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: 21


The tension in the string is 1.87 N

Work Step by Step

We can find the mass of the aluminum as: $M = V~\rho_{Al}$ $M = (100\times 10^{-6}~m^3)(2700~kg/m^3)$ $M = 0.27~kg$ We can find the buoyant force on the aluminum. The buoyant force is equal to the weight of the Ethyl alcohol that is displaced by the volume of the aluminum. Let $\rho_E$ be the density of Ethyl alcohol. Therefore; $F_B = \rho_E~V~g$ $F_B = (790~kg/m^3)~(100\times 10^{-6}~m^3)~(9.80~m/s^2)$ $F_B = 0.7742~N$ The sum of the tension and the buoyant force is equal to the aluminum's weight. We can find the tension in the string as: $T+F_B = Mg$ $T = Mg-F_B$ $T = (0.27~kg)(9.80~m/s^2)- (0.7742~N)$ $T = 1.87~N$ The tension in the string is thus 1.87 N.
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