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

In summary, The conversation revolves around the concept of buoyancy and Archimedes' principle. The formula for buoyant force is Fb = -ρVg, where ρ is the density of the displaced liquid, V is its volume, and g is the gravitational acceleration. The conversation also touches on the difference between a scalar and a vector, and the need for consistent notation when dealing with vectors.
  • #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.
 
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  • #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.
 
  • #51
rudransh verma said:
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.
Expressing a vector using unit vectors is always fine. But you'll eventually have to deal with components. "-mg" is the component of the weight (a vector) using the typical "up is positive" sign convention.
 
  • #52
rudransh verma said:
I think the proper way is to write with unit vectors
For components there's no need to write as vectors. The idea to use components rather than orginal vectors is to simplify things so we use magnitudes along an axis to get equations.

-mg is not a vector, but just part of an equation along a certain axis.
 
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  • #53
Doc Al said:
-mg" is the component of the weight (a vector)
It’s better to say ##-mg\hat j## is component.

Can we now concentrate on the question in OP.
What will be the V displaced?
 
  • #54
rudransh verma said:
It’s better to say ##-mg\hat j## is component.
No it isn't.

rudransh verma said:
Can we now concentrate on the question in OP.
What will be the V displaced?
Just call the volume of the stone V. You don't need to know the actual volume, just the mass of the displaced water. All the info to solve for that is given.
 
  • #55
Doc Al said:
Just call the volume of the stone V. You don't need to know the actual volume, just the mass of the displaced water. All the info to solve for that is given.
W=(-mg+rhoVg)d=-250J
 
  • #56
Hint: The mass of anything = ρV.
 
  • #57
Doc Al said:
Hint: The mass of anything = ρV.
I solved it. -250J
 
  • #58
rudransh verma said:
I solved it. -250J
The work to raise the stone should be positive.
 
  • #59
Doc Al said:
The work to raise the stone should be positive.
How? The net force is downwards and displacement is upwards.
 
  • #60
rudransh verma said:
How? The net force is downwards and displacement is upwards.
I would use the following equation to solve this problem.

##W_{net}= \Delta KE + \Delta PE##, where ##W_{net}## is work done by applied forces (excluding force of gravity)
 
  • #61
rudransh verma said:
How? The net force is downwards and displacement is upwards.
Since the ##PE## of stone increases when it's moved up, so net work done should be postive.
 
  • #62
Excuse me jumping-in but here are a few thoughts.

@rudransh verma, try following these rules:

Put an arrow over the symbol for a vector, e.g. ##\vec V##. (But for unit vectors, use a ‘hat’, e.g. ##\hat i##.)

To represent the magnitude of a vector, use the unadorned letter, e.g. ##V##. Or if required for extra clarity, use ##|\vec V|##.

Never write equations where one side is a vector and the other side is a scalar. It is simply wrong.

I think the above is consistent with what Resnick does in your image in Post #45.

For example, take ##\hat j## as the positive upwards (+y) unit vector. Suppose there are two forces on an object, ##\vec A## up and ##\vec B## down. Then we can write these vector equations:
##\vec F_{net} = \vec A + \vec B##
##\vec F_{net} = A \hat j + B(-\hat j)##
##\vec F_{net} = (A - B)\hat j##
Since ##\vec F_{net}## has only a y-component we can write
##F_{net}\hat j= (A - B)\hat j##

Cancelling out ##\hat j## gives this scalar equation:
##F_{net} = A - B##
________________________

Also worth noting: the original problem in Post #1 should ask for the minimum amount of work. This minimum would be achieved by raising the stone very slowly so that viscous (drag) forces and any increase in kinetic energy can be ignored.

Edit. Typo's corrected. Also, my post is a bit late as the thread has progressed.
 
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  • #63
vcsharp2003 said:
Since the ##PE## of stone increases when it's moved up, so net work done should be postive.
Well why not ##\Delta U=-W##
Since W=-250J so ##\Delta U= 250J## which is +ve as it should be. Meaning U increases.
 
  • #64
rudransh verma said:
Well why not ##\Delta U=-W##
Since W=-250J so ##\Delta U= 250J## which is +ve as it should be. Meaning U increases.
Change in ##PE## is simply ##mgh##. What is change in ##KE##? What you've in above post is wrong since the equation you gave is the relationship between ##PE## and work done by the corresponding conservative force i.e. by mg in this case. You need to use the equation I mentioned and you need to understand the meaning of net work done.
 
  • #65
rudransh verma said:
How? The net force is downwards and displacement is upwards.
What's lifting the stone? That's the force that's doing the work.
 
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  • #66
@rudranch verma, while the stone is being lifted, from your free body diagram, you will see that there are three forces acting (assuming negligible drag):
1) Weight, ##\vec W## downwards
2) Buoyancy ##\vec B## upwards
3) The applied external lifting force, ##\vec L## upwards (e.g. the tension from a piece of string tied to the stone).

You are being asked to find the work done by ##\vec L##, not the work done by ##\vec W+ \vec B##.

Assuming the stone is raised at a constant velocity (no acceleration) note that
##\vec F_{net} = \vec W## + ##\vec B## + ##\vec L## = 0
Or in terms of magnitudes
##F_{net} = -W + B + L = 0##
 
  • #67
Steve4Physics said:
You are being asked to find the work done by L→, not the work done by W→+B→.
Doc Al said:
What's lifting the stone? That's the force that's doing the work.
L=mg-rhoVg (taking g=10)
W=Ld
W=250 J
If it was in air then the applied force would have done W=500J.
As it should be. In air we have to do more work.
vcsharp2003 said:
I would use the following equation to solve this problem.

##W_{net}= \Delta KE + \Delta PE##, where ##W_{net}## is work done by applied forces (excluding force of gravity)
I don't know about this formula. All I know is ##\Delta KE +\Delta U=0## for a closed system.
 
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  • #68
rudransh verma said:
All I know is ΔKE+ΔU=0 for a closed system.
This equation is valid only if no forces other than gravity force act on the object. In your question, the object has an upthrust force as well as an applied force acting as the object is brought up to the surface. Therefore, your equation cannot be applied to the scenario in question.

The equation that I gave is a general form of work energy theorem.
 
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  • #69
vcsharp2003 said:
The equation that I gave is a general form of work energy theorem.
Ok
 
  • #70
rudransh verma said:
Ok
It makes problem solving easier when dealing with problems under Work,Energy and Power chapter.
 
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