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14 September, 22:05

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?

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  1. 14 September, 22:59
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    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°
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