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