Question

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.

Short answer

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

Step-by-step solution

Formulas of the center of a mass on the x-axis and the y-axis

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

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

Given information

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 \({y_h} = \left( {62.2 - 43.1} \right)\;{\rm{units}}\).

Calculate the center of mass of an arm bent at the right angle

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 \(\left( {21.7\;{\rm{cm,7}}{\rm{.6}}\;{\rm{cm}}} \right)\).