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Tension with Different Angles

  1. May 10, 2007 #1
    1. The problem statement, all variables and given/known data

    A climber is suspended between two cliffs and is closer to the left cliff. The left part of the rope makes an angle of 65 to the cliff and the right part an angle of 80 to the cliff. The climber weighs 535N.

    [​IMG]

    2. Relevant equations

    Trigonometry...

    3. The attempt at a solution

    What I did at first was assume that the verticle force on the two sides is equal, then use Trig. to calculate the two tensions, but the answers I got seemed way off...And as I think of it more I think the two Y tensions of the ropes may not be 535/2 as I first thought...So I'm stuck here...

    I keep re-reading my book for clarification, but I can't seem to grasp what I must do.
     
    Last edited: May 10, 2007
  2. jcsd
  3. May 10, 2007 #2

    hage567

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    The tensions will not be the same. You must resolve the forces into their x and y components using trig, then sum them up in each direction to get your two equations with two unknowns.
     
  4. May 10, 2007 #3
    Ahhh, of course! I can't beleive I overlooked that...I was thinking too much, thanks.

    So, after doing all the algebra and trig, I came up with this...

    From the equations for the components:

    T1=Left, T2=Right

    T1 Sin 65 + T2 Sin 80 = 0
    T1 Cos 65 + T2 Cos 80 = 535

    Left string's tension = T1 = 2032.654N
    Right string's tension = T2 = 1870.629N

    Does that look right?

    Thanks again.
     
  5. May 11, 2007 #4

    hage567

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    Don't forget about the force due to the mass of the climber! You must use ALL the forces. That acts in the vertical direction.

    Also, in the horizontal direction, the tension in the first rope acts in the opposite direction of the second rope. So you must subtract instead of add.
     
  6. May 11, 2007 #5
    So, in the vertical part:

    T1 Cos 65 + T2 Cos 80 = 535

    Doesn't the 535 take into account the weight of the climber? Since the two verticle upward forces of each rope must add together to equal the single downward force due to the mass of the climber?
     
  7. May 11, 2007 #6

    hage567

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    Yes, you are right about that. Sorry, I think I misread your earlier post.

    As for the horizontal components of the tension, T1 acts to the left, while T2 acts to the right. So when summing up the forces it will be the difference between the two. Does that make sense?
     
  8. May 11, 2007 #7
    Yeah, I think I get it now, thanks a bunch!
     
  9. May 11, 2007 #8

    hage567

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    Hmmm, when I work through your equations, I get a different answer. One way to check if your that you didn't make a mistake in your algebra is to put your values back into your original equations, and see if they make sense.
     
  10. May 11, 2007 #9
    You're right, I went through it kind of fast and had a wayward "-" sign.

    After fixing that I got:

    T1 = 917.456
    T2 = 844.18

    Which makes a lot more sense, is that close to what you got?
     
  11. May 11, 2007 #10

    hage567

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    Yes, within rounding, those are the values I got.
     
  12. Oct 28, 2009 #11
    hi how would you soolve for t1 and t2?
     
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