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: 24

Answer

The maximum mass that can hang from the sphere is 55.6 kg

Work Step by Step

We can find the volume of the sphere as: $V = \frac{4}{3}\pi~r^3$ $V = \frac{4}{3}\pi~(0.25~m)^3$ $V = 0.06545~m^3$ We can find the mass $M_s$ of the sphere as: $M_s = \rho~V$ $M_s = (150~kg/m^3)(0.06545~m^3)$ $M_s = 9.818~kg$ The buoyant force is equal to the weight of the water that is displaced by the sphere's volume. Let $\rho_w$ be the density of water. We can find the buoyant force on the sphere when the whole sphere is submerged. Therefore; $F_B = \rho_w~V~g$ $F_B = (1000~kg/m^3)(0.06545~m^3)(9.80~m/s^2)$ $F_B = 641.4~N$ To find the maximum mass $M_o$ of an object that can hang on the sphere, we can assume that the whole sphere is submerged and the buoyant force is equal to the total weight of the sphere and the object. $M_s~g+M_o~g = F_B$ $M_o = \frac{F_B-M_s~g}{g}$ $M_o = \frac{(641.4~N)-(9.818~kg)(9.80~m/s^2)}{9.80~m/s^2}$ $M_o = 55.6~kg$ The maximum mass that can hang from the sphere is 55.6 kilograms.
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