Question

A frequently quoted rule of thumb in aircraft design is that wings should produce about 1000 N of lift per square meter of wing. (The fact that a wing has a top and bottom surface does not double its area.) (a) At takeoff the aircraft travels at 63.0 m/s, so that the air speed relative to the bottom of the wing is 63.0 m/s. Given the sea level density of air to be 1.29 kg/m3, how fast (in m/s) must it move over the upper surface to create the ideal lift

Answer 1

v₂ = 63.62 m / s

Step-by-step explanation:

For this exercise in fluid mechanics we will use Bernoulli's equation

P₁ + ρ g v₁² + ρ g y₁ = P₂ + ρ g v₂² + ρ g y₂

where the subscript 1 refers to the inside of the wing and the subscript 2 to the top of the wing.

We will assume that the distance between the two parts is small, so y₁ = y₂

P₁-P₂ = ρ g (v₂² - v₁²)

pressure is defined by

P = F / A

we substitute

ΔF / A = ρ g (v₂² - v₁²)

v₂² =
`(\Delta F)/(A \ \rho \ g) + v_1^2`

suppose that the area of ​​the wing is A = 1 m²

we substitute

v₂² =
`(1000)/(1 \ 1.29 \ 9.8) + 63^2`

v₂² = 79.10 + 3969

v₂ = √4048.1

v₂ = 63.62 m / s