Hydrostatics on plane surface (basic fluid mechanic Q)

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SUMMARY

The discussion focuses on the calculation of hydrostatic forces acting on plane surfaces, specifically addressing the positions of individual forces F1 and F2. The hydrostatic force F1 does not act at the centroid of the square section due to the non-constant pressure distribution with depth, necessitating integration of the pressure distribution to determine the force's magnitude and center of pressure. For F2, the pressure distribution is simpler, being triangular, which allows for straightforward calculations without complex formulas. The distinction between triangular and trapezoidal pressure distributions is crucial for accurate force positioning.

PREREQUISITES
  • Understanding of hydrostatic pressure principles
  • Familiarity with pressure diagrams in fluid mechanics
  • Knowledge of integration techniques for pressure distribution
  • Basic geometric properties of triangles and trapezoids
NEXT STEPS
  • Study the integration of pressure distributions in hydrostatics
  • Learn about calculating the center of pressure for triangular and trapezoidal shapes
  • Explore applications of hydrostatic forces in engineering scenarios
  • Review the derivation of hydrostatic force equations for various geometries
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Students and professionals in fluid mechanics, civil engineers, and anyone involved in the design and analysis of structures subjected to hydrostatic forces.

uzman1243
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I don't understand how they got the "position of individual forces".

For F1, I'm guessing they used a pressure diagram and that's why it's 2/3 from the tip of the triangle. But why wouldn't the force act on the centroid for this square section (thus its 3.5/2)?
?temp_hash=a20ac79e236529c76b2ff2b74bb4e03b.png

But for position of F2 (y2), they used the equation. Can you explain this?
 

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uzman1243 said:
?temp_hash=a20ac79e236529c76b2ff2b74bb4e03b.png


I don't understand how they got the "position of individual forces".

For F1, I'm guessing they used a pressure diagram and that's why it's 2/3 from the tip of the triangle. But why wouldn't the force act on the centroid for this square section (thus its 3.5/2)?

The hydrostatic force F1 doesn't act at the centroid of the square because it's not constant with respect to depth. You must integrate the pressure distribution acting over the area to find the magnitude of the force F1, and then calculate the first moment of this force with respect to the waterline in order to locate the point at which the hydrostatic pressure acts.

Since the pressure distribution is triangular w.r.t. depth, it's easy to calculate the total hydrostatic force and to find the center of pressure, knowing some geometric facts about triangles.
?temp_hash=a20ac79e236529c76b2ff2b74bb4e03b.png

But for position of F2 (y2), they used the equation. Can you explain this?

Since the top of the square is at the waterline for the body, the pressure distribution is pretty simple: it's a triangle. No complicated formulas are required to find the center of pressure.

For the bottom of the body, the bottom is sloping and the pressure distribution is trapezoidal, so there are no easy formulas to apply except the one shown in the solution.
 

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