Boat, Lake and Anchor Question

Responses to yesterday’s post:

Nothing would happen to the water level. When you put the boat into the water, water is displaced depending on the weight of the boat and what is in the boat. IE The water rises when the boat is placed on the lake. When you then drop the anchor overboard, nothing will happen. The boat itself will displace less water, and will float higher. But the anchor still displaces water itself, so the overall displacement of water has not changed. There in regards to the lake system, nothing has changed.

Hopefully this helps.
– Mike

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The level of water will generally rise, but this really depends on the size and shape of your boat. The amount of water displaced is equal to the volume of the body immersed in the water (Archimedes’ Law) – ie. the anchor. The amount displaced by the boat is equal to the volume of the section of boat that is underwater. The size of this section is such that forces pushing the boat down (gravity) and forces pushing it up (buyoancy, which increases the more boat is underwater) are in equilibrium. If your boat is a huge flat-bottomed barge, the amount of water it displaces probably won’t change very much from losing the anchor, so the water level will rise. If the boat is a small rowing boat, the amount of water it ceases to displace when the large anchor is thrown overboard will probably be smaller, so the water level won’t rise as much.

It would be intresting to see some exact calculations to do with this …
– Victor

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I think the water level would go down. When the anchor is on the boat, its weight pushing down on the boat’s hull would cause its own weight in water to be displaced. When the anchor is thrown in the water, its own *volume* in water would be displaced. Since I’m assuming the anchor is steel or iron, the water displaced by the anchor weighs less than the anchor itself, hence the water level would go down.
– Dennis

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It will go down.

When it is in the boat, it displaces its WEIGHT in water, making the level of the lake go up a bit… but when it goes into the lake, it displaces its volume. Because it is more dense that water, its volume is disproportionately small for its weight, and thus the lake would go down… right? i think… anyway… good to see you’re posting again

– Kev

This is my opinion on the matter: The water level would fall if the anchor is thrown overboard.

The answer mainly stems from Archimedes’ Principal, which legend purports was discovered by him trying to figure out if a King’s crown was made out of gold or pyrite. Discovery was, as poetic license has it, celebrated by him
jumping out of his bath and running through the streets of Syracuse butt naked screaming, “Eureka!” But anyway, moving on from promiscuous Greeks…

In our case of the boat, an object that is submerged will displace water equal in volume to the object. If the anchor has a volume of X cm^3, it displaces X cm^3 (X millilitres) of water.

Now, an object that is not submerged, but is floating in water, is held up by two forces – buoyancy and surface tension. The effect of surface tension in this situation is insignificant, so we’ll ignore it, given that boats generally have a keel and are never flat bottomed. By Archimedes’ principal, the weight of water displaced by an object floating in it is equal to the weight of the object floating in it (incidentally the force of buoyancy on
the object is equal to the weight of water displaced). So, the object’s density is important. If the object is denser than water, it will displace a volume of water greater than the volume of the weight. Vice versa for an
object less dense than water. Obviously an anchor is more dense than water (if it’s less dense, it will not sink and that sort of defeats the purpose of an anchor), so lets say it weighs Y kg. Thus, when floating, it displaces
Y kg of water. Volume-wise, more water is required to form Y kg, than is required of the material the anchor is made of. Thus, when the anchor is floating on the boat, a greater volume of water is displaced.

Throw the anchor off, less water is displaced, and the water level falls. It would be rudimentary to mathematically calculate a roughly accurate figure of how much the lake would fall by, if other variables were given.

If we were dealing with objects on a much smaller scale, and of different qualities (perhaps a small pebble lying on a flat sheet of plastic floating in a swimming pool), surface tension would play a part in keeping the object
afloat and then it can’t be conclusively determined if the water level would rise or fall if you chucked the pebble into the pool. I think, anyhow.

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