Question

A jet airplane is in level flight. The mass of the airplane is m=9010kg. The airplane travels at a constant speed around a circular path of radius R=9.77mi and makes one revolution every T=0.129h. Given that the lift force acts perpendicular upward from the plane defined by the wings, what is the magnitude of the lift force acting on the airplane? At what angle is the airplane banked?

Answer 1

Answer:

The magnitude of the lift force L = 92.12 kN

The required angle is ≅ 16.35°

Explanation:

From the given information:

mass of the airplane = 9010 kg

radius of the airplane R = 9.77 mi

period T = 0.129 hours = (0.129 × 3600) secs

= 464.4 secs

The angular speed can be determined by using the expression:

ω = 2π / T

ω = 2 π/ 464.4

ω = 0.01353 rad/sec

The direction $\theta = tan^{-1} ( \dfrac{\omega ^2 R}{g})$

$\theta = tan^{-1} ( \dfrac{0.01353 ^2 \times (9.77\times 1609)}{9.81})$

θ = 16.35°

The magnitude of the lift force  L = mg ÷ Cos(θ)

L = (9010 × 9.81) ÷ Cos(16.35)

L = 88388.1  ÷ 0.9596

L = 92109.32 N

L = 92.12 kN