- #1
Soren4
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- 2
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?)