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.
  • #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.
 
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  • #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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