Re: [MathWithoutBorders] Chapter 4 Ex 4-5 #32
- You are doing it right, and the answer is 36 lb/sq-in. My Teachers' edition also shows 36 as the correct answer. Wherever you are reading 32, it is an error. Is that the answer in the Solution Guide?
The part I would have thought you might wonder about is where the 1.2 comes from. I think it is a factor that compensates for the shape of the heel. If it were square, the formula would simply be P=W/H^2. For a circular heel it would come out ~1.27W/H^2, so it is probably an approximate form for a circular or near circular heel. The area of a circle is pi*R^2, which can be expressed in terms of the diameter, D, as (pi*D^2)/4. The pressure due to a heel with a circular cross section would thus be P= (4W)/(pi*D^2) = ~ (1.27W)/D^2. (The author uses H instead of D.)
--David ChandlerOn Mon, Feb 13, 2012 at 9:16 AM, steitzfitness <steitzfam@...> wrote:
I have a question regarding the Spiked Heel Problem on pg 149. Letter a. answer shows that H squared is 32, but the heel is 3 inches. Am I missing something? Why isn't it 1.2(270)/9 ? Thanks for any enlightenment!!