Find the center of mass of the system when a mass of $90\textrm{ kg}$ is placed at $6$ and a mass of $40\textrm{ kg}$ is placed at $2$.

Solution The system is shown in Figure 67, where the two masses are placed on a weightless seesaw. The center of mass $\bar{x}$ will be at some number where a fulcrum balances the two masses. Then for equilibrium, $$\begin{equation*} \bar{x}=\dfrac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}=\dfrac{40(2)+90(6)}{40+90}=\dfrac{620}{130}\approx 4.769 \end{equation*}$$

The center of mass is at $\bar{x}\approx 4.769$.

Figure 67 At the center of mass $\bar{x}$ the system is in equilibrium.