# Floating block with objects that are thrown in a liquid

• Soren4
In summary, the conversation discusses a block floating on a liquid and the effects of removing objects from the block. In case 1, both the block and the liquid move, with the block moving up and the liquid moving down to fill the space left by the removed objects. In case 2, the liquid moves up as the objects are thrown in, causing a change in the total height of the liquid. The block also moves in both cases, with its motion being dependent on the rate of removal or addition of objects.
Soren4

## 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?

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?)

At fisrt, the objects are removed from the block, Mg<F => the block will move with acceleration F-Mg=Ma. The block moves up, then F reduces the magnitude. So the block will move. While the height of the liquid moves down too.
Actually, the block can oscillate because at equilibrium position it has inertial velocity. If you want it doesn't oscillate, implementation process (the objects are removed from the block) must be very slow.
Sorry because my English isn't good. I am very pleasure if you check my mistake

Last edited:
Soren4
Soren4 said:
In the solution I proposed the block does not move at all, it's the liquid that moves below it.
Consider total volume of liquid and block. Will that change?

Soren4
volume of liquid don't change because we don't pour water on. But the height of liquid changes

Soren4
Hamal_Arietis said:
volume of liquid don't change because we don't pour water on. But the height of liquid changes
As you can see, my question was posed to Soren4. Your post #2 gave too much instruction, effectively providing the answer. The style on these forums is to nudge the student in the right direction, e.g. by asking questions that lead them to understand where their thinking is wrong. That was the aim of my question.

Soren4
Sorry!The first I join this forum so I don't know this. Because I am not American so I worried that my guide makes them don't understand. So I must gave too much instruction, effectively providing the answer.

Soren4
Hamal_Arietis said:
Sorry!The first I join this forum so I don't know this. Because I am not American so I worried that my guide makes them don't understand. So I must gave too much instruction, effectively providing the answer.
That's ok, it's an easy mistake to make. Did you read the forum guidelines?

Soren4
Thanks a lot to both for your suggestions!

So actually both the block and the fluid do move.. Since my aim is to find the variation of fluid level I tried to interpret the phenomenon of case 1 as described by my answer in the question, now that I know that the block moves too. I made a drawing, maybe a little strange, I would like to ask you if that would be correct (as far as the exercise is concerned)

I marked the steps with letters
• ##A##: block and objects on liquid
• ##B## : object is removed. Here i foucused only on the motion of block. The block moved up of a certain height, leaving a blank space below it (again, of course that is not true, is just to understand the exercise). The height the block moved is such that the blank volume is the one called ##\Delta V## in my question.
• ##C## : Here I imagined that I can replace ##\Delta V## with a "slice" of fluid from below (in this way I can relate ##\Delta V## to the level of the liquid in the tank of section ##\mathcal{S}##)
• ##D## : ##\Delta V## is replaced so now there is another blank volume, equal to ##\Delta V##, below
• ##E##: I imagined to move everything, as a rigid body, down, to fill the new blank volume

Total result: the height of the block and the level of the fluid changed both.

As said above this is just an attempt to justify practically my answer, after having learned that the block moves. Would you be so kind as to suggest me if this is the solution I proposed is correct even if the block moves?

Soren4 said:
Thanks a lot to both for your suggestions!

So actually both the block and the fluid do move.. Since my aim is to find the variation of fluid level I tried to interpret the phenomenon of case 1 as described by my answer in the question, now that I know that the block moves too. I made a drawing, maybe a little strange, I would like to ask you if that would be correct (as far as the exercise is concerned)
View attachment 102591
I marked the steps with letters
• ##A##: block and objects on liquid
• ##B## : object is removed. Here i foucused only on the motion of block. The block moved up of a certain height, leaving a blank space below it (again, of course that is not true, is just to understand the exercise). The height the block moved is such that the blank volume is the one called ##\Delta V## in my question.
• ##C## : Here I imagined that I can replace ##\Delta V## with a "slice" of fluid from below (in this way I can relate ##\Delta V## to the level of the liquid in the tank of section ##\mathcal{S}##)
• ##D## : ##\Delta V## is replaced so now there is another blank volume, equal to ##\Delta V##, below
• ##E##: I imagined to move everything, as a rigid body, down, to fill the new blank volume

Total result: the height of the block and the level of the fluid changed both.

As said above this is just an attempt to justify practically my answer, after having learned that the block moves. Would you be so kind as to suggest me if this is the solution I proposed is correct even if the block moves?
Yes, that reasoning all looks correct.
How about case 2?

## 1. How does the floating block stay afloat when objects are thrown in the liquid?

The floating block stays afloat because of the principle of buoyancy. When the objects are thrown into the liquid, they displace a certain amount of liquid, creating an upward force on the block, which counteracts the downward force of gravity.

## 2. What factors affect the buoyancy of the floating block?

The buoyancy of the floating block is affected by the density of the liquid, the density of the block, and the volume of the block. The more dense the liquid, the more buoyant the block will be. Similarly, if the block has a greater density than the liquid, it will sink.

## 3. Can the floating block support objects of any weight?

No, the floating block can only support objects of a certain weight before it sinks. This is because the buoyant force is only strong enough to counteract a certain amount of weight. If the weight of the objects exceeds this limit, the block will sink.

## 4. Why do some objects sink while others float when thrown into the liquid?

Whether an object sinks or floats in a liquid depends on its density. Objects with a greater density than the liquid will sink, while those with a lower density will float. This is why some objects may sink while others float when thrown in the same liquid.

## 5. How can we calculate the buoyant force on the floating block?

The buoyant force on the floating block can be calculated using Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by the object. This means that the more fluid the object displaces, the greater the buoyant force will be.

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