An aeronautical engineer observed that on the upper and the lower surface of the wing of an aeroplane the speed 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 `1.29 kg//m^(3)`.

Explanation

Here, speed of air on the upper surface of the wing of aeroplane , `upsilon_(1) = 120 ms^(-1)`
Speed of air on the lower surface of the wing of aeroplane , `upsilon_(2) = 90 ms^(-1)`,
Area of wing , `A = 3.2 m^(2)`
density of air, `rho = 1.29 kg m^(-3)`
Let `P_(1) and P_(2)` be the pressure on the upper and lower surface of the wing of an aeroplane.
Ac cording to Bernoulli's theorem
`P_(1) + 1/2 rho upsilon_(1)^(2) = P_(2)+1/2 rho upsilon_(2)^(2)`
or `P_(2)-P_(1) = 1/2 rho (upsilon_(1)^(2) - upsilon_(2)^(2))`
`=1/2 xx 1.29 (120^(2) - 90^(2))`
`= 4063.5 Pa`
Left on the wing, `F= (P_(2) - P_(1)) A`
`= 4063.5 xx 3.2 `
`= 13003.2N`.