Properties of matter, film of liquid in ring
I tried a method which was wrong for fairly obvious reasons in hindsight and now I'm confused.
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@Gyro Gearloose
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Consider the tension in a differential portion due to the surface tension Td(theta) which would be equal to 2*pi.rd(theta)
And then apply the formula for young's modulus.
what exactly is alpha and why is it 2?
That is a 2 written badly. 😅 Because of the 2 layers of soap solution tension is 2T.l
Oh alright xD
thats what i thought as well intially but thought that i somehow went wrong
mb for being a lil late lmao
but where did this come from?
the Td(theta) = 2pi*r*d(theta)
also in the picture you've written 2sigma*Rd(theta) ?
i think i dont understand the differential portion
So what i'm getting is $dF = \sigma * 2R d(\theta) \newline$
Integrating theta with limits 0 to $2\pi, F = 2 \sigma R * 2 \pi = 2 \sigma l$
Nimboi [ping if answering]
instead of
dF, you're using F.d(theta) which i guess is why i'm confusedConsider a small portion with F force acting inclined at an angle d(theta) from the horizontal. The resultant force is 2Fsin(d(theta)/2)=Fd(theta)
ohhh
yeah the force direction
brilliant thank you
+solved @SirLancelotDuLac
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