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I recently learned how to draw free body diagrams. Suppose a block is kept on ground and we need to draw FBD of ground. One force would be reaction force by block on ground. Will we also include the gravitational force on ground by block $(= mg )$ as a reaction force to gravitational pull on box by earth?This is what my teachers say is FBD of surface

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  • $\begingroup$ If you have received the answer for many of your Q on physics. Then , do accept them and don’t leave them open. Other users may think that you have still not understood the answer even though you have answers. $\endgroup$ – Srijan M.T May 26 at 8:00
  • $\begingroup$ ok sir I understood $\endgroup$ – lalit tolani May 26 at 14:48
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Yes, when considering the forces acting on the ground, in theory you should include both

  • the downwards normal force by the block and
  • the upwards gravitational force by the block.

In fact the gravitational force felt by the earth due to the object is exactly equal to the gravitational force felt by the object due to earth, just opposite. This Is newton's m 3rd law. The influence in the object is just much greater (it reaches higher acceleration during a fall e.g.) due to its much smaller mass.

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  • $\begingroup$ but sir all of my teachers say that mg is acting on center of earth and therefore we should not include it in our free body diagram of surface of earth $\endgroup$ – lalit tolani May 17 at 6:23
  • $\begingroup$ they say surface is different thing and whole body is different thing $\endgroup$ – lalit tolani May 17 at 6:23
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    $\begingroup$ @lalittolani Perhaps you could upload a picture of the free-body diagram of the ground that your teachers prefer. You cannot have a free-body diagram of the "surface" of anything; "free" means that an isolated object is shown, such as a chunk of Earth (with reaction forces where it meets the rest of the Earth) or the entire Earth. $\endgroup$ – Chemomechanics May 17 at 6:50
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    $\begingroup$ @lalittolani This doesn't make much sense, since if you only consider a small portion of the Earth - such as just the surface, by which I guess you mean a portion of soil or so - then even this small portion of Earth feels a gravitational pull by the rock. It also feels a gravitational pull by the rest of Earth, which is why it doesn't start accelerating upwards towards the rock. $\endgroup$ – Steeven May 17 at 7:56
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    $\begingroup$ @lalittolani Note, though, that gravity does not only act from the centre of Earth. It acts from every single particle with mass. But, from the outside, the effect of Earth's total gravitational force is the same as if Earth was just a point-mass - of this reason we very often average down Earth's gravity or mass into it's centre, since that simplifies a lot of things. But as soon as you start chopping off pieces of Earth and considering them separately, then we should be careful since they naturally carry mass as well and thus experience and exert gravitational forces as well. $\endgroup$ – Steeven May 17 at 7:56
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Suppose you have a block on ground. Then , it’s free body diagram is as follows and as you wrote in your question , you are right!.

enter image description here

mg,N1 and N2 are just action reaction pairs of each other according to the Newton’s third law.

N1=N2.

I hope your question is now clear.

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When a block is put on the ground. The gravitaional force of earth attracts the block with force $mg$ (More precisely, the centre of earth attracts the block and vice-versa). As a result the ground gets pushed and the ground exerts normal reaction on the block ($N$).
So in $y$ direction there are 2 forces on block -
i) Gravitational force by the earth ($mg$) in $-y$ direction.
ii) Normal reaction by ground ($N$) in $+y$ direction
As there is no vertical motion of block, so $N=mg$

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