Register to reply 
Can you prove Newton's Second Law from this experiment? 
Share this thread: 
#19
Jan612, 12:33 PM

P: 153




#20
Jan612, 03:19 PM

Sci Advisor
HW Helper
P: 6,678

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. AM 


#21
Jan812, 09:13 AM

P: 75

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. 


#22
Jan812, 09:58 AM

Homework
Sci Advisor
HW Helper
Thanks
P: 12,873

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 wellhashed 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?) 


#23
Jan812, 09:38 PM

Sci Advisor
HW Helper
P: 6,678

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 noninteracting 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. AM 


#24
Jan812, 09:46 PM

Sci Advisor
HW Helper
P: 6,678

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. AM 


#25
Jan912, 05:22 AM

P: 3,187




#26
Jan912, 05:43 PM

Sci Advisor
HW Helper
P: 6,678

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 twounit 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. AM 


#27
Jan1012, 02:52 AM

P: 3,187




#28
Jan1012, 04:05 AM

P: 75

Why is F=ma^2 inconsistent with the fact that all inertial frames are equivalent?



#29
Jan1012, 04:07 AM

P: 3,187

Note: probably one has to add the assumption that measurements of time and distance are the same in all inertial frames. 


#30
Jan1012, 08:28 AM

Sci Advisor
HW Helper
P: 6,678

AM 


#31
Jan1312, 09:00 AM

Sci Advisor
HW Helper
P: 6,678

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 v_{1} then (using U for a unit of Force, M for a unit body, and t_{1} for a unit of time): (A) 1U to 1M for 2t_{1} → 2v_{1} (B) 2U to 1M for t_{1} → 2v_{1} 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] AM 


#32
Jan1312, 03:25 PM

P: 788




#33
Jan1412, 02:45 AM

Homework
Sci Advisor
HW Helper
Thanks
P: 12,873

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 sidebyside 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 apriori 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 


#34
Jan1412, 10:28 AM

P: 788

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 unitsdependent) but the fact that a direct (linear) proportionality exists at all. 


#35
Jan1412, 12:56 PM

Sci Advisor
HW Helper
P: 6,678

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. AM 


#36
Jan1612, 05:29 AM

P: 3,187




Register to reply 
Related Discussions  
Newton's Thought Experiment  Special & General Relativity  2  
Is there any way to prove Newton's law  General Physics  5  
How to prove the Newton's third law.  Classical Physics  7  
Constructing experiment to prove P=IV  General Physics  1  
Need help with designing an experiment to prove that F=ma  General Physics  6 