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Help with work problem

  1. May 24, 2012 #1
    A 2000lb elevator travels from the ground floor to the 1st floor and on to the 9th floor. At the 1st floor, a 100lb person gets off, on the 2nd floor a 110lb person gets off, on the 3rd floor a 120lb person gets off and so on and so on till the last person gets off on the 9th floor.

    a) How much did the last person to get out of the elevator weigh? What was the total weight to start?
    b) If each floor is 10 feet in height, find the work done by the elevator. (Hint: No integrals needed). Write the Riemann Sum (in terms of i) for the work and find the numerical value of the total work done.

    I know for part a) there is some integration I can use to get the solution. By writing it all out I was able to conclude that the last person to get off was 180lbs, right? I'm just not certain how to conceptually get to that solution through integration...HELP!

    For b) I thought solving it with integrals would be the only possible way and I'm pretty stumped as to how to get the work using Riemann...or I figured I could have used the formula W=F(d) but that's confusing me as well.

    /:
     
    Last edited: May 24, 2012
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  3. May 24, 2012 #2

    Ray Vickson

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    Re: **help** with work problem

    You are supposed to show your work when asking a question here. So, how did you deal with (a)? Show the steps and the calculations.

    As to (b): how much work is done to reach the first floor? How much to reach the second from the first? Etc.

    RGV
     
  4. May 24, 2012 #3
    Re: **help** with work problem

    For a)
    I just wrote it all out to begin with some idea.
    floor 1 - 100 lb
    floor 2 - 110 lb
    floor 3 - 120 lb
    floor 4 - 120 lb + 10 lb = 130 lb
    floor 5 - 130 lb + 10 lb = 140 lb
    floor 6 - 140 lb + 10 lb = 150 lb
    floor 7 - 150 lb + 10 lb = 160 lb
    floor 8 - 160 lb + 10 lb = 170 lb
    floor 9 - 170 lb + 10 lb = 180 lb

    I tried to ∫ 100x from 1 to 9 but that just gave me 4,000...which is definitely not right. Then I tried to ∫ 10x from 1 to 9 and that gave me 400. I've never tried a problem like this before, that's why I'm a little lost...

    As to b)

    W=F×d
    And weight is a force, right?
    Then, each floor is 10 feet in height...so W to reach the first floor would be (the total weight of the elevator, including the people on it) × the distance (10ft)?

    I'm not sure if I'm headed in the right direction, but if I am correct, wouldn't I have to calculate a different force for every floor since the weight of the elevator and the people in it is constantly changing?
     
  5. May 24, 2012 #4

    HallsofIvy

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    That's kind a strange problem. As far as (a) is concerned, you would have to assume that the pattern of weights will continue- which is NOT my experience with elevators.

    Now, for constant weight, w pounds, the work done in moving up h feet is wh "foot pounds". But you are calculating the weights backwards. Initially, everyone was in the elevator. So between the ground floor and first floor the total weight in the elevator was 100+ 110+ 120+ 130+ 140+ 150+ 160+ 170+ 180+ 2000 pounds. Multiply that by the distance between the ground and first floors. Since a person weighing 100 pounds get out at the first floor, the weight between the first and second floors is 110+ 120+ 130+ 140+ 150+ 160+ 170+ 180+ 2000 pounds.
    Multiply that by the distance from the first floor to the second floor to find the work done there.
     
  6. May 24, 2012 #5
    For b) then, would my Riemann sum be
    W=3260(10)+3160(10)+3050(10)+2930(10)+2800(10)+2660(10)+2510(10)+2350(10)+2180(10)+2000(10)? Giving me 269,000 ft-lbs.
     
  7. May 24, 2012 #6

    HallsofIvy

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    You include the work done by the empty elevator going to the 10th floor while the problem talks about the elevator going to the 9th floor. In fact, there is no reason to think the building even has a tenth floor.
     
  8. May 24, 2012 #7
    OH! Oops, got it! So then:

    W=3260(10)+3160(10)+3050(10)+2930(10)+2800(10)+266 0(10)+2510(10)+2350(10)+2180(10)= 249,000 ft-lbs. Right?

    And also, for a), assuming that the patterns of weight is constant, how would my integral look to conclude that the last person to get off the elevator was 180lbs?
     
  9. May 24, 2012 #8

    Ray Vickson

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    Why do you insist on having an integral? It this problem the weight is a discontinuous function of height, so any integral you have will just be a finite sum of separate, easily-computed terms. Integration does not really help here, and it may even be hindering your understanding of the problem.

    RGV
     
  10. May 25, 2012 #9
    I insist on it because there has to be a simpler way to get to the answer without having to write it all out the way I did. I know for sure that an integral can get me the solution, I'm just not sure how to set it up. Are you saying there is no integral that could get me 180 lbs with the information given?
     
  11. May 25, 2012 #10
    The weights are increasing linearly, and so w(f) = 90 + 10f.

    In integral form, w(f) = 90 + ∫10 with your bounds of integration being 0 to f.
    I can see wanting a function, but having an integral in there makes things more complicated.
     
  12. May 25, 2012 #11

    HallsofIvy

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    Using an integral would involve assuming that people got off the elevator in pieces between floors! It will give you an approximation to the sum, not the actual sum. This problem may have been put into a pre-integration section as an introduction to the idea of Riemann sums.

    You can approximate a smooth function by "piecewise" constant functions- that's the idea of "Riemann sums"- but you are trying to approximate a piecewise constant function by a smooth function.
     
  13. May 25, 2012 #12

    Ray Vickson

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    As HallsofIvy has indicated, an integral would be appropriate if the people were made of liquid that seeped out of the elevator at a smooth rate, with a total of 100 lb seeping out during the trip from the ground to the first floor, etc. I don't think that describes the actual situation here. You want to avoid calculating a sum of 9 or 10 fairly simple terms. Why? You could do the whole thing on a hand-held calculator in less than a minute, or in a spreadsheet essentially instantly.

    As I said before, your desire to somehow hammer the problem into a form appropriate for integration is hindering your understanding of the problem!

    You could, of course, formulate the problem as an integration of a discontinuous function, but the only way to _compute_ that integral would be to break it up into pieces and compute each piece separately. That would just lead you back to your starting point.

    RGV
     
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