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Old Nov 1st 2017, 12:50 AM   #1
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Newtons second law for the layman.

Newtons all three laws are valid only in an inertial frame of reference. Here we look at newtons second law. But first, we define inertial frame of reference:

An inertial reference frame is one in which a body at rest stays at rest and a body in motion continues to move at a constant velocity.

So now how can F=ma be true because the formula contains acceleration and in an inertial reference frame there is no acceleration or the acceleration is zero. This is a common doubt we get. Now my explanation is this:

An inertial frame of reference is one in which acceleration can be measured. So what it means is this: Lets take an example: An ball is at rest and you take the ball and throw it horizontally to another person, so initially the body is at rest so acceleration is 0. Now when the ball is thrown it accelerates 3 meters per second. So the force is equal to: say ball has a mass of a kg so F=ma= 1*3= 3 kg m/s^2. So Force is measured when there is acceleration.How?

What this second law states is that initially the body is at rest but later it accelerates but after it accelerates its speed remains constant. From the above example the ball it at rest so acceleration =0. But later it accelerates 3 meters per second and stays at this speed. So this is an inertial reference frame since acceleration is measured and stays constant later.

But if after the ball accelerated and later it decelerated and again it accelerated then newtons second law does not hold and force is not measurable since acceleration is not measurable.

This is in laymans terms, is this correct?
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Old Nov 1st 2017, 02:40 AM   #2
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Originally Posted by avito009 View Post
So now how can F=ma be true because the formula contains acceleration and in an inertial reference frame there is no acceleration or the acceleration is zero. This is a common doubt we get. Now my explanation is this:
Because while velocity is not invariant under a Galilean transformation acceleration is. Do you know calculus? If so then I recommend doing this calculation. Its very instructive. You should also know that Newton didn't use F = ma to define force. He used F = dp/dt.

Originally Posted by avito009 View Post
. So Force is measured when there is acceleration.How?
Not sure what your asking. Are you asking of force is measured or how its calculated? If its the former as you appear to be asking then you should know that force is rarely, if ever, measured directly. In a case like this one would measure the acceleration and then use F = ma to determine the force. Measuring acceleration isn't an easy task either since one would have to measure velocity at two different times and the determine a = dv/dt


Originally Posted by avito009 View Post
What this second law states is that initially the body is at rest but later it accelerates but after it accelerates its speed remains constant.
That's correct. A force is only present when the ball is accelerating.

Originally Posted by avito009 View Post
But if after the ball accelerated and later it decelerated and again it accelerated then newtons second law does not hold and force is not measurable since acceleration is not measurable.
That is quite wrong. Newton's second law always holds true in classical mechanics.

Originally Posted by avito009 View Post
This is in laymans terms, is this correct?
No.
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Old Nov 1st 2017, 02:48 AM   #3
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Changing acceleration

Lets look at changing acceleration. So how would you measure the force if an objects acceleration is changing?

You would have to calculate the force again and again. So what I am saying in the above post is that newtons second law assumes constant motion.

When an object is accelerated at 3 m/s then you calculate the force. Later this same object moving at 3 m/s is accelerated which means force is provided since first law states an object in motion will continue in motion unless a force is applies. So now the acceleration has increased it is accelerating at 6 m/s now so change in acceleration is 6-3 = 3 m/s. So since again force is applied which increases the acceleration force will have to be calculated again.
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Old Nov 1st 2017, 04:38 AM   #4
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Originally Posted by avito009 View Post
Lets look at changing acceleration. So how would you measure the force if an objects acceleration is changing?
I've already explained that.
Originally Posted by avito009 View Post
You would have to calculate the force again and again. So what I am saying in the above post is that newtons second law assumes constant motion.
All that you're doing is talking about a time dependent force rather than a constant one. Force is defined as F = dp/dt. Newton wrote about it as if it was a law of motion but he was wrong about that as any modern textbooks on classical mechanics attest.

Last edited by Pmb; Nov 1st 2017 at 05:35 AM.
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Old Nov 1st 2017, 05:52 AM   #5
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It is utterly pointly having a discussion about a quotation unless you first correctly establish exactly what was said.

One thing Newton most certainly did not say was d something by d (anything).
This notation, which is clearly superior, was due to Leibnitz and a controversy raged for many years between them where Newton steadfastly refused to use it, prefering instead his own.
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Old Nov 1st 2017, 06:37 AM   #6
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Originally Posted by studiot View Post
One thing Newton most certainly did not say was d something by d (anything).
Wrong. You're confusing what I said with the origin of the notation I used. Its unwise to do such a thing. It is in fact quite possible to interpret Newton this way and indeed it often is is. See Concepts of Force by Max Jammer, Dover Pub., page 184.
The second law, likewise, has two possible interpretations. : it may serve as a quantitative definition of force or as a generalization of empirical facts. In modern notation the law, according to Newton, asserts F ~ d(mv).
Max Jammer is a widely known authority on these matters. He was a physicist and philosopher of physics.

From The Principia by Isaac Newton, page 19
LAW II

The alteration of motion is ever proportional to the motive force impression; and is made in the direction of the right line in which that force is impressed.
In order to understand that statement one has to know what Newton meant by "motion" since its not what modern physicists mean by it this context. On page 9 he wrote
DEFINITION II

The quantity of motion is the measure of the same, arising from the velocity and quantity of matter conjunctly.
In modern terms that means "quantity of motion" = mv = momentum. = p. Things have changed since he wrote that since we now consider momentum as a vector whereas Newton didn't.

Originally Posted by studiot View Post
This notation, ...
That's your mistake right there. I didn't make any claim that Newton used the same notation I used whatsoever. All I did was simply explain how that Newton defined force, i.e. as being the time rate of change of momentum. In modern notation that means F = dp/dt. Nowhere did I imply that Newton used that notation.

Please consider such things before you go around making such statements in the future. We're more concerned here with the physics than we are with the origin of any notation.

Last edited by Pmb; Nov 1st 2017 at 06:45 AM.
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Old Nov 1st 2017, 07:58 AM   #7
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PMB

Since you are such an expert on Newton that you can presume to tell me how to phrase a post, perhaps you would like to tell avito why his latest thoughts (about tossing a ball up) on Newton contravene N1
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Old Nov 1st 2017, 08:29 AM   #8
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Originally Posted by avito009 View Post
So now how can F=ma be true because the formula contains acceleration and in an inertial reference frame there is no acceleration or the acceleration is zero.
It's totally fine for inertial frames to contain objects that accelerate. It's accelerating frames (actually... just non-inertial frames) that cause issues with the physics and therefore require additional treatment (e.g. Coriolis terms in rotating frames).

An inertial frame of reference is one in which acceleration can be measured. So what it means is this: Lets take an example: A ball is at rest and you take the ball and throw it horizontally to another person, so initially the body is at rest so acceleration is 0. Now when the ball is thrown it accelerates 3 meters per second. So the force is equal to: say ball has a mass of a kg so F=ma= 1*3= 3 kg m/s^2. So Force is measured when there is acceleration.How?
Well, force is defined as the rate of change of momentum which, for particles with mass, causes acceleration. Therefore, when acceleration is observed, a force can be associated with it. Note that forces can be associated with acceleration regardless of what kind of frame you are in (e.g. the centrifugal force and Coriolis force associated with dynamics in rotating reference frames).

What this second law states is that initially the body is at rest but later it accelerates but after it accelerates its speed remains constant. From the above example the ball it at rest so acceleration =0. But later it accelerates 3 meters per second and stays at this speed. So this is an inertial reference frame since acceleration is measured and stays constant later.

But if after the ball accelerated and later it decelerated and again it accelerated then newtons second law does not hold and force is not measurable since acceleration is not measurable.

This is in laymans terms, is this correct?
No, because what you have here is a measurement of an average acceleration over a given duration and measuring a value of zero for your average acceleration doesn't necessarily mean that the instantaneous accelerations are zero. Even so, measurements of zero don't invalidate laws of physics; they're just zero measurements.

Let's put some numbers to your last point. Let's say that at time t = 0, a 1 kg ball accelerates for 0.001 seconds at 1 m s$\displaystyle ^{-2}$ and then 0.008 seconds later (t=0.009) the ball decelerates for 0.001 seconds at 1 m s$\displaystyle ^{-2}$. At t = 0 and t = 0.01 the ball's speed is zero but between times t = 0.001 and t = 0.008, the ball's speed is 0.001 ms$\displaystyle ^{-1}$. The velocities in the initial and final phases look like linear ramps that connect the 0.001 value to 0.

Now, let's assume we have this particle in an inertial frame and you have an accelerometer that measures the acceleration of an object every 0.01 seconds. The measurement of the average acceleration is made by sampling the velocity at two points in time and then plugging into the following formula:

$\displaystyle a = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{0.01}$

Since the velocities are zero for both times, the measured acceleration is:

$\displaystyle a = \frac{0 - 0}{0.01} = 0 m s^{-2}$

However, remember that a deceleration is the same as a positive acceleration, so the two accelerations can be written down as

$\displaystyle a_1 = 1 m s^{-2}$
$\displaystyle a_2 = -1 m s^{-2}$

So is it surprising that our average acceleration over the interval is 0?

However, let's say we upgraded our accelerometer so it can now make measurements every 0.001 seconds (so it can make 10 measurements in the same that the original accelerometer made 1). For the first 0.001 seconds we get

$\displaystyle a = \frac{0.001 - 0}{0.001} = 1 m s^{-2}$

and for the last 0.001 seconds, we get

$\displaystyle a = \frac{0 - 0.001}{0.001} = -1 m s^{-2}$

so it seems our new accelerometer is now capable of measuring the changes.

Small changes that occur over very short time intervals are called perturbations. Perturbations happen all the time in reality and are considered in physics theory like everything else. In fact, perturbation theory allows for the exploration of what can happen to certain systems if a perturbation does occur.

Now... let's say that our frame of reference is not a accelerating frame (i.e. non-inertial) and it accelerates at a constant 0.001 m s$\displaystyle ^{-2}$. In the period t = 0 to t= 0.001, the object will look stationary in the frame since it's acceleration is the same as the frame of reference. However, in the period t = 0.001 to t = 0.009, the object will appear to accelerate relative to it's inertial frame at -1 m s$\displaystyle ^{-2}$, so the final velocity at t = 0.009 is

$\displaystyle v_2 = v_1 + at = 0 - 1 \times 0.008 = -0.008 m s^{-1}$

The final part will then have the ball accelerate at -2 m s$\displaystyle ^{-2}$ for 0.001 seconds, giving a final velocity of -0.01 m s$\displaystyle ^{-1}$

Okay... now let's look at forces. In the first situation we looked at (inertial frame) we have the forces for the accelerating and decelerating phases as

$\displaystyle F = ma = 1 \times 1 = 1 N$

and

$\displaystyle F = ma = 1 \times -1 = -1 N$

and zero for the constant velocity bit. However, for the second situation (non-inertial frame) we have

$\displaystyle F = ma = 0 N$

for the first phase and

$\displaystyle F = ma = 1 \times -1 = -1 N$

for the second phase. The last phase has an even greater force:

$\displaystyle F = ma = 1 \times -2 = -2 N$

So it is clear that the formula gives incompatible results between inertial and non-inertial frames. The equations that describe mechanics in non-inertial frames does exist though(consider, for example, https://en.wikipedia.org/wiki/Mechan...article_motion)
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Old Nov 3rd 2017, 11:51 AM   #9
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Knowledge of calculus.

Pmb, I know differential and integral calculus. In brief I will explain. Calculus is the study of change. Delta V is acceleration i.e change in velocity is acceleration.

Also derivative of position is velocity. Written at X with one dot above it and derivative of velocity is acceleration. Written as X with two dots above it.
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Old Nov 3rd 2017, 01:01 PM   #10
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Originally Posted by avito009 View Post
Pmb, I know differential and integral calculus. In brief I will explain. Calculus is the study of change. Delta V is acceleration i.e change in velocity is acceleration.

Also derivative of position is velocity. Written at X with one dot above it and derivative of velocity is acceleration. Written as X with two dots above it.
FYI
The word 'calculus' means a theoretical structure which provides a basis and formal procedure for obtaining a desired mathematical result or results.

There are may such calculi.

You are referring to what we now call 'the differential calculus'.

Another example: there is one called the graphical calculus which is a procedure for graphically solving algebraic equations.

Of course the integral calculus in not about change but about 'gathering together'
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