Physics in troubles: the real equation of force is F = ma and not F = dp/dt

[Genady]

Just now, martillo said:

No, in current Physics it applies for constant m only. This is clear in any basic Physics' text. 

Could you show me what exactly that physics text says?

[martillo]

28 minutes ago, swansont said:

They refer to them as forces because they are. They took momentum and applied dp/dt to it. dp/dt is a force, despite your insistence to the contrary.

This is quite clearly spelled out in the derivation.

What they did not do is apply F=ma to anything with variable mass. I don’t know how you can claim that they are.

 

 

 

 

 

Can you show then where and how they derive the Thrust Force from the Thrust Equation (Equation 1.5) applying F = dp/dt?

What I read is just to say (simplified): "the left-hand term is the force as it matches the units".

This is to apply F = mdv/dt = ma. Can't you see it?

24 minutes ago, Genady said:

Could you show me what exactly that physics text says?

Well, no, I don't have a basic Physics text. I have read that in innumerable physics texts all my life. Starting at college. I only have a Blatt's Physics' text but is not so basic, and in Spanish...

I can't believe you are asking me this.

Edited February 21, 2023 by martillo

[Genady]

24 minutes ago, martillo said:

I can't believe you are asking me this.

I doubt it says that the law is not applicable if mass changes.

43 minutes ago, martillo said:

No, in current Physics it applies for constant m only. This is clear in any basic Physics' text. 

Does it apply for constant a only?

[martillo]

13 minutes ago, Genady said:

Does it apply for constant a only?

No, for any acceleration a.

[Genady]

4 minutes ago, martillo said:

No, for any acceleration a.

OK. How it works if the acceleration changes, for example, continuously increases?

[martillo]

I hope you can understand...

6 minutes ago, Genady said:

OK. How it works if the acceleration changes, for example, continuously increases?

F is proportional to a if m is constant. F = ma then m is the constant of proportionality.

Edited February 21, 2023 by martillo

[Genady]

5 minutes ago, martillo said:

F is proportional to a if m is constant. F = ma then m is the constant of proportionality.

My question is, what value of a should be used in F=ma if a is changing in time?

[martillo]

9 minutes ago, Genady said:

My question is, what value of a should be used in F=ma if a is changing in time?

I don't know if I understand your point.

The equation must be solved. a depends on what force is applied to the object. For instance, for a constant force F_^c then a = F_^c/m.

Edited February 21, 2023 by martillo

[Genady]

13 minutes ago, martillo said:

I don't know if I understand your point.

The equation must be solved. a depends on what force is applied to the object. For instance, for a constant force F_^c then a = F_^c/m.

Let's say you observe a body moving with a variable acceleration and you want to know a force.

[martillo]

4 minutes ago, Genady said:

Let's say you observe a body moving with a variable acceleration and you want to know a force.

For a constant mass m just F = ma.

We are going in circles... I think I will take some time out now...

Edited February 21, 2023 by martillo

[Genady]

1 minute ago, martillo said:

For a constant mass m just F = ma.

We are going in circles...

But you need to put some number for a to find F. Which number will you put if a is changing?

[martillo]

2 minutes ago, Genady said:

But you need to put some number for a to find F. Which number will you put if a is changing?

You have one equation with two variables. You must have one determined to find the other one.

Seems you are joking me...

Edited February 21, 2023 by martillo

[Genady]

23 minutes ago, martillo said:

You have one equation with two variables. You must have one determined to find the other one.

Seems you are joking me...

I can't believe you don't get this simple question, but I try again. You have acceleration changing in time, you observe it, you get the values as they change. Mass is known. You need to find force. One equation, one unknown. 

[swansont]

1 hour ago, martillo said:

Can you show then where and how they derive the Thrust Force from the Thrust Equation (Equation 1.5) applying F = dp/dt?

No, because that’s not where it’s derived.

 It’s derived from conservation of momentum, and they divide by delta t, which they make into a derivative by taking the limit as delta t becomes small, just as clearly they explain. equation 1.3 through equation 1.5

 

1 hour ago, martillo said:

What I read is just to say (simplified): "the left-hand term is the force as it matches the units".

Well, that simplification is wrong.They say it has dimension of force. The reason it’s a force is because it was arrived at by taking the time derivative of momentum, which is force, which they explained in equation 1.2.

 

1 hour ago, martillo said:

This is to apply F = mdv/dt = ma. Can't you see it?

No. They didn’t apply F=m dv/dt, they derived it, and identify it as the reaction force

 

[martillo]

1 hour ago, Genady said:

You have acceleration changing in time, you observe it, you get the values as they change.

I can assume a = a(t) as known then, right?

1 hour ago, Genady said:

I can't believe you don't get this simple question, but I try again. You have acceleration changing in time, you observe it, you get the values as they change. Mass is known. You need to find force. One equation, one unknown. 

May be is a joke but fine. I will try to follow you...

Assuming a = a(t) then you will have F = F(t) = ma(t).

Now what? What does this mean for you?

[Genady]

1 minute ago, martillo said:

What does this mean for you?

It means to me that F=ma applies to momentary value of a if a changes in time.

[martillo]

2 hours ago, swansont said:

They didn’t apply F=m dv/dt, they derived it, and identify it as the reaction force

Then you at least agree that they use the equation F = mdv/dt = ma for the motion of the rocket right?

Then there's a problem because in current Physics this equation F = ma is valid for constant values of m only while the rocket has m as a variable mass.

16 minutes ago, Genady said:

It means to me that F=ma applies to momentary value of a if a changes in time.

Well, if you call an instant "momentary value" fine for me, I can handle that, but in current Physics this equation F = ma is valid for constant values of m only as it is stated for instance in the sample photo of the book I posted for you. Is this what I said right for you?

Edited February 22, 2023 by martillo

[swansont]

7 minutes ago, martillo said:

You at least agree that they use the equation F = mdv/dt = ma for the motion of the rocket right?

I don’t know why you would ask this, when you don’t (and can’t) show that they do this in the link. They do not appear to use the equation you cite. I can’t find it in the link. I think you’re making this up. 

 

 

[martillo]

9 minutes ago, swansont said:

I don’t know why you would ask this, when you don’t (and can’t) show that they do this in the link. They do not appear to use the equation you cite. I can’t find it in the link. I think you’re making this up. 

 

 

You said above:

2 hours ago, swansont said:

They didn’t apply F=m dv/dt, they derived it, and identify it as the reaction force

You said this. So you agree then, that it is considered in the link that the force on the rocket is F = ma. 

Edited February 22, 2023 by martillo

[Genady]

36 minutes ago, martillo said:

in current Physics this equation F = ma  it is valid for constant values of m only as it is stated for instance in the sample photo of the book I posted for you. Is this what I said right for you?

Even with my poor Spanish - it is not one of the four languages I've mentioned earlier - I understand that it is not what the book says.

It does not say that F=ma is valid for constant value of m only.

It says, that if a momentum changes only due to a change of velocity while the mass is constant, then from F=dP/dt we can derive a familiar form of the Newton second law, F=ma.

It does not mean, that this form, F=ma, is not valid otherwise. And it is valid. An instant value of force equals an instant value of mass multiplied by an instant value of acceleration. F(t) = m(t) x a(t). Nowhere in textbooks on Newtonian mechanics I ever saw that this is not so.

[martillo]

2 minutes ago, Genady said:

Even with my poor Spanish - it is not one of the four languages I've mentioned earlier - I understand that it is not what the book says.

It does not say that F=ma is valid for constant value of m only.

It says that.

You said:

3 minutes ago, Genady said:

It says, that if a momentum changes only due to a change of velocity while the mass is constant, then from F=dP/dt we can derive a familiar form of the Newton second law, F=ma.

You see, "while the mass is constant"... 

Now:

4 minutes ago, Genady said:

It does not mean, that this form, F=ma, is not valid otherwise. And it is valid. An instant value of force equals an instant value of mass multiplied by an instant value of acceleration. F(t) = m(t) x a(t). Nowhere in textbooks on Newtonian mechanics I ever saw that this is not so.

Well, in this case you are welcome! You then believe that F = ma is valid any case even for variable mass. That is what I'm defending in all this thread!

 

[Genady]

5 minutes ago, martillo said:

Well, in this case you are welcome! You then believe that F = ma is valid any case even for variable mass. That is what I'm defending in all this thread!

There is nothing to defend. You are mistaken when you say that textbooks say otherwise. 

Also in your book. It says, if mass is constant then from F=dP/dt we get F=ma. But it does not say that the mass must be constant.

[martillo]

6 minutes ago, Genady said:

Also in your book. It says, if mass is constant then from F=dP/dt we get F=ma. But it does not say that the mass must be constant.

By defintions:

p = mv then dp/dt = mdv/dt + vdm/dt = ma + vdm/dt

The unique way for dp/dt to be equal to ma is while mass is constant (dm/dt = 0). No other case. Then dp/dt = ma only, and only if m is constant it can be valid.

Edited February 22, 2023 by martillo

[Genady]

13 minutes ago, martillo said:

By defintions:

p = mv

No, this formula holds only for momentum of a system with constant, non-zero mass. Also in your book, they substituted P by mv after they assumed that the mass is constant.

In other cases, the momentum formula is different and depends on specifics of the case.

[martillo]

14 minutes ago, Genady said:

No, this formula holds only for momentum of a system with constant, non-zero mass. Also in your book, they substituted P by mv after they assumed that the mass is constant.

In other cases, the momentum formula is different and depends on specifics of the case.

Momentum p is defined as P = mv. By definition. Mass being constant or not, always.

I never see momentum defined other way.

Edited February 22, 2023 by martillo