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You have been told the problem repeatedly. Please read through what has already been said.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.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?
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.Orodruin said:You have been told the problem repeatedly. Please read through what has already been said.
You are right but some confusions can become a problem of future.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.
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 said:You are right but some confusions can become a problem of future.
Let’s keep the sign convention constant. I always take +ve as up and right as +ve.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.
What book are you using? (Title, author, and edition.)rudransh verma said:because in Resnik the free body diagrams
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 said:So I thought they are vectors.
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.Doc Al said:Sadly, you are making a rather simple problem way too complicated. (And why in the world did you bring up friction?)
Can you post the figure you are talking about in the book by Resnick Halliday?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.
See in FBD they are showing vectors but when solving they are writing like T-mg=-ma.vcsharp2003 said:What's the doubt you have about the free body diagrams you posted?
The equations are for the components of those vectors. (Vectors themselves aren't positive or negative, they just point in some direction.)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.rudransh verma said:See in FBD they are showing vectors but when solving they are writing like T-mg=-ma.
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.Doc Al said:The equations are for the components of those vectors.
That’s what I have been doing , breaking vectors into its components like mgcostheta and mgsintheta.vcsharp2003 said:terms of magnitude of vector components.
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.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.
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.rudransh verma said:I think the proper way is to write with unit vectors
It’s better to say ##-mg\hat j## is component.Doc Al said:-mg" is the component of the weight (a vector)
No it isn't.rudransh verma said:It’s better to say ##-mg\hat j## is component.
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 said:Can we now concentrate on the question in OP.
What will be the V displaced?
W=(-mg+rhoVg)d=-250JDoc 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.
I solved it. -250JDoc Al said:Hint: The mass of anything = ρV.
The work to raise the stone should be positive.rudransh verma said:I solved it. -250J
How? The net force is downwards and displacement is upwards.Doc Al said:The work to raise the stone should be positive.
I would use the following equation to solve this problem.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.rudransh verma said:How? The net force is downwards and displacement is upwards.
Well why not ##\Delta U=-W##vcsharp2003 said:Since the ##PE## of stone increases when it's moved up, so net work done should be postive.
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.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.
What's lifting the stone? That's the force that's doing the work.rudransh verma said:How? The net force is downwards and displacement is upwards.
Steve4Physics said:You are being asked to find the work done by L→, not the work done by W→+B→.
L=mg-rhoVg (taking g=10)Doc Al said:What's lifting the stone? That's the force that's doing the work.
I don't know about this formula. All I know is ##\Delta KE +\Delta U=0## for a closed system.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)
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.rudransh verma said:All I know is ΔKE+ΔU=0 for a closed system.
Okvcsharp2003 said:The equation that I gave is a general form of work energy theorem.
It makes problem solving easier when dealing with problems under Work,Energy and Power chapter.rudransh verma said:Ok