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

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Homework Help Overview

The discussion revolves around the effects of water on the net force acting on a submerged stone, specifically exploring the concepts of buoyancy and gravitational forces. Participants are examining the implications of Archimedes' principle and the relationship between buoyant force and the weight of displaced water.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the buoyant force and its relationship to the weight of the displaced liquid. Questions arise regarding the application of Archimedes' principle, the nature of vectors and scalars in the context of buoyancy, and the correct formulation of related equations.

Discussion Status

The discussion is active, with participants providing insights into the mathematical representation of buoyant force and questioning the consistency of vector notation. Some guidance has been offered regarding the interpretation of equations, but multiple interpretations and clarifications are still being explored.

Contextual Notes

Participants express uncertainty about the law of buoyancy and the principles of fluid mechanics, indicating a need for foundational understanding. There are also references to previous knowledge of physics equations and their application in different contexts.

  • #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 positive.
 
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  • #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 positive.
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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