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Work done by a vector force

  1. Sep 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine the work done by the Force F = yi(hat) - xj(hat) on an object that travels from point A (a,0) to point B (-a,0)

    a) along an elliptical path described by x=acosθ, y=bsinθ
    b) along a straight line from A to B
    c) From these results, can we determine whether or not the force is conservative?

    2. Relevant equations

    W=∫F ° dl
    dl=dxi(hat) + dyj(hat)

    3. The attempt at a solution

    a) W=∫(a to -a) bsinθi(hat)dl - ∫(0 to 0) acosθj(hat)dl
    so that just gives me ∫(a to -a) bsinθi(hat)dl which equals 0. This doesn't seem right to me. What's mainly confusing me is that the y is corresponding to the i(hat) vectors and vice versa for the x. I don't really know what to do with the b.

    b) dl = dxj(hat)
    W = ∫(a to -a) xdy = 0

    c) I know that you can tell if a force is conservative if the work depends on the path. Therefore this is probably not conservative.
     
  2. jcsd
  3. Sep 26, 2012 #2

    gabbagabbahey

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    Use your two relevant equations, what is [itex]\mathbf{F}\cdot d\mathbf{l}[/itex] if [itex]\mathbf{F} = y\mathbf{i} - x\mathbf{j}[/itex] and [itex]d\mathbf{l} = dx\mathbf{i}+ dy\mathbf{j}[/itex]?
     
  4. Sep 26, 2012 #3
    So then I would get ∫ydx - xdy.

    How would I integrate y with dx and x with dy?
    I'm sort of behind on this stuff and this may seem simple and I'm sorry for that.
     
  5. Sep 26, 2012 #4

    gabbagabbahey

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    You have equations for both [itex]x[/itex] and [itex]y[/itex] in terms of a single parameter, [itex]\theta[/itex], use them and remember that [itex]dx =\frac{dx}{d\theta}d\theta[/itex]
     
  6. Sep 27, 2012 #5
    So I would get an equation that looks like ∫bsinθ(dx/dθ)dx - ∫acosθ(dx/dθ)dθ

    which yields:

    -bcosθ - bsinθ

    Would that be my answer? Or are there further steps pertaining to points A and B?
     
  7. Sep 27, 2012 #6

    gabbagabbahey

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    Don't you mean [itex]\int_{\theta(x=a, y=0)}^{\theta(x=-a,y=0)}\left(b\sin\theta \frac{dx}{d\theta} - a\cos\theta \frac{dy}{d\theta} \right)d\theta[/itex]?

    I get something different, can you post your steps?

    Work done is the path integral from A to B, so you should have a definite integral where the limits are the value of [itex]\theta[/itex] at each endpoint.
     
  8. Sep 27, 2012 #7
    Sorry I seemed to have misplaced a dx with what should have been dθ.

    Continuing on:

    I split the integrals:
    ∫bsinθ(dx/dθ)dθ - ∫acosθ(dx/dθ)dθ with the limits a to -a.

    If I integrate bsinθ I get -bcosθ and if I integrate acosθ I get a sinθ (another mistake in my previous answer)

    Therefore the expression would be:

    -bcosθ - asinθ from the limits a to -a

    resulting in:

    (-bcos(-a) - asin(-a)) - (-bcos(a) - asin(a))

    I think I may be integrating the original equation incorectly with the due to confusion by the dθ's and dx's.
     
  9. Sep 27, 2012 #8

    gabbagabbahey

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    What happened to [itex]\frac{dx}{d\theta}[/itex] and [itex]\frac{dy}{d\theta}[/itex]? Why would the limits be a to -a? What is the value of [itex]\theta[/itex] at the point A? What about at point P?
     
  10. Sep 27, 2012 #9
    I made another mistake in writing the final equation down so the final equation from before would be:

    ∫bsinθ(dx/dθ)dθ - ∫acosθ(dy/dθ)dθ with limits a to -a

    This doesn't really change much though.

    Well technically it's θ to θ but since there is no y component wouldn't the limits just be the x components?

    I'm extremely confused at this point on how I would find the θ in point A or B because I have no idea what "a" means.
     
  11. Sep 27, 2012 #10

    gabbagabbahey

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    "a" and "b" are just numbers. Plot out your parametric equations as you range theta from 0 to 2pi... what value of theta gives you x=a and y=0? What value of theta gives you x=-a and y=0?

    As for dx/dθ and dy/dθ, you will need to calculate them...
     
  12. Sep 27, 2012 #11
    Well for sinθ y=0 at 0, ∏ and 2∏ Therefore would "a" be pi/2 and theta be 0, 180 or 360?

    for cosθ y=0 at ∏/2 and 3∏/2 so b would be pi/2 or 3pi/2 and theta be 90 or 270...

    I am just really confused at this point as I feel like I need to review basic geometry however this question is one of the only ones I don't understand in my textbook chapter.
     
  13. Sep 27, 2012 #12

    gabbagabbahey

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    Why ask me? You have [itex]y(\theta)=b\sin\theta[/itex], so what is [itex]y\left( \frac{\pi}{2} \right)[/itex]?

    Huh?:confused:

    You have [itex]x(\theta)=a\cos\theta[/itex], what are [itex]x\left( 0 \right)[/itex], [itex]x\left( \frac{\pi}{2} \right)[/itex], [itex]x\left( \pi \right)[/itex], etc?

    I think you should review linear algebra, especially curves and parameterization.
     
  14. Sep 27, 2012 #13
    Then y=b at theta = pi/2.
    y=0 at theta = 0 as well




    x=a at theta = 0
    x=-a at theta = pi

    Therefore the values that im looking for are theta = 0 and pi.

    I don't understand what the limits for this integral would be though. As well as what im doing with these values of theta.
     
  15. Sep 27, 2012 #14

    gabbagabbahey

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    You are integrating along an eliptical path from point A to point B. When you parametrize that path in terms of [itex]\theta[/itex], you find that [itex]\theta=0[/itex] corresponds to point A and [itex]\theta=\pi[/itex] corresponds to point B, so those values of [itex]\theta[/itex] are your lower and upper limits.
    .
     
  16. Sep 28, 2012 #15
    So if I complete the integral:

    ∫(bsinθdx/dθ - acosθdy/dθ)dθ From 0 to pi

    = -bcosθ - asinθ from 0 to pi

    = b-1

    Would b-1 be my final answer for work? That doesnt seem right to me, did I misplace a number that I can substitute in for b?
     
  17. Sep 28, 2012 #16
    sorry 2b*
     
  18. Sep 28, 2012 #17

    gabbagabbahey

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    No, [itex]\int_0^{\pi} b\sin\theta \frac{dx}{d\theta} d\theta \neq -b\cos\theta[/itex], what happened to [itex]\frac{dx}{d\theta}[/itex]?
     
  19. Sep 28, 2012 #18
    I actually got help from a TA but thank you so much for your help until now.
     
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