How Does Water Affect the Net Force on a Submerged Stone?

AI Thread Summary
The discussion centers on the effects of buoyancy on the net force acting on a submerged stone. Participants explore Archimedes' principle, emphasizing that the buoyant force equals the weight of the displaced water, which reduces the stone's apparent weight in water. There is a debate about the correct notation and representation of vectors versus scalars, particularly regarding gravitational force and buoyancy equations. Misunderstandings arise around the use of negative signs and the distinction between vector and scalar quantities. Ultimately, clarity in notation is deemed crucial for accurately applying the principles of buoyancy and fluid mechanics.
rudransh verma
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Homework Statement
A stone of mass 10kg is lying at the bed of a lake 5m deep. If the relative density of the stone is 2, amount of work to bring it to the top is
Relevant Equations
W=F.d
d=5m
But the force is not mg since there is water so net force must be less. But how much?
 
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What are your thoughts?
 
Orodruin said:
What are your thoughts?
I don’t know the law of baoency.
 
rudransh verma said:
I don’t know the law of baoency.
Then maybe a good place to start is looking that up? Archimedes’ principle is quite fundamental.
 
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Orodruin said:
Then maybe a good place to start is looking that up? Archimedes’ principle is quite fundamental.
##F_b=-rhogV##
This is the formula of baouyant force which i think needs to be applied. But first of all what is this formula telling. There is no vector quantity in right side. So what is -ve doing here. Surely its to tell the sign of force.
 
The gravitational acceleration ##g## is technically a vector. The buoyancy formula tells you in math what Archimedes' principle says: The buoyancy is equal to the weight of the displaced liquid. (##\rho V## being its mass). Now, what is the mass of the displaced liquid if the object has mass 10 kg and relative density 2?
 
Orodruin said:
The gravitational acceleration g is technically a vector.
We always take -g in problems and that is a vector. Oh! you mean ##\vec F_b=-\rho V\vec g## where the value of ##\vec g=-g##. So the ##\vec F_b=F_b## in the upward direction.
Orodruin said:
The buoyancy is equal to the weight of the displaced liquid. (ρV being its mass).
Its obvious that the apparent weight of an object=weight of the object- bouyant force. But how is bouyant force equal to weight of water displaced?
 
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rudransh verma said:
We always take -g in problems and that is a vector. Oh! you mean F→b=−ρVg→ where the value of g→=−g. So the F→b=Fb in the upward direction.
No. ##g## is the magnitude of ##\vec g##. A magnitude can never be equal to a vector.

rudransh verma said:
But how is bouyant force equal to weight of water displaced?
This is Archimedes’ principle. The actual proof requires a bit more math but you can argue for it for any column submerged in a fluid. Basically, assume a column submerged in a fluid and compute the pressure on top and bottom of the column.
 
Orodruin said:
Basically, assume a column submerged in a fluid and compute the pressure on top and bottom of the column.
On top zero
and on bottom mgh/V ,where V is the volume of water and h is height.
Orodruin said:
g is the magnitude of g→. A magnitude can never be equal to a vector.
I mean ##\vec F_b=-\rho V\vec g=>\vec F_b=\rho Vg##
which means Force of bouyancy should be upwards which it is.
 
  • #10
rudransh verma said:
On top zero
and on bottom mgh/V ,where V is the volume of water and h is height.
No, it is not necessarily zero at the top if the column is submerged.

rudransh verma said:
I mean ##\vec F_b=-\rho V\vec g=>\vec F_b=\rho Vg##
which means Force of bouyancy should be upwards which it is.
No. The second equation makes absolutely no sense. The LHS is a vector and the RHS is a scalar.
 
  • #11
Orodruin said:
No. The second equation makes absolutely no sense. The LHS is a vector and the RHS is a scalar.
##\vec F_b=+\rho Vg##
 
  • #12
rudransh verma said:
##\vec F_b=+\rho Vg##
That is still equating a vector with a scalar. You simply cannot write things like that.
 
  • #13
Orodruin said:
That is still equating a vector with a scalar. You simply cannot write things like that.
I have not studied the concept of fluid mechanics. So I don't think i can solve it without your help.
 
  • #14
rudransh verma said:
I have not studied the concept of fluid mechanics. So I don't think i can solve it without your help.
It is not about fluid mechanics, not equating a vector to a scalar is basic vector analysis.

Your first equation in #9 was correct for the buoyancy. After that it is up to you to use that to find the work required.
 
  • #15
Orodruin said:
Your first equation in #9 was correct for the buoyancy.
Ok. But to justify Fb is in upward direction we need to take the ##\vec g=-g##
 
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  • #16
rudransh verma said:
Ok. But to justify Fb is in upward direction we need to take the vector g=-g
No. The minus sign in the correct equation already means that the buoyancy is in the opposite direction of the gravitational field. You can never equate a vector to a scalar. I believe what you are attempting is to resolve the buoyancy into its vertical component. The component of the gravitational field will depend on which direction you choose as positivebin your coordinate system.
 
  • #17
Orodruin said:
No. The minus sign in the correct equation already means that the buoyancy is in the opposite direction of the gravitational field. You can never equate a vector to a scalar. I believe what you are attempting is to resolve the buoyancy into its vertical component. The component of the gravitational field will depend on which direction you choose as positivebin your coordinate system.
is -g not a vector?
 
  • #18
rudransh verma said:
is -g not a vector?
No. Not if you want any sort of consistent notation and use ##\vec g## for the gravitational field.
 
  • #19
Orodruin said:
No. Not if you want any sort of consistent notation and use ##\vec g## for the gravitational field.
Can we write it like this ##\vec F_b=-(-)\rho Vg##
 
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  • #20
rudransh verma said:
Ok. But to justify Fb is in upward direction we need to take the vector g=-g
The only way that g can equal -g is for g to be equal to 0.
rudransh verma said:
is -g not a vector?
The point that @Orodruin repeatedly has been making, and that you aren't getting, is that g (a scalar, a number) and ##\vec g## (a vector) are different things. He is asking you to be consistent with vectors and notation. If the left side of an equation has a vector, the right side must have a vector as well.
rudransh verma said:
Can we write it like this ##\vec F_b=-(-)\rho Vg##
No. This is exactly the same as ##\vec F_b = \rho Vg##
Again, you have a vector on the left side, but a scalar ( g ) on the right side. As written, the equation doesn't make sense.
 
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  • #21
rudransh verma said:
I don’t know the law of baoency.
No such word as "baoency".
rudransh verma said:
This is the formula of baouyant force
Still not a word.
rudransh verma said:
bouyant force
Closer, but the words are buoyant and buoyancy.
 
  • #22
Mark44 said:
o. This is exactly the same as F→b=ρVg
Again, you have a vector on the left side, but a scalar ( g ) on the right side. As written, the equation doesn't make sense.
I have always been putting in suvat eqns a=-g when calculating any of the s,u,v,a,t under gravity. Why is this different now. When doing problems on force and motion I always took -mg as downward weight. Why is this different now.
I believe you want ##\vec F_b=\rho Vg\hat j##
 
  • #23
rudransh verma said:
I have always been putting in suvat eqns a=-g when calculating any of the s,u,v,a,t under gravity. Why is this different now.
The suvat equations generally treat a single component of a vector relationship, not a vector relationship in itself. All of the involved quantities relate to that particular direction. If you want to express yourself in vectors then you meed to use vectors.

You also cannot blindly put a=-g in suvat either as it depends on how you defined your coordinates. If you define your coordinates with increasing coordinate downwards, it would be a=g.
 
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  • #24
Orodruin said:
The suvat equations generally treat a single component of a vector relationship, not a vector relationship in itself. All of the involved quantities relate to that particular direction. If you want to express yourself in vectors then you meed to use vectors.

You also cannot blindly put a=-g in suvat either as it depends on how you defined your coordinates. If you define your coordinates with increasing coordinate downwards, it would be a=g.
Ok
Edit: I suppose ##F_b=\rho Vg## is correct if ##F_b## and ##g## are magnitudes.
I suppose ##F_b=-\rho Vg## is also correct
I suppose ##F_b\hat j =-\rho Vg\hat j ## is same as second one
I suppose ##\vec F_b=-\rho V \vec g## is also correct.
These three eqns above are saying the same thing that ##F_b##=##\rho Vg## and there directions are opposite to each other.
I suppose ##\vec F_b=\rho Vg## is wrong. We need to define ##\vec g=-g\hat j## and put in ##\vec F_b=-\rho V \vec g##
then it becomes ##\vec F_b=\rho Vg\hat j##
 
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  • #25
rudransh verma said:
I suppose Fb=−ρVg is also correct if Fb and g are vectors.
This is a scalar equation that is one of the components of the correct vector equation, not a vector equation. Its correctness depends on how you defined your coordinate directions.

rudransh verma said:
I suppose F→b=−ρVg→ is same as second one.
No. Definitely not. One is a relation between two vectors, the other is a relation between two scalars.

rudransh verma said:
We need to define g→=−gj^
You do not define this. You can pick your coordinate axes such that it is the case however.
 
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  • #26
Orodruin said:
This is a scalar equation that is one of the components of the correct vector equation, not a vector equation.
I defined F_b and g as vectors but if I am wrong please show me what do you mean.
 
  • #27
rudransh verma said:
I defined F_b and g as vectors but if I am wrong please show me what do you mean.
Then your notation is wrong. If you are using vector notation, you need to do it consistently.
 
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  • #28
rudransh verma said:
I defined F_b and g as vectors but if I am wrong please show me what do you mean.
It's best to use a scalar equation for Archimedes principle with the understanding that the buoyant force acts in an upward direction. That way things are kept simple and one can start applying it to problem at hand without unecessary confusion.

So, I would just use the following formula: ##F_b = \rho V g##, where all quantities are positive. ##V## is volume of fluid displaced by the submerged body and ##\rho## is density of fluid. And of course ##g## is acceleration due to gravity.
 
  • #29
Orodruin said:
Then your notation is wrong. If you are using vector notation, you need to do it consistently.
I think you miss understood me. I am trying to write different variations of the same law (law of buoyancy)in vector and magnitude form. It’s not a question about consistency.
I am trying to write correctly and avoid making mistakes with signs which I often make.
 
  • #30
rudransh verma said:
I think you miss understood me. I am trying to write different variations of the same law (law of buoyancy)in vector and magnitude form. It’s not a question about consistency.
I am trying to write correctly and avoid making mistakes with signs which I often make.
If you are trying to avoid making such mistakes then you really need to make sure your notation is on point. You cannot ramdomly assign g to be a vector in some expressions and a vector magnitude in others and on top of everything also introduce ##\vec g##. I am not surprised you make sign errors with such inconsistency in notation.
 
  • #31
Orodruin said:
You cannot ramdomly assign g to be a vector in some expressions and a vector magnitude in others
I didn’t do that intentionally because I know it’s wrong.
So are my variations right in post#24?
Have I got the concept of vectors?
 
  • #32
rudransh verma said:
I didn’t do that intentionally because I know it’s wrong.
So are my variations right in post#24?
Have I got the concept of vectors?
No, as I have already said. Many of them have major notational issues. The first thing you need to do is to take care of those. Then you can start writing down equations.
 
  • #33
Orodruin said:
Many of them have major notational issues.
Like?
We can sort it out one by one.
 
  • #34
rudransh verma said:
Like?
We can sort it out one by one.
No. It cannot be sorted out without the notation being sorted out first.
 
  • #35
Orodruin said:
No. It cannot be sorted out without the notation being sorted out first.
Ok! Let’s sort out the notation.
By the way I wrote what each notation mean in each eqn. Where is the problem?
 
  • #36
rudransh verma said:
Ok! Let’s sort out the notation.
By the way I wrote what each notation mean in each eqn. Where is the problem?
You have been told the problem repeatedly. Please read through what has already been said.
 
  • #37
Orodruin said:
You have been told the problem repeatedly. Please read through what has already been said.
I think I have solved the problem. I have been mixing the scalar and vector versions. g, -g, -mg, Fn(normal force), -Fs(static friction) are not vectors and are used in 1D motion.
Whereas the vectors are defined when we need to describe motion in 2D or 3D motion.
##\vec F_s=F_s\hat i## is correct if static friction is actually in positive direction. We can also write this like ##F_s=F_s=\mu F_N##
##\vec F_s=-F_s\hat i## is same as ##-F_s=-\mu F_N##
We can use the scalar as well as vector form if the motion is 1D. Both will be correct.
I edited the post#24. Please verify it. I hope it’s right now.
In 1D motion we can just use signs to tell the direction. But in2D or 3D we need vectors like ##\vec F_a=a\hat i +b \hat j##
 
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  • #38
vcsharp2003 said:
It's best to use a scalar equation for Archimedes principle with the understanding that the buoyant force acts in an upward direction. That way things are kept simple and one can start applying it to problem at hand without unecessary confusion.
You are right but some confusions can become a problem of future.
 
  • #39
rudransh verma said:
You are right but some confusions can become a problem of future.
Also, remember when you express both sides of the equation in terms of unit vectors in one dimension, you are free to take +ve direction of x-axis pointing downward or upward. Accordingly, your vector equation in one dimension will change.
 
  • #40
vcsharp2003 said:
Also, remember when you express both sides of the equation in terms of unit vectors in one dimension, you are free to take +ve direction of x-axis pointing downward or upwards. Accordingly, your vector equation in one dimension will change.
Let’s keep the sign convention constant. I always take +ve as up and right as +ve.

The problem arised I think because in Resnik the free body diagrams and fig. have vectors indicated. And then they write some thing like -mg+F_N=m(0). So I thought they are vectors.
 
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  • #41
Sadly, you are making a rather simple problem way too complicated. (And why in the world did you bring up friction?)

As the stone is raised, what forces act on it? What are their directions? Then you can choose a sign convention and write an equation for their components.

rudransh verma said:
because in Resnik the free body diagrams
What book are you using? (Title, author, and edition.)
 
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  • #42
rudransh verma said:
So I thought they are vectors.
First draw a free body diagram and then choose your axes system. Then simply write down the equations. This is what probably is done in the book by Resnick Halliday.
 
  • #43
Doc Al said:
Sadly, you are making a rather simple problem way too complicated. (And why in the world did you bring up friction?)
No I am not trying to solve that problem. But I am asking you to clear my doubts on vector and scalar form of any eqn in a general sense. First that needs to be solved and then I will come back to the original problem in OP.
Principles of Physics International student edition.
 
  • #44
rudransh verma said:
Let’s keep the sign convention constant. I always take +ve as up and right as +ve.

The problem arised I think because in Resnik the free body diagrams and fig. have vectors indicated. And then they write some thing like -mg+F_N=m(0). So I thought they are vectors.
Can you post the figure you are talking about in the book by Resnick Halliday?
 
  • #45
vcsharp2003 said:
Can you post the figure you are talking about in the book by Resnick Halliday?
 

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  • #46
What's the doubt you have about the free body diagrams you posted?
 
  • #47
vcsharp2003 said:
What's the doubt you have about the free body diagrams you posted?
See in FBD they are showing vectors but when solving they are writing like T-mg=-ma.
 
  • #48
rudransh verma said:
See in FBD they are showing vectors but when solving they are writing like T-mg=-ma.
The equations are for the components of those vectors. (Vectors themselves aren't positive or negative, they just point in some direction.)
 
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  • #49
rudransh verma said:
See in FBD they are showing vectors but when solving they are writing like T-mg=-ma.
Showing forces in vector notation is entirely upto you. There is no harm in using vector notation in free body diagram. Personally, I don't use vector notation in free body diagram. But, you must note that each vector/force in FBD must be broken into its components along the axes system chosen and then the equations for each axis is written in terms of magnitude of vector components.
That is the process in going from FBD to equations.
 
  • #50
Doc Al said:
The equations are for the components of those vectors.
I think the proper way is to write with unit vectors or just write the scalar eqns and solve. This seems like -mg is a vector.
vcsharp2003 said:
terms of magnitude of vector components.
That’s what I have been doing , breaking vectors into its components like mgcostheta and mgsintheta.
 
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