The leg bone (femur) breaks under a compressive force of about $5 \times 10^{4} \mathrm{N}\( for a human and \)10 \times 10^{4} \mathrm{N}$ for a horse. The human femur has a compressive strength of \(1.6 \times 10^{8} \mathrm{Pa},\) while the horse femur has a compressive strength of $1.4 \times 10^{8} \mathrm{Pa} .$ What is the effective crosssectional area of the femur in a human and in a horse? (Note: since the center of the femur contains bone marrow, which has essentially no compressive strength, the effective cross- sectional area is about \(80 \%\) of the total cross-sectional area.)

Use the formula: Force = Compressive strength * Effective cross-sectional area. For the human femur, Force = \(5 \times 10^{4} \mathrm{N}\) Compressive strength = \(1.6 \times 10^{8} \mathrm{Pa}\) Effective cross-sectional area = \(\frac{Force}{Compressive\, strength}=\frac{5 \times 10^{4} \mathrm{N}}{1.6 \times 10^{8} \mathrm{Pa}} = 3.125 \times 10^{-4} \mathrm{m^2}\) The effective cross-sectional area of the human femur is \(3.125 \times 10^{-4} \mathrm{m^2}\).

Use the same formula as before for the horse femur: Force = \(10 \times 10^{4} \mathrm{N}\) Compressive strength = \(1.4 \times 10^{8} \mathrm{Pa}\) Effective cross-sectional area = \(\frac{Force}{Compressive\, strength}=\frac{10 \times 10^{4} \mathrm{N}}{1.4 \times 10^{8} \mathrm{Pa}} = 7.142857 \times 10^{-4} \mathrm{m^2}\) The effective cross-sectional area of the horse femur is \(7.142857 \times 10^{-4} \mathrm{m^2}\).

We know that the effective cross-sectional area is 80% of the total cross-sectional area. So we can calculate the total cross-sectional area of both the human and horse femur by: For the human femur, Total cross-sectional area = \(\frac{Effective\, cross-sectional\, area}{0.8}\) = \(\frac{3.125 \times 10^{-4} \mathrm{m^2}}{0.8}\) = \(3.90625 \times 10^{-4} \mathrm{m^2}\) For the horse femur, Total cross-sectional area = \(\frac{Effective\, cross-sectional\, area}{0.8}\) = \(\frac{7.142857 \times 10^{-4} \mathrm{m^2}}{0.8}\) = \(8.928571 \times 10^{-4} \mathrm{m^2}\) The total cross-sectional area of the human femur is \(3.90625 \times 10^{-4} \mathrm{m^2}\), and the total cross-sectional area of the horse femur is \(8.928571 \times 10^{-4} \mathrm{m^2}\).