# Couple Newton Questions

1. Jan 15, 2010

### DocZaius

Couple questions regarding Newton:

1) Sort of a historical question...In Newton's time, how did Newton explain the stopping of an object due to friction? I imagine that would have been the first objection to his first law. Could he have said that the earth's momentum had changed due to the stopping of the object? Could his detractors not have said that such a change in the earth's momentum cannot be measured? I am merely curious about how easily his laws, especially in regards to interactions with the earth (where one side of the equation cannot be measured) went over with his contemporaries. I am looking at this purely from a conservation of momentum standpoint, rather than conservation of energy.

2) I often read that Newton's Second Law should be expressed as F = dp/dt rather than F = ma. When mass is treated as a constant (and concerning external forces, it should be), what's the simplest example of a calculation where using F = ma gives the wrong answer (rather than be less convenient), if any? Or is it just that Newton's point in his third law was in regards to momentum, and to speak of momentum rather than acceleration in his second law would be more consistent?

2. Jan 15, 2010

### flatmaster

Your second question requires some math.

Begin with

F=ma

Acceleration is the change in velocity divided by the change in time.

F = m dv/dt

mass does not change with time, so it can be taken inside of the deravitive.

F - dmv/dt

The definition of momentum is.

p=mv.

Thus

F = dp/dt

3. Jan 15, 2010

### DocZaius

You misunderstand my second question. I am asking under what conditions would approaching a problem with the F=ma version lead to the wrong answer. I am familiar with how to get from F=ma to F=dp/dt. If the answer is never, then my next question is why is it always reminded to us that the proper version is F=dp/dt? I then propose that it's to stay consistent with the statement of the third law, but that this proposition could be incorrect.

4. Jan 15, 2010

### Nabeshin

The simplest example I often come across is a rocket. Rockets expend fuel, and thus decrease in mass as they burn their engines, so the simple f=ma is not adequate to describe their motion.

5. Jan 15, 2010

### flatmaster

My guess would be to remind you that newton's third law is a 3D differential equation. F=ma dosen't look like a differential equation. However, F=dp/dt does. As long as you apply F=ma correctly, the two are equivalent. I suppose the correct form of F=ma would be

F=m (d^2x/d^t)
or
F=m x double dot

or F=m (second time derivative of position)

6. Jan 15, 2010

### DocZaius

My understanding is that neither of the forms of the second law are appropriate for systems with variable mass.

http://en.wikipedia.org/wiki/Newton's_laws_of_motion (last paragraphs of the second law section)

Although perhaps your point is that one can only get to the equations for systems with variable masses with the dp/dt version of the second law?

Last edited: Jan 15, 2010
7. Jan 15, 2010

### Cleonis

Well, Newton's formulation in the Principia revolves around the concept of quantity of motion, which today we call momentum.

The Third law focuses on the fact that in an interaction of two masses, exerting a force upon each other, the total momentum is conserved.

When two masses are interacting they are changing each others momentum. Seen in that way I think it's quite natural that Newton formulated the Second law in terms of change of momentum. It ties the Second and Third Law together

Cleonis

8. Jan 16, 2010

### Prologue

I just looked into this for a solid couple hours and basically have gotten the impression that dp/dt is wrong altogether. I won't write up anything now but there are problems if you use dp/dt in the rocket equations, you have to use f=ma and you are OK. So the short answer is (in my opinion), use f=ma and everything works, don't use f=dp/dt.

Although I have some quibbles with the paper, it still spells it out alright. Essentially when m is a function of t you have to be careful with what you actually look at as a 'force'.

f=ma still works in the moving/mass loss cases because the center of mass is still not accelerating. So F = 0. But now you have to use conservation of momentum to solve them, that is where they got the 'new' equation.

Last edited: Jan 16, 2010
9. Jan 16, 2010

### GRDixon

Consider the case where an applied force acts in the same (or opposite) direction as a particle's velocity. What force is required to give the particle a given linear acceleration, a? It is F=(gamma)^3(m0)a and not F=(gamma)(m0)a (where "m0" is the so-called rest mass). Note that both forms work equally well, in the case of radial acceleration. That is, the force needed to give a particle a radial acceleration of a is F=(gamma)(m0)a.

10. Jan 16, 2010

### payumooli

11. Jan 16, 2010

### GRDixon

I believe the answer is because F=dp/dt=d(mv)/dt is relativistically rigorous, whereas F=ma is not (m being relativisitic mass in each case). See reply #9 this thread. It's also interesting that dp/dt transforms identically to the Lorentz force in any given situation. But ma does not. The transformation of dp/dt is identical to the general force transformation (found in numerous texts). Interesting sidelight: the Lorentz force also transforms like the general force transformation. Thus dp/dt and the Lorentz force transform identically.

12. Jan 21, 2010

### ftfaaa

The second question may be easier.
Newton's second law should be expressed as F = dp/dt rather than F = ma,because F=dp/dt may be more universal than F =ma.
When the velocity is far smaller than the speed of light,the mass can be seen as a constant , F=ma and F=dp/dt can be all right. However,when it comes to the velocity that is close to the speed of light ,it will be wrong when you use F= ma ,because this time m is a variable (m=m0 /sqr(1-(v/c)^2) ) .

13. Jan 21, 2010

### Prologue

Yeah, I was talking strictly classical physics in my post. If m is a function of time dp/dt is wrong in classical physics.