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Homework Help: Centroid of a C-Shape

  1. Sep 11, 2012 #1
    1. The problem statement, all variables and given/known data

    http://prntscr.com/fcbm8

    Find the centroid - All dimensions are in mm

    2. Relevant equations

    xbar = (A1X1+A2X2) / (A1+A2)

    Similarly for Ybar I assume

    3. The attempt at a solution

    I got the y co-ordinate to be 20.428mm, and would assume that the x coordinate would be 5mm.

    Is this right?
     
  2. jcsd
  3. Sep 11, 2012 #2

    CWatters

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    You can divide the shape into three rectangular parts several different ways but I used two vertical lines in the obvious places. I assumed the origin is in the bottom left corner.

    xbar = (A1X1+A2X2+A3X3) / (A1+A2+A3)

    = (700*5 + 800*30 +400*30) / (700+800+400)
    = 20.789

    ybar = (A1Y1+A2Y2+A3Y3) / (A1+A2+A3)

    = (700*35 + 800*10 + 400*65) / (700+800+400)
    = 30.789

    Best show your working as my answer is quite different.
     
  4. Sep 11, 2012 #3
    This is what confused me, does this mean the centroid is not on the shape itself?
     
  5. Sep 11, 2012 #4

    CWatters

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    In this case yes. Where would the centroid of a doughnut be?
     
  6. Sep 11, 2012 #5

    HallsofIvy

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    In the center, of course. That's what the "centroid" is- the geometric center. If you were to represent the doughnut as two circled in the in the xy-plane, centered at the origin with radii r and R, and then have other circles as the thickness of the doughnut, the centroid would be at (0, 0, 0).
     
  7. Sep 12, 2012 #6

    CWatters

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    I know. I was using it as an obvious example for the OP to think about. eg a shape that has a centroid that's not on the surface of the shape.
     
  8. Sep 12, 2012 #7
    Yeah, I was probably thinking more about a centroid of a mass, but even then, there's still the donut which proves me redundant.
     
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