Finding downward force on immersed object

[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!

 

[haruspex]

At most, the number of [significant figures] 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.

When a formula involves taking the difference between two positive values, the potential fractional error in the result exceeds the sum of the fractional errors in the operands.
Your formula involves $1-\frac{F_i}{F_g}$. The first data pair in your table is $F_i=0.11, F_g=0.16$. That implies $0.105<F_i<0.115, 0.155<F_g<0.165$, i.e. fractional errors of $\pm 4.5\%, \pm 3.1\%$. The standard rule adds those to produce an error of $\pm 7.6\%$ for $\frac{F_i}{F_g}$, i.e. a value from 0.64 to 0.74. So $0.26<1-\frac{F_i}{F_g}<0.36$, an error of $\pm 15$%. As a result, the answer should be given as $3\cdot 10^3$.

 

[member 731016]

When a formula involves taking the difference between two positive values, the potential fractional error in the result exceeds the sum of the fractional errors in the operands.
Your formula involves $1-\frac{F_i}{F_g}$. The first data pair in your table is $F_i=0.11, F_g=0.16$. That implies $0.105<F_i<0.115, 0.155<F_g<0.165$, i.e. fractional errors of $\pm 4.5\%, \pm 3.1\%$. The standard rule adds those to produce an error of $\pm 7.6\%$ for $\frac{F_i}{F_g}$, i.e. a value from 0.64 to 0.74. So $0.26<1-\frac{F_i}{F_g}<0.36$, an error of $\pm 15$%. As a result, the answer should be given as $3\cdot 10^3$.

Thank you for you reply @haruspex!

I will look into that.

Many thanks!

 

[kuruman]

$\dots~$ but I will come back to this thread when I have more time to think about the physics without being under time pressure.

While you do this consider why you latched on the idea that the downward force exerted by the fluid is important. Here is what is going on.

Let
$F_0$ = force recorded with the mass hanging in air = $mg=\rho_{\text{mass}}Vg.$
$F$ = force recorded with the mass hanging under water = $mg-B=\rho_{\text{mass}}Vg-\rho_{\text{water}}Vg.$

Divide the bottom by the top equation, $$\frac{F}{F_0}=\frac{\rho_{\text{mass}}\cancel{Vg}-\rho_{\text{water}}\cancel{Vg}}{\rho_{\text{mass}}\cancel{Vg}}=1-\frac{\rho_{\text{water}}}{\rho_{\text{mass}}}.$$Then, $$\begin{align}
& \frac{\rho_{\text{water}}}{\rho_{\text{mass}}}=1-\frac{F}{F_0} \nonumber \\
& \rho_{\text{mass}}=\rho_{\text{water}}\left(1-\frac{F}{F_0}\right)^{-1}.
\nonumber\end{align}$$Note that the last equation is the formula that you used to calculate the entries under the heading "ρ(object) [kg/m^3] - using formula in above post", except that I used the symbol $F$ and you used the symbol $F_i$. What these symbols stand for is the same force, namely the reading of the gauge with the object immersed.

Your claim

We need to find the downward force on the immersed object, $F_i~\dots$

is baseless and shows that you lost track of what you set out to determine with this experiment. You need to find the unknown density of the masses in the net by measuring and recording (a) the force $F_0$ (or $mg$) that they exert on the gauge when suspended in air and (b) the force $F$ (or $F_i$) that they exert on the gauge when suspended under water.

I will say no more on this thread because there is nothing left I can say. As someone remarked recently, "You can teach a hungry man how to fish and lead him to the water, but you can't make him fish."

 

[hmmm27]

As someone remarked recently, "You can teach a hungry man how to fish and lead him to the water, but you can't make him fish."

Nono, that's "You can give a horse a fish, but you can't make him swim."

 

[member 731016]

While you do this consider why you latched on the idea that the downward force exerted by the fluid is important. Here is what is going on.

Let
$F_0$ = force recorded with the mass hanging in air = $mg=\rho_{\text{mass}}Vg.$
$F$ = force recorded with the mass hanging under water = $mg-B=\rho_{\text{mass}}Vg-\rho_{\text{water}}Vg.$

Divide the bottom by the top equation, $$\frac{F}{F_0}=\frac{\rho_{\text{mass}}\cancel{Vg}-\rho_{\text{water}}\cancel{Vg}}{\rho_{\text{mass}}\cancel{Vg}}=1-\frac{\rho_{\text{water}}}{\rho_{\text{mass}}}.$$Then, $$\begin{align}
& \frac{\rho_{\text{water}}}{\rho_{\text{mass}}}=1-\frac{F}{F_0} \nonumber \\
& \rho_{\text{mass}}=\rho_{\text{water}}\left(1-\frac{F}{F_0}\right)^{-1}.
\nonumber\end{align}$$Note that the last equation is the formula that you used to calculate the entries under the heading "ρ(object) [kg/m^3] - using formula in above post", except that I used the symbol $F$ and you used the symbol $F_i$. What these symbols stand for is the same force, namely the reading of the gauge with the object immersed.

Your claim

is baseless and shows that you lost track of what you set out to determine with this experiment. You need to find the unknown density of the masses in the net by measuring and recording (a) the force $F_0$ (or $mg$) that they exert on the gauge when suspended in air and (b) the force $F$ (or $F_i$) that they exert on the gauge when suspended under water.

I will say no more on this thread because there is nothing left I can say. As someone remarked recently, "You can teach a hungry man how to fish and lead him to the water, but you can't make him fish."

Thank you for your reply @kuruman!

Your derivation is very helpful!

Now I just trying to understand why the force read on gauge when the object is immersed in equal $F$ is equal to downward force exerted by the water $F_i$ when I compare it to TA's derivation shown in post #9. I think this is the hardest part.

When I go about trying to see this is true $F = mg - B = mg - [F_j - F_i] = mg - F_j + F_i$ it seems very hard.

I don't understand why the professor wrote the equation in terms of $F_i$ when it very hard to see how this is equal to the value read by the force gauge when the object is immersed in the water.

Many thanks!

 

[haruspex]

Now I just trying to understand why the force read on gauge when the object is immersed in equal F is equal to downward force exerted by the water …. I think this is the hardest part.

It's hard to understand because it is not true. So for goodness sake stop trying to understand it.

I think we already established that you introduced a confusion between the force directions: $F_i$ in post #3 is an upward force. But merely correcting that is not enough.
To make the TA's equations work, $F_i$ there must mean the upward force from the net net buoyancy force, i.e. in terms of post #3, $F_i-F_j$.
As I have explained several times in different ways, either your TA is using $F_i$ that way or TA in this case stands for Technical Ass, never to be Trusted Again.
If you wish to determine which, ask the TA whether it is the same as in post #3.

I cannot tell why your prof even bothered to show the diagram in post #3. It really is not very helpful. E.g., change the shape to a sphere. Now you have radially inward forces all over the surface, strongest at the bottom and weakest at the top. The buoyancy force now would have to be computed as an integral. Imagine doing that for the bones.
The power of Archimedes' principle is that you don't have to do the integral if you only need to relate the buoyancy force to the volume.

 

[kuruman]

$\dots~$ TA in this case stands for Technical Ass, never to be Trusted Again.

That would be (TA)².

If you wish to determine which, ask the TA whether it is the same as in post #3.

I wouldn't recommend that. @Callumnc1 is confused enough as it is.

 

[member 731016]

It's hard to understand because it is not true. So for goodness sake stop trying to understand it.

I think we already established that you introduced a confusion between the force directions: $F_i$ in post #3 is an upward force. But merely correcting that is not enough.
To make the TA's equations work, $F_i$ there must mean the net buoyancy force, i.e. in terms of post #3, $F_i-F_j$.
As I have explained several times in different ways, either your TA is using $F_i$ that way or TA in this case stands for Technical Ass, never to be Trusted Again.
If you wish to determine which, ask the TA whether it is the same as in post #3.

I cannot tell why your prof even bothered to show the diagram in post #3. It really is not very helpful. E.g., change the shape to a sphere. Now you have radially inward forces all over the surface, strongest at the bottom and weakest at the top. The buoyancy force now would have to be computed as an integral. Imagine doing that for the bones.
The power of Archimedes' principle is that you don't have to do the integral if you only need to relate the buoyancy force to the volume.

Thank you for your reply @haruspex!

Sorry my prof did not show the diagram in post #3. That was diagram was incorrectly drawn by me.

However, the professor did show the equation that the TA derived:

Maybe I should ask him about it.

Many thanks!

 

[member 731016]

That would be (TA)².

I wouldn't recommend that. @Callumnc1 is confused enough as it is.

Thank you for your reply @kuruman!

 

[member 731016]

I would like to thank you both in alphabetical order (@haruspex and @kuruman) very much for your help! You worked very hard to fix a derivation that was probably wrong in the Manuel for starters. However, all is not wasted, I have learnt a lot from your replies. And I still plan to get back to some of the later replies in the 1st page.

This has also been a long thread and the only way I can see it getting resolved is if I get the professor who wrote that in the image and ask him to derive it himself. That way, hopefully, we can get a fourth opinion on the derivation.

I have emailed the professor and I will post the outcome on this thread when he gets back to me.

Thanks again!

 

[haruspex]

my prof did not show the diagram in post #3… However, the professor did show the equation that the TA derived:
View attachment 323847
Maybe I should ask him about it.

So how did he define $F_i$?

Btw, I was careless in post #41. Please see the edit.

 

[member 731016]

So how did he define $F_i$?

Btw, I was careless in post #41. Please see the edit.

Thank you for your reply @haruspex!

He did not redefine $F_i$

Many thanks!

 

[haruspex]

He did not redefine $F_i$

But you wrote

my prof did not show the diagram in post #3.

So where did the prof define it?

 

[member 731016]

But you wrote

So where did the prof define it?

Thank you for your reply @haruspex!

Sorry but I do not have derivation yet. Apparently Fi = Fg - B (not sure how) then rearrange for B and fellow though with algebra should give the expression in that image.

Where Fi is the downwards force from the water (not redefining)

Many thanks!

 

[member 731016]

But you wrote

So where did the prof define it?

Sorry here is the reasoning for $F_i = F_g - B$:

"$F_i = F_g - B$ (as B pushes up)"

Many thanks!

 

[member 731016]

In response to "You should have F_i = F_g - B (as B pushes up)":

I said:

"How dose F_i = F_g - B?

When I apply Newton II to the immersed object (Assuming T = 0) then B - F_g = 0 so B = mg. Then we try to decompose B into F_i the downward force from the water and F_j the upward force from the water. We get F_j - F_i = mg. However, if F_i = F_g - B as you said then:

F_i = F_g - (F_j - F_i)
-mg + F_i = -F_j+ F_i
mg = F_j

Now I can see a problem. Why dose your relation F_i = F_g - B give the upwards force from the water (F_j) to be equal to the force of gravity (F_g = mg)? It clear that if you derivation is correct then you would be neglecting the downwards force from the water (F_i). Is this a valid assumption?"

I think it clear that their derivation at least not accurate as it could be, would you agree @haruspex and @kuruman?

Many thanks!

 

[haruspex]

Thank you for your reply @haruspex!

Sorry but I do not have derivation yet. Apparently Fi = Fg - B (not sure how) then rearrange for B and fellow though with algebra should give the expression in that image.

Where Fi is the downwards force from the water (not redefining)

Many thanks!

I asked how the prof defined $F_i$. Did the prof actually define it as "the downward force from the water"? If not, what makes you think that is what it represents? You keep saying it is but do not back it up with any evidence (and we keep telling you it cannot be that).
If Fi = Fg - B then Fi+B=Fg, which means Fi must be the upward force the net exerts on the object. It cannot be anything else.

 

[member 731016]

I asked how the prof defined $F_i$. Did the prof actually define it as "the downward force from the water"? If not, what makes you think that is what it represents? You keep saying it is but do not back it up with any evidence (and we keep telling you it cannot be that).
If Fi = Fg - B then Fi+B=Fg, which means Fi must be the upward force the net exerts on the object. It cannot be anything else.

Thank you for your reply! Sorry @haruspex.

Two pieces of evidence:

(1) So it said that Fi is the downward force from the water on the lab script written by the prof:

(2) I said it in email when I was stating my assumptions and the prof had no problem with it:

"Thanks for your email!

B is the net force from the water. If follow the convection written in the main script then F_i is the downward force from the water and F_j is the upward force from the water. Therefore B = F_j - F_i (in the i-hat direction).

I don't have a derivation that match's yours.

Kind regards,
Callum "

Prof replies (with some personal parts excluded):

"
Hi,

You should have Fi = Fg - B (as B pushes up)

Then rearrange for B. Follow through with the algebra.

Hopefully this helps."

Is that enough evidence?

Many thanks!

 

[member 731016]

If Fi = Fg - B then Fi+B=Fg, which means Fi must be the upward force the net exerts on the object. It cannot be anything else.

I agree totally with that statement. But it seems that Fi that they really are defining Fi to be the downward force from the water which leads to my reasoning in post #51. Would you agree with post #52 given that Fi is the downward force from the water?

Many thanks!

 

[kuruman]

So how did he define $F_i$?

A clue to that is the sloppy derivation, item 4, quoted by @Callumnc1 in post #43. We have
Step 1
$\rho_{\text{object}}=\dfrac{m}{V}.$
This is OK.

Step 2
$\dfrac{m}{V_{\text{object}}}=-\dfrac{m \rho_{\text{water}} g}{B}.$
To see where this comes from, note that $B=\rho_{\text{water}} V_{\text{object}} g \implies V_{\text{object}}=\dfrac{B}{\rho_{\text{water}} g}.$
Replacing that in the denominator gives $\dfrac{m}{V_{\text{object}}}= \dfrac{m \rho_{\text{water}} g}{B}.$
There should be no negative sign because that would make the density on the left hand side negative. The buoyant force $B$ cannot be negative because it is equal to $\rho_{\text{water}} V_{\text{object}} g$, which is positive.

Step 3
$\dfrac{m\rho_{\text{water}} g}{B}=\dfrac{F_g}{F_i-F_g}\rho_{\text{water}}.$ (I dropped the negative signs that appear on both sides of the equation.)
Here the buoyant force in the denominator has been replaced by the difference between the weight in air and the mysterious $F_i$. Note that $$\frac{F_g}{F_i-F_g}\rho_{\text{water}}=\left( \frac{F_i-F_g}{F_g} \right)^{-1}\rho_{\text{water}}=\left( \frac{F_i}{F_g}-1 \right)^{-1}\rho_{\text{water}}.$$ This is the negative of what the professor expected to get and explains the negative sign in Step 2. It is a fudge factor to make the answer come right.

When replacing the buoyant force with a difference, that difference should have been $F_g-F_i$ where $F_i$ is the reading of the force gauge with the mass under water, also the tension in the string from which the mass is hanging, also the force that the net exerts on the mass. Where this comes from is shown in post #24, equations (1) and (2).

To @Callumnc1 : Look at my derivation in post #38 and forget the other mess. $F_i$ is the force that you recorded with the masses under water. You don't have to calculate it because you measured it.

 

[member 731016]

A clue to that is the sloppy derivation, item 4, quoted by @Callumnc1 in post #43. We have
Step 1
$\rho_{\text{object}}=\dfrac{m}{V}.$
This is OK.

Step 2
$\dfrac{m}{V_{\text{object}}}=-\dfrac{m \rho_{\text{water}} g}{B}.$
To see where this comes from, note that $B=\rho_{\text{water}} V_{\text{object}} g \implies V_{\text{object}}=\dfrac{B}{\rho_{\text{water}} g}.$
Replacing that in the denominator gives $\dfrac{m}{V_{\text{object}}}= \dfrac{m \rho_{\text{water}} g}{B}.$
There should be no negative sign because that would make the density on the left hand side negative. The buoyant force $B$ cannot be negative because it is equal to $\rho_{\text{water}} V_{\text{object}} g$, which is positive.

Step 3
$\dfrac{m\rho_{\text{water}} g}{B}=\dfrac{F_g}{F_i-F_g}\rho_{\text{water}}.$ (I dropped the negative signs that appear on both sides of the equation.)
Here the buoyant force in the denominator has been replaced by the difference between the weight in air and the mysterious $F_i$. Note that $$\frac{F_g}{F_i-F_g}\rho_{\text{water}}=\left( \frac{F_i-F_g}{F_g} \right)^{-1}\rho_{\text{water}}=\left( \frac{F_i}{F_g}-1 \right)^{-1}\rho_{\text{water}}.$$ This is the negative of what the professor expected to get and explains the negative sign in Step 2. It is a fudge factor to make the answer come right.

When replacing the buoyant force with a difference, that difference should have been $F_g-F_i$ where $F_i$ is the reading of the force gauge with the mass under water, also the tension in the string from which the mass is hanging, also the force that the net exerts on the mass. Where this comes from is shown in post #24, equations (1) and (2).

To @Callumnc1 : Look at my derivation in post #38 and forget the other mess. $F_i$ is the force that you recorded with the masses under water. You don't have to calculate it because you measured it.

Thank you for your reply @kuruman!

Do you please know how $F_i= F_g - B$ if F_i is the downward force from the water on the object (post #51 goes on about this) and Post #53 has the evidence that F_i is the downward force from the water.

Many thanks!

 

[kuruman]

Thank you for your reply @kuruman!

Do you please know how $F_i= F_g - B$ if F_i is the downward force from the water on the object (post #51 goes on about this) and Post #53 has the evidence that F_i is the downward force from the water.

Many thanks!

No, I don't. You have been told numerous times that the downward force from the water is irrelevant to the analysis of this experiment. @haruspex and I have tried to shake you out of this misconception but you refuse to budge. Since I cannot help you, I am out of this thread.

 

[member 731016]

No, I don't. You have been told numerous times that the downward force from the water is irrelevant to the analysis of this experiment. @haruspex and I have tried to shake you out of this misconception but you refuse to budge. Since I cannot help you, I am out of this thread.

Thank you for your reply @kuruman!

Sorry If I was not clear. It is the prof who said that $F_i = F_g - B$. I personally am with you and @haruspex, in that we don't need to decompose the $B$ into $B = F_j - F_i$ to find the density of the object. I was just wondering whether the prof statement that $F_i = F_g - B$ is correct from post #51.

Many thanks!

 

[haruspex]

(1) So it said that Fi is the downward force from the water on the lab script written by the prof:
View attachment 323857

It does not say anything about water; it just says downward force.
(Now, if the object is less dense than water the net will indeed be exerting a downward force, but, from the equation Fi = Fg - B, Fi is measured positive up, so I assume all the objects naturally sink.)
It could be that the prof made a mistake and wrote downward instead of upward, or equivalently, "on the object" instead of "from the object". Either way, it is quite clear from the algebra that Fi is supposed to be the force the net exerts on the object.

It is also clear that this is not quite the same as the force recorded by the sensor, since that force (T?) includes the weight of the net (less any buoyant force on that). This makes me think the TA did not say, or did not mean, that T=0; merely that T is only very slightly different from Fi, so use the sensor reading as Fi and don't worry about T.

 

[haruspex]

Thank you for your reply @kuruman!

Sorry If I was not clear. It is the prof who said that $F_i = F_g - B$. I personally am with you and @haruspex, in that we don't need to decompose the $B$ into $B = F_j - F_i$ to find the density of the object. I was just wondering whether the prof statement that $F_i = F_g - B$ is correct from post #51.

Many thanks!

The professor's statement that $F_i = F_g - B$ proves that $B \neq F_j - F_i$.