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Calc + applications in physics

  1. Oct 6, 2005 #1
    I think I found a way to do this, but it's not working.

    The problem gives a cable weighing 2lb/fot, that is used to lift 800 lb of coal up a mineshaft 500ft deep. find the work

    I looked up the answer, which 650,000 ft-lbs. The weight is the force, so there is no need for the acceleration due to gravity. I tried something that I thought was right, but it didn't yield the correct answer. Am I not integrating correctly or is my set-up wrong?

    [tex] \int_{0 ft}^{500 ft} (800lbs + 2x) x\ dx [/tex]

    I distributed the x and integrated, ending up with [tex] ({800/2})x^2 + ({2/3})x^3 [/tex] from 0 to 500.

    That doesn't come up with the rgith answer, though. Any ideas?
     
    Last edited: Oct 6, 2005
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  3. Oct 6, 2005 #2

    Tom Mattson

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    ecoli, are you continuing a thread here that you started at SFN? Because if you are, I don't think that anyone except me is going to be able to follow you!
     
  4. Oct 6, 2005 #3
    you're right, I forgot to copy the first part of the message... namely the problem. I'm an idiot... sorry guys.
     
  5. Oct 6, 2005 #4
    I couldn't tell from your previous post if you already have an answer, but here's how I would set it up:

    The 800 is constant from the bottom to the top, only the 2x will change. You can write it out as:

    [tex]W=800*500+\int_{0}^{500}2x\,dx[/tex]

    Do you see why?

    Alex
     
  6. Oct 6, 2005 #5
    Where did you get the 500 from... because won't that change when you pull the cable upwards?

    The book gives the answer as 650,000 ft-lbs, so I'm not sure your answer is right...
     
  7. Oct 6, 2005 #6
    Remember for a constant force, the work is W=Fd. You can split the work up into the work for the coal (constant weight=constant force) + work for the cable. The force on the cable is the only thing that changes.

    Alex
     
  8. Oct 6, 2005 #7

    Tom Mattson

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    :rofl: No problem.

    You've got 1 major problem that is screwing the entire thing up, and that is that you don't seem to really get the definition of work.

    It is (for 1D forces and displacements):

    [tex]W=\int_{x_1}^{x_2}Fdx[/tex]

    Now, your integral seems to be taking the definition as:

    [tex]W=\int_{x_1}^{x_2}Fxdx[/tex]

    which is not the same thing. So, drop that extra "x" in the integrand, and you'll be fine.
     
  9. Oct 6, 2005 #8
    the weight of the cable changes. I understand that is equal to the force.

    I see why your answer is correct, but it's not the same answer the book gives... unless I'm integrating incorrectly.
     
  10. Oct 6, 2005 #9

    Tom Mattson

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    If you follow the advice of either apmcavoy or myself, you will get the correct answer of 650,000 ft-lbs.
     
  11. Oct 6, 2005 #10
    I am following your advice, I must be integrating incorrectly. That's making the answer come out wrong, so I thought it was the setup that was wrong. Of course that's not the case.
     
  12. Oct 6, 2005 #11

    Tom Mattson

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    Want to post your steps and have us check them?
     
  13. Oct 6, 2005 #12
    ok, thanks.

    first off, I have
    Fd = (800 + 2x)
    [tex] \int_{0}^{500} 800 + 2x dx [/tex]

    then I pull out the 800 and get [tex] 800 \int_{0}^{500} 2x\ dx [/tex]

    I integrate and get [tex] 800 * x^2 [/tex] from 0 to 500

    which is 200 million, and not the correct answer
     
  14. Oct 6, 2005 #13
    There's your problem, you pulled out the 800! Remember this?

    [tex]\int\left(f(x)+g(x)\right)\,dx=\int f(x)\,dx+\int g(x)\,dx[/tex]

    Alex
     
  15. Oct 6, 2005 #14

    Tom Mattson

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    Right.

    Wrong.

    You're pulling out the 800 as if it were a factor of the integrand. It's not, it's a term of the integrand. So the correct way to split it up would be:

    [tex] \int_0^{500} (800 + 2x) dx = \int_0^{500} 800 dx + \int_0^{500} 2x dx[/tex]

    Now, you can pull 800 out of the first integral and 2 out of the second.
     
  16. Oct 6, 2005 #15
    ahhh... thanks guys, all clear now.
     
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