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

[Andrew Mason]

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 → 2v₁If F = ma^2 \text{ i.e. } \sqrt{F/m} = a then the result in (B) would have to be: \sqrt{2}v_1

AM

 

[olivermsun]

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]

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]

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]

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. I am will result in an acceleration of those unit bodies at the rate a = \frac{n}{i}a1.

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.

AM

 

[harrylin]

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.