Question

the femur bone in a human leg has a minimum effective cross section of 3.05 cm² and an ultimate strength of 1.70×10⁸ n/m². how much compressive force fmax can the femur withstand before breaking? fmax = n

Explanation:

Step1: Convert the cross - sectional area to SI units

The formula for stress is \(\sigma=\frac{F}{A}\), where \(\sigma\) is stress, \(F\) is force, and \(A\) is area.
Given \(A = 3.05\space cm^{2}\). Since \(1\space m=100\space cm\), then \(1\space m^{2}=(100\space cm)^{2} = 10^{4}\space cm^{2}\). So \(A=3.05\times10^{- 4}\space m^{2}\).

Step2: Rearrange the stress formula to solve for force

From \(\sigma=\frac{F}{A}\), we can get \(F=\sigma\times A\).
Given \(\sigma = 1.70\times10^{8}\space N/m^{2}\) and \(A = 3.05\times10^{-4}\space m^{2}\).
Substitute the values into the formula: \(F=(1.70\times 10^{8})\times(3.05\times10^{-4})\).
Using the rule of exponents \(a^{m}\times a^{n}=a^{m + n}\), we have \(F=(1.70\times3.05)\times10^{8+( - 4)}\).
\(1.70\times3.05 = 5.185\) and \(10^{8+( - 4)}=10^{4}\). So \(F = 5.185\times10^{4}\space N\).

Answer:

\(5.185\times 10^{4}\)