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I have asked a question on I pushing the wall,where the force of me is not zero but as per the formula F=m.a,the force is zero and then I got the answer , it is because here the net force is acting on the body.But I a doubt that then a body of mass 5kg and moving with an acceleration 2m/s^2 will experience a force = 10N,Is in this case also the net force acting? Or the net force is acting only when the body doesn't move? If this is correct then my question why?

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  • $\begingroup$ You have posted this question 3 times with different formating in less than an hour. This is not only duplicating but spamming the forum. I think it is rude that even when this was been answered here you still opened the same question again, at least read the answers before you duplicate it man. $\endgroup$ – Swike Jun 23 at 16:04
  • $\begingroup$ No,I am just clearing my doubts $\endgroup$ – user230720 Jun 23 at 16:10
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    $\begingroup$ When clearing your doubts, edit the question that you already asked. Don't open a new question to ask the same question a different way. $\endgroup$ – garyp Jun 23 at 16:18
  • $\begingroup$ Okk I will do that from next time onwards $\endgroup$ – user230720 Jun 23 at 16:20
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Now starting from scratch, what we know from Newton's 2nd law is given below:-$$\vec{F_{net}}=m\vec{a}$$

When we push the wall the forces acting on it are:- 1) Friction in opposite direction of your push. 2)Your pushing force. 3)Normal reaction from earth. 4)Weight. 5)Downward normal reaction from ceiling. As,$$\vec{a_{net}}=0$$ $$\vec{f}+\vec{N_{e}}+m\vec{g}+\vec{N_{c}}+\vec{F_{you}}=0$$ I think these two equations in itself describes the whole mechanics of the problem.

Hope this helps!

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    $\begingroup$ Thank you very much Unique $\endgroup$ – user230720 Jun 23 at 16:22
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In the equation for Newton's second law $F=ma$ the $F$ always represents the net force. It's better to write the equation as $\Sigma F= ma$ or $F_{net}=ma$.

In your example of a 5 kg object experiencing an acceleration of 2 m/s^2 there might be a single 10 N force acting or, more likely, there are two or more forces acting which add to 10 N.

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  • $\begingroup$ In case of force=zero we see a lot of force but in case of a body experiencing force we see a single force why? $\endgroup$ – user230720 Jun 23 at 15:15
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I have asked a question on I pushing the wall,where the force of me is not zero but as per the formula F=m.a,the force is zero and then I got the answer , it is because here the net force is acting on the body.

The net force on a body is the sum of all the forces acting on that body. By sum I don't mean adding the intensity of each force (their absolute values), I mean vector addition.

When you push a wall you are excerting a force on it. Why it doesn't accellerate (starts moving)? because there are more forces at play. The wall is anchored to the floor, this means that when you push the wall the wall pushes on the floor in the same direction. But because of Newton's third law the floor is also pushing on the wall with the same intensity but in the opposite direction. Thus ON the wall there are two forces acting; your force acting on the wall, and the force of the floor reacting to the action of the wall (an action that was created by pure transmission of your force throught the wall to the floor). Thus the net force acting on the wall is the sum of the two forces we have described on this body. Those two forces are equally intense but opposite so they cancel each other out.

When you think of accelleration as the final change in movement of the object Newton's second law can be interpreted as the force been the net force applied on that object. Since the net force is zero the wall does remain at rest.

But I a doubt that then a body of mass 5kg and moving with an acceleration 2m/s^2 will experience a force = 10N

Why? If the body is moving with accelleration $a = 2\; m/s^2$ and the mass of the object is $m = 5\; kg$ then the force has to be $F = 10\; N$. It obeys Newton's second law and also obeys intuition. No object would start to accellerate if there is no force pushing/pulling it. The net force in this case is the same as the force applied since there is no other force at play on the $5\;kg$ body (the sum of 1 thing is just that thing).

In case of force=zero we see a lot of force but in case of a body experiencing force we see a single force why?

This is a plain contradiction. If $F = 0$ then we don't see a lot of $F$. I think you are still playing here with the difference between a force and the net force (the sum of all the forces), but if you use the same word for everything it is impossible to make sense of it. They are not the same concept, even if in the particular scenario when only one force is acting on a body it is the same as the net force logically.


TL;DR;

You seem to have a missconception on what is the difference (or you don't acknowledge that difference at all) between a force acting on an object and the resulting net force acting on an object. The second is the sum of the contributions of any number of the first ones.

You present two scenarios;

Scenario A: I push the wall but the wall doesn't move. Your question is why if $F = something$ I observe $a = nothing$ when $F = ma$ tells me that if $F = something$ then $a = something$? And the answer is that you missed the fact that your push is not the only force acting on the wall, that there is another force; the reaction of the floor (or the thing to which the wall is fixed) against the wall. The net force is thus the sum of both forces and in this scenario it turns out that they cancel each other out. So the answer is that you observe $a = nothing$ because $F = nothing$, which is in perfect concordance to $F=ma$, where $F$ is the net force.

Scenario B: An object (let's say floating in space so we don't have to think of anything else) is pushed by me and I see it accellerating at $a = 2\;m/s^2$. Then how if Newton's laws are correct there should be a net force acting on it. But the net force should be zero like before right? NO! If the object is moving according to $F = ma$ the net force is non-zero. So why this is not the same case as the wall? because the net force, the sum of all the forces acting on the object, is just the same as the only force acting on the object: your push. According to Newton's third law there is a reaction force from the object to you (the pusher), in fact in space when you push the object you would also be flown back (not only the object would be moved but also you in the opposite direction). But this is important, these two forces are acting on two different objects (the action is made by you towards the object and the reaction is applied by the object towards you).

Difference between scenarios A and B I will repeat this: in the Scenario A this is more complex, there are 4 forces at play (your push on the wall, the wall reaction pushing you, the wall pushing the floor (or the object which is fixated to) and the floor pushing on the wall) but only 2 of these 4 forces are acting on the object "wall", the other two are acting on the floor and on you. In the Scenario B there are only 2 forces at play (you pushing the object and the object reaction on you) but only 1 of those 2 forces is applied on the object, the other is applied on you.

The net force acting on the object is not the sum of all the forces at play but only of all the forces acting on the specific object. That's why in Scenario B the object moves and in Scenario A it doesn't, in perfect confirmation of Newton's laws.

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  • $\begingroup$ Thank you very much Swike $\endgroup$ – user230720 Jun 23 at 16:21

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