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Physics and fluids

  1. Sep 6, 2008 #1
    1. The problem statement, all variables and given/known data

    Very small particles moving in fluids are known to experience a drag force proportional to speed. Consider a particle of net weight W dropped in a fluid. The particle experiences a drag force, Fd = kV, where V is the particle speed. Determine the time required for the particle to accelerate from rest to 95% of its terminal velocity, Vt, in terms of k, W, and g.

    2. Relevant equations

    Newtons Second Law of motion, etc.

    3. The attempt at a solution

    I tried to sum forces, etc. but didn't really get anywhere...
     
  2. jcsd
  3. Sep 6, 2008 #2

    Doc Al

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    What forces act on the particle?
     
  4. Sep 6, 2008 #3
    i wrote the question verbatim, so i will assume a drag force (and a force of flowing fluid?)
    i can write the answer if it helps, but it's useless without the method.
     
  5. Sep 6, 2008 #4

    Doc Al

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    Right. An expression for that is given.
    That's the drag force.

    What other force, also given, acts on the particle?
     
  6. Sep 6, 2008 #5
    there is a force on the particle moving it forward and the drag force
     
  7. Sep 6, 2008 #6

    Doc Al

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    Yes. What is that force?
     
  8. Sep 6, 2008 #7
    i dont know.
     
  9. Sep 6, 2008 #8

    Doc Al

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    Hint: It's one of the variables that your answer must be expressed in terms of. :wink:
     
  10. Sep 6, 2008 #9
    gravity.
     
  11. Sep 6, 2008 #10

    Doc Al

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    Of course! Now write an equation using Newton's 2nd law.
     
  12. Sep 6, 2008 #11
    so am i assuming a vertical pipe with fluid in it?

    kV-mg=ma. we don't want it in terms of acceleration so we use a=dV/dt.
     
  13. Sep 6, 2008 #12

    Doc Al

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    It's just a particle placed in some fluid and allowed to fall.
    Good. I would switch the signs around, so that "down" is positive (since you know it's going to fall down).
     
  14. Sep 6, 2008 #13
    mg-kV=m dV/dt. the net weight is W, which is also mg.
    W-kV=m dV/dt
     
  15. Sep 6, 2008 #14

    Doc Al

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    Good. Now just rearrange and integrate.
     
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