Q11-51P - University Physics with Modern Physics

Question

A therapists 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. To comply with these instructions, the patient consults a table of typical mass distributions and finds that both upper leg (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?

Step-by-Step Solution

Answer

The supporting strap should be attached to the cast at a distance of 49.9 cm from the hip joint.

Step 1: The given data

Given that 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. To comply with these instructions, the patient consults a table of typical mass distributions and finds that both upper leg (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.

Mass of each upper leg $m_1=\left(\frac{1}{2}\right)\left(\frac{215}{1000}\right)\left(78\text{ kg}\right)=7.955\text{ kg}$

Position of upper leg $x_1=18\text{ cm}$

Mass of each lower leg $m_2=\left(\frac{1}{2}\right)\left(\frac{14}{100}\right)\left(78\text{ kg}\right)=5.18\text{ kg}$

Position of lower leg $x_2=69\text{ cm}$

Mass of cast $m_3=5.5 \text{kg}$

Position of cast $x_3=78\text{ cm}$

Step 2: Formula used

Center of mass $x=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$

Where are masses and $x_i$’s are position.

Step 3: Find position of supporting strap

$x=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$where x is position of supporting strap

Hence $x=\frac{\left(7.955\text{ kg}\right)\left(18\text{ cm}\right)+\left(5.18\text{ kg}\right)\left(69\text{ cm}\right)+\left(5.5\text{ kg}\right)\left(7.8\text{ cm}\right)}{\left(7.955\text{ kg}\right)+\left(5.18\text{ kg}\right)+\left(5.5\text{ kg}\right)}=49.9\text{ cm}$

Hence, the supporting strap should be attached to the cast at a distance of 49.9 cm from the hip joint.