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

[studiot]

Which is exactly why I asked @martillo to start at the beginning.

2 minutes ago, swansont said:

Right. Because that’s not how they derived the equation.

Since dp/dt is zero (there is no net force) it means momentum is conserved. They solve the problem by applying that principle.

 

When the rocket is standing ready to launch it has exactly zero momentum.

So we can write a conservation of momentum equation right there.

This is usually taken for granted and also what I am referring to when I asked about what are they hiding ?

7 minutes ago, joigus said:

My comment was about maths and definitions, and what you decide to call 'the force on the system.' This requires to define what you mean by 'the system.' If 'the system' is both the rocket, the fuel and the exhaust, then all forces implied are internal and there is no net force. If it's just the rocket with the fuel that's left, well... You see? 

 

Of course but what I am saying is that in order to solve or analyse this problem we start with a complete rocket and fuel as 'the system' and then add further equations coming from the application of mechanics to parts of that system.

I further point out that all of the equations are simultaneous so that one can substitute from one into another
Again a tried and tested technique.

I am also say that there is nothing revolutionary in this, indeed the technique is well used in many simpler systems.

[Lorentz Jr]

8 hours ago, Lorentz Jr said:

F = dp/dt is derived from F = ma.

39 minutes ago, swansont said:

Other way around

I should have said "can be" instead of "is".

Is there something wrong with this, in the context of rockets?

Edited February 21, 2023 by Lorentz Jr

[martillo]

31 minutes ago, Markus Hanke said:

It depends what you mean by “problem”. You were asserting in your OP that F=dp/dt isn’t the correct formulation of Newton’s 2nd law, so I tried to provide you with the bigger picture of where this relation comes from. My point was to show you that it isn’t just made up (and thus potentially wrong), but that it follows from more general principles. It’s essentially a special case of the principle of least action. F=ma then is in turn a special case of F=dp/dt - it works well for many simple systems (which is why we all learn it in school), but its domain of validity is limited. F=dp/dt is more general, and also works in some cases where F=ma doesn’t. And then of course there are systems where neither of these work (as joigus has pointed out), and then you have to use the full Lagrangian formulation of mechanics. 

I understand that.

31 minutes ago, Markus Hanke said:

Either way, just claiming that F=dp/dt is wrong and must always be replaced by F=ma - which is what you seem to be saying - is silly, because that’s manifestly false. 

Could seem silly until analyzing appropriately the presented problem. Everybody (Science, developers, fabrics...) is using F = ma for a rocket which is something with variable mass. It must be generalized then for the general case. 

30 minutes ago, studiot said:

When the rocket is standing ready to launch it has exactly zero momentum.

So we can write a conservation of momentum equation right there.

This is usually taken for granted and also what I am referring to when I asked about what are they hiding ?

Well, fine, the initial total system is the rocket with its fuel...

You want to analyze the problem from the beginning again, fine, I will try for you.

Edited February 21, 2023 by martillo

[Lorentz Jr]

18 minutes ago, martillo said:

Well, fine, the initial total system is the rocket with its fuel...

You want to analyze the problem from the beginning again, fine, I will try for you.

Using F = ma (Newton's second law) for the mass dm_gas of exhaust gasses, we have the internal force of the rocket acting on the gas:

F_int = dm_gas (-v_e/dt) = -v_e dm_gas/dt

And the action-reaction rule (Newton's third law) says the exhaust gasses exert an equal and opposite force on the rocket. Defining F_ext as any external force that might be acting on the rocket, F = ma for the rocket is

F_tot = F_ext + F_int =  F_ext + v_e dm_gas/dt = m_rocket a = m_rocket dv/dt

Defining the final (backwards) speed of the exhaust gasses v_x = v_e - v, and remembering dm_rocket = -dm_gas,

F_ext = -v_e dm_gas/dt + m_rocket dv/dt = -(v + v_x) dm_gas/dt + m_rocket dv/dt = v dm_rocket/dt + m_rocket dv/dt - v_x dm_gas/dt

F_ext = d(m_rocketv)/dt + (-v_x) dm_gas/dt = dp_rocket/dt + dp_gas/dt = dp/dt

You can see there are two contributions to dp/dt: The change in mv of the rocket, plus however much (forward) momentum is carried away by the exhaust gasses.

Edited February 21, 2023 by Lorentz Jr

[Markus Hanke]

13 minutes ago, martillo said:

Could seem silly until analyzing appropriately the presented problem.

No, it’s silly because - as I have already shown you - it directly follows from the principle of least action (as well as other principles, as pointed out by several contributors here), so it makes no sense to claim that it is “wrong”. 

[martillo]

2 minutes ago, Lorentz Jr said:

Using F = ma (Newton's second law)

You see, you are using F = ma and F = dp/dt.

4 minutes ago, Lorentz Jr said:

F = ma for the rocket is

F_tot = F_ext + F_int =  F_ext + v_e dm_gas/dt = m_rocket a = m_rocket dv/dt

Here again.

It is always used F = ma for the rocket,always.

[Lorentz Jr]

5 minutes ago, martillo said:

You see, you are using F = ma and F = dp/dt.

But I'm not solving a rocket problem, martillo. I'm using F = ma to derive F = dp/dt so people can use F = dp/dt to solve rocket problems.

5 minutes ago, martillo said:

It is always used F = ma for the rocket,always.

What do you mean by "used", martillo? Most people use F = dp/dt in their calculations. How are they "using" F = ma?

Edited February 21, 2023 by Lorentz Jr

[martillo]

14 minutes ago, Markus Hanke said:

No, it’s silly because - as I have already shown you - it directly follows from the principle of least action (as well as other principles, as pointed out by several contributors here), so it makes no sense to claim that it is “wrong”. 

The "principle of east action" is a variational principle. As I said, I'm not able to discuss the problem in terms of Variational Calculus". I have no expertisse at all in that.

9 minutes ago, Lorentz Jr said:

What do you mean by "used", martillo? Most people use F = dp/dt in their calculations. How are they "using" F = ma?

Everybody is using F = ma for the rockets, Physics Science, developers, fabrics, and not use F = dp/dt. You can search for this in the web. They just use the Thrust Equation which is derived with F = ma. That's my claim. F = ma is the right equation then. 

Edited February 21, 2023 by martillo

[Genady]

5 minutes ago, martillo said:

The "principle of east action" is a variational principle. As I said, I'm not able to discuss the problem in terms of Variational Calculus". I have no expertisse at all in that.

This is a dishonest response. He doesn't tell you to discuss the problem in terms of Variational Calculus.

He tells you that F=dP/dT follows from more general principles rather than is just made up, and thus it makes no sense to claim that it is wrong, as you do.

Edited February 21, 2023 by Genady

[Lorentz Jr]

5 minutes ago, martillo said:

Everybody is using F = ma for the rockets

What do you mean by "using", martillo? Most people solve rocket problems without every writing down "F = ma". How can they be using F = ma if they never write it down?

[studiot]

8 minutes ago, martillo said:

It is always used F = ma for the rocket,always.

always ??

I'm fed up with members that don't bother to read what's offered, just argue apparantly for the sake of it.

Here is a simple treatment which explains why using f = ma makes life more difficult.

Because m is not necessarily constant.

Professor Sears had a particularly clear way of putting matters.

 

[martillo]

28 minutes ago, Genady said:

This is a dishonest response. He doesn't tell you to discuss the problem in terms of Variational Calculus.

He tells you that F=dP/dT follows from more general principles rather than is just made up, and thus it makes no sense to claim that it is wrong, as you do.

Note dishonest, no. I meant I'm not able to analyze how the principle is applied to derive the equation of motion. Is a matter of VAriational Calculus and I have no expertise at all in this area.

28 minutes ago, studiot said:

always ??

I'm fed up with members that don't bother to read what's offered, just argue apparantly for the sake of it.

Here is a simple treatment which explains why using f = ma makes life more difficult.

Because m is not necessarily constant.

Professor Sears had a particularly clear way of putting matters.

 

Difficult to read but I did. In the third page is written:

This is to apply F = ma. Exactly the same it is said in at the link I provided:

"The right-hand term depends on the characteristics of the rocket and, like the left-hand term, has the dimensions of a force. This force is called the thrust, and is the reaction force exerted on the rocket by the mass that leaves it."

Saying like "the left term is the force..." is to apply F = ma.

28 minutes ago, Lorentz Jr said:

What do you mean by "using", martillo? Most people solve rocket problems without every writing down "F = ma". How can they be using F = ma if they never write it down?

They use the Thrust Force F = -v_edm/dt of the rocket which is derived with F = ma. So they use F = ma at the end. Please read the answer above for @studiot to see better.

Edited February 21, 2023 by martillo

[studiot]

18 minutes ago, martillo said:

This is to apply F = ma. Exactly the same it is said in at the link I provided:

But Sears started, where I and swansont suggested you start  - with conservation of mementum.

Sears also derives the conditions where f = ma is applicable first, before he uses it also as I indicated in my last post.

 

20 minutes ago, martillo said:

Difficult to read but I did. In the third page is written:

Scienceforums offers you the facility to view enlarged vrsions of the attachments, and I specifically uploaded large enough versions for this.

At full enlargement it seems just fine to me.

 

Note yet again I am saying ( as are others) you need both versions for the rocket.

 

You still haven't acknowledged what I said about simultaneous equations.

 

As an electrical engineer try this.

 

V = IR

and

P = IV

 

Are two different equations about different physical quantities.

Because they have some common simultaneous variables you can substitute to find an equation for say power in terms of either voltage or current and resistance.

P = V²/r = I²R

These can be extended further to find energy and other properties.

 

The rocket is an equivalent situation in mechanics.

[martillo]

8 minutes ago, studiot said:

But Sears started, where I and swansont suggested you start  - with conservation of mementum.

Of course it must be started on the equation dp/dt in the total system of constant mass and so dp/dt = 0. 

8 minutes ago, studiot said:

You still haven't acknowledged what I said about simultaneous equations.

Yes I did:

 

2 hours ago, martillo said:

Well, right. We have two equations to apply. One is the Thrust Equation (mdv/dt = v_rel(dm/dt)), the second is the problem, which equation to apply: F = ma or F = dp/dt. F = ma is applied and gives right results for rockets (they work isn't it?). Then follows that the right equation of motion is F = ma and not F = dp/dt, even when mass varies. Don't you get it?

 

Edited February 21, 2023 by martillo

[studiot]

21 minutes ago, martillo said:

dp/dt = 0. 

But I didn't say that and it isn't true all the time.

 

I said

At the start the momentum (note not the rate of change of momentum) is exactly zero.

That is not the same thing.

dp/dt comes later in the analysis when we can say that it starts off at zero and is conserved so the total momentum of the whole system is always zero.

That fact lets us form an equation between parts of the system that do not have zero momentum by themselves.

So we can then say that the upward momentum - the downward momentum must equal zero

Which is a very useful condition.

This equality leads us to consider the third law forces between the parts by way of the second law.

 

The logic of all this is inexorable if you follow it correctly.

But it is easy to leave a step or two out when you are so familiar with it you become bored with it.

[martillo]

10 minutes ago, studiot said:

But I didn't say that and it isn't true all the time.

The initial dP/dt = 0 in the total system is always valid since the total mass is constant. What matters is after. As you said:

10 minutes ago, studiot said:

That fact lets us form an equation between parts of the system that do not have zero momentum by themselves.

The point is that now we have parts with variable masses and so here is where we must pay much attention which formula is used. It is found that it is used F = ma and not F = dp/dt.

This is the point.

 

Edited February 21, 2023 by martillo

[swansont]

1 minute ago, martillo said:

The point is that now we have parts with variable masses and so here is where we must pay much attention which formula is used. It is found that it is used F = ma and not F = dp/dt.

What equation they used does not impact the validity of an equation they didn’t use.

I don’t see PV = nRT anywhere. Is it not valid because they didn’t use it?

[studiot]

1 minute ago, martillo said:

The point is that now we have parts with variable masses and so here is where we must pay much attention which formula is used. It is found that it is used F = ma and not F = dp/dt.

This i the point.

So as you have been told several times

f = ma is used but only in the appropriate place and is not the beginning of the dynamical analysis of a rocket.

 

3 minutes ago, martillo said:

The initial dP/dt = 0 in the total system is always valid since the total mass is constant. What matters is after. As you said:

Only in a simple analysis.

 

[martillo]

41 minutes ago, swansont said:

What equation they used does not impact the validity of an equation they didn’t use.

It is used F = ma on something with variable mass. Don't you see this at least as suspicious?

41 minutes ago, studiot said:

f = ma is used but only in the appropriate place and is not the beginning of the dynamical analysis of a rocket.

F = ma is used in the middle over an object with variable mass. Suspicious or not?

Edited February 21, 2023 by martillo

[studiot]

1 hour ago, martillo said:

It is used F = ma on something with variable mass. Don't you see this at least as suspicious?

F = ma is used in the middle over an object with variable mass. Suspicious or not?

Why should that be suspicious ?

All the equation tells us is that force is the product of two independent quantities.

Independent means that each or both can be varied without reference to the other.

Setting one or both constant just makes for easier maths.

[martillo]

10 minutes ago, studiot said:

Why should that be suspicious ?

All the equation tells us is that force is the product of two independent quantities.

Independent means that each or both can be varied without reference to the other.

Setting one or both constant just makes for easier maths.

Well, if you don't see wrong the applying of F = ma on an object of variable mass then I have nothing else to say...

 

[studiot]

2 minutes ago, martillo said:

Well, if you don't see wrong the applying of F = ma on an object of variable mass then I have nothing else to say...

 

That was a particularly high handed dismissal of my request for you to describe what you see as 'suspicious'

You made the claim, not I, so it is up to you to justify it.

[martillo]

33 minutes ago, studiot said:

That was a particularly high handed dismissal of my request for you to describe what you see as 'suspicious'

You made the claim, not I, so it is up to you to justify it.

It is wrong in current Physics to consider F = ma on an object of variable mass. That's the point. And that is what is done in your book, in my provided link, in Wikipedia and everywhere else. Something wrong I have realized and wanted to share here in the forum.

Edited February 21, 2023 by martillo

[studiot]

9 minutes ago, martillo said:

It is wrong to consider F = ma on an object of variable mass. That's the point. And that is what is done in your book, in my provided link, in Wikipedia and everywhere else. Something wrong I have realized and wanted to share here in the forum.

Just saying it doesn't make it so.

I don't understand why you have this problem as an electrical engineer you must be familiar with the equation

Instantaneous power = instantaneous voltage x instantaneous current.

The analogy is exact

Instantaneous force = instanteous mass x instantaneous voltage.

 

So spit out your objection fully please.

[martillo]

15 minutes ago, studiot said:

Just saying it doesn't make it so.

I don't understand why you have this problem as an electrical engineer you must be familiar with the equation

Instantaneous power = instantaneous voltage x instantaneous current.

The analogy is exact

Instantaneous force = instanteous mass x instantaneous voltage.

 

So spit out your objection fully please.

In current Physics is widely considered F = dp/dt and that in the case, and only in the case of constant mass, the equation F = ma applies. Any Physics' textbook says that. So it would be wrong, in current Physics, to apply it on something with variable mass as it is done for rockets, or not? I do not understand how you don't see this.

If you consider F = dp/dt then F = mdv/dt + vdm/dt which would be equal to mdv/dt only if dm/dt = 0 what would mean constant mass.

Is not for me, for me is very right to apply F = ma always even for variable mass. For me is the right equation of force. That is in the title of the thread.

 

Edited February 21, 2023 by martillo