Does capillary rise violate hydrostatic paradox?












8












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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 $p_A=p_B$?



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fluid-statics capillary-action







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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


















8












$begingroup$


If $p$ is a pressure and $p_A = p_{text{atm}} + hdg,,$ $p_B = p_{text{atm}}$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?



enter image description here






fluid-statics capillary-action







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















8












8








8


1



$begingroup$


If $p$ is a pressure and $p_A = p_{text{atm}} + hdg,,$ $p_B = p_{text{atm}}$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?



enter image description here






fluid-statics capillary-action







share|cite|improve this question











$endgroup$




If $p$ is a pressure and $p_A = p_{text{atm}} + hdg,,$ $p_B = p_{text{atm}}$, is hydrostatic paradox violated, shouldn't $p_A=p_B$?



enter image description here






fluid-statics capillary-action




fluid-statics capillary-action






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edited yesterday









Qmechanic♦

106k121961224




106k121961224










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




















  • $begingroup$
    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












2 Answers
2






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11












$begingroup$

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 $p_{atm}-hdg$.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    $p_A$ 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 $p_A=p_B$.






    share|cite|improve this answer











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      2 Answers
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      2 Answers
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      11












      $begingroup$

      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 $p_{atm}-hdg$.






      share|cite|improve this answer









      $endgroup$


















        11












        $begingroup$

        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 $p_{atm}-hdg$.






        share|cite|improve this answer









        $endgroup$
















          11












          11








          11





          $begingroup$

          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 $p_{atm}-hdg$.






          share|cite|improve this answer









          $endgroup$



          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 $p_{atm}-hdg$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Chet MillerChet Miller

          15.8k2826




          15.8k2826























              4












              $begingroup$

              $p_A$ 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 $p_A=p_B$.






              share|cite|improve this answer











              $endgroup$


















                4












                $begingroup$

                $p_A$ 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 $p_A=p_B$.






                share|cite|improve this answer











                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  $p_A$ 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 $p_A=p_B$.






                  share|cite|improve this answer











                  $endgroup$



                  $p_A$ 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 $p_A=p_B$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited yesterday









                  Sebastiano

                  323119




                  323119










                  answered yesterday









                  himanshuhimanshu

                  604




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