Maximum Overhang for Uniform Blocks

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Homework Help Overview

The discussion revolves around the maximum overhang of uniform blocks stacked at the edge of a table, focusing on the conditions under which the blocks can remain stable without toppling. The subject area includes concepts of center of gravity and equilibrium in mechanics.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the conditions for stability when stacking two identical blocks, questioning the assumptions about the placement of the center of gravity for both blocks. There are attempts to derive the maximum overhang based on the center of gravity calculations.

Discussion Status

Some participants have provided insights into the relationship between the center of gravity of the blocks and their placement, while others are seeking clarification on how to approach the problem from the beginning. There is an ongoing exploration of the implications of the center of gravity's position for both blocks.

Contextual Notes

Participants are discussing the constraints of the problem, including the requirement that the center of gravity of the combined system must be directly over the edge of the table for stability. There is mention of specific assumptions regarding the positioning of the blocks that may need to be reconsidered.

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Overhang of uniform blocks??

Homework Statement


If you put a uniform block at the edge of a table. the center of the hlock must be over the table for the hlock not to fall off. (a) If you stack two identical blocks at the table edge, the center of the top block must be over the bottom block, and the center of gravity of the two blocks together must be over the table. In terms of the length L of each block, what is the maximum overhang(see attached file) possible??

Homework Equations


Xcm= Mx1+Mx2 / (M + M)


The Attempt at a Solution



If i keep the bottom block on the table so that its CG lies at the right edge of the table, then using the above formula if i calculate the CG for both blocks, its CG also lies at the edge of the table...
Taking origin as the left edge of the bottom block...

<br /> \[\begin{array}{l}<br /> L/2 = (M(L/2) + Mx)/(M + M) \\ <br /> x = L/2 \\ <br /> \end{array}\]<br />

kindly help me
 

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Last edited:
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Your problem is that you assume that the bottom block stays where it is, half of it overhanging. If that is the case, you cannot add a second block with any overhang without it toppling over. You need to solve the problem from the beginning with two blocks.
 


can u kindly explain me how to start with this problem?? I 'm still not getting... Like how to find the location of CG of the bottom block??
 


Say the CG of the bottom block is distance x back from the edge. At what distance from the edge must the CG of the top block be so that the CG of the combined system is exactly over the edge?

** Additional explanation on edit **
Note that the CG of the top block must be at the edge of the bottom block for maximum overhang.
 
Last edited:


kuruman said:
Say the CG of the bottom block is distance x back from the edge. At what distance from the edge must the CG of the top block be so that the CG of the combined system is exactly over the edge?

** Additional explanation on edit **
Note that the CG of the top block must be at the edge of the bottom block for maximum overhang.

Thanks ... i got it... :smile:
 

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