Can you prove Newton's Second Law from this experiment?


by titaniumpen
Tags: gravity, newtons second law, proof 2nd law
physwizard
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Jan6-12, 12:33 PM
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Quote Quote by Simon Bridge View Post
You cannot prove the force law ... it the definition of what the Newton means when he uses the word "force". "Force" is defined to be a shorthand word for "the rate of change of momentum".

You are correct that you have to use the force law to determine the force due to the falling weight. Thus the experiment has set up a circular argument.

The best this shows is that falling motion will couple to horizontal motion in a manner consistent with newton's law. So the rules are internally consistent.

What you are seeing is a very common mistake in junior physics classes.

You could also use a newton-meter to drag an object and record the force you use and the acceleration you produced ... no falling weight and you measure force directly... but wait: a Newton-meter actually measures the extension of a spring, converting to force by Hook's Law. The calibration is done by using the 2nd Law, so once again.

I'm sure you can think of others.
thats right. its not really a law, it is the definition of 'force'
Andrew Mason
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Jan6-12, 03:19 PM
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Quote Quote by titaniumpen View Post
As far as I know, you can't prove laws, but anyway...

We got to do an experiment at school. A weight is suspended vertically from a string which is connected to a trolley which is placed horizontally on a flat table. There's a pulley at the edge of the table to reduce friction. Then we let the weight fall due to gravity and pull the trolley across the table. The trolley pulls along a tape as it moves, and the tape has to go through a ticker timer, which automatically dots the tape every 0.2 second.

The set up looks like this:
http://www.mathsrevision.net/alevel/...g%20newton.JPG

Anyway, we studied the tape and tried to measure the acceleration. We know the force exerted by the falling weight, and we also know the acceleration of the trolley from the tape. We also know the mass of the trolley. So if we put the values into F=ma, which is Newton's Second Law of Motion, we should find that both sides of the equation is the same. Which proves that Newton's Second Law is true!

Now, I don't think you can prove the Second Law like that. How do we know the force exerted by the falling weight on the trolley? The force is mg, right? (m=mass of weight, g=9.81m/s^2) But that is determined using Newton's Second Law. We cannot prove a law by using the law itself. Isn't that a circular argument?

Thanks for reading, this got me thinking for some time...
Getting back to the original post here, what you are trying to do is not to prove Newton's second law in the mathematical sense but to merely verify it - to show that it gives a correct prediction of what happens in nature.

This experiment is sometimes used in high school physics to demonstrate the second law, but it has a particular flaw: the tension force on the string pulling the trolley varies not only with the mass of the falling weight but also with mass of the trolley. What you need is the same constant pull on the trolley for different trolley masses. You would need to pull the car with elastics or springs stretched a set amount.

If you do that, you can show that the acceleration is proportional to the number of springs or elastics and inversely proportional to the mass of the trolley, just as the second law predicts. That demonstrates that Newton's second law provides the correct result. And you will not have used Newton's second law in order to produce your data.

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AlonsoMcLaren
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Jan8-12, 09:13 AM
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Quote Quote by Andrew Mason View Post
If you do that, you can show that the acceleration is proportional to the number of springs or elastics and inversely proportional to the mass of the trolley, just as the second law predicts. That demonstrates that Newton's second law provides the correct result. And you will not have used Newton's second law in order to produce your data.

AM
1. How do you define mass?

2. "You will not have used Newton's second law in order to produce your data" But you have to use 2nd law to show that force is additive. That is to say, the force provided by two springs stretched dx each is twice the force provided by one spring stretched dx.

If F=ma^2 (or some other weird stuff), force will not be additive.
Simon Bridge
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Jan8-12, 09:58 AM
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Getting back to the original post here, what you are trying to do is not to prove Newton's second law in the mathematical sense but to merely verify it
+1

The sort of discussion we read in Principia (Newton's, not Mal2s) is showing that the form chosen for the second law - the formal definition of force - is sensible in terms of common understandings of the time. I'd bet that most people would assume that two springs, as described, would provide twice the force without a formal definition.

The formalism, though, is why we have to be careful when we use words like force and work when talking to people unfamiliar with physics.

I think it has already been well-hashed out that the 2nd law can only be demonstrated, not proved.

(That A follows from assumptions B and C would be a logical, analytic, proof ... true, and not empirical - which is important since it is a synthetic statement which is to be proved. I think there's a thread in the philosophy forums for this?)
Andrew Mason
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Jan8-12, 09:38 PM
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Quote Quote by Simon Bridge View Post
I think it has already been well-hashed out that the 2nd law can only be demonstrated, not proved.

(That A follows from assumptions B and C would be a logical, analytic, proof ... true, and not empirical - which is important since it is a synthetic statement which is to be proved. I think there's a thread in the philosophy forums for this?)
It is called mathematics. All I am saying is that the statements:

1. the first law of motion defining inertial frames of reference is true and
2. all intertial frames of reference are equivalent is true.

imply that F = ma.

In my example where I add unit pulls and the same number of unit bodies, if a = (iF)^2/iM = i(constant) then the relationship between Δv and Δt would not be linear. It would be proportional to the number of unit forces or unit bodies that I have added.

The result would be that I can increase the change in velocity of a unit body per unit of time by i+1 times simply by being in the reference frame of i other non-interacting unit bodies each being independently pulled by the same unit of force. The conclusion would be that the change in velocity where the same pull is applied to the same body for the same period of time is not the same in all inertial reference frames, which negates the premise.

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Andrew Mason
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Jan8-12, 09:46 PM
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Quote Quote by AlonsoMcLaren View Post
1. How do you define mass?
In my example I didn't define mass. I defined a unit body. I think it is possible only to logically deduce from the first law and Galilean relativity that the acceleration for the same force will vary inversely as the number unit bodies to which the force is applied.

You would have to do some real experiments to determine the relationship between mass and the number of unit bodies. If it was known that there was a relationship between mass and the number of fundamental particles (protons+electrons or neutrons) a body contains, one could have deduced a general relationship between mass and the number of unit bodies.

2. "You will not have used Newton's second law in order to produce your data" But you have to use 2nd law to show that force is additive. That is to say, the force provided by two springs stretched dx each is twice the force provided by one spring stretched dx.

If F=ma^2 (or some other weird stuff), force will not be additive.
If F = ma^2 (or anything but F=ma) the premises will be contradicted - see my immediately previous post. I don't think that involves any assumptions about adding forces.

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harrylin
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Jan9-12, 05:22 AM
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Quote Quote by AlonsoMcLaren View Post
1. How do you define mass?
Mass is defined differently in different theories. Newton defined mass as the amount of matter, such as (later) the standard kg.
2. "You will not have used Newton's second law in order to produce your data" But you have to use 2nd law to show that force is additive. That is to say, the force provided by two springs stretched dx each is twice the force provided by one spring stretched dx.

If F=ma^2 (or some other weird stuff), force will not be additive.
As discussed earlier (see posts #13, #17), Newton likely used a force definition similar to the later kgf. Force defined like that must be additive just as mass is additive in classical physics. Consequently, if one would find that F=ma^2, then that would disprove Newton's second law.
Andrew Mason
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Jan9-12, 05:43 PM
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Quote Quote by harrylin View Post
Mass is defined differently in different theories. Newton defined mass as the amount of matter, such as (later) the standard kg.
At first this might seem a bit circular. But Newton may have had in mind a primitive concept of atoms - that all bodies consisted of finite numbers of indivisible elements of matter. In that sense his definition is not circular at all. "Quantity of matter" (or however Newton expressed it) is really the essence of what mass is. We know this from our knowledge of atomic structure. The mass of a body is, to a very close approximation, measured by the number of neutrons or protons+electrons contained in the body ie. its "quantity of matter" or "quantity of the individual elements of matter".

As discussed earlier (see posts #13, #17), Newton likely used a force definition similar to the later kgf. Force defined like that must be additive just as mass is additive in classical physics.
Unit forces are necessarily additive the same way that unit masses are if you start with the simple premise that all inertial frames are equivalent.

If I apply one unit of force to each of two equal bodies for the same unit of time, I get the same change in motion of each body. I also get the same result if I applied the same unit force to just one unit body for the same unit of time. Since the two bodies are in the same reference frame at all times, if I join them together to form a 2 unit body I have not changed anything about the physics (ie. I am not applying a net force to the two body system simply by joining them). But now I have two units of force being applied to a two-unit body for a unit of time and I get the same change in motion. Since nothing has changed in the physics, I have to conclude that forces necessarily add together just as the bodies add together eg.. 1 + 1 = 2.

Consequently, if one would find that F=ma^2, then that would disprove Newton's second law.
Certainly. But it would also be inconsistent with the premise that all inertial frames of reference (as defined in accordance with the first law, being the frame of reference of a body on which no net forces act) are equivalent in the sense that no local experiment can be done to differentiate between two inertial frames of reference.

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harrylin
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Jan10-12, 02:52 AM
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Quote Quote by Andrew Mason View Post
At first this might seem a bit circular. But Newton may have had in mind a primitive concept of atoms - that all bodies consisted of finite numbers of indivisible elements of matter. In that sense his definition is not circular at all. "Quantity of matter" (or however Newton expressed it) is really the essence of what mass is. We know this from our knowledge of atomic structure. The mass of a body is, to a very close approximation, measured by the number of neutrons or protons+electrons contained in the body ie. its "quantity of matter" or "quantity of the individual elements of matter".
Yes indeed, thanks for the addition. Newton certainly was thinking along those lines:
The quantity of matter is the measure of the same, arising from its density and bulk conjunctly.

THUS air of double density, in a double space, is quadruple in quantity [..]. The same thing is to be understood of snow, and fine dust or powders [..]. It is this quantity that I mean hereafter everywhere under the name of body or mass. And the same is known by the weight of each body; for it is proportional to the weight, as I have found by experiments on pendulums [..]
Quote Quote by Andrew Mason View Post
Unit forces are necessarily additive the same way that unit masses are if you start with the simple premise that all inertial frames are equivalent.
[..]
the premise that all inertial frames of reference (as defined in accordance with the first law, being the frame of reference of a body on which no net forces act) are equivalent in the sense that no local experiment can be done to differentiate between two inertial frames of reference.
AM
Sure. My point was that in Newton's presentation Galilean relativity follows from the laws of nature; and those laws can be put to the test based on non-circular definitions.
AlonsoMcLaren
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Jan10-12, 04:05 AM
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Why is F=ma^2 inconsistent with the fact that all inertial frames are equivalent?
harrylin
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Jan10-12, 04:07 AM
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Quote Quote by AlonsoMcLaren View Post
Why is F=ma^2 inconsistent with the fact that all inertial frames are equivalent?
I think that Andrew clearly explained that... Did you follow his derivation in post #14 (+ his elaboration in post #23)?

Note: probably one has to add the assumption that measurements of time and distance are the same in all inertial frames.
Andrew Mason
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Jan10-12, 08:28 AM
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Quote Quote by harrylin View Post
I think that Andrew clearly explained that... Did you follow his derivation in post #14 (+ his elaboration in post #23)?

Note: probably one has to add the assumption that measurements of time and distance are the same in all inertial frames.
Yes. That is important. Newton assumed it. The essential difference between Galilean relativity and Special Relativity is that measurements of time (and simultanaeity and, therefore, distance) do not translate equally between frames: t' ≠ t.

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Andrew Mason
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Jan13-12, 09:00 AM
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Quote Quote by AlonsoMcLaren View Post
Why is F=ma^2 inconsistent with the fact that all inertial frames are equivalent?
In looking at this again, I did not specifically address the situation where F/m changes. So I will try to do that here. It can get a little tricky so it is important to identify the principles that follow from the premise (that all inertial reference frames are equivalent).

Since all inertial frames are equivalent then whether I apply a unit of pull to each of two equal bodies simultaneously or sequentially for a unit of time should not matter. It will result in the same change of each body's motion when the pulls end i.e. if they start in the same reference frame they will end up in the same reference frame. (The only difference will be a spatial separation depending on how long I wait between the sequential applications of pull). Let's call this principle Principle 1.

Second, if two bodies are physically equal and at rest in the same reference frame, then one body can be substituted for the other and the same result will be obtained when they are subjected to the same pull for the same amount of time. Let's call this principle, Principle 2.


To make it simple, let the application of one unit of pull to a one unit body for one unit of time result in a speed that we will define as a unit of velocity, v1.

Exp. 1: I apply one unit pull to each of two single unit bodies simultaneously for one unit of time. The result will be a change of v1 for each body. Let's say that the bodies are initially at rest in reference frame i0. They end up at rest in reference frame i1 moving at velocity v1 with respect to i0.

Exp. 2: The same two single unit bodies are initially at rest in i0. I apply one unit of pull to each of the two bodies sequentially, each for one unit of time. By application of Principle 1, this should give the same results as Exp. 1. So the result will be a change of v1 for each body. They both end up in i1 (but separated by a distance).

Exp. 3: This is the same as Exp. 2 except that we start with the first body initially at rest in i0 and the second at rest in i1. Again, the result will be a change of v1 for each body. But in this case, the first body ends up in i1 and the second in i2 traveling at velocity v1 relative to i1 = 2v1 relative to i0.

Exp. 4: This is the same as Exp. 3 except that now we have only one single unit body intially at rest in i0. I apply one unit of pull to the body for one unit of time and the change in velocity is v1 so it is now in frame i1. Then I apply one unit of pull to the SAME body for another unit of time. By application of Principle 2 this will give the same result as Exp. 3: it results in an additional change in velocity of v1 so the body ends up in i2 traveling at velocity v1 relative to i1.


Each of the above four experiments involves the application of a force to a unit body for a unit of time twice. Each application results in the same change of motion of the unit body to which the pull is applied ie. v1. Since, the sequential application of the unit of pull to a unit body results in a change of 2v1, then, by principles 1 and 2, the simultaneous application of the same units of pull for the same unit of time (i.e. 2 units of pull applied to the same body simultaneously rather than sequentially each for a unit of time) will result in the same change of motion. So the unit body must end up in i2 traveling at velocity 2v1 relative to i0.


Letting the standard unit of velocity be the velocity of a unit body after applying a unit of pull for one unit of time be v1 then (using U for a unit of Force, M for a unit body, and t1 for a unit of time):

(A) 1U to 1M for 2t1 → 2v1
(B) 2U to 1M for t1 → 2v1


If [itex]F = ma^2 \text{ i.e. } \sqrt{F/m} = a[/itex] then the result in (B) would have to be: [itex]\sqrt{2}v_1[/itex]

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olivermsun
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Jan13-12, 03:25 PM
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Quote Quote by Simon Bridge View Post
The rubber band is being used as a newtonmeter and has the same problem: you know 2 bands (in parallel) give twice the force because of Newton's laws. However, I'll agree it's a better demonstration ... you can pretend to define force in terms of how much a spring stretches and then show this is proportional to the rate of change of momentum: it'll hold em. (But involve a bit of a rework in how Newtonian physics is usually taught.)
You don't have to use Hooke's law explicitly. You just have to assume that the rubber band pulls with the same force given the same elongation. Two rubber bands pulling with twice the force also doesn't really require you assume Newton's laws. What if you pull two masses in parallel using two rubber bands at fixed elongation and confirm that the distances traveled per time are the same whether the masses are fixed together or allowed to move separately?
Simon Bridge
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Jan14-12, 02:45 AM
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As a proof? Or a demonstration?
Remember you need to prove that the force is equal to the rate of change of momentum - of, F=ma, for a fixed mass.

I like the way you've rigged pulling the masses side-by-side connected or no which makes sure the times are the same also. In fact, notice that you don't need to actually do the experiment - the proportionality logically cannot fail but follow - making a-priori knowledge of newton's law possible (if this method is correct) which tells you that it is not a synthetic truth - eg. not of the World and therefore just a definition.

Looking more closely: Performing the experiment - which would be tricky to say the least - you are measuring F=nf (f= force due to one band/spring and n=1,2,3...) and nm (m is the mass that f pulls at a chosen constant acceleration) - plot F against nm, and get the slope f/m which is equal to independantly measured acceleration and proving f=ma... but wait: you don't know f! You need to know that to show the relation.

How do you find f?

You can show that f is proportional to m ... but you defined that to be the case when you defined force in terms of the extension of the rubber band - in the setup of the experiment... to show Newton's law you need to confirm the constant of proportionality
olivermsun
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Jan14-12, 10:28 AM
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You use the two parallel bands to generate some arbitrary force F and then 2*F without assuming Hooke's law

You don't know a priori that pulling the masses together and separately will result in the same acceleration, which is why you do the experiment.

What you are trying to demonstrate in Newton's law is not the constant of proportionality (since that would be units-dependent) but the fact that a direct (linear) proportionality exists at all.
Andrew Mason
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Jan14-12, 12:56 PM
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Quote Quote by Simon Bridge View Post
As a proof? Or a demonstration?
Looking more closely: Performing the experiment - which would be tricky to say the least - you are measuring F=nf (f= force due to one band/spring and n=1,2,3...) and nm (m is the mass that f pulls at a chosen constant acceleration) - plot F against nm, and get the slope f/m which is equal to independantly measured acceleration and proving f=ma... but wait: you don't know f! You need to know that to show the relation.

How do you find f?

You can show that f is proportional to m ... but you defined that to be the case when you defined force in terms of the extension of the rubber band - in the setup of the experiment... to show Newton's law you need to confirm the constant of proportionality
Olivermsun is correct. f is just an arbitrary unit of "pull". By choosing units you can make the constant of proportionality to be 1.

As we have seen above, because all inertial frames are equivalent (our premise) the application of one unit f to one unit m will result in a constant acceleration a1. Let's define f as the "pull" exerted by a certain stretched elastic such that the motion of an arbitrary unit body, m, will experience an acceleration a1 (ie. a unit change of velocity v1 per unit of time, t1). We can also deduce from the premise that we apply a pull of nf to n unit bodies, ie. nm, the acceleration must be a constant a1.

My previous post was an attempt to show that we can also deduce from the premises (the first law and the equivalence of inertial frames) that n unit pulls, ie nf, applied to i unit bodies ie. im will result in an acceleration of those unit bodies at the rate [itex]a = \frac{n}{i}a1[/itex].

nf would just be a simultaneous application of n unit stretched rubber bands to an aggregation of i unit bodies. We don't have to know anything about Hooke's law or anything else in order to deduce Newton's second law.

If the unit body is one proton+electron or neutron then a gram could be defined as an aggregate of N (Avogadro's number) unit bodies.

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harrylin
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Jan16-12, 05:29 AM
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Quote Quote by olivermsun View Post
You use the two parallel bands to generate some arbitrary force F and then 2*F without assuming Hooke's law

You don't know a priori that pulling the masses together and separately will result in the same acceleration, which is why you do the experiment.

What you are trying to demonstrate in Newton's law is not the constant of proportionality (since that would be units-dependent) but the fact that a direct (linear) proportionality exists at all.
Exactly - as I also discussed in my posts #13, #17 and #25.


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