2
$\begingroup$

As I have understood,

$F(t)=m \cdot a(t)$

can have 2 different meanings:

  • When applying an external force $F$ on a point mass of mass $m$, the resulting acceleration of that mass at time $t$ is $a$.

  • If a point mass is going to be accelerated with $a$ (maybe by another force), the inertial force of the mass is $F$. (Pushing in the opposite direction of $a$).

Which one did Newton mean? Are these two distinct expressions, or am I thinking the wrong way?

|cite|improve this question
$\endgroup$
  • $\begingroup$ By "inertial force" you mean the force pushing back on the accelerating object. This isn't "inertial force" it's called the "reaction force" and it appears whether the body is not moving or moving. $\endgroup$ – Ron Maimon May 9 '12 at 20:20
  • $\begingroup$ No i don't mean the reaction force. What i mean by inertial force is: If i was floating in space along with a satellite in rest; When i push the satellite with a force, the satellite resists to the movement proportional to its mass and its current acceleration. $\endgroup$ – bijan May 9 '12 at 23:32
  • $\begingroup$ And that current acceleration was created by my force. $\endgroup$ – bijan May 9 '12 at 23:34
4
$\begingroup$

Original Newton idea was that

$$\vec{F} = m \cdot \vec{a}$$

meaning in inertial frame of reference (net or total) force equals product of mass of body and its acceleration.

Indeed, as you mentioned similar expression appears in non-inertial frame of reference $$\vec{F'} = - m \cdot \vec{a'}$$

meaning in non-inertial frame of reference you have to add pseudo or fictitious or inertial force $\vec{F'}$ if you wish to use $\vec{F} = m \cdot \vec{a}$.

$\vec{a'}$ in second expression represents the acceleration of the non-inertial frame of reference and not the acceleration of the mass. Also, note minus sign in the second expression, which means that pseudo force is directed in the opposite direction to acceleration of the frame of reference!

By the way, non-inertial frame of reference is usually conveniently chosen in a way that observed body is at rest (position of the body defines that frame of reference), so you end up with $\vec{a} = 0$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thank you very much Pygmalion! Do you maybe have an example or a link to an example? $\endgroup$ – bijan May 8 '12 at 20:28
  • 1
    $\begingroup$ en.wikipedia.org/wiki/Fictitious_force, en.wikipedia.org/wiki/Centrifugal_force, there are even more simple examples I do at my lectures (pendulum in accelerating train/plane), but I cannot find anything similar on Internet. $\endgroup$ – Pygmalion May 8 '12 at 20:29
  • 2
    $\begingroup$ The inertial force is explained by D'Alembert's principle. en.wikipedia.org/wiki/D'Alembert's_principle $\endgroup$ – ja72 May 8 '12 at 20:32
  • $\begingroup$ Ok thank you! So these are really two distinct formulas right? $\endgroup$ – bijan May 8 '12 at 20:36
  • $\begingroup$ Yes, they are. Plainly speaking (sorry Nick), the first formula is "defined" only within inertial frame of reference. But if you want to use the first also in non-intertial frame, you have to use the second too. $\endgroup$ – Pygmalion May 8 '12 at 20:44
3
$\begingroup$

Actually, neither of those is quite correct. The first expression is subtly wrong; the second may be either subtly wrong in the same way, or completely wrong, depending on what you mean by "inertial force."

What Newton's second law in this form really means is that the net (total) force acting on an object at time $t$ is equal to the object's mass times the object's acceleration at that time. It does not apply to any one individual force, only to the net force. For this reason it is most properly written

$$\sum \vec{F} = m\vec{a}$$

with the summation symbol to indicate the sum over all forces.

There are some restrictions on the validity of this equation; in particular, as Pygmalion pointed out, it only works in an inertial frame of reference (or can be taken to define an inertial frame of reference, in Newtonian mechanics). Also, it only works for objects whose mass is constant, and whose velocity is small. (That last one is sort of debatable depending on which definition of force you use) The more general form of Newton's second law (and in fact the way it was originally written) is

$$\sum \vec{F} = \frac{\mathrm{d}\vec{p}}{\mathrm{d}t}$$

This applies for objects of varying mass and high velocity as well.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ As written here wikipedia.org/wiki/Fictitious_force or here www-istp.gsfc.nasa.gov/stargaze/Sframes2.htm, inertial force is synonymous with fictitious/pseudo force $\endgroup$ – Pygmalion May 9 '12 at 19:50
  • $\begingroup$ Interesting, I've never heard that terminology before (at least not in that usage). $\endgroup$ – David Z♦ May 9 '12 at 19:58
  • $\begingroup$ Neither did I. So, I've googled before answering above. $\endgroup$ – Pygmalion May 9 '12 at 20:00
  • $\begingroup$ That's fair. The reason I answered as I did was that I have heard "inertial force" used to represent something that is completely wrong (when I'm teaching students about forces they sometimes invent a force that is due to an object's mass and call it the "force of inertia") so I thought that was what the question mentioned. $\endgroup$ – David Z♦ May 9 '12 at 20:04
  • 1
    $\begingroup$ @DavidZaslavsky After additional brainstorming, checking books and Googling, I added "(net or total)" to my text, leaving the formula the same, as it is actually commonly used. So you are right on that one. However, I think net force applies only in the case of the original Newton's laws. When talking about pseudo forces, it has nothing to do with net force. $\endgroup$ – Pygmalion May 9 '12 at 21:22
2
$\begingroup$

Newton meant both, since action equals reaction. The first is Newton's intended interpretation--- to accelerate a body by a, you need a force equal to ma. The second is the necessary reaction force on the thing that is doing the acceleration. If you push on a body to accelerate it, the body pushes back on you in the opposite direction.

Note that the reaction doesn't care if the body is speeding up or not--- whenever you apply F units of force on a body, the reaction is there, pushing you back by an equal amount.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.