Find the center of mass of the system of objects having masses \(4\), \(6\), and \(9\) kg, located at the points \((-2,1), (3,-2)\), and \((4,3),\) respectively.
460
To find the exact position of the center of mass, we first find the moment of the system about the \(y\)-axis, \(M_{y},\) and the moment of the system about the \(x\)-axis, \(M_{x}\). \begin{align*} M_{y}=\sum\limits_{i=1}^{3} m_{i}x_{i} =4(-2)+6(3)+9(4)=46\\ M_{x}=\sum\limits_{i=1}^{3} m_{i} y_{i} =4(1)+6(-2)+9(3)=19 \end{align*}
Since the mass \(M\) of the system is \(M=4+6+9=19,\) we have \begin{equation*} \bar{x}=\frac{M_{y}}{M}= \frac{46}{19},\qquad \bar{y}=\frac{M_{x}}{M}=\frac{19}{19}=1 \end{equation*}
The center of mass of these objects is at the point \(\left( \dfrac{46}{19},1\right) \). Notice that the center of mass lies in the rectangle \(-2\leq x\leq 4\); \(-2\leq y\leq 3\).