Energy conservation paradox for constrained mass

[euquila]

As sophiecentaur and BruceW pointed out, there is no energy going in or out of this system and therefore no work is done when you move the rail/rod. If the rod had mass, then the work done would be equivalent to the change in strain energy in the rod as you moved the rail.

 

[AlephZero]

But tangential to the rail there are no constraints. In order to neutralize the gravitational force $m*g*sin(\theta)$ tangential to the rail, it would take either friction or some obstacle (i,e, a surface component transverse to the rail).

To get some sort of a solution to the problem, DaleSpam changed the problem, and made the "point" contact between the rod and the rail into two separated points.

That solution works mathematically, but only in the special case where the mass is at the center of the circle. But as a "real world" solution it is impractical, because as $\phi \rightarrow 0$ the two forces both $\rightarrow \infty$ (positive in opposite directions) but the difference betwen them stays finite and equal to $mg$ upwards.

So really, all this is shadow-boxing with something physically unrealistic, and being an engineer not a theoretical physicist, I'm not very interested in believing six impossible things before breakfast every day!

If the rod is a different length and mass is not at the center of the circle, the paradox goes away, because there is no equilibrium solution without friction (even for DaleSpam's change to the original problem) except when the rod is vertical, and the gravitational potential energy of the mass does work as the system moves to an equilibrium position (there are two of them, one stable and one unstable).

DaleSpam's solution only "works" because in the free body diagram of the mass, rod, and two contact points, the three force vectors (two at the rail and the weight of the mass) all pass through the same point (the center of the circle). If you change the length of the rod, this can not happen except when the rod is vertical, and there will always be a moment acting on the mass+rod to rotate it around the circle to its equilibrium position.

You could find an equilibrium solution for one particular angle $\theta$ by replacing $\phi$ by two different angles $\phi_1$ and $\phi_2$, but by symmetry is it clear that there is no solution like that which is correct for all values of $\theta$. When $\theta = 0$, $\phi_1 = \phi_2$.

 

[Dale]

But tangential to the rail there are no constraints. In order to neutralize the gravitational force $m*g*sin(\theta)$ tangential to the rail, it would take either friction or some obstacle (i,e, a surface component transverse to the rail).

No, it doesn't. That is the whole point of what I showed above, and we have discussed similar concepts previously. Because you are constraining both the position and the direction of the arm you can cancel out the gravitational force completely without a tangential force at any point.

Yes please.

See the previous post for a graphic showing the directions of the forces. Using Newton's 2nd law and noting that a=0 we write one expression for the component of the forces parallel to the arm:
$mg \cos(\theta) - F_A \cos(\phi) - F_B \cos(\phi)=0$
and another expression for the component of the forces perpendicular to the arm:
$mg \sin(\theta) + F_A \sin(\phi) - F_B \sin(\phi)=0$

This is two equations in two unknowns, so we can solve for $F_A$ and $F_B$ to get:
$$F_A=\frac{mg}{2}\frac{-\cos(\phi) \sin(\theta)+\cos(\theta) \sin(\phi)}{\cos(\phi)\sin(\phi)}$$ $$F_B=\frac{mg}{2}\frac{\cos(\phi) \sin(\theta)+\cos(\theta) \sin(\phi)}{\cos(\phi)\sin(\phi)}$$
Which simplifies to the above expressions.

 

[Dale]

To get some sort of a solution to the problem, DaleSpam changed the problem, and made the "point" contact between the rod and the rail into two separated points.

I didn't change the problem at all. I analyzed a standard idealized mechanism for providing both a path (force) constraint and a direction (torque) constraint. A point contact cannot do that, and the drawing in the OP clearly shows that a point contact is not assumed. The OP did not specify the mechanism, so I chose the simplest such mechanism without altering the scenario.

Your other comments about the impracticality of such a system are, of course, correct. However, the things you mention were explicit parts of the OP and I maintained them in my analysis precisely to provide a faithful answer to the OP's problem.

 

[Dale]

Hi Fantasist, this comment by AlephZero got me thinking:

as $\phi \rightarrow 0$ the two forces both $\rightarrow \infty$ (positive in opposite directions) but the difference betwen them stays finite and equal to $mg$ upwards.

I may have an alternative explanation which could help. I need to work through it and see if the explanation is valid, but if it is then it may be a different way of looking at the problem which is more similar to your though process and perhaps more understandable to you.

 

[sophiecentaur]

This ihas turned into a futile set of arguments now. Are we not discussing an ideal situation? If we are, then how could any work be done if the height of the mass doesn't change? Whatever maths you care to apply to the situation, if you don't get the answer Zero, then you've done the Maths wrong.

 

[BruceW]

I would agree with sophiecentaur, which is why I've not added any more comments. But I guess the OP wants to see a solution to something that is similar to his original question (and I'm still not sure exactly what that is). So if anyone can give him this, then that's fine with me.

 

[jbriggs444]

I would agree with sophiecentaur, which is why I've not added any more comments. But I guess the OP wants to see a solution to something that is similar to his original question (and I'm still not sure exactly what that is). So if anyone can give him this, then that's fine with me.

I thought SophieCentaur came pretty close in post 14. Here's my try:

It is clear the ideal attachment point is exerting an upward force on the mass. If one views this interaction in isolation then, as the attachment point is moving, some amount of work is done.

But the interaction is not in isolation. The ideal attachment point is also exerting a torque. As it moves, the attachment point also rotates in the direction of the torque. As it rotates, some amount of work is recovered.

As long as the mass is at the center of curvature of the rail, the two effects cancel exactly so that no net work is done.

The paradox is obtained by failing to consider the torque. Which is why Sophie's post #14 (which happens to use the word "moment" as a synonym for torque) seemed to me as if it should have put the matter to rest.

 

[Dale]

Well, maybe this will do it.

Fantasist, in post 20 and 38 I detailed a method of answering the OP using the simplest mechanism for the constraints. However, I have realized that it is actually possible to answer the OP without using any mechanism for the constraints. The key is to realize that the constraints constrain two degrees of freedom, the position and the angle.

So, if you assume some mechanism which provides specific forces then it must provide at least two forces to correspond to the two constrained degrees of freedom. Alternatively, you can simply consider a net force (by whatever mechanism) and a net torque (again, by whatever mechanism). This satisfies the two degrees of freedom without any specification of the mechanism.

Now, as we move the attachment a small distance you want to consider the work done by the constraints (even though we know that constraints do no work, you want to prove it). So, since there are two constraints we get:

$dW=dW_F+dW_T$

where $dW_F$ is the work done by the net force and $dW_T$ is the work done by the net torque at the constraint. Then:

$dW_F=\mathbf{F}\cdot d\mathbf{s}=m\mathbf{g}\cdot d\mathbf{s} = -mg \sin(\theta) ds$
$dW_T=\tau \; d\theta = mg\; r \sin(\theta) d\theta$

and since $r\;d\theta = ds$ we have $dW_T=-dW_F$ so $dW=0$.

Your basic mistake in the OP was that you analyzed only $dW_F$ and neglected $dW_T$. That doesn't provide enough constraints to match what you were trying to do even in an abstract "let's not specify the mechanism" type analysis.

EDIT: I see that jbriggs444 got to the same place that I did, but faster!

 

[Fantasist]

Thanks for all your replies and sorry for the delay with my response, but I had little time last week and wanted to make sure I consider all arguments as thoroughly as possible. I can actually see where those are coming from, but lastly I can't (yet) see them resolving the paradox due to the issues addressed below:

The work-energy principle for a point-like mass is:
$$Work = \int_{t1}^{t2} \vec{F} \cdot \vec{v} dt$$
And in this case, it is zero, because the only significant mass is the one in the middle, which has zero velocity for all time.

But there is no mass in your equation. What it says is that if you move a certain point of a mechanical system against some force over a certain path (or with a certain velocity for a certain time in your formulation), then this will require a certain work i.e. energy. The emphasis is clearly on the point of application here i.e. the point of constraint in this case, and by assumption this moves.

How are you defining "a path that changes the state of the system" ? If you mean something like "the path of the connection point of the rod with the rail", then I challenge the physical meaning to this. Why is such a thing physically meaningful? Why would you choose that path, and not any other arbitrary path? For example, you could choose the path of any other point on the rod, and you would get a different answer for your 'work done'.

Well, if you would do the work at any other point of the rod, you would change the constraint; effectively you would have then a rail with a smaller radius and a shorter rod. This would be part of the paradox.

n order to neutralize the gravitational force tangential to the rail, it would take either friction or some obstacle ...

...or two radial forces at $\theta\pm\phi$ which both have non-zero tangential components w.r.t to the radius at $\theta$.

Only that the forces apply at $\theta\pm\phi$ and not at $\theta$. Translating the vectors from the former to the latter obviously changes the scenario. Consider the following (slightly simpler) example

https://www.physicsforums.com/attachment.php?attachmentid=68144&stc=1&d=1396201366

you have a disk constrained at its center so that it can only rotate. There are two different radial forces $F_A$ and $F_B$ as indicated by the blue vectors. If you translate these vectors both to a common point and add them there, the sum will there have a tangential component $F_T$ (indicated in green), so the two forces should set the disk in rotation. Are you really suggesting the disk will be rotating in this scenario?

To get some sort of a solution to the problem, DaleSpam changed the problem, and made the "point" contact between the rod and the rail into two separated points.
That solution works mathematically, but only in the special case where the mass is at the center of the circle. But as a "real world" solution it is impractical, because as $\phi \rightarrow 0$ the two forces both $\rightarrow \infty$ (positive in opposite directions) but the difference between them stays finite and equal to $mg$ upwards.

Sorry, but for small $\phi$ I make this

$F_A - F_B = - mg/\phi*sin(\theta)$
$F_A + F_B = mg*cos(\theta)$ (which is the radial component parallel to the arm)

In any case, $F_A$ and $F_B$ are clearly defined as radial forces, so they can not produce any net force tangential to the rail that could compensate for the gravitational force (see my drawing above).

Using Newton's 2nd law and noting that a=0 we write one expression for the component of the forces parallel to the arm:
$mg \cos(\theta) - F_A \cos(\phi) - F_B \cos(\phi)=0$
and another expression for the component of the forces perpendicular to the arm:
$mg \sin(\theta) + F_A \sin(\phi) - F_B \sin(\phi)=0$

What you have done here is to set up two equations of constraint, one for the net force along the arm and one for the net force perpendicular to the arm (i.e. tangential to the rail). Effectively, this means you can't move anything anymore, and not surprisingly you won't be able to do any work then. The force components you thus get are just the forces of constraint in the material when you have locked everything in place, but this is not the scenario here.
The error you made here is to assume that the condition a=0 imposes a constraint on the motion perpendicular to the arm as well. It doesn't, as the mass can stay in the same place but still rotate.

Generally speaking, you should not bother about forces of constraint here at all. They may be of interest to material scientists, engineers or architects, but are irrelevant for the dynamical changes of a system.

Now, as we move the attachment a small distance you want to consider the work done by the constraints (even though we know that constraints do no work, you want to prove it). So, since there are two constraints we get:

$dW=dW_F+dW_T$

where $dW_F$ is the work done by the net force and $dW_T$ is the work done by the net torque at the constraint. Then:

$dW_F=\mathbf{F}\cdot d\mathbf{s}=m\mathbf{g}\cdot d\mathbf{s} = -mg \sin(\theta) ds$
$dW_T=\tau \; d\theta = mg\; r \sin(\theta) d\theta$

and since $r\;d\theta = ds$ we have $dW_T=-dW_F$ so $dW=0$.

Your basic mistake in the OP was that you analyzed only $dW_F$ and neglected $dW_T$. That doesn't provide enough constraints to match what you were trying to do even in an abstract "let's not specify the mechanism" type analysis

I don't understand why you are bringing in the torque here. If you apply a force $\mathbf{F}$ to some point of a rigid body and move this point a distance $d\mathbf{s}$ in the process, then the work associated with this is $\mathbf{F}\cdot d\mathbf{s}$, irrespectively of how much of this results in a translation of the center of mass and how much in a rotation due the torque associated with the force. If you say the total work done in the process is zero, then $\mathbf{F}\cdot d\mathbf{s}$ must be zero. But this contradicts your equation $\mathbf{F}\cdot d\mathbf{s}=m\mathbf{g}\cdot d\mathbf{s} = -mg \sin(\theta) ds$.
Like I said already earlier, the important point here is that the work is done at the point of constraint, not directly at the mass.

 

[sophiecentaur]

@Fantasist

I really don't see where you are going with this. If there is a complete disc (massless etc) supporting the mass then there is no work done in rotating it.
If you use number of spokes, there is still no work done.
Two spokes, progressively closer and closer together, will produce no work. If you say there is something special about your single 'rigid arm' with 'non flexing' shoe on the rail then you would need to say what the limiting case is (i.e. when two spokes start behaving as one, only differently). If you are calculating "for small ϕ" and you are getting a non zero result, you have to conclude that you have got the sums wrong. I don't see that, in any way, you have discovered a 'paradox' - you have just produced an error in your derivation. We have all been there and usually look for our error, rather than cry 'paradox'.

 

[Dale]

Sorry, but for small $\phi$ I make this

$F_A - F_B = - mg/\phi*sin(\theta)$
$F_A + F_B = mg*cos(\theta)$ (which is the radial component parallel to the arm)

The variables $F_A$ and $F_B$ in the equations of posts 20 and 38 and are the magnitudes of vectors which point it different directions. It doesn't make any sense to add them or subtract them this way. If you actually calculate the force vectors $\mathbf{F_A}$ and $\mathbf{F_B}$ and add them then you will see, as AlephZero said, that their sum is mg upwards regardless of the size of $\phi$, as it must be.

In any case, $F_A$ and $F_B$ are clearly defined as radial forces, so they can not produce any net force tangential to the rail that could compensate for the gravitational force (see my drawing above).

I already proved the contrary. This is simply false, so please don't repeat it again. Two radial forces together can indeed provide a net force which can compensate for the gravitational force. In general, a radial force at some point A and a radial force at some point B will usually produce a net force which is not radial at a third point C.

What you have done here is to set up two equations of constraint, one for the net force along the arm and one for the net force perpendicular to the arm (i.e. tangential to the rail). Effectively, this means you can't move anything anymore, and not surprisingly you won't be able to do any work then. The force components you thus get are just the forces of constraint in the material when you have locked everything in place, but this is not the scenario here.
The error you made here is to assume that the condition a=0 imposes a constraint on the motion perpendicular to the arm as well. It doesn't, as the mass can stay in the same place but still rotate.

You have this so mixed up I have a hard time knowing where to start.

Newton's 2nd law is ∑F=ma, where ∑F is the sum of all of the forces acting on a body, m is the mass of the body, and a is the acceleration of the center of mass of the body. Because this is a vector equation it does indeed constrain the motion of the center of mass in both the horizontal and vertical direction or equivalently in both the direction parallel to and perpendicular to the arm.

However, the a in the equation is the acceleration of the center of mass. It does not constrain in any way the motion of the rest of the body around the center of mass. In particular, it does not prevent the object, including the massless arm, from rotating around the center of mass. Nothing is being "locked" in place or in any way constrained other than as you specified in the OP.

I don't understand why you are bringing in the torque here.

I didn't bring it in, you brought it in during the OP when you constrained the arm to always point perpendicular to the track. The force of gravity exerts a torque about the connection point, therefore the constraint must exert an equal and opposite torque, simply to maintain the angle of the rod.

If you disagree, then consider what would happen if the rod were hinged rather than being a rigid body. A torque is obviously required.

If you apply a force $\mathbf{F}$ to some point of a rigid body and move this point a distance $d\mathbf{s}$ in the process, then the work associated with this is $\mathbf{F}\cdot d\mathbf{s}$, irrespectively of how much of this results in a translation of the center of mass and how much in a rotation due the torque associated with the force. If you say the total work done in the process is zero, then $\mathbf{F}\cdot d\mathbf{s}$ must be zero. But this contradicts your equation $\mathbf{F}\cdot d\mathbf{s}=m\mathbf{g}\cdot d\mathbf{s} = -mg \sin(\theta) ds$.
Like I said already earlier, the important point here is that the work is done at the point of constraint, not directly at the mass.

You have specified two kinematical constraints: one constraint on the position and one constraint on the direction. A single constraint force is simply incapable of explaining the dynamics of your kinematical constraints. It leads to an underdetermined set of equations. If you disagree then I encourage you to set up the system of equations and see for yourself.

In order to remedy this, you must add an additional dynamical constraint. This can be either in the form of a second force, as shown in posts 20 and 38, or it can be in the form of a torque, as shown in post 44. Either way the work required to move the arm is zero and energy is conserved, as I have shown.

 

[Buckleymanor]

AlphaZero in post 4 and Sophie in post 14 both point out the non paradoxical element of the constrained mass.
To constrain the mass you must have torque at point P where the mass connects and therefore friction.
There is no point in saying it's frictionless at this point try building it, whatever materials used there will be a moment or leverage about this point and friction produced.
You might as well insist it levitates.

 

[jbriggs444]

AlphaZero in post 4 and Sophie in post 14 both point out the non paradoxical element of the constrained mass.
To constrain the mass you must have torque at point P where the mass connects and therefore friction.

The existence of a torque does not require friction.

 

[BruceW]

yeah, and I think this is maybe where Fantasist's main problem is. If the rod was attached to the rail at a point, then the mass cannot be suspended in equilibrium, because the total torque (or moments, I never know the right terminology for this) around the mass, cannot add to zero, while also having the linear forces add to zero.

But, even if the connection point was just finite in size, then it would be possible for the mass to be suspended. For a simpler example, think of a shelf (which is very similar to this problem, where the rod is horizontal). If the shelf was simply attached to the wall at a point, the shelf cannot be in equilibrium. But if you have a shelf bracket, then it is possible to balance the forces and have zero torque. This is possible because the attachment is not just at a point.

 

[A.T.]

To constrain the mass you must have torque at point P where the mass connects and therefore friction.

As DaleSpam has shown no friction is required.

 

[Fantasist]

@Fantasist

I really don't see where you are going with this.

Hopefully to a better understanding of the physics of constrained mechanical systems.

If there is a complete disc (massless etc) supporting the mass then there is no work done in rotating it.
If you use number of spokes, there is still no work done.
Two spokes, progressively closer and closer together, will produce no work. If you say there is something special about your single 'rigid arm' with 'non flexing' shoe on the rail then you would need to say what the limiting case is (i.e. when two spokes start behaving as one, only differently). If you are calculating "for small ϕ" and you are getting a non zero result, you have to conclude that you have got the sums wrong. I don't see that, in any way, you have discovered a 'paradox' - you have just produced an error in your derivation. We have all been there and usually look for our error, rather than cry 'paradox'.

I have indicated above already where the reason of the paradox may lie, namely because of assuming the work is done directly at the mass, when in fact it is done at the point of constraint.

 

[sophiecentaur]

Hopefully to a better understanding of the physics of constrained mechanical systems.



I have indicated above already where the reason of the paradox may lie, namely because of assuming the work is done directly at the mass, when in fact it is done at the point of constraint.

Are you saying that your method of calculation, incontrovertibly shows that work is done "at the point of constraint"? I suggest that the calculation will involve taking limits and that has to be done in a valid way or the answer can't be relied on.
Imo, the way to deal with situations like this is to start with the basic principle and not aim at showing it's wrong. As soon as we start getting into zeros and infinities we depart from the real world so spurious paradoxes start to suggest themselves. "A better understanding" doesn't seem likely when the system being studied is unreal.

 

[Fantasist]

The variables $F_A$ and $F_B$ in the equations of posts 20 and 38 and are the magnitudes of vectors which point it different directions. It doesn't make any sense to add them or subtract them this way. If you actually calculate the force vectors $\mathbf{F_A}$ and $\mathbf{F_B}$ and add them then you will see, as AlephZero said, that their sum is mg upwards regardless of the size of $\phi$, as it must be.

OK, my mistake. I apparently misunderstood AlephZero's statement. Obviously, if you multiply $\sin(\phi)$ to $F_A - F_B$ then this would be equal to $-mg \sin(\theta)$ (as this is what you put in as your assumption in the first place).

Newton's 2nd law is ∑F=ma, where ∑F is the sum of all of the forces acting on a body, m is the mass of the body, and a is the acceleration of the center of mass of the body. Because this is a vector equation it does indeed constrain the motion of the center of mass in both the horizontal and vertical direction or equivalently in both the direction parallel to and perpendicular to the arm.

However, the a in the equation is the acceleration of the center of mass. It does not constrain in any way the motion of the rest of the body around the center of mass. In particular, it does not prevent the object, including the massless arm, from rotating around the center of mass.

I fully agree with this in principle. Only this is not how you set up your equations, where the forces are taken at the constraint, not at the mass. I quote your earlier post:

For concreteness, let's assume that the structure consists of two pairs of frictionless ball bearings, each located an angle of $\phi$ away from the angle of the rod which is itself at an angle $\theta$ from the vertical. Then the forces at the bearings are:
$F_A=-mg \csc(2\phi) \sin(\theta-\phi)$
$F_B=mg \csc(2\phi) \sin(\theta+\phi)$

I already proved the contrary. This is simply false, so please don't repeat it again. Two radial forces together can indeed provide a net force which can compensate for the gravitational force. In general, a radial force at some point A and a radial force at some point B will usually produce a net force which is not radial at a third point C.

Sorry, but you haven't proved anything. You made the assumption that the system is static by postulating the force equations at the constraint as

$mg \cos(\theta) = F_A \cos(\phi) + F_B \cos(\phi)$
$mg \sin(\theta) = F_B \sin(\phi) - F_A \sin(\phi)$

You then calculate from this the forces $F_A$ and $F_B$ that satisfy this assumption. These forces are exactly the internal stress forces in the material when you have constrained not only the radial motion but also the tangential motion along the rail. But it would be a circular conclusion to say that the system will be static without a tangential constraint. I might just as well postulate the second of the equations as

$F_B \sin(\phi) - F_A \sin(\phi) = 0$

and then say I have proven that the system is not static in general.

I don't understand why you are bringing in the torque here. If you apply a force $\mathbf{F}$ to some point of a rigid body and move this point a distance $d\mathbf{s}$ in the process, then the work associated with this is $\mathbf{F}\cdot d\mathbf{s}$, irrespectively of how much of this results in a translation of the center of mass and how much in a rotation due the torque associated with the force. If you say the total work done in the process is zero, then $\mathbf{F}\cdot d\mathbf{s}$ must be zero. But this contradicts your equation $\mathbf{F}\cdot d\mathbf{s}=m\mathbf{g}\cdot d\mathbf{s} = -mg \sin(\theta) ds$.
Like I said already earlier, the important point here is that the work is done at the point of constraint, not directly at the mass.

You have specified two kinematical constraints: one constraint on the position and one constraint on the direction. A single constraint force is simply incapable of explaining the dynamics of your kinematical constraints. It leads to an underdetermined set of equations. If you disagree then I encourage you to set up the system of equations and see for yourself.

I don't know why you assume the system would be underdetermined. Consider the case of tightening a nut with a spanner (wrench): you exert a vertical force $\mathbf{F}$ at a distance r from the nut; when you move this through one rotation you'll do the work $W = 2\pi*r*\mathbf{F}$. There is no additional contribution from the torque (of course, alternatively you could calculate the work from the torque as well, which would give the same result). The present scenario is pretty much identical to this.P.S.: I make this my last post on this subject, as I don't think there is a lot to add anymore here.

 

[Dale]

I fully agree with this in principle. Only this is not how you set up your equations, where the forces are taken at the constraint, not at the mass. I quote your earlier post:

For concreteness, let's assume that the structure consists of two pairs of frictionless ball bearings, each located an angle of $\phi$ away from the angle of the rod which is itself at an angle $\theta$ from the vertical. Then the forces at the bearings are:
$F_A=-mg \csc(2\phi) \sin(\theta-\phi)$
$F_B=mg \csc(2\phi) \sin(\theta+\phi)$

And what is wrong with that? Obviously, the forces on the bearings act at the bearings! Where else would the forces on the bearings act other than at the bearings?

It is completely baffling to me that you would consider this a point of contention. If I push a door then the force of my finger on the door acts where I push on the door and not at the window near the door, and the force of the door on my finger acts on my finger and not on my nose. How can you possibly think otherwise? Why would you think that the forces on the bearings act elsewhere?

Furthermore, why do you think it matters? Look at Newtons second law. Where in the law does the point of application matter? Regardless of where on an object a force is exerted the acceleration of the center of mass of the object responds the same.

Sorry, but you haven't proved anything. You made the assumption that the system is static

Nonsense. YOU made that assumption when you specified the geometry of the problem and "a mass m" and "a rod (with negligible mass)". It is part of your OP that the center of mass did not accelerate, which you even recognize in the OP with "the mass m has always stayed in the same location".

My derivation is simply a consequence of applying Newton's laws to the system that YOU specified. YOU specified that ma=0, not me. As a consequence we know by Newton's laws that ∑f also equals 0, but that is not an assumption I imposed. That is a consequence of your assumption.

I don't know why you assume the system would be underdetermined.

Why don't you actually set up the system of equations and see for yourself.