D'Alembert's principle and the work done by constraint forces

[polytheneman]

From what I understand, constraint forces do no work because they are perpendicular to the allowed virtual displacements of the system. However, if you consider an unbalanced Atwood machine, in which both masses are accelerating in opposite directions, you'll find that the tension force of the wire (a constraint force), which pulls the lighter mass up, is parallel to the displacement, which means it does work (right?).

Now, I understand that the same is true for the other side: the tension force on the heavier mass is parallel to the displacement, but in the opposite direction, so that if you add the work done by the tension force on the heavier side to the work done by the tension force on the lighter side you get zero.

So my question is: would it be correct for me to say that individual constraint forces can do work, but it's the sum of the work done by all the constraint forces which is always equal to zero? If this is true, it's a bit different from the notion I had before, which was that individual constraint forces never do work because they are always perpendicular to the displacement.

 

[Dale]

So, for an Attwood machine analyzed in this way you would typically have generalized coordinates like $h_1$ and $h_2$ for the heights of mass 1 and mass 2 respectively. The constraint would then be $h_1+h_2=C$ for some constant $C$. In the $(h_1,h_2)$ plane the ordinary displacements are along the $(1,-1)$ direction, as one mass goes up the other goes down. The virtual displacements are along the $(1,1)$ direction, both masses go in the same direction. So the virtual displacements are indeed perpendicular to the actual displacements, in the generalized coordinate system.

 

[polytheneman]

So, for an Attwood machine analyzed in this way you would typically have generalized coordinates like $h_1$ and $h_2$ for the heights of mass 1 and mass 2 respectively. The constraint would then be $h_1+h_2=C$ for some constant $C$. In the $(h_1,h_2)$ plane the ordinary displacements are along the $(1,-1)$ direction, as one mass goes up the other goes down. The virtual displacements are along the $(1,1)$ direction, both masses go in the same direction. So the virtual displacements are indeed perpendicular to the actual displacements, in the generalized coordinate system.

Thank you for replying! I apologize, but I had a little trouble visualizing what you meant (I'm self-learning), particularly the part about both masses moving in the same direction. Would you say that from what you have stated here one can conclude that the tension forces of the wire can be considered to have done non-zero work on both masses, but it is the sum of these works that amounts to zero? Or does you analysis lead to the conclusion that the tension forces of the wire really do no work? Thanks again.

 

[polytheneman]

So, for an Attwood machine analyzed in this way you would typically have generalized coordinates like $h_1$ and $h_2$ for the heights of mass 1 and mass 2 respectively. The constraint would then be $h_1+h_2=C$ for some constant $C$. In the $(h_1,h_2)$ plane the ordinary displacements are along the $(1,-1)$ direction, as one mass goes up the other goes down. The virtual displacements are along the $(1,1)$ direction, both masses go in the same direction. So the virtual displacements are indeed perpendicular to the actual displacements, in the generalized coordinate system.

The author of the excellent book where I found this problem (The Lazy Universe) explains in another part of the book:

A surprisingly tricky example is the case of a sliding block which is pushed across a table-top by a force, say, pushed by your finger (we ignore friction). The displacement of the block is anywhere on the surface whereas the reaction-force acts at right-angles to this surface preventing the block from burrowing down into the table. So far, this makes sense. But, hang on, there is also a reaction against your finger, from the block, and this reaction is in line with the block’s displacement. The trick is to appreciate that the block’s displacement due to the finger-push is an actual, not a virtual, displacement. We can hypothetically freeze the block (switch to a different reference frame) and get rid of the distraction of its actual motion. Then we realize that the finger can’t depress the block as if it were so much sponge-cake, as there is a reaction-force of the block against the finger. However, the finger is still allowed, infinitesimally, virtually, to move within the back face of the block, at right-angles to this reaction-force. This is a general result: for any virtual displacement, being ‘harmonious’ is the same thing as being in a direction perpendicular to the reaction forces.

I feel this is the answer to my question, but I can't say I understand how this translates to the Atwood machine problem. This is my guess:

The important thing is not to confuse virtual displacements with actual displacements. Sometimes, the actual physical displacements cannot be chosen as virtual displacements. Virtual displacements occur without the passage of time. If we think of the example above, while time is frozen the finger cannot actually move in the direction the block moves when time is flowing. For it do so, it would have to compress the block, but the block is inelastic. So that's not an allowed virtual displacement. But with things frozen in time, the finger is allowed to move within the face of the block (like up and down across the face), and this motion is at right-angles to the reaction (constraint) force from the block on the finger.

The real constraint in the Atwood machine is that the string has constant length. This means that, like the block in the previous example, the string can't be compressed. Therefore, if we freeze time, we'll find that we can't take the virtual displacement of the blocks to be the same as the actual displacement, i.e, "up and down", i.e, in the direction of the strings. For them to do so with time frozen they would have to compress the strings, and the strings can't be compressed. So we are once again "getting distracted by the actual motions". This is another case where we can't take the actual displacements as the virtual displacements. The virtual displacements allowed for the blocks are actually at right angles to the string, not up and down, but sideways!

Am I onto something?

 

[polytheneman]

On the other hand, this fella is arguing that the whole problem arises from the unnecessary assumption that virtual displacements have to be perpendicular to constraint forces: arxiv.org/pdf/physics/0011004.pdf

 

[Dale]

I had a little trouble visualizing what you meant (I'm self-learning), particularly the part about both masses moving in the same direction

So in the Atwood machine there are two masses which go up and down. The constraint equation requires $h_1+h_2=C$. This means that the allowed motion is where $h_1$ goes up and $h_2$ goes down, or vice versa. The type of motion that is not allowed by the constraint is where both $h_1$ and $h_2$ go down at the same time, since that motion would stretch the string. That is what I meant by moving in the same direction, both masses moving up would shrink the string and both masses moving down would stretch the string. The constraint prevents that type of motion.

Would you say that from what you have stated here one can conclude that the tension forces of the wire can be considered to have done non-zero work on both masses, but it is the sum of these works that amounts to zero? Or does you analysis lead to the conclusion that the tension forces of the wire really do no work?

So, in the Lagrangian formulation there really aren’t two separate masses. There is just one system and the masses are part of that overall system. Also, the coordinates are generalized coordinates and the forces are generalized forces. So you have to be a little more abstract in your thinking.

The system is both masses together and the tension is the generalized constraint force on both masses together, not the individual string tensions on each mass. So the generalized tension does no work on the whole system. The allowed displacements are the up-down displacements, and the constraint prevents up-up displacements. The permitted up-down displacements are orthogonal to the forbidden up-up displacements in a generalized sense.

Am I onto something?

Maybe, but I really didn’t understand that. It didn’t seem like the finger going sideways was consistent with the coordinates, so I didn’t follow the reasoning

 

[polytheneman]

So in the Atwood machine there are two masses which go up and down. The constraint equation requires $h_1+h_2=C$. This means that the allowed motion is where $h_1$ goes up and $h_2$ goes down, or vice versa. The type of motion that is not allowed by the constraint is where both $h_1$ and $h_2$ go down at the same time, since that motion would stretch the string. That is what I meant by moving in the same direction, both masses moving up would shrink the string and both masses moving down would stretch the string. The constraint prevents that type of motion.

So, in the Lagrangian formulation there really aren’t two separate masses. There is just one system and the masses are part of that overall system. Also, the coordinates are generalized coordinates and the forces are generalized forces. So you have to be a little more abstract in your thinking.

The system is both masses together and the tension is the generalized constraint force on both masses together, not the individual string tensions on each mass. So the generalized tension does no work on the whole system. The allowed displacements are the up-down displacements, and the constraint prevents up-up displacements. The permitted up-down displacements are orthogonal to the forbidden up-up displacements in a generalized sense.Maybe, but I really didn’t understand that. It didn’t seem like the finger going sideways was consistent with the coordinates, so I didn’t follow the reasoning

Thank you for your help! I believe I understand this subject better now.