Torque on a square screw full with water

[Hassan2]

Yes that's the formula for the torque due to surface force at that point. For the total torque, we should calculate the force due to all surface forces, includeing the cut srface. Did you have a reason to ignore the forces on the cut surface? Perhaps you think these forces are parallel with r and cause no torque. But if the were paralled with r everywhere, they wouldn't cancel the hydraulic force on the turning thread surfaces and the surface would move in the force direction.

 

[Gh778]

What is the cut surface ? See the message #58 for explain please. The surfaces with water is only up and down surfaces not sides.

 

[Gh778]

Only water where you see blue color. The all system is in air.

 

[Hassan2]

Ok, If you assume the thread to be like what we see in message #58, it has a degree of freedom in directions other than the z- axis. But of course you don:t mean that's all the model. When we consider the thread to be fixed from sides, we are assuming some forces from sides to cancell the lateral focres.

If you assume that the thread is that square only, then the analysis would be different. For a free body, we can define torque?

 

[Gh778]

For a free body, we can define torque?

Because it move up/down and turn ? I don't know how to study this case. For see if an object move I done the sum of all forces. For see if something turn I sum all the moments. If something is force to turn but must move up/down for turn (no force contrary the move up/down) why the thread can't turn ? We know there is no up/down force but I think there is a torque (need calculations for measure it exactly and I try...). Sure like you said the square thread move perpendiculary to the force and this is original, enough for study more. Imagine you're the square thread, you have no up/down force and you have something which want to turn you, what you do ?

 

[Hassan2]

Right, but axis of rotation woud depends on the forces and its won't be the z-axis. The calculation would be different too. Morevover, as the as force cause a small motion perpendicular to the surface, the volumes of water ( or pressurized gas as I assume)would change and the forces become balanced again.

 

[Gh778]

Right, but axis of rotation woud depends on the forces and its won't be the z-axis. The calculation would be different too.

I don't understand could you explain more please ? For me the axis of rotation depends of the mecanical freedom. The square thread can turn around Z axis and can move up/down.

the volumes of water ( or pressurized gas as I assume)would change and the forces become balanced again.

The film of water is pressurized due to the speed ? When the system start maybe, but if we launched the square thread at w rd/s after the water is always at same pressure I think.

 

[Hassan2]

1. Sorry I was wrong. we can calculate torque around any number of arbirary axes.

2. Forget about the gas. Lest assume water only. The forces normal to the surface cause a torque which may not be around the Z axis. Since the thread is locked and can't move normal to the surface, there must be an opposing torque from the reaction of the fixed thread. Since the fixed threads are not directly in contact with the square thread, they apply their forcec through water by increasing the pressure on one side and decreasing on the other side. So when calculating the torque around Z-axis, you should consider these changens in pressure too .

 

[Gh778]

The forces normal to the surface cause a torque which may not be around the Z axis.

Think with a circular thread, apply pressure at up surface and 0 pressure at down surface this will apply a torque of the thread (sure this need energy). Maybe all the torque is not around Z axis but like the thread is forced to move between 2 fixed threads, all others torque have no effect, but this can't delete the part of torque around Z axis. For me the torque around Z axis is like dM=r*P*de*dl*sin(θ), other torque is canceled by axis of the square thread (the drawings don't show the square thread axis but the square thread can have it). The power lossed (or give) by a torque depend of the axis, if we don't turn around the good axis, the torque don't lossed (or give) energy I think.

Since the fixed threads are not directly in contact with the square thread, they apply their forcec through water by increasing the pressure on one side and decreasing on the other side.

I don't understand, for you the fixed thread increase the pressure of water, it's not gravity which create pressure ? The film of water is always fixed to the square thread and never move alone, the volume of the film (the volume where the water is) is constant always. How the pressure can be increase when the volume is constant ? Edit: I understand now what you said ! It's more easy to explain with the high of water, if the thickness of the film change, this change the high of water and this change the pressure, ok. But, with a mecanical system it's possible to adjust the thickness due to no up/down forces I think.

Edit: Don't forget there is no up/down forces, it's important: the thread has a torque, and when the thread turn it need to move up/down, but we can adjust the vertical mouvement freely because we have no up/down force. It's the difference of radius which give torque but vertical movement don't care about radius.

 

[Hassan2]

Sorry for the confusion. Seems I have a different understating of the problem. But it is not the main problem.
1. From message #63 on I thought the thread has no axis.
2. I thought the water films are closed by gasket from bottom and top too.

But these are not the main problem.

With or without axis, if the thread is forced to move with fixed distance from the fixed thread, the force any point is normal and the displacement in normal direction is zero. This means the whole force is canceled somehow. When the whole force is canceled, what part of the force remains to cause another torque? This may be enough to conclude that net torque is zero, but perhaps you want to know the problem in your calculation.

When you calculate the net torque, you should consider other forces too. I mean the forces which keep the thread in the same distance from the fixed thread. You would see that those forces have some components to cancel the torque around Z axis. It's quite simple now.

 

[Gh778]

Sorry for the confusion.

Thanks for your help :)

1. From message #63 on I thought the thread has no axis.

it can, I think this don't change the result. But for simplify the problem consider there is an axis like that all forces which are in the axis is canceled by axis.

2. I thought the water films are closed by gasket from bottom and top too.

Yes all is closed, but the film is always at the same place.

but perhaps you want to know the problem in your calculation.

Yes, I would like to calculate this ! I don't know how, I need help for this.

With or without axis, if the thread is forced to move with fixed distance from the fixed thread, the force any point is normal and the displacement in normal direction is zero.

For me, there is a torque and others forces in the axis of the square thread like you said. The pressure apply forces in 3 directions. Z axis which I see a torque. X and Y, but these axis can't turn due to the Z axis, could you explain how you see the torque is canceled by another forces ?regards

 

[Hassan2]

Seems here is your problem: The axis of the thread is not fixed by a bearing or something. It can't cancel any force or torque. In fact the torque is canceled by the fixed thread through the water films. The reaction forces from the fixed thread is normal to the fixed there ( and also to the square thread) , it has x and y components too.

 

[Gh778]

We can put the square thread like the drawing show. There is an axis but the thread can move free up/down. The sum up/down forces is 0. The X and Y axis can't turn.

 

[Hassan2]

But this has no effect. Since the water films are not compressible, they keep the axis just in the center of the bearing. No radial force then.

 

[Gh778]

Z axis is fixed with square thread. The square thread can only move up/down and turn around Z axis I think. The fixed threads are not fixed with Z axis. Where I wrote "fixed" in the drawing, supports of axis are fixed to fixed referential.

Since the water films are not compressible,

maybe it's the source of misunderstood. The film of water is not compressible, we have only gravity which create pressure.

 

[Hassan2]

Yes I understand you but again its the fixed thread which cancel the torque not the axis of the thread, because the axis of the tread is not pushing against the bearing.

 

[Gh778]

Since the water films are not compressible,

maybe it's the source of conflict. The film of water is not compressible, we have only gravity which create pressure.

 

[Hassan2]

That's why I prefer to fill the space between the threads with pressured gas.

 

[Gh778]

Ah, ok, but with gas, like I said before, I agree: there is no torque. For me, the torque appear with the change of pressure due to the gravity. I don't pressure the film of water, there is only pressure from gravity. And like you see with the last drawing, the axis of the square thread prevent the square thread to move X and Y axis and prevent to move around X and Y axis. The only liberty is move up/down and rotate around Z axis. Like that the square thread never move on X and Y, the film of water is always with gravity pressure only.

For me, now I see only one solution: the torque is the same everywhere in the thread but you said my calculation of the torque is good.

 

[Hassan2]

Another thing is that, for closed gasket, the pressure is not the gravity only. The pressures are determined by equilibrium equations. I think you consider the pressures as if their top was open.

 

[Gh778]

Another thing is that, for closed gasket, the pressure is not the gravity only.

But if you pressurized water, the pressure is added at each point and you add a fixed constant to the phase angle torque. The phase angle exist always I think.

But I give a drawing of the square thread with open gasket at top if you think it's important. The water can't move down.

 

[Hassan2]

You can even leave the lower gap filled with air. Only the upper film is filled with water. Yo can even fill it to a certain level. Again the axis cancels the torque.
Added: Sorry, in this case, the thread sits on the fixed thread.

 

[Gh778]

You can even leave the lower gap filled with air. Only the upper film is filled with water.

I don't understand can you explain ? Up and down gaps are filled with water not air.

Added: Sorry, in this case, the thread sits on the fixed thread

Not at all, there is no up/down force, so a mechanical system can easily adjust the altitude of the square thread. The mechanical system support the weight of the system (weight move up/down => energy=0). Again, films of water (up AND down) are always the same, the volume is constant. Water is under gravity pressure only.

Are you sure about the formula : dM=r*P*de*dl*sin(θ) ?

 

[Hassan2]

dM=dF \times r

It is a vector, when you integrate, you can't threat is like an scalar. the magnitude of this vector is

|dM|=PdA r sinθ , where θ is the angle between the force and r , not the angle between the r and Z axis. Due to the curvature of the thread, the normal force is not along the Z axis.

dA: an infinitesimal area around the point.

 

[Gh778]

So, for you the formula is good ?

 

[Hassan2]

Seems fine. In this problem, θ seems to be constant at any phase angle.

 

[Gh778]

θ change with the radius, it is the slope of the thread. The pitch is the same at every radius but the length is bigger when the radius is increasing. With pitch = 1, when we pass from r=1 to r=3 we pass from sin(θ)=0.707 to sin(θ)= 0.316. The torque is not the same at each point of the thread, bigger if the radius is bigger.

 

[Hassan2]

I can't imagine the slope of the thread in radial direction correctly. What are the slopes at r=0 and r=infinity?

The torque is bigger mainly due to increase in r itself.

 

[Gh778]

r->0, the slope -> 90°
r->infinity, the slope ->0°
See last message for the drawing, the slopes are visible on the circular thread.

 

[Hassan2]

I don't get it. Normal to surface a any point makes an angle with Z axis. does the angle depends on r?