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Center of pressure calculation

  1. Apr 8, 2015 #1
    1. The problem statement:

    A removable plate is mounted in the end of a swimming pool. The plate is 60cm square with the top edge 30cm below surface of water. Compute the force on plate and distance to centre of pressure below surface of water.

    2. Relevant equations.
    ImageUploadedByPhysics Forums1428511297.905568.jpg

    3. The attempt at solution :

    ImageUploadedByPhysics Forums1428511756.434984.jpg

    Spent a lot of research and time on cop to no avail, so this is my last resort!

    Kind regards ,

  2. jcsd
  3. Apr 8, 2015 #2
    Just I've Notices that you've written F = pgh, In fact it's the change in pressure that equals pgh, I don't know,maybe you can the force out of that!
    Try to modelize your situation by defining the limits ;)
    Hope that helps :)
  4. Apr 8, 2015 #3


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    Your calculation for yp is OK except for one detail. y-bar for the plate must be measured from the surface of the water, not the top edge of the plate.

    Remember, since the pressure increases with depth, the center of pressure must be lower than the centroid of the plate. :wink:
  5. Apr 8, 2015 #4
    Thanks , still can't get the answer though!
    Tried using y bar as 0.9 and 0.45.
    Do I use the water line as the reference line and measure y bar up to...where? Centroid, the bottom plate part, see what I mean?
    And Yep, my badly drawn diagram shows COP lower than centroid :)
  6. Apr 8, 2015 #5
    Solved ! I see what you now mean, y bar is 0.6, distance between centroid and water line, that makes more sense, thank you!!!

    Just lastly if possible, what is that last equation in my original first post, with the integral sign? A more complex version of the Yp formula I used initially? Thanks!
  7. Apr 10, 2015 #6
    I think, in order to calculate the centre of pressure you can use the same formula as for the centre of mass(you can simply look it up in wikipedia https://en.wikipedia.org/wiki/Center_of_mass). Just replace the total mass with the total force and the volume density with the pressure and integrate over the whole square.
  8. Jan 18, 2018 #7
    Its simple, use your x bar to be 0.6 I.e distance from centre of gravity to free surface,that way you would get the answer 0.65
  9. Jan 18, 2018 #8
    And that's what you rightly denoted as y bar.......
    So,y bar= 0.3 +0.3
    Which is the addition of the distance of centre of gravity and the distance of the square from the free surface....
  10. Jan 18, 2018 #9
    You realize that this thread is almost 3 years old, right? The OP has not been seen for almost 3 years.

    Anyway, as SteamKing correctly pointed out in post #3, the center of pressure is below the center of mass of the plate.

    This thread is hereby closed.
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