# Fluid Pressure

1. Feb 10, 2014

### blahraptors

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

2. Relevant equations
fluid pressure = fluid density * depth * g

3. The attempt at a solution
My understanding is that fluid pressure at any two points at the same depth is equal, for a fluid at rest. But because one end is open, the pressure at point C is less than B and equal to A? I don't understand intuitively how this works.

Any help would be appreciated - thanks!

2. Feb 10, 2014

### Simon Bridge

Well at point A - what is exerting the pressure?

Would this be a high or a low pressure compared with the rest?

Remains to consider B and C ... what is exerting the pressure in each case?
Is there anything else that would contribute to the pressure?

3. Feb 11, 2014

### blahraptors

At point A, we have atmospheric pressure, as does point D, so they are equal I think. B has the weight of the water molecules above it, so it experiences more pressure than A and D. I'm just not sure why C is < B when they are at the same depth. Is it because molecules at C are moving but not at B?

4. Feb 11, 2014

### Simon Bridge

That's the one - notice that B is placed conspicuously far away from the entrance to the pipe that C is in.

You could also look at it like this: The pressure has to drop as you approach D or A.
C is closer to the low pressure region than B is.

5. Feb 11, 2014

### Staff: Mentor

What's happening is that the fluid that is approaching the outlet pipe has to get accelerated from the very low velocity away from the pipe (e.g., at point B ) to the higher velocity at point C. To accelerate it, you need an imbalance of force. This is supplied by a pressure variation, from higher pressure away from the pipe entrance to lower pressure within the pipe.