# Variation of pressure with depth

1. Nov 19, 2012

### whynot314

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$_{0}$ + $\rho$gh variation with pressure
P=F/A

3. 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 ammount $\rho$gh?

2. Nov 19, 2012

### 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 $\Delta x$?

3. Nov 19, 2012

### 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?

4. Nov 19, 2012

### 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.