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Tension! (and torque?)

  1. Nov 12, 2007 #1
    1. 'A rope of negligible mass is stretched horizontally between two supports that are 3.44 meters apart. When an object of weight 3160 N is hung at the center of the rope, the is observed to sag by 35.0 cm. What is the tension in the rope?'

    2. Does this have anything to do with torque?!?!

    3. ATTEMPTS:

    I divided 3.44 by 2 (1.72) to find the horizontal length from one support to the object. From that I found that the length of the rope from the object to one of the supports is 1.72^2 + .35^2 = X^2 (it's a triangle), x=1.75

    I can assume because the object is in the center of the rope that the tension on both sides of the rope is equal. Can I also assume the forces sum to 0 because its stagnant? BAH I'm so confused, I don't know what to do!
    The correct answer is 7920. Help!
  2. jcsd
  3. Nov 12, 2007 #2
    it's not torque. draw a free-body diagram of the rope and label all the forces.
  4. Nov 12, 2007 #3
    I drew a free-body diagram and I'm still confused. I really have no idea how to do this. How can I solve for tension!? I don't know the tension along the x axis or the y axis ? I know the angle - 11.5 degrees.
  5. Nov 12, 2007 #4
    you DO know the downward force on the rope. you also know that the forces at either side should do what? is the rope moving? you have an angle and a side, you can get the other side, provided you draw a right triangle with the vectors.
  6. Nov 12, 2007 #5
    ahhh i see, the rope is not moving so forces in the x and y direction are balanced.. i think i understand. so EFy = (<-- supposed to be sigma) = 0 = -W + 2Frope,y .. so Frope,y = 1580 on each side and the angle is 11 degrees .. so 1580*sin11=7925 .. about the right answer .. but that is the tension in the rope for only one side ? Oh dear, well thank you anyway !
  7. Nov 12, 2007 #6
    remember that sin = opposite/hypotenuse
    Last edited: Nov 12, 2007
  8. Nov 12, 2007 #7
    i think that sin = opposite / hypotenuse .. o/a = tan
  9. Nov 12, 2007 #8
    oops, i'm tired :(

    besides the original poster still used the sin funtion to calculate the hypotenuse by multiplying by the opposite if i follow correctly.
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