Mechanical Advantage in absence of resistance or load

[UMath1]

I want to understand what happens to the energy inputted into a simple machine if there is no resistance or load. For example, let's say you have a U-shaped tube of water with greater surface area on the left side. If a piston applies pressure to the left side and there is no piston on the right side, what does the water apply pressure to on the left side? Where does the energy go?

 

[NascentOxygen]

With no load, it will be to the water that the input energy is transferred; the water will be accelerated outwards, exiting the manometer.

 

[UMath1]

But what if the water level simply rises on the other side, like shown in this video at 8:09 () ?

What about in the case of a simple lever, what happens if I apply force to the short end and raise the long end of the lever, where is the energy transferred? To the lever? If so, how come when a block is placed on the long end, we say that all the energy is transferred to the block?

 

[Ronie Bayron]

I want to understand what happens to the energy inputted into a simple machine if there is no resistance or load. For example, let's say you have a U-shaped tube of water with greater surface area on the left side. If a piston applies pressure to the left side and there is no piston on the right side, what does the water apply pressure to on the left side? Where does the energy go?

The energy input (load) is exerted for the water level to rise at certain height at the otherside. If load is applied at the side where the height is elevated, then simultaneously the input load is increased by that much. There are no energy loss in the system. Energy conservation is still there (w/with out restriction)

 

[UMath1]

What do you mean by the input load being increased? If the same volume of water is raised on the other side, there is no change in the potential energy of the water, correct?

And what about the lever case?

 

[NascentOxygen]

In a simple lever, such as a seesaw, if you push one end down, with no load on the other end all your input energy gets transferred to the lever bar itself, causing it to spin up and over and clock you if there is nothing to stop it. You've seen comedy skits of someone stepping on the upright tines of a garden rake?

 

[sophiecentaur]

I want to understand what happens to the energy inputted into a simple machine if there is no resistance or load. For example, let's say you have a U-shaped tube of water with greater surface area on the left side. If a piston applies pressure to the left side and there is no piston on the right side, what does the water apply pressure to on the left side? Where does the energy go?

If there is no Load then the MA is zero (Load / Effort). There is another quantity - Ideal Mechanical Advantage or Velocity Ratio, which is the ratio of distance moved by the effort / distance moved by the load. That ignores resistance and dead weight.
That simple hydraulic system will consume finite energy, just sloshing the water about, of course and, with no load, the efficiency is zero. Efficiency = MA/VR

 

[UMath1]

Yeah..but then it stops spinning once the side you apply force on touches the ground. At that point, where has the energy been transferred? It can't be friction because had there been a load, you would say the energy would be transferred to the load.

 

[Ronie Bayron]

What do you mean by the input load being increased? If the same volume of water is raised on the other side, there is no change in the potential energy of the water, correct?

Wrong concept, potential energy is = mgh, even that the other side elevation has minimal change because of bigger cross section, see that mass-m is increased hence m=ρV.

 

[NascentOxygen]

Circuses have seesaws that don't hit on the ground. If the end of a seesaw hits the ground, the hole it makes accounts for the energy loss.

 

[UMath1]

But the same mass water has lost elevation on the other side, therefore there is no net change.

 

[Ronie Bayron]

But the same mass water has lost elevation on the other side, therefore there is no net change.

I suggest you need more study on hydrostatic pressure.

 

[UMath1]

But the same mass water has lost elevation on the other side, therefore there is no net change.

Circuses have seesaws that don't hit on the ground. If the end of a seesaw hits the ground, the hole it makes accounts for the energy loss.

What about if no hole is made? When there is a load, it still hits the ground, yet the energy is transferred to the load not the hole.

 

[sophiecentaur]

But the same mass water has lost elevation on the other side, therefore there is no net change.

No change in what? Work out the mgh on each side before you make assertions like that.

This is a pretty futile discussion. The Energy in any system like this can always be accounted for, one way or another. Things get hot and deformed. That always accounts for any suggested deficit one can calculate.

 

[UMath1]

In potential energy. If the volume is the same, then Mg is the same. The only different quantity is height. However, on the side that loses elevation, some water molecules are at higher elevations than other. So the net change in potential energy should be 0...or am I missing something?

 

[sophiecentaur]

or am I missing something?

You are missing the am[litudes of the changes in mgh. One goes down by less than the other goes up but the mass transferred is the same. Always do the sums when possible, before coming to intuitive conclusions. Those numbers are what counts.

 

[UMath1]

Ok..so the potential energy decreases because the water moves less height upwards on the side with no load?

So where does the energy go?

 

[Dale]

If you have a mechanical device with no load and you input energy, then you will raise the internal energy of the device, usually to internal KE and PE, vibrations, and heat. The vibrations and heat can then dissipate the energy to the environment, sometimes after breaking the device.

 

[sophiecentaur]

The minimum Potential Energy is (of course) when the two water columns are leverl. To change that situation requires work to be done. Work out (in proper detail) what happens to the centre of mass of the whole mass of the water. Either do it with algebra (preferable) or for a specific example (which isn't a proof - its just an indication).
PS - did you actually work it out or are you just repeating what I wrote?

 

[UMath1]

I worked it out. I found that M, and g, are same on both sides and that -h, height water displaced downwards, is greater than +h, height water displaced upwards. Isn't that sufficient to say potential energy is decreased?

If you have a mechanical device with no load and you input energy, then you will raise the internal energy of the device, usually to internal KE, vibrations, and heat. The vibrations and heat can then dissipate the energy to the environment, sometimes after breaking the device.

So essentially the water increases in temperature? What would happen in the lever case if no hole is created?

 

[Dale]

I worked it out. I found that M, and g, are same on both sides and that -h, height water displaced downwards, is greater than +h, height water displaced upwards. Isn't that sufficient to say potential energy is decreased?

Please show your work in detail. You made a mistake, the PE increases.

 

[russ_watters]

I want to understand what happens to the energy inputted into a simple machine if there is no resistance or load. For example, let's say you have a U-shaped tube of water with greater surface area on the left side. If a piston applies pressure to the left side and there is no piston on the right side, what does the water apply pressure to on the left side? Where does the energy go?

You already asked this question in a slightly different form (a massless lever with nothing on it) and received the answer: if there is no resistance/load, there is no force and therefore no energy input (however, in the manometer example, there is a force and energy input...so you failed to construct an example that meets your requirements).

Oh, wait, there it is again:

What about in the case of a simple lever, what happens if I apply force to the short end and raise the long end of the lever, where is the energy transferred? To the lever? If so, how come when a block is placed on the long end, we say that all the energy is transferred to the block?

You really need to do a better job remembering and applying what you've learned from one week to the next. These concepts you are stumbling over are not difficult, but you seem to be getting stuck with a wrong idea in your head that you are unwilling to let go of.

Part of it, as Dale says, is your lack of showing your work that is preventing you from figuring out the exact answers on your own. For example, in your lever example you are being too vague to answer it properly: you haven't said which side you are pushing down on. In your previous version of the question you asked about pushing down on the long side, which, of course, you can't do if there is nothing sitting on the short side (it just falls under its own weight).

 

[UMath1]

Heres my calculation for the manometer

 

[UMath1]

Heres a diagram for the lever.

 

[russ_watters]

Heres a diagram for the lever.View attachment 91468

What is the mass of the lever? This will determine how much force can be/has to be applied to move it.

 

[UMath1]

Lets say it has a nonzero mass, M.

 

[russ_watters]

Lets say it has a nonzero mass, M.

Great! So, can you use geometry to calculate the force required to hold the lever stationary and level or move it?

Hints:
1. The force to hold it stationary or move it is the same if you move it slowly.
2. The force doesn't change if we keep the rotation angle small.

By the way, just to make sure we're clear: you do recognize now that the starting premise of the thread, "no resistance" is false, right? Every scenario you've constructed includes resistance.

 

[sophiecentaur]

I worked it out. I found that M, and g, are same on both sides and that -h, height water displaced downwards, is greater than +h, height water displaced upwards. Isn't that sufficient to say potential energy is decreased?

You have shown that net work is needed to disturb the levels from the same height. That's good enough because the system will tend to return. Aamof, the net GPE will have Increased - so the system will 'fall' to a lower energy state as the levels go back to where they were.
In a machine with no load then any energy / work put in must end up as heat, distortion etc or Kinetic energy.

 

[UMath1]

I am not sure if I did this right, but if I assume that all of the energy becomes potential energy..then this would be the force. But I don't know whether that is correct, because if there had been a load, the displacement would still be the same but the energy of the lever would not change.

You have shown that net work is needed to disturb the levels from the same height. That's good enough because the system will tend to return. Aamof, the net GPE will have Increased - so the system will 'fall' to a lower energy state as the levels go back to where they were.
In a machine with no load then any energy / work put in must end up as heat, distortion etc or Kinetic energy.

Can you explain how it will have increased based on the manometer diagram?

 

[sophiecentaur]

Can you explain how it will have increased based on the manometer diagram?

Work has been done to alter the levels from the equilibrium heights. If not, the system would not be stable and the water would go shooting out of the U, one way or another. That's the philosophy behind it. The reason that you get a negative will be because of the sign you have chosen for the directions used. The calculation should show that the work put in is greater than zero (making the PE of the system less negative).

 

[russ_watters]

I am not sure if I did this right, but if I assume that all of the energy becomes potential energy..then this would be the force. But I don't know whether that is correct, because if there had been a load, the displacement would still be the same but the energy of the lever would not change.
View attachment 91474

Wow, there are several ways to do this and I would have avoided the integral and trig, but it looks correct (though I'm not completely clear on all your constants). If you sum the moments about the fulcrum it just becomes simple ratios.

In any case, do you see now that because the lever has mass, there is a force and a distance on each side and therefore contrary to your title premise, it is not a "no load" situation?

[edit]
I'm not sure you answered your own question though. You should probably add the forces of gravity to the diagram (in terms of F_in and M) to show that you've accounted for everything and understand where the force/energy in is "going".

 

[UMath1]

So we would then say that the lever is a load and that all of the energy is transferred to the lever(which is the load)? But what about the problems in which a separate load is placed on the long side, how can we completely neglect the energy transfer to the lever? Why do these problems assume that the entirety of the energy is transferred to the weight? In both cases, when there is a weight and when there is no weight, the change in position of the lever is exactly the same..so in both cases the change in energy of the lever should be the same.

 

[UMath1]

Work has been done to alter the levels from the equilibrium heights. If not, the system would not be stable and the water would go shooting out of the U, one way or another. That's the philosophy behind it. The reason that you get a negative will be because of the sign you have chosen for the directions used. The calculation should show that the work put in is greater than zero (making the PE of the system less negative).

Intuitively that makes sense, but I don't quite understand it with the diagram. Doesn't the potential energy decrease on the left side and increase on the right side?

 

[russ_watters]

So we would then say that the lever is a load and that all of the energy is transferred to the lever(which is the load)?

Yes. Indeed, if you just wanted to do this in terms of energy instead of force, you could ignore the force and just calculate the change in mgh on each side, like you did for the water. That would make the math simpler!

But what about the problems in which a separate load is placed on the long side, how can we completely neglect the energy transfer to the lever? Why do these problems assume that the entirety of the energy is transferred to the weight? In both cases, when there is a weight and when there is no weight, the change in position of the lever is exactly the same..so in both cases the change in energy of the lever should be the same.

When people are learning these things, it is often best to simplify the problems. So the lever is usually assumed to be massless.

 

[Dale]

View attachment 91467 Heres my calculation for the manometer

You are thinking about this wrong. Instead of just doing a calculation involving delta h, try calculating the PE before and the PE after. Only calculate the change in PE afterwards.

I can show this later, but if you set up the problem right then you will immediately see that the change in PE is positive regardless of the differences in A.

Concentrate on the height of the volume, not the surface

 

[UMath1]

Ok that makes sense. Can you explain the flaw in using delta h though? I intuitively understand its wrong, but I don't know why.

As for the OP, my understanding now is that the simple machine is considered to be a load, and that the inputted energy becomes potential energy of the simple machine. If a sepearte load is present, some of the inputted energy still goes to the machine, but it is neglected?

 

[russ_watters]

As for the OP, my understanding now is that the simple machine is considered to be a load, and that the inputted energy becomes potential energy of the simple machine. If a sepearte load is present, some of the inputted energy still goes to the machine, but it is neglected?

Sometimes yes, sometimes no. It depends on the requirements of the specific problem.

 

[UMath1]

The sometimes yes and no is for the neglecting of the energy of the machine, correct?

 

[Dale]

Ok that makes sense. Can you explain the flaw in using delta h though? I intuitively understand its wrong, but I don't know why.

The delta H is the surface. The surface doesn't have any mass. You need to look at what happens to the mass.

Draw a before and after drawing and look at the center of gravity of the mass of fluid in each case.

 

[russ_watters]

The sometimes yes and no is for the neglecting of the energy of the machine, correct?

Yes. It is typically when the energy of the load is much greater than the energy of the machine that you can safely neglect the energy of the machine.

 

[Dale]

Can you explain the flaw in using delta h though? I intuitively understand its wrong, but I don't know why.

OK, look at the attached drawing. This shows the manometer in the initial condition at equilibrium and in the final condition after the piston is pushed down. The volume is divided up into 3 segments, the volume in the box on the left, the volume in the box on the right, and the volume outside either box.

The volume outside either box begins filled with fluid and ends filled with fluid, so its PE does not change.

The volume in the box on the left begins with a PE of $-mgh_1/2$, and ends with PE of $0$, so the change in PE for the volume in the left is positive $mgh_1/2$.

The volume in the box on the right begins with a PE of 0, and ends with a PE of $mgh_2/2$, so the change in PE for that volume is positive $mgh_2/2$.

So the total change in PE is $mg(h_1+h_2)/2$, which is always positive (regardless of the differences in the area). Essentially, you are moving fluid uphill from the volume on the left to the volume on the right.