Larry -

Glad you made it through Isaac. I've seen a lot of gators, but none ever swimming around my yard!!

Good luck with the cleanup and getting the power back on.

Tom

To ALL,

This is the official closing of the TBZED challenge entry window.

If you have already messaged me, good building!

I answer question each morning Central time USA.
Good luck, and get wet!
Wayne Travis

Hi all,


I have one question see picture below.


Simplifications:
1. air is not compressible, always keeps volume
2. red riser have infinitely small weight

Step 1 is initial situation. Riser have no lift, there are 4 units of water in left and 4 in right side. Pressure inside of riser is the same as outside = 1atm.
Step 2. Adding 2 units of water. Since air is not compresible this action will push water in column 2 and 11 down by one unit and also raise columns 1 and 12 by one.

Question: what is the lifting force F after step 2? Equivalent weight of 4 units of water or 20units?

Marcel

Should that have read "4 units or 2 units" instead of 20 units?

Second question: Through what distance will that lifting force act? That is, how high will the riser rise before the water levels in and out are again equal?

(I like the simplification of incompressible air. It makes it a bit easier to see what's happening.)

But wait a minute.... you must have the riser locked down at first, otherwise it will simply rise up from the air pressure and the water columns will remain the same height. The only way the pressure from the input water can be transferred to the water columns is if the riser can't move, and if it's weightless then it must be restrained.
So in your experiment then, you are unlocking the riser after injecting the water... under pressure..... and then asking with what force it will try to float up, right?
And I'm asking how high it will float before the water levels are equal again.

So the water columns will be equal again once the weightless and free riser has risen, right? And since volume is conserved, the two units of volume added to the central column's eight units will raise the riser's 10 unit width by 1/5 unit, I think. So you will have essentially the lift buoyancy of 4 units of displaced water initially, but the riser only needs to rise 1/5 unit to equalise everything again.

Ok, tear me up, how am I going wrong?

(I think I see where the "20" came from now... but I don't think that's a right answer.)