# Work and Fluid Force, lifting water out of a triangular prism tank

1. Feb 20, 2011

### schlynn

1. The problem statement, all variables and given/known data
A vertical cross section of a tank is shown. Assume the tank is 16 feet long and full of water. ($$\delta=62.4$$, 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.

2. Relevant equations
Not sure that there are any.

3. 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=$$\frac{5}{2}-\frac{5}{16}y$$. 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$$\delta$$ factor too, but you just bring that outside of your integral. After I simply the integrand and anti differentiate I got $$40y-\frac{15}{4}y^{2}+\frac{5}{48}y^{3}$$ evaluated from 0 to 8, again, with the factor of 16$$\delta$$. 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?

Last edited: Feb 20, 2011
2. Feb 21, 2011

### lanedance

how about using conservation of energy and consider the centre of mass's

3. Feb 21, 2011

### schlynn

Because this is a calculus 2 class, not a physics class, I don't know how to do it that way. I know how to find the centroid of an area that has uniform density, but that's not how we are supposed to do it.

4. Feb 21, 2011

### lanedance

ok well i would find w(h)

then infinetsimal vol element
dV = w(h).L.dh

think of the work dW required to get this infinitesimal element to teh reuired hieght and then integrate over h

its all the same thing though

Last edited: Feb 21, 2011
5. Feb 21, 2011

### schlynn

That's what I'm having trouble with. I can't find the infitesimal volume, I can't get the width as a function of the height. Could you walk me through it?

6. Feb 21, 2011

### lanedance

changed notation in last post

ok so you know
w(0) = 5
w(8) = 0

and as its a triangle, its width will vary linearly in between...

so basically you have two points (0,5) and (8,0) find the equation of the line that connects them, and that will be w(h)