Understanding the Force of Two Cubes

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The pressure inside the solid cube is zero, while the external pressure is atmospheric at 1.01 × 10^5 Pa. The force exerted by the pressure on one cube is calculated as F = pA, resulting in 6.3 × 10^3 N using the area of one face of the cube. The discussion raises a question about why only one cube's area is considered, suggesting that both cubes should be accounted for, similar to tension in a rope. However, the focus remains on the forces acting on a single cube in the direction normal to its missing face. Understanding these forces clarifies the application of pressure in this scenario.
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Homework Statement
Two hollow cubes of side 25 cm with one face missing are placed
together at the missing face. The air inside the solid
formed is pumped out. Determine the force that is necessary to
separate the cubes.
Relevant Equations
P=F/A
The solution is :

The pressure inside the solid is zero and outside it equals atmospheric pressure, 1.01 × 10^5 Pa.

Thus, the force is given by: F =pA = 1.01 × 10^5 × (0.25)^2 = 6.3 × 10^3 N

I do not understand why only the area of one cube is used in the solution, should it not be the area of both cubes e.g. Area of 2 cubes = 2 x (0.25)^2?
 
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This is analogous to a rope being pulled with force F at each end. Is the tension F or 2F?
Just consider one cube. What forces act on it in the direction normal to the missing face?
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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