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Tension in one of two ropes of a hung sign

  1. Oct 7, 2012 #1
    1. The problem statement, all variables and given/known data

    A sign hangs precariously from your prof's office door. Calculate the magnitude of the tension in string 1, if theta1 = 35.67°, theta2 = 62.43°, and the mass of the sign is 3.3 kg. 1262751-1315-setDynamics_no_friction-prob8--prob29.gif

    2. Relevant equations


    3. The attempt at a solution
    Ft1=cos35.67t1 + cos117.57t2=0
    Fxt1=0.57t2

    Fyt2=sin35.67(0.5t2)+sin117.57t2=-32.34(from gravity and mass of sigh)
    in the end it gave me t2= 27.5 and t1= 15.675
    ... which was wrong
    What do I need to do
     
  2. jcsd
  3. Oct 7, 2012 #2

    TSny

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    Don't know if you just had a typo there, but the .5 should be .57 . Should you have a negative sign on the right side of the last equation (in front of 32.34)?
     
  4. Oct 7, 2012 #3
    thank-you! yes .57 was a typo, but I think I forgot the (-) sign in front of 32.34.
     
  5. Oct 7, 2012 #4

    TSny

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    Good. But just to make sure, there should not be a negative sign in front of the 32.34 on the right hand side of the equation.
     
  6. Oct 7, 2012 #5
    I calculated the answer with the negative sign, but then took it out in the final answer because it was the magnitude they were looking for. Does it give the same answer because I ended up with the right answer anyway?
     
  7. Oct 7, 2012 #6

    TSny

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    A good way to think about it is that the sum of the vertical components of all of the forces must equal zero (for equillibrium). So, you have an equation with the sum of vertical components of all forces on the left side and zero on the right. If you take upward as the positive vertical direction, then the vertical component of weight will be negative. Then, if you take this term over to the other side of the equation, it will be positive there.
     
    Last edited: Oct 7, 2012
  8. Oct 7, 2012 #7

    TSny

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    Symbolically: [itex]\sum t_1y + t_2y + W_y = 0[/itex]
    Since ##W_y = -mg##,
    [itex]\sum t_1y + t_2y -mg = 0[/itex] or [itex]\sum t_1y + t_2y = mg[/itex]

    (no negative sign in front of mg on the right)
     
  9. Oct 7, 2012 #8
    Ohh, ok I get it! Thank-youu!!
     
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