Hi stpmmaths,
i have another approach. Think about the http://en.wikipedia.org/wiki/Center_of_mass" of each block! That means, that each block can be imagined as a point mass, at a certain Point.
So the x-component (see sketch below) of the center of mass of a bunch of blocks must be on the edge of the block below (acutally an infinitesimal piece forward the edge, to be in stable equilibrium, otherwise the "point mass" falls down!)
Cause my english is not so well, I've made a sketch:
http://go-krang.de/physicsforums.com%20-%20thread%20-%202339260.png
So considered from the origin of the coordinate system i have chosen, i declare the x component of the center of mass of the blocks with
- Block 1 : r^{^{(cm)}}_{x,1}
- Block 2 : r^{^{(cm)}}_{x,2}
- Block 3 : r^{^{(cm)}}_{x,3}
with R_x = \frac{1}{\sum_i m_i} \, \sum_i m_i \cdot r^{^{(cm)}}_{i,x} it yields
R_x = \underbrace{L}_{\mbox{edge of the lowest block}} = \frac{1}{3\cdot m_{block}} \cdot m_{block} \cdot \left( \underbrace{\left(z+\frac{L}{2}\right)}_{ \mbox{center of mass B3} } + \underbrace{\left(y+z+\frac{L}{2}\right)}_{ \mbox{center of mass B2} }+ \underbrace{\left(x+y+z+\frac{L}{2}\right)}_{ \mbox{center of mass B1} } \right) = \frac{L}{2} + \frac{x}{3} + \frac{2y}{3} + z
if you put the answers for x,y,z in the above equation it results L, like I've promised ;)
But my equation above has infinity results (Cause my equation treats the 3 blocks as one continuum...) , so you must think about what constraints are on x and on y!
TIP: work top down through the blocks
with best regards and i hope i could help