A fishermen drops a small heavy object into the pond. Does the level of the lake rise or fall?

Question

A fisherman who has a small but a very heavy object in his boat. He is fishing in a pond which has no inlet or outlet (for the problem's sake). Accidentally he drops this small but very object into the pond. As a result of this careless act, would the level of the water of the pond rise,fall or remains the same?

Answer 1

The water level is going to fall. While the small object is in the boat, the small but heavy object pushes the boat down (although not able to completely sink the boat), which causes a greater dislplacement in the water because of the size of the boat which goes below the water surface. If the small object were to be dropped, then a considerable volume of the boat would rise above the surface, as the weight that is pushing it down disappears. The small and heavy object would definitely have a smaller volume displaced, even though it sinks completely.

Answer 2

When he takes the small but heavy object out of the boat, then the weight of the boat decreases, resulting in a reduced bouyant force from the water, meaning that the boat displaces less water. Therefore the waterlevel of the lake drops, according to some independent measure.

When the small object is dropped into the water, it will displace an amout of liquid exactly equal to it's own volume. However, this will be small in comparison to the volume displaced by the boat.

Overall, I think that there would probably be a minute decrease in the water level.

Answer 3

Remains the same
Any object immersed in water is provided with a buoyant force that counters the force of gravity, appearing to make the object less heavy. If the overall density of the object exceeds the density of water, the object sinks. If the overall density is less than the density of water, the object rises until it floats on the surface.
Archimedes' principle
The principle that the net fluid force on a body submerged (or floating) in a stationary fluid is an upward force equal to the weight of the fluid displaced by the body. This concept, perhaps the oldest stated principle in fluid mechanics, was first put forth by Archimedes in the third century B.C.

In a static fluid, the weight of the fluid causes an increase in pressure with depth. Thus, at the surface of the fluid, the pressure is atmospheric pressure (p0 = 14.7 lb/in.2 = 101 kilonewtons/m2), while at a depth h the pressure has a larger value of p1, given by Eq. (1), where γ is the specific weight
1.
of the fluid (weight/volume). The difference in pressure force between the bottom and the top of a water column is therefore given by Eq. (2), where h′ and A are the height and area
2.
of the column, and pb and pt are the pressures at the bottom and top of the column. This difference is precisely equal to the weight W of the water within the column, given by Eq. (3). If
3.
the water column were replaced with a solid object, the pressure forces on the object would be the same as on the original water column. That is, the net hydrostatic pressure force on the object, termed the buoyant force, would be equal to the weight of the water displaced (which is the statement of Archimedes' principle). The same concept holds for a body of arbitrary shape, which can be thought of a consisting of many small vertical columns fastened together. Archimedes' principle is valid for submerged or floating bodies in liquids or gases. See also Buoyancy; Specific gravity.

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Answer 4

Stay the same.

The object would have already displaced the water when it was inside the boat, so dropping it in wont cause any more displacement than that.

Although to be honest I dont really know the correct answer to this, but that would be my guess. Everyone on here has given different answers so good luck working it out!

Answer 5

The water level would fall. The heavy object, while it is in the boat, is already displacing an amount of water with the same mass as the object. It will displace an equal VOLUME of water once it is dropped, which will be smaller than the volume of water it was displacing while in the boat, therefore the water level will fall.

Answer 6

Wow this could be an extremely large question and actual have been given my thinking... the respond is the preliminary point, x, would be greater than the result point, y. think of of it in ranges... a million. while the rock is interior the boat, that's floating, so, to boot to what the boat might displace devoid of the rock, it additionally displaces the quantity of water with a weight equivalent to the rock. 2. while the stone is hurled interior the air -- so as that that's out of the boat yet no longer yet interior the water -- the boat is relieved of the rock's weight and displaces much less water. subsequently, the water point interior the pool drops. 3. while the rock enters the water, it displaces a quantity of water equivalent to its length. because of the fact the comparable quantity of water weighs much less (maybe in basic terms 3 kilos) than the comparable quantity of rock (say 10 kilos), the denser rock sinks to the backside. The water point rises back, yet no longer as extreme because it replaced into while the rock replaced into interior the boat in view that's now displacing much less water. the internet results of throwing the rock interior the pool, then, is a reducing of the water point. wish this clarification is obvious for you!

Answer 7

Indiana... is right. The water level goes down. Sometimes solving problems is easiest if you look at an extreme example that you can do in your head.
Imagine the fisherman has a black hole (or something REALLY heavy but small.)

The boat is very heavy, so is very low in the water, which would make the water level high. He drops the black hole. The boat comes up a lot, the water level goes down a lot. The black hole enters the water, but it's so tiny, the water level hardly changes at all.

So, answer is "level goes down." As I said, some other responders already gave you the correct answer. But do you see how taking an extreme case sometimes makes it easier to see the solution?
-Cheers!

Answer 8

The displacement of the hull would require more area with the weight, bot launching in would make surface area rise. Heavy object in the vessel tossed over would decrease the area of displacement required to make the boat boyaunt Vessel would rise, water would also. It is a Volume Factor.

Answer 9

it would stay the same because once the boat went on the water ( with the object in the boat) any displacement already took place, so by the object going into the water will make no difference at this point.

Answer 10

Displacement would make the water level rise.
However, if the object was so heavy that it embedded itself into the pond base, the level would remain the same.