Identical blocks each of mass `M` and length `L` are placed one above the other such that each extends out by a maximum length as shown in figure. Find the maximum extension of the `n^(th)` block from the top, so that the blocks will not fall.

Explanation

For a two block system, the centre of mass `(C_(1))` of upper block should be at the edge of lower block i.e., at `l/2` distance. But if centre of mass of upper block is not resting on the lower block then, the upper block falls down because of unbalanced torque created by gravitational force.

If a third block `(EF)` is arranged below the two blocks then

The centre of mass `(C_(2))` of `(AB)` and `(CD)` block system lie on the edge `E` of third block. to find `x_(2)` consider `C` as origin. then
`x_(2)=(M(0)+M(l/2))/(2M)=l/4`
`x_(2)=l/4` So, centre of mass of upper two blocks is at `l/4` distance from edge of lower block. also, if another block `(GH)` is placed below the three blocks in equilibrium, then

The centre of mass `(C_(3))` of the upper three blocks must lie on the edge of the lower fourth block i.e., at `G`. to find `x_(3)` consider `E` as origin.
`x_(3)=(2M(0)+M(l/2))/(3M)=l/6 :. x_(3)=l/6`
Similarly `x_(4)=l/8, x_(5)=l/10,.......`
for `n^(th)` block `x_(n)=l/(2n)`