all of u are wrong
hey That is an easy problem
I am solving the problem in terms of the variables involved in the problem then the solution is as follows
Assuming the length of the brick is L:
since the bricks are homogenous the point of application of the weight of each brick will lie half the way along its length. Consequently, the top block will still be on equilibrium with respect to the second when the center of gravity is arranged above the edges of the second brick: the maximum length of the free edge of the first brick will be L/2
the center of gravity of the first and second bricks taken together would be at a distance L/4 from the edge of the second block. This is precisely the length by which the second brick may be displaced with respect to a third. Once again applying the same principle we get the center of gravity of the first three top blocks to be at a distance at L/6. This essentially means that the maximum overhanging part of the third block is l/6
taking the length of the block to 1 as per the question :maximum overhanging parts of the blocks are as follows :
the first top block : ½ units
the second block : ¼ units
the third block : 1/6 units
Drawing the figure really helps you to understand the problem
I hope everybody will give me a salute for every line I have written
Moderation Team
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