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Resistive force in fluids

  1. Dec 8, 2007 #1
    heheh...I need some help understanding something. If there was an object, such as a sphere, in water, gravity would be pushing it downwards, while a resistive force R = -bv would be pushing it upward (b as a constant).

    that would imply that the net Force Fy would be
    Fy = mg - bv = ma = m(dv/dt)

    dv/dt = g - (b/m)v

    How do I come up with

    v = (mg/b)(1 - e^(-bt/m))
     
  2. jcsd
  3. Dec 8, 2007 #2

    Shooting Star

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    Have you learned integration? That's how you get the final eqn.

    The force R is a resistive force, acting against the direction of motion, and not necessarily pushing it up. But if the body is falling vertically downward, of course it is acting upward.
     
  4. Dec 8, 2007 #3
    I see...I've learned integration, but I'm still just tipping it.

    I was just thinking that using dv/dt = g - (b/m)v could be rearranged for v = (mg/b)(1-(a/g)), in which dv/dt = acceleration = a. After that, I would somehow have to state that (a/g) = e^(-bt/m)....which I apparently didn't do.

    Actually...nevermind...I get whats being said. The terminal velocity (when the net force is 0N) is only approached, not touched, in which the terminal velocity = (mg/b). So, the equation changes a bit.
     
    Last edited: Dec 8, 2007
  5. Dec 8, 2007 #4

    Shooting Star

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    I'm glad you got the essence of it. In practice, the actual velocity gets indistinguishably close to the terminal velocity within a very short time, depending, of course on b/m. The higher this ratio is, the faster it happens.
     
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