(Figure 1) shows the foot and lower leg of a person standing on the ball of one

Q: Figure 1) shows the foot and lower leg of a person standing on the ball of one

A) apply net torque about Ankle joint = 0 n*11.6 - T*4.55*sin(90-25) = 0 T = n*11.6/(4.55*sin(65)) = 897*11.6/(4.55*sin(65)) = 2523 N

ID: 1496386 • Letter: #

Question

(Figure 1) shows the foot and lower leg of a person standing on the ball of one foot. Three forces act to maintain this equilibrium: the tension force T in the Achilles tendon, the contact force F c at the ankle joint, and the normal force n that supports the person's 897-Nweight. The application points for these forces are shown in the figure. The person's center of gravity is directly above the contact point with the ground, and you can treat the mass of the foot itself as being negligible.

Part A

Find the magnitude of the tension in the Achilles tendon.

Part B

Find the magnitude of the contact force at the ankle joint.

Explanation / Answer

A) apply net torque about Ankle joint = 0

n*11.6 - T*4.55*sin(90-25) = 0

T = n*11.6/(4.55*sin(65))

= 897*11.6/(4.55*sin(65))

= 2523 N <<<<-----------------Answer

B) now Apply net torque about achille tendon = 0

n*(11.6+4.55) - Fc*4.55*sin(90-25) = 0

Fc = n*(11.6+4.55)/(4.55*sin(65))

= 897*(11.6+4.55)/(4.55*sin(65))

= 3513 N <<<<-----------------Answer

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(Figure 1) shows the foot and lower leg of a person standing on the ball of one

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