# Think of a body which is at rest but not in equilibrium

• SSG-E

#### SSG-E

Homework Statement
Think of a body which is at rest but not in equilibrium. Give explanation as well as figure/diagram.
Relevant Equations
Σ F = x (x is not equal to 0)
When a ball is thrown upward it becomes at rest at maximum height, at this it is not in equilibrium although it is at rest. It is not at equilibrium because force of gravity is acting on it? Still I cannot find good explanation from exam point of view.I also cannot find the figure/diagram.

sysprog

When a ball is thrown upward it becomes at rest at maximum height, at this it is not in equilibrium although it is at rest. It is not at equilibrium because force of gravity is acting on it? .
Right.

A ball thrown vertically upward will be momentarily at rest even though a net force acts on it (and thus it's not at equilibrium).

scottdave, hutchphd and sysprog
A pendulum and an oscillating spring-mass are two more examples.
Both cases show points of minimum velocity, which are as well the points at which maximum force and acceleration occur.

Perhaps not completely correct, but I would say that a body that immediately accelerates after being in repose, if no additional force is acting, has received impulse from a resultant force; therefore, it has not been in equilibrium immediately before the magnitude of its velocity starts growing.

scottdave, SSG-E and sysprog
Right.

A ball thrown vertically upward will be momentarily at rest even though a net force acts on it (and thus it's not at equilibrium).
Nevertheless, the ball may, in some discourse, be said to have been at rest at its perigee for 'zero' time, despite it not having never been at rest there.

I think that 'momentarily' mathematically would mean for an infinitesimal duration, and that physically it would in the posited instance mean for a duration consistent with speed of propagation of gravitational force, and that saying that it was there for zero time is inconsistent with saying that it was ever there.

As far as I know, we don't yet know whether gravity acts continuously, or in discrete pulses of unknown but extremely high frequency.

hutchphd and Delta2
How about a rubber ball as it hits the ground, right before it pops back up, again.

Lnewqban
A pendulum and an oscillating spring-mass are two more examples.
Both cases show points of minimum velocity, which are as well the points at which maximum force and acceleration occur.

Perhaps not completely correct, but I would say that a body that immediately accelerates after being in repose, if no additional force is acting, has received impulse from a resultant force; therefore, it has not been in equilibrium immediately before the magnitude of its velocity starts growing.

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Its correct because it is momentarily at rest and not completely at rest. This satisfies Newton's law.

Lnewqban
Its correct because it is momentarily at rest and not completely at rest. This satisfies Newton's law.
OK.
I would describe it as being momentarily at rest but still accelerating.

Lnewqban
Its correct because it is momentarily at rest and not completely at rest. This satisfies Newton's law.
As you can see in the animation of the pendulum, the vector velocity is constantly changing magnitude and direction.

Its correct because it is momentarily at rest and not completely at rest. This satisfies Newton's law.
OK.
I would describe it as being momentarily at rest but still accelerating.

Both can be incorporated to clarify the connection.
"It's correct because it is momentarily at rest but will not remain at rest because it is being accelerated. This satisfies Newton's law."

scottdave, Lnewqban and Doc Al
As you can see in the animation of the pendulum, the vector velocity is constantly changing magnitude and direction.
At each extreme of the pendulum the velocity is equal to zero therefore the magnitude of velocity vector is also equal to zero. But the gravitational acceleration is constant at every position of the pendulum. Therefore, the object is not completely at rest but momentarily at rest as it will move after an instant. In other words the object passes through the rest frame but isn't continuously at rest as it doesn't have time to rest. Words are tricky in physics. They sometimes project "mirages" in our minds...

Lnewqban
passes through the rest frame
What are these words supposed to mean? A rest frame is not something that you can "pass through".

phinds and etotheipi
What are these words supposed to mean? A rest frame is not something that you can "pass through".
At maximum height, the velocity of the ball is equal to 0, but it doesn't continuously remain at zero for a continuous time period, even not for the millionth part of second….but the ball simply passes through rest.
Look at the velocity-time graph in the picture below. At point B , when the velocity graph crosses the axis v = 0. It is just for "one instant". Therefore, by passing through rest I mean the ball isn't at rest for a continuous time period. It is at rest only for a particular instant. The velocity starts increasing after that.

At maximum height, the velocity of the ball is equal to 0, but it doesn't continuously remain at zero for a continuous time period, even not for the millionth part of second….but the ball simply passes through rest.
Thank you. This time you've removed the problematic word "frame". The ball passes through a condition of being at rest. Not through a condition of being in a particular frame.

What are these words supposed to mean? A rest frame is not something that you can "pass through".

I guess we could speak of the rest frame of the particle at any given point in time. And at the instant it is at rest w.r.t. the observer, the rest frame of the particle is the rest frame of the stationary observer.

I'd agree though it's bad to say it "passes through" a rest frame. It would seem better to say "the particle instantaneously shares a rest frame with the observer".

But why we'd want to invoke such a notion here is questionable

SSG-E
I guess we could speak of the rest frame of the particle at any given point in time. And at the instant it is at rest w.r.t. the observer, the rest frame of the particle is the rest frame of the stationary observer.

I'd agree though it's bad to say it "passes through" a rest frame. It would seem better to say "the particle instantaneously shares a rest frame with the observer".

But why we'd want to invoke such a notion here is questionable
Thank you. This time you've removed the problematic word "frame". The ball passes through a condition of being at rest. Not through a condition of being in a particular frame.
Can anyone tell the meaning of frame? Just for confirmation.

Can anyone tell the meaning of frame? Just for confirmation.
For the purposes of classical mechanics, one can think of "frame of reference" as a synonym for "coordinate system" without going too badly wrong. It is a tool for analysis -- lines drawn on a piece of paper. Not anything physical.

One can distinguish frame of reference from coordinate system because the frame of reference is really more a standard of rest while the coordinate system gets into the nuts and bolts of putting numbers on positions.

etotheipi
I like to think of it in classical physics as a set of observers who share a certain rigid body motion. And in any frame of reference you can choose how to position and orient your origin and coordinate axes, so a frame of reference corresponds to a set of coordinate systems connected via constant translations and rotations.

But you'll often see the terms frame of reference and coordinate system used interchangeably, which isn't too big of a deal for the most part since it's usually fairly evident from context what's being discussed.

jbriggs444
For the purposes of classical mechanics, one can think of "frame of reference" as a synonym for "coordinate system" without going too badly wrong. It is a tool for analysis -- lines drawn on a piece of paper. Not anything physical.
By saying passing through rest frame, I didn't mean that the object got through another physical body but now, I too agree that it's bad to say it "passes through" a rest frame. It would seem better to say "the particle instantaneously shares a rest frame with the observer".

By saying passing through rest frame, I didn't mean that the object got through another physical body but now, I too agree that it's bad to say it "passes through" a rest frame. It would seem better to say "the particle instantaneously shares a rest frame with the observer".
Yes. Though it seems like too much effort for a simple concept. Rather than saying that "the frame of reference within which an object is always at rest is itself instantaneously at rest with respect to the frame of reference where the observer is always at rest", just say "the object is instantaneously at rest".

etotheipi
Yes. Though it seems like too much effort for a simple concept. Rather than saying that "the frame of reference within which an object is always at rest is itself instantaneously at rest with respect to the frame of reference where the observer is always at rest", just say "the object is instantaneously at rest".

That is, unless one wants to show off

jbriggs444
That is, unless one wants to show off
Or, outright be wrong : if I throw a ball into the air, watching it isn't what makes it momentarily stop. A frame of reference need not be occupied, nor interactive.

Or, outright be wrong : if I throw a ball into the air, watching it isn't what makes it momentarily stop. A frame of reference need not be occupied, nor interactive.

I don't understand your point; that's not what was said. At the instant the ball reaches the highest point in its trajectory, the rest frame of the ball is the rest frame of the person who threw the ball - i.e. they both occupy the same rest frame for an instant.

It's perhaps not a helpful fact, but it's a true one, nonetheless.

nor interactive.

Also I'm not too sure what this means. A frame of reference cannot "interact", it's the notion of a set of points (observers) in the Euclidian space that share a common motion, with which you can define a system of measurement (e.g. coordinate system).

The choice of reference frame of course must not influence the phenomena we are trying to measure.

At maximum height, the velocity of the ball is equal to 0, but it doesn't continuously remain at zero for a continuous time period, even not for the millionth part of second….but the ball simply passes through rest.
Look at the velocity-time graph in the picture below. At point B , when the velocity graph crosses the axis v = 0. It is just for "one instant". Therefore, by passing through rest I mean the ball isn't at rest for a continuous time period. It is at rest only for a particular instant. The velocity starts increasing after that.
View attachment 264538
How might we without inconsistency envisage instantaneity? Is there physically such a thing as an 'instant'? If there is, does it have a duration of zero, or is its duration positive?

How might we without inconsistency envisage instantaneity? Is there physically such a thing as an 'instant'? If there is, does it have a duration of zero, or is its duration positive?
Physically? That's for experiment to decide. No experiment I can imagine can detect whether there is such a thing.

Edit: This article by @john baez indicates that there is no evidence that gives a conclusive result either way.

Mathematically? Sure. A course in real analysis will provide a great deal of insight. It's duration is zero.

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etotheipi
Physically? That's for experiment to decide. No experiment I can imagine can detect whether there is such a thing.

Edit: This article by @john baez indicates that there is no evidence that gives a conclusive result either way.

Mathematically? Sure. A course in real analysis will provide a great deal of insight. It's duration is zero.
I understand that to be conventional usage. In my opinion, it should be called something that doesn't also mean nonexistent ##-## it seems to me to be a misappropriation of the term 'zero'. I prefer 'infintesimal'.

I understand that to be conventional usage. In my opinion, it should be called something that doesn't also mean nonexistent ##-## it seems to me to be a misappropriation of the term 'zero'. I prefer 'infintesimal'.
As I said, a course in real analysis would not be amiss here. Zero or, in particular, zero duration is not a synonym for non-existent.

I don't understand your point; that's not what was said.
Perhaps not, my reading of this thread was a bit sketchy : it just seemed that somebody was correlating observation with occurrence.

As I said, a course in real analysis would not be amiss here. Zero or, in particular, zero duration is not a synonym for non-existent.
I demur from some of the usages of 'zero' that are commonly presented in real analysis courses.

In ordinary English, if someone asks "for exactly how much time was thing '##a##' at location '##b##'", and the response is "for zero time", that means that '##a##' was never at '##b##', i.e. that the posited eventuality of '##a##' being present at '##b##' did not occur.

Mathematically, one could reply that '##a##' was at '##b##' "for infinitesimal time", without violating the common meaning of 'zero'.

I demur from some of the usages of 'zero' that are commonly presented in real analysis courses.

In ordinary English, if someone asks "for exactly how much time was thing '##a##' at location '##b##'", and the response is "for zero time", that means that '##a##' was never at '##b##', i.e. that the posited eventuality of '##a##' being present at '##b##' did not occur.

Mathematically, one could reply that '##a##' was at '##b##' "for infinitesimal time", without violating the common meaning of 'zero'.
Zero is an infinitesimal. The only infinitesimal in the standard reals. When I say "zero duration" I do not mean a non-zero infinitesimal duration. Nor do I mean that something never happened.

etotheipi
Zero is an infinitesimal. The only infinitesimal in the standard reals. When I say "zero duration" I do not mean a non-zero infinitesimal duration. Nor do I mean that something never happened.
I think that this highlights an inadequacy of the 'standard reals'. In my view, ##\epsilon \in {}^*\mathbb{R}: \forall r [(|\epsilon| < r)\wedge(r>0)]##. That entails that ##\epsilon## is not a real number in the standard reals, as ##0## is, and it also restricts the infinitesimal to occupying 'the leftmost place' to the right of zero on the number line. I know that real analysis will immediately raise the objection that there is no such place on the real number line; however, such a place is nameable, albeit naming something that does not exist among the standard reals, wherefore the resorting to hyperreals.

etotheipi
I think that this highlights an inadequacy of the 'standard reals'. In my view, ##\epsilon \in {}^*\mathbb{R}: \forall r [(|\epsilon| < r)\wedge(r>0)]##. That entails that ##\epsilon## is not a real number in the standard reals, as ##0## is, and it also restricts the infinitesimal to occupying 'the leftmost place' to the right of zero on the number line. I know that real analysis will immediately raise the objection that there is no such place on the real number line; however, such a place is nameable, albeit naming something that does not exist among the standard reals, wherefore the resorting to hyperreals.
But remember the point of the exercise.

If we allow for hyper-real intervals we should allow for hyper-real velocities. The hyper-real interval during which the velocity is infinitesimally different from zero is infinitesimal. But the hyper-real interval during which the velocity is zero is still exactly zero.

sysprog and etotheipi
jbriggs444 said:
If we allow for hyper-real intervals we should allow for hyper-real velocities. The hyper-real interval during which the velocity is infinitesimally different from zero is infinitesimal. But the hyper-real interval during which the velocity is zero is still exactly zero.
I model the zero-velocity interval to be infinitesimal. Calling it an interval is calling it non-zero. I like your succinct positure, and I see its appeal; however, a velocity is a ratio of a directed distance per duration, while a duration is a positive interval. If I model the zero velocity as having occurred for an infinitesimal duration, I avoid division by zero, leaving the velocity as zero, and the duration of the zero velocity as infinitesimal. In my view, an 'instant' cannot be of zero duration, but must be of minimally greater than zero duration.

I don't know much about the technicalities, but here I think @jbriggs444's approach seems more natural. If the body comes to rest instantaneously at ##t_0##, then ##v(t_0+\epsilon) \neq 0## and ##v(t_0-\epsilon) \neq 0## no matter how arbitrarily small you make ##\epsilon##. So I think it would make sense to say that the body is at rest for zero time.

The Baez article is really interesting, but from a pure maths perspective all I can say is that the trajectory intersects the ##v=0## axis at one point, and a "point" in Euclidian space doesn't have any "width" (indeed such a notion doesn't even make sense!).

sysprog
a velocity is a ratio of a directed distance per duration
A velocity is the limit of the ratio of a directed distance over a duration as the duration and distance approach zero. It is a limit, not a ratio.

If you are careful, you can use the transfer principle to turn this statement about velocity in the reals as a limit into a statement about a velocity in the reals as a ratio of infinitesimals. But that does not help make the point you need to be trying to make.

The set of points where velocity is zero is a degenerate interval consisting of a single point. The measure of that set is zero.

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sysprog, etotheipi and Lnewqban