Finding downward force on immersed object

[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 $B \neq F_j - F_i$.

 

[member 731016]

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.

Thank you for you reply @haruspex!

I had a chat with the Professors today. Apparently, Fi is defined to be the upward force exerted by the force sensor and it not actually the downward force exerted by the water. They are going to change the lab script so that the definition of Fi will no longer say 'downward force on immersed object' which they agreed could imply the downward force from the water above the object. You are correct that $B ≠ F_j - F_i$.

$F_i + B - mg = 0$ and rearranging gives their expression.

Thank you and @kuruman very much for your help!

 

[member 731016]

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

Thank you for you help @haruspex!

 

[malawi_glenn]

Thank you for your reply @kuruman!

Thank you for you help @haruspex!

What did we say about making posts just to say "thanks"?