An airplane is flying in a horizontal circle at a speed of 480 km/h (Fig. 6-42). If its wings are tilted at angle θ = 40° to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an “aerodynamic lift” that is perpendicular to the wing surface.
Fnet = ma
a = v^2 / R
The Attempt at a Solution
Converted g[9.8 m/s^2] to km/h^2 to get 35.28 km/h^2
I designed two free-body diagrams: one for the plane's angle [40 degrees] and for the perpendicular "aerodynamic lift" [50 degrees] on a normal cartesian x,y coordinate.
For the free body diagram for the plane [40 degrees] I got: [x is noted as theta]
x: -FnCosx = ma
y: FnSinx - Fg = 0
Solved for Fn in the y and got Fn = Fg/Sinx and substituted Fn in x to get:
(mg/sinx)cosx = m(v^2/R)
Through algebra I got R = [(V^2)(Tan(40))] /g.
The answer was far too large and far from the answer.