Forces 2) Left: Anatomical diagram of a patient standing on one leg during slow

Question

Forces

2) Left: Anatomical diagram of a patient standing on one leg during slow walking (i.e., assume static equilibrium). Right: Free body diagram of the lower leg. Assume the patient has a body mass of 65 kg. The vertical (i.e., Normal) ground reaction force (N) is equal to the patient's total body weight (Wb). Assume the weight of their lower leg is 16% of Wb. a) Compute the magnitude (Newtons) of the hip abductor muscle force (M). b) Compute the magnitude (Newtons) of the net joint reaction force (R). c) Express your answers for both M and R as a percentage of Wb. Do your answers seem reasonable? Why or why not?

Explanation / Answer

As person is in static equilibrium, so net force and net torque will be zero.

so, Wb = N= mg = 65*9.8=637N (where m is the mass of the body)

weight of the lower leg = 16/100 *65 * 9.8 = 101.92Newton

(a) net torque about joint,

M*sin70 * 7 - N*10.8 + W*3.2 = 0 (where W is the weight of the lower legthat is 16% of tatal body weight)

M = N*10.8- W*3.2/sin70* 7

M = 637*10.8-326.144/6.573 = 997.02 Newton

(b) Ry = N - W - M sin70 = 404.81 Newton

Rx = Mcos70 = 341 Newton

the net reaction force, R =( R2x + R2y)1/2 = (116281 + 163871.13)1/2 = (280152.13)1/2 = 509.29 Newton

(c) increase of force in M from Wb = 997.02 - 637 = 360.02Newton

% increase in M = 360.02 * 100/ 637 = 56.51

% decrease in R = (637 - 509.29)* 100 / 509.29 =25

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