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Hydr. force on a triang. tank (force one end & pumping out)

  • Thread starter leo255
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  • #1
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Homework Statement


[/B]
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.

Homework Equations



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.

For b:

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
 

Answers and Replies

  • #2
SteamKing
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Homework Statement


[/B]
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.

Homework Equations



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.
But what's your final answer?

For b:

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).
You can't use 9.8 m/s2 here. Why?

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
 
  • #3
57
2
Using a calculator, I got 656.25 lbs of total force for A. Yeah, you can't use meters because the triangle is measured in feet - that was a dumb mistake. Would I just replace that with 32.15 ft/sec^2?

Using that for B, I would get approximately 316,476.56. Not sure exactly which unit of measure to use for this.
 
Last edited:
  • #4
SteamKing
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Using a calculator, I got 656.25 lbs of total force for A. Yeah, you can't use meters because the triangle is measured in feet - that was a dumb mistake. Would I just replace that with 32.15 ft/sec^2?
That's OK. g is usually taken as 32.2 ft/s2 in the imperial units.

leo255 said:
Using that for B, I would get approximately 316,476.56. Not sure exactly which unit of measure to use for this.
If you are calculating work, then you must have units for work (force * distance, eh?)

Going back to your derivation of the hydrostatic force, you used the depth as (3 - y) and the width wi as (3 - 2y/3). When y = 3 feet, the depth is zero, but wi must be 2 feet, the width of the trough at the top. Similarly, when y = 0, wi must be zero, since we are at the bottom of the trough.
 
Last edited by a moderator:
  • #5
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Would it work if I made it (6 - y) instead?
 
Last edited:
  • #6
SteamKing
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Would it work if I made it (6 - y) instead?
You can answer that question yourself. Try drawing a sketch of the end of the trough.
 

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