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 438: 50

Answer

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Work Step by Step

The cylinder is in static equilibrium, so the net force exerted on it is zero. $$\sum F_y=F_{B1}+F_{B2}-mg=ma_y=m(0)=0$$ where $F_{B1}$ is the buoyant force exerted by the upper liquid and $F_{B2}$ is the buoyant force exerted by the second liquid, as we see in the figure below. $$F_{B1}+F_{B2}=mg $$ where $m$ is given by the density law $m=\rho V=\rho A l$ $$F_{B1}+F_{B2}=\rho A lg\tag 1 $$ Recall that the buoyant force is given by Archimedes's principle. which states that the buoyant force equals the weight of the fluid displaced by the object So, $F_{B1}=m_1g=\rho_1V_1g=\rho_1y_1Ag$ and $F_{B2}=m_2g=\rho_2V_2 g=\rho_2y_2Ag$ Plugging into (1); $$\rho_1y_1\color{red}{\bf\not} A\color{red}{\bf\not} g+\rho_2y_2\color{red}{\bf\not} A\color{red}{\bf\not} g=\rho \color{red}{\bf\not} A l\color{red}{\bf\not} g $$ we need to find $y_2/l$, so we need to replace $y_1$ by $y_1=l-y_2$ $$\rho_1 (l-y_2)+\rho_2y_2=\rho l$$ $$\rho_1 l-\rho_1 y_2 +\rho_2y_2=\rho l$$ $$(-\rho_1 +\rho_2)y_2=\rho l-\rho_1 l$$ $$( \rho_2-\rho_1 )y_2=l(\rho -\rho_1 )$$ Therefore, $$\dfrac{y_2}{l}=\dfrac{\rho -\rho_1}{\rho_2-\rho_1}$$ $$\boxed{ y_2 =\left[\dfrac{\rho -\rho_1}{\rho_2-\rho_1}\right]l}$$
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