# Question about buoyancy force affecting mass measurement

Consider weighing a container fulled to the brim with water and a solid ball next to it to obtain $m_1$. That mass is made up from the balls mass, the containers mass and from the mass of the water.

Now place the ball in the container with water, so that a volume $V$ of water overflows the container and the ball sinks to the bottom. Now measure the mass again to obtain $m_2$. Now obviously $m_2$ is going to be less than $m_1$ for the mass of the overflowed water which establishes the relation $m_2=m_1-\rho V$ where $\rho$ is the density of water.

However, my question is this. If we are measuring mass by measuring weight via the normal force, then when we put the ball inside the water container there will be a buoyancy force that will decrease the normal force on the ball itself. Therefore, apart from the overflowed water, shouldn't there also be an apparent loss of mass (decrease of weight) because of the buoyancy force? I have heard that "the buoyancy force will come in a action-reaction pair that has no effect on the scale". However, why doesn't it affect the scale?

Simply put, why the buoyancy force exerted by a fluid on bodies inside it (floating or not) does not affect mass measurement of the fluid with the bodies in it?

Now let's use Newton's second law for the ball, that is, taking in account only forces that are acting upon the ball. $$mg=F_N+F_B$$ where $F_N$ is the normal force and $F_B$ is the buoyancy force. There is no problem with this. However, if we limit our observation of the case to the ball only, then one might ask: "Well obviously from that equation we can deduce that the normal force is less than usual, so why doesn't the buoyancy force affect mass/weight measurement?" And that's exactly what confused me. The simple solution to this is to recall that the weight of the ball not only acts upon the ball itself but also upon the container since the ball is in contact with it. Therefore, if the container is resting on a weighing scale, the sum of the forces must be zero, that is, there must exist a normal force that counters the weight of the ball pushing the container down. And that force, as now is obvious cannot be anything but the weight of the ball itself.