Asking for help with a physical problem that uses a piston to push water

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SUMMARY

The discussion centers on calculating the force exerted by a piston system involving water, specifically addressing the relationship between the surface areas of the pistons (S1, S2, and S) and the force (F) applied. The correct formula for the force is established as F = -ρgHS1, where ρ is the fluid density, g is the acceleration due to gravity, and H is the height of the water column. The participants clarify that the pressure of the water drives the piston and that the areas S1 and S2 are coupled, affecting the overall force calculation. The conversation emphasizes the importance of understanding hydraulic principles and the conservation of energy in these systems.

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  • Familiarity with Pascal's Law
  • Knowledge of pressure calculations in hydraulic systems
  • Basic concepts of force and motion in physics
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Engineers, physics students, and anyone involved in designing or analyzing hydraulic systems will benefit from this discussion, particularly those looking to deepen their understanding of piston dynamics and fluid pressure calculations.

  • #31
As far as I am concerned, the inflow and outflow of water refers to the outflow of water from the tube to the place in contact with air, S and S1 meet this condition, while S and S2 do not meet this condition, because the water at the mouth of S2 has no contact with air.
 
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  • #32
I think I've figured out how to do it, and I can do it very quickly with PASCAL's law. 1. F=-ρgSH; 2. F=-(1/2)ρgSH; 3. F=-(1/2)ρgS(H1+H2); 4. F =-ρgS(H1+H2).
 
  • #33
Baluncore said:
2. The area of S2 exposed to atmosphere cancels an equal area of S exposed to the atmosphere.
I can't understand what you said here. I'm not sure if it's because I misunderstood the translation.
If S2 and S are equal to the air contact surface, how to understand figure 2? Does it also result in F =-ρgHS1?
 
  • #34
Chestermiller said:
Except for the sign.
You told me what the answer was, but you didn't tell me how it was calculated. There are several possible ways to compute this answer, and I don't know which one is the most accurate and can be applied to other graphs, so I assumed a few graphs and tried to learn how to compute it from more calculations.
 
  • #35
vxiaoyu18 said:
I can't understand what you said here. I'm not sure if it's because I misunderstood the translation.
If S2 and S are equal to the air contact surface, how to understand figure 2? Does it also result in F =-ρgHS1?
@Baluncore is saying to consider the area S as made of two areas S1' and S2' side by side, the first of area S1 and the second of area S2. The pressure on S2' is the same as that on S2, and they have the same area, so the forces on them are equal and opposite, so cancel.
 
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  • #36
haruspex said:
@Baluncore is saying to consider the area S as made of two areas ...
Can you think of it this way, that if you don't change the height, no matter what S2' prime does, it doesn't matter what S1' prime does, and the value of F stays the same? Is F =-ρgHS1.
 
  • #37
vxiaoyu18 said:
I think I've figured out how to do it, and I can do it very quickly with PASCAL's law.
1. F=-ρgSH;
2. F=-(1/2)ρgSH;
3. F=-(1/2)ρgS(H1+H2);
4. F =-ρgS(H1+H2).
Thanks for the answer, and I think I'm pretty sure now that the answer is correct, which is to use PASCAL's law.
 
Last edited:
  • #38
Baluncore said:
The sign of F will depend on convention. F = - ρ⋅g⋅H⋅S1
Watch out for fires in your national forests. Be safe. Hope you get through this soon.
 

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