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Now through my understanding, there is a weight force that points downward on block A, also another weight force that points downward on block B.

Between block A and block B, there is two normal forces equal in magnitude and opposite in direction right? (contact force). I believe the reason is because of electrostatic thingy between the two surfaces.

Then, isn't there two normal forces equal in magnitude and opposite in direction as well between block A and the surface? Since they're in contact of each other as well?

The left diagram is what I think the free-body diagram is. The right one is what I usually see. Which one is correct?

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    $\begingroup$ Many authors suggest that a formal "free body diagram" include only that single body and the forces acting on it. (That is you would draw your sketch with two blocks and the surface under them and separately a FBD or block A and a FDB of block B totaling three different sketches.) It is very common for people to include the forces on the main sketch when the situation is simple, but as in this case, doing so can make for a confusing diagram when there are several bodies involved. $\endgroup$ – dmckee --- ex-moderator kitten Jul 4 '15 at 2:55
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I think that the diagram should be somewhat simpler than previously posted. Here is my drawing of this system: enter image description here

Note that I separated the forces by a horizontal distance on both blocks to make the drawing easier to read. Normally, I would place a dot in the middle of each rectangle, and each force would originate from the dot in such a way that they were obviously equal and opposite, and such that the forces could not produce a torque on either block.

Block B has a weight, so it is being pulled down by the earth. It puts a force on block A. Since block B is not falling through block A, block A has to put an equal and opposite normal force on block B, denoted as Fn(A-B) in the drawing, where the order shown in the parentheses indicates that block A is putting a force on block B.

For block A, it also has a weight, and it has to support the weight of block B, so it is transmitting a force equal to its weight plus the weight of block B, onto the floor. Because blocks A and B are not falling through the floor, the floor has to put a normal force on block A that is equal to the weight of both blocks and opposite in direction.

The situation whereby the normal forces involved are equal and opposite to the weights involved satisfies Newton's 3rd law without the need for additional forces.

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I would recommend that you use different lengths to show different sizes of force; and that you ask yourself whether the entire system is in equilibrium, or whether it is accelerating. If the system is in equilibrium, then forces need to be balanced (no net force). Neither of your pictures seems to provide that. I would draw the diagram like this:

enter image description here

You start with two forces of gravity: labeled 1 and 2 in the diagram.

Then you add the reaction forces. Block B is resting on block A, so it applies force 1a to A. In return, A applies equal and opposite force 1b to B. Now there is no net force on B, and it remains stationary.

Looking at A, it has the force of gravity 2 acting on it. This force results in a force on the supporting surface 2a, and a reaction force 2b. At the same time, force 1a needs to be balanced: it is transmitted to the supporting surface as 1c, with reaction force 1d.

So for A as well, all the forces balance, and A does not move.

Note I displaced the force vectors horizontally to make it easier to draw / distinguish them. In reality they will be close to / on top of each other.

Maybe this is the same as your drawing on the left... it's a bit hard to tell.

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  • $\begingroup$ Hmm.. Understood. I guess mine is just missing the forces (action-reaction)transmitted from the top block. So my original question is true right? That there is also 2 normal forces equal and opposite to each other between the surface AND the lower block. $\endgroup$ – Zhi J Teoh Nov 11 '14 at 7:45
  • $\begingroup$ That's how it looks - but the arrows are thicker so it's hard to tell... $\endgroup$ – Floris Nov 11 '14 at 7:46
  • $\begingroup$ This is a poor way to represent forces, since it looks like you have unbalanced forces. In fact, 1 and 1a represent the same force (the same with 2 and 2a). This is the only way to have balanced forces and the objects are not accelerating. $\endgroup$ – LDC3 Jun 1 '15 at 3:06
  • $\begingroup$ @LDC3 I obviously didn't make it clear what I was trying to do. The forces acting on the top block are 1 and 1a - and they cancel (after you displace horizontally which I explicitly stated in the answer). Similarly the forces 2 and 2b cancel, as do the forces 1a and 1d. You are left with "unbalanced forces" 2a and 1c which are just the transmission of the force of gravity from the blocks to the support. What I am not showing is the fact that 1 and 2 are attracting the earth in the same way that the earth is attracting them - if I did, you could see that the earth is also not being accelerated. $\endgroup$ – Floris Jun 1 '15 at 3:40
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First of all you must find your system. Suppose we want to study the system B. What kind of forces exert on it? Obviously the earth exerts $W_B$ (weight). But since the system B can't penetrate somethings that are below it, so we say it must exist another force that exerts on it. We call it $N_B$ (or normal force in literature).

Now consider system A. Again we have the weight ($W_A$) and the normal force ($N_A$) by the same reasoning. But there is something above system A and that pushes it. We call this force $T_{AB}$.

Now the question is can we find it in terms of other known forces? According to Newton's third law of mechanics, the way that system B pushes system A, is as the same way as system A pushes system B. It means, $\vec T_{AB}=-\vec N_B$.

And nothing else remains. We found all forces that exerts on both systems.

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