Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 15 - Fluids and Elasticity - Exercises and Problems - Page 436: 13


The minimum diameter of the hose is 3.5 cm

Work Step by Step

The pressure difference between the pressure inside the ideal vacuum cleaner and the atmospheric pressure is $1.01\times 10^5~N/m^2$. The force exerted on the dog is equal to this pressure difference multiplied by the area $A$ of the hose. To find the minimum radius of the hose, we can let this force equal the dog's weight. Therefore; $F = (1.01\times 10^5~N/m^2)~A = Mg$ $(1.01\times 10^5~N/m^2)~(\pi~R^2) = Mg$ $R^2 = \frac{Mg}{(1.01\times 10^5~N/m^2)~(\pi)}$ $R = \sqrt{\frac{Mg}{(1.01\times 10^5~N/m^2)~(\pi)}}$ $R = \sqrt{\frac{(10~kg)(9.80~m/s^2)}{(1.01\times 10^5~N/m^2)~(\pi)}}$ $R = 0.01757~m = 1.757~cm$ We then find the minimum diameter of the hose. $d = 2R = (2)(1.757~cm) = 3.5~cm$ Therefore, the minimum diameter of the hose is 3.5 cm.
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