Suppose we have a ball of volume V and a block of the same volume V whith same density. We submerge the ball into some kind of liquid so that $\frac{1}{2}V$ is submerged into the liquid. We do the same thing with the block. Now $\frac{1}{2}V$ of both block and ball is submerged into the liquid. At this moment we can say that the buoyant force on both objects is the same. If we now push both the ball and the block by little bit into the liquid by the same volume, the force acting on the bottom of the ball would increase (because it is now a little deeper in the liquid) and so will the force acting on the bottom of the block. If we now look at the geometry of the objects, wouldn't there be additional force on the ball pushing it downward because of the curvature of the ball which wouldn't appear on the block? Can we still say the buoyant force is the same on both objects? I added a picture and highlighted the area where I think the force would act in red.Red

  • $\begingroup$ All that really needs to be proven is that, based on the pressure distribution on the submerged surface of the sphere, the buoyant force is equal to the submerged volume times the density of water times g (in agreement with Archimedes principle). Would you like me to prove that? $\endgroup$ – Chet Miller Mar 26 '19 at 18:55
  • $\begingroup$ Thank you, but you don't have to prove the Archimedes principle. $\endgroup$ – ToTheSpace 2 Mar 26 '19 at 20:12

The buoyant force will remain the same, The buoyant force is the net force that the liquid exerts on the body, this includes the force which acts on the bottom and the top(the red portion).

We can prove this by using Archimedes Principle, which states that the force of buoyancy is equal to the weight of fluid displaced by an object. Here, both the sphere and cube displace the same amount of liquid and hence will have same force of buoyancy acting on it.

  • $\begingroup$ Proving it by manually finding the force of buoyancy (using integration) for the sphere will be a quite troublesome process ,which is why I kept it simple and used Archimedes Principle. $\endgroup$ – Vaishakh Sreekanth Menon Mar 24 '19 at 12:37
  • $\begingroup$ Yes, but the OP seems to know that already. The question that the OP is asking is precisely about the apparent mismatch between the result that we know from the Archimedes principle to be true and the result that the OP seems to get when they apply the first-principles arguments to the specific case that they are discussing. $\endgroup$ – Dvij Mankad Mar 24 '19 at 13:13
  • $\begingroup$ Oh, my bad. Shall I delete this answer? $\endgroup$ – Vaishakh Sreekanth Menon Mar 24 '19 at 13:21
  • $\begingroup$ Well, I am not sure. I would wait to see what the OP thinks. $\endgroup$ – Dvij Mankad Mar 25 '19 at 1:51
  • $\begingroup$ Thank you for giving me a general explanation, but as Dvij Mankad said, I already knew Archimedes principle. I wanted to see how other people can come up with an apprehensible explanation as to why the buoyancy is the same although it can't be directly concluded. $\endgroup$ – ToTheSpace 2 Mar 26 '19 at 14:44

This diagram might help you understand what is going on?

enter image description here

The initial position of the sphere is shown in grey and the new position of the sphere is shown in orange.

The pressure on all parts of the sphere below $XY$ increases by $h\rho g$ when in the new position $X'Y'$ and those parts all contribute to an increase in the net force upwards.

The section of the sphere $X''Y''Y'X'$ which was originally in the air is now in the liquid with the change in pressure on that surface ranging from $0$ along $X''Y''$ to $h\rho g$ along $X'Y'$ and the surface area which has become immersed is much less than the surface area already in the liquid.
This newly submerged part of the sphere thus contributed a net downward force which is less than the increase in the upward force contributed by the permanently submerged part of the sphere.

  • $\begingroup$ Thank you for your explanation. Although you didn't express equivalence in buoyant force between sphere and block, I see what you are saying. $\endgroup$ – ToTheSpace 2 Mar 26 '19 at 14:53

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