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shivam_bhalla
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My question is what is the pressure felt by the wall
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 2 years ago
 2 years ago
shivam_bhalla Group Title
See the question in diagram My question is what is the pressure felt by the wall Wait for diagram
 2 years ago
 2 years ago

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shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
dw:1335593779148:dw P>Pressure applied on wall of cylinder Assume one side of the cylinder is filled with liquid as in figure i) Does P vary with depth of cylinder ii)How do you get P?
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
@Aditya790 . Any tip?
 2 years ago

mos1635 Group TitleBest ResponseYou've already chosen the best response.0
are we looking for pressure or total force???
 2 years ago

Aditya790 Group TitleBest ResponseYou've already chosen the best response.2
@shivam_bhalla Pressure increases linearly with depth, the exact relation being P=pgh. In this problem, the density is same and gravitational constant is also same. The only thing that varies the pressure down the wall is how deep we go. From the figure, the wall is h units high and 2r units in width. At the top of the wall, the pressure is 0 (p*g*0) and at the bottom it is pgh. Since it increases linearly, we can use the average value of pressure that is pgh/2 as the pressure on the entire wall. The area of the wall is 2rh. So the force on the wall is (pgh/2)*(2rh), this is pressure times area. Alternately, you can divide the wall into horizontal strips of height dh and length 2r. the force on a particular segment at a depth of h is 2rpghdh. You can find the total force by integrating this expression from 0 to H. I wouldn't recommend this though because it is a longer method. Remember in general that for a quantity changing linearly, you can use the average value to represent the quantity at any instance. We do this in constantly accelerated motion where u is initial velocity and v is final velocity. We can compute distance by (u+v)/2 * time.
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
At the top of the wall, the pressure is 0 (p*g*0) >how, what about surface tension??
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
We have the force, but the area would be negligigble? Am I right @Aditya790
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
And so pressure due to surface tension>Almost absent
 2 years ago

Aditya790 Group TitleBest ResponseYou've already chosen the best response.2
@shivam_bhalla Yes. Compared to the pressure, surface tension is a very small value. At 20 degrees centigrade, the force due to suface tension is on the top is 0.0727*2r newton, which is very slight. Nice point though, I never considered surface tension in such a case. Since the surface tension acts only on the top, it causes a torque inward. To weak to cause a steel/glass vessal to break. But if you pour water into a nonrigid container like a plastic cover, the cover kind of closes up on the top and the body of the cover becomes spherical in shape (sphere has minimum surface area for a given volume) as water tries to minimise its surface area. Thanks for bringing my attention to it.
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
@Aditya790 , You are a computer or what :P . Great work :)
 2 years ago

Aditya790 Group TitleBest ResponseYou've already chosen the best response.2
@shivam_bhalla Not really, but thanks. Always feels good to be appreciated.
 2 years ago
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