With no stress the femur is $L_0=500mm$ long. When a $70-kg$ person is standing, the weight is distributed over the femur. Given the area of the femur to be $5\mathrm{.}2{10}^{-4}{m}^2$ and the Young's Modulus of the femur to be $1\mathrm{.}6{10}^{10}\frac{N}{m^2},$ how much does the femur shorten $($in $mm)$ when a person is standing on one foot?