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The Peculiar Pillar (Posted on 2003-04-15) Difficulty: 4 of 5
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?

  Submitted by Bryan    
Rating: 4.0000 (6 votes)
Solution: (Hide)
As stated in the problem, since the base of the pillar has 4 sides and the top has 3, two of the edges running the length of the pillar must meet at one of the corners of the triangle. Since the edge of the triangle opposite this corner is parallel to one of the edges of the bottom, every cross section of the pillar between the top and bottom is a trapezoid. To find the volume, we can integrate the trapezoidal cross sections along the height of the pillar.
For a trapezoid with parallel sides B1 and B2 and height H, its area is

A=.5(B1 + B2)H

At the square base, B1 = B2 = H = 20cm
At the top, B1 = 16, B2 = 0, & H = 8*sqrt(3)

All three values change linearly from the base of the pillar to its top 200cm above. Let y = 0 at the base and y = 200 at the top. Then

B1 = f(y) = 20 – y*4/200 = 20 – y/50
B2 = f(y) = 20 – y*20/200 = 20 – y/10
H = f(y) = 20 – y*20/200 + y*8*sqrt(3)/200 = 20 – y/10 + y*sqrt(3)/25

A = f(y) = .5(20 – y/50 + 20 – y/10)(20 – y/10 + y*sqrt(3)/25)
A = y^2*(15 – 6*sqrt(3))/2500 + y*(4*sqrt(3) – 16)/5 + 400

To find volume V:

V = integral[A dy] from 0 to 200
V = 1/3 *y^3*(15 – 6*sqrt(3))/2500 + .5*y^2*(4*sqrt(3) – 16)/5 + 400y ] from 0 to 200
V = 200^3*(15 – 6*sqrt(3))/7500 + .5*200^2*(4*sqrt(3) – 16)/5 + 80000
V = 48627.7 cm^3

Therefore the weight W of the pillar is

W = 48627.7*2.8 = 136157.6 g = 136.16 kg

Note: Whether the triangular top of the pillar is centered above the square base or not will have no effect on the pillar’s volume.
v

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionAnother SolutionP C2004-02-04 20:36:39
Hmmm...Silly Jilly2003-05-12 02:25:30
I do not recall...Tim Axoy2003-05-11 13:15:21
I do not recall...Tim Axoy2003-05-11 13:14:24
Oh My God !!!!Ravi Raja2003-04-19 07:56:01
SolutionsolutionTim Axoy2003-04-18 12:54:46
SolutionA TwistCharlie2003-04-15 05:02:49
No SubjectCharlie2003-04-15 05:02:05
Solutiona solutionHank2003-04-15 04:22:19
re: solutionRobin Gatter2003-04-15 04:01:05
My solutionRobin Gatter2003-04-15 03:52:34
SolutionsolutionCharlie2003-04-15 03:44:41
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