A well-known  problem  is the following  (see, for example, Scientific American,  November  1964, p. 128): Uniform bricks are placed one upon another in such a manner as to have the maximum  offset. This is accomplished  by having the center of gravity of the top brick directly above the edge of the brick below it, the center of gravity of the two top bricks combined directly above the edge of the third brick from the top, and so on. (a)  Justify  this criterion  for maximum  offset;  find the largest  equilibrium  offsets for four bricks.  (b) Show that, if the process is continued  downward, one can obtain as large an offset as one wants. (Martin Gardner, in the article referred to above, states: "With 52 playing  cards, the first placed so that its end is flush with a table edge, the maximum  overhang is a little more than   cardlengths ") (c) Suppose  now, in- stead, one piles up uniform bricks so that the end of one brick is offset from the one below it by a constant fraction,  l/n, of a brick  length  L. How many  bricks,  N, can one use  in this process before the pile will fall over? Check the plausibility of your answer for n = 1, n = 2, n = ∞.

Jitender Pal
9 years ago

The center of gravity of the three bricks is given as:

It is important to note that this distance is measured relative to the other edge of the third brick at the bottom.

Therefore, the total offset of the bricks is equal to the sum of the offset of the first brick, the second brick and the third brick as:

the brick beneath it (n is the number of bricks counted from the top).
The total offset of the bricks for n bricks from the top form the series:

From the series above, it can be seen that by increasing the value of n , one can increase the total offset by the corresponding amount. However, at larger values of n, the increase in offset will be insignificant.