Biophysics Problem 38

You should realize that this is a problem in scaling. In both cases, we have to cover a surface.

It is important notice that in both sweaters, the thickness of material will be the same. Thus the wool covering one square metre of a child will have the same mass as the wool covering one square metre of a man.

It should be obvious that:

\(\frac{\text{mass of wool for man}}{\text{area of man}}=\frac{\text{area of man}}{\text{area of child}}\)

We do not need to know the absolute area of either the child or the man; all we need is the ratio. Since the child and the man have the same density, then the ratio of their volumes will be the same as the ratio of their masses.

How many times is the volume of the man than that of the child?

You should have found that:

\(\frac{\text{volume of man}}{\text{volume of child}}= \frac{\text{mass of man}}{\text{mass of child}}= \frac{120}{15}= 8\)

If the linear dimensions of the man are greater than those of the child by a factor of \('L',\) then the volume of the man will be greater than the child by a factor of \(' L^3.'\)

So, \(L^3 = 8.\) This gives us a scaling factor of what? (i.e. \(L = ?\))

Since \(L^3 = 8,\) then \(L = 2.\)

We are now in a position to calculate the ratio of the area of the man relative to that of the child.

Calculate this ratio.

Remember, areas are two dimensional, so scale as \(L^2.\) Thus the area of the man is 4 times that of the child.

Earlier we wrote down:

\(\frac{\text{mass of wool for man}}{\text{area of man}}=\frac{\text{area of man}}{\text{area of child}}\)

Now calculate the mass of wool needed for the man's sweater.

Because the man has four times the area, it will take four times as much wool to cover him.

Therefore, \(4 \times 100\; g = 400 \;g\) of wool needed to cover the man.