<p>What is the lift (in newtons) due to Bernoulli\'s principle on a wing of area

Question

<p>What is the lift (in newtons) due to Bernoulli's principle on a wing of area <span>90</span>&#160;<img src="http://session.masteringphysics.com/render?units=m%5E2" alt="m^2" align="middle" /> if the air passes over the top and bottom surfaces at speeds of <span>300</span>&#160;<img src="http://session.masteringphysics.com/render?units=m%2Fs" alt="m/s" align="middle" /> and <span>180</span>&#160;<img src="http://session.masteringphysics.com/render?units=m%2Fs" alt="m/s" align="middle" />, respectively?</p>
<p>I would really appreciate any help with this problem.&#160; Thanks!</p>

Explanation / Answer

The area of the wing span is A = 90 m^2 The speed of the air at top v_top = 300 m/s The speed of the air at bottom is v_bottom = 180 m/s The density of the air rho = 1.29 kg/m^3 According to Bernouli's equation P_top + 0.5*rho*v_top^2 = P_bottom + 0.5*rho*v_bottom^2 The pressure difference P_bottom - P_top = 0.5*rho*v_top^2 - 0.5*rho*v_bottom^2 DELTA P = 0.5*1.29*(300^2 - 180^2) DELTA P = 37152 The lift force experienced by the wing is F = DELTA P*A F = 37152*90 F = 3343680 N

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