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Say we have a container of water full to the brim and some object with volume $V$. We measure the mass of the container with the object by its side to obtain $m$. Now we put the object inside the container and it sinks, and we measure the mass of the container with the object in it to obtain $m_1$. Now, since the container is full to the brim, when we put the object inside, it will push out a volume $V$ of water, but also because of the buoyant force the objects weight will be less for the mass of the displaced water so we can conclude that $m-m_1=2\rho gV$ where $\rho$ stands for water density. Is this correct?

What happens if the object floats? Will then the measured mass be $m-\rho g \Delta V$ where $\Delta V$ is the volume of the object submerged/volume of displaced (pushed out) water?

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  • $\begingroup$ Please explain why the down vote? $\endgroup$
    – bonehead
    Commented Dec 8, 2016 at 14:02

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It's not correct, there's no "2" in the first expression, if I understand the setup. It sounds like you are saying that you put a beaker filled to the top on a scale, and zero the scale. Then you put an object next to the beaker, and the scale reading gives you the weight of the object. Then you put the object in the beaker, discard the overflow, and measure the weight again. If that's what you mean, that second weight is less than the first one only by the weight of the overflow that was discarded. The buoyant force on the object is an internal force to that system, so will come in an action-reaction pair that has no effect on the scale reading.

Thus the difference if the object sinks, or floats, is only in the weight of the discarded overflow. If the object sinks, the overflow is the weight of water that fits in the object's volume (no "2"). If the object floats, the overflow is the weight of the object itself. So the scale will read zero if the object floats and the overflow is discarded. If the overflow is not discarded, there is no difference in the weight when the object is placed in the beaker rather than next to it, regardless of whether the object sinks or floats, either way the scale will read the weight of the object.

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  • $\begingroup$ What would be the reaction force to buoyancy? Normal force of the object itself preventing it from being crushed by the water? $\endgroup$
    – bonehead
    Commented Dec 8, 2016 at 14:29
  • $\begingroup$ Yes, that's it exactly. Note there are actually two such forces that are relevant, top and bottom, because the buoyant force is also the difference between two forces, top and bottom. They are pressure forces, and the body resists them by not compressing. This also means that a body that compresses without resistance is not buoyant. $\endgroup$
    – Ken G
    Commented Dec 8, 2016 at 17:45
  • $\begingroup$ I understand. Though, consider when a ball is resting on the bottom of the container. There's weight acting downwards, buoyancy and normal force acting upwards. If we were to put a weighing scale on the bottom wouldn't it measure the weight indirectly by measuring the normal force which would be $W-FB$ ? $\endgroup$
    – bonehead
    Commented Dec 9, 2016 at 10:36
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    $\begingroup$ Ah, now you mean an underwater scale. Yes, if the scale is itself underwater, then it will measure the weight of the object minus the buoyancy force. You can tell this must be true because of the "object" is itself pure water, the scale (say, an underwater spring) will not register anything at all. $\endgroup$
    – Ken G
    Commented Dec 10, 2016 at 11:44
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First of all, your first conclusion is incorrect because you have incorrectly included the buoyancy force. The mass difference $m-m_1=\rho V$. But it is right for floating case $m-\rho \Delta V$.

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