What will happen to the water level of the swimming pool?

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Discussion Overview

The discussion revolves around the effects on the water level of a swimming pool when a pebble is thrown from a boat into the pool. Participants explore the implications of buoyancy, density, and displacement in this scenario, considering both theoretical and practical aspects.

Discussion Character

  • Exploratory
  • Debate/contested
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that when the pebble is in the boat, it contributes to the boat's weight, causing the boat to displace water based on the total weight of the boat and the pebble.
  • Others argue that once the pebble is thrown into the pool, it displaces its own volume of water, potentially leading to the same water level as before.
  • A participant introduces the idea of considering extreme cases, such as a weightless boat and a pebble with infinite density, to simplify understanding of the problem.
  • Some participants discuss the role of buoyancy and how the pebble's position (in the boat versus in the water) affects the overall displacement of water.
  • There are claims that the boat's presence adds complexity to the problem, with suggestions that the question could be simplified by focusing solely on the pebble's volume and displacement.
  • One participant emphasizes the importance of understanding the base math of the problem, suggesting that additional information may be extraneous.
  • Disagreements arise regarding the interpretation of buoyancy and displacement, with some participants questioning the reasoning of others without providing explicit corrections.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the outcome of the water level change. Multiple competing views remain regarding the effects of the pebble's displacement and the role of the boat in the scenario.

Contextual Notes

Participants express various assumptions about buoyancy, density, and the mechanics of displacement without resolving these assumptions. The discussion includes hypothetical scenarios that may not align with practical applications.

  • #31
Try some example numbers, like the mass weighs 1,000 kg, and has a volume of .05 m^3 (1/20th cubic meter). When the mass is in the boat, how much water is displaced? When the mass is resting at the bottom of the pool, how much water is displaced?
 
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  • #32
hsdrop said:
...but you're saying that if the mass was submerged the water level would stay the same no matter what the volume of the mass is(provided it would sink and not float on it own)??

Check again. Did he really say that?
 
  • #33
ok now I'm starting to see what the water level would do. it just the way the question was given had me a little mixed up
also i am a veary slow at read (i have a savear reading disablety with decoding) and i just now got to the boddom of the https://en.wikipedia.org/wiki/Archimedes'_principle page that gives a detaled explanchen on how and why the principle works witch help emancely
 
  • #34
CWatters said:
Check again. Did he really say that?
and you're right i looked and no he did not say that

thank you everyone for being so patient with me
 
  • #35
Interesting problem. I don't know if the OP's question ever got answered, but when the rock is in the boat it will cause an extra displacement of water equal to its weight. If the density of the rock is greater than the density of water, the amount of water displaced by the rock when it is in the boat will be greater than the volume of the rock. When the rock is thrown in the water, it sinks to the bottom and the rock only displaces its volume. Thereby the water level of the pool holding the boat drops a small amount. Alternatively, if the rock is less dense than water, it will float (it won't be entirely submerged) when tossed out of the boat and the amount of water displaced by the rock will be the same whether it was in the boat or out of the boat.
 
  • #36
Now what I understood reading the complete thread. Boat is assumed weightless and stone is of finite density & some finite volume. Now only weight of stone will contribute in displacing certain volume of water. The proportion of density between water and pebble will decide how much volume (multiple of stone's volume) water will be displaced. As the weight of stone will be equal to the weight of displaced water I.e. Archimedes principle. Let's say the density of stone is 1000 times that of water. So when pebble is in the boat it will displace 1000 times more volume compared to the pebble sinking to the bottom of the pool. Because when pebble is sinking it is only displacing its own volume. So level of swimming pool will fall after throwing the pebble.
 
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