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Hydrostatic force problem

  1. Nov 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the hydrostatic force against one side of the plate by Reimann sum. Then express the force as an integral and evaluate it.


    2. Relevant equations
    F=pg*Ad
    pg do not need to be solved. they can be left like that.
    F=force
    p= density
    g=gravity
    a= area
    d= distance

    3. The attempt at a solution
    so far i have the equation
    in the attachment. However, I do not know where to put the 1 ft that lies above the triangle.
    i have the area as (4/3)(3-y)
     

    Attached Files:

    Last edited: Nov 15, 2011
  2. jcsd
  3. Nov 15, 2011 #2
    That one foot needs to be depth as a function of y. For instance, if you put the top of the triangle as y=0, increasing positively downward, then d(y) = (y+1), because at y=0, the depth will then be 1, as expected. At y=3, the depth will be 4, and so on.

    You are really computing the volume of water above a particular area (more specifically, the force it causes), as a function of depth more or less. To solve these problems you can always think "How can I express depth/length/etc as a function of my independent variable"?
     
  4. Nov 15, 2011 #3
    would this integral be correct? i can never understand the whole concept about which way y is
     

    Attached Files:

  5. Nov 15, 2011 #4
    Where are you calling zero? Can you label your axis.

    Just think of it like this:
    [tex]F_{H} = δ \int_{a}^{b}L(x) D(x) dx[/tex]
     
  6. Nov 15, 2011 #5
    y= 0 starts at the very bottom of the triangle for this integral.
     
  7. Nov 15, 2011 #6
    Okay, so you have (4-y) for your D(y), when y = 1, is the depth at that point 3? Can you test the length function L(y) in the same manner against the diagram?
     
  8. Nov 15, 2011 #7
    if by L(x) you are talking about the area that needs to be integrated, then the area is (4/3)(3-y) since y is that unknown length above the dy section.
     
  9. Nov 15, 2011 #8
    Yes, the area. L(y) dy is the area of the strip/rectangle. Using your axis, where y=0 is at the bottom of the triangle. When y increases (going closer to the surface), does your function 4/3(3-y) represent the length of the strip at that point? When y = 0, that L(y) function evaluates to 4/3(3-0) = 4, does that sound right to you?
     
  10. Nov 15, 2011 #9
    no. it doesn't sound right.
    ok i changed the area formula to A(y)= (4/3)y
    now i think it's correct. thank you
     
  11. Nov 15, 2011 #10
    That sounds right!

    Usually, an easier way to do these problems is to put y=0 at the surface of the water, that way your depth function D(x) is just x. Then you simply need to find a way to get out the Length at some particular x value. This can greatly simplify your integrals in most cases that I have seen. If the object is circular, putting the y=0 at the center of the circle can do the same thing.
     
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