0
$\begingroup$

Lets keep it short:
I am confused regarding the domain over which buoyancy force due to a liquid acts.
If I have a hollow sphere and I dip it in water, I observe that it floats partially with certain part dipped in the water.
Now I cut a negligibly small hole on the outer part so that the system is NOT affected. Through this hole I release small masses (having finite volume) or maybe pebbles and seal it off quickly. Of course, due to a larger mass, the sphere will sink (maybe fully till the bottom or maybe not).
The Doubt : Does buoyancy force act on the small weights ?

Essentially, does buoyancy force act on an object inside an object ?

Well, if it doesn't, then the object HAS to sink till the bottom due to weight being more than upthrust at any instant.**

**I may be wrong here because I assumed that the body without additional masses floats partially

$\endgroup$
1
  • $\begingroup$ Hint: When you add a few pebbles, and then re-seal your vessel, you effectively have increased the average density of the vessel. $\endgroup$ Sep 8, 2021 at 19:30

2 Answers 2

2
$\begingroup$

If I have a hollow sphere and I dip it in water, I observe that it floats partially with certain part dipped in the water.

If it floats partially submerged the density of the sphere is less than the density of water.

Now I cut a negligibly small hole on the outer part so that the system is NOT affected.

As long as you cut out a hole, the amount of the sphere submerged will decrease since the mass is less. Naturally, the smaller the hole the less the effect.

Through this hole I release small masses (having finite volume) or maybe pebbles and seal it off quickly. Of course, due to a larger mass, the sphere will sink (maybe fully till the bottom or maybe not).

The buoyancy force depends on the amount of water displaced by the sphere. The amount of water displaced by the sphere depends on its density compared to that of the water. It it is less dense, it will float partially submerged, as in your first instance.

As long as adding the pebbles does not result in the density of the sphere, which is its total mass (empty sphere plus pebbles) divided by the total sphere volume (assumed fixed) being greater than water, it will still float.

It will float partially submerged if its density is less that of water, and float totally submerged if its density exactly equals the density of water. It will sink if its density is greater than water.

The Doubt : Does buoyancy force act on the small weights ?

Essentially, does buoyancy force act on an object inside an object ?

The buoyancy force acts on the exterior of the sphere and is based on the overall density of the sphere (total mass divided by total volume). It does not individually act on the small weights. The only effect the small weights may have is on which part of the surface of the sphere is submerged (the part of the surface where the weights accumulate.

Well, if it doesn't, then the object HAS to sink till the bottom due to weight being more than upthrust at any instant.**

**I may be wrong here because I assumed that the body without additional masses floats partially

I'm having trouble following you here. But like I said, the sphere will not sink as long as its density is not greater than water.

Hope this helps.

$\endgroup$
2
  • $\begingroup$ yep got it ! so we need to just check density and do stuff where we take total mass/total volume, right? And we don't really bother with whether the two materials are different while calculating the densities ? $\endgroup$ Sep 9, 2021 at 15:43
  • $\begingroup$ @omjoglekar That is correct. The distribution of the mass can effect what part of the object is submerged while floating because object is stable when the center of mass is lowest. But the volume of portion of the object that is submerged is strictly determine by the relative density of the object vs the liquid. $\endgroup$
    – Bob D
    Sep 9, 2021 at 15:50
2
$\begingroup$

Bouyant force exerted by a fluid on a body is equal to the weight of the fluid displaced by the body.

In the case you described, the hollow sphere would initially displace some liquid and would attain equilibrium when the weight of the hollow sphere would be balanced by the buoyant force. Let us assume that the sphere is still floating.

Now when you drill a hole in the sphere and add some mass, the weight of the sphere would increase. This would cause a force imbalance and the sphere would be submerged to a greater extent. If the weight of the sphere plus the added masses exceeds the weight of the water displaced, the sphere would sink completely.

So, in a nutshell, the buoyant force only depends upon the weight of the fluid displayed.

$\endgroup$
2
  • $\begingroup$ so will it act on a body in a body ? $\endgroup$ Sep 9, 2021 at 15:43
  • $\begingroup$ The buoyant force will only depend on the weight of the fluid displaced. Adding more mass to the inside of the sphere causes a force imbalance as stated in the answer which results in the body being submerged to a greater extent until the new increased magnitude of buoyant force balances the increased weight. I suggest doing some problems on buoyancy from standard textbooks. You will get the idea. $\endgroup$
    – Mechanic
    Sep 9, 2021 at 16:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.