A pillar 2m tall stands on a square base, 20cm on a side. The peculiar thing about this pillar is that its top, which is parallel to the base, is an equilateral triangle 16cm on a side.
The four edges running the length of the pillar are linear. Two of these edges meet at one corner of the triangle, and the edge of the triangle opposite this vertex is parallel to two edges of the square. All horizontal cross sections have straight edges.
If the pillar is made of basalt (density = 2.8 g/cm^3), what is its total mass?
Let's assume that if for example the triangle's vertex that lies to the north is the one that has two edges joined to opposite ends of an edge of the square, and that it connects to both ends of the north edge of the square.
Each horizontal slice will be a trapezoid (that is, in the U.S.; in England it would be called a trapezium) with two parallel edges. At the extremes they are degenerate, as the square, with two sets of parallel edges, and the triangle where one of the parallel edges has length zero.
First we need to find the volume. To simplify the math, let x range from zero at the bottom to 1 at the top. The differential of volume is therefore 200 dx (measured in centimeters).
The larger (south in our setup) side of each trapezoid will be 20-4x cm, while the smaller (north) side will be 20-20x cm. These average to 20-12x cm. The altitude of each trapezoid will be 20-(20-8√3)x cm.
The area of each trapezoid will be (20-12x)(20-(20-8√3)x). We need to integrate this from zero to one and multiply that by 200 to get the volume.
This comes out to 200 ∫{0 to 1} (400 - (640-160√3)x + (240-96√3)x²)dx
We get 200[400x - ((640-160√3)/2)x² + ((240-96√3)/3)x³]{0 to 1}
which evaluates to 48627.68775...
We need to multiply by the 2.8 g/cm³ density. Blindly multiplying we'd get 136157.5... grams, but 2.8 g/cm³ was given to only 2 significant figures. Allowing for the fact that the answer begins with a lower digit, we can stretch and allow 3 significant digits in the answer, and conveniently say 136 kg.
|
Posted by Charlie
on 2003-04-15 03:44:41 |