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Static equilibrium mass problem

  1. Mar 15, 2009 #1
    1. The problem statement, all variables and given/known data
    A uniform 0.122 kg rod of 0.90 m length is used to suspend two masses as shown.
    At what distance x should the 0.20 kg mass be placed to achieve static equilibrium?
    4010113.gif


    2. The attempt at a solution

    F1=1.96N
    F2=4.9N

    torque = 4.9 x 0.25 = 1.225 Nm.cw

    1.23 = 1.96r

    r=0.63m


    answer is actually 0.5

    I think its because of the force of 1.96 (0.20kg) being to the right of the end of the rod (which has a mass)


    could someone please help me.
     
    Last edited: Mar 15, 2009
  2. jcsd
  3. Mar 15, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi brentwoodbc! Welcome to PF! :smile:
    What about F3 (for the rod)?

    (And please don't multiply everything by 9.8 …

    just call it g … all the g's wil cancel in the end, anyway :rolleyes:)
     
  4. Mar 15, 2009 #3
    Re: Welcome to PF!


    thanks, ya Im trying to figure out how to factor in the force of the rod but we were not given any example where the rod had a mass.
     
  5. Mar 15, 2009 #4

    tiny-tim

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    ok … the rod has a weight …

    where does that weight act? :smile:
     
  6. Mar 15, 2009 #5
    Re: Equalibrium

    on the centre of rotation. and the sum of all forces = zero. What direction is the 3rd force though? against the other two?
     
  7. Mar 15, 2009 #6

    tiny-tim

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    (sorry, i'm not following you)

    the weight of a rod (or indeed anything else) acts through its centre of mass
     
  8. Mar 15, 2009 #7
    Re: Equalibrium

    so whats the equation going to look like?

    F1+F2+F3=0?

    Im lost on this question.
     
  9. Mar 15, 2009 #8

    tiny-tim

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    No, torque1 + torque2 + torque3 = 0 :smile:
     
  10. Mar 15, 2009 #9
    Re: Equalibrium

    ok, but I keep getting -0.35

    .5gr+.2gr+.122gr=0

    .2gr=-(.122g(.45 -.25)+.5(.25))

    cancel g

    divide .2

    r = 0.75?
     
  11. Mar 15, 2009 #10

    tiny-tim

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    curiously, that seems to be the right answer …

    but what was the (.45 -.25) supposed to be? :confused:
     
  12. Mar 15, 2009 #11
    Re: Equalibrium

    half of distance of beam is 0.45metres

    so minus the 0.25metres is the distance of the centre of mass from the centre of rotation so (r for F3)
     
  13. Mar 15, 2009 #12
    Re: Equalibrium

    I noticed that the correct answer "0.5"is my answer "0.75" minus the 0.25 to the right of the centre of rotation... hmmm.
     
  14. Mar 15, 2009 #13

    tiny-tim

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    ah … got it!

    that's fine then :smile:
     
  15. Mar 15, 2009 #14
    Re: Equalibrium

    Im not 100% but I think that since the force with mass.5 is clockwise "in direction" it is negative. the other two are in a counter clockwise direction. So I made the 0.25metres negative and I got 0.5.

    Seems to be fine. Thank you very much.:biggrin:
     
  16. Mar 15, 2009 #15

    tiny-tim

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    oh, I missed the minus in your
    that does make it .5 :redface:
     
  17. Mar 15, 2009 #16
    Re: Equalibrium

    pfft. I'm so smart lol :cool:

    thanks again though. Being spring break I cant get help and I have tests a couple days after I go back. And there are more question s in my homework similar to this so I should be cool now.
     
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