# Static Equilibrium in Fluids: Pressure and Depth

1. Dec 7, 2007

### smichels

A cylindrical container 1.2 m tall contains mercury to a certain depth, d. The rest of the cylinder is filled with water. If the pressure at the bottom of the cylinder is 1.9 atm, what is the depth d?

Does anyone have any ideas on how to approach this problem, better yet,solve it!?

2. Dec 7, 2007

### Shooting Star

We are not here to solve problems, until you show us that you have tried your best and shown us the attempts. Then we'll guide you as best as we can.

3. Dec 7, 2007

### stewartcs

Hint: Potential Energy

4. Dec 8, 2007

### smichels

Sure, I understand. Here is what I have done so far:
The pressure at the bottom of a cylinder is equal to the force at the bottom divided by the Area, or

[P(bottom) = P(atmosphere) + density*gravity*height.

because we are dealing with water and mercury, do I need to equate this formula to:

P(at) + dens(water)*gravity*Height(cylinder)=P(at)+dens(mercury)*gravity*height(x). Where we solve for the height (x)

Am I on the right track?..

5. Dec 8, 2007

### Shooting Star

Some corrections.

P at bottom = P_atm + P(due to mercury) + P(due to water)
= P_atm + dens(mercury)*(height of mercury)*g + dens(water)*(height of water)*g.

Now you can put h of Hg as d and h of water as 1.2-d, ans solve. (Whether you have to neglect atm pressure depends on whether that has been mentioned in the problem.)