A 70.0 kg man supports 62.0 kg of his weight on one leg, modeled as a uniform rod 40.0 cm long and 2.00 cm in radius, with a compression Young's modulus for bone of 9 times ten to the ninth newtons per square meter. Using g = 9.80 m/s squared, roughly how much does the bone shorten?

1. The supported force is F = mg = (62.0 kg)(9.80 m/s squared), which is 607.6 N. 2. The cross-sectional area is pi times the radius squared, with radius 2.00 cm, giving about 1.257 times ten to the minus 3 square meters. 3. The change in length is delta L equal to (1/Y)(F/A)(L0), using the compression modulus for bone. 4. Substituting these values gives a change in length of about 2 times ten to the minus 5 meters. 5. Such a tiny shortening is consistent with the everyday experience that bones feel completely rigid under normal loads. 6. The other options are off by several orders of magnitude, describing either a huge deformation or a far smaller one than the calculation gives. _Source: OpenStax College Physics (CC BY 4.0), Ch 5 "Further Applications of Newton's Laws: Friction, Drag, and Elasticity", section 5.3 Elasticity: Stress and Strain_