A therapist tells a 80.0 {kg} patient with a broken leg that he must have his le

Question

A therapist tells a 80.0 {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 .) 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 { m cm} from the hip joint. The cast has a mass of 5.50 {kg}, and its center of mass is 78.0 { m cm} from the hip joint A therapist tells a 80.0 {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 .) 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 {rm cm} from the hip joint. The cast has a mass of 5.50 {kg}, and its center of mass is 78.0 {rm cm} from the hip joint 1.How far from the hip joint should the supporting strap be attached to the cast? (in cm)

Explanation / Answer

Mass of one thigh is (Mass of person)(0.215)(0.5) = 8.6 kg Mass of one lower leg is (Mass of person)(0.140)(0.5) = 5.6 kg So, we can consolidate this system into a coordinate system with the hip at x=0. The CM of the thigh at x1, CM of lower leg at x2, CM of cast at x3. CM of Leg/Cast system: [(x1)(8.6)+(x2)(5.6)+(x3)(5.5)]/(8.6+5.6+5.5) = (970.2)*kg/(19.7kg)= 49.2 Place support at 49.2 cm from hip joint

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