Question

g 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 61.1 m/s, so that the air speed relative to the bottom of the wing is 61.1 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

The value is
`u = 72.69 \ m/s`

Step-by-step explanation:

From the question we are told that

The amount of force a square meter of an aircraft wing should produce is
`F = 1000 \ N`

The air speed relative to the bottom of the wing is
`v = 61.1 \ m/s`

The air level density of air is
`\rho_s = 1.29\ kg/m^3`

Gnerally this force per square meter of an aircraft wing is mathematically represented as

`F = (1)/(2) * \rho_s * A * [ u^2 - v^2 ]`

Here u is the speed air need to go over the top surface to create the ideal lift

A is the area of a square meter i.e
`A = 1 \ m^2`

So

`1000 = (1)/(2) * 1.29 * 1 * [ u^2 - 61.1 ^2 ]`

=>
`u = 72.69 \ m/s`