What is the center of gravity of the perimeter of a triangle (as when a piece of wire is bent into triangular form)?

If I had a triangular piece of card and attempted to balance it on a needle point, the C of G would be at the point of bisection of the three sides.

Since there is no solid interior, if I suspended the wire object (and the wire is uniform in cross-section and density) by one vertex then the C of G is directly beneath that vertex in terms of gravity.  If you could 'draw' that vertical line, hold it, and then perform the same exercise using a different vertex the C of G would still lie on that line.

Such a "physical" experiment only mirrors what one would really derive from a triangle in the plane.

I had used the wrong term at the end of the last paragraph, not only that I was also incorrect in my opening paragraph. 

Edited on May 19, 2008, 9:26 pm

Posted by brianjn on 2008-05-19 10:27:52