A man with a mass of 71.2 kg stands on one foot. His femur has a cross-sectional area of 8.00 cm$^2$ and an uncompressed length of 51.7 cm. Young’s modulus for compression of the human femur is 9.40 × 10$^9$ N/m$^2$.

a. How much shorter is the femur when the man stands on one foot? (Answer in cm)

b. What is the fractional length change of the femur when the person moves from standing on two feet to standing on one foot?

The man's femur becomes about 3.77 cm shorter when he stands on one foot, which also represents a fractional length change, or strain, of 7.31%.

a. Young's modulus (Y) is defined as the ratio of stress (σ) to strain (ε). Stress is the pressure applied (Force/Area), and strain is the resulting change in length over original length (∆L/L).

When the man stands on one foot, the weight of his entire body applies a pressure (stress) on that femur.

This stress is given by σ = Force/Area = mg/A = (71.2 kg * 9.8 m/s2) / (8.00 cm2 * (1 m/100 cm)2) = 0.687 x 108 N/m2.

Solving for strain, we get ε = σ / Y = (0.687 x 108 N/m2) / (9.4 x 109 N/m2) = 0.0731.

Then, the change in length of the femur (∆L) is given by ∆L = ε*L = 0.0731 * 51.7 cm = 3.77 cm. So the femur becomes about 3.77 cm shorter when the man stands on one foot.

b. The fractional length change of the femur is simply the strain, which is 0.0731 or 7.31%.

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