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Hydrostatic Equilibrium and Forces

  1. Feb 11, 2010 #1
    1. The problem statement, all variables and given/known data

    Two Piston cylinders are connected by a pipe with radius (Rp). Piston 1 which is located at height (h1) with an area (a1) is pushed down with a force (F1). In the opposite cylinder, Piston 2 is at a height (h2) with an area (a2), with force (F2) pushing up on the piston due to pascal's law. Piston system is filled with water with a viscosity (p).

    Givens :
    F1 = 100 N
    Rp = 6 m
    h1 = 5 m
    h2 = 10 m
    a1 = 10m^2
    a2 = 20m^2
    p = 1000kg/m^3

    Given the information, what opposing force must be applied to piston 2, to make piston system stay in hydrostatic equilibrium?

    2. Relevant equations
    P = F/A
    delta h = h2 - h1

    3. The attempt at a solution

    I initially looked at this and thought that pascal's law would be able to work, however upon inspection, cannot be used due to the different liquid height.

    With that in mind I started to take the Summation of the forces
    F = P * A
    F = rho*gravity*height
    height of 1 : h2-h1 = 5m = h1
    height of 2 : 10m

    Sum F : F2 + F1 = 0

    This is where I'm unsure, we have a pipe that connects the both and I know the velocity of the liquid accelerates in the area of the small tube, so it would make a difference in the equation.

    F1 = P1 A1 + Ppipe Apipe
    F2 = P2 A2 + Ppipe Apipe

    this is where I'm stuck
  2. jcsd
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