# Would the upward force be equal to weight of immersed body

1. Sep 4, 2015

### SMPS-PHYSICS

I was just reading about Archimedes principle. Suppose a cylindrical body (attached figure) is immersed in liquid of density d. if the weight of body outside liquid is Fout then can we say that this body will immerse in liquid when released to a depth where upward force F2 = Fout. And then it will stay there forever.

2. Sep 4, 2015

### Titan97

There are three possible cases. If density if the object < density of liquid, the object remains at rest with some part outside the liquid.
If density if the object = density of liquid, the object is fully immersed and it remains at rest. (If its released inside the liquid from a depth d, it remains at rest)
If density if the object > density of liquid, the object sinks and hits the floor

3. Sep 4, 2015

### SMPS-PHYSICS

@Titan97 Being specific can you describe your answer in terms of Fout and F2?

4. Sep 4, 2015

### Staff: Mentor

Assuming the object is fully submerged, the upward buoyant force on the object will equal F2 - F1. That has nothing directly to do with the weight of the object (what you are calling Fout).

5. Sep 4, 2015

### SMPS-PHYSICS

Then which forces will determine that how deep an object will stay inside liquid?

6. Sep 4, 2015

### davenn

Titan97 answered that in post #2

Dave

7. Sep 4, 2015

### SMPS-PHYSICS

I think you all are confused for replying to equality F2 = Fout

8. Sep 4, 2015

### davenn

you have already been told that what you are calling Fout is irrelevant

we can only go on what you have told us ... if you don't understand post #2
maybe you need to reword your query to better explain your problem so that a clearer solution can be explained

9. Sep 5, 2015

### SMPS-PHYSICS

I think I got it. The buoyant force B is that of liquid in upward direction. If we consider downward weight of fluid of same cylindrical shape Wfl then B = Wfl and this is in equilibrium and that piece of liquid will not move from its place in static rest of fluid. However if we replace the same cylindrical volume of fluid with some solid cylinder then now B may not equal to weight of cylinder and so it can move up or down. However if they are same then it will remain at that position. Now B is always same at all depths as B = F2-F1 would be same. So if we release object (for which B = Fout) at some deeper depth then it will remain there. If we release that object at some smaller depths again it will remain static there if fluid is not moving.

10. Sep 5, 2015

### Merlin3189

I'm not sure what is going on in this thread. When I read OP, I just thought that F1 was missing, ie. F2-F1 = Fout when the cylinder is in equilibrium. This all seemed to be sorted by #4 and I can't see why Dave & Doc say that the weight of the object is irrelevant?

When Starter in #5 asks, what determines the depth at which equilibrium is reached, then it gets interesting.
Generally we assume the liquid is incompressible, so that F2-F1 is constant. (As #2 & #9 seem to say) But if it were compressible, then I think the OP might be correct (except for the F1 .) Density would increase with depth, as would F2-F1, so there could be a specific depth at which equilibrium were reached.

Of course that then assumes that the object is incompressible, or at least less compressible than the liquid. I've no idea whether that is possible?

There is the Cartesian diver (and for that matter, human divers!), where the object is more compressible than the liquid. Then I think there is no stable equilibrium depth: if it rises, it becomes more buoyant and if it sinks it becomes less buoyant.

11. Sep 5, 2015

### Staff: Mentor

The weight of the object is irrelevant in calculating the buoyant force. The weight is certainly relevant when determining if the object floats, sinks, or remains in neutral buoyancy.

Only if the fluid (or object) were compressible would the buoyant force depend on depth.

12. Sep 5, 2015

### sophiecentaur

Doesn't the Archimede's Principle we were taught in school say it all? The upthrust is (always) equal to the weight of the water displaced (relates to the volume displaced). This applies to every case you can think of - floating, sitting on the bottom, half way up or dipped into the surface. Apply that principle rigorously and the answer consistently comes out for you. That applies to pretty much all the classical Physics questions that people bring up, aamof; it's usually just a matter of complexity.
Note: This thread has to ignore peripheral effects like surface tension and finite air density etc. but it needs to be sorted out at this level before the smart alec exceptions get introduced.

13. Sep 13, 2015

### Buckleymanor

Is it necessary to relate to volumes or density.
Could you not just say if the weight of the water displaced by the object is greater than the objects weight then the object will float.
If the displaced waters weight is less than the object's weight then the object will sink.
If they are both the same then the object will remain stationary submerged within the liquid when placed.

14. Sep 13, 2015

### sophiecentaur

The up thrust is not directly affected by the density or weight of the object. Just the weight of fluid displaced. Sinking or floating comes later - depending on the net force.
You can state Archie's principle in several ways but I reckon the basic form is as good as any other. As with all these Classical Physics problems, if you follow the basic rules you can predict the outcome.

15. Sep 13, 2015

### rcgldr

If the object is less compressible than water, and it's density is close to the density of water at some depth, then the object would tend to rise or sink to a point where the water's density equals the objects density due to compression.

16. Sep 13, 2015

### sophiecentaur

Compressibility is another issue. The result of both object and fluid being compressible will depend on the particular conditions. AP will still apply, decide on the densities after deciding what the modifications give you and you have your answer. It could entail modifying AP into an Integral equation.

17. Sep 14, 2015

### ogg

Of course the same principles apply to objects in our ocean of air (we call the atmosphere). What is needed for Archimedes' to work is a low viscosity fluid and an object, both with "well behaved" (defined) density. In some wet analytical chemistry work I've done, I had to factor in both the volume of the container and the barometric pressure I was weighing it (and its contents) in; in other words I had to factor in the buoyancy of the glassware in air. In other work, surface tension, temperature, density gradients, compression, all factor in. In working with large systems, gravitational gradients, tidal forces need be considered.
Anyway, Fout is a result of the air-object system, and as said isn't directly related to the water-object system. (I assume Fout was measured in air, rather than in vacuum). Also, nothing is "forever" - in our Universe: "indefinitely" is more accurate. But most of these kinds of problems are based on simplifications of real world into some idealized (thermodynamic) system. The reason why a denser object sinks isn't because its density is less, its because its density is less, its in a (low viscosity) fluid, and its in a gravitational field. Hence mass isn't totally irrelevant. Archimedes allows us to ignore mass because our systems are very small relative to the planet (gravitational field) we're working on (in).