Let's say
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the cylinder has a diameter equal to its height of $25\;\mathrm{cm}$
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the cylinder is sealed at the top and filled with sand
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the cylinder is also buried in the sand at sea at a depth of $10\;\mathrm{m}$ with only its top exposed
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the top has a handle for you to pull on
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the cylinder's weight is negligible ($0.0\;\mathrm{kg}$)
How much force is needed to pull it out (excluding friction)?
I just need the formulas to figure out how big of a cylinder I need to anchor a buoy. Precision is up to you.
Please correct my computations if they are wrong.
Assuming the density of wet sand is $1920\;\mathrm{kg\;m^{-3}}$:
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Volume of cylinder $= \pi r^2 h = 3.1416 \times 0.015625\;\mathrm{m^2} \times 0.25\;\mathrm{m} \approx 0.012272\;\mathrm{m^3}$
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Mass of sand in cylinder $= 0.012272\;\mathrm{m^3} \times 1920\;\mathrm{kg\;m^{-3}} = 23.56\;\mathrm{kg}$
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Weight of sand in cylinder $= 23.56\;\mathrm{kg} \times 9.8\;\mathrm{m\;s^{-2}} = 231\;\mathrm{N}$
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Water pressure at $10\;\mathrm{m}$ depth $= 2\;\mathrm{atm} = 200000\;\mathrm{N\;m^{-2}}$
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Surface area of top of cylinder $= \pi r^2 = 3.1416 \times (0.125\;\mathrm{m})^2 = 0.049\;\mathrm{m^2}$
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Pressure force on top of cylinder $= 0.049\;\mathrm{m^2} \times 200000\;\mathrm{N\;m^{-2}} = 9800\;\mathrm{N}$
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Total force $= 231\;\mathrm{N} + 9800\;\mathrm{N} = 10031\;\mathrm{N}$
Edit: Converted to newtons ($\mathrm{N}$). Thanks @KvdLingen!
Note: This is not homework. I'm a freediver. I am literally thinking of burying a plastic cylinder into a patch of sand at a $10\;\mathrm{m}$ depth. At the top of the cylinder, a one way check valve will be attached to let water get out when burying and a release valve so I can pull it out. I will attach to the "handle" a rope and a buoy to the other end of the rope. The cylinder will serve as a temporary anchor. A real anchor can be easily pulled up by vertical forces when there are waves acting on the buoy. An anchor can also damage corals in dive sites–thus having a need to secure it in a patch of sand.
I've seen this technique used in construction of underwater tunnels and oil rigs to fix the structure in place. It is called a vacuum-anchor (e.g. Troll A platform). I want to apply it in a smaller scale.
Best Answer
First, water at seabed pressure will also be between the sand grains in the bucket, so the water pressure does not add up to the force that will oppose your pull. This is already at equilibrium. So away go your 9800 N.
I'm not sure either about the 231 N, as the sand will probably mostly stay with the seabed as you pull out the bucket.
So your hope is in actual suction, that is, how much force is necessary to let water in when you pull the bucket out. First, as noted by User58220, this will not resist a continuous pull, as a flow of water will start as soon as there is any pull up. So unless the cylinder is denser than water and expels it slowly after a pull, it will eventually come out after some number of pulls.
If you assume that the boundary condition is really that the sand around the cylinder wall is as packed as the rest of the seabed, then you need to estimate the permeability $k$ of the sand there and from that apply D'Arcy's law to get the flow rate as a function of force, wikipedia, with the pressure drop equal to the pulling force divided by cylinder area. You're interested in $T=Ad/Q$, the time to pull out by a distance $d$ of the cylinder, as a function of the force, $T\simeq 2\mu Ahd /(kF)$. You can work out the time for it to settle back under its weight (this is true only if $d$ is small enough that the sand did not move).
What I fear is that you'll have a detachment between the packed sand and the walls of your cylinder, and flow will be much easier there. Cylinder should be very rigid and have rough walls at the scale of sand grains to try to prevent this (gluing sand on it is used in related experiments)