Maximum Overhang for Uniform Blocks

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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...

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

kindly help me
 

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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: