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 288: 52

Answer

The lift on the wing is $3.2\times 10^6~N$

Work Step by Step

We can use Bernoulli's equation to find the pressure difference below the wing and above the wing. Let $P_1$ be the pressure below the wing and let $P_2$ be the pressure above the wing. $P_1 + \frac{1}{2}\rho~v_1^2 = P_2 + \frac{1}{2}\rho~v_2^2$ $P_1-P_2 = \frac{1}{2}\rho~(v_2^2-v_1^2)$ $P_1-P_2 = \frac{1}{2}(1.29~kg/m^3)~[(280~m/s)^2-(150~m/s)^2]$ $P_1-P_2 = 3.6\times 10^4~N/m^2$ We can find the upward force exerted on the wing from the pressure difference. $F = (P_1-P_2)~A$ $F = (3.6\times 10^4~N/m^2)(88~m^2)$ $F = 3.2\times 10^6~N$ The lift on the wing is $3.2\times 10^6~N$.
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