Solve for Work in Triangular Vat (H:3m, B:4m, L:6m, Water H:2m)

[darthxepher]

Ugh. Ok so I have been working on this problem for 3 days and haven't gotten anywhere wiht it. Ok so you have a triangular vat. the height of each triangle on the ends is 3 m and the base of each triangle is 4 m. The length of the vat is 6 m. The water inside the vat is 2 m high. Now I need to find out the work needed in order to push the 2m's of water out. I come up wiht all these different set ups of integrals but htey don't work out in the end. Can someone come up with one?

 

[Defennder]

Ok, so maybe I have a poor imagination, but I'm not able to visualise what the vat looks like. Is there a picture of it?

 

[HallsofIvy]

Sounds to me like a trough with triangular ends.

Are we to presume that the vertices of the triangles are up or down? And are we to presume that the triangles are isosceles?

Assuming the triangles are isosceles with vertices down, start by drawing a picture of a triangular end. Draw a horizontal line across the triangle at height x. That represents one "layer" of water to be lifted at height 3- x to the top of the trough. Work= force times distance so the work done to lift that "layer" is its weight times 3- x. Weight is density times volume. Density is a constant but the volume depends upon x. Each "layer" a rectangle with length 6 m and width equal to the length of that horizontal line (call that w for now). We can take the thickness of the "layer" to be dx. Now the volume of that "layer" is 6wdx and its weight is 6\delta w dx where \delta is the density of water. The work done in lifting that "layer" to the top of the trough is 6\delta w (3- x)dx. The work done in lifting all of the water in the trough out of the trough is
6\delta\int_0^2 (3-x)w dx[/itex]<br /> <br /> Now, to find w as a function of x, use &quot;similar triangles&quot;. The triangle of water has height x and base width w and is the same shape as the entire end of the trough which has height 3 and base width 4. w/x= 4/3.