Why doesn't a hydraulic lever violate conservation of energy? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) 2019 Moderator Election Q&A - Question CollectionHow is energy transferred from one incompressible fluid to another?Pascal's Principle and hydraulic liftMicroscopic source of pressure in an incompressible fluidIs work done on a fluid in a communicating vessel necessarily equal to work done by the fluid inside it?Force amplification and Newton's third lawEnergy paradox in fluid mechanicsAircraft lift theory vs energy conservationConfusion about Conservation of energyWhy the excess pressures are equated in hydraulic press?Why Pascal's Law is true and what is the mechanism for force amplification at molecular level?
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Why doesn't a hydraulic lever violate conservation of energy?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
2019 Moderator Election Q&A - Question CollectionHow is energy transferred from one incompressible fluid to another?Pascal's Principle and hydraulic liftMicroscopic source of pressure in an incompressible fluidIs work done on a fluid in a communicating vessel necessarily equal to work done by the fluid inside it?Force amplification and Newton's third lawEnergy paradox in fluid mechanicsAircraft lift theory vs energy conservationConfusion about Conservation of energyWhy the excess pressures are equated in hydraulic press?Why Pascal's Law is true and what is the mechanism for force amplification at molecular level?
$begingroup$
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
edited Apr 12 at 17:15
knzhou
47.1k11127226
asked Apr 12 at 11:53
Sawan KumawatSawan Kumawat
465
$endgroup$
add a comment |
$begingroup$
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
edited Apr 12 at 17:15
knzhou
47.1k11127226
asked Apr 12 at 11:53
Sawan KumawatSawan Kumawat
465
$endgroup$
18
$begingroup$
Your statement “displacement on both sides is same” is incorrect.
$endgroup$
– Farcher
Apr 12 at 12:22
3
$begingroup$
do you think the levers also violate energy conservation?
$endgroup$
– user8718165
Apr 12 at 12:41
2
$begingroup$
displacement means "volume", right?
$endgroup$
– JEB
Apr 12 at 13:58
7
$begingroup$
@JEB hits the point. Displacement here means a distance moved and not the volume displaced.
$endgroup$
– JimmyB
Apr 12 at 14:44
add a comment |
$begingroup$
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
edited Apr 12 at 17:15
knzhou
47.1k11127226
asked Apr 12 at 11:53
Sawan KumawatSawan Kumawat
465
$endgroup$
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
edited Apr 12 at 17:15
knzhou
47.1k11127226
asked Apr 12 at 11:53
Sawan KumawatSawan Kumawat
465
edited Apr 12 at 17:15
knzhou
47.1k11127226
asked Apr 12 at 11:53
Sawan KumawatSawan Kumawat
465
edited Apr 12 at 17:15
knzhou
47.1k11127226
edited Apr 12 at 17:15
knzhou
47.1k11127226
edited Apr 12 at 17:15
knzhou
47.1k11127226
47.1k11127226
asked Apr 12 at 11:53
Sawan KumawatSawan Kumawat
465
asked Apr 12 at 11:53
Sawan KumawatSawan Kumawat
465
asked Apr 12 at 11:53
Sawan KumawatSawan Kumawat
465
465
18
$begingroup$
Your statement “displacement on both sides is same” is incorrect.
$endgroup$
– Farcher
Apr 12 at 12:22
3
$begingroup$
do you think the levers also violate energy conservation?
$endgroup$
– user8718165
Apr 12 at 12:41
2
$begingroup$
displacement means "volume", right?
$endgroup$
– JEB
Apr 12 at 13:58
7
$begingroup$
@JEB hits the point. Displacement here means a distance moved and not the volume displaced.
$endgroup$
– JimmyB
Apr 12 at 14:44
add a comment |
18
$begingroup$
Your statement “displacement on both sides is same” is incorrect.
$endgroup$
– Farcher
Apr 12 at 12:22
3
$begingroup$
do you think the levers also violate energy conservation?
$endgroup$
– user8718165
Apr 12 at 12:41
2
$begingroup$
displacement means "volume", right?
$endgroup$
– JEB
Apr 12 at 13:58
7
$begingroup$
@JEB hits the point. Displacement here means a distance moved and not the volume displaced.
$endgroup$
– JimmyB
Apr 12 at 14:44
18
18
$begingroup$
Your statement “displacement on both sides is same” is incorrect.
$endgroup$
– Farcher
Apr 12 at 12:22
$begingroup$
Your statement “displacement on both sides is same” is incorrect.
$endgroup$
– Farcher
Apr 12 at 12:22
3
3
$begingroup$
do you think the levers also violate energy conservation?
$endgroup$
– user8718165
Apr 12 at 12:41
$begingroup$
do you think the levers also violate energy conservation?
$endgroup$
– user8718165
Apr 12 at 12:41
2
2
$begingroup$
displacement means "volume", right?
$endgroup$
– JEB
Apr 12 at 13:58
$begingroup$
displacement means "volume", right?
$endgroup$
– JEB
Apr 12 at 13:58
7
7
$begingroup$
@JEB hits the point. Displacement here means a distance moved and not the volume displaced.
$endgroup$
– JimmyB
Apr 12 at 14:44
$begingroup$
@JEB hits the point. Displacement here means a distance moved and not the volume displaced.
$endgroup$
– JimmyB
Apr 12 at 14:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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), $Delta V=A_1x = A_2y$.
$$y=fracA_1xA_2$$
Since, $A_2 >A_1$, clearly, $y<x$.
answered Apr 12 at 12:25
KV18KV18
1,177516
$endgroup$
add a comment |
$begingroup$
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 $d_1$.
edited Apr 12 at 12:33
MarianD
278129
answered Apr 12 at 12:22
BrolyBroly
499215
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.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
oldest
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oldest
votes
$begingroup$
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), $Delta V=A_1x = A_2y$.
$$y=fracA_1xA_2$$
Since, $A_2 >A_1$, clearly, $y<x$.
answered Apr 12 at 12:25
KV18KV18
1,177516
$endgroup$
add a comment |
$begingroup$
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), $Delta V=A_1x = A_2y$.
$$y=fracA_1xA_2$$
Since, $A_2 >A_1$, clearly, $y<x$.
answered Apr 12 at 12:25
KV18KV18
1,177516
$endgroup$
add a comment |
$begingroup$
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), $Delta V=A_1x = A_2y$.
$$y=fracA_1xA_2$$
Since, $A_2 >A_1$, clearly, $y<x$.
answered Apr 12 at 12:25
KV18KV18
1,177516
$endgroup$
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), $Delta V=A_1x = A_2y$.
$$y=fracA_1xA_2$$
Since, $A_2 >A_1$, clearly, $y<x$.
answered Apr 12 at 12:25
KV18KV18
1,177516
answered Apr 12 at 12:25
KV18KV18
1,177516
answered Apr 12 at 12:25
KV18KV18
1,177516
answered Apr 12 at 12:25
KV18KV18
1,177516
1,177516
add a comment |
add a comment |
$begingroup$
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 $d_1$.
edited Apr 12 at 12:33
MarianD
278129
answered Apr 12 at 12:22
BrolyBroly
499215
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.
$endgroup$
add a comment |
$begingroup$
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 $d_1$.
edited Apr 12 at 12:33
MarianD
278129
answered Apr 12 at 12:22
BrolyBroly
499215
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.
$endgroup$
add a comment |
$begingroup$
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 $d_1$.
edited Apr 12 at 12:33
MarianD
278129
answered Apr 12 at 12:22
BrolyBroly
499215
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.
$endgroup$
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 $d_1$.
edited Apr 12 at 12:33
MarianD
278129
answered Apr 12 at 12:22
BrolyBroly
499215
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.
edited Apr 12 at 12:33
MarianD
278129
edited Apr 12 at 12:33
MarianD
278129
edited Apr 12 at 12:33
MarianD
278129
278129
answered Apr 12 at 12:22
BrolyBroly
499215
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 Apr 12 at 12:22
BrolyBroly
499215
answered Apr 12 at 12:22
BrolyBroly
499215
499215
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.
add a comment |
add a comment |
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18
$begingroup$
Your statement “displacement on both sides is same” is incorrect.
$endgroup$
– Farcher
Apr 12 at 12:22
3
$begingroup$
do you think the levers also violate energy conservation?
$endgroup$
– user8718165
Apr 12 at 12:41
2
$begingroup$
displacement means "volume", right?
$endgroup$
– JEB
Apr 12 at 13:58
7
$begingroup$
@JEB hits the point. Displacement here means a distance moved and not the volume displaced.
$endgroup$
– JimmyB
Apr 12 at 14:44