# Speed of Force or 'Transfer of Momentum'

Discussion in 'Physics & Math' started by hansda, Feb 14, 2013.

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1. ### hansdaValued Senior Member

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This is a vague statement, unless you identify a specific statement which you thought to be wrong.

In this energy transfer who should move? The swimmer or water.

3. ### originTrump is the best argument against a democracy.Valued Senior Member

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It really does seem like all your effort is to NOT understand this. The swimmer moves through the water. The water also moves of course because it is is a liquid. The point remains that the water is not transfering energy to the swimmer.

5. ### hansdaValued Senior Member

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Doesn't the swimmer have kinetic energy when he is moving forward? If the swimmer transfers his kinetic energy to the water, how the swimmer is gaining kinetic energy for forward movement.

What is the "reactive force"? Is it "reaction-force" or "frictional-force" or "some other force without any energy"?

When a net force is applied on the shoulder, a net energy is also applied on the shoulder.

What is so difficult in this? It is simply "conservation of energy" in "Newton's Third Law of Motion".

Here, we are only discussing action-reaction as per Newton's Third Law of motion.

7. ### hansdaValued Senior Member

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You mean to say, water is transferring force but not energy. Isn't this violation of Newton's Second Law of Motion?

8. ### originTrump is the best argument against a democracy.Valued Senior Member

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Yes.

The friction of the water trys to stop him so his KE is transfered to the water through friction. He is changing his chemical potential energy to kenetic energy in his arms and legs and the movement of his limbs propels him through the water, he is production enough KE, to overcome the frictional forces.

For every action there is an equal an opposite reaction, but you are asking why the swimmer moves. Frictional forces resist the movement of his limbs which results in a force on his limbs which is in the opposite direction of the friction, which results in a net force moving him forward

9. ### hansdaValued Senior Member

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Doesn't the "reaction-force" work on the swimmer? Do you mean "reaction-force" is same as "friction force"?

"Friction force" just opposes any movement. It will oppose action-force as well the reaction-force.

Swimmer moves due to reaction force from water.

"Friction-force" does not cause any motion. It only opposes motion. "Friction-force" only absorbs energy and does not supply any energy.

[note: Consider there is no frictional loss in this case of swimming]

10. ### originTrump is the best argument against a democracy.Valued Senior Member

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The swimmer moves due to the resistence (friction) of the water.

Come on. On a completely fricionless surface you could not walk - you wouldn't go anywhere. Does friction cause motion? Of course not, but your foot moving against the ground propels you because of the friction of the surface. Swimming is the same thing the friction or resistence of the water is what propels you.

You must be trying to not understand this. Why, what is the point?

11. ### hansdaValued Senior Member

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Well, can you clarify my question in post #280?

12. ### originTrump is the best argument against a democracy.Valued Senior Member

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I can't clarify your question, only you can. But it the energy transfer is in the form of a force, such as a pool ball hitting another pool ball for instance, the answer is Option 2.

13. ### hansdaValued Senior Member

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At-least you have clarified your views for option-2. It is also my view that, it is option-2. Option-2 is also matching with all the examples of action-reaction pair as per Newton's Third Law of Motion.

In action-reaction pair energy transfers in the form of force only. As energy transfers from one mass to another mass, it also generates a force as "F=dE/dx" or "F=ma".

Let us consider your example of pool-ball. Say there are two pool-ball A and B. Pool-ball A strikes pool-ball B. So, pool-ball A imparts some kinetic-energy(E) and force(F) to pool-ball B. This energy "E" should be conserved in the action-reaction of these two ball, as it is the external input energy.

Energy transfer in action-force(F) from pool-ball A to pool-ball B is "E".

Energy transfer in reaction-force(F) from pool-ball B to pool-ball A is "E".

If action and reaction are happening simultaneously, we get total energy "2E" in the system due to input energy "E"(from A to B). This is contradicting "conservation of energy" principle. So, this may not be true.

14. ### originTrump is the best argument against a democracy.Valued Senior Member

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1. Fine lets use F=ma. Lets assume that these are perfect pool balls and it is a completely elastic collision.

The masses of the balls are identical. m1 = m2
The velocity of ball A before the collision is equal to the velocity of the ball B after the collision. V1 = V2

For ball A $m_1 (-\frac{dV_1}{dt}) = -F$ The force is in the negative direction because the ball is decelerating.

For ball B $m_2 (\frac{dV_2}{dt}) = F$ The force is in the positive direction because the ball is accelerating.

The positive and negative convention is not important what is important is that the forces are in opposite directions.
This change in forces between the balls does not occur instantaneously but it does occur simultaneously.

At each instant in time as the ball A decelerates there is a force in the negative direction, simultaneously there an acceleration of ball B resulting in a force in the positive direction. Equal and opposite forcese. The KE of ball A decrease as the velocity slows due to the deceleration and an increase in the KE of ball B as the velocity increases due to the acceleration.

There is no 2X energy.

In the less idealized case the deceleration and acceleration are partially expressed as deformation of the body and not bulk movement of the balls.

15. ### hansdaValued Senior Member

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You have considered here two mass m1 and m2 as equal.

Now consider mass m2 is much much greater than mass m1. Say mass m2 is "a wall" and mass m1 is "a ball" or "a man pushing the wall". How do you think the force equations will be in this case?

Also calculate energy-conservation here.

You can refer following examples for action-reaction pair.

1. http://www.physchem.co.za/OB11-mec/law3.htm

2. http://physics.tutorvista.com/motion/newton-s-third-law-of-motion.html

3. http://www.qrg.northwestern.edu/pro...2-every-action-has-an-equal-and-opposite.html

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17. ### hansdaValued Senior Member

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That's why you are still not able to clarify my question in post #280.

18. ### eramSciengineerValued Senior Member

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No, #280 wasn't vague, it was nonsense.

19. ### originTrump is the best argument against a democracy.Valued Senior Member

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No thanks, I can only go around so many times on this merry-go-round. You have my permission to believe what ever you want, if you do not understand something feel free to substitute any sort of nonsense you like.

20. ### hansdaValued Senior Member

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The way you have explained "Reaction-Force" here; it is always decelerating and can never cause an acceleration.

21. ### eramSciengineerValued Senior Member

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What is that supposed to mean?

I don't even know how to start, because your understanding is too poor at the moment.

Last edited: Apr 13, 2013
22. ### hansdaValued Senior Member

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What is your difficulty in understanding that. The way you are projecting a "Reaction-Force", it can only cause deceleration and no acceleration.

Do you know that, a "Reaction-Force" can also cause "propulsion".

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