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Variable force? and work

  1. Apr 5, 2010 #1
    1. The problem statement, all variables and given/known data
    A force F=[(3x)[tex]\hat{}x[/tex]-2[tex]\hat{}y[/tex]]N acts on an object. How much work is done by this force as it move the object from the origin to the point r=(2[tex]\hat{}x[/tex]+ 3[tex]\hat{}y[/tex])m?


    2. Relevant equations

    W=[tex]\int[/tex]F*d[tex]\bar{}r[/tex]

    3. The attempt at a solution
    W=[((3/2)x^2)[tex]\hat{}x[/tex]-(2y)[tex]\hat{}y[/tex]] * (2[tex]\hat{}x[/tex]+ 3[tex]\hat{}y[/tex])m
    =6-6=0

    I am a little confused on how to do this particular integral and we never really discussed it in class...
     
  2. jcsd
  3. Apr 5, 2010 #2

    vela

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    Were you given a path to integrate along?
     
  4. Apr 5, 2010 #3

    tiny-tim

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    Welcome to PF!

    Hi smhippe ! Welcome to PF! :smile:

    (erm :redface: … you have to put x or y inside the hat-box!! :wink:)

    No, it's not W = ∫ F*dr, it's W = ∫ F "dot" dr

    try again :smile:
     
  5. Apr 5, 2010 #4

    vela

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    Let me elaborate a bit more. You need to evaluate the dot product inside the integral first. Depending on what you're given, you may be able to turn the integrand into a "normal" one with respect to just one variable and then just integrate like you usually would.
     
  6. Apr 5, 2010 #5
    Another path to take, is if you know what a conservative field is, and how to take the curl of a vector field, then you can show that this force is conservative, and that therefore it does not matter what path you take. (Which should hold for this simple case, if I understood what a 'simply connected region' means sufficiently well)

    And seeing how a specific path was not given, apparently the aim of the question is that you prove that the field is conservative, in whatever way you may know, so that you can take any path you'd like.

    Having established that, you can integrate along any path you find convenient to reach your answer.
     
    Last edited: Apr 5, 2010
  7. Apr 5, 2010 #6
    Okay...so not too much of any of that made sense to me. I was not given a specific path, what I have posted is exactly what was given to me.
    I understand that it is the dot product...I couldn't find any symbol that would represent the dot product and I figured that would be implied. Computers aren't my thing so sorry about the whole "hat" thing being out of line.
    I honestly don't know what to do with the integral of force and the distance vector.
    Thanks royalcat for the input....unfortunately that is a little over my head. Just a simple explanation of how to do this integral would be amazing.
    -Thanks guys
     
  8. Apr 5, 2010 #7
    Well, you could just outright go and say that the force is conservative and that therefore any path is valid. From there on, it's just a question of choosing the most convenient path (I suggest going along the x direction, and then along the y direction) and taking the integral of the product [tex]\vec F \cdot \vec dr =F_x dx + F_y dy + F_z dz[/tex]
     
    Last edited: Apr 5, 2010
  9. Apr 5, 2010 #8

    vela

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    Suppose you're given a curve [itex]\vec{r}(t)=x(t)\hat{x}+y(t)\hat{y}[/itex]. In a time dt, [itex]\vec{r}[/itex] changes by the amount [itex]d\vec{r}=dx\hat{x}+dy\hat{y}=x'(t)dt\hat{x}+y'(t)dt\hat{y}[/itex]. Now use this to simplify the integrand:

    [tex]\vec{F}\cdot d\vec{r} = F_x dx + F_y dy = F_x x'(t) dt + F_y y'(t) dt = (F_x x'(t)+F_y y'(t)) dt[/tex]

    So your integral becomes

    [tex]W = \int_{t_0}^{t_1} (F_x x'(t)+F_y y'(t)) dt[/tex]

    where [itex]\vec{r}(t_0)[/itex] is your starting point and [itex]\vec{r}(t_1)[/itex] is your endpoint. It's just a matter now of rewriting everything in terms of t and integrating.

    Suppose instead you're told to integrate along the path given by y=x2. In this case, you can write dy=2x dx, and your integrand becomes

    [tex]\vec{F}\cdot d\vec{r} = F_x dx + F_y dy = F_x dx + F_y (2x dx) = (F_x+2xF_y) dx[/tex]

    The integral is then done wrt to x:

    [tex]W = \int_{x_0}^{x_1} (F_x+2xF_y) dx[/tex]

    So you write everything in terms of x this time, and then integrate.

    For this problem, the path you choose doesn't matter. You'll always get the same answer, so just pick one that's easy to work with.
     
  10. Apr 6, 2010 #9

    tiny-tim

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    hi smhippe! :smile:

    (just got up :zzz: …)
    erm :redface:i always use a dot (.) ! :biggrin:
    ok, go along the x-axis first, from (0,0) to (2m,0) …

    F is (3x,-2) and dr is (1,0)dx,

    then go up the line x = 2m, from (2m,0) to (2m,3m) …

    F is still (3x,-2) but dr is now (0,1)dy …

    what do you get? :smile:

    (btw, it does work if you go direct, but then dr = (2,3)mds/√13, and you have to rewrite 3x in terms of s, where s is distance in the (2,3) direction, but that's much more difficult! :rolleyes:)
     
  11. Apr 6, 2010 #10

    ehild

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    Let me contribute with an other method. As this is a conservative force, potential V(x,y) exist, and the force is the negative gradient of V:

    [tex]F_x=-\frac{\partial V(x,y)}{\partial x}[/tex]

    and

    [tex]F_y=-\frac{\partial V(x,y)}{\partial y}[/tex].

    Fx=3x, Fy=-2. Integrate -Fx with respect to x to get V(x,y), (now it is an indefinite integral) but note that the integration constant can contain the other variable, y.

    V(x,y) = -3x^2/2+f(y)

    Take the partial derivative of V with respect to y: f'(y) =-Fy = 2 . Integrate with respect to y : you get f(y) = 2y +C. So your potential function is

    V(x,y) = -3/2 x^2 + 2 y +C.

    The work is negative difference between V(2,3) and V(0,0)

    ehild
     
  12. Apr 6, 2010 #11
    Awesome!
    ...so if I understand correctly the final answer should be 0 if I understand the math correctly. We're just starting to talk about some of the things that I'm seeing you guys use. I know this is kinda simple...but I would rather understand the basics now and save myself some headache later on. I really hope I'm not still missing anything! Thanks again!
     
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