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Tension on two cables.

  1. Sep 25, 2009 #1
    1. The problem statement, all variables and given/known data

    A chandelier with mass m is attached to the ceiling of a large concert hall by two cables. Because the ceiling is covered with intricate architectural decorations (not indicated in the figure, which uses a humbler depiction), the workers who hung the chandelier couldn't attach the cables to the ceiling directly above the chandelier. Instead, they attached the cables to the ceiling near the walls. Cable 1 has tension T1 and makes an angle of [tex]\theta[/tex]1 with the ceiling. Cable 2 has tension T2 and makes an angle of [tex]\theta[/tex]2 with the ceiling.

    Find the forces in the x direction.

    Find an expression for T1, the tension in cable 1, that does not depend on T2.

    2. Relevant equations
    FT = ma


    3. The attempt at a solution
    Would I be correct in saying that the total Ftx = mgsin[tex]\theta[/tex]1+mgsin[tex]\theta[/tex]2? If so, I would take that and add it to mgcos[tex]\theta[/tex]1+mgcos[tex]\theta[/tex]2 for the total.

    P.S This is my first post so I apologize if anything is in the wrong place.
     
  2. jcsd
  3. Sep 25, 2009 #2

    kuruman

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    Gold Member

    Hi gb14, welcome to PF.

    No, you would not be correct. The sum of the vertical components of the two tensions must be equal to the weight of the chandelier. That's because there is no unbalanced force in the vertical direction and you need to say that with an equation.
    Again, no. Forces are vectors and you cannot add the vertical components to the horizontal components in any meaningful way. You need to say that the two horizontal components of the tension are equal and opposite. That's because there is no unbalanced force in the horizontal direction either.

    When you write the vector equation

    [tex]\vec{F}_{net}=m \vec{a}[/tex]

    that's really two equations in one. One equation for the horizontal direction and a separate equation for the vertical direction.
     
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