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

[Orodruin]

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.

 

[rudransh verma]

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 $\vec F_a=a\hat i +b \hat j$

 

[rudransh verma]

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.

 

[vcsharp2003]

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.

 

[rudransh verma]

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.

 

[Doc Al]

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.

because in Resnik the free body diagrams

What book are you using? (Title, author, and edition.)

 

[vcsharp2003]

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.

 

[rudransh verma]

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.

 

[vcsharp2003]

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?

 

[rudransh verma]

Can you post the figure you are talking about in the book by Resnick Halliday?

 

[vcsharp2003]

What's the doubt you have about the free body diagrams you posted?

 

[rudransh verma]

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.

 

[Doc Al]

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.)

 

[vcsharp2003]

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.

 

[rudransh verma]

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.

terms of magnitude of vector components.

That’s what I have been doing , breaking vectors into its components like mgcostheta and mgsintheta.

 

[Doc Al]

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.

 

[vcsharp2003]

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.

 

[rudransh verma]

-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?

 

[Doc Al]

It’s better to say $-mg\hat j$ is component.

No it isn't.

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.

 

[rudransh verma]

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

 

[Doc Al]

Hint: The mass of anything = ρV.

 

[rudransh verma]

Hint: The mass of anything = ρV.

I solved it. -250J

 

[Doc Al]

I solved it. -250J

The work to raise the stone should be positive.

 

[rudransh verma]

The work to raise the stone should be positive.

How? The net force is downwards and displacement is upwards.

 

[vcsharp2003]

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)

 

[vcsharp2003]

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.

 

[Steve4Physics]

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.

 

[rudransh verma]

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.

 

[vcsharp2003]

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.

 

[Doc Al]

How? The net force is downwards and displacement is upwards.

What's lifting the stone? That's the force that's doing the work.

 

[Steve4Physics]

@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
$F_{net} = -W + B + L = 0$

 

[rudransh verma]

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

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.

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.

 

[vcsharp2003]

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.

 

[rudransh verma]

The equation that I gave is a general form of work energy theorem.

Ok

 

[vcsharp2003]

Ok

It makes problem solving easier when dealing with problems under Work,Energy and Power chapter.