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Hanging a deer from a pole - find tension

  1. Nov 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Ricky Bobby hangs a deer from a uniform horizontal pole as shown below. The Deer's mass is 57.6 kg, the pole has a mass of 12.1kg, and the length of the pole is 5.36m. The Deer is hung 1.92m from the point where the pole is attached to the verticle beam, and the angle of the support wire at the end of the pole 21.3 degrees. What is the tension in the wire? what is the direction and the magnitude of the force of the verticle support against the pole at the point where they meet?


    2. Relevant equations
    T=<T<145degrees>
    W=<57.6<270degrees>
    Weight of pole=<1.92<270degrees>
    force on pole=<F<21.3>


    3. The attempt at a solution
    <T<145>+<57.6<270>+<1.92<270>+<F<21.3>=0 I don't know if im doing this right can you help me please
     
  2. jcsd
  3. Nov 11, 2008 #2

    tiny-tim

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    Hi ldbaseball16! :smile:

    Yes, that's certainly right in principle …

    you've correctly applied good ol' Newton's second law for the forces on the end of the pole …

    but your (r,θ) notation is useless for adding the vectors … you'll need to convert to (x,y) …

    and you'll also need to take moments about the other end of the pole. :smile:
     
  4. Nov 11, 2008 #3
    Re: Deer

    yea but how does 5.36m of the pole and 12.1 kgs tie into the equation???? thats what throws me off
     
  5. Nov 12, 2008 #4

    tiny-tim

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    Hi ldbaseball16! :smile:

    The 5.36m comes in when you take moments about the other end of the pole. :wink:
     
  6. Mar 16, 2009 #5
    Re: Deer

    i don't understand you mean like this ????? <T<145>+<1.92<270>+<12.1<180>+<5.36<180>+<57.6<270>+<F<21.3>=0 and then convert to (x,y)??????? right
     
  7. Mar 16, 2009 #6

    tiny-tim

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    Re: Deer

    wow, that was three months ago …

    that deer must be smelling pretty niffy by now :bugeye:

    sorry, but I can't follow your equation …

    can you please convert it to x and y first, and then ask me? :smile:
     
  8. Mar 16, 2009 #7
    Re: Deer

    Tcos(145)+1.92cos(270)+12.1cos(180)+5.36cos(180)+57.6cos(270)+Fcos(21.3)=0

    Tsin(145)+1.92sin(270)+12.1sin(180)+5.36sin(180)+57.6sin(270)+Fsin(21.3)=0
     
  9. Mar 16, 2009 #8

    tiny-tim

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    uhh? :confused:

    where did 145º come from?

    and please write it out properly, with all angles < 90º, and putting cos or sin = 0 or 1 where appropriate :smile:
     
  10. Mar 16, 2009 #9
    Re: Deer

    Tcos<145>+1.92cos<270>+12.1cos<180>+5.36cos<180>+5 7.6cos<270>+Fcos<21.3>=0

    Tsin<145>+1.92sin<270>+12.1sin<180>+5.36sin<180>+5 7.6sin<270>+Fsin<21.3>=0


    the <145> is the angle of the missing tension of the wire
     
  11. Mar 16, 2009 #10

    tiny-tim

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    no it isn't …
    and what is sin180º, for example?
     
  12. Mar 16, 2009 #11
    Re: Deer

    Tcos(145)+1.92cos(270)+12.1cos(180)+5.36cos(180)+5 7.6cos(270)+Fcos(21.3)=0

    Tsin(145)+1.92sin(270)+12.1sin(180)+5.36sin(180)+5 7.6sin(270)+Fsin(21.3)=0

    T+0-12.1-5.36+0+F.9316912276=0

    T-1.92+0+0-57.6+F.3632512305=0

    so it just stays as T and no <145>??????
     
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