mspeedwagon: I apologize for butting in like that. Schenker was explaining and I just distracted you further by offering a quick fix. That was not polite. Schenker please correct me if I misstate anything.....
Solution 1: So center of mass is an abstraction where we assume the sum of all forces are concentrated and act. In this problem let us assume that the 16 kg is the origin O of x axis. The 22 kg acts at a distance of 2 m. Furthermore, let point C on the yoke be the center of mass which is at a distance 'x' from the origin O or 16 kg mass. So the center of mass calculation becomes, Total Mass*x = 16*0 + 22*2. The 16 kg is at the origin so therefore I multiply it by 0. So we solve this eqn. (16+22)*x = 44 so x = 44/38 so x is a little more than 1 i.e., x = 1.15 approx. So now we know that the center of mass is around 1.15 m from the Origin O which is at the 16 kg end. Which means it is about .15 m closer to the 22 kg end from the center of the yoke, which is what we expect. It does not matter whether you take origin at 16 kg or 22 kg end. Either way the math will work out. If you take the origin at 22 kg end, (16+22)*x = 22*0 + 16*2 so x = 32/38 so approx 0.85 m away from 22 kg end or 1.15 m from 16 kg end. Intuitively speaking think of a baseball bat. One end is heavier and the other end is lighter. If u balance that bat on a finger, the finger will be closer to the heavier end of the bat, right. Same case here. Center of mass is where the sum of all forces are assumed to act and so placing a support there (a finger) we can balance the whole yoke.
Solution 2: I use Torques. Makes it extremely simple. The center of mass has to be on the yoke so just assume a place on the yoke to be center of mass, say C. Place a fulcrum there. Now the Torques or moments about that fulcrum have to balance for C to be center of mass. So let us assume C to be at a distance x from 16 kg mass. So clockwise moment = anticlockwise moment to balance out the moments. Which means 16*x = 22(2-x), solving for x = 44/38 = 1.15. So the center of mass C will be 1.15 from the 16 Kg mass which is what we expect meaning slightly closer to the heavier end of 22 kg. It does not matter from which mass I take that distance x. You can take x to be from 22 kg mass as well. So now, if I had chosen x to be the distance from 22 kg, in that case 22*x = 16(2-x) so x = 32/38 which is approx 0.85 which means that x is 0.85 m away from 22 kg mass meaning, x is 1.15 m from 16 kg mass. Once again, the same result as the first case. So either way we know the center of mass to be 1.15 m from 16 kg mass. Hope this helps....
