(I) Determine the CM of an outstretched arm using Table 7–1.

The center of mass of an extended object is that point at which the net force can be assumed to act for determining the object's translational motion as a whole.

From table 7-1,

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

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

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

The diagram of an outstretched arm is shown below.

The center of mass of an outstretched arm can be calculated asshown below:

\({x_{{\rm{CM}}}} = \frac{{{m_{{\rm{UA}}}}{x_{{\rm{UA}}}} + {m_{{\rm{LA}}}}{x_{{\rm{LA}}}} + {m_{\rm{H}}}{x_{\rm{H}}}}}{{{m_{{\rm{UA}}}} + {m_{{\rm{LA}}}} + {m_{\rm{H}}}}}\)

Substitute the values in the above equation.

\(\begin{array}{l}{x_{{\rm{CM}}}} = \frac{{\left( {6.6} \right)\left( {81.2 - 71.7} \right) + \left( {4.2} \right)\left( {81.2 - 55.3} \right) + \left( {1.7} \right)\left( {81.2 - 43.1} \right)}}{{\left( {6.6} \right) + \left( {4.2} \right) + \left( {1.7} \right)}}\\{x_{{\rm{CM}}}} = \frac{{\left( {6.6} \right)\left( {9.5} \right) + \left( {4.2} \right)\left( {25.9} \right) + \left( {1.7} \right)\left( {38.1} \right)}}{{\left( {6.6} \right) + \left( {4.2} \right) + \left( {1.7} \right)}}\\{x_{{\rm{CM}}}} = 18.9\; \approx 19\;{\rm{units}}\end{array}\)

Thus, the CM of an outstretched arm is 19% of the person’s height, along the line from shoulder to hand.