Archimedes Problem ```Name: Tom M. Status: other Age: 40s Location: N/A Country: N/A Date: 2000-2001 ``` Question: Do I understand Archimedies principal correctly? If I build a 5'x5'x1' plywood frame (waterproof of course)it would displace 25 cubic feet of sea water @ 64 pounds per cubic foot the vessel would be able to support 2600 pounds total of vessel weight and cargo? Seems way to high? I was too embarassed to ask my daughter. Replies: Tom, Your assessment of the situation is correct. The box would displace 25 cubic feet of water at about 60+ pounds / cubic foot. Amazing, but true. Regards, ProfHoff 316 You have the basic principle down correctly. A vessel that displaces 25 cubic feet of seawater can support or float with about 2600 pounds of total weight. The reason that this seems too high to you is that you could not build a boat of this size and put 26000 pounds of cargo in it, because it would quickly sink. In a boat, the top is open, so water can spill over the side. In order to support 2600 pounds, the frame would have to be perfectly balanced, the cargo would have to be perfectly balanced, and the water would have to be perfectly calm. In reality none of these situations can be acheived. If the box had a sealed top, then you the total weight could be 2600 pounds and it would float at or near the surface, depending on the temperature and salt content of the seawater. But, boats, without a sealed top, have to account for stability under conditions where the load (people and cargo), are not perfectly balanced and are shifting around, and where there are waves and other disturbances in the water. In fact, with the simple frame you have described, one person could stand in a corner of the boat, and force the corner under the water. Then the boat would fill up with water and sink. So while your frame could (in theory) support over 2000 pounds, you would find that it really would not be safe to attempt more than a few hundred pounds in the frame, and even then you would need to be careful so it did not sink. I hope this helps. Eric Tolman Hi, Tom!! I did not understand quite well your question and may be you could bring more light on it. But, it seems to me that you will build a 5 ft x 5 ft x 1 ft plywood, right? Its volume reaches V = 5 x 5 x 1 = 25 ft3. If 1 ft3 of seawater has a mass of 64 lb, then it results (25 ft3) x (64 lb/ft3) =1600 lb plus cargo. You mentioned 2600 lb and said it seems way to high. Is it clear to you now? Or what more do you wish to know? What are you going to do with this vessel? Alcir Grohmann Beschaffung SAM Embarassment not necessary, a lot of people do not get this right. Archimedes principle states that the bouyant force exerted on a body is equal to the force exerted by the mass of the volume of DISPLACED FLUID (say water). The important point here is this: The volume of the displaced fluid depends both upon the shape and mass of the object. You can see this by doing the following "thought experiment". Take two identical pieces of aluminum foil. Crumple the first into a ball. Form the second into the shape of a shallow "boat". Place both on the surface of water. The ball sinks, but the boat floats. The reason is the ball does not displace much water, but the boat because of its shape displaces a much larger amount of water even though it has the same mass as the ball. A side note: The capacity of a ship is often expressed in terms of its "displacement", which is the amount of water the ship displaces. That is why flat barges are used to carry heavy cargo. Vince Calder Your initial numbers are correct, but 25 ft^3 x 64 lb/ft^3 = 1600 lb, not 2600. This is still a fair amount of weight. It doesn't mean that your "craft" would be sea-worthy, though. One wave over the gunwales, and the whole thing would go down like a rock if you've loaded it to capacity. Richard E. Barrans Jr., Ph.D. Assistant Director PG Research Foundation, Darien, Illinois No, you are absolutely correct (except for one detail -- see below). These numbers are roughly consistent with my experience building rafts on the Wisconsin River as a young man, although I mostly used waterlogged ties. Do ask your daughter! I am sure she would be delighted to discuss it with you. You might want to build a diver, as my father did with me and I did for my daughter. Get a wide mouthed bottle and fill it with water. Put in it a small bottle, mouth down, almost filled with water so it is just floating. Then stretch a sheet of rubber or similar across the mouth and secure with a rubber band. Now with a little effort you can make the diver sink by pressing on the rubber and rise when you release it. Why is that? Detail: But before you discuss with your daughter, note that 25 * 64 = 1600 (not 2600). Have fun! Best, Dick Plano You have the concepts right but the answer is incorrect. I get 25 cubic feet X 64 lbs per cubit foot = 1600 lbs This, of course, is an upper limit. Any additional weight would sink the craft to the bottom, so a safer approach would be to leave about 6 inches above water, resulting in a maximum "safe" capacity of 800 lbs. Greg Bradburn You understand right, but you have a multiplication error. Tim Mooney 25 * 64 = 1600 lbs, not 2600 lbs Remember that rafts are normally made up of round logs. Say each log is 6" in diameter. A raft that is 25 cubic feet would take be about 8' x 8'. I can imagine that a raft that large would be capable of holding 10 small men. -Wil Lam Tom, Correcting the calculator error produces a value of 25x16=1600 pounds of vessel weight and cargo. What this does not take into account is the fact that 1600 pounds is just enough to sink the vessel in perfectly calm water. It would push the top of the box down to the water surface. A bug landing on the box would push it under. To accomodate things such as small waves and bouncing, I would recommend limiting it to 800 pounds. This is not unreasonable. A 5'x5' raft of solid wood could easily support six adults. Dr. Ken Mellendorf Illinois Central College Click here to return to the Physics Archives

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