Question

A 75 kg person is crouching with his weight evenly distributed on both tiptoes. The forces on the skelet part of foot are shown in Fig. 12.9 below. In this position, the Achilles tendon experiences a force F., making an angle a = 59.2° with the horizontal. The tendon acts on the ankle a horizontal distance d= 9.66 cm from the point where foot touches the floor. The tibia acts on the foot with magnitude | F and makes an angle B with the vertical. This force acts on the ankle at a horizontal distance d = 5.41 cm from the point where the foot touches the floor. Ignoring the weight of the foot and taking B = 60°, calculate the magnitudes of the tension in Achilles tendon, F. and F, in Newtons. FH 59.2° 5) 609 N = 223 N 0 4.25 cm 9.66 cm Fig. 12.9 Let's first mention that we are not counting the weight force (37.5 kg x 9.8 m/s2 on each foot) as that has been taken care of by Nin the figure. Therefore, FH sin59.2° +223 = F, cos 60°.

Answer 1

FH 1 N= 223N S9 2 P g. 66cm G 4.2518 moment balance aboute FJX sintox4.25 = 223X(9.66+4.25) 52 842-77 N fut sin (59 21+ 223= dj aggo 842 223096 fu= 230-96N