What Height of Water in a Long Arm Risks Popping the Seal?

[Tater]

Homework Statement

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?

Homework Equations

P = Po + pgh
P = F / A

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!

 

[Redbelly98]

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.

 

[tomcue]

Good start! You are correct in recognizing that at the equal heights, the pressure is in equilibrium. However, there are a few things to consider in order to find the total height of the water in the long arm that will put the seal on the verge of popping.

First, we need to determine the pressure at the bottom of the long arm, as this will be the maximum pressure that the seal can withstand before popping off. This can be found using the equation P = Po + pgh, where P is the pressure at the bottom, Po is the atmospheric pressure, p is the density of water, g is the acceleration due to gravity, and h is the height of the water in the long arm.

Next, we can use the equation P = F/A to relate the pressure to the force acting on the seal. In this case, the force is the weight of the water in the long arm, which can be found using the density of water and the volume of water in the long arm (which can be calculated using the cross-sectional area and the height of the water).

Finally, we can set these two equations equal to each other and solve for the height h, which will give us the total height of the water in the long arm that will put the seal on the verge of popping.

I hope this helps! Remember to always consider the relevant equations and variables, and think about how they are related to each other. Good luck!