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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?Lets say it has a nonzero mass, M.
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.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?
Can you explain how it will have increased based on the manometer diagram?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.
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).Can you explain how it will have increased based on the manometer diagram?
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.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
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?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).
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!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)?
When people are learning these things, it is often best to simplify the problems. So the lever is usually assumed to be massless.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.
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.View attachment 91467 Heres my calculation for the manometer
Sometimes yes, sometimes no. It depends on the requirements of the specific problem.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?
The delta H is the surface. The surface doesn't have any mass. You need to look at what happens to the mass.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.
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.The sometimes yes and no is for the neglecting of the energy of the machine, correct?
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.Can you explain the flaw in using delta h though? I intuitively understand its wrong, but I don't know why.