Use Table 7–1 to calculate the position of the CM of an arm bent at a right angle. Assume that the person is 155 cm tall.

The formula for the location of the center of mass on the x-axis is

\({{\bf{x}}_{{\bf{cm}}}}{\bf{ = }}\frac{{{{\bf{m}}_{\bf{u}}}{{\bf{x}}_{\bf{u}}}{\bf{ + }}{{\bf{m}}_{\bf{l}}}{{\bf{x}}_{\bf{l}}}{\bf{ + }}{{\bf{m}}_{\bf{h}}}{{\bf{x}}_{\bf{h}}}}}{{{{\bf{m}}_{\bf{u}}}{\bf{ + }}{{\bf{m}}_{\bf{l}}}{\bf{ + }}{{\bf{m}}_{\bf{h}}}}}\)

The formula for the location of the center of mass on the y-axis is

\({{\bf{y}}_{{\bf{cm}}}}{\bf{ = }}\frac{{{{\bf{m}}_{\bf{u}}}{{\bf{y}}_{\bf{u}}}{\bf{ + }}{{\bf{m}}_{\bf{l}}}{{\bf{y}}_{\bf{l}}}{\bf{ + }}{{\bf{m}}_{\bf{h}}}{{\bf{y}}_{\bf{h}}}}}{{{{\bf{m}}_{\bf{u}}}{\bf{ + }}{{\bf{m}}_{\bf{l}}}{\bf{ + }}{{\bf{m}}_{\bf{h}}}}}\)

The height of the person is \(h = 155\;{\rm{cm}}\).

From table 7-1,

The percent mass of the upper arm is \({m_{\rm{u}}} = 6.6\;{\rm{units}}\).

The percent mass of the lower arm is \({m_{\rm{l}}} = 4.2\;{\rm{units}}\).

The percent mass of the hand is \({m_{\rm{h}}} = 1.7\;{\rm{units}}\).

The distance of the upper arm from the center of mass on the x-axis is \({x_{\rm{u}}} = \left( {81.2 - 71.7} \right)\;{\rm{units}}\).

The distance of the lower arm from the center of mass on the x-axis is \({x_{\rm{l}}} = \left( {81.2 - 62.2} \right)\;{\rm{units}}\).

The distance of the hand from the center of mass on the x-axis is \({x_h} = \left( {81.2 - 62.2} \right)\;{\rm{units}}\).

The distance of the upper arm from the center of mass on the y-axis is \({y_{\rm{u}}} = \left( 0 \right)\;{\rm{units}}\).

The distance of the lower arm from the center of mass on the y-axis is \({y_{\rm{l}}} = \left( {62.2 - 55.3} \right)\;{\rm{units}}\).

The distance of the hand from the center of mass on the y-axis is \({y_h} = \left( {62.2 - 43.1} \right)\;{\rm{units}}\).

Consider the shoulder as the origin. The location of the center of mass of the system is given by its coordinates, \({x_{{\rm{cm}}}}\) and \({y_{{\rm{cm}}}}\).

The \({x_{{\rm{cm}}}}\) coordinate can be calculated as shown below:

\(\begin{array}{l}{x_{{\rm{cm}}}} = \frac{{{m_{\rm{u}}}{x_{\rm{u}}} + {m_{\rm{l}}}{x_{\rm{l}}} + {m_{\rm{h}}}{x_{\rm{h}}}}}{{{m_{\rm{u}}} + {m_{\rm{l}}} + {m_{\rm{h}}}}}\\{x_{{\rm{cm}}}} = \frac{{\left( {6.6} \right)\left( {81.2 - 71.7} \right) + \left( {4.2} \right)\left( {81.2 - 62.2} \right) + \left( {1.7} \right)\left( {81.2 - 62.2} \right)}}{{\left( {6.6} \right) + \left( {4.2} \right) + \left( {1.7} \right)}}\\{x_{{\rm{cm}}}} = 13.9\;{\rm{units}} \approx {\rm{14}}\;{\rm{units}}\end{array}\)

The \({y_{{\rm{cm}}}}\) coordinate can be calculated as shown below:

\(\begin{array}{l}{y_{{\rm{cm}}}} = \frac{{{m_{\rm{u}}}{y_{\rm{u}}} + {m_{\rm{l}}}{y_{\rm{l}}} + {m_{\rm{h}}}{y_{\rm{h}}}}}{{{m_{\rm{u}}} + {m_{\rm{l}}} + {m_{\rm{h}}}}}\\{y_{{\rm{cm}}}} = \frac{{\left( {6.6} \right)\left( 0 \right) + \left( {4.2} \right)\left( {62.2 - 55.3} \right) + \left( {1.7} \right)\left( {62.2 - 43.1} \right)}}{{\left( {6.6} \right) + \left( {4.2} \right) + \left( {1.7} \right)}}\\{y_{{\rm{cm}}}} = 4.92\;{\rm{units}}\end{array}\)

The actual distances can be obtained using the person’s height on the x-axis.

\(\begin{array}{l}{x_{{\rm{a,cm}}}} = \left( {14\% } \right)\left( h \right)\\{x_{{\rm{a,cm}}}} = \left( {14\% } \right)\left( {155\;{\rm{cm}}} \right)\\{x_{{\rm{a,cm}}}} = 21.7\;{\rm{cm}}\end{array}\)

The actual distances can be obtained using the person’s height on the y-axis.

\(\begin{array}{l}{y_{{\rm{a,cm}}}} = \left( {4.92\% } \right)\left( h \right)\\{y_{{\rm{a,cm}}}} = \left( {4.92\% } \right)\left( {155\;{\rm{cm}}} \right)\\{y_{{\rm{a,cm}}}} = 7.6\;{\rm{cm}}\end{array}\)

Thus, the position of the center of mass of an arm bent at the right angle is \(\left( {21.7\;{\rm{cm,7}}{\rm{.6}}\;{\rm{cm}}} \right)\).