Question

The area of an airplane wing is \(A=4\,m^{2}\). Air flows with velocity \(v_{1}=80\,m/s\) above the wing and \(v_{2}=60\,m/s\) below it. The density of air is \(1.2\,kg/m^{3}\). Find the pressure difference \((\Delta P)\) between the upper and lower surfaces of the wing.

• A. \(1200\;Pa\)
• B. \(1680\;Pa\)
• C. \(2400\;Pa\)
• D. \(3600\;Pa\)

Hint:

For airplane wing problems, remember: \[ \boxed{\Delta P=\frac{1}{2}\rho\left(v_{\text{fast}}^{2}-v_{\text{slow}}^{2}\right)} \] Greater fluid velocity corresponds to lower pressure according to Bernoulli's principle. If the lift force is asked, \[ \boxed{F=\Delta P\times A.} \] Always check whether the question asks for pressure difference or lift force.

Answer 1

The Correct Option is B

By Bernoulli's theorem, the pressure difference between the wing surfaces is $\Delta P = \frac{1}{2}\rho(v_1^2 - v_2^2) = \frac{1}{2}(1.3)(6400 - 3600) = 1820$ Pa. The lift force is $F = \Delta P \times A = 1820 \times 4 = 7280$ N.