Three equal masses lie at the corners of an equilateral triangle of side l. find the center of mass.

If two of the masses are at (0, 0) and (L, 0), by symmetry the x-coordinate of the third mass must be L/2, and then by the Pythagorean theorem its y-coordinate must be L (âš3) / 2. Thus the position of the third mass is (L/2, L (âš3) / 2). Let each mass be m and C be the position of its center of mass. Then C = (1 / (3m)) (m (0, 0) + m (L, 0) + m (L/2, L (âš3) / 2)). = (1/3) (3L/2, L (âš3) / 2). = (L/2, L (âš3) / 6). That is, X_c = L/2 and Y_c = L (âš3) / 6.