College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 9 - Problems - Page 363: 52


The magnitude of the force on the roof is $1.02\times 10^9~N$

Work Step by Step

We can convert the wind speed to units of m/s: $v_2 = 150~mph\times \frac{1609~m}{1~mi} \times \frac{1~hr}{3600~s} = 67.0~m/s$ Let $P_1$ be the pressure inside the house. We can use Bernoulli's equation to find the pressure difference between the inside and outside of the house: $P_1 + \frac{1}{2}\rho~v_1^2 = P_2+\frac{1}{2}\rho~v_2^2$ $P_1 + 0 = P_2+\frac{1}{2}\rho~v_2^2$ $P_1 - P_2 = \frac{1}{2}\rho~v_2^2$ $P_1 - P_2 = (\frac{1}{2})(1.225~kg/m^3)(67.0~m/s)^2$ $P_1 - P_2 = 2750~Pa$ We can find the force that this pressure difference exerts on the roof: $F = \Delta P~A$ $F = (2750~N/m^2)(2000~ft^2)(\frac{1~m}{3.28~ft})^2$ $F = 1.02\times 10^9~N$ The magnitude of the force on the roof is $1.02\times 10^9~N$.
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