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$_{0}$ + $\rho$gh 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 $\rho$gh?

 

[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.

 

[Nickie]

As a scientist, it is important to carefully consider all variables and equations involved in a problem. In this case, the equation P = P_{0} + \rhogh represents the variation of pressure with depth, where P is the pressure at a certain depth, P_{0} is the atmospheric pressure, \rho is the density of the liquid, g is the acceleration due to gravity, and h is the depth. This equation is derived from the hydrostatic equation and takes into account the atmospheric pressure at the surface.

In the given problem, the pressure gauge is being lowered into water in a lake, and the piston is moving by a distance of 0.0075m. This change in depth is causing a change in pressure, which is measured by the gauge. Since the problem does not specify the depth of the lake or the depth at which the gauge is initially placed, it is not possible to calculate the pressure at the new depth without considering the atmospheric pressure.

Therefore, it is important to include the atmospheric pressure in the equation and consider it as a variable in the problem. It is possible that the error in the initial attempt was due to a miscalculation or incorrect use of the equation, rather than the inclusion of the atmospheric pressure. As a scientist, it is important to double check calculations and consider all relevant variables before concluding a solution.