Suppose that in Example 7–14 (Fig. 7–28), \({{\bf{m}}_{{\bf{II}}}}{\bf{ = 3}}{{\bf{m}}_{\bf{I}}}\).

(a) Where then would \({{\bf{m}}_{{\bf{II}}}}\) land?

(b) What if \({{\bf{m}}_{\bf{I}}}{\bf{ = 3}}{{\bf{m}}_{{\bf{II}}}}\)?

The CM of any object is the point that runs as if all of the mass were gathered there, and all the external forces were employed there.

(a)

It is given that \({m_{{\rm{II}}}} = 3{m_{\rm{I}}}\).

From examples 7-14, the center of mass of the system is located at \({x_{{\rm{CM}}}} = 2d\) , and the position of part I from the CM is \({x_{\rm{I}}} = d\).

The location of part II from the center of mass can be calculated asshown below:

\(\begin{array}{c}{x_{{\rm{CM}}}} = \frac{{{m_{\rm{I}}}{x_{\rm{I}}} + {m_{{\rm{II}}}}{x_{{\rm{II}}}}}}{{{m_{\rm{I}}} + {m_{{\rm{II}}}}}}\\2d = \frac{{{m_{\rm{I}}}d + 3{m_{\rm{I}}}{x_{{\rm{II}}}}}}{{{m_{\rm{I}}} + 3{m_{\rm{I}}}}}\\{x_{{\rm{II}}}} = \frac{7}{3}d\end{array}\)

Thus, the location of part II from the center of mass is \(\frac{7}{3}d\).

(b)

In this case, \({m_{\rm{I}}} = 3{m_{{\rm{II}}}}\). So, the location of part II from the center of mass can be calculated asshown below:

\(\begin{array}{c}{x_{{\rm{CM}}}} = \frac{{{m_{\rm{I}}}{x_{\rm{I}}} + {m_{{\rm{II}}}}{x_{{\rm{II}}}}}}{{{m_{\rm{I}}} + {m_{{\rm{II}}}}}}\\2d = \frac{{3{m_{{\rm{II}}}}d + 3{m_{{\rm{II}}}}{x_{{\rm{II}}}}}}{{3{m_{{\rm{II}}}} + {m_{{\rm{II}}}}}}\\{x_{{\rm{II}}}} = 5d\end{array}\)

Thus, the location of part II from the center of mass is \(5d\).