Question about mass measurement and Archimedes law

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?

• Please explain why the down vote? – ahra Dec 8 '16 at 14:02

• 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 $W-FB$ ? – ahra Dec 9 '16 at 10:36
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 $m-\rho \Delta V$.