# Hydrostatic pressure (basic basic stuff)

1. Dec 14, 2006

### mbrmbrg

hydrostatic pressure (basic, basic stuff)

1. The problem statement, all variables and given/known data

In Figure 14-31 (attatched), an open tube of length L = 1.8 m and cross-sectional area A = 4.6 cm2 is fixed to the top of a cylindrical barrel of diameter D = 1.2 m and height H = 1.8 m. The barrel and tube are filled with water (to the top of the tube).
(a)Calculate the ratio of the hydrostatic force on the bottom of the barrel to the gravitational force on the water contained in the barrel.
(b)Why is that ratio not equal to 1.0? (You need not consider the atmospheric pressure.)

2. Relevant equations

P=F/A

$$P=P_0+\rho gh$$

$$mg=\rho Vg$$

3. The attempt at a solution

If it please the honored members of the site, their humble servant shall not blabber on with the method to the solution. For part (a) I got 2, which is the correct answer.

But part (b) has me stumped--my current answer reads "why would the ratio be 1?" I think I might be able to give a slightly more mature answer if I knew what logical fallacy I am supposed to be disproving...

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Last edited: Dec 14, 2006
2. Dec 14, 2006

### billiards

Just elaborate on your answer. In science it helps to be clear what you mean, so you could say "Why would the ratio be one? Clearly ........."

I can't help you any more than that without giving you the answer? Think along these lines

y/x will only equal 1 if y=x.

3. Dec 14, 2006

### mbrmbrg

So do you think a good answer would be:
"Because the hydrostatic force (which=$$\rho gh\pi r^2$$) does not equal the weight of the water (which=$$\rho h(\pi r^2+\pi R^2)g$$)."
or is there some blindingly obvious simple satement that I'm just missing?

4. Dec 14, 2006

### billiards

I'm not sure if that pi*r2 should be there for you hydrostatic force. And I'm not sure your other equation is right either.

Other than that I think it's a decent answer, I would elaborate and give the condition under which the hydrostatic force/weight of water does equal one(start by saying hydro force=weight). And then maybe say that this situation doesn't meet those conditions.

5. Dec 14, 2006

### mbrmbrg

Nope, the equations are fine (after all, they giveth me the correct anthwer to part a ); I just totally failed to explain them in my blurb.

Thenk you very much!

6. Dec 14, 2006

### OlderDan

You could delve into the forces acting that cause the local force on a bit of area at the bottom of a part of the drum that is not under the tube to be greater than the weight of the water above it.