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## Homework Statement

I'm confused about the following kind of situation. Consider a block of density ##\rho_b##, mass ##M_b## and section ##S_b## that floats on a liquid of density ##\rho_l##, in a tank of section ##\mathcal{S}##. On the block there are some objects (all equal), of density ##\rho_o## and mass ##m_0##.

Now consider two cases

1. The objects are removed from the block

2. The objects are thrown in the liquid

What does happen to the system?

## Homework Equations

Bouyancy

## The Attempt at a Solution

__Case 1__

Initially the block is immerged for an height ##h_1## such that $$(h_1 S_b) \rho_l= M_b + \sum m_{o}$$

When objects are removed the height of the immerged part is ##h_2## such that

$$(h_2 S_b)\rho_l= M_b$$

From here I can calculate the variation of immerged volume $$\Delta V= S_b (h_1-h_2)$$

My main question is: who does move? The liquid, the block, or both?

As far as I understood **only the liquid moves** going down, of an height ##h^*## given by

$$\Delta V=(h_1-h_2) S_b= \mathcal{h^*} \mathcal{S}$$

But I'm not convinced because maybe also the block moves, precisely it may go up. In the solution I proposed the block does not move at all, it's the liquid that moves below it. Is that correct?

__Case 2__

After the height ##h^*## calculated in case 1, the fluid goes up of an height ##\bar{h}## because the objects are (at least partially) immerged in the fluid.

$$ \sum m_o =\rho_l \bar{h} \mathcal{S}$$

Summing up the total change i height of the fluid should be $$\Delta h= h^*- \bar{h}$$

As said above all of this assumes that only the fluid moves, while the block does not. So my question is: is that assumption correct? (And, possibly, why?)