The Lee-Tseung Lead Out Theory

[chrisC]

Am I confusing all the forum members who have no formal training in integral calculus???

No. Tseung. There is nothing wrong with real mathematics and real physics. There is everything wrong with your theory of Physics. cCan you really be so STUPID? Don't you get it? Is your Physics that poor?

cheers

chrisC

[Top Gun]

@ Top gun:

No, but you are confusing some of us by indicating that you are going to pull a rabbit out of your ass. (mathematically speaking)  I can't wait to see the numbers on this one.

Bill

Dear Pirate88179,

Numbers are easy if you have the formulae and the Excel spread sheet program.

Let us make the same assumptions in slide 3/Pendulum08.jpg on pendulum12.jpg :

Mass of Bob = 60 Kg
Acceleration due to gravity = 9.8 m/s/s
Force due to the bob is Mg = 588 Newtons
External force F = 1/6 Mg = 98 Newtons
Length of string is 1 m
The final angle a is given by Tan(a) = F/Mg, thus a = 9.46 degrees
The vertical displacement dH = L(1-cos(a)) = 0.013599 meters
The horizontal displacement dX= Lsin(a) = 0.164359 meters

The increase in potential energy of the bob Mg x dH
= 588 x 0.013599 = 7.99646 joules
The work done by the external force F x (dH + dX) = F x (0.177958)
= 98 * 0.177958 = 17.94993 joules

It is clear that Mg x dH is not equal to F x (dH+dX).
(7.99646 is not equal to 17.94993)

Thank you to utilitarian for correcting a mathematical error (see his post that follows)

Can any one find errors in the calculation now???

[utilitarian]

Can any one find errors in the calculation???

Yes, almost immediately.  Some Top Gun you are.  1 kg = 9.8 N.  So the bob exerts a 588 N force downward, not 58.8.

I suggest you ask some university students to explain this to you.  Or maybe the wikipedia.

http://en.wikipedia.org/wiki/Kilogram-force

[utilitarian]

The increase in potential energy of the bob Mg x dH
= 588 x 0.013599 = 7.99646 joules
The work done by the external force F x (dH + dX) = F x (0.177958)
= 98 * 0.177958 = 17.94993 joules

Can any one find errors in the calculation now???

Yes, there is another error.

You are correct in calculating increase in potential energy.  Indeed, 60kg * 9.8 * .013599 = 7.996471 J

You err in calculating the work done by the external force.  As you had done previously, lets reduce this to an inclined plane problem.  What we have is essentially a right triangle.  The long horizontal side is 0.164358 meters.  The short vertical side is 0.013599 meters.  This makes the hypotenuse 0.164921 meters.  Also, the angle of incline, per a simple trig calculation, is 4.73 degrees.

The formula for calculating work done along an inclined plane, with no friction, is as follows:

mass * g * distance * sin (theta), where theta is the angle of incline

So, let's calculate.

60 kg * 9.8 * 0.164921 * sin (4.73) = 7.996471 J

So everything matches.  No extra energy.

[Top Gun]

Yes, there is another error.

You are correct in calculating increase in potential energy.  Indeed, 60kg * 9.8 * .013599 = 7.996471 J

You err in calculating the work done by the external force.  As you had done previously, lets reduce this to an inclined plane problem.  What we have is essentially a right triangle.  The long horizontal side is 0.164358 meters.  The short vertical side is 0.013599 meters.  This makes the hypotenuse 0.164921 meters.  Also, the angle of incline, per a simple trig calculation, is 4.73 degrees.

The formula for calculating work done along an inclined plane, with no friction, is as follows:

mass * g * distance * sin (theta), where theta is the angle of incline

So, let's calculate.

60 kg * 9.8 * 0.164921 * sin (4.73) = 7.996471 J

So everything matches.  No extra energy.

Dear utilitarian,

I am glad that we can discuss physics without the insults.

You are perfectly correct if you push a load up an inclined plane.  There is no string and there is no Lead Out energy.  Thus the incline plane is not capable of performing the Lead Out energy function.

However, we are considering a totally different simple machine in the case of a pendulum under Lee-Tseung Pull.  We have to consider the forces, displacement, work and energy in this particular machine.  As I stated before:

(1)   We know the beginning Position A in Pendulum08.jpg.  It is at rest.  We use it as reference.  The potential energy is zero.  There is no velocity.  The kinetic energy is zero.

(2)   We know the maximum displaced Position B in Pendulum08.jpg.  The pendulum system under the three forces at equilibrium is also at rest.  There is no velocity.  The kinetic energy is zero.  We want to find out how much work has been done (or energy spent) to get the bob from Position A to Position B.

(3)   The vigorous physics and mathematics tell us that we can use different forces on different paths to achieve the end result of going from Position A to Position B.  Some will require all the energy to be supplied by the external force.  Your example of the inclined plane is one of them.  My example in Pendulum10.jpg where the tension in the string is zero is another.

(4)   However, whenever the tension of the string is involved, we have to consider whether (a) does the tension in the string represent a force? (b) does this force has displacement in direction of the force?  Gravity does work and imparts energy via the tension in the string.

(5)   Once you take away the string, gravity cannot do work via the string and thus cannot impart energy.  Thus your inclined plane example is perfectly correct in its own right but it is not a pendulum under Lee-Tseung Pull.  It will not Lead Out gravitational energy.