Moving Stonehenge
A look at the fluid mechanics questions that are raised by the Stonehenge 'bluestones'.
Problem
Stonehenge is formed from massive stones made from Dolerite, which is has a density of $2.9$ tons $m^{-3}$. Interestingly, the quarries where such stones are found are a great distance from Stonehenge itself. It is usually assumed that the massive blocks were transported by water.
If we assumed that rafts were used to transport the blocks, what is the minimum length of trees, of diameter 20cm, that would have been required to transport each stone? An indicative density of wood is $\frac{3}{5}$ that of water, and assume that a typical Stonehenge stone can be approximated as a cuboid 5m high with 60cm square side length.
How would things change if barges were used instead of rafts?
You might want to experiment with wooden rods and kilogram weights to see how much weight a small raft can actually support. In order to do this, you will need to work out an estimate for the actual density of your wooden rods. Do your results tie in with the theory? What other mathematical and modelling issues arise?
Getting Started
Archimedes' Principle says that when you displace some volume of fluid, you experience an upward force equal to the weight of the fluid displaced.
The problem boils down to finding out how much weight a metre of log can support.
Many physics A-Level syllabuses cover Archimedes' Principle, which is covered in more depth in most 1st year general engineering courses.
Student Solutions
If transporting the stones dry, the minimum volume of wood required:
$(V_{stone}\rho_{stone} + V_{wood}\rho_{wood})g = V_{wood}\rho_{water}g$
$\therefore V_{wood}(\rho_{water} - \rho_{wood}) = V_{stone}\rho_{stone}$
$\therefore V_{wood} = \frac{V_{stone}\rho_{stone}}{\rho_{water} - \rho_{wood}} = 13.05m^3$
$V_{wood}/(\pi r_{tree}^2) = length_{tree} = 415.4m$
That's nearly half a kilometer of sizeable trees!
If the stones could be transported wet, which would of course require a river about 2 feet deeper, then less wood would have been required:
$(V_{stone}\rho_{stone} + V_{wood}\rho_{wood})g = (V_{wood} + V_{stone})\rho_{water}g$
$\therefore V_{wood}(\rho_{water} - \rho_{wood}) = V_{stone}\rho_{stone}$
$\therefore V_{wood} = \frac{V_{stone}(\rho_{stone} - \rho_{water})}{\rho_{water} - \rho_{wood}} = 8.55m^3$
$V_{wood}/(\pi r_{tree}^2) = length_{tree} = 272.2m$
That's still a lot of trees, but considerably fewer, maybe 25 large trees. That is the absolute minimum value though, at which the object will have neutral buoyancy, i.e. will have the overall density of water.