Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 14 - Fluids - Problems - Page 409: 44b

Answer

$2.01\ kg$

Work Step by Step

In this case, the wood has a part of its volume in the water, and the lead has all of its volume in the water. Volume of water displaced = 0.900(volume of the wood)+(volume of the lead). $(V_{wood}=\displaystyle \frac{m_{wood}}{\rho_{wood}}, \quad V_{lead}=\frac{m_{lead}}{\rho_{lead}})$ The combined mass of the lead and wood is floating, so there is no vertical acceleration $\Rightarrow$ resultant net force is zero. Forces acting: weight $F_{g}$ (downward) and buoyant force $F_{b}$(upward), and, their magnitudes are equal: \begin{align*} (m_{wood}+m_{lead})g&=\rho_{water}[(0.900V_{wood})+V_{lead}]\cdot g\\\\ m_{wood}+m_{lead}&= \displaystyle \rho_{water}\cdot (\frac{0.900m_{wood}}{\rho_{wood}}+\frac{m_{lead}}{\rho_{lead}})\\ \\ m_{lead}- \displaystyle \rho_{water}(\frac{m_{lead}}{\rho_{lead}})&=\displaystyle \rho_{water}\cdot (\frac{0.900m_{wood}}{\rho_{wood}})-m_{wood}\\ \\ m_{lead}(1-\displaystyle \frac{\rho_{water}}{\rho_{lead}})&=\displaystyle \rho_{water}\cdot (\frac{0.900m_{wood}}{\rho_{wood}})-m_{wood}\\ \\ m_{lead}&=[(1000\displaystyle \frac{kg}{m^{3}})\cdot (\frac{0.900(3.67kg)}{(600\frac{kg}{m^{3}})})-(3.67kg)]\div(1-\frac{(1.00\times 10^{3}\frac{kg}{m^{3}})}{(1.13\times 10^{4}\frac{kg}{m^{3}})})\\ &=2.01kg \end{align*}
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