Why doesn't a hydraulic lever violate conservation of energy?












3












$begingroup$


Suppose I apply some force on one side of Hydraulic lift where area is less, and the fluid in the lift raises some heavier object on the other side where area is more, Now work done is $Forcetimes displacement$ and displacement on both side is same (incompressible liquid) but force on one side is less, so we get more energy on other side. Then why doesn't the law of Conservation of energy fail here.






newtonian-mechanics fluid-dynamics pressure energy-conservation







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$endgroup$








  • 12




    $begingroup$
    Your statement “displacement on both sides is same” is incorrect.
    $endgroup$
    – Farcher
    7 hours ago






  • 1




    $begingroup$
    do you think the levers also violate energy conservation?
    $endgroup$
    – user8718165
    7 hours ago






  • 2




    $begingroup$
    displacement means "volume", right?
    $endgroup$
    – JEB
    5 hours ago






  • 2




    $begingroup$
    @JEB hits the point. Displacement here means a distance moved and not the volume displaced.
    $endgroup$
    – JimmyB
    5 hours ago
















3












$begingroup$


Suppose I apply some force on one side of Hydraulic lift where area is less, and the fluid in the lift raises some heavier object on the other side where area is more, Now work done is $Forcetimes displacement$ and displacement on both side is same (incompressible liquid) but force on one side is less, so we get more energy on other side. Then why doesn't the law of Conservation of energy fail here.






newtonian-mechanics fluid-dynamics pressure energy-conservation







share|cite|improve this question











$endgroup$








  • 12




    $begingroup$
    Your statement “displacement on both sides is same” is incorrect.
    $endgroup$
    – Farcher
    7 hours ago






  • 1




    $begingroup$
    do you think the levers also violate energy conservation?
    $endgroup$
    – user8718165
    7 hours ago






  • 2




    $begingroup$
    displacement means "volume", right?
    $endgroup$
    – JEB
    5 hours ago






  • 2




    $begingroup$
    @JEB hits the point. Displacement here means a distance moved and not the volume displaced.
    $endgroup$
    – JimmyB
    5 hours ago














3












3








3





$begingroup$


Suppose I apply some force on one side of Hydraulic lift where area is less, and the fluid in the lift raises some heavier object on the other side where area is more, Now work done is $Forcetimes displacement$ and displacement on both side is same (incompressible liquid) but force on one side is less, so we get more energy on other side. Then why doesn't the law of Conservation of energy fail here.






newtonian-mechanics fluid-dynamics pressure energy-conservation







share|cite|improve this question











$endgroup$




Suppose I apply some force on one side of Hydraulic lift where area is less, and the fluid in the lift raises some heavier object on the other side where area is more, Now work done is $Forcetimes displacement$ and displacement on both side is same (incompressible liquid) but force on one side is less, so we get more energy on other side. Then why doesn't the law of Conservation of energy fail here.






newtonian-mechanics fluid-dynamics pressure energy-conservation




newtonian-mechanics fluid-dynamics pressure energy-conservation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









knzhou

47k11127226




47k11127226










asked 7 hours ago









Sawan KumawatSawan Kumawat

334




334








  • 12




    $begingroup$
    Your statement “displacement on both sides is same” is incorrect.
    $endgroup$
    – Farcher
    7 hours ago






  • 1




    $begingroup$
    do you think the levers also violate energy conservation?
    $endgroup$
    – user8718165
    7 hours ago






  • 2




    $begingroup$
    displacement means "volume", right?
    $endgroup$
    – JEB
    5 hours ago






  • 2




    $begingroup$
    @JEB hits the point. Displacement here means a distance moved and not the volume displaced.
    $endgroup$
    – JimmyB
    5 hours ago














  • 12




    $begingroup$
    Your statement “displacement on both sides is same” is incorrect.
    $endgroup$
    – Farcher
    7 hours ago






  • 1




    $begingroup$
    do you think the levers also violate energy conservation?
    $endgroup$
    – user8718165
    7 hours ago






  • 2




    $begingroup$
    displacement means "volume", right?
    $endgroup$
    – JEB
    5 hours ago






  • 2




    $begingroup$
    @JEB hits the point. Displacement here means a distance moved and not the volume displaced.
    $endgroup$
    – JimmyB
    5 hours ago








12




12




$begingroup$
Your statement “displacement on both sides is same” is incorrect.
$endgroup$
– Farcher
7 hours ago




$begingroup$
Your statement “displacement on both sides is same” is incorrect.
$endgroup$
– Farcher
7 hours ago




1




1




$begingroup$
do you think the levers also violate energy conservation?
$endgroup$
– user8718165
7 hours ago




$begingroup$
do you think the levers also violate energy conservation?
$endgroup$
– user8718165
7 hours ago




2




2




$begingroup$
displacement means "volume", right?
$endgroup$
– JEB
5 hours ago




$begingroup$
displacement means "volume", right?
$endgroup$
– JEB
5 hours ago




2




2




$begingroup$
@JEB hits the point. Displacement here means a distance moved and not the volume displaced.
$endgroup$
– JimmyB
5 hours ago




$begingroup$
@JEB hits the point. Displacement here means a distance moved and not the volume displaced.
$endgroup$
– JimmyB
5 hours ago










2 Answers
2






active

oldest

votes


















18












$begingroup$

enter image description here
Displacement in both sides is not same. If on one side of lift the area is $A_1$, and on other side it is $A_2$, and we apply a force $F_1$ on one side to distance $d_1$ then volume decreased in one side is $=A_1 times d_1$



Equal amount of volume will raise in the other side.



So $$A_1 times d_1=A_2 times d_2$$



$A_1 not= A_2$, so $d_1 not=d_2$.



Actually, we need to apply the little force $F_1$ for a greater distance $d_1$.






share|cite|improve this answer










New contributor




Broly is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$





















    4












    $begingroup$

    The displacement produced is not the same. That is why, energy is conserved.



    When you apply force on one side of the opening (with smaller $A$, i.e. $A_1$), the displacement in the piston that does the work on the water is say, $x$. The displacement on the other side of the lift with $A_2$ where $A_2>A_1$, has a displacement smaller than $x$, which we'll call $y$.



    What happens here is that the water absorbs energy from the piston and sends it straight to the lift on the other end with area $A_2$. The volume of water remains the same. But the displacements need not be the same.



    Consider the work done $W=PDelta V$ where $Delta V$ is the change in volume. Since the first and the second openings are subjected to the same pressure (from the piston to the water, and from something that lifts the object in the larger opening), $Delta V=A_1x = A_2y$.



    $$y=frac{A_1x}{A_2}$$



    Since, $A_2 >A_1$, clearly, $y<x$.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

      votes









      18












      $begingroup$

      enter image description here
      Displacement in both sides is not same. If on one side of lift the area is $A_1$, and on other side it is $A_2$, and we apply a force $F_1$ on one side to distance $d_1$ then volume decreased in one side is $=A_1 times d_1$



      Equal amount of volume will raise in the other side.



      So $$A_1 times d_1=A_2 times d_2$$



      $A_1 not= A_2$, so $d_1 not=d_2$.



      Actually, we need to apply the little force $F_1$ for a greater distance $d_1$.






      share|cite|improve this answer










      New contributor




      Broly is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$


















        18












        $begingroup$

        enter image description here
        Displacement in both sides is not same. If on one side of lift the area is $A_1$, and on other side it is $A_2$, and we apply a force $F_1$ on one side to distance $d_1$ then volume decreased in one side is $=A_1 times d_1$



        Equal amount of volume will raise in the other side.



        So $$A_1 times d_1=A_2 times d_2$$



        $A_1 not= A_2$, so $d_1 not=d_2$.



        Actually, we need to apply the little force $F_1$ for a greater distance $d_1$.






        share|cite|improve this answer










        New contributor




        Broly is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        $endgroup$
















          18












          18








          18





          $begingroup$

          enter image description here
          Displacement in both sides is not same. If on one side of lift the area is $A_1$, and on other side it is $A_2$, and we apply a force $F_1$ on one side to distance $d_1$ then volume decreased in one side is $=A_1 times d_1$



          Equal amount of volume will raise in the other side.



          So $$A_1 times d_1=A_2 times d_2$$



          $A_1 not= A_2$, so $d_1 not=d_2$.



          Actually, we need to apply the little force $F_1$ for a greater distance $d_1$.






          share|cite|improve this answer










          New contributor




          Broly is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$



          enter image description here
          Displacement in both sides is not same. If on one side of lift the area is $A_1$, and on other side it is $A_2$, and we apply a force $F_1$ on one side to distance $d_1$ then volume decreased in one side is $=A_1 times d_1$



          Equal amount of volume will raise in the other side.



          So $$A_1 times d_1=A_2 times d_2$$



          $A_1 not= A_2$, so $d_1 not=d_2$.



          Actually, we need to apply the little force $F_1$ for a greater distance $d_1$.







          share|cite|improve this answer










          New contributor




          Broly is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          share|cite|improve this answer



          share|cite|improve this answer








          edited 7 hours ago









          MarianD

          242128




          242128






          New contributor




          Broly is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered 7 hours ago









          BrolyBroly

          401113




          401113




          New contributor




          Broly is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          New contributor





          Broly is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          Broly is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.























              4












              $begingroup$

              The displacement produced is not the same. That is why, energy is conserved.



              When you apply force on one side of the opening (with smaller $A$, i.e. $A_1$), the displacement in the piston that does the work on the water is say, $x$. The displacement on the other side of the lift with $A_2$ where $A_2>A_1$, has a displacement smaller than $x$, which we'll call $y$.



              What happens here is that the water absorbs energy from the piston and sends it straight to the lift on the other end with area $A_2$. The volume of water remains the same. But the displacements need not be the same.



              Consider the work done $W=PDelta V$ where $Delta V$ is the change in volume. Since the first and the second openings are subjected to the same pressure (from the piston to the water, and from something that lifts the object in the larger opening), $Delta V=A_1x = A_2y$.



              $$y=frac{A_1x}{A_2}$$



              Since, $A_2 >A_1$, clearly, $y<x$.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                The displacement produced is not the same. That is why, energy is conserved.



                When you apply force on one side of the opening (with smaller $A$, i.e. $A_1$), the displacement in the piston that does the work on the water is say, $x$. The displacement on the other side of the lift with $A_2$ where $A_2>A_1$, has a displacement smaller than $x$, which we'll call $y$.



                What happens here is that the water absorbs energy from the piston and sends it straight to the lift on the other end with area $A_2$. The volume of water remains the same. But the displacements need not be the same.



                Consider the work done $W=PDelta V$ where $Delta V$ is the change in volume. Since the first and the second openings are subjected to the same pressure (from the piston to the water, and from something that lifts the object in the larger opening), $Delta V=A_1x = A_2y$.



                $$y=frac{A_1x}{A_2}$$



                Since, $A_2 >A_1$, clearly, $y<x$.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  The displacement produced is not the same. That is why, energy is conserved.



                  When you apply force on one side of the opening (with smaller $A$, i.e. $A_1$), the displacement in the piston that does the work on the water is say, $x$. The displacement on the other side of the lift with $A_2$ where $A_2>A_1$, has a displacement smaller than $x$, which we'll call $y$.



                  What happens here is that the water absorbs energy from the piston and sends it straight to the lift on the other end with area $A_2$. The volume of water remains the same. But the displacements need not be the same.



                  Consider the work done $W=PDelta V$ where $Delta V$ is the change in volume. Since the first and the second openings are subjected to the same pressure (from the piston to the water, and from something that lifts the object in the larger opening), $Delta V=A_1x = A_2y$.



                  $$y=frac{A_1x}{A_2}$$



                  Since, $A_2 >A_1$, clearly, $y<x$.






                  share|cite|improve this answer









                  $endgroup$



                  The displacement produced is not the same. That is why, energy is conserved.



                  When you apply force on one side of the opening (with smaller $A$, i.e. $A_1$), the displacement in the piston that does the work on the water is say, $x$. The displacement on the other side of the lift with $A_2$ where $A_2>A_1$, has a displacement smaller than $x$, which we'll call $y$.



                  What happens here is that the water absorbs energy from the piston and sends it straight to the lift on the other end with area $A_2$. The volume of water remains the same. But the displacements need not be the same.



                  Consider the work done $W=PDelta V$ where $Delta V$ is the change in volume. Since the first and the second openings are subjected to the same pressure (from the piston to the water, and from something that lifts the object in the larger opening), $Delta V=A_1x = A_2y$.



                  $$y=frac{A_1x}{A_2}$$



                  Since, $A_2 >A_1$, clearly, $y<x$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 7 hours ago









                  KV18KV18

                  950415




                  950415






























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