In Problem 8.41, we found that the force of the tibia (shinbone) on the ankle joint for a person (of weight 750 N) standing on the ball of one foot was 2800 N. The ankle joint therefore pushes upward on the bottom of the tibia with a force of 2800 N, while the top end of the tibia must feel a net downward force of approximately 2800 N (ignoring the weight of the tibia itself). The tibia has a length of 0.40 m, an average inner diameter of 1.3 cm, and an average outer diameter of 2.5 cm. (The central core of the bone contains marrow that has negligible compressive strength.)
(a) Find the average cross sectional area of the tibia.
(b) Find the compressive stress in the tibia.
(c) Find the change in length for the tibia due to the compressive forces.

Transcribed Image Text:

2800 N 20S0N 750N 1.3 cm 0.40 m 2800 N 2.5 cm Cross section 2050 N 750 N 2800 N 2800 N