A jet airplane is in level flight. The mass of the airplane is m = 8970 kg . m=8970 kg. The airplane travels at a constant speed around a circular path of radius R = 8.45 mi R=8.45 mi and makes one revolution every T = 0.112 h . T=0.112 h. Given that the lift force acts perpendicularly upward from the plane defined by the wings, what is the magnitude of the lift force acting on the airplane?

F=92838 N

θ = 18.5°

Explanation:

F: lift force

θ: banking angle of the airplane, clockwise from vertical

W: weight of the airplane

v: velocity of the airplane

m = 8970 kg

R = 8.45 miles = 13598.96 meters

T = 0.112 hours = 403.2 seconds

g = 9.81 m/s^2

Fy = F*cos(θ)

Fx = F*sin(θ)

W = m*g

v = 2*pi*R/T    (velocity = distance/time) =2*pi*13598.96m/403.2 sec

v=211.8 m/s

As the airplane is in level flight,

ΣFy = 0 = Fy - W

> Fy = m*g =(8970*9.81)N

   Fy=87995.7

F_cent = m*v^2/R = Fx =8970 kg*211.8 m/s^2 /13598.96m

Fx = m*(2*pi*R/T)^2 / R = 29592N

> Fx = 4*pi^2*m*R/(T^2)

The magnitude of the lift force is

F = sqrt(Fx^2 + Fy^2)

F=sqrt(29592 N^2+87995.7^2)

>> F = 92838 N

The angle of the banking will be

θ = atan(Fx/Fy)

θ=atan(29592N/87995.7N)

>> θ = 18.5°

the answer is 12 as i may remember

: )