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I'm having difficulty understanding how to find Hydrostatic Force on various shapes. One problem is a right triangle with a height of 3ft and a width of 7ft. Its submerged 1ft below the surface with the 7ft side facing the surface. What do I do?

I understand the idea. My problem is more about setting up the integral.

so I don't need to take the integral? There is no thickness. the way we've been doing this in class is that dF=pgdA where F=force on the triangular plate, pg=62.5 and dA=is the change in area that somehow includes dx OR dy.

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Did my answer help? Please provide some feedback, either by clicking the checkmark if it worked for you, or by expanding your question if not. – Ross Millikan Feb 23 '11 at 4:58
up vote 3 down vote accepted

The force is simply the weight of the water displaced. You should have a theorem that proves that. So integrate to find the volume and multiply by the density of water. But if you have a triangle, the volume is zero unless it has some thickness. Then it's easy.

Added: if it is completely submerged, the force is simply the volume of the shape. For a right triangle with legs of 3 and 7, the area is 21/2 and the volume would be $\frac{21t}{2}$, where $t$ is the thickness. Then the force would be $\frac{21t}{2}62.4$ pounds upward.

Added2: If there is no thickness, you are considering everything to have unit thickness. The 62.5 is the density of water in lb/ft^3 (though it is really 62.43 max unless you have salt water). The reason you don't need an integral is that you said it was submerged 1 foot below the surface. It's like if you asked in integral calculus for the area under $y=1-x$ from 0 to 2. If the upper limit were below 1, I might do an integral, but with limit above 1 I would just find the area of the triangle.

But maybe you meant the lower corner is 1 foot down, so the triangle is partially submerged. You still need the area of what is in the water and your expression integrating $dA$ is to do exactly that. You can either use similar triangles to see that the submerged shape has height 1 foot and width $\frac{7}{3}$ feet for an area of $\frac{7}{6}$ foot, or integate $\int_0^1\frac{7}{3}xdx$ get the same as the volume element is $dx$ high and $\frac{7}{3}x$ wide (determined by the same similar triangles I used in the first).

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