Question

Four identical bricks are placed on top of each other at the edge of a table. Is it possible to
slide them horizontally across each other in such a way that the projection of the topmost one
is completely outside the table? No marks for guessing. Hint: Start the process from the top.
The general strategy is to move each sub-pile of bricks until its combined center of gravity is
just above the edge of the brick below it.​

Answer 1

Answer:

ANSWER

Let the weight of each brick be W and length a . As bricks are homogeneous, the centre of gravity of each brick must be at the midpoint. Therefore, the topmost brick will be in equilibrium if its centre of gravity lies at the edge of brick below it, i.e., II brick. Thus the topmost brick can have maximum equilibrium extension of a/2 .

C

1

is the centre of mass of the top two bricks which lies on the edge of the third brick.

C

2

is the centre of mass of the top three bricks which lies on the edge of the fourth brick.

Thus, the maximum overhanging length to top from the edge of bottom brick is

2

l

+

4

l

+

6

l

=

12

11

a