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Homework Help: Engineering Mechanics - Resultant of || forces

  1. Nov 17, 2007 #1
    1. The problem statement, all variables and given/known data
    The 16-ft wing of an airplane is subjected to a lift w/c varies from zero at the tip to 360lb per ft at the fuselage according to [tex]w = 90x^{\frac{1}{2}}[/tex] lb per ft where
    x is measured from the tip. Compute the resultant and its location from the wing tip.

    2. Relevant equations

    3. The attempt at a solution

    Is this what the problem illustrates? This is my own drawing btw
    first the drawing looks like this:

    then i minus 360lb/ft to 90x^(1/2)lb/ft to make the intensity at x = 0 to 0

    then i used a portion of dx and a portion of dR to calculate its area....
    then im gonna use ratio and proportion to compute its resultant force in pounds in my

    The latex cant be edited so regard 360ft - lb as 360ft.lb
    By ratio:
    [tex]\frac{y}{x} = \frac{360ft.lb - 90x^{\frac{1}{2}}lb}{16ft}[/tex]

    [tex] y = \frac{360ft.lb - 90x^{\frac{1}{2}}lb}{16ft} [/tex]

    Since dR = ydx integrate
    [tex]R = \int_{0}^{12} \frac{360ft.lb - 90x^{\frac{1}{2}}lb}{16ft} dx [/tex]
    (from 0 to 16)

    then i integrated it,
    my answer seems to be 576 lb....

    The books answer is R = 3840 lb....
    can u tell me what is wrong with my solution?
    Last edited: Nov 17, 2007
  2. jcsd
  3. Nov 17, 2007 #2

    Shooting Star

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    Homework Helper

    Integral(from 0 to 16) [90*sqrt(x)] = 3840, if you do the integral.

    Can you find the location of the resultant?
  4. Nov 17, 2007 #3
    R . d = [tex] \int_{0}^{16} 90\sqrt{x}(x) [/tex]dx
    3840lb d = 36864lb-ft
    d = 9.6ft from the tip of the wing

    hey thanks alot... one question... how did you find out the resultant is the integral of 90sqrt(x)? that means its in the fuselage?.... and how about the 360ft/lb?
    thank you shooting star
    Last edited: Nov 17, 2007
  5. Nov 17, 2007 #4

    Shooting Star

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    The resultant is the sum of all the parallel forces here. Because the forces are not discrete but continuously varying, you have to integrate over its profile.

    The value at the junction of wing and fuselage is not supposed to be given if the force as a function of distance is given. Perhaps the given 360 lb/ft acts on the fuselage itself, not on where the wing meets the fuselage. It seems a bit extraneous to the problem.
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