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This question already has an answer here:

I have a doubt on the formula, $$F=m\cdot a$$ In pushing the wall, as per formula, $F= m\cdot 0=0$ as there is no acceleration hence zero. Therefore force become zero. But we have applied some force which is non zero.

How can then the force be zero?

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marked as duplicate by stafusa, Jon Custer, Kyle Kanos, M. Enns, Aaron Stevens Jun 27 at 16:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ it’s only for resultant force. $\endgroup$ – Ubaid Hassan Jun 23 at 14:16
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    $\begingroup$ Possible duplicate of Explanation on Newton's second law $\endgroup$ – Swike Jun 23 at 15:59
  • $\begingroup$ No duplicate I have just asked that question $\endgroup$ – user230720 Jun 23 at 16:02
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    $\begingroup$ I'm willing to upvote this question given that you change its title to make it easier to find for other people who might have the same doubt. $\endgroup$ – corcholatacolormarengo Jun 24 at 16:33
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Here F does not mean the force that you apply, but it is the sum of all the forces that are being applied. The wall is also applying an equal opposite force F back. This force is applied by the rigidity of the wall and by the its contact with the rest of the room.

$$ F_{ext}=M*a_{CM}$$

And even more fundamental is $$ F_{ext}=\frac{dp}{dt}$$ where P is the momentum.

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The total force on the wall includes the part provided by your hand, and also the part provided by the floor and other things to which the wall is attached. This total force on the wall all adds up to zero.

Similarly, the total force on yourself includes the part on your hand and also the friction between your feet and the floor. All this adds to zero in total (if you are not accelerating).

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  • $\begingroup$ Force of any object which we will calculate through this formula.will it give the net force? $\endgroup$ – user230720 Jun 23 at 14:14
  • $\begingroup$ @user230720 That is correct. $\endgroup$ – SmarthBansal Jun 23 at 14:17
  • $\begingroup$ Then a body having mass suppose 5kg and moved with an acceleration of 2m/s^2 then the force= 10N.In this case, Is resultant force acting? $\endgroup$ – user230720 Jun 23 at 14:26
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The correct formula is $\sum \vec F = m\vec a$. FOr using this formula first you need to define a system. In your case the system is you and the wall. And as net force is zero so acceleration becomes zero.

The net force is zero because, in reaction to the force that you applied on the wall, the wall also applied the same force but in the opposite direction.

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Now starting from scratch, what we know from Newton's 2nd law is given below:- $$\vec{F}_{net}=m\vec{a}_{net}$$

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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