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Center of mass of right triangle

  1. Sep 6, 2012 #1
    I've been reading that the center of mass of a right triangle - the coordinates of the COM, is (1/3b,1/3h)- I can't for the life of me figure out why this is. Is there some sort of clear proof I can take a look at?
    I don't really know what to integrate..
     
    Last edited: Sep 6, 2012
  2. jcsd
  3. Sep 6, 2012 #2
    The center of mass of a triangle is at the intersection of its medians. Why? Because each median divides the triangle into two equal area (which means equal mass) subtriangles, so all those subtriangles balance each other out.
     
  4. Sep 6, 2012 #3
    i was actually talking about how to get the answer from an integration.
     
  5. Sep 6, 2012 #4

    ehild

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  6. Sep 7, 2012 #5
    xcmƩmi=Ʃmixi
    xcmM=Ʃ(dm)ixi ........(1)

    Taking a triangle lying on x-axis with point of hypotenuse and adjacent at origin
    Then you slice vertiacally the triangle into small as possible until each slice resembles rectangular piece with each area equal to f(x)dx......(2)

    Let the length of adjacent is a and opposite length is b.
    Now the density related to area is total mass divided by total area
    ρ=Mass/Area=2M/ab
    Thus dm/dA= 2M/ab
    dm=dA(2M/ab)
    Subt.(2)

    dm=f(x)dx (2M/ab)
    f(x)=(b/a)x
    Thus
    dm=x(b/a)dx(2M/ab)
    dm=2M/a2 (xdx)

    Subt. in (1)
    xcmM=Ʃ(2M/a2 (xdx))ixi

    [itex]x=\frac {2}{a^2}\int_0^a \! x^2 \, \mathrm{d} x[/itex]
     
  7. Sep 8, 2012 #6
    that pdf is wonderful, ehild. And thanks azizlwl- very helpful. I was having trouble reading the integrals because they looked complicated, but I think I got it now.
     
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