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Vector calculus and force field

  1. Jun 29, 2011 #1
    1.10.1 The force field acting on a two-dimensional linear oscillator may be described by
    F=−ˆxkx − ˆyky.
    Compare the work done moving against this force field when going from (1, 1) to (4, 4)
    by the following straight-line paths:
    (a) (1, 1)→(4, 1)→(4, 4)
    (b) (1, 1)→(1, 4)→(4, 4)
    (c) (1, 1)→(4, 4) along x = y.
    This means evaluating

    (4,4)
    (1,1)
    F · dr
    along each path.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 29, 2011 #2

    lanedance

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    hi tibphysic, welcome to pf, the idea is to have a try

    do you know about conservative vector fields?
     
  4. Jul 1, 2011 #3
    yes i do.. conservative vectors fields does not depend on the path taken.. just on the initial and the final point.
     
  5. Jul 1, 2011 #4

    lanedance

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    could be useful....
     
  6. Jul 1, 2011 #5

    LCKurtz

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    I'm not familiar with that notation. Is that <-kx, -ky> or something else? Are ^x and ^y notations for i and j?
     
  7. Jul 1, 2011 #6

    LCKurtz

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    True, but perhaps the intent of the exercise is to actually work them...
     
  8. Jul 1, 2011 #7
    im assuming his notation ^x , read x hat. where the carrot should be over the x.
    is equivalent to i , j , k . they use x hat and y hat and z hat in physics more.
     
  9. Jul 1, 2011 #8

    SammyS

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    For instance:

    Along the path (1, 1)→(4, 1), [itex]d\hat{r}=\hat{x}\,dx\,.[/itex]

    So, find [itex]\displaystyle\int_{(1,1)}^{(4,1)}\,dW=\int_{x=1}^{x=4}{\vec{F}\,|_{y=1}\cdot\hat{x}}\,dx[/itex]
     
    Last edited: Jul 1, 2011
  10. Jul 1, 2011 #9

    lanedance

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    yeah I thought the notation was a little strange but took them as unit vectors, so the potentials were of harmonic oscillator form
     
  11. Jul 2, 2011 #10

    HallsofIvy

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    True, but if the force field is conservative, as lanedance suggested in the first response, the problem is much easier. You don't need to do all those integrals.
     
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