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Fluid Force

  1. Jun 20, 2010 #1
    2. A plate shaped as in the figure [picture.jpg is attached] is submerged vertically in a fluid as indicted. Find the fluid force on the plate if the fluid has weight density 62.4 lb/ft^3

    The integral I set up was the limits being -5 to -1 and the integral being (5-y)(-7/4y - 7/4) dy.


    [Also on this one I was getting help on it yesterday on another thread...but the tutor seems to offline right now, and im hoping someone else can assist me to get this problem done]

    I was told that the (-7/4y - 7/4) is correct but the (5-y) is not.

    After many posts on the last thread I still can't figure out the depth.

    Help?
     

    Attached Files:

  2. jcsd
  3. Jun 20, 2010 #2

    HallsofIvy

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    Actually, it is the other say around- the "5- y" is correct, it is the "(7/4)- (7/4)y" that is incorrect. You want (7/4)(4- y)= 7- (7/4)y.

    Let "y" be the height above the base line of the triangle. Then the depth of water above that line is 5- x. Also, by "similar triangles" we have
    [tex]\frac{x}{7}= \frac{4-y}{4}[/tex]
    where "x" is the width of the triangle at height y.
    (The triangle above height y has height 4- y, width x and is similar to the entire triangle with height 4 and width 7)
    [tex]x= \frac{7}{4}(4- y)= 7- (7/4)y[/tex]

    So a thin rectangle, at height y, with thickness dy, would have area (7- (7/4)y)dx. Taking [itex]\delta[/itex] as the density of water, the weight of water above that rectangle is the area times the height above y (the volume of water above that rectangle) times the density of water: [itex]((7-(7/4)y)dx)(5- y)\delta[/itex]

    Integrate that from y= 0 to y= 4.
     
    Last edited: Jun 20, 2010
  4. Jun 20, 2010 #3
    So wait, What is it that im integrating? (5-y)(7-7/4y) dy

    From the limits [0,4]

    ?
     
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