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Center mass

  1. Jul 14, 2013 #1
    1. The problem statement, all variables and given/known data

    An 800 g steel plate has the shape of the isosceles triangle shown in the figure. What are the x and y coordinates of the center of mass?
    https://www.physicsforums.com/attachment.php?attachmentid=13559&d=1208292919

    2. Relevant equations

    x=1/M ∫ x dm

    3. The attempt at a solution[/b
    I have no idea how to start the question
     
    Last edited: Jul 14, 2013
  2. jcsd
  3. Jul 14, 2013 #2

    SteamKing

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    The image gives you a big ole hairy HINT. Did you try following it?
     
  4. Jul 15, 2013 #3
    what hint is that ? the triangle is a smmetry and I put xcm=1/M∫xd(ρ/AH)
    and the height is y=2/3 x as =mx+C. and should I put h = 2/3 x ?
     
  5. Jul 16, 2013 #4

    haruspex

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    That's a pretty confusing mix of variable names. What's AH, C, the m in mx+C? Why switch from y to h?
    That said, y (=h) = 2x/3 looks right. What do you get for the integral?
     
  6. Jul 16, 2013 #5
    xcm=1/M∫xd(ρ/AH) , A is the area and H is the height , I need to find h and I use h=mx+c to find the height and the slope is 2/3 . Question is to find the center of mass and only this integration could help me find the answer .
     
  7. Jul 16, 2013 #6

    haruspex

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    ρ/AH is not going to give you mass. And... the area and height of what, exactly?
    You want the mass, dm, of an element of width dx and height y = 2x/3. It's a 2-dimensional lamina, so the density is mass/area. So what is dm equal to?
     
  8. Jul 16, 2013 #7

    SteamKing

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    More importantly, you have a plate with a simple shape composed of a material which has a constant density. Do you really need an integral to determine the location of the centroid?
     
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