Center of Mass of a Triangle (uniform)

AI Thread Summary
The center of mass of a uniform triangle, specifically an equilateral triangle, is located at the centroid, which is found two-thirds of the way along the median from the vertex. This centroid is determined by averaging the coordinates of the triangle's vertices. In an equilateral triangle, the centroid lies on the line connecting the midpoints of two sides. A geometric approach using right triangles can confirm that the centroid remains consistent regardless of the triangle's orientation. Understanding these principles helps clarify the location of the center of mass in triangular shapes.
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Homework Statement



Where exactly is the center of mass of a triangle ?
the file attached shows a diagram of a triangle which is equilateral, and the blue spots are the mid-points of each side.



Homework Equations



a distance = 2/3 from the vertex (along the median)


The Attempt at a Solution



I think the Midpoint will be equal to (2h)/3 from the top vertex.
But here is the real problem, is this Center of mass gonnah lie on the Yellow line passing through the mid points of two sides of the triangle ?
 

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hms.tech said:
I think the Midpoint will be equal to (2h)/3 from the top vertex.
But here is the real problem, is this Center of mass gonnah lie on the Yellow line passing through the mid points of two sides of the triangle ?

It does, the centroid is an average of the three coordinates, ( (stigma x) /3, (sigma y)/3). since its an equilateral triangle, the angles are all the same, and will all have the same terminal vector of 30 degrees (~0.5 rad) with equal length

A good way to see it is (using a triangle of 2 unit length sides)

cut the equilateral triangle into a right triangle (results in legs of 1, sqrt(3) and a hypotenuse of 2), taking the arc-tangent (opposite over adjacent) of 30 degrees, and rearranging to solve for the opposite (height) with an adjacent (run) length of 1 give you the height. Then rotate the equilateral triangle 120 degrees (360/3) and again bisect to create the right triangle. Do the same thing after rotating the triangle another 120 degrees. The lines should intersect at the same location.


Hope it helps
Joe


an external link to visualize it:
http://www.easycalculation.com/analytical/learn-centroid.php
 
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