Repeat Problem 59 assuming the body bends at the hip joint by about 15°. Estimate, using Fig. 7–27 as a model.

The CM of a system is that point in the system that follows the law of physics for a single particle.

In a uniform gravitational field, the CM is at the same place as the center of gravity.

The FBD for the given situation is shown below:

In the above situation, the upper legs are parallel to the bar, and the feet are hanging down vertically. The head, trunk, and neck are tilted downward by \(15^\circ \).

From table 7-1,

The percent mass of the trunk and neck is \({m_{{\rm{TN}}}} = 46.1\;{\rm{units}}\).

The percent mass of the head is \({m_{{\rm{Head}}}} = 6.9\;{\rm{units}}\).

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

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

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

The percent mass of the full body is \({m_{{\rm{Full}}\;{\rm{body}}}} = 100\;{\rm{units}}\).

The percent mass of the lower legs is \({m_{{\rm{LL}}}} = 9.6\;{\rm{units}}\).

The percent mass of the feet is \({m_{\rm{F}}} = 3.4\;{\rm{units}}\).

The distance of the trunk and neck from the center of mass is

\({y_{{\rm{TN}}}} = \left( {71.1 - 52.1} \right)\sin \left( {15^\circ } \right) = 4.92\;{\rm{units}}\).

The distance of the head from the center of mass is

\({y_{{\rm{Head}}}} = \left( {93.5 - 52.1} \right)\sin \left( {15^\circ } \right) = 10.72\;{\rm{units}}\).

The distance of the shoulder from the center of mass is

\({y_{\rm{S}}} = \left( {81.2 - 52.1} \right)\sin \left( {15^\circ } \right) = 7.53\;{\rm{units}}\).

The distance of the upper arm from the center of mass is

\({y_{{\rm{UA}}}} = \left\{ \begin{array}{l}\left( {{y_{\rm{S}}}} \right) + \left( {81.2 - 71.7} \right)\cos \left( {15^\circ } \right)\\ = \left( {7.53} \right) + \left( {81.2 - 71.7} \right)\cos \left( {15^\circ } \right) = 16.71\;{\rm{units}}\end{array} \right\}\).

The distance of the lower arm from the center of mass is

\({y_{{\rm{LA}}}} = \left\{ \begin{array}{l}\left( {{y_{\rm{S}}}} \right) + \left( {81.2 - 55.3} \right)\cos \left( {15^\circ } \right)\\ = \left( {7.53} \right) + \left( {81.2 - 55.3} \right)\cos \left( {15^\circ } \right) = 32.55\;{\rm{units}}\end{array} \right\}\).

The distance of the hands from the center of mass is

\({y_{\rm{H}}} = \left\{ \begin{array}{l}\left( {{y_{\rm{S}}}} \right) + \left( {81.2 - 43.1} \right)\cos \left( {15^\circ } \right)\\ = \left( {7.53} \right) + \left( {81.2 - 43.1} \right)\cos \left( {15^\circ } \right) = 44.33\;{\rm{units}}\end{array} \right\}\).

The distance of the lower legs from the center of mass is

\({y_{{\rm{LL}}}} = \left( {28.5 - 18.2} \right) = 10.3\;{\rm{units}}\).

The distance of the lower legs from the center of mass is

\({y_{\rm{F}}} = \left( {28.5 - 1.8} \right) = 26.7\;{\rm{units}}\).

The vertical location of CM can be calculated as shown below:

\(\begin{array}{l}{y_{{\rm{cm}}}} = \frac{{\left\{ \begin{array}{l}{m_{{\rm{TN}}}}{y_{{\rm{TN}}}} + {m_{{\rm{Head}}}}{y_{{\rm{Head}}}} + {m_{{\rm{UA}}}}{y_{{\rm{UA}}}}\\ + {m_{{\rm{LA}}}}{y_{{\rm{LA}}}} + {m_{\rm{H}}}{y_{\rm{H}}} + {m_{{\rm{LL}}}}{y_{{\rm{LL}}}} + {m_{\rm{F}}}{y_{\rm{F}}}\end{array} \right\}}}{{{m_{{\rm{Full}}\;{\rm{body}}}}}}\\{y_{{\rm{cm}}}} = \frac{{\left\{ \begin{array}{l}\left( {46.1} \right)\left( {4.92} \right) + \left( {6.9} \right)\left( {10.72} \right) + \left( {6.6} \right)\left( {16.71} \right)\\ + \left( {4.2} \right)\left( {32.55} \right) + \left( {1.7} \right)\left( {44.33} \right)\\ + \left( {9.6} \right)\left( {10.3} \right) + \left( {3.4} \right)\left( {26.7} \right)\end{array} \right\}}}{{\left( {100} \right)}}\\{y_{{\rm{cm}}}} = 8.1\;{\rm{units}}\end{array}\)

Thus, the center of mass is 8.1% of the full body height, below the torso’s median line. For a person of height 1.7 m, this will be:

\(\begin{array}{l}{y_{{\rm{cm}}}} = \left( {8.1\% } \right)\left( {1.7\;{\rm{m}}} \right)\left( {\frac{{100\;{\rm{cm}}}}{{1\;{\rm{m}}}}} \right)\\{y_{{\rm{cm}}}} = 13.7\;{\rm{cm}}{\rm{.}}\end{array}\)

The thickness of the torso from the median line is about 10 cm. Therefore, the CM is well outside the body.