Buoyancy force on object between three liquids I have a thought experiment.
Suppose you have 3 layers of liquid resting on each other in a beaker. A long cylinder of some material is placed in the beaker so that it passes through all layers of liquid. My question is if the middle layer of liquid exerts a buoyancy force on the cylinder.
My guess is that it wouldn't because the cylinder doesn't have an upper and lower surface exposed to the middle liquid layer since it passes through the middle layer. So it can't experience a force caused by the pressure difference. 
On the contrary, according to Archimedes principle, there should be a force on the cylinder exerted by the middle layer since the cylinder displaces liquid of the middle layer.
So what would be the correct answer?
 A: It does have an effect, but it is not direct, unlike the pressure acting on the horizontal faces.  
It contributes to the total force that the cylinder will experience at it's bottom face, because the weight of that middle fluid contributes to the pressure in the bottom fluid.
Because of this, when you analyze the whole system, the amount of the middle fluid displaced will still have a contribution to the overall force balance of the system.  When thinking of Archimedes' principle, it's easy to see that, as you mention.  
When analyzing the forces on the top and bottom faces, it is less apparent.  If you consider the case of just two liquids, you can determine the pressure on each face and find the force balance.  When you replace some of the liquid with one of a density in between, the pressure due to the fluid column changes, because the $\rho$ in $F_B = \rho g h$ changes.  This is how the third liquid changes the force balance of the system, and why Archimedes' Principle still agrees with this.
A: The fact that the buoyant force is equal to the volume displaced is a mathematical coincidence, but the buoyant force comes from a difference in pressure at the top and bottom of an object. Let's consider one fluid and a cylinder of height $h$ and radius $r$ placed upright. The sides can't be pushed up or down, only the faces can. The pressure sry the top is equal to 
$$P_{atm} + \rho g h_1(\pi r^2)$$
and at the bottom,
$$P_{atm} + \rho g h_2(\pi r^2)$$
Subtracting the two gives simply
$$\rho g(h_2 - h_1)A = \rho g V$$
Now let's consider the cylinder on its side. The pressure on the top is
$$P_{atm} + \rho gh\int\limits_0^\pi (h_0-r\sin\theta)\sin\theta\ rd\theta$$
And for the bottom just change the limits to $\pi$ to $2\pi$.  Subtract and you'll get the same result, $\rho g V$. I don't know how to prove the general result, but the important thing is, just calculate the pressure on top and on bottom. The middle fluid still has an effect because it will add pressure to the bottom liquid. 
