How Do You Calculate the Work Required to Excavate a Rectangular Prism Well?

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Homework Help Overview

The problem involves calculating the work required to excavate soil from a well shaped like a rectangular prism, with specific dimensions and weight density of the soil. The original poster presents their approach to the problem, including their equations and results, while expressing confusion over the correctness of their solution.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formulation of the work equation and the interpretation of the distance variable in the context of the problem. There is debate over whether to use 'x' or '30-x' for the distance in the integral, with some participants questioning the implications of these choices on the final answer.

Discussion Status

Multiple interpretations of the problem setup are being explored, particularly regarding the coordinate system and the formulation of the work integral. Some participants have offered guidance on potential errors in the original poster's calculations, while others express confusion about the differing approaches.

Contextual Notes

There is a mention of the original poster's assumption about the shape of the excavation, with some participants considering the implications of the term "prism" in their interpretations. The discussion reflects uncertainty about the correct application of the formulas involved.

Sarangalex
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Homework Statement



A well is dug in the shape of a rectangular prism. It is 30ft deep and has a base with area of 40ft^2. Assuming that the soil weighs 150 lbs/ft^3, calculate the work W required to raise the soil to ground level.


Homework Equations



W = ∫dW
dW = ρ(area)(distance)dx


The Attempt at a Solution



I had dW = ρ40(30-x)dx

Then I said W = 150∫[from 0 to 30] 1200-40x dx

Solved to get 2700000ft/lb, which is wrong. I don't really understand what I did wrong here. Isn't this just like pumping water out of a tank or something?
 
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I get the same answer. I guess we are both assuming "rectangular prism" means the same thing as "rectangular parallelepiped" or "rectangular box".
 
Sarangalex said:

Homework Statement



A well is dug in the shape of a rectangular prism. It is 30ft deep and has a base with area of 40ft^2. Assuming that the soil weighs 150 lbs/ft^3, calculate the work W required to raise the soil to ground level.


Homework Equations



W = ∫dW
dW = ρ(area)(distance)dx


The Attempt at a Solution



I had dW = ρ40(30-x)dx

Then I said W = 150∫[from 0 to 30] 1200-40x dx

Solved to get 2700000ft/lb, which is wrong. I don't really understand what I did wrong here. Isn't this just like pumping water out of a tank or something?


I think your answer is too large by a factor of 30. Go back to your dW formula to see why.

RGV
 
I don't understand why it would just be x for the distance rather than 30-x. Could someone explain this to me?
 
It doesn't matter whether you use x or 30 - x. I think you are correct and I am puzzled what RGV sees that we don't, since his answers are usually spot on.
 
But x and 30-x give you totally different answers...
 
Sarangalex said:
But x and 30-x give you totally different answers...

No they don't.
 
LCKurtz said:
It doesn't matter whether you use x or 30 - x. I think you are correct and I am puzzled what RGV sees that we don't, since his answers are usually spot on.

The cross-sectional area at depth x is 40 at x = 0, but the formula given by the OP gives 40*30. I think the correct formula should be dW = ρ*40*(30-x)/30 dx.

RGV
 
Ray Vickson said:
The cross-sectional area at depth x is 40 at x = 0, but the formula given by the OP gives 40*30. I think the correct formula should be dW = ρ*40*(30-x)/30 dx.

RGV

But the OP has just chosen his coordinates measuring depth from the other end; that's the 30-x, which varies from 30 to 0. You could measure from the other end and use x as I mentioned earlier. The cross section is a constant 40.

We are both thinking of a hole shaped by a rectangular box, right?
 
  • #10
Oh, I'm sorry. I see they are the same now, I just don't understand how.
 
  • #11
Sarangalex said:
Oh, I'm sorry. I see they are the same now, I just don't understand how.

It is just a matter of which end you put your coordinate system and whether x is measured
positive up or down.
 
  • #12
No, I was thinking of a wedge-shaped excavation (because of the word "prism" in the original post). In that case I would be wrong anyway, because (my) dW is weight, not work. For a wedge-shaped hole the answer would be
Work = int rho * 40 * x*(1 - x/30) dx = 900000 ft-lb.

RGV
 
Last edited:

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