# Pressure Measurements

1. Apr 7, 2009

### science.girl

1. The problem statement, all variables and given/known data
The spring of a pressure gauge has a force constant of 1250 N/m, and the piston has a radius of 1.20 cm. As the gauge is lowered into water, what change in depth causes the piston to move in by 0.750 cm?

2. Relevant equations
F = -k$$\Delta$$x

$$\frac{F}{A}$$ = Y$$\frac{\Delta L}{L 0}$$

$$\Delta$$P = -B$$\frac{\Delta V}{V}$$

$$\rho$$ = $$\frac{M}{V}$$

P = $$\frac{F}{A}$$

P2 = P1 + $$\rho$$gh

3. The attempt at a solution
I'm not sure exactly how to begin.
I know that P = F/A, so would the pressure of the water on the gauge be:

P = F/A
P = F/($$\pi$$r2)
P = F/(1.44$$\pi$$)

But wouldn't another equation involve the force constant, because that would be a force acting against the water pressure?

Any guidance would be appreciated.

2. Apr 8, 2009

### Carid

To compress that spring a distance of one metre you'd need to exert a force of 1250 newtons. But we are only going to compress it by 0.75 centimetres. So how much force is required?

Now you've found the force, and we know its acting on a circular area of radius 1.2 cm, what pressure (force per unit area) is exerted on the piston?

Now in a liquid the static pressure is always equal to depth * density.

So equate the required pressure to the static liquid pressure and discover the depth.

3. Apr 8, 2009

### science.girl

F = -k$$\Delta$$x
F = (-1250 N/m)(-0.0075 m)
F = 9.375 N

P = $$\frac{F}{A}$$

P = $$\frac{9.375 N}{\pi(1.20 cm) 2}$$

P = 2.07 N/cm2

So, is this equivalent to the equation: P = P$$_{0}$$ + $$\rho$$gh?

And would the density be the density of the fluid? (In which case calculation would be unnecessary, and I would just use the standard density of water?)

By the way, thank you for your help thus far.

Last edited: Apr 8, 2009
4. Apr 8, 2009

### science.girl

I think I have it:

P = F/A = (9.375 N)/(pi * 0.012^2) = 20723.3 Pa

P = $$\rho$$gh
20723.3 Pa = (1000 kg/m^3)(9.8)(h)
h = 2.11 m

So... h = 2.11 m

Is this correct?

5. Apr 9, 2009

### Carid

Yes that looks right to me. I haven't checked all the units though.

We ignore the atmospheric pressure because the spring was already subjected to that at the beginning of the experiment and the movement of the spring is entirely due to immersion in the liquid.

6. Apr 9, 2009

### science.girl

Oh, now I understand. Thank you for mentioning that last tidbit. I appreciate your help, Carid!