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Curvilinear Integral

  1. Apr 2, 2010 #1
    1. The problem statement, all variables and given/known data
    A 750N man raises a 200N bucket of water 30m up a helicoidal staircase. If we know that the staircase has a 10 meter radius and does exactly 4 complete turns in the 30m span, what was the total work done by the man against gravity?

    Also, in the same scenario, what is the work required if the bucket had a small hole so that 50N leaked out at a constant rate while it was raised up the stairs?

    2. Relevant equations
    [tex]\oint F dr[/tex]

    3. The attempt at a solution
    Althought this can be solved with conservation of energy (750N+200N)*30m = 28,500J, I'd like to solve it using a line integral. I've solved the first part, but what's giving me a hard time is the 50N lost at a constant rate. Without the line integral, we can simply figure it out as [tex]\int \frac{50N}{30M}t[/tex] with t from 0 to 30, which represents the work with the variable force.

    However, how would this be done with the line integral?

    For the first part, I found the equation of the staircase:

    x=10cos(t) dx=-10sin(t)
    y=10sin(t) dy=10cos(t)
    z=(30t)/8*Pi dz= 15/(4pi)
    (As we want z=30 @ 8pi (4 complete turns))

    I obtain: [tex]\oint 950dx + 950dy +950dz [/tex], t from 0 to 8pi

    I can't seem to figure out how to with the variable mass.

    Thanks for the help!

  2. jcsd
  3. Apr 2, 2010 #2


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    Homework Helper

    few points hopefully, they're helpful..
    first [itex]\oint F dr[/itex] represents an integral over a closed path,

    that isn't the case here, otherwise the wok would be zero, so you should use [itex]\int F dr[/itex]

    that said, shouldn't the integrand be a vector dot product? [itex]\int \vec{F} \bullet \vec{dr}[/itex]

    This means the translation contirbuting to the work is only the component of the translation in the direction of the force.

    You should check your first intergal gives the correct value as compared to conservation of energy & should probably be able to be reduced to the form you quote
    Last edited: Apr 2, 2010
  4. Apr 2, 2010 #3


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    Gold Member

    No it can't because the force varies with t.


    So you have

    [tex]\vec R(t) = \langle 10\cos(t),10\sin(t),\frac {15}{4\pi}t\rangle[/tex]

    which looks correct. Now you just need to write the force vector:

    [tex]\vec F(t) = \langle ?,?,?\rangle[/tex]

    Here are some hints:

    1. What direction is the force vector in?
    2. If the water is 200N when t = 0 and 150N when [itex]t = 8\pi[/itex], and it loses weight at a constant rate, can you figure out an equation for W(t)?
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