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Fluid Force problem

  1. Jan 30, 2007 #1
    I would appreciate any help I can get w/the following problem:
    A square plate of side 'a' ft is dipped in a liquid of weight density 'p' lb/ft^3. Find the fluid force on the plate if a vertex is at the surface and a diagonal is perpendicular to the surface.

    I understand the surface has a diamond shape--I have tried dividing it into two triangles and finding the fluid force for each and adding them but have been unable to obtain the textbook's answer.

    The two integrals I get are:
    fluid force 1 = integral (limits: a*sqrt2/2 and 0) p * 2h^2 * dh
    fluid force 2 = integral (limits: a*sqrt2 and a*sqrt2/2) p * ( h + (a*sqrt2)/2) * (2h) * dh

    The book's answer is (a^3*p)/sqrt2.

  2. jcsd
  3. Jan 31, 2007 #2


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    Since this is clearly homework, I am moving it.

    Yes, dividing into two triangles is probably the best way to go. If we take a coordinate system with (0,0) at the upper corner, positive y-axis along the vertical diagonal, then the sides of the upper triangle are given by y= x (x> 0) and y= -x (x< 0). The horizontal diagonal is at y= [itex]a\sqrt{2}/2[/itex], half the length of the diagonal. A thin (thickness dy) horizontal rectangle, at given y will have length y-(-y)= 2y and so area 2ydy. The depth is y so the pressure at that depth is py. The total force on that thin rectangle is py2dy. Integrate that from y= 0 to [itex]a\sqrt{2}/2[/itex] to get the total force on the upper triangle.

    The lower triangle can be done in a similar way. Now, let the origin, (0, 0) be at the lower vertex, the positive y axis upward. The calculation for the area is exactly the same: 2ydy. However, the depth of that vertex at (0,0) is now [itex]a\sqrt{2}[/itex] and the depth of the horizontal line at height y above the vertex is [itex]a\sqrt{2}-y[/itex] so you must integrate [itex]2py(a\sqrt{2}-y)[/itex] from 0 to [itex]a\sqrt{2}/2[/itex].
    Hmmm- if you are clever, the hard part of those two integrals will cancel!
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