You are given a triangular trough (base of triangle on top) that is 15 feet long, 2 feet wide, and 3 feet tall.
a. Find the force of the water on one triangular end.
b. Find the work required to pump all the water over the top.
Just a note: This was question that I got wrong on a previous exam. I am doing it over now to study for the final, and don't know the answer, so my work could be correct or incorrect. Please feel free to be critical of any mistakes/errors that I make. I try to explain my reasoning in my attempt at the solution.
Water weighs 62.4 lb/ft^3.
The Attempt at a Solution
For a: I believe I will need area * depth * water weight (62.4 lb/ft^3)
The height is 3 feet, so I made the height of wi to be (3-y). To get the width of wi, I used similar triangles, and ended up with 3 - 2y/3. So, I have the depth (3-y) * the area of wi (3 - 2y/3 dy) * the water weight (62.4).
I integrated from 0 to 3 of 62.4 (3 - y)(3 - 2y/3) dy.
I feel I can use the above integral, and add to it (the area of the thin strip will still be wi * 3-y * the water weight. Since we're using the entire tank, I need to use the 15 feet length, and since this is work required to empty the tank, I feel that I will need to introduce the gravitational constant (9.8 m/s^2).
So, I integrate, from 0 to 3, of 15 * 9.8 * 62.4 * (3-2y/3)(3-y) dy
= 9172.8 * the integral, from 0 to 3, of (3 - 2y/3)(3-y) dy