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How is this system in equilibrium?

  1. Oct 14, 2012 #1
    [diagram attached]

    Assume some friction between M and the surface.

    The forces acting on the connector point are the forces of friction of the M blocks and the weight of m, but they are in different directions; how do they manage to cancel out?
     

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  2. jcsd
  3. Oct 15, 2012 #2

    CWatters

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    By being in opposite directions.

    Aside: If the string (?) joining the blocks really was horizontal the tension would be infinite so it must sag a bit. Assume it sags to an angle θ with the horizontal. Then look at the horizontal and vertical components. The vertical components adds to the normal force on the blocks. The horizontal force is opposed by friction etc.
     
    Last edited: Oct 15, 2012
  4. Oct 15, 2012 #3
    But the weight of m is perpendicular to the frictional forces
     
  5. Oct 15, 2012 #4

    CWatters

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    There are other vertical forces.

    See my earlier reply that I edited.
     
  6. Oct 15, 2012 #5

    CWatters

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    Can I check the line joining the blocks is indeed something like string and not a rigid rod?
     
  7. Oct 15, 2012 #6
    Yes. Sorry I thought that was implied.

    Ah, I see now. It is physically impossible for the strings to be 100% perpendicular, correct?
     
  8. Oct 15, 2012 #7

    CWatters

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    Yes because the Tension in the string is T = mg/sin(θ) and sin(θ) approaches zero. Perhaps draw it with the string forming a V and work out all the forces.
     
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