1. The answer is 5 cm. The answer is 2.5 cm. For every tile you add
on the overhang will have to be halved to have equal balance with
both of the other tiles.
Yuliang Djanogly City Academy identified
that the best overhang in the three tile case is 6.6cm over the
base tile.The explanation for the
sort of reasoning for this was given by Jason, by using centre of
mass considerations
Whatever amount of tiles there are, when they are placed on another
tile, half the mass must be on the tile for them to just balance.
So, for the first two, the top tile may sit 10cm on and 10cm off
the bottom one. This gives a pivot point for the mass at the
halfway point, ie. 15cm. ergo these two may be placed 15cm on and
15cm off the bottom one. Now the pivotal point of the mass of these
three tiles is about 16.6cm.
An algebraic extension was given by Lindsay
from Collyer's Sixth Form College
So we've established that the maximum overhang for two tiles is
10cm.
Now we can consider three tiles: The centre of mass of the first
two tiles must sit on the edge of the third tiles to achieve
maximum overhang.
Using the idea of centre of mass we can produce the following
equation, where $x$ is the new centre of mass
$$10m + 20m = 2m\times x$$
. Rearranging this gives
$$ \frac{10m + 20m}{2m} = x = 15cm$$
The two tiles can therefore be placed 5cm from the end of the
third, giving an additional overhang of 5cm and an overall overhang
of 10 + 5 = 15cm.
This process can be repeated to add a fourth tile: Centre of mass
of three tiles
$$
\frac{10m + 20\times 2m}{3m} = \frac{50}{3} = 16 \frac{2}{3}
$$
So they can be placed $20 - 16\frac{2}{3} = 3\frac{1}{3}$cm from
the end of the fourth tile, creating an overall overhang of $10 + 5
+ 3\frac{1}{3} = 18\frac{1}{3}cm$.
There seems to be a pattern in these results, which becomes clearer
once we tabulate them. This can be generalised for n tiles in the
following way:
$$
10m + 20\times(n-1)m = nmx, \mbox{ where } x \mbox{ is the centre
of mass}
$$
This rearranges to $\frac{10+20(n-1)}{n} = x$. Because the edge of
the bottom tile must be directly below the common centre of mass of
the stack of n tiles above, x represents the total overhang of a
stack of n tiles.