A pile of ore has a rectangular base, 60 feet wide and 500 feet long. If the sides of the pile are all inclined 45degrees to the horizontal, and the ore weighs 110 lb. per cu.ft. Find the number of tons of ore in the pile. ( 1ton = 2000 lb).
If possible, use the formula for solving the volume of prismatoid, (as this is what our topic is all about)
V = ((b + B 4M)/6)*h
b = area of upper base; B = area of lower base; M = area of midsection; h = height
Thanks!
This is the formula for a general prismatoid that's bounded by horizontal surfaces on both sides. It works by quadratically interpolating the three areas, much like Simpson's rule. Since you don't specify a height nor any dimensions for the upper horizontal surface, I presume that the pile extends all the way up until the width becomes zero. In that case you can use a simpler formula, which you can derive from this formula by noting that in this case the midsection has one fourth of the area of the lower base, so the volume is
$$ V=\frac{b+B+4M}6h=\frac{0+B+4\frac B4}6h=\frac{Bh}3\;. $$
In your case the lower base area is $60\def\ft{\,\mathrm{ft}}\ft\cdot500\ft=30000\ft^2$ and the height is half the width, $15\ft$, so the volume is $30000\ft^2\cdot15\ft/3=150000\ft^3$. Mutliplying this by the density $110\def\lb{\,\mathrm{lb}}\lb/\mathrm{\ft}^3$ yields $16500000\lb=8250\,\mathrm t$.
