The discussion revolves around a physical problem involving pistons and water dynamics, specifically focusing on the forces acting on a piston system with multiple surface areas (S1, S2, and S) and their relationship to water pressure and movement. Participants explore the implications of these relationships in a hydraulic context.
The discussion is ongoing, with multiple interpretations of the problem being explored. Some participants have offered guidance on considering the pressure dynamics and the role of the water's height in force calculations. There is recognition of differing opinions on the correct sign convention for the force, indicating a productive exchange of ideas.
Participants express uncertainty regarding the assumptions made about the system, including the fixed areas of the pistons and the conditions under which the force is calculated. The discussion also highlights the need for clarity on the definitions of the forces involved and their respective directions.
I can't understand what you said here. I'm not sure if it's because I misunderstood the translation.Baluncore said:2. The area of S2 exposed to atmosphere cancels an equal area of S exposed to the atmosphere.
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.Chestermiller said:Except for the sign.
@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.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?
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.haruspex said:@Baluncore is saying to consider the area S as made of two areas ...
Thanks for the answer, and I think I'm pretty sure now that the answer is correct, which is to use PASCAL's law.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).
Watch out for fires in your national forests. Be safe. Hope you get through this soon.Baluncore said:The sign of F will depend on convention. F = - ρ⋅g⋅H⋅S1