91Ó°ÊÓ

Given that to keep the calculations fairly simple but still reasonable, we model a human leg that is 92 cm long (measured from hip joint) by assuming that the upper leg and lower leg (which includes the foot) have equal lengths and are uniform. For a 70kg person, the mass of the upper leg is 8.6 kg, while that of the lower leg (including the foot) is 5.25 kg.

$m_1=8.6$ kg

$m_2=5.25$kg

Length of lower leg = 46 cm

Length of upper leg = 46 cm

Center of mass is$x=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$ and$y=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}$

$$\begin{gathered} x_1=23 \\ {x}_2=69 \end{gathered}$$ and $$\begin{gathered} y_1=0 \\ {y}_2=0 \end{gathered}$$

Center of mass isrole="math" localid="1668078807053" $x=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$ androle="math" localid="1668078826127" $y=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}$

role="math" localid="1668078839425" $x=\frac{\left(23\right)(8.6)+\left(69\right)(5.25)}{8.6+5.25}=40.4$cm and y =0

Hence, the center of mass is at a horizontal distance of 40.4 cm from the hip.

$$\begin{gathered} x_1=23 \\ {x}_2=46 \end{gathered}$$ and$$\begin{gathered} y_1=0 \\ {y}_2=23 \end{gathered}$$

Center of mass is$x=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$ and$y=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}$

$x=\frac{\left(23\right)(8.6)+\left(46\right)(5.25)}{8.6+5.25}=31.7$cm and$y=\frac{0+\left(23\right)(5.25)}{8.6+5.25}=8.7$ cm

Hence, center of mass is at a vertical distance of 8.7 cm below the hip and a horizontal distance of 31.7 cm from the hip.