# 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?

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?

\(\frac{\Delta \ell}{\ell} \times 100 = 0.1 \\ \frac{\Delta \ell}{\ell} = 0.001\)

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 \(D = 2.1 \times 10^{-3}\; m.\)