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I Is my approach for pascal's law wrong?

  1. Mar 4, 2017 #1
    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
     

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  2. jcsd
  3. Mar 4, 2017 #2

    mfb

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    Staff: Mentor

    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.
     
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