Variation of pressure with depth

  • Thread starter whynot314
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  • #1
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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[itex]_{0}[/itex] + [itex]\rho[/itex]gh variation with pressure
P=F/A

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?
 

Answers and Replies

  • #2
191
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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]?
 
  • #3
76
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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
191
8
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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