Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 10 - Fluids - Problems - Page 287: 34

Answer

(a) The buoyant force on the chamber is $7.39\times 10^5~N$ (b) The tension in the cable is 9880 N

Work Step by Step

(a) We can find the volume $V$ of the chamber as: $V = \frac{4}{3}\pi~r^3$ $V = \frac{4}{3}\pi~(2.60~m)^3$ $V = 73.6~m^3$ We can find the buoyant force which is equal to the weight of seawater that is displaced by the chamber. Let $\rho$ be the density of seawater. $F_B = \rho~V~g$ $F_B = (1.025\times 10^3~kg/m^3)(73.6~m^3)(9.80~m/s^2)$ $F_B = 7.39\times 10^5~N$ The buoyant force on the chamber is $7.39\times 10^5~N$. (b) The sum of the tension in the cable and the chamber's actual weight is equal to the buoyant force. $T + M~g = F_B$ $T = F_B - M~g$ $T = (7.39\times 10^5~N) - (74,400~kg)(9.80~m/s^2)$ $T = 9880~N$ The tension in the cable is thus 9880 N.
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