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?
Best Answer
I think that the diagram should be somewhat simpler than previously posted. Here is my drawing of this system:
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.