Fluid mechanics problem, need explanation please

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Discussion Overview

The discussion revolves around a fluid mechanics problem, specifically addressing the forces acting on a plug in a fluid system. Participants seek clarification on the treatment of pressure forces and the role of atmospheric pressure in the solution.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant questions why the plug was not treated as a vertical wall for calculating force, suggesting that the force should be the integral of pressure times area.
  • Another participant responds that the integral was indeed taken, but due to the small size of the plug, the height can be treated as a constant, simplifying the integral.
  • A participant inquires about the atmospheric pressure, asking where it is accounted for in the problem.
  • Responses indicate that atmospheric pressure acts on both sides of the plug and can be ignored in the calculations since it cancels out.

Areas of Agreement / Disagreement

Participants appear to have differing views on the treatment of atmospheric pressure and the simplifications made in the calculations. The discussion remains unresolved regarding the implications of these assumptions.

Contextual Notes

There are assumptions regarding the size of the plug and the treatment of atmospheric pressure that may affect the validity of the simplifications made in the discussion.

Bassel
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Hi please check the highlighted question in the attachment and the solution and see the problem, i need help please. I have two questions about it:

1- why didn't we take the plug as a vertical wall and thus the force would be integral of pressure x area?

2-WHERE DID THE FORCE DUE TO ATMOSPHERIC PRESSURE GO?

Thnx for your help.
 

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Hi Bassel! Welcome to PF! :smile:
Bassel said:
1- why didn't we take the plug as a vertical wall and thus the force would be integral of pressure x area?

You did.

You did ∫ ρgA(H+h) dh, but the plug is so small that you can take H+h to be a constant, H …

so that integral is just ∫ ρgAH dh :wink:
2-WHERE DID THE FORCE DUE TO ATMOSPHERIC PRESSURE GO?

it nipped round the other side when you weren't looking, and started pushing back :smile:

(so it's the same on both sides, and you can ignore it)
 
by the other side you mean from outside ??
 
yes, the plug has two sides :smile:
 

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