Does capillary rise violate hydrostatic paradox?
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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 yesterday
Qmechanic♦
106k121961224
asked yesterday
Lelouche LamperougeLelouche Lamperouge
834
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add a comment |
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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 yesterday
Qmechanic♦
106k121961224
asked yesterday
Lelouche LamperougeLelouche Lamperouge
834
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$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
yesterday
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
yesterday
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 yesterday
Qmechanic♦
106k121961224
asked yesterday
Lelouche LamperougeLelouche Lamperouge
834
$endgroup$
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 yesterday
Qmechanic♦
106k121961224
asked yesterday
Lelouche LamperougeLelouche Lamperouge
834
edited yesterday
Qmechanic♦
106k121961224
asked yesterday
Lelouche LamperougeLelouche Lamperouge
834
edited yesterday
Qmechanic♦
106k121961224
edited yesterday
Qmechanic♦
106k121961224
edited yesterday
Qmechanic♦
106k121961224
106k121961224
asked yesterday
Lelouche LamperougeLelouche Lamperouge
834
asked yesterday
Lelouche LamperougeLelouche Lamperouge
834
asked yesterday
Lelouche LamperougeLelouche Lamperouge
834
834
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
yesterday
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
yesterday
add a comment |
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
yesterday
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
yesterday
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
yesterday
$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
yesterday
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
yesterday
$begingroup$
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
yesterday
add a comment |
2 Answers
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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 yesterday
Chet MillerChet Miller
15.8k2826
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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 yesterday
Sebastiano
323119
answered yesterday
himanshuhimanshu
604
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add a comment |
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2 Answers
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2 Answers
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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 yesterday
Chet MillerChet Miller
15.8k2826
$endgroup$
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 yesterday
Chet MillerChet Miller
15.8k2826
$endgroup$
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 yesterday
Chet MillerChet Miller
15.8k2826
$endgroup$
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 yesterday
Chet MillerChet Miller
15.8k2826
answered yesterday
Chet MillerChet Miller
15.8k2826
answered yesterday
Chet MillerChet Miller
15.8k2826
answered yesterday
Chet MillerChet Miller
15.8k2826
15.8k2826
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 yesterday
Sebastiano
323119
answered yesterday
himanshuhimanshu
604
$endgroup$
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 yesterday
Sebastiano
323119
answered yesterday
himanshuhimanshu
604
$endgroup$
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 yesterday
Sebastiano
323119
answered yesterday
himanshuhimanshu
604
$endgroup$
$p_A$ 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 yesterday
Sebastiano
323119
answered yesterday
himanshuhimanshu
604
edited yesterday
Sebastiano
323119
edited yesterday
Sebastiano
323119
edited yesterday
Sebastiano
323119
323119
answered yesterday
himanshuhimanshu
604
answered yesterday
himanshuhimanshu
604
answered yesterday
himanshuhimanshu
604
604
add a comment |
add a comment |
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$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
yesterday
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
yesterday