# Biophysics Problem 31

When standing erect, a person's weight is supported chiefly by the larger of the two leg bones. Assuming this bone to be a hollow circular tube of $2.5 \;cm$ internal diameter and $3.5 \;cm$ external diameter, what compressive load must this bone (in each leg) carry in the case of a $70 \;kg$ person?

#### First Step

First of all, you must make a diagram of the cross-section of the bone. Draw a diagram on a piece of paper.

#### Diagram

Remember that the $70\;kg$ represents the person's mass, and thus is not a force.

Calculate the downward force.

#### Calculations

$\text{downward force} = 70 \;kg \times 9.8\; m/s^2 = 686\; N$

Now calculate the cross-sectional area of the bone.

Remember that the area of a circle is: $A = \pi r^2$

Cross-sectional area, we'll call $'A'$is:

$A = \pi (3.5/2)^2 - \pi (2.5/2)^2 \\ = 9.62 - 4.91 \\ = 4.71\; cm^2 \\ = 4.71 \times 10^{-4}\; m^2$

Now calculate the compressive load -- that is, the force per unit area of bone. Remember that there are two bones, each carrying an equal share of the load.

Compressive load, we'll call $'CL'$ is:

$CL = \frac{force}{unit\;area\;of\; bone} \\ = \frac{686\; N}{2 \times 4.71 \times 1-^{-4}} \\ = 7.3 \times 10^5 \; Nm^2$