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

In summary: So the work done in lifting all of the water in the trough out of the trough is [tex]6\delta\int_0^2 (3-x)w dx[/itex]In summary, the water in a triangular vat must be lifted out using work done on "layers" of water.
  • #1
darthxepher
56
0
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?
 
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  • #2
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?
 
  • #3
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 [itex]6\delta w dx[/itex] where [itex]\delta[/itex] is the density of water. The work done in lifting that "layer" to the top of the trough is [itex]6\delta w (3- x)dx[/itex]. The work done in lifting all of the water in the trough out of the trough is
[tex]6\delta\int_0^2 (3-x)w dx[/itex]

Now, to find w as a function of x, use "similar triangles". 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.
 

1. What is the formula for finding the volume of a triangular vat?

The formula for finding the volume of a triangular vat is V = (1/2) x b x h x l, where b represents the base length, h represents the height, and l represents the length of the vat.

2. How do I solve for the volume of a triangular vat with specific measurements?

To solve for the volume of a triangular vat with specific measurements, plug the given values into the formula V = (1/2) x b x h x l. In this case, b = 4m, h = 3m, and l = 6m. So, V = (1/2) x 4m x 3m x 6m = 36m³.

3. What is the purpose of finding the volume of a triangular vat?

The purpose of finding the volume of a triangular vat is to determine the amount of space inside the vat. This information is important for various applications, such as filling the vat with a specific amount of liquid or solid materials.

4. How does the height of the water affect the volume of the triangular vat?

The height of the water does not affect the volume of the triangular vat. The volume of the vat is determined by the dimensions of the vat itself, not the amount of liquid inside.

5. Can this formula be used for all types of triangular vats?

Yes, this formula can be used for all types of triangular vats as long as the measurements of the base, height, and length are known. It is a general formula for calculating the volume of any triangular prism.

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