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Does capillary rise violate hydrostatic paradox?
The Next CEO of Stack OverflowQuestion on the hydrostatic paradoxIt's about capillary rise of waterForces causing capillary riseIf a hole is drilled at the bottom of a vessel, why is the pressure of the liquid leaving the vessel equal to atmospheric pressure?Is hydrostatic pressure independent of temperature?Pressure on horizontal levels same?About hydrostatic pressure affecting measured weight on a scaleIs Pascal's law incorrect?Hydrostatic pressure in a gasDerivation of height of a liquid in a capillary tube
$begingroup$
If $p$ is a pressure and $p_A = p_textatm + hdg,,$ $p_B = p_textatm$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?
fluid-statics capillary-action
edited Mar 19 at 21:50
Qmechanic♦
107k121981230
asked Mar 19 at 16:26
Lelouche LamperougeLelouche Lamperouge
905
$endgroup$
add a comment |
$begingroup$
If $p$ is a pressure and $p_A = p_textatm + hdg,,$ $p_B = p_textatm$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?
fluid-statics capillary-action
edited Mar 19 at 21:50
Qmechanic♦
107k121981230
asked Mar 19 at 16:26
Lelouche LamperougeLelouche Lamperouge
905
$endgroup$
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
Mar 19 at 16:41
2
$begingroup$
You must have meant hydrostatic principle rather than hydrostatic paradox. ;) A paradox cannot be "violated"--rather, it appears to violate principles. And true principles resolve the paradox rather than violating it. :P
$endgroup$
– Dvij Mankad
Mar 19 at 17:08
add a comment |
$begingroup$
If $p$ is a pressure and $p_A = p_textatm + hdg,,$ $p_B = p_textatm$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?
fluid-statics capillary-action
edited Mar 19 at 21:50
Qmechanic♦
107k121981230
asked Mar 19 at 16:26
Lelouche LamperougeLelouche Lamperouge
905
$endgroup$
If $p$ is a pressure and $p_A = p_textatm + hdg,,$ $p_B = p_textatm$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?
fluid-statics capillary-action
fluid-statics capillary-action
edited Mar 19 at 21:50
Qmechanic♦
107k121981230
asked Mar 19 at 16:26
Lelouche LamperougeLelouche Lamperouge
905
edited Mar 19 at 21:50
Qmechanic♦
107k121981230
asked Mar 19 at 16:26
Lelouche LamperougeLelouche Lamperouge
905
edited Mar 19 at 21:50
Qmechanic♦
107k121981230
edited Mar 19 at 21:50
Qmechanic♦
107k121981230
edited Mar 19 at 21:50
Qmechanic♦
107k121981230
107k121981230
asked Mar 19 at 16:26
Lelouche LamperougeLelouche Lamperouge
905
asked Mar 19 at 16:26
Lelouche LamperougeLelouche Lamperouge
905
asked Mar 19 at 16:26
Lelouche LamperougeLelouche Lamperouge
905
905
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
Mar 19 at 16:41
2
$begingroup$
You must have meant hydrostatic principle rather than hydrostatic paradox. ;) A paradox cannot be "violated"--rather, it appears to violate principles. And true principles resolve the paradox rather than violating it. :P
$endgroup$
– Dvij Mankad
Mar 19 at 17:08
add a comment |
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
Mar 19 at 16:41
2
$begingroup$
You must have meant hydrostatic principle rather than hydrostatic paradox. ;) A paradox cannot be "violated"--rather, it appears to violate principles. And true principles resolve the paradox rather than violating it. :P
$endgroup$
– Dvij Mankad
Mar 19 at 17:08
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
Mar 19 at 16:41
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
Mar 19 at 16:41
2
2
$begingroup$
You must have meant hydrostatic principle rather than hydrostatic paradox. ;) A paradox cannot be "violated"--rather, it appears to violate principles. And true principles resolve the paradox rather than violating it. :P
$endgroup$
– Dvij Mankad
Mar 19 at 17:08
$begingroup$
You must have meant hydrostatic principle rather than hydrostatic paradox. ;) A paradox cannot be "violated"--rather, it appears to violate principles. And true principles resolve the paradox rather than violating it. :P
$endgroup$
– Dvij Mankad
Mar 19 at 17:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The pressures at A and B are indeed equal. However, the pressure in the fluid immediately below the curved meniscus is equal to $p_atm-hdg$ as a result of surface tension. So the pressure at A is $$p_A=p_atm-hdg+hdg=p_atm=p_B$$That is, there is a discontinuous change in pressure across the meniscus as a result of the surface tension in combination with the curvature. The pressure on the upper side of the interface is $p_atm$ and the pressure on the lower side of the interface is $p_atm-hdg$.
answered Mar 19 at 17:09
Chet MillerChet Miller
16k2826
$endgroup$
add a comment |
$begingroup$
$p_A$ is equal to $p_B$ here. The disparity is arising due to the fact that pressure just outside the meniscus is greater than the pressure inside. This is due to the curvature of the meniscus and surface tension.
This difference is compensated by $hdg$ to make $p_A=p_B$.
edited Mar 19 at 21:28
Sebastiano
344119
answered Mar 19 at 17:04
himanshuhimanshu
804
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
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active
oldest
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active
oldest
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$begingroup$
The pressures at A and B are indeed equal. However, the pressure in the fluid immediately below the curved meniscus is equal to $p_atm-hdg$ as a result of surface tension. So the pressure at A is $$p_A=p_atm-hdg+hdg=p_atm=p_B$$That is, there is a discontinuous change in pressure across the meniscus as a result of the surface tension in combination with the curvature. The pressure on the upper side of the interface is $p_atm$ and the pressure on the lower side of the interface is $p_atm-hdg$.
answered Mar 19 at 17:09
Chet MillerChet Miller
16k2826
$endgroup$
add a comment |
$begingroup$
The pressures at A and B are indeed equal. However, the pressure in the fluid immediately below the curved meniscus is equal to $p_atm-hdg$ as a result of surface tension. So the pressure at A is $$p_A=p_atm-hdg+hdg=p_atm=p_B$$That is, there is a discontinuous change in pressure across the meniscus as a result of the surface tension in combination with the curvature. The pressure on the upper side of the interface is $p_atm$ and the pressure on the lower side of the interface is $p_atm-hdg$.
answered Mar 19 at 17:09
Chet MillerChet Miller
16k2826
$endgroup$
add a comment |
$begingroup$
The pressures at A and B are indeed equal. However, the pressure in the fluid immediately below the curved meniscus is equal to $p_atm-hdg$ as a result of surface tension. So the pressure at A is $$p_A=p_atm-hdg+hdg=p_atm=p_B$$That is, there is a discontinuous change in pressure across the meniscus as a result of the surface tension in combination with the curvature. The pressure on the upper side of the interface is $p_atm$ and the pressure on the lower side of the interface is $p_atm-hdg$.
answered Mar 19 at 17:09
Chet MillerChet Miller
16k2826
$endgroup$
The pressures at A and B are indeed equal. However, the pressure in the fluid immediately below the curved meniscus is equal to $p_atm-hdg$ as a result of surface tension. So the pressure at A is $$p_A=p_atm-hdg+hdg=p_atm=p_B$$That is, there is a discontinuous change in pressure across the meniscus as a result of the surface tension in combination with the curvature. The pressure on the upper side of the interface is $p_atm$ and the pressure on the lower side of the interface is $p_atm-hdg$.
answered Mar 19 at 17:09
Chet MillerChet Miller
16k2826
answered Mar 19 at 17:09
Chet MillerChet Miller
16k2826
answered Mar 19 at 17:09
Chet MillerChet Miller
16k2826
answered Mar 19 at 17:09
Chet MillerChet Miller
16k2826
16k2826
add a comment |
add a comment |
$begingroup$
$p_A$ is equal to $p_B$ here. The disparity is arising due to the fact that pressure just outside the meniscus is greater than the pressure inside. This is due to the curvature of the meniscus and surface tension.
This difference is compensated by $hdg$ to make $p_A=p_B$.
edited Mar 19 at 21:28
Sebastiano
344119
answered Mar 19 at 17:04
himanshuhimanshu
804
$endgroup$
add a comment |
$begingroup$
$p_A$ is equal to $p_B$ here. The disparity is arising due to the fact that pressure just outside the meniscus is greater than the pressure inside. This is due to the curvature of the meniscus and surface tension.
This difference is compensated by $hdg$ to make $p_A=p_B$.
edited Mar 19 at 21:28
Sebastiano
344119
answered Mar 19 at 17:04
himanshuhimanshu
804
$endgroup$
add a comment |
$begingroup$
$p_A$ is equal to $p_B$ here. The disparity is arising due to the fact that pressure just outside the meniscus is greater than the pressure inside. This is due to the curvature of the meniscus and surface tension.
This difference is compensated by $hdg$ to make $p_A=p_B$.
edited Mar 19 at 21:28
Sebastiano
344119
answered Mar 19 at 17:04
himanshuhimanshu
804
$endgroup$
$p_A$ is equal to $p_B$ here. The disparity is arising due to the fact that pressure just outside the meniscus is greater than the pressure inside. This is due to the curvature of the meniscus and surface tension.
This difference is compensated by $hdg$ to make $p_A=p_B$.
edited Mar 19 at 21:28
Sebastiano
344119
answered Mar 19 at 17:04
himanshuhimanshu
804
edited Mar 19 at 21:28
Sebastiano
344119
edited Mar 19 at 21:28
Sebastiano
344119
edited Mar 19 at 21:28
Sebastiano
344119
344119
answered Mar 19 at 17:04
himanshuhimanshu
804
answered Mar 19 at 17:04
himanshuhimanshu
804
answered Mar 19 at 17:04
himanshuhimanshu
804
804
add a comment |
add a comment |
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$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
Mar 19 at 16:41
2
$begingroup$
You must have meant hydrostatic principle rather than hydrostatic paradox. ;) A paradox cannot be "violated"--rather, it appears to violate principles. And true principles resolve the paradox rather than violating it. :P
$endgroup$
– Dvij Mankad
Mar 19 at 17:08