I say it stays the same, a friend says it rises, and a coworker says it lowers. Which is it?!
You’re in a row boat, which is in a large tank filled with water. You have an anchor on board, which you throw overboard (the chain is long enough so the anchor rests completely on the bottom of the tank). Does the water level in the tank rise or fall?
Who puts an anchor in a tank of water in the first place? Theres your problem right there
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The water level increases, because when an object is dropped in water, the water around it rises slightly. Edit: Also, try dropping ice cubes in water and look at the water level.
Well, I think we all know that. The problem is sorting out the matter of whether the boat's submersion with all that added mass would be equal, less, or greater than the mass of the anchor by itself.
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It rises. Having the anchor in the water, makes the mass of the water increase. The fact that the anchor isn't imaginary, it has mass just like every existing thing in this world.
Your sitting in class before every one else and its 10 minutes before the bell rings. With 20 students in the class originally you witness every student gradually enter the room as time goes on. With every student entering the room, the equivalent of their mass of air leaves the room with them. The more people in a room, less air there is to breathe. Ever have several friends in your room playing video games and it's really hot? Whereas throughout the week it's normal? That's why.
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When the anchor is dropped into the water, the water level is displaced, causing it to rise slightly. This is because the anchor is more dense. An object with more mass has more weight, and thus more density.
It rises. Having the anchor in the water, makes the mass of the water increase. The fact that the anchor isn't imaginary, it has mass just like every existing thing in this world.
As WW mentioned, though, the question isn't whether the anchor would displace water on its own. It certainly would. It's a question of whether or not that displacement would be greater than, less than or equal to the displacement it's already causing by weighing down the boat.
I haven't decided on how to obtain the correct, reliable answer, but no one here seems to be paying attention to the real problem (EDIT: except Chicken, of course ).
Here's an excerpt from Wikipedia to get us started, with a couple of points highlighted:
More tersely: Buoyancy = weight of displaced fluid
Archimedes' principle does not consider the surface tension (capillarity) acting on the body,[4] but this additional force modifies only the amount of fluid displaced, so the principle that Buoyancy = weight of displaced fluid remains valid.
The weight of the displaced fluid is directly proportional to the volume of the displaced fluid (if the surrounding fluid is of uniform density). In simple terms, the principle states that the buoyancy force on an object is going to be equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational constant, g. Thus, among completely submerged objects with equal masses, objects with greater volume have greater buoyancy.
The water displaced depends not on the mass of the object but on its volume. That changes everything.
So, the question is this: with the anchor in the boat, does the volume of the boat that is submerged increase more than the volume of water displaced by the anchor alone?
Also, thanks: now I'm going to have to try this when I get home, using a heavy, dense object siting inside a small, boatlike object floating in my bathtub. In other words, a typical Monday evening.
Twitter is a good place to throw your nonsense. Wii FC: 8378 9716 1696 8633 || "How can mushrooms give you extra life? Get the green ones." -
Also, thanks: now I'm going to have to try this when I get home, using a heavy, dense object siting inside a small, boatlike object floating in my bathtub. In other words, a typical Monday evening.
Dammit, I cheated and managed to find a source on this.
We assume that there's a water level gauge of great precision somewhere, perhaps at the boat dock, to measure the lake level. This puzzle, and the next two puzzles, use Archimedes' principle: "The Buoyant force on an object submerged or floating is equal to the weight of the displaced liquid." They also use Newton's law: "The net force on a body in equilibrium is zero."
The anchor in the boat adds to the boat's weight, and the presence of the anchor there causes a displacement of an amount of water that weighs as much as the anchor. (The upward buoyant force due to the displaced water has the same size as the weight of the anchor, by Newton's law.) I.e., the anchor in the boat displaces a weight of water equal to its own weight. When the anchor is at the bottom of the lake it displaces an amount of water equal to its own volume. Its weight is greater than the weight of an equivalent volume of water (that's why it sank), so it displaces more water when in the boat than when at the bottom. The water level of the lake drops when the anchor is thrown overboard.
Dammit, I cheated and managed to find a source on this.
We assume that there's a water level gauge of great precision somewhere, perhaps at the boat dock, to measure the lake level. This puzzle, and the next two puzzles, use Archimedes' principle: "The Buoyant force on an object submerged or floating is equal to the weight of the displaced liquid." They also use Newton's law: "The net force on a body in equilibrium is zero."
The anchor in the boat adds to the boat's weight, and the presence of the anchor there causes a displacement of an amount of water that weighs as much as the anchor. (The upward buoyant force due to the displaced water has the same size as the weight of the anchor, by Newton's law.) I.e., the anchor in the boat displaces a weight of water equal to its own weight. When the anchor is at the bottom of the lake it displaces an amount of water equal to its own volume. Its weight is greater than the weight of an equivalent volume of water (that's why it sank), so it displaces more water when in the boat than when at the bottom. The water level of the lake drops when the anchor is thrown overboard.
Dammit, I cheated and managed to find a source on this.
We assume that there's a water level gauge of great precision somewhere, perhaps at the boat dock, to measure the lake level. This puzzle, and the next two puzzles, use Archimedes' principle: "The Buoyant force on an object submerged or floating is equal to the weight of the displaced liquid." They also use Newton's law: "The net force on a body in equilibrium is zero."
The anchor in the boat adds to the boat's weight, and the presence of the anchor there causes a displacement of an amount of water that weighs as much as the anchor. (The upward buoyant force due to the displaced water has the same size as the weight of the anchor, by Newton's law.) I.e., the anchor in the boat displaces a weight of water equal to its own weight. When the anchor is at the bottom of the lake it displaces an amount of water equal to its own volume. Its weight is greater than the weight of an equivalent volume of water (that's why it sank), so it displaces more water when in the boat than when at the bottom. The water level of the lake drops when the anchor is thrown overboard.
Well, when you remove the anchor from the boat, the boat would rise, as there is not as much weight on the boat, meaning the water level would then lower. When you drop the anchor into the water, some mass is replaced, making the water rise. It really depends on how much mass there is in the boat with the anchor, and how much there is when the anchor is dropped. So, it really depends. This coming from someone in 8th grade science
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