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Fluids - bringing together pressure, depth and friction

  1. Dec 4, 2005 #1
    Fluids -- bringing together pressure, depth and friction

    Hi Again!

    More troubles with fluids.

    I was pretty sure I had this one.

    There is a
    freshwater pond
    density, D=1000 kg/m^3
    depth = 15.1 m
    one of the sides is blocked off by a cliff.
    A nearly horizontal tube
    diameter = 3.17 cm
    depth below the pond's surface is 6.2 m
    eroded to the other side of the cliff
    a rock blocks cuts off the flow of the water.

    The question asks for the frictional force between the tube's wall and the rock blocking the exit.

    I figured it would be fairly easy, since
    F = P*A
    and we can find out the hydrostatic pressure.
    so
    P = Patm + D*g*h
    P = 101300 + 1000*9.8*6.2
    P = 162060 Pa

    A = pi*r^2 = pi* (0.01585)^2 = 7.89e-4 m^2

    and therefore,

    F = P*A = 127.9 N is the force on the rock blocking the exit

    because there is no acceleration, the system is at equilibrium

    so Fnet = F(ontherock) - F(friction) = 0
    so F(on the rock) = F(friction)
    so then my answer would be 127.9 N

    Could you point out where I've made an error, and push me in the right direction?
     
  2. jcsd
  3. Dec 5, 2005 #2

    Tide

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    Homework Helper

    I think you didn't take into account the pressure of the air pushing back on the rock from the other side!
     
  4. Dec 5, 2005 #3
    LOL i just posted a question EXACTLY like this. Anyways, I understand arctans arguement, but I dont know where you would add the pressure pushing back from the other side . . .
     
  5. Dec 5, 2005 #4

    ahh good point!

    :biggrin:

    So It'd be:

    Fnet = F(airpushingonrock) + F(friction) - F(water) = 0

    Am I correct?
     
  6. Dec 5, 2005 #5

    Tide

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    Science Advisor
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    That looks fine - just be careful with the signs of the forces.
     
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