# Homework Help: Pull a ring out of water

1. Sep 23, 2007

### iskar

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

Problem from Fluid Mechanics (Streeter, McGraw Hill, 9th ed.)

1.70 A method used to calculate surface tension of a liquid is finding the force necessary to pull a platinum ring out of the surface of that liquid. Calculate the force necessary to pull a ring with a 20mm diameter out of water at 20 degrees Celsius.

surface tension of water at given state: 0.074 N/m

2. Relevant equations

pressure = surface tension / radius
force = pressure * surface area

3. The attempt at a solution

I used the previous formulas and the surface tension for water at 20 Celsius.
Pressure = 7.4 Pa so the Force = 0.00232 N

1. This answer does not match the one in the book. 0.00919 N

2. The problem does not provide any other info about the size of the ring, but even if it was, I dont think the weight of the ring would have anything to do with the surface tension. This has more to do with the size of the ring and the properties of the liquid at the given temperature.

I am lost, and this is my first post, I dont know if I am posting this correctly.

Last edited: Sep 23, 2007
2. Sep 24, 2007

### Gokul43201

Staff Emeritus
The problem does tell you the size of the ring, which is important. The diameter is given to be 20mm.

The only thing you need to know about surface tension to solve this problem is its definition. The work done in creating an extra surface area dA is given by: dW = SdA, where S is the surface tension of the liquid with respect to its surroundings.

How much additional surface area is created when you pull a ring out of the water through some height h?

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