Why doesn't a hydraulic lever violate conservation of energy? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How 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?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How 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 yesterday
knzhou
47k11127226
asked yesterday
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 yesterday
knzhou
47k11127226
asked yesterday
Sawan KumawatSawan Kumawat
465
$endgroup$
18
$begingroup$
Your statement “displacement on both sides is same” is incorrect.
$endgroup$
– Farcher
yesterday
3
$begingroup$
do you think the levers also violate energy conservation?
$endgroup$
– user8718165
yesterday
2
$begingroup$
displacement means "volume", right?
$endgroup$
– JEB
yesterday
7
$begingroup$
@JEB hits the point. Displacement here means a distance moved and not the volume displaced.
$endgroup$
– JimmyB
yesterday
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 yesterday
knzhou
47k11127226
asked yesterday
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 yesterday
knzhou
47k11127226
asked yesterday
Sawan KumawatSawan Kumawat
465
edited yesterday
knzhou
47k11127226
asked yesterday
Sawan KumawatSawan Kumawat
465
edited yesterday
knzhou
47k11127226
edited yesterday
knzhou
47k11127226
edited yesterday
knzhou
47k11127226
47k11127226
asked yesterday
Sawan KumawatSawan Kumawat
465
asked yesterday
Sawan KumawatSawan Kumawat
465
asked yesterday
Sawan KumawatSawan Kumawat
465
465
18
$begingroup$
Your statement “displacement on both sides is same” is incorrect.
$endgroup$
– Farcher
yesterday
3
$begingroup$
do you think the levers also violate energy conservation?
$endgroup$
– user8718165
yesterday
2
$begingroup$
displacement means "volume", right?
$endgroup$
– JEB
yesterday
7
$begingroup$
@JEB hits the point. Displacement here means a distance moved and not the volume displaced.
$endgroup$
– JimmyB
yesterday
add a comment |
18
$begingroup$
Your statement “displacement on both sides is same” is incorrect.
$endgroup$
– Farcher
yesterday
3
$begingroup$
do you think the levers also violate energy conservation?
$endgroup$
– user8718165
yesterday
2
$begingroup$
displacement means "volume", right?
$endgroup$
– JEB
yesterday
7
$begingroup$
@JEB hits the point. Displacement here means a distance moved and not the volume displaced.
$endgroup$
– JimmyB
yesterday
18
18
$begingroup$
Your statement “displacement on both sides is same” is incorrect.
$endgroup$
– Farcher
yesterday
$begingroup$
Your statement “displacement on both sides is same” is incorrect.
$endgroup$
– Farcher
yesterday
3
3
$begingroup$
do you think the levers also violate energy conservation?
$endgroup$
– user8718165
yesterday
$begingroup$
do you think the levers also violate energy conservation?
$endgroup$
– user8718165
yesterday
2
2
$begingroup$
displacement means "volume", right?
$endgroup$
– JEB
yesterday
$begingroup$
displacement means "volume", right?
$endgroup$
– JEB
yesterday
7
7
$begingroup$
@JEB hits the point. Displacement here means a distance moved and not the volume displaced.
$endgroup$
– JimmyB
yesterday
$begingroup$
@JEB hits the point. Displacement here means a distance moved and not the volume displaced.
$endgroup$
– JimmyB
yesterday
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 yesterday
KV18KV18
1,043516
$endgroup$
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 yesterday
MarianD
258128
answered yesterday
BrolyBroly
476214
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
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 yesterday
KV18KV18
1,043516
$endgroup$
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 yesterday
KV18KV18
1,043516
$endgroup$
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 yesterday
KV18KV18
1,043516
$endgroup$
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 yesterday
KV18KV18
1,043516
answered yesterday
KV18KV18
1,043516
answered yesterday
KV18KV18
1,043516
answered yesterday
KV18KV18
1,043516
1,043516
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 yesterday
MarianD
258128
answered yesterday
BrolyBroly
476214
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 yesterday
MarianD
258128
answered yesterday
BrolyBroly
476214
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 yesterday
MarianD
258128
answered yesterday
BrolyBroly
476214
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 yesterday
MarianD
258128
answered yesterday
BrolyBroly
476214
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 yesterday
MarianD
258128
edited yesterday
MarianD
258128
edited yesterday
MarianD
258128
258128
answered yesterday
BrolyBroly
476214
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 yesterday
BrolyBroly
476214
answered yesterday
BrolyBroly
476214
476214
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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-energy-conservation, fluid-dynamics, newtonian-mechanics, pressure
18
$begingroup$
Your statement “displacement on both sides is same” is incorrect.
$endgroup$
– Farcher
yesterday
3
$begingroup$
do you think the levers also violate energy conservation?
$endgroup$
– user8718165
yesterday
2
$begingroup$
displacement means "volume", right?
$endgroup$
– JEB
yesterday
7
$begingroup$
@JEB hits the point. Displacement here means a distance moved and not the volume displaced.
$endgroup$
– JimmyB
yesterday