Center of mass of right triangle

AI Thread Summary
The discussion centers on understanding the center of mass (COM) of a right triangle, specifically its coordinates at (1/3b, 1/3h). Participants seek clarity on the proof behind this formula and the role of integration in deriving it. The center of mass is identified as the intersection of the triangle's medians, which divide it into equal area subtriangles. A detailed integration approach is provided, involving slicing the triangle into vertical sections to calculate mass distribution. The conversation concludes with appreciation for the shared resources that clarify the integration process.
mariexotoni
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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..
 
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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.
 
i was actually talking about how to get the answer from an integration.
 
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

x=\frac {2}{a^2}\int_0^a \! x^2 \, \mathrm{d} x
 
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