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Moments, calculating torque ,

  1. Dec 12, 2008 #1
    1. The problem statement, all variables and given/known data

    I am stuck on question 6c on this paper. Can anyone have a look at it please. I dont know were to start.

    do I place the rock 'x' meters away from one of the strings and set one of the strings tension to 40g?


    2. Relevant equations



    3. The attempt at a solution
     

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  3. Dec 13, 2008 #2

    Doc Al

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    Yes, that will work.

    Find the tension in the strings as a function of 'x'.
     
  4. Dec 13, 2008 #3
    Okay I have done this, but I don't get the right answer.

    heres my working,

    taking moments about A [tex] (490 \times 4) + (196 \times x) = 392 \times 8 [/tex]

    [tex] 1960 + 196x = 3136 [/tex]

    [tex] 196x = 1176 [/tex]

    [tex] x = 6 [/tex]

    is my diagram correct?
    http://www.mathhelpforum.com/math-h...h/9244d1229179227-mechanics-help-untitled.jpg

    the right answer is [tex] \frac{1}{2} [/tex]
     
  5. Dec 13, 2008 #4

    Doc Al

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    Your work and diagram are perfectly correct. You just need to translate your answer into the form that they want. They asked for the fraction of the board, not the position of the mass. :wink:
     
  6. Dec 13, 2008 #5
    I dont know how to work that out....
     
  7. Dec 13, 2008 #6

    Doc Al

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    What does your answer (x = 6) mean? What happens for x > 6? You did torques about point A, what if you did the same thing about point B?

    (Hint: In your diagram, mark each portion of the board where placing the mass would lead to a string breaking.)
     
  8. Dec 13, 2008 #7
    x=6 means the distance the weight of the rock acts from point A?

    if x is greater than 6 than one of the strings will break right?
     
  9. Dec 13, 2008 #8

    Doc Al

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    Right. So how much of the board is "off limits" for the string not to break? (Consider both ends.)
     
  10. Dec 13, 2008 #9
    8-6=2?
     
  11. Dec 13, 2008 #10

    Doc Al

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    Yes, viewed from side A. Now view from the other side.
     
  12. Dec 13, 2008 #11
    it would be the same wouldn't it?

    I dont think I am going to ever understand this hehe

    so thanks for you help!

    edit: is it 4?
     
  13. Dec 13, 2008 #12

    Doc Al

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    That's right.

    Yes. Analzying things form side A tells you that placing the mass between 6 and 8 meters from A will cause string B to break. A similar analysis applies from side B. That means 2 m on both ends of the board are off limits--a total of 4 m out of 8. What fraction is that? :wink:
     
  14. Dec 13, 2008 #13
    OH MY GOD I GET IT NOW!!

    thanks for patiently helping me
     
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