If I bury a cylinder (top end sealed) in sand, how much force do I need to pull it out?

Let's say

• the cylinder has a diameter equal to its height of $25\;\mathrm{cm}$

• the cylinder is sealed at the top and filled with sand

• the cylinder is also buried in the sand at sea at a depth of $10\;\mathrm{m}$ with only its top exposed

• the top has a handle for you to pull on

• 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 $1920\;\mathrm{kg\;m^{-3}}$:

• Volume of cylinder $= \pi r^2 h = 3.1416 \times 0.015625\;\mathrm{m^2} \times 0.25\;\mathrm{m} \approx 0.012272\;\mathrm{m^3}$

• Mass of sand in cylinder $= 0.012272\;\mathrm{m^3} \times 1920\;\mathrm{kg\;m^{-3}} = 23.56\;\mathrm{kg}$

• Weight of sand in cylinder $= 23.56\;\mathrm{kg} \times 9.8\;\mathrm{m\;s^{-2}} = 231\;\mathrm{N}$

• Water pressure at $10\;\mathrm{m}$ depth $= 2\;\mathrm{atm} = 200000\;\mathrm{N\;m^{-2}}$

• Surface area of top of cylinder $= \pi r^2 = 3.1416 \times (0.125\;\mathrm{m})^2 = 0.049\;\mathrm{m^2}$

• Pressure force on top of cylinder $= 0.049\;\mathrm{m^2} \times 200000\;\mathrm{N\;m^{-2}} = 9800\;\mathrm{N}$

• 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.

• Force of gravity is 9.8 times mass – KvdLingen Mar 17 '14 at 16:16
• Hi @KvdLingen, correct me if my assumption is wrong, I think I also need to consider the water pressure (2atm) at 10 meter depth since one end is sealed. It's like slowly raising an inverted glass out of water. If you keep the open end submerged, you are not only using force equivalent to the weight of the glass and the water trapped in the vacuum within the glass. You are also pulling onto 1atm distributed at the submerged opening of the glass. Is that pressure negligible (or maybe different a different force is acting on the submerged cylinder)? – John Tolentino Mar 18 '14 at 5:47
• I might have misunderstood but you stated that you miss the acceleration part, furthermore you use kg as unit for weight and kg/m2 for pressure, where it should read N and N/m2. So I gave you the relationship for the weight. You can apply it also for the pressure as that is caused by the weight of the water. – KvdLingen Mar 18 '14 at 6:48
• @KvdLingen, You're right. I'm not really good with Physics and I don't totally understand this problem. What I'm hoping for is to use the negative pressure and the weight of the sand being pulled by that negative pressure to resist the vertical pull. I'm just guessing that an inverted bucket buried in the ground is harder to pull out than a right-side-up bucket buried in the ground. Was hoping to get confirmation if a vacuum exists and formulas for the forces acting on the cylinder from experts here. – John Tolentino Mar 18 '14 at 7:02

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)

• Thanks for the answer Joce. I'll read up on the D'Arcy's law. Thanks for the link. I'll be making the suction anchor out of PVC pipe with an end cap and some release valve. I'm sure it will be very rigid. So how does suction anchor/caissons work if it's all gravity? Please see en.wikipedia.org/wiki/Suction_caisson – John Tolentino May 6 '14 at 3:30
• Good point about the grains on the walls, by the way. I'm thinking though that it will be harder to embed the cylinder if it does not have a smooth sides due to friction. – John Tolentino May 6 '14 at 5:13
• @John: It kind of makes sense that the easier it is to put in, the easier it'll get out, I guess... In this case, I believe rough walls will increase pulling-out resistance more than the driving-in one. If you're using PVC, maybe you can abrase the surface to get something rough, it's another thing that's done in experiments with sand flow. – Joce May 7 '14 at 13:20
• @John: I haven't looked in the literature about it, but I believe this is as I say: suction increases greatly resistance for short pulls, but then in order to recover the caisson needs its weight. If you'd pull continuously on it with a force stronger than its weight, it would eventually come out. Therefore it should be denser than seabed. – Joce May 7 '14 at 13:26
• I see your point. I was thinking that the waves can be considered short pulls. But you are right. I should put some weights to set it back into the ground or attach a continuously running suction pump. – John Tolentino May 9 '14 at 1:53

Consider the forces at work on the cylinder, ignoring friction - you have the weight of the cylinder (and containing sand) and the weight of the water on top of the cylinder pushing the cylinder down. The total force of the water on the cylinder is equal to the pressure multiplied by the area of the top of the cylinder. Therefore the force required to pull it out is any force greater than the total force pushing the cylinder down. So to get the minimum force required, total the forces pushing it down. Any force greater than this will pull the cylinder out, with the greater the force, the faster the pulling-out.

I haven't checked your figures but they seem sensible!

• How is the sand "contained"? – DJohnM Mar 17 '14 at 17:42
• @User582220, please see my comments to your answer. Thank you! – John Tolentino Mar 18 '14 at 6:33

I think the answer depends on how quickly you try to lift the cylinder.

Note that the cylinder is sealed at the top! To my mind, that implies something like an inverted drinking glass.

If you pull up with a force equal to the apparent weight of the submerged cylinder alone, and in the absence of friction, the cylinder will begin to rise, and the sand will stay where it is. There is no external upward force on the grains of sand to counteract gravity and make them rise up. Water will percolate through the sand and fill the space at the top of the cylinder created as the cylinder leaves the sand behind.

Look at it this way: Suppose the cylinder were open at both ends. There would be no doubt that the cylinder would lift and leave the sand behind. So, what's the difference between water getting in through the top of the cylinder, or just through the bottom through the sand?

• Water will not be able to pass through the sand as fast as you pull on the cylinder. A temporary vacuum will be created inside the cylinder, unless you pull it very slowly. This is the force that I'm interested in--the pressure difference between the exposed part of the cylinder and negative pressure inside the cylinder. This is similar to the effect of a suction cup. Although a suction cup is light, it is hard to pull out because of the vacuum underneath it. – John Tolentino Mar 18 '14 at 1:14
• On the other hand, you presented a good point which I wasn't able to consider: the rate of flow of water through the sand given the volume of the cylinder. I'll search for those figures or formula. Thank you! :) – John Tolentino Mar 18 '14 at 1:22
• Check Google for "sand boil" to see what happens when there's a pressure difference across some submerged sand that happens to be supporting a flood levee.. Also, "sand filter". – DJohnM Mar 18 '14 at 1:45
• Thanks @User58220. I've been reading on those subjects. I think I have to use the Hydraulic conductivity of sand... except that the value varies (where K = 1 to 0.000001). See en.wikipedia.org/wiki/Hydraulic_conductivity. – John Tolentino Mar 18 '14 at 5:36

I think the other answers are making this too complicated. Yes, the force pulling the bucket down is due to pressure differences, but ultimately the thing driving these pressure differences is the weight of the sand that's being lifted up when you try to lift the bucket.

Therefore, a rough estimation of the force the bucket can exert can be found by assuming the tube is sealed at the bottom. Then we just need to calculate the weight of the wet sand in the tube, minus the weight of water that it would displace. According to this table, wet sand has a density of around $2000\:\mathrm{kg/m^3}$. Water's density is about $1000\:\mathrm{kg/m^3}$, so that leaves us with a nice round $1000\:\mathrm{kg/m^3}$ of relevant mass in the bucket: the force exerted will be roughly the same as the force it would exert if it were filled with water and surrounded by air.

In your case the volume is $\pi*(0.125)^2*(0.25)\approx 0.012\:\mathrm{m^3}$, meaning the force exerted is $g*1000*0.012 \approx 120 \:\mathrm{N}$. I imagine this is pretty small compared to the vertical force exerted by even quite a small wave, but I can't seem to find a good figure for the latter. I guess it depends on the size of your buoy.

The actual force could be more or less than this. It could be more because you're not just lifting the sand inside the bucket, you're lifting a fair bit of the sand underneath it as well. It could be less because water finds a way to get in and replace the sand somehow (I suspect the likelihood of this will be strongly dependent on the consistency of the sand), or because a vacuum forms in the top of the bucket (but I imagine the bucket would probably break first). The only way to know for sure is to test it, but I think this calculation should give you a good idea of the magnitude to expect.

Incidentally I think the only reason to use an inverted bucket rather than a right-way-up one is that it's easier to bury - I'm pretty certain there's no means by which an open-bottomed tube could exert more force than a closed-bottomed one. On such a small scale it might be easier to use a right-way-up bucket (or a bag, or just a large sheet of plastic with the corners tied together), and fill it with sand without bothering to bury it at all.

• Thanks @Nathaniel for your answer (still digesting it). Yes, I've considered just having a bucket right side up. However, it will be lighter compared to lead weights. We've been using weights as an anchor but a buoy can easily lift a fair amount of weights. That's why I'm thinking of using physics to find a better solution, hence using suction. My logic is, it's like having your shoe stuck in the mud. Your shoe isn't really "that" heavy, it's just that air can't pass through the sides of your shoe fast enough to counteract the air pressure pushing inside your shoe. Does that make sense? – John Tolentino Mar 18 '14 at 8:30
• Well, the stuck shoe thing is ultimately for the same reason: when you try to lift it, the sand tries to come up as well (because the air/water can't get in, as you say), and so you're actually trying to lift a whole load of mud as well as your shoe. That example does suggest that the weight you're lifting can be a lot more than the sand inside the bucket, which means my answer's probably a massive underestimate. – Nathaniel Mar 18 '14 at 8:47
• Simpler answer: I cannot just rely on gravity. Lead is heavier than sand and the buoy can easily lift it. I need an alternative to gravity. :) – John Tolentino Mar 18 '14 at 8:51
• It's still gravity that's holding it down ultimately. It's just that you you want to take advantage of the sand below the device, which is heavier than any lead weight would be. – Nathaniel Mar 18 '14 at 8:53
• Just thinking... perhaps easier than an inverted bucket might be a rigid flat surface, like a sheet of plywood weighed down with weights. As long as it's flat against the sand (e.g. if you bury it under a small amount of sand), it should behave like the sole of the stuck shoe, sticking to the sand below it. (Of course it would move around horizontally, but I'm sure you could solve that.) That's just an idea - I don't have the practical experience to tell if it would work - but it sounds more plausible than the bucket to my naïve imagination, because of its greater surface area. – Nathaniel Mar 18 '14 at 8:53