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Variation of pressure with depth

  1. Nov 19, 2012 #1
    1. The problem statement, all variables and given/known data
    the spring of the pressure gauge. has a force constant of 1250 N/m, and the piston has a diameter of .012m. As the gauge is lowered into water in a lake, what change in depth causes the piston to move by .0075 m


    2. Relevant equations
    P= P[itex]_{0}[/itex] + [itex]\rho[/itex]gh variation with pressure
    P=F/A

    3. The attempt at a solution
    I did this problem including P[itex]_{0}[/itex] (atmospheric pressure) and it was wrong, then did it without atmospheric pressure and got it right. I am confused as to why atmospheric pressure is not included. because Isn't the pressure at P at a depth H below a point in the liquid at which pressure is P[itex]_{0}[/itex] by an ammount [itex]\rho[/itex]gh?
     
  2. jcsd
  3. Nov 19, 2012 #2
    The piston moves due to a change in the pressure, doesn't it?

    So, above surface the pressure is already [itex]P_0[/itex], which corresponds to a certain compression [itex]x_0[/itex]. Under the surface your equation holds and gives a compression [itex]x(P)=x_0+\Delta x[/itex]. What is [itex]\Delta x[/itex]?
     
  4. Nov 19, 2012 #3
    Ah, so when I am calculating the force to move the spring F=kx 1250 X .0075, I can think atmospheric pressure is included in this calculation. Then I just end up with P=ρgh?
     
  5. Nov 19, 2012 #4
    Yes, that's right. [itex]\Delta x[/itex] only depends on [itex]\rho g h[/itex] (and vice versa). By the way, this is why we often call this pressure (i.e. absolute pressure minus atmospheric pressure) gauge pressure.
     
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