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Dynamics Question involving Tension (Work Provided)

  1. Jul 24, 2007 #1
    1. The problem statement, all variables and given/known data

    During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of 137 N on the wire. The left section of the wire makes an angle of 15° relative to the horizontal and sustains a tension of 447 N. Find the magnitude and direction of the tension that the right section of the wire sustains.

    2. Relevant equations



    3. The attempt at a solution

    The question is, where do I go from here. If the y component of the tension is the same for both sides of the wire, is there an equation i can set up to find the tension on right side? I wondered this, but I don't have an angle for the right side of the wire.
     

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    Last edited: Jul 24, 2007
  2. jcsd
  3. Jul 24, 2007 #2

    Doc Al

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    You don't need the angle: solve for the components of the unknown tension. (Once you have the components, then you can figure out the angle.)

    Since the branch is in equilibrium, what must the net force be on it? That's the equation you need.
     
  4. Jul 24, 2007 #3
    so the sum of all the forces must equal 0 right? So in the y axis, 137N = 0

    T=W ?
     
    Last edited: Jul 24, 2007
  5. Jul 24, 2007 #4
    I thought about what you said for a long time. I still can't figure it out. Can I have some more help?
     
  6. Jul 24, 2007 #5

    mgb_phys

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    The vertical force on the wire must equal the weight of the branch (137N). The horizontal forces on each side must be the same because the branch is not moving sideways.

    What is the horizontal force on the side you do know the angle of?

    The tension in the wire isn't necessarily the same on both sides if the branch is not smooth.
     
  7. Jul 24, 2007 #6
    That would mean that the magnitude of the tension on both sides is equal, according to what you are saying. Is this true? Is there something I'm missing? Is this a trick question?
     
  8. Jul 24, 2007 #7

    mgb_phys

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    Yes I think your right, I was assuming that if the branch had friction with the wire then the tension could be different but the horizontal tension must be the same and so the overall magnitude must be the same.

    You will have to write the angle in terms of the relative lengths of the two lengths of wire.
     
  9. Jul 24, 2007 #8

    Doc Al

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    The magnitude of the tension on both sides is not necessarily equal--but the horizontal components of the tensions must be equal (and opposite).

    Now you know the horizontal component of the right wire tension. All that's left is to figure out the vertical component: Set up the equation making the sum of the vertical components of all three forces equal to zero and solve for it.

    This is not a trick question, just a standard application of the conditions for equilibrium.
     
  10. Jul 24, 2007 #9
    cool thanks, got the answer
     
  11. Jul 24, 2007 #10

    mgb_phys

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    Sorry - the TOTAL downward force on the wire is 137N.
    The proportion of this force on the two halfs of the wire is of course different.

    This is why I should always draw diagrams - even for simple homework problems!
     
  12. Jul 25, 2007 #11

    andrevdh

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    One can solve for the tension using the cosine rule.
     

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  13. Feb 18, 2010 #12
    I didnt understand any of that. Are the horizontal components of the tension equal? Or are the vertical components of the tension equal?
     
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