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Did I solve this correctly? (Calculating forces at rest)

  1. Feb 12, 2014 #1
    1. The problem statement, all variables and given/known data
    dXURK4v.gif
    A beam (red line) has mass m and length L.
    3 chains are attached to the beam to keep it at rest. C1 which is perpendicular to the roof, C2 which is perpendicular at the center of mass on the beam, and C3 which is perpendicular to the side wall.

    Questions:
    Draw a free body diagram.
    Calculate the forces from the 3 chains.

    2. Relevant equations
    Newton's law, torque.


    3. The attempt at a solution
    FBD:
    GKya3ke.gif

    To find the forces from the 3 chains, I set up 3 equations. Using Newton's 1st law for forces on the x-axis and again Newton's 1st law for forces on the y-axis:

    [itex]\rightarrow \sum F_x=0: C2 \cdot \cos \theta - C3=0[/itex]

    [itex]\uparrow \sum F_y=0: C1 - C2 \cdot \sin \theta - mg=0[/itex]

    For the 3rd equation, I use the torque on southern most point of the beam (where C3 is attached) Counter-clockwise:

    [itex]C1 \cdot L \sin \theta - C2 \cdot \frac{L}{2}=0[/itex]

    Now I have 3 equations which is enough to find the values for C1, C2, and C3.
    What do you guys think of my method? Did I do this correctly?
     
  2. jcsd
  3. Feb 12, 2014 #2

    Simon Bridge

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    Resolving forces to horizontal and vertical can hide some subtleties.
    Why did you choose that particular pivot point? Did you check the result against other pivot points?
    Using that pivot - doesn't the weight also provide a torque?

    I'd resolve the forces to those going through the com and those acting perpendicularly to the line joining their point of action and the center of mass ...
     
  4. Feb 13, 2014 #3
    Oh yes, you're correct, I forgot about the weight in my torque equation.

    I'm not really sure why I chose that pivot point. I always choose the point with most forces going through the point (to cancel them out since the "arm" between the force and the torque point becomes zero).

    I'm not sure exactly what you mean by the center of mass? Would you choose the center of the mass on the beam to be your torque point?

    I guess that does make sense because we have 4 forces, C1, C2, C3, and the weight. Using a pivot point at the the center of mass will eliminate both C2 and the weight as opposed to only eliminating C3 in my previous attempt. Thus the torque equation when using the center of mass will only include C1 and C3. Would that be the best solution?
     
  5. Feb 13, 2014 #4

    Simon Bridge

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    When you draw a free-body diagram, the body is drawn with no physical support and everything has to be modeled by forces. If you apply an arbitrary force to a free object, that object will generally accelerate as well as rotate. The point that it will naturally rotate about is the center of mass.

    It is generally easier to exploit the symmetry of the problem where you can.
    Stick a Cartesian coordinate system on the diagram - doesn't it seem "natural" to line the axes up with the rod and put the origin on the com?
     
  6. Feb 13, 2014 #5
    Why would you put the origin of the coordinate system at the com? I don't see how that would help. Why would the origin of the coordinate system matter?
     
  7. Feb 13, 2014 #6

    haruspex

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    Putting the torque reference point where some lines of force intersect allows you to omit those. There are 5 such points here, but only two intersect at any point. Some are hard to locate, but the CoM is easy, and the perpendicular distances to the remaining two forces are easy.
     
  8. Feb 13, 2014 #7

    Simon Bridge

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    Why would anyone not? I can tell from your approach to the problem above that you don't default to using symmetry when you do calculations. It can take a while to get this attitude. Most people seem to manage it by post-grad where it's pretty much all anyone talks about.

    Try it and see for yourself ;)

    Since there is no motion, there is no physical pivot. So you are free to sum the torques about any point, so you pick the point where the math is easiest. It's also good to pick the point where the physics is clearest - so it is easiest to troubleshoot when things go wrong and easiest to defend your work should it be challenged. You'll see what I mean when you try it - especially for more complicated problems.

    However - you can get an inkling by comparing with the horizontal beam/distributed load problems - this is a generalization of those - where you usually need to sum the torques about several (usually two) potential pivot points to get the effect of the load at the supports.

    Aligning the coordinate system with the beam simplifies the problem - you can put the origin anywhere along the beam you like - but, in general, more symmetry means easier maths.

    Mind you - doing things just for the symmetry is not always optimal:
    https://xkcd.com/403/

    After a while you find that you can get a bit obsessive about symmetry and particularly structure:
    http://cpsievert.github.io/projects/615/xkcd/
     
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