Question

11.55 ** bio supporting a broken leg. a therapist tells a 74 kg patient with a broken leg that he must have his leg in a cast suspended horizontally. for minimum discomfort, the leg should be supported by a vertical strap attached at the center of mass of the leg - cast system (fig. p11.55). to comply with these instructions, the patient consults a table of typical mass distributions and finds that both upper legs (thighs) together typically account for 21.5% of body weight and the center of mass of each thigh is 18.0 cm from the hip joint. the patient also reads that the two lower legs (including the feet) are 14.0% of body weight, with a center of mass 69.0 cm from the hip joint. the cast has a mass of 5.50 kg, and its center of mass is 78.0 cm from the hip joint. how far from the hip joint should the supporting strap be attached to the cast?

Explanation:

Step1: Calculate the mass of each part

• Mass of both thighs: \(m_{thighs}=0.215\times74\space kg = 15.91\space kg\). Since there are two thighs, mass of one thigh \(m_1=\frac{15.91}{2}=7.955\space kg\)
• Mass of both lower - legs: \(m_{lower - legs}=0.14\times74\space kg=10.36\space kg\). Mass of one lower - leg \(m_2 = \frac{10.36}{2}=5.18\space kg\)
• Mass of the cast \(m_3 = 5.50\space kg\)

Step2: Use the formula for the center of mass

The formula for the \(x\) - coordinate of the center of mass \(x_{cm}=\frac{\sum_{i = 1}^{n}m_ix_i}{m_{total}}\)
The total mass of the leg - cast system \(m_{total}=m_1 + m_2+m_3\)
\(m_{total}=7.955+5.18 + 5.50=18.635\space kg\)
Let \(x_1 = 18.0\space cm\), \(x_2 = 69.0\space cm\), \(x_3 = 78.0\space cm\)
\(x_{cm}=\frac{m_1x_1+m_2x_2+m_3x_3}{m_{total}}\)
Substitute the values:

$$
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$$

Answer:

\(x_{cm}\approx49.9\space cm\)