Rene Dekker said:
I wasn't, I was speaking of objects that are lighter than water, that are held to the floor through a water-tight seal. Although it does not matter for the principle. Your rubber ball example shows exactly what I was pointing out. There is a buoyancy force on the ball, that is counteracted by the force in the string tying it down to the floor. The same is true for an object that is held to the floor with a water tight seal. There is the same buoyancy force on the object, that is counteracted by a force from the floor. The same is true for an object that is welded to the floor. There is the same buoyancy force on the object, that is counteracted by the force through the weld.
In all cases, the buoyancy force is still there, it does not matter whether there is water under the object. It does not suddenly disappear when you replace the rope with a weld, or with a vacuum.
No, Rene! You are making the mistake that Haruspex pointed out in a recent reply. What do you know about hydrostatics? Look at the following drawing.
There is a pool , with a vertical wall on the left, and an inclined wall on the right. The reason for the inclined right wall is only the make it clear that the (relative) pressure , on horizontal planes, is the same. So, in points A (left) and B ( right) we have the same pressure because the depth under the free surface is the same : h_A = H_B . The diagram on the left is known as “pressure diagram” , and shows the linear increase of pressure with depth.
An important issue to understand is that pressure is NOT a vector quantity , but a scalar. But this would be too long and difficult to explain for me, because I have some problems with English, which isn’t my mother tongue. So let me go on.
Consider the ball , kept tied to the bottom by a rope. The buoyant force acting on the ball is due to vertical components of all elementary forces acting on elementary surfaces of the ball, which are to be "summed up” (to be mathematically correct: integrated) all over the surface. You will easily understand that the pressure at point U is less than the one at point V , which is vertically under U at a distance of 2R (R being the radius of the ball) . So you have a difference of pressure , which increases with depth, as already said. The length of the rope has no importance, it can be as short as a single chain link, connected by a hook to a fixed half-ring bolted to the bottom. There is still buoyancy here, because the geometry hasn’t changed, and the water is still “pushing up” the ball.
Now, consider a cubic box made of steel , 5 mm thickness of each wall and 1 square meter of each face. When put free into water, it floats, because it is lighter than an equal volume of water; its mean density, mass/volume, is less than that of water . Push it until it reaches the bottom of the pool, and weld it all around the base perimeter ( suppose the pool too is made of steel) . After having welded it this way , no pressure acts beneath , because no water is there. The pressure acts on the four vertical walls and the horizontal upper surface.
No buoyant force acts on the cube.
Of course, the water level of the pool has increased , but this means only an increase of max pressure at the bottom, as well as at any surface of objects in fixed position in the pool.
