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: 20


The tension in the string is 43.9 N

Work Step by Step

We can find the volume of the rock as: $V = \frac{M}{\rho}$ $V = \frac{5.0~kg}{4800~kg/m^3}$ $V = 1.04\times 10^{-3}~m^3$ We can find the buoyant force on the rock. The buoyant force is equal to the weight of the water that is displaced by half the volume of the rock. Let $\rho_w$ be the density of water. Therefore; $F_B = (\rho_w)~(\frac{V}{2})~(g)$ $F_B = (1000~kg/m^3)~(\frac{1.04\times 10^{-3}~m^3}{2})~(9.80~m/s^2)$ $F_B = 5.1~N$ The sum of the tension and the buoyant force is equal to the rock's weight. We can find the tension in the string as: $T+F_B = Mg$ $T = Mg-F_B$ $T = (5.0~kg)(9.80~m/s^2)- 5.1~N$ $T = 43.9~N$ The tension in the string is thus 43.9 N.
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