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Tension of hammock problem

  1. Jun 7, 2005 #1
    hi I have this hw problem where I don't know how to approach it.

    A person weight 150lbs lies in a hammock. the rope at the person's head makes an angle of 40 with the horizontal, while the rope at the persons feet makes an angle of 20. find the tension in the two ropes.
  2. jcsd
  3. Jun 7, 2005 #2


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    You can model this problem as three strings and a weight hanging from the central string with the other two tied off at the specified angles. The details of the curvature of the hammock under the person are of no consequence to computing the tension in the end ropes. Can you do the simpler looking problem?
  4. Jun 7, 2005 #3
    ok i had that drawn. this is my first time posting so i should have been a little bit more specific where i was stuck. so what is the mathematical relationship between the weight of the person and the two angles made. do i break it down into the vertical components?
  5. Jun 7, 2005 #4


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    You're going to end up with two equations with two unknowns. Break the tension forces into vertical and horizontal components. Once you get the equations for the component directions, substitute the term for the overall tension back in to get your two unknowns and then solve.
  6. Jun 7, 2005 #5
    ok i did that i tried to do a relationship that was similar in another problem that I had and I tried to do it like this

    for example i took the rope with the angle that was 20 degrees and tried this
    T2Sin20 - T1 = 0 = T2Sin20 - W

    so T2sin20 = W
    T2 = w/sin20 = 2.92 W

    but that was nowhere near the answer....this worked when one of the ropes was in the horizontal.

    I think i need some more guidance on this problem :-/
  7. Jun 7, 2005 #6


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    We need to agree on notation. It looks like you are calling T2 the tension in the string at 20 degrees. Then let T1 be the tension in the string at 40 degrees. Both of these have vertical components, and the sum of their vertical components must support the weight. What can you say about their horizontal components?
  8. Jun 7, 2005 #7
    i am thinking that the sum of the horizontal components would equal zero because they cancel one another out.

    so T1cos40 + T2Cos40 = 0

    or am I wrong?

    how do i apply these concepts mathematically?
  9. Jun 7, 2005 #8


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    You are correct.

    Almost. They'll be in opposite directions so their signs will be opposite.

    You just did.
  10. Jun 7, 2005 #9


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    Like OlderDan said, you got T2 in terms of the vertical component, but you didn't the same for T1.
  11. Jun 7, 2005 #10
    ok ok things are starting to come together...kinda

    so with my mathematical relationship for the vertical components of

    T2 = W/sin20
    T1 = w/sin40

    would that be the answer of the tension OR just the tension in the y direction

    because when I calculate this
    T2 = 438
    T1 = 233
    and the answers should be 132 and 162 lb....so that doesn't look right at all

    ...do i have to find out what the tension is in the horizontal component and then find the resultant of the two using pythageoreans theorem...or am i just getting off track now :).

    also i'm a bit confused to do with my other relationship of

    T2Cos20 - T1Cos40 = O
  12. Jun 7, 2005 #11
    please i'm still stuck...more help would be appreciated!
  13. Jun 7, 2005 #12


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    This is a classic in Statics. It's 2 equations and 2 unknowns problem, simply do the sum of forces in y and x, or you could use the parellogram and the law of sin and solve it geometrically and faster.

    Solving it with sum of forces.

    [tex] \sum F_{x} = T_{2} \cos 20 - T_{1} \cos 40 = 0 [/tex]

    [tex] \sum F_{y} = T_{2} \sin 20 + T_{1} \sin 40 - W_{g} = 0 [/tex]
  14. Jun 7, 2005 #13

    Doc Al

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    This is incorrect. Instead, realize that in order for the man to be in balance, the vertical components of tensions must add up to be equal to the weight.

    What is the vertical component of T2? of T1? Then write the condition for vertical equilibrium (which I just stated above).

    This will give you an equation in terms of T1 and T2. In order to solve for T1 and T2, you'll need another equation. The one for the horizontal components.

    That's the second equation you'll need.
  15. Jun 7, 2005 #14
    alright guys got that ...makes sense

    but i think its the having of the 2 unknowns that is really killing me..I haven't used algebra in a while.

    so my equations for the vertical is

    .34T2 + .64T1 = Wg

    how now do i manipulate this to find the values of T1 and T2?

    oh wait...do i solve this as a system of equations and use like the 2nd equaiton of
    T2Cos20 - T1Cos40 = O and re-arrange that in terms of T1 and T2 and use that to substitute into the first equation??
  16. Jun 7, 2005 #15

    I solved it!!
    thanks everyone!
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