Question

The speed of air flow on the upper and lower surfaces of a wing of an aeroplane are $v_1$ and $v_2$ respectively. If A is the cross section area of the wing and $\rho$ is the density of air, then the upward lift is :

• A. $\frac{1}{2}\rho A(v_1-v_2)$
• B. $\frac{1}{2}\rho A(v_1+v_2)$
• C. $\frac{1}{2}\rho A(v_1^2-v_2^2)$
• D. $\frac{1}{2}\rho A(v_1^2+v_2^2)$

Answer 1

The correct option is C $\frac{1}{2}\rho A(v_1^2-v_2^2)$

Applying Bernoulli's Equation at the two surfaces of the wing gives

$⟹P_u+\frac{1}{2}\rho v_1^2=P_l+\frac{1}{2}\rho v_2^2$

$⟹P_l-P_u=\frac{1}{2}\rho (v_1^2-v_2^2)$

Thus, upward force $=(P_l-P_u)A=\frac{1}{2}\rho A(v_1^2-v_2^2)$