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#40 by whitelancer64 on 12 Mar, 2019 22:23
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*snip* there will be an unbalance of momentum, and with that and enough force, it can move the hole center of mass.
Where is the unbalanced momentum?QuoteBy the way, the piping or better the flexible garden pipe, will be affected by the same forces as in the other pipe, going opposite direction, Newton will be there, but that dont affect the moving containers. You dont see it ehhh?
BY THE WAY THE THIRD LAW OF NEWTON THAT RELATES TO THE CENTER OF MASS, EXPLAIN THAT NO INTERNAL FORCE CAN MOVE THE CENTER OF ITS OWN MASS BECAUSE OF ACTION-REACTION, BUT IN THIS DEVICE HOW IT LOSES MASS THE CONTAINER B, IT HAS NOTHING TO DO WITH NEWTON.
You are quite right, no internal force can move a center of its own mass. Your device is very much proving this. Newton is still there. -
#41 by esposcar on 12 Mar, 2019 22:30
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This I found it in the net:
Why does a system need an external force to shift its center of mass?
It's a consequence of conservation of momentum, or alternatively, of Newton's third law. Any force that the boy exerts on the raft is countered by an equal but opposite force that the raft exerts on the boy. Those internal forces cancel, and therefore play zero role in the motion of the center of mass relative to an inertial frame.
Well, if one of the Containers lose mass in the transfer and no momentum will be transmitted to the stationary container, then container A will lose momentum because of mass. And the fluid will land as it departed, in total rest. And this have nothing to do with center of mass itself. If the force have not done reaction it works, and if it loses mass it loses momentum. -
#42 by esposcar on 12 Mar, 2019 22:31
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*snip* there will be an unbalance of momentum, and with that and enough force, it can move the hole center of mass.
Where is the unbalanced momentum?QuoteBy the way, the piping or better the flexible garden pipe, will be affected by the same forces as in the other pipe, going opposite direction, Newton will be there, but that dont affect the moving containers. You dont see it ehhh?
BY THE WAY THE THIRD LAW OF NEWTON THAT RELATES TO THE CENTER OF MASS, EXPLAIN THAT NO INTERNAL FORCE CAN MOVE THE CENTER OF ITS OWN MASS BECAUSE OF ACTION-REACTION, BUT IN THIS DEVICE HOW IT LOSES MASS THE CONTAINER B, IT HAS NOTHING TO DO WITH NEWTON.
You are quite right, no internal force can move a center of its own mass. Your device is very much proving this. Newton is still there.
Have you viewed the pics? I think its clear, in the transfer of fluid from container A to B.
Teletransporting mass literally as I show, does not do any force but moving the piston but changes the momentum of the container. I post you again the pic, and you tell me what happens with the stationary container. First view how the device works and then you can critic me, without generalities, that you can not show in the device, with all my respects.
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#43 by meberbs on 12 Mar, 2019 23:09
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I find difficult to find equations that describes how the mass is transfered. Do you have any clue?.
It seems that everyone in this thread except you has a clue on this. I posted a link to a video on how to calculate force through a bent pipe. Others pointed out Bernoulli's equation for fluid motion, which is also relevant. (See the other link I had provided that points out that fluid flow speeds would be different between the inside and outside curves of a curved pipe.)Step one goes in Newton, step two also, step three of course, the only thing that in my idea makes it reactionless is the transfer of fluid, once the container have a final velocity, say 5 metres per second or whatever, the fluid aboard is at rest. NEVER IN ACCELERATION the transfer of fluid must happen.
It is reactionless when you transfer fluid, because you simply are choosing to ignore what everyone here is telling you: there would be a force present. You are saying that the fluid starts in container A which is moving away from container B. Somehow you claim that the fluid will end up in container B, and also be at rest. By definition the velocity of the fluid is different between the initial and final conditions. This cannot happen if there is no acceleration. The velocity of the fluid changed therefore it must have accelerated this is simply a definition.Then what happens if one of the container pulls the system with more force than the other after the transfer, because have more fluid in it?
Your question is invalid. You are assuming the result you want (unbalanced forces). The if part of the question is phrased wrong. the question you should be asking is "what would happen?"the center of mass, wont move? I really see it difficult that stays the same. Then Newton third law would not work.
According to Newton's laws, the center of mass of a system at rest will remain at rest unless an external force acts on the system. There are no external forces on the system. (If the center of mass is not at rest, which is the case for some of the diagrams you posted, then the center of mass instead moves with constant velocity, but you can simplify things by moving to the center of mass frame so my previous statement holds.)Since when two opposite mass forces, both tied up, the opposite object with more weight, will not able to pull the other object and the system to its side, if both have the same speed? its not logic. Explain me visually how it would be arranged, that no force win. What would be the momentum of one deposit full of fluid respect to the other deposit going light?
No clue what any of what you are saying here is supposed to mean.Can you show what you say with an equation? So it will move container B to the same direction as container A has. Sorry but you cannot show that.
Asserting that others can't show something is not going to help you.
Here is some math for you:
Mc = mass of empty cart (both carts are the same)
Mf = Mass of fluid that fits in 1 cart
Vi = initial velocity of cart A.
rho = fluid density
Total initial momentum of the system:
(assume the hose is made of some relatively super lightweight material, so we can ignore that part of it is moving, and part isn't)
Ptot = (Mc+Mf)*Vi
Let us assume that it is a long flexible tube as you suggest. For some length of hose, if the direction and velocity of fluid flow going in and out of the tubing is the same then, (ignoring fiction losses) there is no net momentum change in the fluid, so we can just ignore that section. To take advantage of this, assume a 90 degree bend fixed to the top of each cart and pointing to each other. Any net forces experienced by the piping will therefore be only from those bends. (You could drop this assumption and get the same results I am about to show, but then you would have to take into account the changing shape of the hose as the carts move.)
assume the fluid flow during the transfer happens at a constant rate dM/dt and the cross sectional area of all piping including the 90 degree bends is a constant area, A.
The cart will empty in time T = Mf / (dM/dt). The velocity of the fluid in the pipe is vt = dM/dt /(rho*A).
The fluid enters the pipe bend with upwards velocity, but leaves with velocity pointed to the right. Ignoring the upwards component of the force (which wouldn't end up relevant in this setup), then in the reference frame of cart A, the fluid after the bend must be moving to the right at vt, so it is carrying momentum to the right away at a rate of vt*dM/dt. By Newton's second law, that means it had a force being continuously exerted on it by the pipe bend of F1 = vt*dM/dt pointed to the right. by Newton's third law, it exerts a force on the pipe equal in magnitude to F1, but pointed to the left.
This is a force effectively applied to cart A through the pipe bend during the transfer. the final velocity of cart A is therefore:
vi + F1*T/Mc =vi + vt*dM/dt * Mf/ (dM/dt) /Mc = vi + vt*Mf/Mc
In other words, as the mass in the first cart decreases, the velocity of the cart actually increases.
More work can be done to show that a comparable force is applied to cart B, but making it move to the right so when you add everything together, the net momentum of the system is unchanged. Also, there are details about what happens when the cart is partially full, so to actually get everything correct, you would need to account for the variable effective mass of each cart with time (Go look up a derivation of the rocket equation for details). Plus, there are some details I glossed over about the hose. Since it is attached to 2 carts moving at different velocities, it will move itself, along with all of the fluid in it, and this will be balanced out when you consider that if the velocity of the fluid relative to the tubing is the same at either end, but the ends of the tubing are moving at different speeds, then the fluid must be accelerating (speeding up or slowing down) relative to the other frames, and therefore actually applying forces, as well the changing velocities of the carts dragging the hose with them.Have you viewed the pics? I think its clear, in the transfer of fluid from container A to B.
Too bad pipes don't operate by teleporting, but by having fluid flow through them and apply forces.
Teletransporting mass literally as I show, does not do any force but moving the piston but changes the momentum of the container. -
#44 by esposcar on 12 Mar, 2019 23:57
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I will look with detail all what you wrote, bit please understand that the fluid inside the containers will be calm, the pipe will move also in a moment at constant velocity, and as the containers gets rid of fluid, ok, the pipe will not push further the empty container to compensate, becuase it travels already at 100 m/s and it cannot give more speed with the same speed of 100 m/s, because it ends are pulled also at 100 m/s and all the pipe structure in a moment will have 100 m/s, but the container will not accelerate, because the pipes cannot acellerate over 100 m/S. I dont know if I am clear at this? And about the water flow in a closed system ruled by Pascal, the forces are the same everywhere, no flow exist in Pascal equation. The links that you showed me before dont apply to Pascal principle, they talk about flow, but flow like river, I dont know what flow will do an input piston into the fluid.
The two links that you sent me, talks about hydrodynamics, like constant flow, and dont apply to Pascal laws, that define hydrostatic. An interesant name, that you want to put flow in it
By the way, I need advices from experts, not people like me and honestly, very few I saw here. Experts come with proves to discredit somebody, but in this case it did not happen in any away, just moving on with topics. Maybe it is what happens when you arrive in new phisics topics Lol
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#45 by esposcar on 13 Mar, 2019 00:14
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I find difficult to find equations that describes how the mass is transfered. Do you have any clue?.
It seems that everyone in this thread except you has a clue on this. I posted a link to a video on how to calculate force through a bent pipe. Others pointed out Bernoulli's equation for fluid motion, which is also relevant. (See the other link I had provided that points out that fluid flow speeds would be different between the inside and outside curves of a curved pipe.)Step one goes in Newton, step two also, step three of course, the only thing that in my idea makes it reactionless is the transfer of fluid, once the container have a final velocity, say 5 metres per second or whatever, the fluid aboard is at rest. NEVER IN ACCELERATION the transfer of fluid must happen.
It is reactionless when you transfer fluid, because you simply are choosing to ignore what everyone here is telling you: there would be a force present. You are saying that the fluid starts in container A which is moving away from container B. Somehow you claim that the fluid will end up in container B, and also be at rest. By definition the velocity of the fluid is different between the initial and final conditions. This cannot happen if there is no acceleration. The velocity of the fluid changed therefore it must have accelerated this is simply a definition.Then what happens if one of the container pulls the system with more force than the other after the transfer, because have more fluid in it?
Your question is invalid. You are assuming the result you want (unbalanced forces). The if part of the question is phrased wrong. the question you should be asking is "what would happen?"the center of mass, wont move? I really see it difficult that stays the same. Then Newton third law would not work.
According to Newton's laws, the center of mass of a system at rest will remain at rest unless an external force acts on the system. There are no external forces on the system. (If the center of mass is not at rest, which is the case for some of the diagrams you posted, then the center of mass instead moves with constant velocity, but you can simplify things by moving to the center of mass frame so my previous statement holds.)Since when two opposite mass forces, both tied up, the opposite object with more weight, will not able to pull the other object and the system to its side, if both have the same speed? its not logic. Explain me visually how it would be arranged, that no force win. What would be the momentum of one deposit full of fluid respect to the other deposit going light?
No clue what any of what you are saying here is supposed to mean.Can you show what you say with an equation? So it will move container B to the same direction as container A has. Sorry but you cannot show that.
Asserting that others can't show something is not going to help you.
Here is some math for you:
Mc = mass of empty cart (both carts are the same)
Mf = Mass of fluid that fits in 1 cart
Vi = initial velocity of cart A.
rho = fluid density
Total initial momentum of the system:
(assume the hose is made of some relatively super lightweight material, so we can ignore that part of it is moving, and part isn't)
Ptot = (Mc+Mf)*Vi
Let us assume that it is a long flexible tube as you suggest. For some length of hose, if the direction and velocity of fluid flow going in and out of the tubing is the same then, (ignoring fiction losses) there is no net momentum change in the fluid, so we can just ignore that section. To take advantage of this, assume a 90 degree bend fixed to the top of each cart and pointing to each other. Any net forces experienced by the piping will therefore be only from those bends. (You could drop this assumption and get the same results I am about to show, but then you would have to take into account the changing shape of the hose as the carts move.)
assume the fluid flow during the transfer happens at a constant rate dM/dt and the cross sectional area of all piping including the 90 degree bends is a constant area, A.
The cart will empty in time T = Mf / (dM/dt). The velocity of the fluid in the pipe is vt = dM/dt /(rho*A).
The fluid enters the pipe bend with upwards velocity, but leaves with velocity pointed to the right. Ignoring the upwards component of the force (which wouldn't end up relevant in this setup), then in the reference frame of cart A, the fluid after the bend must be moving to the right at vt, so it is carrying momentum to the right away at a rate of vt*dM/dt. By Newton's second law, that means it had a force being continuously exerted on it by the pipe bend of F1 = vt*dM/dt pointed to the right. by Newton's third law, it exerts a force on the pipe equal in magnitude to F1, but pointed to the left.
This is a force effectively applied to cart A through the pipe bend during the transfer. the final velocity of cart A is therefore:
vi + F1*T/Mc =vi + vt*dM/dt * Mf/ (dM/dt) /Mc = vi + vt*Mf/Mc
In other words, as the mass in the first cart decreases, the velocity of the cart actually increases.
More work can be done to show that a comparable force is applied to cart B, but making it move to the right so when you add everything together, the net momentum of the system is unchanged. Also, there are details about what happens when the cart is partially full, so to actually get everything correct, you would need to account for the variable effective mass of each cart with time (Go look up a derivation of the rocket equation for details). Plus, there are some details I glossed over about the hose. Since it is attached to 2 carts moving at different velocities, it will move itself, along with all of the fluid in it, and this will be balanced out when you consider that if the velocity of the fluid relative to the tubing is the same at either end, but the ends of the tubing are moving at different speeds, then the fluid must be accelerating (speeding up or slowing down) relative to the other frames, and therefore actually applying forces, as well the changing velocities of the carts dragging the hose with them.Have you viewed the pics? I think its clear, in the transfer of fluid from container A to B.
Too bad pipes don't operate by teleporting, but by having fluid flow through them and apply forces.
Teletransporting mass literally as I show, does not do any force but moving the piston but changes the momentum of the container.
Here I got from Physycs stack exchange. Please have it clear about what we are talking.
https://physics.stackexchange.com/questions/434907/why-does-pascals-principle-apply-to-a-hydraulic-jack-but-not-to-stream-lines-in -
#46 by meberbs on 13 Mar, 2019 00:50
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I will look with detail all what you wrote, bit please understand that the fluid inside the containers will be calm, the pipe will move also in a moment at constant velocity, and as the containers gets rid of fluid, ok, the pipe will not push further the empty container to compensate, becuase it travels already at 100 m/s and it cannot give more speed with the same speed of 100 m/s, because it ends are pulled also at 100 m/s and all the pipe structure in a moment will have 100 m/s, but the container will not accelerate, because the pipes cannot acellerate over 100 m/S. I dont know if I am clear at this?
The one thing that is clear in what you just wrote is that you are completely wrong and have no clue what you are talking about. Imagine if the hose wasn't there, just a 90 degree bend. Pushing the piston up would force water up through the bend and then out the back, accelerating the cart. Attaching a hose changes nothing about these forces. other forces may act on the hose itself, or whatever is on the other end, and dragging the hose increases the mass the cart has to deal with, but, the cart most certainly can accelerate like that.And about the water flow in a closed system ruled by Pascal, the forces are the same everywhere, no flow exist in Pascal equation. The links that you showed me before dont apply to Pascal principle, they talk about flow, but flow like river, I dont know what flow will do an input piston into the fluid.
Again, there are flows happening in the system you described, so you need hydrodynamics, not hydrostatics, and your blind application of Pascal is wrong.
The two links that you sent me, talks about hydrodynamics, like constant flow, and dont apply to Pascal laws, that define hydrostatic. An interesant name, that you want to put flow in itBy the way, I need advices from experts, not people like me and honestly, very few I saw here. Experts come with proves to discredit somebody, but in this case it did not happen in any away, just moving on with topics.
You appear to be defining an expert as someone who agrees with you. You have been given multiple proofs, starting in my first post where I pointed out that your question was essentially contradictory.
I have lost track of how many ways people have tried to explain to you that the fluid must move to get from one box to the other. So many simple proofs of it:
-the boxes are moving at different velocities, so the fluid must accelerate to get from one to the other
-the fluid starts moving away from the second box, so to get to the second box, it must move towards it at some point.
-etc. (re-read this thread if you need to see some others)
If you can't even accept such simple proofs, then there is no point in discussing anything with you, and there is really no point in going through a detailed proof of basic physical facts such as the equivalence between Newton's laws and conservation of momentum.Here I got from Physycs stack exchange. Please have it clear about what we are talking.
In addition to what is stated in the answer to that question, the fluid can't be moving for Pascal's law to apply. Please stop ignoring this statement. (And the fluid is moving in your device, stop denying that.)
https://physics.stackexchange.com/questions/434907/why-does-pascals-principle-apply-to-a-hydraulic-jack-but-not-to-stream-lines-in -
#47 by esposcar on 13 Mar, 2019 01:05
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Have it clear, in a closed hydraulic system there is no flow, NO FLOW, ask and hydraulic engineer or go in the net. I found a link just in a second. NO flow. You have got it wrong.
Second, the impulse of the container up to its final velocity will move the pipe structure, so if the container loses weight, the container will not accelerate by the pipe structure, because the momentum was donated by the container itself in the acceleration and final speed. So what can happen in any case it that will slow down.
the transfer ocurrs under pascal laws, and no flow happen. Show me a good link, no flow, o. If there is flow, it will rebound back and forward inbetween the pistons, so even in that fantasy case, it would be useless flow. No flow, closed system. If you are not able to understand this then I cannot do anything for you. You dont have any idea how pascal laws operates by insisting in flows.
And so about the pic, if you have a bumper machine, and you install it in a car and you live the end of the pipe in earth, a big long pipe of 400 hundred meters and you begin to bump the water from the track. If there is somebody in earth holding the pipe, it will jump the water, back at 100 km per hour as it goes out of the pipe? come on man Lol, it will go out as if the bumper was next to you hahaha -
#48 by donaldp on 13 Mar, 2019 01:07
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Quote
Here I got from Physycs stack exchange. Please have it clear about what we are talking.
https://physics.stackexchange.com/questions/434907/why-does-pascals-principle-apply-to-a-hydraulic-jack-but-not-to-stream-lines-in
I'm not sure why I'm bothering to post this as you are clearly either not listening, ignoring or not understanding anything that is being posted. The information you have linked to states for Pascal's principle that:
"the pressure applied at one point in an enclosed fluid under equilibrium conditions is transmitted equally to all parts of the fluid"
Your system is clearly not under equilibrium conditions as you required the fluid to move from one place to another. By definition that is not in equilibrium. What the statement above says is that if you apply pressure at one point and there is no movement of the fluid in the system then the pressure is the same at all points.
In other words Pascal's principle does not apply to your system. It does not apply when fluid is moving from one place to another. For there to be movement there has to be a pressure difference. In the case of a hydraulic jack Pascal principle only applies when there is no movement of the fluid.
It would seem that you are not clear about what you are talking about. -
#49 by esposcar on 13 Mar, 2019 01:15
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Quote
Here I got from Physycs stack exchange. Please have it clear about what we are talking.
https://physics.stackexchange.com/questions/434907/why-does-pascals-principle-apply-to-a-hydraulic-jack-but-not-to-stream-lines-in
I'm not sure why I'm bothering to post this as you are clearly either not listening, ignoring or not understanding anything that is being posted. The information you have linked to states for Pascal's principle that:
"the pressure applied at one point in an enclosed fluid under equilibrium conditions is transmitted equally to all parts of the fluid"
Your system is clearly not under equilibrium conditions as you required the fluid to move from one place to another. By definition that is not in equilibrium. What the statement above says is that if you apply pressure at one point and there is no movement of the fluid in the system then the pressure is the same at all points.
In other words Pascal's principle does not apply to your system. It does not apply when fluid is moving from one place to another. For there to be movement there has to be a pressure difference. In the case of a hydraulic jack Pascal principle only applies when there is no movement of the fluid.
It would seem that you are not clear about what you are talking about.
In constant speed they are in equilibrium the liquids settled in the container in constant velocity. This I have repeated thousands of times, and you insist in no equilibrium, and I tell you why not, and you insist one and another, get it, I have explained it but no way. I repeat read about Pascal because you dont have it at all, in Pascal systems. The jack hydraulic operates in the same way, there is no flow inside, in a compressor system, there is a pipe that have liquid and move it, to make it work. Please have it clear.
http://aplusphysics.com/courses/honors/fluids/Pascal.html
dont mix up continuity of fluids with statical fluids. Remeber the pipe is always full, no new flow arrives, just flow levels go up and down in the containers, view it like that
http://aplusphysics.com/courses/honors/fluids/continuity.html
Hey you know the difference between constant velocity and acceleration? we can begin by there. -
#50 by esposcar on 13 Mar, 2019 01:30
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https://www.myodesie.com/wiki/index/returnEntry/id/2986 here explains on the milimeter how it works. And to the next one that comes again with the equilibrium, just to tell him, that sitting in an airplane is nearly like sitting at home
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#51 by Coastal Ron on 13 Mar, 2019 02:05
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Remeber the pipe is always full, no new flow arrives, just flow levels go up and down in the containers, view it like that
You seem to be contradicting yourself. In order for mass to arrive in the containers, mass must flow through the pipe. See the animation on the "continuity" website.
Also, as mass flows through the pipe there is resistance. That means there is less force at the outflow than it took to push the mass into the pipe. A net loss in energy.
And again, you don't seem to want to gain knowledge from the NASASpaceFlight members, so I'm not sure what your goal is here. No one seems to be impressed by what you are proposing, especially since it seems like some sort of perpetual motion machine, and science has proven that those don't exist.
My $0.02 -
#52 by esposcar on 13 Mar, 2019 02:21
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Remeber the pipe is always full, no new flow arrives, just flow levels go up and down in the containers, view it like that
You seem to be contradicting yourself. In order for mass to arrive in the containers, mass must flow through the pipe. See the animation on the "continuity" website.
Also, as mass flows through the pipe there is resistance. That means there is less force at the outflow than it took to push the mass into the pipe. A net loss in energy.
And again, you don't seem to want to gain knowledge from the NASASpaceFlight members, so I'm not sure what your goal is here. No one seems to be impressed by what you are proposing, especially since it seems like some sort of perpetual motion machine, and science has proven that those don't exist.
My $0.02
The members of a forum, dont make it sound like god. No specialist, have been talking with me, because they acuse me of not having maths proves, and they insist on a thing that is not true. Since when they have becomed specialist of Pascal law? come on, lets be honest.
Between two pistons that there is only fluid, if you apply force to one piston it pushes the other piston and inbetween, force will be in all place the same. What kind of flow we are talking about? in a closed hydraulic system that works through Pascal laws, have the same charachteristics as the transfer of fluid in my device does. There is no lose, the force of the piston will be translated through all the fluid in the way of pression instantanly.
Science have not proven that it dont exists, again you mistake, because why that interest by the EM Drive and all that, for something that you say that is proved, as if you say is the law Lol. Be accurate and show me wrong empirically if not, well do another activity -
#53 by esposcar on 13 Mar, 2019 02:27
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I will look with detail all what you wrote, bit please understand that the fluid inside the containers will be calm, the pipe will move also in a moment at constant velocity, and as the containers gets rid of fluid, ok, the pipe will not push further the empty container to compensate, becuase it travels already at 100 m/s and it cannot give more speed with the same speed of 100 m/s, because it ends are pulled also at 100 m/s and all the pipe structure in a moment will have 100 m/s, but the container will not accelerate, because the pipes cannot acellerate over 100 m/S. I dont know if I am clear at this?
The one thing that is clear in what you just wrote is that you are completely wrong and have no clue what you are talking about. Imagine if the hose wasn't there, just a 90 degree bend. Pushing the piston up would force water up through the bend and then out the back, accelerating the cart. Attaching a hose changes nothing about these forces. other forces may act on the hose itself, or whatever is on the other end, and dragging the hose increases the mass the cart has to deal with, but, the cart most certainly can accelerate like that.And about the water flow in a closed system ruled by Pascal, the forces are the same everywhere, no flow exist in Pascal equation. The links that you showed me before dont apply to Pascal principle, they talk about flow, but flow like river, I dont know what flow will do an input piston into the fluid.
Again, there are flows happening in the system you described, so you need hydrodynamics, not hydrostatics, and your blind application of Pascal is wrong.
The two links that you sent me, talks about hydrodynamics, like constant flow, and dont apply to Pascal laws, that define hydrostatic. An interesant name, that you want to put flow in itBy the way, I need advices from experts, not people like me and honestly, very few I saw here. Experts come with proves to discredit somebody, but in this case it did not happen in any away, just moving on with topics.
You appear to be defining an expert as someone who agrees with you. You have been given multiple proofs, starting in my first post where I pointed out that your question was essentially contradictory.
I have lost track of how many ways people have tried to explain to you that the fluid must move to get from one box to the other. So many simple proofs of it:
-the boxes are moving at different velocities, so the fluid must accelerate to get from one to the other
-the fluid starts moving away from the second box, so to get to the second box, it must move towards it at some point.
-etc. (re-read this thread if you need to see some others)
If you can't even accept such simple proofs, then there is no point in discussing anything with you, and there is really no point in going through a detailed proof of basic physical facts such as the equivalence between Newton's laws and conservation of momentum.Here I got from Physycs stack exchange. Please have it clear about what we are talking.
In addition to what is stated in the answer to that question, the fluid can't be moving for Pascal's law to apply. Please stop ignoring this statement. (And the fluid is moving in your device, stop denying that.)
https://physics.stackexchange.com/questions/434907/why-does-pascals-principle-apply-to-a-hydraulic-jack-but-not-to-stream-lines-in
There is no flow, show me a real link that shows that, because again your two other links, in any video I saw a mention of Pascal by the way. Proves, I show you usefull links to confirm what I say, do the same. Flow does not exist
So you agree that bumping fluid from a car on 100 km/h, if I hold the pipe at earth, the water will go back at 100 km/h on the direction of the car, as the water drops go out of the pipe? if you tell me yes, then sorry, but you have discovered today Pascal laws -
#54 by esposcar on 13 Mar, 2019 02:57
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For the transfer of fluids, you only need this equation to move the fluid.
P1=P2
F1/A1 = F2/A2
When you push the piston you Only need to know this. No flow in the equation to have in consideration, so no flow exists in this system, not at least continous flow. Get it clear everybody. And the fluid is stable at a constant speed, so stop arguing about something that is not true. -
#55 by esposcar on 13 Mar, 2019 03:11
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A definition of what means water at rest in a confined area under Pascal law. This let clear that then Pascal law applies and a lot of people here will have to reconsider their positions.
https://en.wikiversity.org/wiki/Hydrostatics:_Fluids_at_Rest
A fluid in rest, is a fluid that is not moving, so this applies to my device. The fluid dont goes out anywhere, its contained, so its at rest, as what it means. Bring me if not a link that contradict what I said, and then reconsider your error. Who was right
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#56 by hop on 13 Mar, 2019 03:13
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There is no flow, show me a real link that shows that, because again your two other links, in any video I saw a mention of Pascal by the way. Proves, I show you usefull links to confirm what I say, do the same. Flow does not exist
You claim in your very first post that fluid transfers from A to B. Yet you also claim "flow does not exist" between them, and in the original diagram, you claim the fluid is stationary.
In other words, you are claiming the volume of fluid inside A and B changes without fluid moving moving into or out of them. This is just utter nonsense. There's no need for further analysis or consideration, you can send your device directly to the museum.
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#57 by meberbs on 13 Mar, 2019 05:00
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There is no flow, show me a real link that shows that, because again your two other links, in any video I saw a mention of Pascal by the way. Proves, I show you usefull links to confirm what I say, do the same. Flow does not exist
How about you actually read the links that have already been posted in this thread, here is one you posted:https://www.myodesie.com/wiki/index/returnEntry/id/2986 here explains on the milimeter how it works. And to the next one that comes again with the equilibrium, just to tell him, that sitting in an airplane is nearly like sitting at home
It says:QuoteIn the operation of hydraulic systems, there must be a flow of fluid. The amount of flow will vary from system to system. To understand power hydraulic systems in action, it is necessary to become familiar with the elementary characteristics of fluids in motion.
Emphasis added.
So as everyone has been trying to tell you, you are wrong.So you agree that bumping fluid from a car on 100 km/h, if I hold the pipe at earth, the water will go back at 100 km/h on the direction of the car, as the water drops go out of the pipe? if you tell me yes, then sorry, but you have discovered today Pascal laws
I honestly have no clue what this setup is supposed to look like. by "bumping" I suppose you mean pumping. I have no clue what "hold the pipe at earth" means, but the answer is that Pascal's law does not apply because you are pumping fluid, and the exit velocity depends on the pump.For the transfer of fluids, you only need this equation to move the fluid.
P1=P2
F1/A1 = F2/A2
When you push the piston you Only need to know this. No flow in the equation to have in consideration, so no flow exists in this system, not at least continous flow. Get it clear everybody. And the fluid is stable at a constant speed, so stop arguing about something that is not true.
Completely gibberish. You have been given at least half a dozen definitions of Pascal's law. Every single one states that Pascal's law does not apply if there is any motion, flow, or transfer of fluid. It only applies to hydraulic systems that are stationary. You cannot conclude that there is no flow just because you are trying to apply Pascal'slaw in a situation where it does not apply. That amounts to simply assuming the thing you want to prove, which fails because your assumption is wrong.A definition of what means water at rest in a confined area under Pascal law. This let clear that then Pascal law applies and a lot of people here will have to reconsider their positions.
There is no definition of what "at rest" means on that page, it just again says that Pascal's law only applies when the fluid is at rest which it is not true in your device.
https://en.wikiversity.org/wiki/Hydrostatics:_Fluids_at_Rest
A fluid in rest, is a fluid that is not moving, so this applies to my device. The fluid dont goes out anywhere, its contained, so its at rest, as what it means. Bring me if not a link that contradict what I said, and then reconsider your error. Who was rightNo specialist, have been talking with me, because they acuse me of not having maths proves, and they insist on a thing that is not true. Since when they have becomed specialist of Pascal law? come on, lets be honest.
You are new here, so here is a warning: insults are not an accepted form of communication on this site. You claim that no specialists have been communicating with you, despite one poster having listed a bunch of relevant engineering courses they have taken. (And I can assure you from past post history of many of the posters on this thread, quite a few of them are far more of an expert in this field than you are.) On the other hand, you have had to do research just to understand Newton's laws which are taught in any introductory physics class.
The only one insisting on things that are not true here is you. The fact that every single poster in this thread besides you is in agreement should be a clue for you. The real question is where did you decide that you know the what is "true" despite being contradicted repeatedly by the sources you yourself have provided. -
#58 by ChrisWilson68 on 13 Mar, 2019 07:07
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esposcar: I believe you are entirely wrong. I believe that virtually every word you have said is wrong. So does every single other poster who has responded here.
Your responses haven't convinced anyone, they just make us think you continue to be out-of-touch with reality. And nothing anyone else has said has convinced you of anything.
Is there any point in continuing? I don't see any point.
Lets just agree to disagree and all move on.
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#59 by aceshigh on 13 Mar, 2019 23:58
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The second half of your question can be reworded and has the exact same meaning as the following:
Too many new posters who present ideas like yours react poorly to being told "no" for me to want to spend much time on a device specific answer, at least before having some evidence that you are interested in listening. It would do you more good to work it out yourself. (Hint: at some point to get from A to B, the fluid must have momentum that is at least partially pointed to the right.)
You nailed it.
And just after that, you contradicted yourself, because you have posted a zillion times, wasting your precious time, with someone CLEARLY not interested in listening.
Can this thread please be closed before Meberbs loses more of his time (because he feels compelled to correct Espocar's nonsense) for no good reason at all?
