FBDs of Stacked Boxes: Understanding Force Distribution in Connected Objects

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The discussion focuses on understanding force distribution in two stacked boxes using free body diagrams (FBDs). The original poster struggled to visualize the forces acting on the boxes, particularly the action-reaction forces as described by Newton's third law. They noted that box A experiences an upward force from box B, which in turn feels a downward force from A, complicating the analysis. The ground must provide a force to support both boxes, accounting for the weight of box A on top of box B. Ultimately, the poster found clarity in the problem after considering the interactions between the forces involved.
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To help me understand how a force is distributed over multiple connected objects, I have been trying to draw the FBDs of two stacked boxes which are sitting on the ground. I haven't been able to solve the problem however, and looking for the solution with Google did not turn up anything useful. I'm aware of Newton's third law, but I still have not been able to solve this... I'm at a loss, which is why I am posting here.

Here is an image of the problem:
fbd_problem.jpg


One attempt at solving it using Newton's third law ended up as (what appeared to be) an infinite loop of action-and-reaction forces going up and down the stack. I tried thinking of the two objects as a single object, but that just ended in guesswork without anything real concrete.

What would the FBD of A and the FBD of B look like, with all involved forces?
 
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So A and B each feel a gravitational force (also resultant in reactionate forces meaning you have to draw two upward vectors in the Earth's core, but i think we're ignoring them here). Since A is not moving, it has to experience an equal (in magnitude) force upward: this can only come from B, and this force's reactionate brother is the force B experiences from A, the latter obviously pushing down. This force makes B extra heavy and thus the ground under B must deliver a force not only for B itself, but also the load B is carrying (this is indirectly what allows B to hold A up in its place). This last force from the ground to B also has a reactionate brother pushing from the blocks onto the ground (which is the force that would make for a weighing scale placed under B to react)

I hope this helps?
 
That solved the problem, thanks :)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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