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Tension Question

  1. Apr 27, 2009 #1
    1. The problem statement, all variables and given/known data

    A hungry bear weighing 720 N walks out on a beam in an attempt to retrieve a basket of food hanging at the end of the beam. The beam is uniform, weighs 200 N, and is 5.40 m long; the basket weighs 80.0 N.

    a) When the bear is at x = 1.00 m, find the tension in the wire and the components of the force exerted by the wall on the left end of the beam.
    N (tension)
    N ( Fx)
    N ( Fy)



    3. The attempt at a solution

    Okay well I sort of understood this question and created a balanced torque equation. After doing this I got that:

    Tension = 345.66
    Fx = 172.83

    But I need help with finiding Fy Please.
    I only have like one more chance to answer the question on my homework so I want to make sure I can get the right answer.
    I know I have to write out some equation like maybe:

    Rsin(theta) = 720 + 200 + 80 - 345.66sin(60)

    Is this correct? And if its not, am I on the right track?
     
    Last edited: Apr 27, 2009
  2. jcsd
  3. Apr 27, 2009 #2
    is the beam supported by a horizontal string on each end?
     
  4. Apr 27, 2009 #3
    Need more data on how the beam is supported.

    Sounds like a cantilever beam with a tension wire at the other end, but is this tension wire vertical or anchored to the wall.
     
  5. Apr 27, 2009 #4
  6. Apr 27, 2009 #5

    tiny-tim

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    Hi yb1013! :smile:

    I think you've calculated the y-component of the tension, instead of the y-component of the force at the wall. :redface:
     
  7. Apr 27, 2009 #6
  8. Apr 27, 2009 #7
    would this be closer to the y-component for the force?

    Rsin(theta) = 720 + 200 + 80 - 345.66sin(60)
     
  9. Apr 27, 2009 #8
    what a dumb bear...

    Anyways, I broke the componets into x and y directions. The y dir. is Tcos(theta) -mg =0 due to thier being no motion.

    also, thier are 3 masses. bear, basket, and beam.
     
  10. Apr 27, 2009 #9

    tiny-tim

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    He's smarter than the average bear!! :biggrin:
    Yup … if the tension is right, then that looks right too! :smile:
     
  11. Apr 27, 2009 #10

    So would it go like 345.66cos(60) - 1020 = 0 ??
     
  12. Apr 27, 2009 #11
    Ok well the numbers in that question aren't actually my exact numbers, but with my numbers I come out with:

    Rsin(theta) = 720.65

    would that mean that its:

    Rsin(60) = 720.65
    R = 832.13

    So 832.13 would be the force in the y-component??
    I only have one more guess left on my homework, so I wanna make sure lol
     
  13. Apr 27, 2009 #12

    tiny-tim

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    uhh? :confused:

    perhaps I misunderstood your Rsin(theta) …

    I assumed you meant that to be the y-component of the reaction force at the wall:

    Ry = 720 + 200 + 80 - 345.66sin(60)

    (ooh, and it's 1000 not 1020 :wink:)
     
  14. Apr 27, 2009 #13
    Yea forget what I just said, I had a brain fart for a second there lol

    Its just 720, sorry about the confusion.

    Thanks again!
     
  15. Apr 27, 2009 #14

    tiny-tim

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    better check that …
     
  16. Apr 27, 2009 #15
    well its right, I stated in one of my other posts that the numbers were exactly what my actual problem contained. the 720 N of the bear is actually 740, which is why when I added all the numbers up I got 1020 instead of 1000.. So in the end the force is equal to 720

    Sorry for all the confusion
     
  17. Apr 28, 2009 #16

    tiny-tim

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    stuffing down more goodies than the average bear …

    ah :rolleyes:heavier than the average bear!! :biggrin:
     
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