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Question about force on a curve

  1. Dec 6, 2008 #1
    i know that the formula of work is W=F*X

    but when i saw how they describe the work of a constant force applied on a mass moving at a curve it looks very different (integrals,differentials)

    i cant see the logic of these equations??

    http://img177.imageshack.us/img177/7741/64565014ea9.gif
     
  2. jcsd
  3. Dec 6, 2008 #2

    Hootenanny

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    The expressions can be derived from the general definition of work done by a force:

    [tex]W = \int_\gamma \bold{F}\left(\bold{x}\right)\bold{\cdot}d\bold{x}[/tex]

    Where the integral is taken over the path [itex]\gamma[/itex] which the force acts and x is the spacial variable (vector).
     
    Last edited: Dec 6, 2008
  4. Dec 6, 2008 #3

    Doc Al

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    That's OK as a basic definition, but it only works if the force is constant and in the same direction as the displacement. A more general definition is:
    [tex]dW = \vec{F}\cdot d\vec{s}[/tex]

    [ah... Hoot beat me to it while I was getting coffee!]
     
  5. Dec 6, 2008 #4
    so how do i apply this formula (dot product of force and distance)
    to my case

    [tex]
    W = \int_\gamma \bold{F}\left(\bold{x}\right)\bold{\cdot}d\bold{x}
    [/tex]?
     
  6. Dec 6, 2008 #5

    Doc Al

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    Why don't you describe the problem you are trying to solve. (It's not obvious from your diagram.)
     
  7. Dec 6, 2008 #6

    Hootenanny

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    For the straight section of the path, between A & B:

    F(x) = F0 [itex]\hat{i}[/itex]

    And

    dx = dx1 [itex]\hat{i}[/itex]

    So

    [tex]\bold{F}\left(\bold{x}\right)\bold{\cdot}d\bold{x} = F_0 dx_1 \underbrace{\left(\hat{i} \cdot \hat{i} \right)}_{1}[/tex]

    Hence,

    [tex]W_{AB} = \int^B_A F_0 dx_1 = F_0\int_B^A dx_1 = F_0\left[x_1\right]^B_A[/tex]

    [tex]W_{AB} = F_0\left(B-A\right) = F_0\alpha[/tex]

    Do you follow?

    Note that I am using the notation [itex]\bold{x} = x_1\hat{i} + x_2\hat{j}[/itex]
     
  8. Dec 6, 2008 #7
    i want to build the equations of work
    from point B to D of my diagram

    the equation that i was given as solution are odd
    i cant follow their logic on the diagram ??
     
  9. Dec 6, 2008 #8

    Doc Al

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    But you haven't told us why you can't follow the logic or what you think the answer should be.

    If I assume, like Hoot did, that the force F is constant and in the +x direction, then I disagree with the answer. Note that under this assumption no integration is needed to get the answer. (But it makes a good exercise.)
     
  10. Dec 6, 2008 #9
    i cant understand how they use angle theta
    ??
     
  11. Dec 6, 2008 #10

    Doc Al

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    First things first: Is it true that the force is constant and directed toward +x?

    In finding the work done, you must take the component of the force in the direction of the displacement (or vice versa). That involves the angle theta.
     
  12. Dec 6, 2008 #11
    the force is constant

    i need to build an integral of force with small displacement
    the displacement cannot be an angle

    its suppose to be a small distance
    ??
     
  13. Dec 6, 2008 #12

    Doc Al

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    Assuming that curved section is part of a circle, a small displacement (ds) can be written in terms of a small angle (dθ) via ds = Rdθ.
     
  14. Dec 6, 2008 #13
    Rsinθ can meen a distance
    but Rdθ is radius times angle
    that is not a small distance
    ??
     
  15. Dec 6, 2008 #14

    Doc Al

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    When the angle is measured in radians, Rθ is an arc length. When θ is small, as in dθ, the length (Rdθ) is small. (Review the concept of radian and arc length.)
     
  16. Dec 6, 2008 #15
    theta is in radians

    so Rθ=Rsinθ
    ??
     
  17. Dec 6, 2008 #16

    Doc Al

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    No. (That would imply that θ=sinθ, which is not true in general.)
     
  18. Dec 14, 2008 #17
    can you recommend a manual for that kind of questions
    explaining the integral thing with the use of angles

    for me an integral is a way to calculate an area
    i cant link it to work
    even more if they use angles
    ??
     
  19. Dec 15, 2008 #18

    turin

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    Have you taken multivariable calculus? You should find this stuff in a calculus text. BTW, the work integral can be thought of as an area, but you have to imagine that it is out of the page. It is sort of like the area of a ribbon that you are looking edge on, so it can curve along the path and all you see is a line. (But this analogy only works in 2 dimensions.) Check the Wikipedia article on "line integral". Wikipedia even has the work integral as the prototypical example in the introduction.
     
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