How Much Force to Pull a Platinum Ring from Water?

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SUMMARY

The discussion focuses on calculating the force required to pull a platinum ring with a 20mm diameter out of water at 20 degrees Celsius, utilizing the surface tension of water, which is 0.074 N/m. The initial calculation yielded a force of 0.00232 N, which was incorrect compared to the expected value of 0.00919 N. The key formulas used include pressure = surface tension / radius and force = pressure * surface area. The importance of the ring's diameter and the properties of the liquid at the specified temperature are emphasized as critical factors in the calculation.

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Homework Statement



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

Homework Equations



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

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 don't 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 don't know if I am posting this correctly.
 
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iskar said:

Homework Statement



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

Homework Equations



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

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 don't 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 don't know if I am posting this correctly.
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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