Question about something i just read

[Johnny5]

I just read this site:

 

Deductive proof of Newton's third law

 

Look where he says:

Fnet = ma, tells us that the net (vector sum) of all forces acting on a body is equal to the product of the body's mass and its vector acceleration.

 

I think the verbal formulation of this is wrong.

 

Draw a circle. Then pick a point on the circumference.

Draw the tangent line. Draw the normal line.

 

If a force F is applied along the normal line, then all of that force will go into translating the object, because the force is aimed right at the center of mass.

 

But suppose the contact force isn't along that line, suppose that the angle of incidence is [math] \theta [/math]

 

Then the amount of force that goes into translating this circular shell, is going to be:

 

[math] F cos \theta [/math]

 

And the rest of the applied force is going to give it rotation. The rest of the applied force is:

 

[math] F sin \theta [/math]

 

 

So suppose that two forces act upon a spherical ball, in a plane containing the center of the ball.

 

Let the forces have equal magnitude but opposite direction.

 

But suppose that the points of contact are 180 degrees apart.

 

And let the angles of incidence be 89.9999999999 degrees.

 

So both applied forces will operate collectively to spin the ball. The ball will start spinning around an axis of rotation, but the center of inertia wont move from where it was. Hence there will be no acceleration of the center of inertia relative to the center of the universe.

 

Now, as this person has expressed Newton's second law, he just says that the sum of the forces on an object will equal that objects inertial mass, times the acceleration of that object.

 

But here the vector sum will be zero, since both forces have the same direction, but are anti-parallel.

 

So am I making an error, or is he?

 

Thank you

 

Regards

[swansont]

So both applied forces will operate collectively to spin the ball. The ball will start spinning around an axis of rotation' date=' but the center of inertia wont move from where it was. Hence[b'] there will be no acceleration of the center of inertia[/b] relative to the center of the universe.

 

Now, as this person has expressed Newton's second law, he just says that the sum of the forces on an object will equal that objects inertial mass, times the acceleration of that object.

 

But here the vector sum will be zero, since both forces have the same direction, but are anti-parallel.

 

So am I making an error, or is he?

(emphasis added)

 

Where is the contradiction?

[Johnny5]

(emphasis added)

 

Where is the contradiction?

 

Mmm I guess there is none the way I have it set up huh?

 

Well what if only one force acts, but the angle is 90 degrees relative to the normal, then what?

 

Regards

[swansont]

If the force is at 90 degrees, there is no acceleration of the center of mass. The second law still works.

 

You usually break things up into center-of-mass motion, in which case you assume point masses, and rotational systems.

[Johnny5]

If the force is at 90 degrees' date=' there is no acceleration of the center of mass. The second law still works.

 

You usually break things up into center-of-mass motion, in which case you assume point masses, and rotational systems.[/quote']

 

Correction: Yesterday I said angle of incidence, but I was referring to 90 - angle of incidence, but I can see you knew which angle I meant, because of this answer.

 

At any rate, you say that "if the force is at 90 degrees, there is no acceleration of the center of mass" <---- Yes, this statement is true. But what is the net force acting upon the object? It cannot be zero, can it?

 

Regards

[5614]

For an object to accelerate or decelerate the forces acting upon it must be unbalanced.

 

If an object is not accelerating or decelerating (so it's still or moving at a constant speed) the net force acting on it is 0.

[Johnny5]

For an object to accelerate or decelerate the forces acting upon it must be unbalanced.

 

If an object is not accelerating or decelerating (so it's still or moving at a constant speed) the net force acting on it is 0.

 

I understand your comment but something is still bugging me.

 

Does Newton's second law (as usually stated verbally at least) properly handle cases where the applied force at the point of contact, is not aimed right at the center of mass of the body?

 

To state this another way, Newton's second law doesn't verbally say anything about the importance of the direction of a contact force in relation to the point of contact, and that matters since a portion of the force will go into translating the object, and the rest will go into rotating it. I am having a problem with the verbal translation of the law, not the law itself.

 

 

Regards,

[swansont]

You usually break things up into center-of-mass motion, in which case you assume point masses, and rotational systems.

 

.

[Johnny5]

Ok Swanson, how do you verbally express the second law?

 

Regards

[ydoaPs]

"Lex II. Mutationem motus proportionalem esse vi motrici impressae, & fieri secundum lineam rectam qua vis illa imprimatur."

[Johnny5]

"Lex II. Mutationem motus proportionalem esse vi motrici impressae, & fieri secundum lineam rectam qua vis illa imprimatur."

 

Translation: The change in motion is essentially proportional to the impressed force, and (unintelligible) the straight line (unintelligible).

 

Huh?

 

Regards

[Skye]

The second law still works for rotational systems, you just assume that there are lots of individual masses.

 

Eg. http://theory.uwinnipeg.ca/physics/rot/node5.html

[Johnny5]

The second law still works for rotational systems' date=' you just assume that there are lots of individual masses.

 

Eg. http://theory.uwinnipeg.ca/physics/rot/node5.html[/quote']

 

You misunderstand my question. My question is not about the mathematical applicability of the law, my question is about translating it coherently, into English. As you can see, there are still modern questions as to its meaning, by the very article which I sited. In fact, you cannot prove that Newton's first law comes from his second, as is stated there. All the second Law is going to give you, is conservation of momentum.

 

In usual expressions of it, no reference is made to the point of contact. I have a problem with that is all.

 

Regards

[swansont]

You misunderstand my question. My question is not about the mathematical applicability of the law' date=' my question is about translating it coherently, into English. As you can see, there are still modern questions as to its meaning, by the very article which I sited. In fact, you cannot prove that Newton's first law comes from his second, as is stated there. All the second Law is going to give you, is conservation of momentum.

 

In usual expressions of it, no reference is made to the point of contact. I have a problem with that is all.

 

Regards[/quote']

 

That's because it is a law, i.e. a short mathematical expression (or an equivalent). F=ma

 

It doesn't tell you how and when to apply it. That is separate knowledge. As in, you use this incarnation for center-of-mass motion. If you have an off-axis force, i.e. a torque, there is an analogous expression, ^{[math] \tau = I\alpha [/math]}

[Johnny5]

That's because it is a law' date=' i.e. a short mathematical expression (or an equivalent). F=ma

 

It doesn't tell you how and when to apply it. That is separate knowledge. As in, you use this incarnation for center-of-mass motion. If you have an off-axis force, i.e. a torque, there is an analogous expression, [sup'][math] \tau = I\alpha [/math][/sup]

 

Follow me here. I am interested in a proper translation of Newton's second law into English.

 

I will now quote you:

 

"(F=ma) doesn't tell you how and when to apply it... you use this incarnation for center-of-mass motion"

 

Let us suppose what you say is true. Then it is you who have now added the stipulation that F=ma can only be used for "center of mass motion."

 

That phrase came from you, not Newton. Nor is it anywhere to be found in customary translations of F=ma into English, nor is it even clear what it means.

 

Must that phrase be added to the customary English translation??? Tell me your answer in the next post. I predict this. You are going to run into a logical problem, the moment you try to convert the meaning of "F=ma" into English, simply because your translation won't take into account where the force is applied.

 

Let me put it to you my way, so that you clearly understand what it is I am trying to say:

 

F=ma works when the applied contact force points right at the center of inertia. You can then treat the motion of the center of inertia as a single particle with mass M, where M is the inertial mass of the entire object you pushed. BUT, and here is the whole point...

 

If the direction of the contact force is not aimed right at the center of mass, then customary English translations of F=ma do not explain what to do, so either the meaning of F=ma is unclear, or not enough information is being given, in order to express the meaning.

 

And another point, is also that it isn't always clear whether or not Newton's laws are to be taken for particles only, bodies, or both.

 

The laws get inferred from the motion of bodies, but in most textbooks, they are expressed as laws for particle motion.

 

So my point, which I still maintain, is that "F=ma" does not get translated into English correctly, and the source of the problem has to do with the point of contact of an applied force on a BODY, not a particle.

 

If I am not making myself clear, I can be more mathematical.

 

Regards

 

PS: I think what you want to say is something like this.

 

F=ma always applied to particles.

F=ma sometimes applies to bodies. It applies to bodies only when the the applied force points at the center of inertia.

If an external force does not point right at the center of inertia, then the portion of the applied force which points at the center of inertia will go into translating the center of inertia, and the portion of the applied force which points perpendicular to the normal drawn through the point of contact will go into rotating the object about an axis. And for rotation the following mathematical statement expresses the law of physics:

 

[math] \tau = I \alpha [/math]

 

Where I is what is called the moment of inertia, and alpha is the angular acceleration. The moment of inertia can be calculated precisely, only when the shape of the object is perfectly known, like a sphere, a cube, cylinder, etc, and when the density of the object is either uniform, or a simple function which can be expressed in the rest frame of the body.

 

Etc etc, on and on.

 

In other words, rather than 'F=ma' expressing the law of physics, there is a loophole where a whole bunch of neverending garbage has to come out. This is unacceptable.

 

Regards

[J.C.MacSwell]

Follow me here. I am interested in a proper translation of Newton's second law into English.

 

I will now quote you:

 

"(F=ma) doesn't tell you how and when to apply it... you use this incarnation for center-of-mass motion"

 

Let us suppose what you say is true. Then it is you who have now added the stipulation that F=ma can only be used for "center of mass motion."

 

That phrase came from you' date=' not Newton. Nor is it anywhere to be found in customary translations of F=ma into English, nor is it even clear what it means.

 

Must that phrase be added to the customary English translation??? Tell me your answer in the next post. I predict this. You are going to run into a logical problem, the moment you try to convert the meaning of "F=ma" into English, simply because your translation won't take into account where the force is applied.

 

Let me put it to you my way, so that you clearly understand what it is I am trying to say:

 

F=ma works when the applied contact force points right at the center of inertia. You can then treat the motion of the center of inertia as a single particle with mass M, where M is the inertial mass of the entire object you pushed. BUT, and here is the whole point...

 

If the direction of the contact force is not aimed right at the center of mass, then customary English translations of F=ma do not explain what to do, so either the meaning of F=ma is unclear, or not enough information is being given, in order to express the meaning.

 

And another point, is also that it isn't always clear whether or not Newton's laws are to be taken for particles only, bodies, or both.

 

The laws get inferred from the motion of bodies, but in most textbooks, they are expressed as laws for particle motion.

 

So my point, which I still maintain, is that "F=ma" does not get translated into English correctly, and the source of the problem has to do with the point of contact of an applied force on a BODY, not a particle.

 

If I am not making myself clear, I can be more mathematical.

 

Regards

 

PS: I think what you want to say is something like this.

 

F=ma always applied to particles.

F=ma sometimes applies to bodies. It applies to bodies only when the the applied force points at the center of inertia.

If an external force does not point right at the center of inertia, then the portion of the applied force which points at the center of inertia will go into translating the center of inertia, and the portion of the applied force which points perpendicular to the normal drawn through the point of contact will go into rotating the object about an axis. And for rotation the following mathematical statement expresses the law of physics:

 

[math'] \tau = I \alpha [/math]

 

Where I is what is called the moment of inertia, and alpha is the angular acceleration. The moment of inertia can be calculated precisely, only when the shape of the object is perfectly known, like a sphere, a cube, cylinder, etc, and when the density of the object is either uniform, or a simple function which can be expressed in the rest frame of the body.

 

Etc etc, on and on.

 

In other words, rather than 'F=ma' expressing the law of physics, there is a loophole where a whole bunch of neverending garbage has to come out. This is unacceptable.

 

Regards

 

F=ma

 

It does not matter if it acts through the centre of mass. It will accelerate the centre of mass of that "body" in the direction of the force. It does not matter what shape, rigidity or "coherence" the body, gas cloud or "deformable object" is getting "accelerated". The centre of mass will accelerate according to F=ma, in the direction of the force except for relativistic effects

 

If it does not act through the centre of mass it will also impart a torque or moment on the body.

[Johnny5]

F=ma

 

It does not matter if it acts through the centre of mass. It will accelerate the centre of mass of that "body" in the direction of the force. It does not matter what shape' date=' rigidity or "coherence" the body, gas cloud or "deformable object" is getting "accelerated". The centre of mass will accelerate according to F=ma, in the direction of the force [b']except for relativistic effects[/b]

 

If it does not act through the centre of mass it will also impart a torque or moment on the body.

 

What if the applied force is 90 degrees relative to the normal drawn through the point of contact? What then?

[J.C.MacSwell]

What if the applied force is 90 degrees relative to the normal drawn through the point of contact? What then?

 

Then you will accelerate the centre of mass according to F=ma

 

If you do this to a billiard ball in "outer space" (no table) you will also impart rotational motion.

 

Careful with the point of contact and direction of force as they will "want" to change if we are not careful.

[Johnny5]

Then you will accelerate the centre of mass according to F=ma

 

Mmm in what direction?

 

Regards

[J.C.MacSwell]

Mmm in what direction?

 

Regards

 

Only in the direction of the force.

[Johnny5]

Suppose you have a line mass distribution

 

..........................................

 

 

and something pushes the left most particle, leftmost side. How does the center of mass move?

[J.C.MacSwell]

Suppose you have a line mass distribution

 

..........................................

 

 

and something pushes the left most particle' date=' leftmost side. How does the center of mass move?[/quote']

Only in the direction of the force and according to F=ma

[Johnny5]

Only [/b'] in the direction of the force and according to F=ma

Ok lets try something.

 

Suppose the mass of each particle is 1.67 x 10^-27 kilograms, and that there are 100 such particles at rest with respect to each other, but held together firmly, so that the line mass is rigid.

 

Then there is some kind of interaction with the leftmost particle, and the force there has magnitude F, and the direction is up (perpendicular to the line charge).

 

Describe for me the path which the leftmost particle takes, as a function of time.

 

I understand that you are going to take some of the applied force, a fraction of F, and say that this portion of F acts as if it pushed the center of mass upwards. Therefore, the center of mass of the system will have some upwards motion.

 

The rest of the force will go into causing some rotation.

 

But here is where I am going with this. I want to watch your solution to the problem, and see if you use the concept of an inertial wave.

 

Regards

[J.C.MacSwell]

Ok lets try something.

 

Suppose the mass of each particle is 1.67 x 10^-27 kilograms' date=' and that there are 100 such particles at rest with respect to each other, but held together firmly, so that the line mass is rigid.

 

Then there is some kind of interaction with the leftmost particle, and the force there has magnitude F, and the direction is up (perpendicular to the line charge).

 

Describe for me the path which the leftmost particle takes, as a function of time.

 

I understand that you are going to take some of the applied force, a fraction of F, and say that this portion of F acts as if it pushed the center of mass upwards. Therefore, the center of mass of the system will have some upwards motion.

 

The rest of the force will go into causing some rotation.

 

But here is where I am going with this. I want to watch your solution to the problem, and see if you use the concept of an inertial wave.

 

Regards[/quote']

All of the force will move the center of mass in the upward direction

[Johnny5]

All [/b'] of the force will move the center of mass in the upward direction

 

Why?