Randy Scalise generated this in response to a e-mail question,
but we though it would be of use to include on the web page.
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Dear Mr. Willis,
This is Randy Scalise from SMU Physics. I was the one who performed
the cantilevered blocks demonstration.
Use 7 identical blocks. The trick is to start at the top.
Align each block (BBBBBBBB) with the edge of the table (TTTTTTTT).
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
TTTTTTTTTTTT
The top block can be pulled half way off the second block
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
TTTTTTTTTTTT
The top two blocks can be pulled as a unit 1/4 off the third block
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
TTTTTTTTTTTT
This is a great time to ask the class to guess the next fraction in the
pattern. About half will guess "1/8" seeing the pattern as inverse
powers of two. The correct answer is "1/6" and the pattern then emerges
as inverse even integers.
The top three blocks can be pulled as a unit 1/6 off the fourth block
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
BBBBBBBBBBBB
TTTTTTTTTTTT
Continue. Move the top four blocks as a unit 1/8 off the fifth block.
Move the top five blocks as a unit 1/10 off the sixth block.
Move the top six blocks as a unit 1/12 off the seventh block.
Move all the blocks as a unit 1/14 off the table.
1/2 + 1/4 + 1/6 + 1/8 + 1/10 + 1/12 + 1/14 = 1.29643 > 1
In theory, four blocks would be enough (1/2 + 1/4 + 1/6 + 1/8 = 1.04 > 1),
but in practice at each stage you must back off from the ideal overhang
to avoid collapse.
Now is a good time to ask the class to calculate in theory how many
blocks would be required to get the top TWO blocks off the edge of the
table (the answer is 31).
You can also introduce the harmonic series (1 + 1/2 + 1/3 + 1/4 + ...)
at this point. Our series is one half times the harmonic series.
The harmonic series diverges, that is, the sum is infinite. As a
consequence, one can cantilever any number of blocks past the edge of
the table, but the total number of blocks in the vertical stack gets
large VERY quickly. To cantilever three blocks past the edge of the
table requires a stack of 227 blocks. To cantilever four blocks past
the edge of the table requires a stack of 1674 blocks.
Feel free to write or call if I can help further.
--Dr. Randall J. Scalise
Senior Lecturer in Physics
,____________________________________________________________________.
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