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Representing Free body diagrams

  1. Oct 22, 2015 #1
    I have a simple question. I know that the objects that we work with in introductory physics are point particles. Thus, say we have a picture frame that is put on a wall. There is a wire holding it up, where the wire is attached to the top right and the top left corners of the frame. The passes through a nail so that the whole picture frame stays up. the wire makes a triangular shape with the picture frame. My question is how do we represent the tension forces on the frame in a free-body diagram if the picture frame is not a point particle, since the tension forces are acting on the corners of the frame.
     
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  3. Oct 22, 2015 #2

    mfb

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    You can reduce the picture frame to a beam between the two points where the wire is attached. Symmetry allows to ignore rotations of the beam, so you can simplify the problem sufficiently to get back to point-like objects.
     
  4. Oct 22, 2015 #3

    Doc Al

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    Well, not always, as your example illustrates. You can certainly draw an extended free body diagram where the points of application of the forces are indicated and the distances between them are shown.
     
  5. Oct 22, 2015 #4
    So it's not possible to represent the situation I am talking about with a point-like free-body diagram?
     
  6. Oct 22, 2015 #5

    Doc Al

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    It would be somewhat odd to represent the picture as a point object. Why would you want to? (Depending upon your purpose, the location of the forces may not matter.)
     
  7. Oct 22, 2015 #6
    Well I am trying to find at which angle the wires approaches makes with the horizontal makes the tension approach a maximum. It would seem I would have to use Newton's 2nd law on a point particle to find out.
     
  8. Oct 22, 2015 #7

    Doc Al

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    Don't get hung up on the diagram. Newton's 2nd law applies to extended objects just as it does to point particles.
     
  9. Oct 22, 2015 #8
    So if I were to apply F = ma to the system I have described, it would be something like ##\displaystyle \sum \vec{F} = \vec{T} + \vec{T} + \vec{W} = \vec{0}##, which would mean ##\displaystyle T = \frac{mg}{2\sin \theta}##?
     
  10. Oct 22, 2015 #9
    In general, if an extended object is acted upon by forces applied at different points of the object, and is in equilibrium, the particle model fails, and you have to satisfy conditions for translational equilibrium (sum of forces = 90), and rotational equilibrium (sum of torques = 0).
     
  11. Oct 23, 2015 #10

    Doc Al

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    Sure. (Assuming symmetry, of course.)
     
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