Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 14 - Fluids and Elasticity - Exercises and Problems - Page 384: 22


The length of steel that is above the surface is 8.4 cm

Work Step by Step

Let $V_c$ be the volume of the cylinder. We can find the mass of the cylinder as: $M = V_c~\rho_{steel}$ $M = (\pi)(0.05~m)^2(0.20~m)(7900~kg/m^3)$ $M = 12.4~kg$ The buoyant force on the cylinder is equal to the cylinder's weight $Mg$. The buoyant force is equal to the weight of the mercury that is displaced. Let $\rho_m$ be the density of mercury. We can find the volume $V_m$ of mercury that is displaced. Therefore; $F_B = Mg$ $\rho_m~V_m~g = Mg$ $V_m = \frac{M}{\rho_m}$ $V_m = \frac{12.4~kg}{13.6\times 10^3~kg/m^3}$ $V_m = 9.12\times 10^{-4}~m^3$ We can find the height $h$ such that the bottom part of the cylinder which is submerged has the volume $V_m$ as: $\pi~r^2~h = V_m$ $h = \frac{V_m}{\pi~r^2}$ $h = \frac{9.12\times 10^{-4}~m^3}{(\pi)~(0.05~m)^2}$ $h = 0.116~m = 11.6~cm$ The length of steel that is below the surface is 11.6 cm. Therefore, the length of steel that is above the surface is 20 cm - 11.6 cm which is 8.4 cm.
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