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Mechanics problem involving a Differential Equation

  1. Sep 23, 2011 #1
    1. The problem statement, all variables and given/known data

    Fnet = Ffriction + Fairresistance = μmg + cv2 = ma

    Essentially I just need to get this in a v(x) form. Prof said that we need to go about it by doing this:

    F = m[itex]\ddot{x}[/itex] = m [itex]\dot{x}[/itex] * d[itex]\dot{x}[/itex]/dx

    and then integrating to get v(x)

    3. The attempt at a solution

    so...

    μmg + cv2 = m [itex]\dot{x}[/itex] * d[itex]\dot{x}[/itex]/dx

    And this is where I just can't figure it out. I'm terrible with DEs.

    I thought of moving the 1/dx to the other side and then integrating both sides, but I get stuck, especially since v2 can be written as [itex]\dot{x}[/itex]2, right? And that messes with my sleep-deprived brain a little, lol. Any nudge in the right direction would be greatly appreciated!
     
  2. jcsd
  3. Sep 23, 2011 #2
    Hi!
    If I recall correctly, the v^2 proportionality law(of drag) applies to high speeds, only; Are you sure that you need to have v^2 there? Stoke's law, in this case, I suppose, is a better fit(i.e k*v);
    How did you arrive at the following relation:?[itex] {\frac{d}{dx}\dot{x}} = 0 [/itex], I can't see its use here...
    We'll work on it,
    Daniel
     
  4. Sep 23, 2011 #3
    The question unfortunately doesn't give much of a hint as to how fast this might be traveling, so I'm not sure I could cancel out the v2. I also can't see where I typed d[itex]\dot{x}[/itex]/dx = 0. :S
     
  5. Sep 23, 2011 #4
    It's not so much that you typed it, it's what it is... :)
    Meaning:
    [itex] \frac{d}{dx}{dx}{dt} = \frac{d}{dt}\cdot\frac{dx}{dx} = 0 [/itex] Meaning that it can't be used...
    The first thing to do, in my view would be to rewrite it, such that:
    [itex] \Large mx''(t) = mk(x'(t))^2 + C [/itex] Where C is a constant carrying your constant friction force.
    Then one would lean towards differentiating again(to get rid of the C) with respect to t:
    [itex] \large mx'''(t) = 2mkx'(t)x''(t) [/itex] and dividing by m, and substituting z(t) = x'(t), and for simplicity, taking k=1:
    [itex] \large z'' = zz', \frac{d^2z}{dt^2} = z\frac{dz}{dt} [/itex].
    Multiplying by dt:
    [itex] \large zdz = (\frac{d^2z}{dt^2})dt [/itex].
    Now one must integrate, and continue the process further.
    Try it,
    Hope that works for you,
    Daniel
    P.S
    Here's some "cheating"(:D), just to verify your steps:
    http://www.wolframalpha.com/input/?i=-x''+==+x*x'"
     
    Last edited by a moderator: Apr 26, 2017
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