Please help with the set up and possibly a discription A person with 100kg mass
ID: 1331655 • Letter: P
Question
Please help with the set up and possibly a discription
A person with 100kg mass is standing erect on the tip of both feel. Note that each foot is subject to the normal force, equal to half of his or her weight. Find the contact force C exerted by the bones, tibia and fibula, on the ankle joint, and the muscle force F on the tendon of calcareous at 37 with the horizontal. In this posture, the distance between the application points of the contact force and the muscle force in a longitudinal cut is 5 cm, and between the application points of the contact force and the normal force, it is 6 cm.Explanation / Answer
N = mg/2 = 100*9.8/2 = 490 N
net torque = 0
F*sin37*0.05 = N*0.06
F*sin37*0.05 = 490*0.06
F = 977 N
net force = 0
along vertical
F*sin37 + N = C*cosalfa
(977*sin37)+490 = C*cosalfa
c*cosalfa = 1078
along horizantal
F*cos37 = c*sinalfa
C*sinalfa = 780
c = sqrt(1078^2+780^2)
C = 1331 N <<<<------answer