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Tension of ropes when laying on a hammock

  1. Feb 12, 2012 #1
    1. The problem statement, all variables and given/known data

    A 50.0 kg person takes a nap in a backyard hammock. Both ropes supporting the hammock are at an angle of 15° above the horizontal. Find the tension in the ropes.

    2. Relevant equations

    T=mg, F=ma, T= (Tcos15°)+(Tsin15°)


    3. The attempt at a solution

    What I did was I started with X and Y coordinates and I figured the coordinates in the X direction would cancel out to be 0 because in one X direction it is positive and the other is negative so the total tension would be the tension in the y direction multiplied by two. What I did was I got the total tension as ƩF= 2Tsin15° which is nowhere near the correct answer in the answer key. The correct answer is 948N which would be the cos of 15 degrees and not the sin. I don't understand how you know which one to us because if the ropes are 15 degrees above the horizontal I figured the x coordinates would cancel out.
     
  2. jcsd
  3. Feb 12, 2012 #2

    PeterO

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    What I did was I got the total tension as ƩF= 2Tsin15°

    seems OK.

    You do need to transpose to find T of course.

    T = ƩF/2sin15° where ƩF is of course the weight of the person: 50*9.8
     
  4. Feb 12, 2012 #3
    Thank you so much. I don't know where I went wrong with that one.
     
  5. Feb 12, 2012 #4

    PeterO

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    Note that the first line of my previous post was a "cut and paste" from your original post, so you started out OK.
     
  6. Feb 12, 2012 #5
    Yes but I figured out the answer with that specific equation. I guess I just didn't think of dividing it out. Error on basic Algebra.
     
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