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Digging a Well (Work Problem)

by Sarangalex
Tags: calculus, work
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Sarangalex
#1
Nov20-11, 06:16 PM
P: 9
1. The problem statement, all variables and given/known data

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.


2. Relevant equations

W = ∫dW
dW = ρ(area)(distance)dx


3. 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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LCKurtz
#2
Nov20-11, 10:15 PM
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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".
Ray Vickson
#3
Nov21-11, 02:10 AM
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Quote Quote by Sarangalex View Post
1. The problem statement, all variables and given/known data

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.


2. Relevant equations

W = ∫dW
dW = ρ(area)(distance)dx


3. 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

Sarangalex
#4
Nov21-11, 11:24 AM
P: 9
Digging a Well (Work Problem)

I don't understand why it would just be x for the distance rather than 30-x. Could someone explain this to me?
LCKurtz
#5
Nov21-11, 11:31 AM
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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.
Sarangalex
#6
Nov21-11, 11:46 AM
P: 9
But x and 30-x give you totally different answers...
LCKurtz
#7
Nov21-11, 11:50 AM
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Quote Quote by Sarangalex View Post
But x and 30-x give you totally different answers...
No they don't.
Ray Vickson
#8
Nov21-11, 11:57 AM
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Quote Quote by LCKurtz View Post
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
LCKurtz
#9
Nov21-11, 12:06 PM
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Quote Quote by Ray Vickson View Post
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?
Sarangalex
#10
Nov21-11, 12:08 PM
P: 9
Oh, I'm sorry. I see they are the same now, I just don't understand how.
LCKurtz
#11
Nov21-11, 01:03 PM
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Quote Quote by Sarangalex View Post
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.
Ray Vickson
#12
Nov21-11, 02:12 PM
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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


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