Buoyancy Puzzler - just for fun

[Doc Al]

Here's a little puzzle that may provide some amusement. (I hope it's not too trivial!) I have attached a (rough) diagram to accompany the following description.

Consider an empty tank with a circular hole in its bottom. (The tank is supported by a table that also has a hole it.) Also consider a steel cylinder that just perfectly fits within the hole, making a frictionless, watertight seal. The cylinder (which has mass m, area A, length L) is suspended by a thin, massless cord attached to the ceiling. The cylinder extends half-way through the hole in the (empty) tank. The tension in the rope is equal to mg, the weight of the cylinder.

Question: What happens to the tension in the cord as water is slowly added to the tank?

 

[NateTG]

Unless there is some mechanical interaction between the water and the cylinder other than pressure, you would get no change in force until the the water level reached the top of the cylinder. until then, the tension in the string remains mg.

Then, as the water level rises above the top of the cylinder, and we assume that water has a density of 1 the force of pressure on the top of the cylinder wouldt have magnitude A*(H-L/2) where H is the height of the water in the tank, so the tension in the string would be mg+(A*(H-L/2)) from that point forward.

This makes a couple of reasonable approximations regarding the ideal behavior of water and air as fluids.

 

[Moose352]

Perhaps I haven't understood the question or Nate's explanation, but I would think that as water is added the tension would increase as the water level would increase since the water above the cylinder would exert a greater pressure, thus adding another downward force.

(Edit: I figured out that Nate's post is the same as mine. I just assumed that the cylinder Nate referred to was the tank of water, not the steel bar. Also, when looking at the diagram, I assumed that the dotted line was the water level. Given that the bottom part of the tank is empty, wouldn't that contradict the assumption that the tension on the cord is equal to mg?)

 

[way2go]

Originally posted by NateTG
Unless there is some mechanical interaction between the water and the cylinder other than pressure, you would get no change in force until the the water level reached the top of the cylinder. until then, the tension in the string remains mg.

Then, as the water level rises above the top of the cylinder, and we assume that water has a density of 1 the force of pressure on the top of the cylinder wouldt have magnitude A*(H-L/2) where H is the height of the water in the tank, so the tension in the string would be mg+(A*(H-L/2)) from that point forward.

This makes a couple of reasonable approximations regarding the ideal behavior of water and air as fluids.

Yes, this explanation is right, I think. Buoyance doesn't apply here, because there is no pressure of the water from below the cylinder.

 

[Doc Al]

Yes, of course NateTG's answer is correct. (I would expect no less )

The (admittedly trivial) point of the exercise is as an example where the "buoyant" force does not equal the weight of the displaced fluid, as is often blindly assumed by some. Another (equivalant) example I like to use is this. Imagine a deep lake with a telephone pole buried halfway into the lakebed. What is the buoyant force tending to pull the pole out from the lakebed? The answer, of course, is that there is no "buoyant" force. On the contrary, the water exerts a net downward force on the pole.

 

[NateTG]

Originally posted by Doc Al
Yes, of course NateTG's answer is correct. (I would expect no less )

The (admittedly trivial) point of the exercise is as an example where the "buoyant" force does not equal the weight of the displaced fluid, as is often blindly assumed by some.

Actually, there is a buoyant force due to the air displaced by the cylinder, but it's negligible. A slightly more amusing example would involve a lighter-than-air cylinder and a cord coming up from the floor. Eventually, the water would push the block out of the bottom of the tank.

 

[LittleNose]

NateTG, without a surface to act upon, there cannot be a positive or negative bouyancy force.
The only surface available is perpendicular to the action required, and thus it surely cannot act upon it ina buoyant manner.
However, if the water pressure was enough to deform the cylinder, then you would get the effect as you described.

 

[Doc Al]

Originally posted by LittleNose
NateTG, without a surface to act upon, there cannot be a positive or negative bouyancy force.
The only surface available is perpendicular to the action required, and thus it surely cannot act upon it ina buoyant manner.
However, if the water pressure was enough to deform the cylinder, then you would get the effect as you described.

I'm not sure what statement of NateTG you are responding to. As far as I can see, everything NateTG said (in both of his posts) is correct as is.

In my examples (and also in NateTG's) there is certainly a surface to act on once the top of the cylinder (or pole) is submerged. The result is a negative "buoyant" force.

 

[LittleNose]

Don't you just wish that you could time travel sometimes and take back all the clangers that you have dropped along the way?

My apologies NateTG

 

[NateTG]

Originally posted by LittleNose
Don't you just wish that you could time travel sometimes and take back all the clangers that you have dropped along the way?

My apologies NateTG

No worries. I never take anything like this personally anyway.

If you were feeling less honorable (and perhaps more savy) you could have hit the edit link on your prior post.