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Old Dec 20th 2018, 12:32 AM   #1
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Lightbulb surface tension

There is horizontal film of soap solution.On it a thread is placed in the form of a loop.The film is punctured inside the loop and the thread becomes a circular loop of radius R.If the surface tension of the soap solution is T,then the tension in the thread will be?

Plz help me out.
Plz provide the solution also with ur answer.
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Old Dec 20th 2018, 01:23 AM   #2
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So this is the situation .....



there is a circular hole in the soap bubble.

Surface tension is measured in N/m

we need the length of the loop, the circumference .... 2 pi R

T (surface tension in N/m) = t/ 2 pi R ......(t is tension in thread in N)

t = T 2 pi R
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Old Dec 20th 2018, 06:50 AM   #3
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I also thought the same. But the answer as given in the book turned out to be 2rT.
r-radius
T-surface tension constant
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Old Dec 20th 2018, 02:05 PM   #4
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There is an error in our calculations !!

the answer should be t= 2T 2pi R .......... t = 4 T pi R

we forgot there are two surfaces to the soap film

but this t is the force pulling outward on the thread , to get the tension in the thread we have to divide by 2 pi.

This 4 min video answers this exact question
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Old Dec 21st 2018, 01:46 AM   #5
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Thanks.

But I still don't get why dividing by 2pi
Plz help.
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Old Dec 21st 2018, 05:40 PM   #6
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I'm glad you asked that mak .... I'll have to walk myself through to make sure I get it myself ......

Here's a diagram which will suit our purpose.... It relates to the pressure in a tube and the stresses that produces , but the calculation is the same for our situation.



the diagram on the left shows the thread , all the arrows are the forces from the surface tension ... to find the tension in the thread we have to add all the components of the small forces together.

In the diagram on the right , T is the tension we need to find . The 2 small arrows pointing up and down will not impart any force because they are at right angles ... the small arrow pointing right will transmit all of it's force ... and all the arrows in between just some of their force ... you have to multiply by cos the angle to find out how much each arrow contributes .....

And when you add all these little forces together by integration it comes out to be 1/2pi
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