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Problem with the weigth of the moment arm

  1. Mar 9, 2014 #1
    Hi
    A rod comprises a counterweight (a) a handle and a hook which hangs mass (m), the rods mass is 400g counterweight mass (a) = 1600g and the rod is 0.8m long. Equilibrium exists when the handle is at position X = 12.0cm

    How I did

    1: calculated the length of the handle to (m) which was 0.68m
    2: figured kg from 1.6 kg * 0.12 = 0.192 kg
    3 0.192 = 0.68*m
    but how do you do with the rod weighs 400g how do I calculate it??
    The question is to calculate the total mass of (m) to obtain equilibrium
    hope you understand the language and metrics, used google translate, did work out that great, but I´m loosing time

    regards fredrik
     
  2. jcsd
  3. Mar 9, 2014 #2

    haruspex

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    Consider a uniform horizontal rod length L pivoted at one end. Using calculus:
    An element length dx at distance x from the pivot has mass ρdx. What is its moment about the pivot? Can you integrate that to get the total moment?
    Or, without calculus, where is the mass centre of the rod?
     
  4. Mar 9, 2014 #3
    Oki now I see, its in its center then You just calculated it to the handle..
    0.8/2-0.12=0.28
    0.4kg*0.28=0.112kg

    Thanx for your help!!


    Sent from my iPhone using Physics Forums
     
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