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Calculation Help Needed

  1. Apr 7, 2007 #1

    Air

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    1. The problem statement, all variables and given/known data
    Find the mass:
    [​IMG]

    2. Relevant equations
    Not Sure.


    3. The attempt at a solution
    I don't know where to start as I believe that there is less information.

    All that I have got upto is Tan-1(3/8) but I don't know if this is needed to find the mass? :confused:
     
  2. jcsd
  3. Apr 7, 2007 #2

    Hootenanny

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    I'm assuming the frame is massless, pivoted bout the right angle of the triangle and the system is in equilibrium, am I correct?
     
  4. Apr 7, 2007 #3

    Air

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    Yes, You are correct.
     
  5. Apr 7, 2007 #4

    Hootenanny

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    So, we know that since the system is in equilibrium, the sum of the moments about any point should be zero. However, lets consider the forces; if the system is not translating (i.e. moving up or down) what must be the vector sum of the vertical forces?
     
  6. Apr 7, 2007 #5

    Air

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    Erm...Not sure but guessing zero. :redface:
     
  7. Apr 7, 2007 #6

    Hootenanny

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    Sounds good to me :smile: and where will this reation force be applied?
     
  8. Apr 7, 2007 #7

    Air

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    Either at the angle at the end of Y or near the mass?
     
  9. Apr 7, 2007 #8

    Hootenanny

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    Normal forces are applied at the point where the system pivots. Just checking, is the side x attached to a wall of some kind?
     
  10. Apr 7, 2007 #9

    Air

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    Nope. The pivot is at the corner of x and y.

    x and y dimensions are given if to calculate the angle (I think).
     
  11. Apr 7, 2007 #10

    Hootenanny

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    Okay, let us consider the torques about the pivot point;

    [tex]0.08T\sin\left(\arctan\left(\frac{3}{8}\right)\right) = 0.4mg[/tex]

    Where T is the tension in the diagonal member. Do you follow?
     
    Last edited: Apr 7, 2007
  12. Apr 7, 2007 #11

    Air

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    Yes, I do. :smile:
     
  13. Apr 7, 2007 #12

    Air

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    Can Tension be calculated without mass? :confused:
     
  14. Apr 7, 2007 #13

    arildno

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    You still have not set up the equalibrium of FORCES, only of torques..

    Also, for your information,
    [tex]\sin(\arctan(x))=\frac{x}{\sqrt{1+x^{2}}}[/tex]
     
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