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?

First I'd like to congratulate Charley and Tim Axoy whom I believe have obtained the correction weight of 136.15752 Kg.
A note for Hank: If you still haven't found where your problem was, just think that a solid pillar 20x20x200 cm3 weighs only 224 kg, less than the 236.957 kg that you proposed.

What I am posting is an arithmetic method for calculating volumes, a universal formula that is unfortunately not very well known. The formula goes as follows:

V=H*(A1+4A2+A3)/6

For those (e.g. Charlie) who are familiar with numerical integration using Simpson's formula, you will see the similitude.

Here A1 is the area at one end of the solid, A2 is the cross section area in the middle section, and A3 is the area at the other end, and H is the total length or height of the solid.

Take the case of a rectangle, A1=A2=A3=A, the formula reduces to
Vr=H*6A/6=H*A.
For a cone, A1=0, A2=πr²/4, A3=πr²
Vc=H*(0+4*πr²/4+πr²)/6
=πr²H/3
For a sphere, A1=0, A2=πr², A3=0
Vs=2*r*(0+4*πr²+0)/6
=4πr³/3
Now for our familiar pillar,
A1=64√3 (top section)
A3=20*20=400 (bottom section)
A2 is a trapezium, top side is (0+20)/2=10, bottom side is (16+20)/2=18, and the height is the average of 20 and the height of the triangle, equal to 10+4√3, therefore
A2=(10+18)*(10+4√3)/2
The volume is thereore:
Vp=(400+4(140+56√3)+64√3)*200/6
=32000+9600√3
Mass=2.8*Vp
=2.8*(32000+9600√3)
=136.15752 Kg.

Hope you all go out and enjoy this little know formula.

Best regards.

Posted by P C on 2004-02-04 20:36:39