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Formula of force

  1. May 26, 2016 #1
    1. The problem statement, all variables and given/known data
    A model boat 1/100 size of its prototype has 0.12N of resistance when stimulating a speed of 5m/s of the prototype , what is the corresponding resistance in the prototype ? water is the fluid in both cases and frictional forces can be neglected.
    Why the author need to transform the force into ρ(L^2)(v^2) ?
    I know the unit of force is kg(m)(s^-2) , so , IMO , F is directly proportional to L only , right . but , not (L^3)
    Fr = Fp / Fm
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    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. May 26, 2016 #2

    haruspex

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    That's for an accelerating mass, the mass being constant. Here, it's the water that is being moved, and the mass of that is related to the cross-sectional area (L2) and the velocity.
     
  4. May 26, 2016 #3
    since we do not know the mass of water ( we only know the density of water) , so we use (rho)(L^3 ) to find the mass of water ?
     
  5. May 26, 2016 #4

    haruspex

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    That would be true if we were considering a cube of water of side L, but L is the characteristic dimension of the model/prototype. How does the mass of water displaced in time t relate to L, t, ρ and v?
     
  6. May 26, 2016 #5
    ρ(L^3) [(L) / (T^-2) ] , which is ρ(L^3)v what are you trying to say ?
     
  7. May 26, 2016 #6

    haruspex

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    No, that's not it. The object moves at speed v. Think of it as a box of side L. What volume of water has to move aside to make way for it in time t?
     
  8. May 26, 2016 #7
    L^3 ???
    i dont really understand what you are trying to say , can you explain further ??
     
  9. May 26, 2016 #8

    haruspex

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    How far does the box move in time t? What volume does the leading face of the box sweep through as the box advances that distance?
     
  10. May 26, 2016 #9
    the box will move by L in time , t , am i right ? the volume that the leading face of the box sweep through as the box advances that distance is (L^3) ?
     
  11. May 27, 2016 #10

    haruspex

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    It is moving at speed v, not L/t.
     
  12. May 27, 2016 #11
    so , what are you trying to say , i didnt get you
     
  13. May 27, 2016 #12

    haruspex

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    If a car length L moves at speed v for time t, how far does it go?
     
  14. May 27, 2016 #13
    car move by vt
     
  15. May 27, 2016 #14
    is it correct ??
     
  16. May 27, 2016 #15

    haruspex

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    Yes, so try answering my post #8 again.
     
  17. May 27, 2016 #16
    distance moved = vt , volume of the leading face of the box sweep through as the box advances that distance is vt + L
     
  18. May 27, 2016 #17

    haruspex

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    How do you get vt+L? Maybe it would help you to think about a more solid analogy. You are digging a tunnel. The excavator is a cube LxLxL. If it advances a distance vt, how much soil does it have to excavate? Draw a picture.
     
  19. May 27, 2016 #18
    Still vt + L
     

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  20. May 27, 2016 #19
    sorry , the volume of the leading face of the box sweep through as the box advances that distance is (L^3) , what are you trying to say ?
     
  21. May 27, 2016 #20

    haruspex

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    No.
    Think about this... if it went for twice as long, 2t, still at speed v, would it still be the same volume?
     
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