Does capillary rise violate hydrostatic paradox?












6












$begingroup$


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







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$endgroup$












  • $begingroup$
    So P(A) need not be equal to P(B)???
    $endgroup$
    – Lelouche Lamperouge
    11 hours ago






  • 1




    $begingroup$
    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
    10 hours ago


















6












$begingroup$


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
    11 hours ago






  • 1




    $begingroup$
    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
    10 hours ago
















6












6








6





$begingroup$


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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 6 hours ago









Qmechanic♦

106k121961223




106k121961223










asked 11 hours ago









Lelouche LamperougeLelouche Lamperouge

754




754












  • $begingroup$
    So P(A) need not be equal to P(B)???
    $endgroup$
    – Lelouche Lamperouge
    11 hours ago






  • 1




    $begingroup$
    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
    10 hours ago




















  • $begingroup$
    So P(A) need not be equal to P(B)???
    $endgroup$
    – Lelouche Lamperouge
    11 hours ago






  • 1




    $begingroup$
    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
    10 hours ago


















$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
11 hours ago




$begingroup$
So P(A) need not be equal to P(B)???
$endgroup$
– Lelouche Lamperouge
11 hours ago




1




1




$begingroup$
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
10 hours ago






$begingroup$
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
10 hours ago












2 Answers
2






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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 $p_{atm}-hdg$.






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
      2






      active

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      active

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      8












      $begingroup$

      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









      $endgroup$


















        8












        $begingroup$

        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









        $endgroup$
















          8












          8








          8





          $begingroup$

          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 10 hours ago









          Chet MillerChet Miller

          15.8k2825




          15.8k2825























              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 6 hours ago









                  Sebastiano

                  318119




                  318119










                  answered 10 hours ago









                  himanshuhimanshu

                  603




                  603






























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