red}{\bf 784}\;\rm N$$
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