# Biophysics Problem 35

Suppose you are supported by a single steel wire. (Ouch!) What must be the diameter of the wire if the elongation is not to exceed $0.1$ percent?

#### First Step

The answer given assumes a mass of $70\; kg.$ (So thoughtful of them to tell you in the question!)

You should realize that you have to use Hooke's Law. Young's modulus for steel (from page 103 in the textbook) is:

$Y = 2 \times 10^{11} \;N\; m^{-2}$

Many students get the wrong answer because they use the wrong value for $\ell / \ell.$

What do you think this value should be?

#### Calculations

$\frac{\Delta \ell}{\ell} \times 100 = 0.1 \\ \frac{\Delta \ell}{\ell} = 0.001$

You should then be able to calculate the area $'A'$ of the wire, and then its diameter.

Calculate $A.$

You should be using:

$\frac{F}{A}= Y\frac{\Delta \ell}{\ell}$

Substituting, you should find that:

$A = 3.43 \times 10^{-6}\; m^2$

Now that you have the area, it should be simple to find the diameter of the wire.

Remember we found area to be:

$A = 3.43 \times 10^{-6}\; m^2$

Further,

$A = \pi r^2 = \pi D^2/4$

where $r$ is the radius, and $D$ is the diameter.

A simple subsitution should give you $D = 2.1 \times 10^{-3}\; m.$