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

  1. Sep 11, 2016 #1
    if the person sitting in the boat throws a pebble to the swimming pool. Pebble was initially contained inside the boat and of course it has higher density than water.
     
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  3. Sep 11, 2016 #2

    CWatters

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    What do you think will happen and why?
     
  4. Sep 11, 2016 #3
    The density of boat will become low after throwing the pebble. Therefore boat will rise decreasing the level of the water in the swimming pool but at the same time stone is inside swimming pool again displacing some volume that means water level will be the same.
     
  5. Sep 11, 2016 #4

    rcgldr

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    The boat displaces water based on the boats total weight and density of water. The pebble is denser than water.
     
  6. Sep 12, 2016 #5
    So what's the conclusion
     
  7. Sep 12, 2016 #6

    rcgldr

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    Lost an edit to my last post. Posting here. Think of this as another clue.

    When the pebble is in the boat, all of it's weight is supported by the buoyant force of displaced water. When the pebble is resting on the bottom of the pool, some of it's weight is supported directly by the bottom of the pool.
     
  8. Sep 12, 2016 #7

    billy_joule

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    Cosider the extreme case and the problem is much easier to understand;
    The boat is weightless and the pebble has close to infinite density.
     
  9. Sep 12, 2016 #8
    That means initially boat and pebble was drowned, displaced volume is (volume of boat + volume of pebble). When pebble is thrown, boat will rise to the top and boat's bottom surface will just kiss the water surface and pebble is drowned only; so this time only displaced volume is volume of the pebble. Am I correct in the interpretation of weightless boat and pebble of infinite density?
     
  10. Sep 13, 2016 #9

    billy_joule

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    A boat can support near infinite weight without sinking, it just has to be very large.
    The boat will only sink if we constrain its size in some way, if we let the boat sink the question is very different, and IMO losses the interesting aspect.

    The weightless boat and very dense pebble has the same result as the question in your OP, it's just that the extreme case makes the outcome much more dramatic/easier to understand.
     
  11. Sep 13, 2016 #10

    CWatters

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    Perhaps focus on the very dense pebble part.
     
  12. Sep 13, 2016 #11

    rcgldr

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    The second part is correct. For the first part (pebble in boat), the displaced volume is the volume of water equal to the weight of the boat and the pebble.
     
  13. Sep 13, 2016 #12

    A.T.

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    How did you get from "some" to "same"?
     
  14. Sep 13, 2016 #13
    It seems to me that the boat is what confusing. All the boat does in this case is add bounce to the rock. Allowing it to float which means that some of the total surface of the rock is out of the water at this point setting the level of the water. Then you take away the boat or at least take away the bounce of the rock. The rock sinking to the bottom getting completely covered in water making the level of the water go up. You can also look at the question like this. Say you have a ball that you can add and subtract sand from. when the ball is empty the ball floats taking up less space in the water. When you fill the ball makes it sink. The ball takes up more space in the water making the water level go up. From my point of view the question is one of volume and all the rest of info is just there for continuity of speech. Real life or word problems always have extra info. You just have to figure out what the base math is in the problem and everything else is useless info
     
  15. Sep 13, 2016 #14

    A.T.

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    When in the boat, the entire rock surface is dry, but the rock can still be completely below the water level.
     
  16. Sep 13, 2016 #15
    you know i could wipe the rock in cellophane. keeping the rock dry at the bottom of the water Then call the cellophane the boat the question does not ask if the rock gets wet or not it asks what the level of the water when the rock is at the bottom of the water compared to floating on top
     
  17. Sep 13, 2016 #16

    A.T.

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    That would be a submarine though.
     
  18. Sep 13, 2016 #17
    lol very true
    by definition: to float on or in water the sum of the parts have to be less dense than the water. Everything that floats has some part is out of the water and that would lead you to think that when pushed down (either by a finger or grative) the water would have to rise ??? you did say the boat is wireless which means that the boat does not displace any water on its own. rock bears the total load of the boat and rock when in the water so you take away the waterless boat (the upward force on the rock making it float) and the rock sinks i'm just looking at the boat as a way to lesson the density of the rock
     
  19. Sep 17, 2016 #18
    It doesn't have to be. The volume of the pebble has to be very, very small to sustain the wieght of a "normal" pebble.
    As long as it's not below ##2GM/c^2## or it will evaporate all the water in the swimming pool, and hopefully not the entire city. :smile:
     
  20. Sep 17, 2016 #19

    A.T.

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    If that's your conclusion, then something is wrong with your reasoning.
     
  21. Sep 17, 2016 #20

    olivermsun

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    To float on or in the water the parts must have density less than or equal to the water.

    Equal (neutral buoyancy) is just the limiting case where the object must displace all the water it can possibly displace to achieve a balance between gravity and buoyancy.
     
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