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Tension in two ropes at angles

  1. Oct 12, 2008 #1
    This question really confuses me, I have got answers before but they dont fit with the answer in the book and it's due tomorrow

    1. The problem statement, all variables and given/known data

    A large helium balloon is attached to the ground by two fixing ropes. Each rope makes an angle of 50 degrees with the ground. There is a force F vertically upwards of 2.15x10^3 N. The total mass of the balloon and it's basket is 1.95x10^2 kg

    Find the tension in either rope

    2. Relevant equations

    I'm not sure what equation to use, in my book theres only upthrust = weight + tension which doesn't work with angles, and the equilibirum equations are explained properly in my text book

    3. The attempt at a solution

    upthrust = weight + tension

    tension = upthrust - weight
    =2.15x10^3 - 1911
    =239
    239/2 = 119.5

    but the answer in the book says it should be 131 N
    Please help.
     
  2. jcsd
  3. Oct 12, 2008 #2
    Sum up the forces in each plane (x and y) and make them equal zero.

    So you have, upthrust = weight + 2tensiony
    and tensionx - tensionx = 0

    Solve the simultaneous eqn
     
  4. Oct 12, 2008 #3
    How would I work out the tension in x, as there is no stated force going along horizontally?
     
  5. Oct 12, 2008 #4
    Well you've done it yourself!!!!
    Look, nothing is moving in the x direction, however both ropes have x components, that means that the x components must be equal and opposite, to cancel each other out.
     
  6. Oct 12, 2008 #5
    erm I'm still really confused so the tensiony = 119.5, right?
    I dont see how the x part comes into it, or how to acheive it
    I know they cancel each other out but for all I know it could be +1 -1 :confused:
     
  7. Oct 12, 2008 #6
    Uh, sorry I have this terrible habit of not reading the question properly haha, it didn't click to me that their angles are the same.

    Well the y tension is 119.5, so you just need to calculate the magnitude of tension in the rope, however this won't give 131.. closer to 150 by my guess, perhaps try it incase my estimate is wrong
     
  8. Oct 12, 2008 #7
    I can see the problem. They take gravity as 10ms-1
    Finding the y component of tension again, then finding the total tension will yield 130.5N
     
  9. Oct 12, 2008 #8
    ahh I did think the books answer may be wrong
    so the magnitude of the tension in the rope would be sin 50 x 199.5 = 152.8?
     
  10. Oct 12, 2008 #9
    No... Perhaps it would help you if you drew a diagram.

    The hypotenuse is what you're trying to solve, and we have the opposite length.

    so it will be [tex] \frac{200}{2sin(50)} = 130.5N [/tex]
     
  11. Oct 12, 2008 #10
    ohh, right thank you
    I forgot the way of re arranging the trig function
     
  12. Oct 12, 2008 #11
    No worries! I hope it all makes sense.
     
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