- #1
Dileep Ramisetty
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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
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