Airplane Flying

  • #1

Homework Statement



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.

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c06/fig06_39.gif

Homework Equations



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.
 

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
10
Interesting; I didn't think there was enough information but it did work out.
I think you have mixed up the angle. In the y direction, Fn*cos(theta) - mg = 0.
I converted to meters and seconds - much easier! Got a little over 2 km for the radius.
 
  • #3
So
x: -FnSin(theta) = m(-a)
y: FnCos(theta) - mg = 0 ?
 
  • #4
This is how my free body diagram looks like:

Where theta is the angle of the plane [40 degrees]

l_eb18de4951a8477dbbc9fd693c0d498e.jpg
 
  • #5
Delphi51
Homework Helper
3,407
10
plane.jpg

Yes, agree with
x: -FnSin(theta) = m(-a)
y: FnCos(theta) - mg = 0 ?
where theta is 40 degrees.
 
  • #6
thanks a bunch!

* I just messed up on the conversion of my g from m/s^2 to km/h^2 silly me
 

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