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Homework Help: Find the hydrostatic force on one end of the trough

  1. Feb 19, 2007 #1
    I have a quiz coming up that includes this, so thanks for any help.

    1. The problem statement, all variables and given/known data
    A trough is filled with a liquid of density 840 kg/m^3. The ends of the trough are equilateral triangles with sides 8 m long and vertex at the bottom. Find the hydrostatic force on one end of the trough.

    2. Relevant equations

    3. The attempt at a solution
    I'm really not sure how to do this problem except that hydrostatic force = integral of pgAd (p = pressure, g = gravitational constant in SI units, A = area, d = distance). My biggest problem is coming up with the area of the approximating strip, so please explain carefully! Thanks
  2. jcsd
  3. Feb 19, 2007 #2


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    Science Advisor

    You are asked to find the pressure on the triangular end. Lets say that the trough has height h and length of the top edge of the triangle is w.

    There are two ways to approach the problem (essentially the same). One way is to set up a coordinate system so that the vertex of the triangle is at (0,0) and the ends of the top are at (-w/2, h), (w/2,h). It is easy to see that the equation of the line from (0,0) to (w/2,h) is y= (2h/w)x (and, of course, the other line is y= (-2h/w)x). Solving for x, x=(w/2h)y. For any given y, then the length of the horizontal line (the width of an infinitesmal rectangle) would be (w/2h)y- (-w/2h)h= (w/h)y. The area of that infinitesmal rectangle is (w/h)ydy.

    The other method is to use "similar triangles". The triangle formed by the vertex and a horizontal line at height y is, of course, similar to the triangle formed by one end of the trough. In particular, the ratio of the base of each triangle to its height is the same: x/y= (w/h) so x= (w/h)y (x here is twice the "x coordinate" above) and so the area of the infinitesmal rectangle is (w/h)ydy as before.
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