Finding downward force on immersed object

[member 731016]

Homework Statement

Pls see below

Relevant Equations

Pls see below

For this task, we are meant to immerse objects (held by a mesh) in water to measure the buoyant force and deduce the object densities.

We can measure the weight force of the object, $F_g$, and tension T and buoyancy force. We need to find the downward force on the immersed object, $F_i$

However, I am unsure how to find the downward force $F_i$ on the immersed object. Applying Newton II in the y-direction

$T + F_j - F_i -F_g = 0$ where $T$ and $F_g$ and $B = F_j - F_i$ are knowns.

And solving, $T + F_j - F_g = F_i$, however, how do we solve this since $F_j$ and $F_i$ are unknowns.

EDIT: I think we can solve this actually since $F_j$ can be measured using a pressure sensor in the water.

Many thanks!

 

[haruspex]

Homework Statement:: Pls see below
Relevant Equations:: Pls see below

For this task, we are meant to immerse objects (held by a mesh) in water to measure the buoyant force and deduce the object densities.

We can measure the weight force of the object, $F_g$, and tension T and buoyancy force. We need to find the downward force on the immersed object, $F_i$

However, I am unsure how to find the downward force $F_i$ on the immersed object. Applying Newton II in the y-direction

$T + F_j - F_i -F_g = 0$ where $T$ and $F_g$ and $B = F_j - F_i$ are knowns.

And solving, $T + F_j - F_g = F_i$, however, how do we solve this since $F_j$ and $F_i$ are unknowns.

EDIT: I think we can solve this actually since $F_j$ can be measured using a pressure sensor in the water.

Many thanks!

I would need a diagram to understand why there are so many forces involved.

 

[member 731016]

I would need a diagram to understand why there are so many forces involved.

Thank you for your reply @haruspex!

I think my original confusion is solved, but here is a diagram with the force notation that I'm using superimposed.

Many thanks!

 

[haruspex]

Thank you for your reply @haruspex!

I think my original confusion is solved, but here is a diagram with the force notation that I'm using superimposed.
View attachment 323673
Many thanks!

I fail to see how that relates to the task described in post #1. That only has gravitational force, the restraining force from some mesh, and a deduced buoyancy force.

 

[member 731016]

I fail to see how that relates to the task described in post #1. That only has gravitational force, the restraining force from some mesh, and a deduced buoyancy force.

Thank you for your reply @haruspex! I will draw a new diagram. Turns out it more confusing that it looks since for the experiment we are meant to assume that the tension from the mesh is zero while calculating the downward force only from the water (they call that $F_i$).

Many thanks!

 

[member 731016]

Here is the diagram @haruspex,

Where $F_j - F_i = B$. The black line is a force sensor so we assume T = 0 then use the formula

To find the object density.

Many thanks!

 

[member 731016]

Apparently when the object is floating, the the value of read by the force sensor is equal to the downward force from the water. Would you please know how that is so?

This lead to data below. Please let me know if my question dose not make sense.

Many thanks!

Fg(N)	Fi	ρ(object) [kg/m^3] - using formula in above post
0.16​	0.11​	3190.4​
2.05​	1.75​	6812.833333​
0.75​	0.355​	1893.037975​
0.28​	0.17​	2537.818182​
0.15​	0.04​	1359.545455​
0.2​	0.14​	3323.333333​
1.34​	1.18​	8349.875​

 

[haruspex]

Here is the diagram @haruspex,

View attachment 323713
Where $F_j - F_i = B$. The black line is a force sensor so we assume T = 0 then use the formula

View attachment 323714
To find the object density.

Many thanks!

I cannot see the point of splitting the buoyancy into a downward force F_i on the top of the object and an upward force F_j on the bottom. In general, it would be very difficult to separate it like that; only for objects with horizontal surfaces top and bottom and vertical sides. Even then, you’d have to measure the depth at top and bottom. You don't need to that to solve the problem, just handle it as B.

Next, the last equation above seems like nonsense if the terms have the meanings you have stated. First, where did the minus sign come from? It appears B is positive up, so the equation would give a negative density.
Secondly, how did B turn into F_i-F_g?

You remark that T is taken to be zero, but since you have not defined T I cannot comment on that.

Here's how it should work:
You know the tension in the cable supporting the net. I will call that T. $B+T=mg$.
$\rho_{object}=\rho_{water}\frac{mg}B=\rho_{water}\frac{mg}{mg-T}$.

From that, it seems to me that your equation is using $F_i$ for the tension in the cable and $F_j$ to mean the same as $F_g$, which is the same as mg.

 

[member 731016]

I cannot see the point of splitting the buoyancy into a downward force F_i on the top of the object and an upward force F_j on the bottom. In general, it would be very difficult to separate it like that; only for objects with horizontal surfaces top and bottom and vertical sides. Even then, you’d have to measure the depth at top and bottom. You don't need to that to solve the problem, just handle it as B.

Next, the last equation above seems like nonsense if the terms have the meanings you have stated. First, where did the minus sign come from? It appears B is positive up, so the equation would give a negative density.
Secondly, how did B turn into F_i-F_g?

You remark that T is taken to be zero, but since you have not defined T I cannot comment on that.

Here's how it should work:
You know the tension in the cable supporting the net. I will call that T. $B+T=mg$.
$\rho_{object}=\rho_{water}\frac{mg}B=\rho_{water}\frac{mg}{mg-T}$.

From that, it seems to me that your equation is using $F_i$ for the tension in the cable and $F_j$ to mean the same as $F_g$, which is the same as mg.

Thank you for your reply @haruspex!

Sorry I define T to be the tension.

This is how the TA derived the equation I gave. The only assumption he said was that tension was zero.

Where $F_i$ is the downward force from the water.

I am also somewhat confused how when the object is in water when the net is slack (TA said assume tension = 0) that the force sensor reads the downward force from the water. I though it would read $F_j = F_i + mg$ where $F_j$ is the upwards force from the water.

Many thanks!

 

[haruspex]

Thank you for your reply @haruspex!

Sorry I define T to be the tension.

This is how the TA derived the equation I gave. The only assumption he said was that tension was zero.

Where $F_i$ is the downward force from the water.

I am also somewhat confused how when the object is in water when the net is slack (TA said assume tension = 0) that the force sensor reads the downward force from the water. I though it would read $F_j = F_i + mg$ where $F_j$ is the upwards force from the water.

Many thanks!

I remain bewildered by your explanations.

What do you mean by "downward force from the water"? We cannot know and do not care what the force from water pressure is on top of the objects. Is it possible that the TA is using $F_i$ to mean something else?

The only way the TA's algebra works is if $F_i$ is the upward force applied by the net. Even then, it takes two mistakes that cancel … unless B is defined as positive down.

If the net is slack then what is holding the objects up? Seems they must be less dense than water and are floating, in which case we cannot determine their density because some of their volume is not displacing water.

Putting all that aside, can you tell me how you’re measuring the $F_i$ in your table?

 

[kuruman]

Putting all that aside, can you tell me how you’re measuring the $F_i$ in your table?

What @Callumnc1 calls $F_i$ is what you call the tension when the objects are under water. The experiment apparently involves weighing different irregular objects (the red globs inside the mesh, post #6) and finding their density. The so called tension, which @Callumnc1 says is zero, is the reading of the force gauge. It is not zero unless the objects float.

Here is my interpretation of the column headings
$F_g$ in the data is the reading of the force gauge in air and includes the weight of the mesh.
$F_i$ in the data is the reading of the force gauge with the objects immersed.

Neglecting the buoyancy of the air when measuring $F_g$ and the buoyancy of the mesh when measuring $F_i$, the relevant equation is as presented in your post #8. You have $$\rho_{object}=\rho_{water}\frac{mg}B=\rho_{water}\frac{mg}{mg-T}.\tag{1}$$The column in the data labeled "ρ(object) [kg/m^3] - using formula in above post" is calculated using $$\rho_{object} =\rho_{water}\left(1-\frac{F_i}{F_g}\right)^{-1}.\tag{2}$$If you substitute $F_i=T$ and $F_g = mg$ and rearrange equation (2), you get equation (1).

I think @Callumnc1's confusion here likely arises from his belief that "the tension is zero" (post #5) and from his incomplete understanding of the FBD for an immersed object.

 

[member 731016]

I remain bewildered by your explanations.

What do you mean by "downward force from the water"? We cannot know and do not care what the force from water pressure is on top of the objects. Is it possible that the TA is using $F_i$ to mean something else?

Thank you for your reply @haruspex!

The downward force from the water is equal to the $P_iA$ where $P_i$ is the pressure acting on the top of the immersed object. Also, since the objects were completely immersed in water, $P_j > P_0$. We assume that area of the object is the same on each side.

The only way the TA's algebra works is if $F_i$ is the upward force applied by the net. Even then, it takes two mistakes that cancel … unless B is defined as positive down.

I am not sure. I believe they defined B to be positive up.

If the net is slack then what is holding the objects up? Seems they must be less dense than water and are floating, in which case we cannot determine their density because some of their volume is not displacing water.

Sorry I think I wrote this part wrong. The net is actually not slack for most of the objects we immersed in the water, however, we are ignoring the tension since the TA said that it would only change the answer by something in the order of ±0.01%

Putting all that aside, can you tell me how you’re measuring the $F_i$ in your table?

So the apparatus we used was a stand with the force sensor attached hanging from the top. The object that we were trying to measure the density of would be attached to the force sensor using a net.

Each time we ran the experiment, we first took one reading for the F_g(N) column which was the reading from the sensor when the object was hanging from the force sensor without being in the water. This part we assume that the tension is non-zero so T = mg.

For the F_i column the TA said that when the object is completely submerged in the water, then we assume T is zero (even when the net is not slack in the case where the object is denser than water), and the force sensor measures $F_i$. That is the very confusing part. I thought the sensor would read $B - mg = F_j - F_i - mg$.

Many thanks!

 

[member 731016]

What @Callumnc1 calls $F_i$ is what you call the tension when the objects are under water. The experiment apparently involves weighing different irregular objects (the red globs inside the mesh, post #6) and finding their density. The so called tension, which @Callumnc1 says is zero, is the reading of the force gauge. It is not zero unless the objects float.

Thank you for your reply @kuruman!

Sorry, I did not specify, only one of the objects was irregular (it was a knuckle bone), the rest were rectangular and spherical in shape.

Here is my interpretation of the column headings
$F_g$ in the data is the reading of the force gauge in air and includes the weight of the mesh.
$F_i$ in the data is the reading of the force gauge with the objects immersed.

You interpretation is correct. However, my confusion is how $F_i$ can be the reading on the force gauge when the objects are immersed. It seems strange since I thought the sensor would read $B - mg$.

Neglecting the buoyancy of the air when measuring $F_g$ and the buoyancy of the mesh when measuring $F_i$, the relevant equation is as presented in your post #8. You have $$\rho_{object}=\rho_{water}\frac{mg}B=\rho_{water}\frac{mg}{mg-T}.\tag{1}$$The column in the data labeled "ρ(object) [kg/m^3] - using formula in above post" is calculated using $$\rho_{object} =\rho_{water}\left(1-\frac{F_i}{F_g}\right)^{-1}.\tag{2}$$If you substitute $F_i=T$ and $F_g = mg$ and rearrange equation (2), you get equation (1).

I think @Callumnc1's confusion here likely arises from his belief that "the tension is zero" (post #5) and from his incomplete understanding of the FBD for an immersed object.

It will actually the TA's assumption that tension is zero. I thought we could do it with the tension using the pressure sensor since we know the area of most of the objects.

Many thanks!

 

[haruspex]

What @Callumnc1 calls $F_i$ is what you call the tension when the objects are under water. The experiment apparently involves weighing different irregular objects (the red globs inside the mesh, post #6) and finding their density. The so called tension, which @Callumnc1 says is zero, is the reading of the force gauge. It is not zero unless the objects float.

which is what I wrote at the end of post #8, but your confirmation is welcome.

 

[kuruman]

However, my confusion is how $F_i$ can be the reading on the force gauge when the objects are immersed. It seems strange since I thought the sensor would read $B - mg$.

The sensor reads whatever force is pulling on its end. It doesn't know that there is an immersed object at he other end. If the object is immersed in a fluid, it would display $mg -B$, but if there is no fluid, it would display just $mg$. It is you, not the sensor, that interprets the displayed value as $mg-B$ when the mass is immersed and as $mg$ when it is not.

 

[kuruman]

which is what I wrote at the end of post #8, but your confirmation is welcome.

So you did. I lost track of it, what with the $F_i~$, $F_j$ and the upward and downward forces exerted by the fluid mentioned in post #10.

 

[member 731016]

The sensor reads whatever force is pulling on its end. It doesn't know that there is an immersed object at he other end. If the object is immersed in a fluid, it would display $mg -B$, but if there is no fluid, it would display just $mg$. It is you, not the sensor, that interprets the displayed value as $mg-B$ when the mass is immersed and as $mg$ when it is not.

Thank you for your reply @kuruman!

I now think when the TA said ignore tension, he was referring to the tension in the net. For the force sensor-object system to remain in vertical equilibrium while the object is immersed in the water, then the force exerted $F$ by the force sensor must be $F = mg - B$ from Newton II: $F + B - mg = 0$. I assume that the value displayed by force sensor is $F$.

If we write the $B$ in terms of upwards force from the water $F_f = P_fA$ and the downwards force from the water $F_i = P_iA$ then, $F = mg - (F_f - F_i)$ which gives $F = mg - F_f + F_i$.

According to the last equation, the force sensor only reads the downward force from the water $F_i$ when $F_f = mg$ which gives $F = F_i$. However, is $F_f = mg$ a valid assumption to make for each object we immerse?

Many thanks!

 

[haruspex]

my confusion is how Fi can be the reading on the force gauge when the objects are immersed. It seems strange since I thought the sensor would read B−mg.

No, the reading would be mg-B. Without the water there is no buoyancy, B=0, and the reading is mg.
I have no idea why the TA is naming the reading $F_i$, or why you think it has anything to do with the diagram in post #3. Nothing in the print image refers to $F_i$ or $F_j$. That seems to be a confusion of your own.
That image is entirely irrelevant to your experiment for reasons I already gave.

the TA said that it would only change the answer by something in the order of ±0.01%

That suggests to me that the TA is distinguishing the upward force the net exerts on the objects (and naming that $F_i$) from the force reading on the sensor, T. The difference would be the weight of the cable and net, less the buoyant force on the submerged part of the net. It's not that T=0, but that the difference between T and $F_i$ would be extremely small. Hence, you are to use the reading as though it is $F_i$.

Turning to your table of results, I presume that each row corresponds to a different object or collection of objects, so the densities can be quite different. That being the case, I see no way to evaluate the experimental results. There ought to be another column identifying the object.
You should not show so many significant figures in the density column. At most, the number of them should be the lower of those in the first two columns; mostly that's three.
But given the subtraction in the formula, even that may be too many. I'll explain why in a later post.

 

[member 731016]

No, the reading would be mg-B. Without the water there is no buoyancy, B=0, and the reading is mg.
I have no idea why the TA is naming the reading $F_i$, or why you think it has anything to do with the diagram in post #3. Nothing in the print image refers to $F_i$ or $F_j$. That seems to be a confusion of your own.
That image is entirely irrelevant to your experiment for reasons I already gave.

Thank you for your reply @haruspex!

Sorry here is a free body diagram once the object is immersed in the water (whilst been inside the net):

Many thanks!

 

[member 731016]

That suggests to me that the TA is distinguishing the upward force the net exerts on the objects (and naming that $F_i$) from the force reading on the sensor, T. The difference would be the weight of the cable and net, less the buoyant force on the submerged part of the net. It's not that T=0, but that the difference between T and $F_i$ would be extremely small. Hence, you are to use the reading as though it is $F_i$.

Thank you for your reply @haruspex!

Sorry I don't understand how the TA would define the upward force the net exerts on the objects as $F_i$. The Professor already defined $F_i$ to be is the downwards force from the water.

I think the TA (please see post #17), was referring to the upwards force $T$ from the net to be zero when he was talking about the tension from the net.

Many thanks!

 

[member 731016]

Turning to your table of results, I presume that each row corresponds to a different object or collection of objects, so the densities can be quite different. That being the case, I see no way to evaluate the experimental results. There ought to be another column identifying the object.
You should not show so many significant figures in the density column. At most, the number of them should be the lower of those in the first two columns; mostly that's three.
But given the subtraction in the formula, even that may be too many. I'll explain why in a later post.

Thank you for your reply @haruspex!

You are correct. We had a column of objects names. We first calculated the density theoretically using dimensions and mass and then used the immersion experiment and compared the two results.

Many thanks!

 

[haruspex]

Thank you for your reply @haruspex!

Sorry here is a free body diagram once the object is immersed in the water (whilst been inside the net):
View attachment 323748
Many thanks!

where $F_i$ and$F_j$ are what?

If you are going to say they are the vertical forces exerted by the water on the top and bottom of the objects then a) what about the net? b) how do you have any hope of figuring out what their values are when the object is irregular in shape?

Forget the diagram in post #3, it is irrelevant to the thread.

 

[haruspex]

Sorry I don't understand how the TA would define the upward force the net exerts on the objects as $F_i$. The Professor already defined $F_i$ to be is the downwards force from the water.

Did the professor do that in the context of this problem or in the context of the diagram in post #3? Unless you have a solid reason to believe that the TA means $F_i$ to be the same as that marked on the diagram post #3, don’t assume it is.
If that is what the TA means then he/she is a nincompoop and would be better ignored.

Btw, according to your notes on that diagram, $F_i$ is the upward force from the water on the bottom of the object. This conflicts with your FBD. It also creates the possibility that the TA is using it to mean the net upward force, B, from the water, i.e. corresponds to $F_i-F_j$ in the diagram.

If you will not accept what @kuruman and I are telling you then I do not see how we can assist further.

 

[kuruman]

$\dots~$ then I do not see how we can assist further.

Being an incorrigible optimist, I will try one more time.
To @Callumnc1 :
When an object is under the influence of forces exerted on it by entities in the environment, one draws a free body diagram (FBD). In that diagram puts all the forces that the entities exert on the object. It is very important to note that there is a single force corresponding to each entity because if an entity exerted two forces on the same object then the two can always be combined to their resultant which is one. With me so far?

First consider the mass suspended in water and supported by a mesh. There are three and only three entities that exert forces on the mass. These are

1. Earth. It exerts a force $W$ down. It is equal to the weight of the mass $mg$.
2. Fluid. It exerts force $B$ (the buoyant force) up. It is equal to the weight of the displaced water.
3. Mesh. It exerts a (normal) force $N$ up. It is the force that keeps the mass from sinking all the way to the bottom.

The mass does not accelerate. This means that the sum of all the forces acting on it is zero: $$-mg+B+N=0 \implies N=mg-B.\tag 1$$ Now consider a second system consisting of the mesh and the mass. The entities acting on this two-component system are

1. Earth. It exerts a force $W$ down. It is equal to the weight of the mass $mg$ assuming that the mesh has negligible mass.
2. Fluid. It exerts force $B$ (the buoyant force) up. It is equal to the weight of the displaced water but the mass assuming that the mesh displaces not water.
3. String. It is tied to the mesh and exerts a (tension) force $T$ up. It is the force that keeps the mesh and its contents from sinking all the way to the bottom. It is also the force recorded by the gauge.

The system does not accelerate. This means that the sum of all the forces acting on it is zero: $$-mg+B+T=0 \implies T=mg-B.\tag 2$$ Comparison of equations (1) and (2) shows that $T=N$.

Do you see now how the forces and FBDs are put together here? If so, I recommend that you abandon $F_i$, $F_j$ etc.

 

[member 731016]

where $F_i$ and$F_j$ are what?

If you are going to say they are the vertical forces exerted by the water on the top and bottom of the objects then a) what about the net? b) how do you have any hope of figuring out what their values are when the object is irregular in shape?

Forget the diagram in post #3, it is irrelevant to the thread.

Thank you for your reply @haruspex!

I am very sorry about diagram in post #3. I just realized the labels for the force are the wrong way round. My bad. I will forget about that diagram. We will use the FBD from now on which agrees with the course readings.

$F_i$ = downward force exerted from water on top of immersed object

$F_j$ = upward force exerted from water on bottom of immersed object

a) The net of the forces from the water acting on the object is the buoyancy force. $B = F_j - F_i$.

b) I am not sure. We only had one irregular object.

Many thanks!

 

[member 731016]

Did the professor do that in the context of this problem or in the context of the diagram in post #3? Unless you have a solid reason to believe that the TA means $F_i$ to be the same as that marked on the diagram post #3, don’t assume it is.
If that is what the TA means then he/she is a nincompoop and would be better ignored.

Btw, according to your notes on that diagram, $F_i$ is the upward force from the water on the bottom of the object. This conflicts with your FBD. It also creates the possibility that the TA is using it to mean the net upward force, B, from the water, i.e. corresponds to $F_i-F_j$ in the diagram.

If you will not accept what @kuruman and I are telling you then I do not see how we can assist further.

Thank you for your reply @haruspex!

The Professor did in context of this problem. The diagram in post #3 is wrong. Thank you letting me know about the diagram.

Many thanks!

 

[member 731016]

Being an incorrigible optimist, I will try one more time.
To @Callumnc1 :
When an object is under the influence of forces exerted on it by entities in the environment, one draws a free body diagram (FBD). In that diagram puts all the forces that the entities exert on the object. It is very important to note that there is a single force corresponding to each entity because if an entity exerted two forces on the same object then the two can always be combined to their resultant which is one. With me so far?

First consider the mass suspended in water and supported by a mesh. There are three and only three entities that exert forces on the mass. These are

1. Earth. It exerts a force $W$ down. It is equal to the weight of the mass $mg$.
2. Fluid. It exerts force $B$ (the buoyant force) up. It is equal to the weight of the displaced water.
3. Mesh. It exerts a (normal) force $N$ up. It is the force that keeps the mass from sinking all the way to the bottom.

The mass does not accelerate. This means that the sum of all the forces acting on it is zero: $$-mg+B+N=0 \implies N=mg-B.\tag 1$$ Now consider a second system consisting of the mesh and the mass. The entities acting on this two-component system are

1. Earth. It exerts a force $W$ down. It is equal to the weight of the mass $mg$ assuming that the mesh has negligible mass.
2. Fluid. It exerts force $B$ (the buoyant force) up. It is equal to the weight of the displaced water but the mass assuming that the mesh displaces not water.
3. String. It is tied to the mesh and exerts a (tension) force $T$ up. It is the force that keeps the mesh and its contents from sinking all the way to the bottom. It is also the force recorded by the gauge.

The system does not accelerate. This means that the sum of all the forces acting on it is zero: $$-mg+B+T=0 \implies T=mg-B.\tag 2$$ Comparison of equations (1) and (2) shows that $T=N$.

Do you see now how the forces and FBDs are put together here? If so, I recommend that you abandon $F_i$, $F_j$ etc.

Thank you for your reply @kuruman!

I see how the forces are put together. That is quite interesting since I have never seen that before. However, why would abandon $F_i$ since that is what we are trying to measure.

Many thanks!

 

[kuruman]

However, why would abandon $F_i$ since that is what we are trying to measure.

OK, don't abandon $F_i$, you can give it any name you wish. However, since you see how the forces are put together, please tell me which entity exerts this force $F_i$. I want to make sure that you truly understand. You have three choices
(A) Earth
(B) Fluid
(C) Mesh (or net)

 

[member 731016]

OK, don't abandon $F_i$, you can give it any name you wish. However, since you see how the forces are put together, please tell me which entity exerts this force $F_i$. I want to make sure that you truly understand. You have three choices
(A) Earth
(B) Fluid
(C) Mesh (or net)

Thank you for your reply @kuruman!

(B) Fluid.

Many thanks!

 

[kuruman]

(B) Fluid.

If it's the force exerted by the fluid, then it must be the buoyant force. However, the buoyant force is directed up. If I look at your drawing in post #19, I see that you have drawn $F_i$ down. Here are more choices in view of your answer and that drawing:
(A) $F_i$ should have been drawn up instead of down in order to represent the buoyant force correctly.
(B) $F_i$ is not the buoyant force and has no place in the FBD.
(C) $F_i$ is correctly drawn but should be identified as the weight because the Earth is the only entity that exerts a down force on the mass.

 

[member 731016]

If it's the force exerted by the fluid, then it must be the buoyant force. However, the buoyant force is directed up. If I look at your drawing in post #19, I see that you have drawn $F_i$ down. Here are more choices in view of your answer and that drawing:
(A) $F_i$ should have been drawn up instead of down in order to represent the buoyant force correctly.
(B) $F_i$ is not the buoyant force and has no place in the FBD.
(C) $F_i$ is correctly drawn but should be identified as the weight because the Earth is the only entity that exerts a down force on the mass.

Thank you for your reply @kuruman!

Ok so it is definitely not (A). I don't think it (B) either since $F_i$ is not a buoyant force but one of the two forces from the water that add to form a resultant force we call the buoyant force.

So it must be (C). It seems strange, I am not use to the concept. How could $F_i$ be identified as a weight?

Many thanks!

 

[haruspex]

Ok so it is definitely not (A).

Why not?

 

[kuruman]

Ok so it is definitely not (A). I don't think it (B) either since $F_i$ is not a buoyant force but one of the two forces from the water that add to form a resultant force we call the buoyant force.

So it must be (C). It seems strange, I am not use to the concept. How could $F_i$ be identified as a weight?

As you contemplate answering @haruspex's question in post #32, also think about your BFD in post #19. It shows two forces directed down and one force directed up. In post #27 you claimed

I see how the forces are put together.

If that is indeed the case, then you should agree that the correct FBD should show two forces directed up (mesh and fluid) and one directed down (Earth.) Remember, that you should draw one arrow for each entity exerting a force. So let's see you post a correct FBD with 3 arrows, 2 up and 1 down labeled with the names of the entities that exert the forces represented by these arrows.

To help guide your thinking, I show below a FBD of the block of fluid in the figure from post #3. There are two entities exerting a force on this block, Earth (down) and Fluid (up). It says that "water floats on water" and that the two forces have equal magnitudes and opposite directions. If I label the force exerted by the Earth $mg$ and the force exerted by the fluid $BF$, it follows that $BF=mg.$ Note that, for this particular situation, the force exerted by the fluid is the sum of an up force exerted by the fluid at the bottom of the block, $F_B=p_BA$ and a down force exerted by the fluid at the top of the block, $F_T=-p_TA.$ The single arrow representing the net force exerted by the fluid on the block is the sum of the two which is called the buoyant force, $BF=p_BA-p_TA.$

For some reason that I do not understand, your professor and/or your TA impressed upon you that $F_T$ is important in the analysis of your experiment. In my opinion it is not and I think @haruspex agrees with me. I recommended that you abandon $F_i$ and $F_j$ because they only serve to confuse you and prevent you from seeing clearly what is going on. Their usefulness extends no further than the derivation of the Archimedes principle, namely that the buoyant force is equal to the weight of the displaced fluid. That's all you need to draw correct FBDs related to floating or immersed objects as in the case of your experiment.

 

[haruspex]

Their usefulness extends no further than the derivation of the Archimedes principle,

Not even that; it merely demonstrates the principle in one simple case. The derivation of the principle was surely a thought experiment.
Archimedes must have imagined replacing a floating body with water that would fill its submerged portion. That parcel of water would obviously float and be subject to the same forces from the surrounding water at every point. Therefore the buoyant force equals the weight of that parcel.

 

[member 731016]

Why not?

As you contemplate answering @haruspex's question in post #32, also think about your BFD in post #19. It shows two forces directed down and one force directed up. In post #27 you claimed

If that is indeed the case, then you should agree that the correct FBD should show two forces directed up (mesh and fluid) and one directed down (Earth.) Remember, that you should draw one arrow for each entity exerting a force. So let's see you post a correct FBD with 3 arrows, 2 up and 1 down labeled with the names of the entities that exert the forces represented by these arrows.

To help guide your thinking, I show below a FBD of the block of fluid in the figure from post #3. There are two entities exerting a force on this block, Earth (down) and Fluid (up). It says that "water floats on water" and that the two forces have equal magnitudes and opposite directions. If I label the force exerted by the Earth $mg$ and the force exerted by the fluid $BF$, it follows that $BF=mg.$ Note that, for this particular situation, the force exerted by the fluid is the sum of an up force exerted by the fluid at the bottom of the block, $F_B=p_BA$ and a down force exerted by the fluid at the top of the block, $F_T=-p_TA.$ The single arrow representing the net force exerted by the fluid on the block is the sum of the two which is called the buoyant force, $BF=p_BA-p_TA.$
View attachment 323780
For some reason that I do not understand, your professor and/or your TA impressed upon you that $F_T$ is important in the analysis of your experiment. In my opinion it is not and I think @haruspex agrees with me. I recommended that you abandon $F_i$ and $F_j$ because they only serve to confuse you and prevent you from seeing clearly what is going on. Their usefulness extends no further than the derivation of the Archimedes principle, namely that the buoyant force is equal to the weight of the displaced fluid. That's all you need to draw correct FBDs related to floating or immersed objects as in the case of your experiment.

Not even that; it merely demonstrates the principle in one simple case. The derivation of the principle was surely a thought experiment.
Archimedes must have imagined replacing a floating body with water that would fill its submerged portion. That parcel of water would obviously float and be subject to the same forces from the surrounding water at every point. Therefore the buoyant force equals the weight of that parcel.

Thank you for your replies @haruspex and @kuruman!

I have got assignments to do at the moment, but I will come back to this thread when I have more time to think about the physics without being under time pressure.

Many thanks!