Question

The velocity of the wind over the surface of the wing of an aeroplane is $80$ and under the wing $60$ if the area of the wing is $4$ the dynamic lift experienced by the wing is density of air.

Answer 1

Given, $v_1=80m/s{v}_2=60m/s$

A= $4m^2$

Density of air, $\rho =1.3kg{m}^{-3}$ (From internet)

From Bernoulli's theorem,$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$

$P_2-P_1=\frac{1}{2}\rho (v_1^2-v_2^2)$-------- (1)

Lift on the wings=$(P_2-P_1)A$

From (1),

=$\frac{1}{2}\rho (v_1^2-v_2^2)A$

= $\frac{1}{2}\times 1.3\times ({80}^2-{60}^2)\times 4$

= 7280 N