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Calculating work

  1. Jul 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Given the attached picture, Calculate the work required to go around the contour shown if the force is [tex]\vec{F} = y\hat{x} - x^{2} \hat{y}[/tex]


    First by Contour integration:
    Work = [tex]\oint \vec{F} \bullet \vec{dl} = \int_{a}^{b} (\vec{F} \bullet \hat{x})dx + \int_{b}^{c} (\vec{F} \bullet \hat{y})dy - \int_{c}^{d} (\vec{F} \bullet \hat{x})dx - \int_{d}^{a} (\vec{F} \bullet \hat{y})dy [/tex]

    [tex]= \int_{1}^{2} (2) dx + \int_{2}^{3} (4)dy - \int_{2}^{1} (3)dx - \int_{3}^{2} (1)dy = 0 [/tex]

    I think the answer above is suppose to be -4 not 0.


    Now by Stokes Theorem:
    When this was done with "Stokes theorem", we get the following:
    Work = [tex]\oint \vec{F} \bullet \vec{dl} \equiv \int \int (\nabla \times \hat{F})\bullet \vec{ds} = - \int \int (2x + 1)( \hat{z} \bullet \hat{z})ds = \int_{2}^{3} \int_{1}^{2} (2x + 1)dxdy = -4[/tex]
    Note: [tex]( \hat{z} \bullet \hat{z}) = 1[/tex] since [tex]( \hat{z} \bullet \vec{ds}) = ds [/tex]

    Question:
    When I look at this, Stokes method seems apparently correct, but the method before it seems wrong. Can someone help me find my error?


    Thanks,


    JL
     

    Attached Files:

    Last edited: Jul 10, 2009
  2. jcsd
  3. Jul 11, 2009 #2
    Your attachment is still pending approval, but if we're speaking of the work done around a closed path, it's certainly not zero. This is due to the fact that the force field is not conservative. While the force is position dependent (as is required by a conservative force), it's curl, [tex]\nabla \times F[/tex], is not the zero vector.
     
  4. Jul 11, 2009 #3

    HallsofIvy

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    The contour is the square with corners at (1, 2), (2, 2), (2, 3), and (1, 3).


    Why do you have minus signs on the last two integrals? Assuming that your contour goes from point a to point b, then to point c, then to point d, then back to point a, these should all be "+". (Although the integrals themselves might be 0.)

    On the first leg, with x= t, y= 2, the integral is [itex]\int_1^2 2 dx[/itex] as you say. On the second leg, with x= 2, y= t, the integral is [itex]\int_2^3 (-4) dt[/itex]. Did you forget the "-" on "[itex]-x^2\vec{j}[/itex]"? On the third leg, with x= t, y= 3, the integral is [itex]\int_2^1 3dt= -\int_1^2 3dt[/itex]. You have the wrong sign. On the fourth leg, with x= 1, y= t, the integral is [itex]\int_3^2(-1) dt= \int_2^3 dt[/itex]. That is what you have, but, I think, because of two canceling sign errors!. Evaluating, the integral is 2- 4+ (-3)+ (1)= -4.

     
  5. Jul 11, 2009 #4
    Thanks Halls.
     
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