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A jet transport with a landing speed of 200 km/h reduces its speed to 60 km/h with a negative thrust R from its jet thrust reversers in a distance of 425 m along the runway with constant deceleration. The total mass of the aircraft is 140 Mg with mass center at G. Compute the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking. At lower speed, aerodynamic forces on the aircraft are small and may be neglected.

Respuesta :

codedmog101

Answer:

257 kN.

Explanation:

So, we are given the following data or parameters or information in the following questions;

=> "A jet transport with a landing speed

= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"

= > The distance = 425 m along the runway with constant deceleration."

=> "The total mass of the aircraft is 140 Mg with mass center at G. "

We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"

Step one: determine the acceleration;

=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.

=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.

Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).

= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).

= 257 kN.

batolisis

The reaction N under the nose wheel B towards the end of the braking interval =  257 kN

Given data :

Landing speed of Jet = 200 km/h

Distance = 425 m

Total mass of aircraft = 140 Mg  with mass center at G

Determine the reaction N under the nose of wheel B

  • First step : calculate the value of the Jet acceleration

  Jet acceleration = 1 / (2 *425) * (200²  - 60² ) *  1 / (3.6)²

                              = 3.3 m/s²

  • Next step : determine the reaction N under the nose of Wheel

Reaction N = Total mass of aircraft * jet acceleration* 1.2 = 15N - (9.8*2.4* 140).   ----- ( 1 )

∴ Reaction N = 140 * 3.3 * 1.2 = 15 N - ( 9.8*2.4* 140 )  

 Hence Reaction N = 257 KN

                     

We can conclude that the The reaction N under the nose wheel B towards the end of the braking interval =  257 kN

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