In Problem \(8.41,\) we found that the force of the tibia (shinbone) on the ankle joint for a person (of weight \(750 \mathrm{N})\) standing on the ball of one foot was \(2800 \mathrm{N}\). The ankle joint therefore pushes upward on the bottom of the tibia with a force of \(2800 \mathrm{N},\) while the top end of the tibia must feel a net downward force of approximately \(2800 \mathrm{N}\) (ignoring the weight of the tibia itself). The tibia has a length of $0.40 \mathrm{m},\( an average inner diameter of \)1.3 \mathrm{cm},$ and an average outer diameter of \(2.5 \mathrm{cm} .\) (The central core of the bone contains marrow that has negligible compressive strength.) (a) Find the average crosssectional area of the tibia. (b) Find the compressive stress in the tibia. (c) Find the change in length for the tibia due to the compressive forces.

Step by step solution

01

Calculate the average cross-sectional area of the tibia.

First, we need to find the average radius of the inner and outer diameters of the tibia. After that, we will calculate the difference in the areas of the outer and inner circle to find the average cross-sectional area. The radius of the inner circle (r1) = \(\frac{inner diameter}{2} = \frac{1.3}{2} = 0.65cm\) The radius of the outer circle (r2) = \(\frac{outer diameter}{2} = \frac{2.5}{2} = 1.25cm\) The area of inner circle (A1) = \(\pi \times r1^2 = \pi \times (0.65 \times 10^{-2})^2 = 1.33 \times 10^{-4}m^2\) The area of the outer circle (A2) = \(\pi \times r2^2 = \pi \times (1.25 \times 10^{-2})^2 = 4.91 \times 10^{-4}m^2\) The average cross-sectional area (A) = A2 - A1 = \(4.91 \times 10^{-4}m^2 - 1.33 \times 10^{-4}m^2 = 3.58 \times 10^{-4}m^2\)

02

Calculate the compressive stress in the tibia.

To find the compressive stress in the tibia, we will use the force and the cross-sectional area: Compressive stress (σ) = \(\frac{Force}{Area} = \frac{2800 N}{3.58 \times 10^{-4}m^2} = 7.82 \times 10^6 N/m^2\)

03

Calculate the change in length for the tibia.

To find the change in length, we will use the formula: \(\Delta L = \frac{Force \times Length}{Area \times E}\) In this problem, we will use the approximate value for the modulus of elasticity (E) for bone, which is around \(1.5 \times 10^{10} N/m^2\). \(\Delta L = \frac{2800 N \times 0.40 m}{3.58 \times 10^{-4}m^2 \times 1.5 \times 10^{10} N/m^2} = 1.09 \times 10^{-4}m\) So, the change in the length of the tibia is approximately \(1.09 \times 10^{-4} m\).

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