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Buoyancy and Weight

  1. Jan 20, 2012 #1
    This is more of a conceptional question.

    Let's say you have 1L of water in a bucket on a weighing scale, which currently reads 50N. If a couple of ice cubes were added, would the weighing scale read 50N + weight of 2 ice cubes?

    Side-Notes:

    density of the ice < density of water
    ice cubes are floating

    At first glance, it looks reasonable, but since the floating ice cubes are in equilibrium, wouldn't the buoyancy force cancel out the gravitational pull of the ice cubes, essentially making the ice cubes weightless?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 20, 2012 #2

    DaveC426913

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    Where does the bouyancy come from? It comes from the water which, in pushing up the ice cubes, also pushes down on the beaker with equal and opposite force.
     
  4. Jan 20, 2012 #3
    So the scale reading would be 50N + buoyancy force of the 2 ice cubes?
     
  5. Jan 20, 2012 #4

    DaveC426913

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    Gold Member

    Well, 50N plus the weight of 2 ice cubes.

    Remove the confusion around bouyancy.

    Imagine a zero mass beaker full of 100 glass marbles. Each marble weighs 10g.
    The full beaker weighs 1000g.
    But wait. One of the glass marbles has a bubble in it. It is the same size but only weighs 8g.
    The one marble is less dense than the others. Technically, if given the opportunity, it will rise to the top. (Like any other less dense substance, it is bouyant.)
    Why would its position in the beaker change the total weight of the beaker of marbles?
    Though it is on top, it contributes 8g of weight to the marble below it, and on through to the beaker's bottom.
    The full beaker weighs 998g.
     
    Last edited: Jan 20, 2012
  6. Jan 20, 2012 #5
    Drawing the Free-Body diagram can really help in this case. This always helps me with physics problems.

    If you draw the entire beaker with the water, before the ice is added, as a single body, the force acting downward would be 50 N.

    When you add the ice cubes to the beaker, the cubes are now part of the beaker along with the water. When you add anything, be it ice cubes, a cookie, or a balloon full of helium to the beaker of water (assuming you could keep it from floating out by tying it down to the sides of the beaker) you also have to add the weight of that object to the total weight you had before.

    Buoyancy only comes to play if you were to model the ice cubes as one object and the beaker of water as another object. The ice cubes would have a weight that acts downward on the water, and the water would have a force acting upward on the ice due to its lower density.
     
  7. Jan 21, 2012 #6
    ok. thanks for the input, you two.

    I have one final question. If the ice cubes started from the bottom of the water-filled beaker, it would rise to the top. During its ascent, would the ice cubes' "weight" (according to the weighing scale) be any different than as if they were on the top, floating?

    I came up with 3 possible answers:

    1. the weight of the ice cubes would be simply its gravitational pull

    2. there are two forces on the cubes:

    - the upward buoyancy force
    - the downward gravitational pull

    the net force would be upwards since the ice is moving up. then if so, the "weight" of the ice cubes would be the opposite-reaction force of the net force.

    3. the weight of the cube would be its downward gravitational pull, plus the downward opposite-reaction force of the upward buoyancy force.

    ---

    I am leaning towards answer #3. Answer #2 doesn't seem reasonable because it wouldn't work for the previous question.
     
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