Constants A therapist tells a 72 kg patient with a broken leg that he must have
ID: 2031533 • Letter: C
Question
Constants A therapist tells a 72 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. (See the figure below (Figure 1).) In order 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 77.0 cm from the hip joint. Figure 1 of 1 Supporting strap Hip joint CastExplanation / Answer
The problem can be solved by splitting the leg up into 3 parts: the thigh, the lower leg, the cast
Mass(thigh) = (1/2)(0.215)(72.0kg)
= 7.74 kg
Mass(lower leg) = (1/2)(0.14)(72.0kg)
= 5.04 kg
Mass(cast) = 5.50 kg
com(center of mass) (thigh) = 18.0 cm
COM(lower leg) = 69.0 cm
COM(cast) = 77.0 cm
Now use the expression of COM -
x = (m1x1 + m2x2 + m3x3) / (m1 + m2 + m3)
= (7.74*18 + 5.04*69 + 5.50*77) / (7.74 + 5.04 + 5.50)
= 910.58 / 18.28 = 49.81 from the hip joint.