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Diff. paths of a Force in xy-plane

  1. Nov 17, 2007 #1
    [SOLVED] diff. paths of a Force in xy-plane

    1. The problem statement, all variables and given/known data

    A force acting on a particle in the xy-plane is given by

    [tex]\vec{F} = (2yi + x^2j)[/tex] where x and y are in meters.

    The particle moves from the origin to a final position having coordinates

    [tex]x = 5.00m[/tex] and [tex]y = 5.00m[/tex]

    calculate the word done by [tex]F[/tex] on the particle as it moves along

    a) [tex]OAC[/tex] ------------- solved
    b) [tex]OBC[/tex] ------------- solved
    c) [tex]OC[/tex] ---------------- need help

    Here is the xy-plane graphed with each point ( ignore the white ) :

    B|....................C ( 5.00m , 5.00m )
    O.....................A ( 5.00m , 0.00m )

    2. Relevant equations

    [tex]W = |F| \Delta r cos\theta[/tex]

    [tex]W_{x} = \int_a^b F_{x} dx[/tex]

    [tex]W = \vec{F}_{x} \cdot \Delta \vec{r}[/tex]

    3. The attempt at a solution

    [tex]W_O_A = <2y, x^2> \cdot <5,0> = 10y = 10(0) = 0[/tex]
    [tex]W_A_C = <2y, x^2> \cdot <0,5> = 5x^2 = 5(5)^2 = 125[/tex]
    [tex]W_O_A_C = 0 + 125 = 125 J[/tex]

    [tex]W_O_B = <2y, x^2> \cdot <0,5> = 5x^2 = 5(0)^2 = 0[/tex]
    [tex]W_B_C = <2y, x^2> \cdot <5,0> = 10y = 10(5) = 50[/tex]
    [tex]W_O_B_C = 0 + 50 = 50J[/tex]

    I attempted this using 3 different equations, all of which gave me different answers and none of which were the correct answer.

    Correct answer: 66.7 J

    1. [tex]W_O_C = <2y, x^2> \cdot <5,5> = 10y + 5x^2 = 10(5) + 5(5)^2 = 50 + 125 = 175J[/tex]

    2. [tex]W_O_C = |F| \Delta r cos \theta = \sqrt{4y^2 + x^4}(7.04)(cos0) = 7.07 \sqrt{4(5)^2 + (5)^4} = 7.07 sqrt{725} = 190.34 J[/tex]

    3. [tex]W_O_C = \int_0^5 x dx = \frac{1}{2} x^2 |_0^5 = \frac{1}{2} (125) = 12.5 J [/tex]

    I can find logical faults in each of my approaches, but I do not know what else to do. I would highly appreciate hints/answers to the right approach, as well as why my above approaches are inappropriate.

    One way I would try to solve this, using Calc 3 (which should not necessary for this level physics) would be something like this:

    [tex]\int_0^5 \int_0^5 F_x_y dy dx = \int_0^5 \int_0^5 \sqrt{4y^2i + x^4j } dy dx[/tex]

    But at first look it seems pretty difficult.

    Thank you very much,

  2. jcsd
  3. Nov 17, 2007 #2


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    EDIT: ooops, i hit the post button by mistake.... here is my complete answer

    You did a verygood job presenting your question very clearly.

    here is a general approach. You may always write [itex] \vec{r} [/itex] as [itex] dx \hat{i} + dy \hat{j} [/itex]

    So the work is the integral of [itex] F_x dx + F_y dy [/itex].

    Now, the path will determine what you can say about dx and dy. For example, if y is constant (so dy=0) , the above reduces to the integral of

    [tex] F_x (evaluated~at ~the~given ~value~of~y)~ dx [/tex].

    Then you must integrate the result. If your F_x happens to be independent of x, then the result is just the value of x times the displacement [itex] \Delta x [/itex]

    In your last case, you are going along a straight line, so y and x are related by [itex] y = m x + b [/itex]. Actually, your case is simple because m=1 and b=0. because of this, y = x and dx = dy. Then you can convert the entire thing into a single integral over x (or if you prefer over y).

    try that.
    Last edited: Nov 17, 2007
  4. Nov 17, 2007 #3
    Thank you for the information, nrqed.

    I tried this in my 3rd attempt at problem (c):

    [itex] \int_0^5 x dx = 1/2 x^2 |_0^5 = 1/2 (5)^2 = 12.5[/itex] J

    correct answer: 66.7 J
  5. Nov 17, 2007 #4

    I would say:


    Project all your F vectors on the bissectrice: y=x.

    Use matrices:

    transpose[(2y, x^2)] is your vector.

    Use formula: F=|F|(Fu.Vu)Vu. F is the projection of all your F vectors on the bissectrice.

    Where Vu = unit vector of(transpose[(1,1)]), your bissectrice vector. (I'm not sure about this either...)

    Then you take |F|=sqrt(F.F)
    So x=y and you'll have F(x) in function of x.

    Now for every x position you now know the absolute amount of force in the direction of OC.

    And then I'm stuck...
    How do you do the integral? Because you have to project the x axis to the bissectrice to do the integral...

    Or maybe it's just: W = int(O->A)(F(x)dx)
    Maybe: int(0->A)(F(x)cos(45).x.dx?
    Last edited: Nov 17, 2007
  6. Nov 17, 2007 #5


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    I am not sure you got this. Start from
    [tex] W = \int (f_x dx + f_y dy) = \int (2 y dx + x^2 dy) [/tex]
    where I used your force in the second step. Now use y =x for your path (so that dx=dy). replace everything in terms of x (or of y, it does not matter). Now integrate. You will get the correct answer.
  7. Nov 17, 2007 #6

    THANK YOU very much, nrqed.

    If you would like help with drums, fighting, DHTML or elementary linux, I'd be happy to return the favour.


    Do you understand it as well, katchum?
  8. Nov 17, 2007 #7
    I understand, but can someone help me with my other technique problem? The difficult solution?

    I mean, it is right to say that W = (the displacement) X (the force in the direction of the displacement)?
  9. Nov 17, 2007 #8

    It is not the cross product, it is the dot product:

    [tex] W = \Delta \vec r \cdot \vec F [/tex]
  10. Nov 17, 2007 #9


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    You are every welcome. But please don't fight!! :frown:
    Last edited: Nov 17, 2007
  11. Nov 17, 2007 #10


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    yes, you are correct (if you X means an ordinary product and by displacement you mean the magnitude of the displacement vector). The difficulty is to set up things when the force is not constant (or the displacement is not a straight line). Your earlier post seemed to be heading in the right direction but it was too vague. If you would actually work out explicitly all the steps it would get back what I wrote.
    Last edited: Nov 17, 2007
  12. Nov 17, 2007 #11
    Aaargh! I quit!

    I tried to calculate my things there but it was way too difficult to do. Let's keep it simple and shove this alternate solution somewhere in a corner... :)

    I mean I got square roots everywhere, and machs to the number of 4 and all. And we're not even close to the integral...
    Last edited: Nov 17, 2007
  13. Nov 17, 2007 #12
    Heheheh, point taken :) .

    Katchum, it was difficult for me to understand your alternate method. If you aren't worried about it, neither am I :cool: . If you want to figure it out, I'm willing to help out as well but I'd need you to repost with more literal equations. Take care! :approve:
    Last edited: Nov 17, 2007
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