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Another Mechanics problem. (Equilibrium)

  1. Aug 4, 2011 #1
    HAHA. Here I am again. Now I know that most of you this is just a piece of cakeee..
    Well for me, the cake is a lie. It's HAAARRDD. Most especially if you're a newbie. HAHA


    "A girl weighing 620N sits in the middle of a 2.8m long hammock which sags 1.2m below the point of support. What force is exerted by each of the ropes supporting the hammock?"

    I need some guidance. AGAIN. Thanks.
     
  2. jcsd
  3. Aug 4, 2011 #2
    Drawing a diagram showing the length, sag and forces can be beneficial. Ultimately you want to express the condition for equilibrium; you need an angle for that, which can be found from the diagram.
     
  4. Aug 4, 2011 #3
    Yup, I did that but I disregarded the sag. I only calculated the tensions on both ropes which is I think is wrong.
     
  5. Aug 4, 2011 #4
    Can you show your diagram?
     
  6. Aug 4, 2011 #5

    PeterO

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    Homework Helper

    You must include the sag.
     
  7. Aug 4, 2011 #6
    No diagram is given and I haven't made one yet.

    But I think using the first condition of equilibrium is the one to be used.

    [itex]\Sigma[/itex]Fx = 0
    [itex]\Sigma[/itex]Fy = 0

    Summation of all forces must be equal to zero, amirite?

    I just don't know how to pull this off.

    Our teacher never taught us about the sags or they way it will sag because of the forces acting on it.
     
  8. Aug 4, 2011 #7

    SteamKing

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    I guess your teacher never showed you how to sit in a hammock, either.
     
  9. Aug 4, 2011 #8
    Use your common sense; which direction can a rope sag in if you sit on it?
     
  10. Aug 4, 2011 #9
    The ropes will sag on the center. Then what now?
     
  11. Aug 4, 2011 #10
    Right, it sags in the center, but which direction? (Up or down?)
     
  12. Aug 4, 2011 #11
    Down. obviously.
     
  13. Aug 5, 2011 #12
    Draw a diagram with all this information, and show us.
     
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