Mechanical Advantage in absence of resistance or load

[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.