Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Airplane Flying

  1. Mar 18, 2009 #1
    1. The problem statement, all variables and given/known data

    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.


    2. Relevant equations

    Fnet = ma

    a = v^2 / R

    3. 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.
  2. jcsd
  3. Mar 18, 2009 #2


    User Avatar
    Homework Helper

    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.
  4. Mar 18, 2009 #3
    x: -FnSin(theta) = m(-a)
    y: FnCos(theta) - mg = 0 ?
  5. Mar 18, 2009 #4
    This is how my free body diagram looks like:

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

  6. Mar 18, 2009 #5


    User Avatar
    Homework Helper

    Yes, agree with
    x: -FnSin(theta) = m(-a)
    y: FnCos(theta) - mg = 0 ?
    where theta is 40 degrees.
  7. Mar 19, 2009 #6
    thanks a bunch!

    * I just messed up on the conversion of my g from m/s^2 to km/h^2 silly me
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook