Consider the tibia (shinbone) for a person of weight 750 N standing on the ball of one foot as in the following figure. The ankle joint 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 of the tibia due to the compressive forces.

Transcribed Image Text:

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