# Find the work required to pump all the water

1. Feb 10, 2009

### nameVoid

the ends of an 8 foot long water trough are equilateral triangles having sides of length 2 feet. If the trough is full of water, find the work required to pump all the water over the top.

2. Feb 10, 2009

### nameVoid

Re: Work

possible trapazoid section? area?

3. Feb 10, 2009

### HallsofIvy

Staff Emeritus
Re: Work

I presume the trough has the vertex of the triangle down! Don't just try to remember formulas- think about what work means. The work done to lift a "piece of water" out of the trough is the weight of the piece times the height it must be lifted: the distance from the top base of the triangle to the piece. We can handle all "pieces of water" that have that same distance at once by thinking of a "layer of water" at a given distance from the top base. That will be a rectanglar layer having length 8 feet, depth dx, and width w, depending on what x is. That is, its volume will be 8w(x)dx and so its weight will be $\gamma (8w(x)dx)$ where $\gamma$ is the density of water (look it up) in pounds per cubic foot. That layer has to be lifted x feet to the top of the trough so the work done is $8\gamma w(x)x dx$. Integrate that from 0 to the height of the triangle.

Now, determining w(x). Draw a picture: an equilateral triangle with vertex at the bottom, base at the height and a horizontal line x units below the base. The length of that line is w(x). "Similar triangles" is a good way to find its length. Another way would be to find the equations of the two lines, both through (0, 0) and one through $(1, \sqrt{3})$, the other through $(-1, \sqrt{3})$. Be careful what you call "x" and what "w". Do you see where I got the $\sqrt{3}$?