What happens to buoyancy when a non-uniform force is applied to the liquid?

[some bloke]

TL;DR Summary

If there is a liquid medium which has an additional force exerted on it, how does this affect the buoyancy and positioning of a floating ball?

This is a thought experiment which I have had rattling around in my head for a while, and I think that it is too specific to find any answers for by google-fu alone, so I'm hoping someone here can help me to understand this!

I have a fair understanding of buoyancy. The medium in which an object sits is pulled downwards by gravity, as is the object. If the object is pulled less than the body of medium it displaces, then it floats upwards.

My next thought is the results of a non-uniform force being applied to the liquid, but which doesn't affect the object. Imagine, if you will, a liquid which can be pulled by magnetism, and a floating ping-pong ball. (I don't know if such a liquid exists, ferrofluid doesn't seem to remain fluid near magnets!).

The fluid is in a tub. The ball is floating in the centre. If I place a magnet directly below the ball, it will float higher, as the fluid is pulled down with more force than before. What if I place the magnet in the corner? The fluid will be pulled towards it, but the ball will not. Would it create a slope? would the ball roll down it? If the ball were held in place with a force-meter attached, which direction would the force act?

Now, what if there is just enough magnetic pull to cancel out the weight of the fluid, and the tub is inverted - will the ball float up, or fall down, when fully immersed in the fluid?

In microgravity, what happens to buoyancy? Does the ball sit on the outside of the floating sphere of fluid, or is it pulled into the centre by the surface tension? What if the fluid was encased in a sphere (to prevent it from moving), and a magnetic field turned on - I think the ball would float away from the magnetic pull, is this right?

Another question: If you put said sphere containing the liquid and the ball onto the arm of a centrifuge and span it up to, say 5G's, the ball would float inwards as the liquid is forced outwards. If you then had a magnetic field which pulled the liquid inwards which canceled the force from the centrifuge, what forces are being exerted on the ball at this point? Is there a way to make the ball float freely (IE neutral buoyancy) in the liquid, whilst it is still being spun at the equivalent of 5G? What sort of forces might a sensor inside said ball register (assuming it was immune to the magnetic fields effects)? would it be akin to free-falling inside a lift - where it feels weightless - or would the sensor continue to experience 5G's of acceleration from the centrifuge (which I suspect, as it is still being spun!)Thankyou for your time with this! I'm full of questions and can seldom find anything which helps me to understand, and have little resources with which to do experiments (if they are even possible)!

 

[JT Smith]

Spend a little time learning how to derive Archimedes' principle. It's very straightforward. That should give you what you need to answer your questions.

 

[kuruman]

And while you are at it, consider that in a non-inertial accelerating frame of reference, the system's acceleration $\vec a$ appears as a fictitious force $=-m \vec a$ that gets added to all the other forces in the system. So if $\vec w=m\vec g_0$ is a vector describing the weight of a system when the system is not accelerating, then $\vec w_{eff}=m\vec g_0+(-m \vec a)$ is the effective weight of the system. So when you write Archimedes's law in an accelerating system. you should replace good old $g$ with the effective value $\vec g_{eff}=\vec g_0-m \vec a$ (note the vector addition).

 

[Michael Price]

Spend a little time learning how to derive Archimedes' principle. It's very straightforward. That should give you what you need to answer your questions.

Excellent advice. Archimedes' Principle is arguably the oldest and first physical insight in history. My favourite version reads like this:
A floating object in a liquid displaces its own weight of liquid. Imagine a bath full to the brim, and you introduce and float a rubber duck on the water. The weight of water that spills over has the same weight as the rubber duck.
Eureka!

 

[some bloke]

Thank-you all for your replies. I've had a good read of Archimedes principle and it's not told me anything I didn't already know, though at least it has confirmed that what I know is correct.

The issue I have found is that the equation, sensibly, is built on the assumption that the forces acting on both the liquid and the buoyant object are acting in the same direction - namely gravity, towards the centre of the Earth - and that both the object and the liquid are exclusively being acted upon by gravity and each other.

I'll start hashing out a basic one, to see if I get it right. I'll assume G=10ms-2 for ease of numbers.
A 1mx1mx1m hollow steel cube weighing 500kg sits in a pool of water large enough to accommodate it, and displaces 500kg of water to reach an equilibrium, where it is floating half into the water. Then we turn on an electromagnet on the bottom of the pool which exerts a force of 2500N on the cube, directly downwards. The cube will, logically, sink to sit 750mm under the water, as the force exerted is the same as if it weighed 250kg more.

Would it be correct to assert then that the downward force acting upon the volume of displaced water must equal the downward force acted upon the object in order for it to float?

As such, in a standard floating object in water in uniform gravity (apologies in advanced for not knowing how to make formulae):
m1 = mass of object
m2 = mass of displaced water
p = density of liquid
v = volume of displaced water
f1 = additional force applied to object in same direction as gravity
f2 = additional force applied to displaced liquid in same direction as gravity

m1g = m2g
m2 = pv
therefore m1g = pvg

now include the additional forces:
m1g + f1 = pvg + f2

pvg = m1g+f1-f2
v = (m1g+f1-f2)/pg

from which we can establish the depth at which an object floats. Plugging in the figures above (to test):

v = (500x10+0-0)/(1000x10) = 5000/10000 = 1/2.
now with the extra force:
v = (500x10+2500-0)/1000x10 = 7500/10000 = 3/4

which is right, first example displaces half a cubic meter, and the second displaces 3/4 of a cubic meter.

Applying a force on the liquid will instead decrease the immersion depth & increase the buoyancy. in all instances the final net force is zero as it has achieved equilibrium.

Am I right so far? I'll start thinking about my earlier questions, and see what I can make of them.

Cheers!

 

[GreatestPhysician99]

Buoyancy is not a real force . It is caused by the difference in pressure between 2 areas of the submerged item . It is the result of Pascal's principle for submerged objects.Pascal's principle says in simple that as we go down to the container of a fluid , pressure exerted by the fluid is increased.Fluids exert pressure to all directions . So if we submerge a can the top side of the can will feel less pressure (so less force) than the bottom side of the can.So the total force would push the can to the surface of the fluid.