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Multivariable Calculus: Force along Line Integral

  1. Dec 6, 2007 #1
    1. The problem statement, all variables and given/known data

    Part 1: A 160lb man carries a 25lb paint can up a spiral staircase, which has radius 20 feet, completes 3 revolutions, and has final height 90 feet. What is the work done?

    Part 2: This time, the man's paint can leaks at a constant rate such that he loses 9lbs of paint from bottom to top.

    2. Relevant equations

    I know that the integral of Force F along curve C has a couple equations.

    Integral of...
    F[tex]\bullet[/tex]dr
    F[tex]\bullet[/tex]Tds
    F(r(t))[tex]\bullet[/tex]r'(t)dt



    3. The attempt at a solution


    I tried to make a parametrization but got stuck on that for part 1. So I just used that Work is Scalar product of F and Dispacement vectors.

    This gave me 185lb x 90 feet x sin(90) because he goes vertical 90 feet from where he started. The path he takes t get there is irrelevant.

    Part 2 is where I'm having trouble.

    I've made equations for r(t) (I think)
    (Note, pi is being multiplied, not raised as an exponent)

    x = 20 cost
    y = 20 sint
    z = 15t/[tex]\pi[/tex]
    t[0, 6[tex]\pi[/tex]]

    The force he has to apply at a given t has to change because he's losing paint.

    F(0) = 185
    F(6[tex]\pi[/tex]) = 185 - 9 = 176

    F(t) = 185 - 9t/6[tex]\pi[/tex]

    I know now that I'm going to have to do an integration, but I'm not sure if what I've done is right, for I have an F(t), not F(r(t)) or F(x,y,z). Any help would be appreciated :)
     
  2. jcsd
  3. Dec 6, 2007 #2

    Avodyne

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    Science Advisor

    Because force is still vertical, only vertical motion matters.
     
  4. Dec 6, 2007 #3
    If the man moves around the staircase at a constant rate, then his movement in the z direction is also at a constant rate.

    Remember, the only movement that matters in this problem is movement parallel to the force he applies (in the z direction).
     
  5. Dec 6, 2007 #4
    So I have the integral from t[0, 6pi] of

    <0,0,185-(9t/6pi)> dot <-20sint, 20cost, 15/pi>dt

    = 2775/pi - (45 x pi x t) / 2

    Which gives me 4,100 foot lb.

    Is the difference in the work done really that large? 166500 - 4100?
     
  6. Dec 6, 2007 #5
    Nvm, no it's not that big. I put a pi in the wrong place when writing it out. The work is slightly less, 16245 foot -lb
     
  7. Dec 6, 2007 #6
    Your most recent answer is correct, but there's a simpler way to get it. You know that

    F(t) = 185lb - 9lb*t/T

    Where T is the total time he takes to climb to the top. Because he moves at a constant vertical velocity, you also know:

    t/T = z/90feet
    ==> F(z) = 185lb - 9lb*(z/90feet)

    Which you can integrate to get the force. No [tex]\pi[/tex]'s involved :smile:
     
  8. Dec 6, 2007 #7

    HallsofIvy

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    Staff Emeritus
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    You don't even need to integrate. Since the total weight changes linearly, the average weight is the average of weight at the top and the bottom. Find the average weight and multiply by the height.
     
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