College Physics (4th Edition)

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

Chapter 9 - Problems - Page 363: 53


The upward force exerted on each wing is $1.91\times 10^5~N$

Work Step by Step

Let $P_1$ be the pressure below the wing and let $P_2$ be the pressure above the wing. We can use Bernoulli's equation to find the pressure difference below the wing and above the wing: $P_1 + \frac{1}{2}\rho~v_1^2 = P_2+\frac{1}{2}\rho~v_2^2$ $P_1 - P_2 = \frac{1}{2}\rho~(v_2^2-v_1^2)$ $P_1 - P_2 = (\frac{1}{2})(1.3~kg/m^3)~[(190~m/s)^2-(160~m/s)^2]$ $P_1 - P_2 = 6825~N/m^2$ We can find the force that this pressure difference exerts on each wing: $F = \Delta P~A$ $F = (6825~N/m^2)(28~m^2)$ $F = 1.91\times 10^5~N$ The upward force exerted on each wing is $1.91\times 10^5~N$.
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