Variation of pressure with depth

[whynot314]

Homework Statement

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

Homework Equations

P= P_{0} + \rhogh variation with pressure
P=F/A

The Attempt at a Solution

I did this problem including P_{0} (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_{0} by an amount \rhogh?

 

[Hypersphere]

The piston moves due to a change in the pressure, doesn't it?

So, above surface the pressure is already P_0, which corresponds to a certain compression x_0. Under the surface your equation holds and gives a compression x(P)=x_0+\Delta x. What is \Delta x?

 

[whynot314]

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?

 

[Hypersphere]

Yes, that's right. \Delta x only depends on \rho g h (and vice versa). By the way, this is why we often call this pressure (i.e. absolute pressure minus atmospheric pressure) gauge pressure.