CP BIO Stress on the Shin Bone. The compressive strength of our bones is important in everyday life. Young’s modulus for bone is about $\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{Pa}$. Bone can take only about a 1.0%change in its length before fracturing.

1. What is the maximum force that can be applied to a bone whose minimum cross-sectional area is $\mathbf{3}\mathbf{.}\mathbf{00}{\mathbf{cm}}^{\mathbf{2}}$? (This is approximately the cross-sectional area of a tibia, or shin bone, at its narrowest point.)
2. Estimate the maximum height from which aman could jump and not fracture his tibia. Take the time between when he first touches the floor and when he has stopped to be 0.030 s, and assume that the stress on his two legs is distributed equally.

Consider the formula for Newton’s law of motion:

${\mathbf{v}}^{\mathbf{2}}\mathbf{=}{\mathbf{v}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{gH}$ (1)

Here, vis final velocity,${\mathbf{v}}_{\mathbf{0}}$ is initial velocity, g is gravitational acceleration, H is height.

Here we have, Young’s modulus of bone is$\mathrm{Y}=1.4\times {10}^{10}\mathrm{Pa}$

Cross-sectional area is $\mathrm{A}=3\times {10}^{-4}{\mathrm{m}}^2$

Change in length is 1.0%.

(a)

Derive the formula for the force as:

$\begin{array}{r}Y=\frac{F_{\perp }{I}_0}{A\mathrm{\Delta }I}\\ F_{\perp }=YA\frac{\mathrm{\Delta }I}{I_0}\end{array}$ (2)

Consider the percentage by which the length is changed is 1.0% .

So, $\frac{∆l}{l_0}=1.0\%$

Substitute the values in equation (2) and solve as:

$\begin{array}{r}{F}_{\perp }=\left(1.4\times {10}^{10}\mathrm{Pa}\right)\left(3\times {10}^{-4}{\mathrm{m}}^2\right)\left(1.0\mathrm{\%}\right)\\ =\left(1.4\times {10}^{10}\mathrm{Pa}\right)\left(3\times {10}^{-4}{\mathrm{m}}^2\right)\left(0.01\right)\\ =42\times {10}^3\mathrm{N}\end{array}$

Hence, the maximum force that can be applied to a bone whose minimum cross-sectional area is$3.00{\mathrm{cm}}^2\mathrm{is}42\times {10}^3\mathrm{N}$

(b)

Simply the equation (1) as follows:

$\begin{array}{r}{v}^2=v_0^2+2gH\\ v^2=2gH(\because v_0=0)\\ H=\frac{v^2}{2g}\left(3\right)\end{array}$

Determine the expression for velocity from Newton’s second Law:

$\begin{array}{r}2F_{max}=\frac{\mathrm{\Delta }p}{\mathrm{\Delta }t}\\ 2F_{max}=\frac{mv}{\mathrm{\Delta }t}\\ V=\frac{2F_{max}\mathrm{\Delta }t}{m}\end{array}$

Substitute the values and solve as:

$\begin{array}{r}{\mathrm{v}}_{max}=\frac{2\left(42\times {10}^3\mathrm{N}\right)(0.030\mathrm{s})}{\left(70\mathrm{kg}\right)}\\ =36\frac{\mathrm{m}}{\mathrm{s}}\\ \mathrm{Now},\mathrm{to}\mathrm{find}\mathrm{maximum}\mathrm{height}\mathrm{put}\mathrm{the}\mathrm{value}\mathrm{of}{\mathrm{v}}_{\max }\mathrm{in}\mathrm{equation}\left(3\right).\\ {\mathrm{H}}_{max}=\frac{{\mathrm{v}}_{max}^2}{2\mathrm{g}}\\ =\frac{{\left(36\frac{\mathrm{m}}{\mathrm{s}}\right)}^2}{2\left(9.8\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)}\\ =65\mathrm{m}\end{array}$

Hence, the maximum height from which a 70 kg man could jump and not fracture his tibia is 65 m .