Steven W. answered 07/31/19
Physics Ph.D., college instructor (calc- and algebra-based)
A very good question! I think the best way to look at this is to compare it with the same container with the water in it but NOT the cube. The origin of the buoyant force is that some force has to hold up a given parcel of fluid against gravity, if we are in a gravitational field, such as around the Earth's surface. Starting from that idea, we can either call it the buoyant force (F_B), equal to the weight of the parcel of fluid, or call it the difference in pressure-created forces applied to the top and bottom of the parcel. These are equivalent ways of describing the same phenomenon.
For a parcel of fluid equal in size to the cube, but whose bottom touches the bottom of the container, this upward force is still needed. There is still a pressure-created force on top, but the force on the bottom is created by contact with the bottom of the container. However, that bottom force is EQUAL to what would be provided if there were fluid below this parcel instead of the bottom of the container.
Now, we have the iron cube there instead (by the way, assuming we pour in the same amount of water, there IS displacement of fluid, since the water will rise marginally higher inside the container with the iron cube in, compared to when it is out). The cube feels the same pressure on the top, but it takes more force from the bottom to hold it up than it would to hold up the corresponding amount of water, since it is heavier. This means the upward force has to counter the same pressure-created downward force as when water was in the cube space, but also the heavier weight of the iron cube compared to the water.
Since the upward force is a contact force (normal force) provided by the container, we presume that (unless the container is very weak), the structural integrity of the container is great enough to provide the extra force needed to hold up the iron cube. But the contact force of the bottom of the container on the cube will INCREASE.
I guess you could say that, when the water was originally there, the container was providing the upward force on water, and that level of force would be the "buoyant force" on the iron cube as it sits there. Then the additional force to hold up the cube is the "normal force." That is a bit pedantic, but since the cube still feels a "buoyant force" equal to the weight of the water which would have been there, it fits with Archimedes' principle. Or, if we are sticklers for the idea that the buoyant force can only be provided by liquid, we could -- with equal sticklishness -- say that the object is not "immersed" in the fluid if it is in contact with a solid on any side, and so Archimedes' principle does not directly apply. Either way, Archimedes' principle holds.
However, it may be an unsatisfying hold, since we are used to the buoyant force *reducing* the apparent weight of objects. If we were holding up a heavy rock underwater, we would expect to have to support it with less force than we would have to support it in air.
So let's suppose again that the iron cube is on the bottom of the empty container. We can say that the support force of the container holds up the cube, and call that support force the normal force of the container on the cube. However, if we instead call part of that support force a "buoyant force," which would support the weight of an equal volume of water, and only call the part of the support force ABOVE the "buoyant force" the "normal force," then the "normal force" would still be less than the entire support force needed without the water (which we could call normal force in the absence of any fluid).
The main difference with this case and the hand under the rock is that we usually assume the hand under the rock covers relatively little area, and fluid covers the rest underneath. In this case, though, the container does it all, and artificially calling part of its support force a "buoyant force" seems odd and arbitrary, perhaps. But that is how, I think, to make the most direct comparison.