In our example the total mass of the system is 22 kg. Where should we place this single mass? If we place this mass d meters to the right of O, we would have a moment of M = 22d kgm. So, M 75 d =: 22 ~ 3.4 m So we see that the center of mass is 3.4 meters from O. 3.4 m 22 kg 2-dimensional Case For the 2 dimensional case, we imagine that all of the masses are set upon a co-ordinate axis, and we will imagine that this co-ordinate axis can only rotate about the x-axis and the y-axis. The tendency to rotate about the y-axis will be called the moment about the y-axis and will be denoted My. Clearly, if a mass is placed on the y-axis it will not cause the plane to rotate about the y-axis, but if the mass is placed to the left or right of the y-axis it would induce such a rotation. In this way the moment about the y-axis is only affected by the x-coordinate of the place where the mass was placed. So if masses m1, m2, ..., mn are placed at positions (x1. )1). (X2. )2), ... (Xn. )n) then the moment about the y-axis is given by My = m1 . x1 + m2 . X2 + ... + Mn. Xn Symmetrically, we can define the tendency to rotate about the x-axis to be the moment about the x-axis which will be denoted My. Arguments like the ones made above give us Mx = m, . y1 + m2 ' yz + ... + mn ' In The center of mass of the system is the point where we can place all of the mass and obtain the same moments. The center of mass is the point (x, y) given by X = My y = Mx m m where m is the total mass of the system. Exercises 1) If point masses of 12, 15, and 20 kg are placed 3 m to the left, 2 m to the right and 8 m to the right of the origin, find the center of mass. 2) If the masses 5, 4, 3, and 6 are located at the points (-4, 2), (0, 5), (3,2) and (1, -2), find the center of mass