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Grade: 10

                        

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 = ∞.

5 years ago

Answers : (1)

Jitender Pal
askIITians Faculty
365 Points
							235-651_1.PNG
The center of gravity of the three bricks is given as:
235-1778_1.PNG
It is important to note that this distance is measured relative to the other edge of the third brick at the bottom.
235-1330_1.PNG
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:
235-521_1.PNG 235-2457_1.PNG
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:
235-199_1.PNG
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.

235-1671_1.PNG
5 years ago
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