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

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Homework Help Overview

The problem involves determining the maximum overhang distance of two identical uniform bricks stacked over the edge of a horizontal surface without falling. The context relates to concepts of center of gravity and stability in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster questions whether the problem is centered around the concept of center of gravity and seeks guidance on how to set it up. Other participants reference previous discussions and explore the conditions under which the bricks remain stable, particularly focusing on the center of mass.

Discussion Status

The discussion includes various interpretations of the conditions necessary for stability. Some participants have offered insights into the relationship between the center of mass of the stacked bricks and the edge of the surface, while others express confusion about the implications of those conditions. There is no explicit consensus reached yet.

Contextual Notes

Participants are navigating assumptions about the physical arrangement of the bricks and the implications of their center of mass in relation to the edge of the surface. The original poster and others are working within the constraints of the problem as posed, without additional information or definitive solutions provided.

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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