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schlynn
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Homework Statement
A vertical cross section of a tank is shown. Assume the tank is 16 feet long and full of water. ([tex]\delta=62.4[/tex], and that the water is to be pumped to a height of 8 feet above the top of the tank. Find the work done in emptying the tank. The tank is a triangular prisim with base=5ft and height=8ft.
Homework Equations
Not sure that there are any.
The Attempt at a Solution
First, I am supposed to find the force of the water. It says to find the width as a function of the height, and the book is unclear how to do this very well. From what I can gather it's the height divided by half the base equals the distance to pump the water divided by W, W being the width. So I solved and got W=[tex]\frac{5}{2}-\frac{5}{16}y[/tex]. And then since work is force times height, you just multiply that by the distance it has to be lifted, that's 16-y. And that's your integrad, but you have a 16[tex]\delta[/tex] factor too, but you just bring that outside of your integral. After I simply the integrand and anti differentiate I got [tex]40y-\frac{15}{4}y^{2}+\frac{5}{48}y^{3}[/tex] evaluated from 0 to 8, again, with the factor of 16[tex]\delta[/tex]. Fundamental theorm it and I got 132787 rounded to the nearest whole number. The answer is wrong, and I'm fairly certain I know how to do all of this except finding the width as a function of the height. The book says it has to do with similar triangles. But I don't get what they are saying. Can someone shed some light on this for me?
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