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Find angles between two rope and the ceiling

  1. Sep 20, 2011 #1
    ***Excuse the pluralization error in the title***

    1. The problem statement, all variables and given/known data

    I attached a picture below...


    2. Relevant equations

    sin^2(theta)+cos^2(theta)=1

    a^2+b^2=c^2

    Vector properties

    Trig identities

    3. The attempt at a solution


    I am familiar with trig identities and vector properties, but I can't get anywhere with this. I tried using the 697N weight as the y-component vector for either side, but that didn't work.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

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  3. Sep 20, 2011 #2

    PeterO

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    resolve each tension into vertical and horizontal components.
    The horizontal components have to balance each other,
    The vertical components together support the mass.
     
  4. Sep 20, 2011 #3
    I understand that part. The part I don't understand is how to separate each vector into its components. I only have the magnitude with no given angles. Looking at the hint, I need to play around with some trig. But I can't see how to incorporate the identity it hints at.

    And I can't use 697N as the shared vertical component, right? Because that would make this much easier, but it did not work for me.
     
    Last edited: Sep 20, 2011
  5. Sep 20, 2011 #4

    PeterO

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    You have two unknowns, theta 1 and theta 2

    Vertical considerations will give one equation involving them

    Horizontal considerations will give a second equation connecting them

    Two equations in two unknowns should mean a simultaneous equations solution is possible.

    EDIT: you could also use the cosine rule to find the angles.
     
    Last edited: Sep 20, 2011
  6. Sep 20, 2011 #5
    Ah, I see what you mean. Thank you so much!
     
  7. Sep 20, 2011 #6

    PeterO

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    Note my edit about the cosine rule.
     
  8. Sep 20, 2011 #7
    Witch cosine rule? The Law of Cosines?
     
  9. Sep 20, 2011 #8

    PeterO

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    yes: a2 = b2 + c2 - 2bc.cos(A)
     
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