Calculate Surface Tension of Blood Plasma: Help Needed

AI Thread Summary
To calculate the surface tension of blood plasma, a vertical force of 1.90×10-2 N is needed to lift a wire ring with a radius of 2.07 cm. The surface tension formula is surface tension = force/length, where length is the circumference of the wire loop. The correct length to use is twice the circumference, leading to a revised calculation. After correcting the approach, the final surface tension value obtained was 0.0731 N/m, which matched the expected answer. This demonstrates the importance of considering both sides of the wire loop when calculating surface tension.
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can anyone help me out with this question:In order to lift a wire ring of radius 2.07 cm from the surface of a container of blood plasma, a vertical force of 1.90×10-2 N greater than the weight of the ring is required. Calculate the surface tension of blood plasma from this information.
I can't figure it out it seemed so easy at first but then i got the wrong answer.
 
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Details? This looks like homework --- forum policy is that you show us your work, or state the difference between your answer and what is expected.
 
surface tension is equal to force over length so what i though is that since the force and radiua are given that length L would be the radius times two and to get surface tension the force that is given would be divided by this number, but that's not right because it was the wrong answer.
 
You got half the number expected?

Too late at night, I meant you got a surface tension "twice" what was expected.
 
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do you mean that i should take my answer and divide by two or multiply by two?
 
Post your answer --- we'll go from there.
 
surface tension= F/L
2.07 cm= .0207m
L= .0207m x 2= .0414m

1.90×10-2 N/ .0414m= .458 N/m
where do I go from here?
 
Got bumped off for a minute --- Okay, now we get somewhere:
1) remember the formula for circumference of a circle? 2πr

2) when you're pulling the loop from the surface of the liquid, you noticed that the surface "draped" BOTH directions from the wire? Does it make sense to you that you're actually pulling against surface tension exerted against twice the circumference of the wire loop? Once around for the outside of the loop, and once more for the inside?

3) What I come up with on the back of the envelope, and I'm truncating --- you'll have to take it beyond the single significant digit, is 0.07 N/m.

I thought at first that you were hung up on just the trick of doubling the circumference of the loop for this type of measurement --- that's why I said "half" mistakenly, and corrected things to guess that you were getting "twice" the value you were expecting, but it's more like 6 times larger than what your answer key is looking for.

This work for you?
 
Thanks alot! You really helped me Igot the answer of 0.0731 N/m, and it was right thanks again.
 
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