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Center of Mass of a Hollow 2D Triangle

  1. Apr 3, 2009 #1
    1. The problem statement, all variables and given/known data

    What is the center of mass a hollow 2D right triangle having legs of length A and B. The sides of the triangle are also extremely thin. Assume uniform density.

    2. Relevant equations

    [tex]\overline{x}[/tex]=∫xmdx/∫mdx

    3. The attempt at a solution

    [tex]\overline{x}=\int\frac{ax^2dx}{b}/\int\frac{axdx}{b}[/tex]<-something like that
    my final answer was ([tex]\frac{a^2}{b}[/tex],[tex]\frac{b^2}{a}[/tex])
     

    Attached Files:

  2. jcsd
  3. Apr 3, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi datdo! Welcome to PF! :smile:

    (i can't see your picture yet, but …)

    Wouldn't it be simpler to replace each side by an equal mass at its centre? :wink:
     
  4. Apr 4, 2009 #3
    hmm.... i didn't think of that...
    A:(.5A,0)
    B:(0,.5B)
    hypotenuse:(.5A,.5B)
    [tex]\overline{x}=\frac{A*.5A+B*0+\sqrt{A^2+B^2}*.5A}{A+B+\sqrt{A^2+B^2}}=\frac{.5A^2+\sqrt{A^2+B^2}*.5A}{A+B+\sqrt{A^2+B^2}}[/tex]
    [tex]\overline{y}=\frac{A*0+B*.5B+\sqrt{A^2+B^2}*.5B}{A+B+\sqrt{A^2+B^2}}=\frac{.5B^2+\sqrt{A^2+B^2}*.5B}{A+B+\sqrt{A^2+B^2}}[/tex]

    Oh but I had to use integrals. I should have put in the problem.
     
    Last edited: Apr 4, 2009
  5. Apr 4, 2009 #4

    tiny-tim

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    Hi datdo! :smile:
    yeeees :rolleyes:

    ok, then for the x-coordinate divide the triangle into "vertical" slices of thickness dx …

    then for the bottom side just integrate xdx, and for the slanty side integrate … ? :smile:
     
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