# Fluids at Rest: U-tube model

1. Sep 13, 2009

### Tater

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

The plastic tube has a cross-sectional area of 5.00 cm². The tube is filled with water until the short arm (of length d = 0.800 m) is full. Then the short arm is sealed and more water is gradually poured into the long arm. If the seal will pop off when the force on it exceds 9.80 N, what total height of the water in the long arm will put the seal on the verge of popping?

2. Relevant equations

P = Po + pgh
P = F / A

3. The attempt at a solution
Well I know that at their equal height, their pressure is in equilibrium (equal at that point).

H = h + d
Fpo = 9.80 N

(From free body diagram):
F2 - F1 = Fpo
PtA - PoA = Fpo

Pt = Po + pgh

This is where I'm stuck. I don't know what to do from here. Maybe i'm doing it completely wrong or there's an easier way. Can someone please help!

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2. Sep 19, 2009

### Redbelly98

Staff Emeritus
Don't know if you're still working on this one, but here are my thoughts.

When the short tube is first sealed off, there is zero net force acting on the seal because the pressure is Po both above it (from the air) and below it (from the water).

So the net force on the seal can be thought of as entirely due to the water added afterwards.