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Homework Help: Force on the pilot

  1. Oct 15, 2009 #1
    1. The problem statement, all variables and given/known data

    The plane is travelling at a constant speed of 800 ft/s along the curve y = 20(10^-6)x^2 + 5000, where x and y are in feet. If the pilot has a weight of 180 lb, determine the normal and tangential components of the force the seat exerts on the pilot when the plane is at its lowest point.


    2. Relevant equations

    [tex]\sum F = ma[/tex]

    [tex]\rho = \frac{[1 + (dy/dx)^{2}]^{3/2}}{|d^{2}y/dx^{2}|}[/tex]

    3. The attempt at a solution



    [tex]\uparrow\sum F_{n} = ma_{n}[/tex]

    [tex]N - 180lb = (180lb/32.2ft/s^{2})((800ft/s)^{2}/\rho)[/tex]

    [tex]\rightarrow\sum F_{t} = ma_{t}[/tex]

    [tex]0 = (180lb/32.2ft/s^{2})a_{t}[/tex]

    [tex]a_{t} = 0[/tex]

    [tex]\rho = \frac{[1 + (4x10^{-4}x)^{2}]^{3/2}}{|4x10^{-4}|}[/tex]

    x = 0

    [tex]\rho = 2500 ft[/tex]

    [tex]N - 180lb = (180lb/32.2ft/s^{2})((800ft/s)^{2}/2500ft)[/tex]

    [tex]N = 1611.06 lb[/tex]

    Attached Files:

  2. jcsd
  3. Oct 15, 2009 #2

    Andrew Mason

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    Find the algebraic solution and plug in the numbers. Otherwise it is very hard to follow your reasoning.

    The force on the seat is the weight of the pilot + the centripetal force. I think that is what you have shown in your diagram.

    To find the centripetal force you have to find the radius of curvature. Since this is a parabolic curve, you have to find the radius of the circle that approximates the parabola at the vertex. You seem to have worked that out to be 2500 ft. I don't think that is right.

    I think the radius of curvature at the vertex is the distance from the parabola to the focus. The focus is p where x^2 = 4py. Since in this case, 1/4p = 20 x 10^-6 = 1/50,000, the radius, p, should be 50,000/4.

  4. Oct 15, 2009 #3
    thanks for the help.

    I checked the back of the book and the answer is F_n = 323 lb so I looked over my work I made a mistake on the p its 25000 ft not 2500 ft I messed up my derivative of y = 20(10^-6)x^2.
  5. Oct 15, 2009 #4

    Andrew Mason

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    Ok, so the radius of curvature of the circle at the vertex is 2p not p. I stand corrected.

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