An aeronautical engineer observes that on the upper and the lower surface of the wing of an aeroplane the speeds of the air are 120 `ms^(-1)` and 90 `ms^(-1)` respectively during flight. What is the lift on the wing of aeroplane if its area is 3.2 `m^(2)`? Given density of air is `1.29kg//m^(3)`.

Explanation

Here, speed of air on the upper surface of the wing of aeroplane, `V_(1)=120ms^(-1)`
Speed of air on the lower surface of the wing of aeroplane, `V_(2)=90ms^(-1)`
Area of using, `A=3.2m^(2)`,
density of air, `rho=1.29kgm^(-3)`
Let `P_(1)andP_(2)` be the pressure on the upper and lower surface of the using of an aeroplane.
According to Bernoulli's theorem,
`P_(1)+(1)/(2)rhoV_(1)^(2)=P_(2)+(1)/(2)rhoV_(2)^(2)`
(or) `P_(2)-P_(1)=(1)/(2)rho(v_(1)^(2)-v_(2)^(2))`
`=(1)/(2)xx1.29(120^(2)-90^(2))`
`=4063.5Pa`
Lift on the wing,
`F=(P_(2)-P_(1))A`
`=4063.5xx3.2`
`=13003.2N`