Density versus Pressure -- Balancing the bouyancy of a floating capsule

[mrapple]

[Mentors’ note: no template as this thread had initially been misplaced in the technical forums and was moved here]

Summary:: Enclosed Cubic foot capsule passing between two bodies of different densities but questioned pressure.

You have a tub of fresh water 32 feet high sharing a wall with a pressurized container of air. The container of pressurized air is the same size as tub of water; the difference being, it's sealed. There is a hole a foot squared 30-31 feet below the surface of the water. There is an enclosed cubic foot shaped capsule of unpressurized air. The capsule is filling the foot squared hole. Half of the capsule is in the water and half is in the container of pressurized air. Disregarding friction, what should the container of air be pressurized to in order for the capsule to not move? How much should the air container be pressurized in order for the capsule to move?

 

[Ibix]

What do you think and why?

Could you explain the relevance of the capsule being full of unpressurised air. Is the capsule simply being a stopper between your pressurised air container and your water container? If so, is it rigid?

 

[jrmichler]

Is this homework? Or a take home quiz problem? If so, we can move it to a homework forum.

 

[mrapple]

What do you think and why?

Could you explain the relevance of the capsule being full of unpressurised air. Is the capsule simply being a stopper between your pressurised air container and your water container? If so, is it rigid?

Yes, everything is rigid. There is no relevance of the capsule being filled with unpressurized air. Yes it is a stopper.

 

[Ibix]

Ok. So what's your thinking?

 

[mrapple]

Given equal pressure even though the densities are different the capsule won't move.

 

[mrapple]

Is this homework? Or a take home quiz problem? If so, we can move it to a homework forum.

Yes but I'd like to make the question simpler. If an Enclosed Cubic foot capsule passing between two bodies of different densities but equal pressure there will be no resistance right?

 

[Ibix]

Given equal pressure even though the densities are different the capsule won't move.

Yes. This is just Newton's second law - if there are equal pressures on equal areas then the forces are equal and opposite and the net force is zero. So the acceleration is zero.

Note that the water pressure will vary by about 0.03 bar over the height of your stopper, while the pressurised air pressure won't vary much. So you'll be setting the air pressure to match the average water pressure.

 

[mrapple]

Yes, everything is rigid. There is no relevance of the capsule being filled with unpressurized air. Yes it is a stopper.

Yes. This is just Newton's second law - if there are equal pressures on equal areas then the forces are equal and opposite and the net force is zero. So the acceleration is zero.

Note that the water pressure will vary by about 0.03 bar over the height of your stopper, while the pressurised air pressure won't vary much. So you'll be setting the air pressure to match the average water pressure.

Thank you!

 

[Lnewqban]

Yes but I'd like to make the question simpler. If an Enclosed Cubic foot capsule passing between two bodies of different densities but equal pressure there will be no resistance right?

Any resistance should come from the buoyancy force acting on the liquid side, increasing friction and inducing an asymmetric moment respect to the square hole (making more difficulty the free sliding of the cube).
Sliding the cube into the container of pressurized air should be easier than in the opposite direction.

 

[mrapple]

Thanks. To follow up, I have a container of water 35 feet deep. If I push a cubic foot container of air into the side of the tank of water 30 feet below the surface I use hydrostatic pressure not water pressure to figure out resistance correct?

 

[jbriggs444]

hydrostatic pressure not water pressure

What distinction do you make between those two terms?

 

[mrapple]

I would say hydrostatic pressure is only important for a closed system such as a pipe where a denser fluid rises on the other side when pushing a less dense fluid. I would say regular water pressure would be the force required to plug a hole because it is from the side and not dependent upon gravity? But, what if I am inserting the object at an angle?

 

[jbriggs444]

I would say hydrostatic pressure is only important for a closed system such as a pipe where a denser fluid rises on the other side when pushing a less dense fluid. I would say regular water pressure would be the force required to plug a hole because it is from the side and not dependent upon gravity? But, what if I am inserting the object at an angle?

I would say that the terms are synonymous for the situation at hand.

"hydrostatic" just means that we are considering a situation in which velocity is not contributing substantially to pressure. So we do not have to worry about Bernoulli, for instance.

"water" just means that the working fluid is water rather than air or hydraulic fluid.

 

[mrapple]

When using a lever the movement is in the shape of an arch. Do I use cos (for example) to calculate the force for both the output and the input? If not, why? Thanks

 

[jbriggs444]

When using a lever the movement is in the shape of an arch. Do I use cos (for example) to calculate the force for both the output and the input? If not, why? Thanks

So you are mating a lever moving in a circular arc with a cube-shaped lump which is moving linearly. If you have a rack and pinion arrangement, a constant torque would work nicely. Can you supply a drawing?

 

[mrapple]

Any resistance should come from the buoyancy force acting on the liquid side, increasing friction and inducing an asymmetric moment respect to the square hole (making more difficulty the free sliding of the cube).
Sliding the cube into the container of pressurized air should be easier than in the opposite direction.

The buoyant object is being pushed into a hole at the bottom a tub of water 40 feet below the surface of the water. This object is being pushed in at a 45 degree angle downward. What equation do I use to find out the force needed to push the object into the tub of water neglecting friction?

 

[mrapple]

Yes. This is just Newton's second law - if there are equal pressures on equal areas then the forces are equal and opposite and the net force is zero. So the acceleration is zero.

Note that the water pressure will vary by about 0.03 bar over the height of your stopper, while the pressurised air pressure won't vary much. So you'll be setting the air pressure to match the average water pressure.

Also, a float with 5 pounds of water displacement is resting on the bottom of a pool 10 feet deep. The float is tethered to the bottom of the pool via a string that is pulled tight lengthwise and so is also laying on the bottom of the pool. The string is 4 feet long. If the float is released the float must travel in an arch because of the string tether. What equation do I use to figure out the resulting upward force of the float if it is traveling in an arch?

 

[berkeman]

Thread closed temporarily for Moderation...

 

[berkeman]

Also, a float with 5 pounds of water displacement is resting on the bottom of a pool 10 feet deep. The float is tethered to the bottom of the pool via a string that is pulled tight lengthwise and so is also laying on the bottom of the pool. The string is 4 feet long. If the float is released the float must travel in an arch because of the string tether. What equation do I use to figure out the resulting upward force of the float if it is traveling in an arch?

Please start a new thread with your new question, and be sure to show your work on the solution. That is how we handle schoolwork-type questions at PF. Thank you.

Thread will remain closed now.