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Simple pressure problem

  1. Nov 16, 2007 #1
    1. The problem statement, all variables and given/known data
    A spherical diving bell containing a camera is in the ocean at a depth of 147 m. It has a flat, transparent, circular port with a diameter of 12.7 cm. Find the total force on the port (use ρsea water = 1025 kg/m3).


    2. Relevant equations
    p = po + pgd
    P = F/A
    F = pA


    3. The attempt at a solution
    So do I find the total pressure and then multiply it be the area? I get a really big number (2.00 x 10 ^ 8) which doesn't seem right?
     
  2. jcsd
  3. Nov 16, 2007 #2

    mgb_phys

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  4. Nov 16, 2007 #3
    No I don't think so cause it is worded differently and in our class they are all worded the same just different numbers...but thats weird its almost identical!
     
  5. Nov 16, 2007 #4
    yea that answer isn't right...I just tried it :(
     
  6. Nov 16, 2007 #5
    does anyone know how to do this?
     
  7. Nov 16, 2007 #6

    mgb_phys

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    Sorry - I thought your comment meant you had solved it!
    The area of the window is A = pi r^2
    There is a column of water on top of it 147m high
    So the volume of water sitting on the window is V = pi r^2 * depth
    Add the density of water and you get the mass of water on the window.

    V = pi * (0.127/2)^2 * 147 = 1.89m^3
    Density is 1025 kg/m^3 so mass of water is 1900kg
    Total force is F = mass * g = 18.7KN

    Note there is also an extra 1 atmosphere of pressure pushing down on the surface of the water, but there is also an atmosphere of pressure inside the bell pushing out which cancel out.
     
  8. Nov 16, 2007 #7
    thats ok...but its telling me thats incorrect?
     
  9. Nov 16, 2007 #8
    this is the msg it gave me...
    Calculate the total pressure at the given depth by adding up the atmospheric pressure (1.01e5 N/m2) plus the pressure due to the sea water. This assumes that there is no air in the bell, (probably a bad assumption).
     
  10. Nov 16, 2007 #9

    mgb_phys

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    Alternate method - the weight of the column of water = m g h = 1.47MN
    Area A= pi r^2 = 0.0127m^2
    Force = pressure * area = 1.47MN/m^2 * 0.0127m^2 = 18.67N

    EDIT - then just add atmospheric pressure to the water pressure, answer should be 20KN.
     
    Last edited: Nov 16, 2007
  11. Nov 16, 2007 #10
    I'm having deja vu and amnesia at the same time....I think I've forgotten this once before.
     
  12. Nov 16, 2007 #11
    LoL...
     
  13. Nov 16, 2007 #12
    I don't understand I enter 0.02 N (it doesn't recognize units like KN or MN) and it says incorrect??
     
  14. Nov 16, 2007 #13

    mgb_phys

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    20 KiloNewtons ie 20,000N or 2E4 N
     
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