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Differential equation for grav, boyant and drag force

  1. Oct 14, 2015 #1
    1. The problem statement, all variables and given/known data
    So there is a falling object, you have to take into account the boyant force, the pull of gravity and the drag force
    A time dependent distance equation is what we're looking for

    2. Relevant equations
    Fd=CdApav2/2
    Where
    Fd is the drag force
    Cd is the drag coefficient
    A is the area exposed to the fall
    pa the air density
    v the immediate velocity

    Fb=mgpa/pc
    Fb is the boyant force
    mg is the weight of the object
    pc is the object's density
    Note ( this equation is found from the original equation Fb=Volume submerged x air density x gravity; where the submerged volume is m/pc)

    Fg=mg
    3. The attempt at a solution
    ma=mg - mgpa/pc - CdApav2/2
    Or
    dv/dt = A - Bv2
    Where
    A=g( 1 - pa/pc)
    B=CdApa/2m

    I solve for v

    v (t) = (A/B)1/2 (1 + C1e-2(BA)1/2t)/( 1 - C1e-2(BA)1/2t)

    Where C1 is some arbitrary constant

    Integrating we get the distance formula:

    X (t) = (A/B)1/2t+(1/B)ln( 1 - C1e-2 (BA)1/2t) + C2
    I don't know wether it's correct or not. I've used techniques i found on the internet for the integration. http://www.freemathhelp.com/forum/threads/47073-integral-(1-(e-x-1))-dx-Using-Partial-Fractions
     
  2. jcsd
  3. Oct 15, 2015 #2

    Geofleur

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    Science Advisor
    Gold Member

    See if it gives the right answers in special cases. For example, what does your result say for the case of no buoyant force and no drag force? What happens as ## t \rightarrow \infty ##, and what would you expect physically in this case? There are lots of things like this to look into.

    Oh, and can you check that the resulting equation is dimensionally correct?
     
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