A uniform bar, of mass M , with seven evenly spaced holes is held by sliding the bar over a horizontal peg through one of the seven holes. The peg passes through hole D, and a cylinder hangs from a hook placed through hole G as shown above. The mass of the bar is the same as the mass of the cylinder, and the location of the center of mass of the bar is at the center of hole D. In this configuration, the bar-cylinder system is not in rotational equilibrium and is free to rotate about the peg in hole D. Frictional forces acting on the bar are negligible Two additional cylinders, identical to the one shown, can each be hung on the bar by hooks from any of the open holes so that the bar will be in rotational equilibrium when all three cylinders are hanging from it. From which two holes should the two cylinders be hung? Briefly explain your reasoning.