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Work done by Three Dimensional Inverse Square Field

  1. Apr 10, 2016 #1
    1. The problem statement, all variables and given/known data

    Find the work done by the three-dimensional inverse-square field

    ## F(r) = \frac {1} {||r||^3} r ##

    on a particle that moves along the line segment from P(6, 2, 3) to Q(4,2,4)

    2. Relevant equations

    ## \int_C F \bullet dr = \int_a^b f(h(t), g(t)) \sqrt {(\frac {dx} {dt})^2 + (\frac {dy} {dt})^2} = \int_a^b f(h(t),g(t)) ||\vec r'(t)|| dt##

    3. The attempt at a solution

    I cannot figure out where to start. I've been generally confused by line integrals the entire section. The only relationship I can see is with the 3rd equation I provided in the relevant equations where we have a magnitude ##||\vec r'(t)||##. But then what would be the function F? Is it the ##\frac {1} {||r||^3} r##? But that is a function of F(r), so I'm not sure how this changes the dynamics of the problem...? The most I could thing of would be along the lines of

    ## \int_C F \bullet dr = \int_a^b f(h(t),g(t)) ||\vec r'(t)|| dt##

    ## \int_C F \bullet dr = \int_a^b \frac {r} {||r||^3} ||\vec r'(t)|| dt ##

    but now we have a function of r AND t inside this integral, but the integral is with respect to t, so could I just pull the function of r out?

    ## \int_C F \bullet dr = \frac {r} {||r||^3} \int_a^b ||\vec r'(t)|| dt ##

    but then what is the function for r(t), I don't understand parameterization very well, and I'm not sure if I'm supposed to make some kind of parametric substitution?
     
  2. jcsd
  3. Apr 10, 2016 #2

    BvU

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    What did you do with the dot product ? In your first equation you pretend ##\vec F## and ##d\vec r## are collinear -- they are not !

    And your ##
    F(r) = \frac {1} {||r||^3} r## should be ##
    \vec F(\vec r) = \frac {1} {||r||^3} \vec r## or
    ##
    \vec F(\vec r) = \frac {1} {\vec r\cdot\vec r} \hat r##
     
  4. Apr 10, 2016 #3
    I'm not sure I follow. Since F is a function of r in this case, unlike it is usually a function re-parameterized, I'm not sure how to deal with F as a function of r, in general.
     
  5. Apr 10, 2016 #4

    BvU

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    A force is a vector. Your inverse-square forces point away from the origin. So the force vector at position ##(x,y,z)## has a magnitude that only depends on ##|\vec r| = \sqrt{x^2+y^2+z^2}## but a direction that depends on each of x, y and z.

    You path is from ##(6,2,3)## to ##(4,2,4)##. I could draw the path as a line segment in the plane ##y = 2##, but that doesn't help much. Your job (as I understand it *) is to find a way to express ##\vec F\cdot d\vec s## ( I use ##\vec s## for the path direction, to distinguish from the position vector ##\vec r##) and integrate.

    *) there are other more indirect ways to calculate this work, but they depend on establishing that ##\vec F## is conservative and finding the corresponding potential field; my impression is that you are supposed to actually do the integral in this exercise.
     
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