# Fisherman boat and sea level

A fisherman rowing his boat on a very small lake throws his anchor into the water. Does the water level of the lake rise, fall or stay the same?

• Here's a non-trivial twist: What if the anchor doesn't go all the way to the bottom? The line is too short. Now the anchor is suspended at some depth and is pulling down on the boat. – Engineer Toast Apr 22 '15 at 2:13
• @EngineerToast First the water level will raise, as long as the anchor is sinking after being thrown in. If the chain is not long enough, the anchor will then pull the boat down again and the water level will be the same as in the beginning, since boat+anchor is a floating object with the same mass as before. – Falco Apr 22 '15 at 9:25
• Clarification required: is the anchor resting on the lakebed or is it still supported (partially or totally) by the boat via the rope/chain? – David Richerby Apr 22 '15 at 10:03
• @Falco Will the net force on the boat+anchor be the same? Is the buoyant force on boat+anchor more than or less than the sum of the buoyant force on the boat and the buoyant force on the anchor? – Engineer Toast Apr 22 '15 at 12:24
• @EngineerToast it is of course the same! Since the boat will rise as much as the anchor is displacing water volume, when the anchor is dropped into the water. – Falco Apr 22 '15 at 12:27

Assuming that the anchor will actually sink and reach the bottom of the very small lake, the water level of the lake will

fall

Reason:

The anchor, because it sinks, is more dense than the water. While it was in the boat, because the boat was floating an amount of water equal to the weight of the boat plus the weight of the anchor was being displaced. After the anchor is thrown into the water, the weight of the boat plus the volume of the anchor is being displaced, but the volume occupied by the anchor is less than the volume occupied by an amount of water with the weight of the anchor.

More intuitive reasoning:

The boat's going to be riding higher in the water without the anchor in the boat.

• I don't follow. If the same weight (i.e amount of water) is being displaced both before and after deploying the anchor, shouldn't the level stay the same? – dberm22 Apr 22 '15 at 11:24
• @dberm22 jonc101.tripod.com/brainteaser/key/… explains the same way – leoll2 Apr 22 '15 at 12:22
• @leoll2 It seems this example works assuming the anchor is small and heavy, what about a large light(just dense enough to sink) anchor? – Eamonn McEvoy Apr 22 '15 at 13:58
• @EamonnMcEvoy You've just read the "In other terms" part. Read the whole solution, including hints, and everything will be clear. – leoll2 Apr 22 '15 at 14:03
• @leoll2 Ahh, I see. In the boat, all the downward force is reacted by the water, so the displaced water is equal to the mass of the anchor (plus the boat, obviously). When the anchor is deployed, and on the bottom of the lake, the lake bed reacts some of the mass of the anchor, so the amount of displaced water is less. Thanks for the link. – dberm22 Apr 22 '15 at 14:39

Two thinking outside the box answers:

1. If the very small lake is shaped as a glass of water and there is water all the way up, then the water level will not rise, since any increase in water level will run over. More likely is that the throwing of the anchor (assuming it even fits the lake) will cause splashing, and the water will splash over the edge and by that decreasing the water level.

2. The boat is touching the bottom of the very small lake. By that, throwing it in will increase the water displacement will not be any different no matter how much you put in the boat.

• Your first point isn't as crazy as you make it sound. :-) If the lake has any rivers flowing out of it then any increase in water level would indeed float away. Your second point assumes that throwing the anchor doesn't decrease the weight of the boat enough for it to start floating. – David Richerby Apr 22 '15 at 21:01
• Yes, you are correct :) – Viktor Mellgren Apr 24 '15 at 12:44

The following answer is wrong, and I am just leaving it here for anyone who has the same idea.

It depends on the size and weight of the anchor in relation to the size of the boat.

Reason:

When the anchor is in the boat, water will be displaced by the boat, with the weight of the anchor contributing to this. When the anchor is thrown overboard the boat will displace less water but now the anchor in the water is displacing some. If the anchor is very large and light the water level will rise, if it is small and very heavy it will go down.

• No i was thinking the same but: "the anchor weighs the boat down which will result in the anchor effectively displacing a volume of water that would weigh as much as the anchor." the volume of water displaced by it weight will always be higher than the volume displaced by the size of the anchor, if the seize of the anchor would displace more water than it's weight it would float. – Vincent Apr 22 '15 at 12:27
• @VincentAdvocaat - strictly speaking, the question never established that it was a regular anchor. It might have been a fat TV news anchor. More seriously, it might have been a sea anchor (en.wikipedia.org/wiki/Sea_anchor). – Glen O Apr 22 '15 at 13:05
• Consider two extremes for the anchor. The first is very small and heavy, say 1cm^3 and 1000kg (made from some super dense material), when in the boat this will cause the boat to displace a large amount of water, but when thrown overboard it will only displace 1cm^3 of water, therefore the water level will drop. The second anchor is large and light 1000cm^3 and 1.1kg (just dense enough to sink) it will have little effect while in the boat but displace 1000cm^3 in the water causing the water level to rise. – Eamonn McEvoy Apr 22 '15 at 13:09
• it will displace more water than the before mentioned anchor but it will not displace more water as when inside the boat. Here i asked it on Math.SE for a rigid proof, hopefully that will clear things up: math.stackexchange.com/questions/1246597/displacement-of-water/… – Vincent Apr 22 '15 at 14:04
• I am now convinced :) – Eamonn McEvoy Apr 22 '15 at 14:32

The level of water will:

Remain the same

Reason:

The Archimedes' principle states that "when a body is immersed in a fluid fully or partially in a fluid, the magnitude of the weight of the fluid displaced is equal to the magnitude of the weight of the body". So as the lake was barring the weight of both the anchor and the boat in both the cases, i.e., when the anchor was not yet thrown and the anchor was thrown, the amount of water displaced will remain the same, hence the water level remains the same.

• Partially true. That is the correct statement of the Archimedes principle IF the object is floating. The true statement of the principle is that the magnitude of the weight of the fluid displaced is equal to the buoyant force exerted on the object. The net force on the anchor was 0 when it was floating with the boat. After thrown in, the net force on the anchor is now negative, since it's sinking (it wouldn't be a very good anchor if it didn't). Since the net force on the anchor has changed, the displaced water has changed. It's now displacing the water based on its volume, not its weight. – Duncan Apr 22 '15 at 4:10
• @Duncan please see puzzling.stackexchange.com/a/12554/5871 – user5871 Apr 22 '15 at 5:08
• @Switch: Before throwing the anchor, the water indeed needs to support the weight of the anchor through the volume of water displaced by the boat, but after it's thrown, it's (supposedly) not touching the ground and hence supported by the ground. The water still gives force to the anchor, but only equal to the volume of water displaced, which is now less, since the anchor cannot displace more water than its volume. Since there are less water displaced, the water level will fall. (btw this is similar to what Duncan said, just added the "ground" factor for easier understanding) – justhalf Apr 22 '15 at 5:20
• Your statement of Archimedes' principal is incorrect. An object that is fully underwater cannot possibly displace more than its own volume. – David Richerby Apr 22 '15 at 10:02