Is my approach for pascal's law wrong?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 1K views
Dileep Ramisetty
Messages
3
Reaction score
0
In hydraulic lift utilising pascal law, larger piston area is A1 and the smaller piston area is A2
mass in = mass out and volume in = volume out (in-compressible),so
A1*x1=A2*x2 (let x1 and x2 are displacements of respective pistons)
A1*V1 =A2*V2 (on differentiation gives continuity eq)
A1 * acceleration1=A2*acceleration2 (on differentiating again)
as mass flow rates m are equal,
m*A1*acceleration1=m*A2*acceleration2
A1*F1=A2*F2 (is giving me inverse of the pascal law)
please clarify...Thank you
 

Attachments

  • doubt.png
    doubt.png
    1.9 KB · Views: 552
Physics news on Phys.org
m*a1 is not the force on plate 1, no matter what "m" is.
m*a2 is not the force on plate 2, no matter what "m" is.

In particular, you can have a force without any acceleration.

Typically those problems assume that the acceleration is small and can be neglected. Otherwise Pascal's law needs modifications.
 
  • Like
Likes   Reactions: Dileep Ramisetty