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A meterstaff

  1. Oct 11, 2005 #1
    A meterstaff who weights 0,12 kg is moveable around of it's ends. In the other staff it hangs a weight of 0,50 kg. The staff is hold in a line by a dynamometer, who is fastened in the middle of the staff. What does the dynamometer show?

    I take 0,12*0,50*6,67*10^-11 =4,002*10^-12 / 1^2 =4,002*10^-12

    Which is totaly wrong this should be -11

    tell me how to do this im totaly lost
     
  2. jcsd
  3. Oct 11, 2005 #2

    Integral

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    I am also lost. I got that way trying to read your problem, Could you look for a better translation?
     
  4. Oct 11, 2005 #3

    HallsofIvy

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    Actually, not bad English- not compared to my (put whatever language you like here!).

    You have a meter stick pinned to a wall by one end (so that end cannot go up or down but the stick can swing about that point). There is a mass of 0.5 kg at the other end and there is a support at the center of the stick, which itself has mass 0.12 kg. What is the force exerted on the stick by that support?

     
  5. Oct 11, 2005 #4
    Gahh

    You know I was using the the the F=Gm1m2/r2 and G is Constant of Gravitation = 6,667*10^-11
    Yeah now that I calulated it I also got 0,56 because i thought in another way... but how do I resolv F I'm like this :confused: :confused: :confused: it stands nothing in my book!

    I added a pic to those who dont understand my english !!!

    Thx
     

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  6. Oct 11, 2005 #5
    Is this the way

    The Force for the weight is 0,56nM and then the force must be utilised on the other side to so 0,56+0,56 * 9,82 =10,984=11nM
     
  7. Oct 11, 2005 #6

    HallsofIvy

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    Yes, exactly.

    (It would be more correct (in English, at least) to say the "torque" rather than the force, is 0,56 Nm- in any language you shouldn't use the word "force" with two different meanings. {But who am I to talk, we use "pounds" for both mass and force!})

    The formula I gave you was 0,5F= 0,56g so that F= 0.56g/0,5= (0,56)(9,81)/0,5. That gives the same thing.

    I hope you realize that you are NOT calculating F by (Gm1m2)/r2. You certainly don't need that for things happening on the surface of the earth!
     
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