The Lee-Tseung Lead Out Theory

[Top Gun]

Dear Kul_ash and Tseung,

You have stimulated the artist in me.  In the diagram below, I showed a possible path keeping the tension in the string equal to 0 until the very end.  In this particular scenario, all the work done in moving the bob from position A to B is from the external force F alone.

However, any other scenario showing tension in the string will show a different result.
Work = Force x displacement (vector arithmetic).  If there were tension in the string (force) and displacement, then some work must have been by the string.  Or to be more exact, work must be done by gravity through the string.

If you need me to draw more diagrams to bring out more clarity, I shall be happy to do so.

[chrisC]

Dear Kul_ash and Tseung,

You have stimulated the artist in me.  In the diagram below, I showed a possible path keeping the tension in the string equal to 0 until the very end.  In this particular scenario, all the work done in moving the bob from position A to B is from the external force F alone.

However, any other scenario showing tension in the string will show a different result.
Work = Force x displacement (vector arithmetic).  If there were tension in the string (force) and displacement, then some work must have been by the string.  Or to be more exact, work must be done by gravity through the string.

If you need me to draw more diagrams to bring out more clarity, I shall be happy to do so.

Tseung, you don't need to draw any more fancy diagrams. You need to take a pre-'O' level Physics class
in middle school.

cheers
chrisC

[Kul_ash]

Dear Kul_ash,

Looks like we still cannot get to a common point to continue the discussion.

I have prepared the following diagram which is effectively an O-level Physics experiment.

Please look at the diagram carefully.  Do you agree with the angles and the scale readings shown?

Obviously, the direction of F can be different.  However, the experiment shows a possible outcome that obeys both Physics and Mathematics Rules.  Please comment on whether you agree with the diagram?

I do not agree simply because:
1. If 2nd fig is final position of the bob, then there is no more Horizontal force F. Its consumed. Because if you are still pulling it at the same force, that will not be its final position, isn't it? Your applied force has lifted the bob to say height h, then at final position it will have no velocity but only PE = mgh.
2. Simpe trignometary tells me that actual angle between the bob and support would be definately lesser than the angle between support and point of force application. Do you agree?
3. There will be always Tension in the vertical part of the string below the point of force application which would be equal to mg. Where have you considered that?
4. If you are calling it as a final position, then simply point of force application becomes point of support. There will be no tension in string between two suppots but the total tension will be takn by the remaining vertical part of string and will bring you to the initial postion of T = mg. When you let go pendulum from this position, pendulum will never swing. Because suddenly the slack portion of sting will experience the force and it will induce equal and opposite force in remaining part and try to bring the system in equilibrium. You will not be able to swing it like the way you could when the complete string is in system. Try it !
5. So in all applying horizontal force in between will not serve any problem but will induce lots of others.

I urge you again that do not unnecessarily go in 0 level physics when you have simple normal physics available to you. Because no matter what Physics is going to be physics and all the levels and it will not allow you to defy the laws of nature.

[Top Gun]

I do not agree simply because:

1. If 2nd fig is final position of the bob, then there is no more Horizontal force F. Its consumed. Because if you are still pulling it at the same force, that will not be its final position, isn't it?.....

Dear Kul_ash,

I hope that we both are examining the same diagram pendulum08.jpg in reply 2156.  That diagram represents three forces at equilibrium.  These three forces are Mg, T1 and F.  From the parallelogram of forces and trigonometry, the following relationships MUST hold:

(1)   tan(a) = F/Mg = 10/60 and thus the angle a = 9.46 degrees.
(2)   T1cos(a) = Mg = 60 units
(3)   T1sin(a) = F = 10 units
(4)   T1 = Mg/cos(a) = 60.83 units

F must be present to achieve this equilibrium.  In other words, I am still pulling with a horizontal force F of 10 units at this point of the analysis.  In particular, T has changed to T1 (from 60 units to 60.83 units).  At equilibrium, even if I am still pulling, there will be no motion because F is balanced by the equal and opposite T1sin(a) from the tension of the string.   This can easily be confirmed by experiment.  Hopefully it answers your question. 

Please confirm your understanding before we carry on to the other juicy parts of the scientific discussion.

[Kul_ash]

I do not agree simply because:

1. If 2nd fig is final position of the bob, then there is no more Horizontal force F. Its consumed. Because if you are still pulling it at the same force, that will not be its final position, isn't it?.....

Dear Kul_ash,

I hope that we both are examining the same diagram pendulum08.jpg in reply 2156.  That diagram represents three forces at equilibrium.  These three forces are Mg, T1 and F.  From the parallelogram of forces and trigonometry, the following relationships MUST hold:

(1)   tan(a) = F/Mg = 10/60 and thus the angle a = 9.46 degrees.
(2)   T1cos(a) = Mg = 60 units
(3)   T1sin(a) = F = 10 units
(4)   T1 = Mg/cos(a) = 60.83 units

F must be present to achieve this equilibrium.  In other words, I am still pulling with a horizontal force F of 10 units at this point of the analysis.  In particular, T has changed to T1 (from 60 units to 60.83 units).  At equilibrium, even if I am still pulling, there will be no motion because F is balanced by the equal and opposite T1sin(a) from the tension of the string.   This can easily be confirmed by experiment.  Hopefully it answers your question. 

Please confirm your understanding before we carry on to the other juicy parts of the scientific discussion.

Nopes! Absolutely incorrect. Because the moment you are going to introduce your horizontal force from any where above the bob i.e in the string, the whole system is going to change. Weigh of bob, mg, will be always balanced by the straight portion of the string. So the straight part of the string will always have T = mg equation. Because your point of horizontal force application is always going to act as point of support.

The moment you introduce any more new support, they system will look totally different! If you only want to consider three forces i.e. mg, T and horizontal force and if you want to consider this as a homogeneous system with only one point of support, then the force has to be given on the bob and not on string.
Again these are basics. Because the moment you introduce support in between there are going to be different forces. Namely, tension in straight part o string, tension in slack part of string, mg and horizontal force. e.g. Consider I hod the string in between and lift it vertically, the moment I do that, tension in string above the point where i am holding is going to be zero. All the tension now is going to be on the vertical part of string. Now if I move it horizontally, tension in that slack part is always going to be zero. Because the system essentially have reduced down to my hand support, straight vertical part of string and bob. So when I move my hand horizontally, its like moving support of pendulum horizontally. Till the moment you are going to pull the bob itself, the system is not going to be pendulum system.

Really man, please I request you to do your analysis based on simple physics first.