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Homework Help: Buoyancy: tossing Archimedes' rock out of a boat

  1. May 27, 2010 #1
    You're floating in a boat in a pool with a 100 pound rock in the boat.

    You carefully measure the water level of the pool.

    Then you toss the rock out of the boat into the water. It sinks to the bottom.

    Does the water level of the pool go up, down or stay the same?

    Quantitative explanations would be very helpful.
     
  2. jcsd
  3. May 27, 2010 #2

    Doc Al

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    Staff: Mentor

    What do you think?
     
  4. May 27, 2010 #3
    I think that my subjective intuitions are not compelling to me even if they happen to be correct.

    I do not have sufficient facility with the relevant equations to prove the answer. That's what I'm looking for help with.

    (I think I'm also trying to figure out what kind of question formulation will elicit the most thoughtful responses on this forum. My lengthy elaboration of several simple buoyancy experiments in my "Archimedes Wooden Spheres" thread may not have been the best approach.)
     
  5. May 27, 2010 #4
    Try this: When it sinks to the bottom, the normal force from the ground acts on it. Previously, there was no normal force.
     
  6. May 27, 2010 #5
    Not sure why this thread was moved here. It's got nothing to do with homework nor am I in any course.

    Whatever.

    I think the water level goes down. That is based upon a comparison of the water displaced by the rock when in the boat (its weight) vs. the water displaced by the rock when submerged (its volume), and the fact that the rock is denser than water. That is, the rock will displace more water when in the boat than when it is submerged.

    However, I don't know how to quantify the analysis more than that, even if it is correct. That's why I am soliciting others.

    I had not considered the normal force of the ground, nor do I even have a guess why that may be relevant.
     
  7. May 27, 2010 #6
    This is just a guess, but I would assume it depends on the volume of the rock.
     
  8. May 27, 2010 #7
    I think you're right, for the rock to displace more water than if it were on the boat, it would have to be less dense than water, which means it wouldn't sink in the first place.
     
  9. May 27, 2010 #8
    I don't think so, not as long as the rock isn't so massive that it sinks the boat.

    Assuming that, no matter the volume of the rock, the weight of the denser-than-water rock will displace more water (raising the level of the pool) when it is "artificially" buoyed up in the boat than it would displace if allowed to sink.

    Hoping for quantification to confirm or disprove my hypothesis.
     
  10. May 27, 2010 #9
    Yes, those are my intuitions. Meaning, if the the rock were less dense than water, it would displace the same amount of water (its weight) whether it was in the boat or tossed out to float -- in which case the water level wouldn't change if it were tossed overboard.

    Not positive about this intuition. Hence still hoping for equations.
     
  11. May 27, 2010 #10
    Here's what I did, not sure if it's right though:

    In the case where it is boat + rock:

    Fb = Fg ; Fb = pVg is the buoyant force and V is the water displaced

    In the case where it is just the boat:

    Fb1 = Fg1; Fb1 = pV1g is the buoyant force keeping just the boat up

    If we say the mass of the rock is m then:

    Fb1 = Fb - mg and Fg1 = Fg - mg

    The volume of water that the rock displaces is mr/d where d is the density of the rock

    Assume that the rock displaces more water while submerged than if it was in the boat, then

    Vr (volume of water that the rock displaces submerged) > V - V1

    mr/d > Fb/(pg) - (Fb1 - mg)/(pg) -> p is the density of water
    mr/d > (mb + mr)g/pg - (mb - mr)g/pg -> mb is the mass of the boat + person
    mr/d > mb/p + mr/p - mb/p + mr/p
    1/d > 2/p
    so d < 2d < p must hold for the rock to displace more water while submerged but if this were the case then the rock wouldn't sink at all
     
  12. May 27, 2010 #11

    Doc Al

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    Perfectly correct on all counts.

    You can "quantify" it just by translating what you said above into mathematical terms.

    Weight of rock: mrock*g
    Volume of rock: Vrock = mrock/densityrock
    Vout = Volume of water displaced when rock is outside the boat = Volume of rock

    Weight of water displaced when rock is in the boat: mrock*g
    Vin = Volume of water displaced when rock is in the boat: mrock/densitywater

    Compare Vout to Vin:
    mrock/densityrock < mrock/densitywater

    Since the density of rock is greater than the density of water.

    It's not.
     
  13. May 27, 2010 #12
    This might help your intuition: to lift the rock off the bottom, attach it to a balloon. You have to inflate the balloon (raising the water level) to get the rock to lift. The open boat is the same as a balloon that is neutrally bouyant.
     
  14. May 28, 2010 #13
    Doc Al, here's why the normal force is relevant.

    When the boat is holding the rock, the buoyant force necessary equals the weight of the boat + rock. Therefore, the volume of the water displaced is proportional to this weight.

    When the rock has sunk to the bottom, the buoyant force necessary is less than the weight of boat + rock because there is an additional normal force holding the rock up. So the weight of the water displaced has to be less, and thus the water level has to go down.
     
  15. May 28, 2010 #14

    Doc Al

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    There's certainly nothing wrong with including it in your analysis, but to determine whether the level rises or falls, all you need to know is that the rock sinks. (Once the rock is out of the boat, we don't care how it's supported, just that it displaces an amount of water equal to its volume, not its weight.)
     
  16. May 28, 2010 #15
    Good point! :blushing:
     
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