Blocks stacked on an edge - max dist. b4 they fall?

AI Thread Summary
The discussion revolves around determining the maximum overhang distance (x) of two stacked bricks without them falling off a ledge. The key concept is the center of gravity, which must remain over the edge of the surface to prevent the bricks from toppling. The correct calculation shows that the maximum distance x is 13.5 cm, derived from the combined center of mass of the two bricks. Clarifications were made regarding the conditions for stability, emphasizing that while the center of mass must be above the bottom brick, it cannot extend beyond the edge. Understanding these principles resolves the initial confusion about the overhang distance.
lizzyb
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The question is "Two identical uniform bricks of length 18 cm are stacked over the edge of a horizontal surface with the maximum overhand possible without falling."

Code:
|<-- 18 cm -->|
+-------------+
|    block 1  |
+-------------+
      +-------------+
      |    block 2  |
      +-------------+
|<-- x -->|+-----------------
           |XXXXXXXXXXXXXXXXX
           |XXXXX ledge XXXXX
           |XXXXXXXXXXXXXXXXX

find the maximum distance x. answer in cm.

Is this a center of gravity question? if so, then how do i set it up? thank you.
 
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But in that thread, OlderDan wrote: "The condition for one book to not fall off the table is the same condition for the second book to not fall off the first book."

Why then is the answer not 18? The condition of the second book to not fall off of the first is that it's center of gravity is over the first book, hence .5L = 9, and the first book must have it's COG at 9 too. hmmm
 
lizzyb said:
But in that thread, OlderDan wrote: "The condition for one book to not fall off the table is the same condition for the second book to not fall off the first book."

Why then is the answer not 18? The condition of the second book to not fall off of the first is that it's center of gravity is over the first book, hence .5L = 9, and the first book must have it's COG at 9 too. hmmm
OlderDan can't remember anything he did that long ago. The condition is the same, but that does not mean the distance is the same. If you put one book on top of another, the center of mass of the top book has to be somewhere above the bottom book. If you take the combination of the two books arranged as close as possible to this limiting arrangement, the center of mass of the two books combined has to be somewhere over the table, not past the edge. Find the center of mass of the two books and see how far that is from the farthest edge of the top book.
 
ok i think i understand it now. thanks!
 
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