A therapist tells a 76 kg patient with a broken leg that he must have his leg in
ID: 1464628 • Letter: A
Question
A therapist tells a 76 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 5.60 kg , and its center of mass is 80.0 cm from the hip joint.
How far from the hip joint should the supporting strap be attached to the cast?
Explanation / Answer
mass of thigh each, M = 0.215*76/2 = 8.17 kg
mass of each lower leg, m = 0.14*76/2 = 5.32 kg
mass of cast = 5.6 kg
let x is the distance where the suuporting strap should be placed.
when theleg is in equilibrium, net force and net torque acting on it must be zero.
Let T is the tension n the string.
Apply, Fnety = 0
T - (8.17 + 5.32+5.6)*9.8 = 0
T = (8.17 + 5.32+5.6)*9.8
T = 187 N
Apply net torque about hip joint = 0
8.17*9.8*18 + T*x - 5.32*9.8*69 - 5.6*9.8*-80 = 0
8.17*9.8*18 + 187*x - 5.32*9.8*69 - 5.6*9.8*80 = 0
x = (5.32*9.8*69 + 5.6*9.8*80 - 8.17*9.8*18)/187
= 35 cm <<<<<<<<<---------------Answer
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