Does capillary rise violate hydrostatic paradox?
$begingroup$
If $p$ is a pressure and $p_A = p_{text{atm}} + hdg,,$ $p_B = p_{text{atm}}$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?
fluid-statics capillary-action
edited 2 days ago
Qmechanic♦
106k121961224
asked Mar 19 at 16:26
Lelouche LamperougeLelouche Lamperouge
905
$endgroup$
add a comment |
$begingroup$
If $p$ is a pressure and $p_A = p_{text{atm}} + hdg,,$ $p_B = p_{text{atm}}$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?
fluid-statics capillary-action
edited 2 days ago
Qmechanic♦
106k121961224
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 $p_A = p_{text{atm}} + hdg,,$ $p_B = p_{text{atm}}$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?
fluid-statics capillary-action
edited 2 days ago
Qmechanic♦
106k121961224
asked Mar 19 at 16:26
Lelouche LamperougeLelouche Lamperouge
905
$endgroup$
If $p$ is a pressure and $p_A = p_{text{atm}} + hdg,,$ $p_B = p_{text{atm}}$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?
fluid-statics capillary-action
fluid-statics capillary-action
edited 2 days ago
Qmechanic♦
106k121961224
asked Mar 19 at 16:26
Lelouche LamperougeLelouche Lamperouge
905
edited 2 days ago
Qmechanic♦
106k121961224
asked Mar 19 at 16:26
Lelouche LamperougeLelouche Lamperouge
905
edited 2 days ago
Qmechanic♦
106k121961224
edited 2 days ago
Qmechanic♦
106k121961224
edited 2 days ago
Qmechanic♦
106k121961224
106k121961224
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
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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
15.9k2826
$endgroup$
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
329119
answered Mar 19 at 17:04
himanshuhimanshu
664
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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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
15.9k2826
$endgroup$
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
15.9k2826
$endgroup$
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
15.9k2826
$endgroup$
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
15.9k2826
answered Mar 19 at 17:09
Chet MillerChet Miller
15.9k2826
answered Mar 19 at 17:09
Chet MillerChet Miller
15.9k2826
answered Mar 19 at 17:09
Chet MillerChet Miller
15.9k2826
15.9k2826
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
329119
answered Mar 19 at 17:04
himanshuhimanshu
664
$endgroup$
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
329119
answered Mar 19 at 17:04
himanshuhimanshu
664
$endgroup$
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
329119
answered Mar 19 at 17:04
himanshuhimanshu
664
$endgroup$
$p_A$ 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
329119
answered Mar 19 at 17:04
himanshuhimanshu
664
edited Mar 19 at 21:28
Sebastiano
329119
edited Mar 19 at 21:28
Sebastiano
329119
edited Mar 19 at 21:28
Sebastiano
329119
329119
answered Mar 19 at 17:04
himanshuhimanshu
664
answered Mar 19 at 17:04
himanshuhimanshu
664
answered Mar 19 at 17:04
himanshuhimanshu
664
664
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