As it is clear that the sum of forces are not zero at these situations, i'm assuming that your issue is about the restness of the body at these points (please correct me if I am wrong).

The restness at these points are very quick (in fact, lasts an infinitesimal time), so it's not a "stable rest", but just a particular point in a velocity curve. Saying that the particle is at rest at those points is equivalent to say that it is at constant velocity at the other points - in other words, treat a particular point as a curve behavior. If this is your concern, I agree with your discomfort.

But I think that the pedagogical reason to use such examples are to clarify for the students that a stopped particle is not necessarily at equilibrium. However, using the word "rest" can be tricky.

At rest means velocities vanish; at equilibrium means forces (and torques) vanish. Newton's second law implies that IF forces (and torques) vanish, velocities don't change, NOT that they vanish; so if the velocities are zero they stay zero, if they're non-zero, they remain non-zero.

That's why they're distinct issues.

This is a homework assignment in elementary mechanics...

This question, although from elementary mechanics, is relevant and interesting both.

At the outset, let us recapitulate that both the terminologies "rest" and "equilibrium" are relative words.

It depends absolutely on the Examiners or Observers how they want to define these terms without or with "relativity".

You are always at rest, if comoving. Now, comoving is a precise term in the corpus of physics. It means same velocity and same position. For example, comoving observers never experience length contraction or time dilation, no matter what speed, acceleration, or arbitrary motion. This is quite different from the physics concept of equilibrium.

Yes, "co-" leads to "rest", such as "comoving" or "corotating"; but, with what? It refers to "co-" relative to one object only, right? That, is, "co-" lies in the "source-observer" frame coupling, isn't it?

1) a stationary box with a gas which initially only fills only half the volume.

2) a mixture of certain chemicals in a jar which just keep churning and churning around. This exists in chemistry.

@ Rehman, It looks like your are confused between stable equilibrium and astable (or unstable) equilibrium. Both types of equilibrium satisfy both of these conditions as stated by you above.

(i) Sum of forces acting on body= 0

(ii) Resultant torque = 0

But, a key property of stable equilibrium behavior is that if the object is disturbed from its equilibrium point (not by a very large amplitude), then the object should return back to its stable equilibrium position. A good example is the simple pendulum at rest. On disturbing it, the dissipative frictional forces restore the pendulum back to its previous position equilibrium position.

An example of unstable equilibrium could be yourself sitting at "rest" at the "tip" of Mt. Everest, or a particle at rest at the peak of a hump. One small nudge to you or the particle pushes you into a region from which you or the particle cannot come back to its previous equilibrium position.

Another example of unstable equilibrium is an egg in a vertically upright "standing" position though it requires some effort to put the egg in such a sensitive equilibrium position. Just push the egg a little and it loses its balance and falls back into a more regular position, not the vertical upright one.

In unstable equilibrium positions, objects are necessarily at rest, but not in a REAL equilibrium state when compared to objects in stable equilibrium.

Hope this explains!

Marcos Souza Stam Nicolis Ed Gerck P. K. Karmakar Juan Weisz Vikash Pandey thanks all of you for participating in discussion. Actually I want very short and simple answer. I teach. And I'm having issue to answer this question exactly. I need simple and short answer from examination point of view.

Question is, does there exist any example of a body which is at rest but not equilibrium. From my point of view, no such body exists. But to some other teachers and according to some helping notes by other teachers, it is said that yes there exist examples of such bodies. Two examples I have quoted in my question above.

Now let's discuss these examples. No doubt, a stone will be at rest momentarily when thrown upward against gravity and it reached at maximum height. Admitted. But how will it be in equilibrium? According to some teachers and according to some helping notes by other teacher, this stone after reaching at maximum height will be at rest but not in equilibrium. They present the logic that at this position only once force i.e. gravity acts on the body/stone, so a body under one force is not in equilibrium. BUT according to me, this explanation is not correct. I say that at maximum height, body/stone is at rest and in equilibrium too. At that position, gravity is being considered by other teachers BUT weight of the body is being ignored. At the same time ,weight of the body balancing the gravity. So sum of the forces is zero. Condition of equilibrium is fulfilled. NOW WHO IS AT THE RIGHT? I or other teachers? Please comment.

@Muhammad Safi Ur Rehman , what do you mean by "at that position, gravity is being considered by other teachers BUT weight of the body is being ignored. At the same time ,weight of the body balancing the gravity. So sum of the forces is zero"? I didn't understand. Is not weight just an approximation of gravitational force at earth surface?

Sir Marcos Souza i mean that other teacher NOT ME, say that the stone/body at its maximum height when thrown upward, is at rest BUT NOT IN EQUILIBRUM. While I say that it is at rest and in equilibrium as well. They say no, because only once force GRAVITY acts on stone, no body is in equilibrium under one force. But I say that weight of the body balances gravity. So stone is in equilibrium. Further I present the example of a book lying on table is at rest and in equilibrium too. Same is the case of this stone. What do you think?

With your stone at maximum height , the stone has no speed, but the force is not zero and neither is the acceleration.

ie the sum of forces is not zero. The statment weight balances gravity is nonsensical, weight is due to gravity.

W=mg

The acceleration is the acceleration of gravity g, W and g point downward.

Dear All,

I, as an RG participant and learner, would like to thank you all for your active participation in the discussion thematized as

"Body (Rest, Nonequilibrium)".

Is it now possible to see the overall brief summary (concise point-wise) about the absolute reality?

Could someone please still throw further light into "Body (Rest, Nonequilibrium)"?

With best wishes and regards,

PK Karmakar

Muhammad Safi Ur Rehman , weight is the gravity force itself, approximate for the Earth's surface, so there is only one force acting, instead of two, like you are proposing. Think about it: if the sum of forces is zero, the body will preserve its velocity, so, if it is at rest, it will remain at rest. Thus, if your proposal were correct, the body would never fall down, don't you think?

And the example of the table is different. in that case, the equilibrium is due the normal force between the body and the table.

Look in Internet for the "Norton dome". An object on the top of the dome is at rest but in an unstable equilibrium. A blow of wind, and the object falls. On top of the dome the object is at rest, in equilibrium, but the equilibrium is unstable.

A body at the highest point when thrown upward against gravity, and a pendulum at its extremes, are not in equilibrium but under the action of the force of gravity.

in this world all systems tend to be in the steady state but any exited to its state make some dynamical movement to return back to this steady state and all our papers prove that