I understand that an object, say a closed plastic cube full of air, stuck to the bottom of an aquarium with a smooth glass bottom will not float if you firmly place it such that no water can get underneath. What if you change the shape such that you add surfaces that water CAN get under to provide lift? Take the cube and change it into a T shape for instance. Now the horizontal part at the top DOES have area under which water can provide lift. Will this break the suction holding the bottom of the T to the bottom of the aquarium and cause it to float?
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$\begingroup$ It comes down to can the buoyancy force overcome the adhesion force. What's the magnitude of each? What's causing the item to be stuck & what is the area of contact? Deep embeddment into sticky mud is going to take a lot of buoyancy to overcome. $\endgroup$– FredNov 7, 2020 at 14:25
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$\begingroup$ Thank you for responding, I appreciate it. So are you saying adding a horizontal member under which water could provide lift would at least to some extent add to the net upwards force? Assuming that it was large enough in relation to the part stuck in the mud or flush against the bottom such that no water could get under the touching part, it could then overcome that bond with the bottom and lift the entire piece? It seems that it could. $\endgroup$– doctorremulac3Nov 7, 2020 at 14:59
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$\begingroup$ OK, so while a: | shape would be held fast to the bottom, the: T shape would float because of the area at the bottom of the horizontal member, the cross on the top of the T right? $\endgroup$– doctorremulac3Nov 7, 2020 at 20:18
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3 Answers
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The object will flat if sum of the upward forces is more than the sum of downward forces. When the sum of the upward forces equal to the sum of the download forces, the object is in the "limiting equilibrium" condition, instability and floating will follow.
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In the majority of the scenarios (material type, thickness of walls, width of T flannges etc) the difference on pressure in the flanges will be greater that the adhesive force keeping it to the bottom.
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let's annotate the following
- m = weight of the object
- H = height of the object's flange
- D is the depth of the water to the top of the object
- rho.g.D is the pressure of water on the top
- rho.g.(D+H)= the pressure on the bottom of the flange
- L= width of the object
- we assume the thickness of the object as 1 for now.
The trapezoid areas of the pressure on the sides cancel out.
when $ \rho.g.(D+H)2B \geq m+\rho .g.D.L$ the buoyancy will float the object.
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