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: 49

Answer

The weight of the roof is $1.2\times 10^5~N$

Work Step by Step

We first convert the wind speed to units of m/s: $v = (180~km/h)(\frac{1000~m}{1~km})(\frac{1~hr}{3600~s})$ $v = 50~m/s$ We then use Bernoulli's equation to find the pressure difference below the roof and above the roof. $P_1 = P_2 + \frac{1}{2}\rho~v^2$ $P_1-P_2 = \frac{1}{2}\rho~v^2$ $P_1-P_2 = \frac{1}{2}(1.29~kg/m^3)~(50~m/s)^2$ $P_1-P_2 = 1612.5~N/m^2$ We then find the upward force exerted on the roof from the pressure difference. $F = (P_1-P_2)~A$ $F = (1612.5~N/m^2)(6.2~m)(12.4~m)$ $F = 1.2\times 10^5~N$ Since the roof comes off the house, we can assume that the upward force exerted on the roof is equal to the weight of the roof. The weight of the roof is $1.2\times 10^5~N$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.