Finding the Center of Mass of a System of Objects on a Line

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 \(\bar{x}\approx 4.769\).