Static vs. Neutral Equilibrium of a Mesh

[Shoelace Thm.]

Suppose a square metal mesh sheet is hung off-center from a string and you are charged with the task of hanging various weights so that the mesh will be in static equilibrium, roughly horizontal. Why is it that this task must be done with masses that can exert torques large enough to flex the sheet, otherwise the sheet will be in neutral equilibrium, stable at a wide variety of angles with the horizontal?

I don't quite understand why flexing would have anything to do with this. However, I can see why smaller torques could cause problems because very slight deviations from the horizontal (which are almost unavoidable) would be relatively large and so the system would not have a preferred orientation.

 

[mfb]

However, I can see why smaller torques could cause problems because very slight deviations from the horizontal (which are almost unavoidable) would be relatively large and so the system would not have a preferred orientation.

That is the reason - if the setup is perfectly rigid, the whole setups is in a plane, and tilting this plane does not change the net torque, so you don't have restoring forces.

If the mesh has to flex a bit (downwards, where masses are), you have some pendulum-like setup, and a deviation from the horizontal orientiation gives a net force in the opposite direction.

 

[Shoelace Thm.]

What is the "net force in the opposite direction"? The way I see it is that in both cases the net force on the mesh equals 0 at all angles and so somehow flexing the mesh provides a restoring torque.

Also why does tilting the plane when the mesh is flexed change the net torque from 0?

 

[Shoelace Thm.]

Can someone resolve these issues?

 

[haruspex]

Any weights suspended from the mesh act, as far as the mesh is concerned, as though attached as point masses right on the mesh. To get stable equilibrium, you need the effective C of G of the mesh (with attached point masses) directly and strictly below the point of attachment of the string (S). Clearly that will involve some flexing. E.g., hang a weight at each corner; the flexing will displace the C of G below the plane tangent to the mesh at S. An appropriate combination of such weights will make that plane horizontal.

 

[Shoelace Thm.]

I agree, but I am more interested in why the mesh must flex in order to be stable horizontally. In particular, what is the restoring force/torque acting in the opposite direction that mfb mentions?

 

[256bits]

I agree, but I am more interested in why the mesh must flex in order to be stable horizontally. In particular, what is the restoring force/torque acting in the opposite direction that mfb mentions?

Use geometry to see how the horizontal components of the ends AS and BS of a straight line AB rotating about S do not vary from the ratio AS/BS.
For a line bent at S the ratio AS/BS does not remain constant.

The change in the ratio AS/BS for forces attached to the ends gives the restoring torque.

 

[haruspex]

I agree, but I am more interested in why the mesh must flex in order to be stable horizontally. In particular, what is the restoring force/torque acting in the opposite direction that mfb mentions?

If it does not flex then, as I said, the effective C of G is still in the plane of the mesh. If it is not at the point, S, from which it is suspended then it will hang vertically. If it is right at S then the C of G has no moment about S, so the mesh can sit at any angle.

 

[Shoelace Thm.]

According to your reasoning, even if the C of G is strictly below S (mesh is flexed) there is no restoring torque because the angle between moment arm (about S) and force of gravity is 0.

 

[mfb]

There is no restoring force in equilibrium, but there is a restoring force if the object is not in that single equilibrium point.

 

[Shoelace Thm.]

Ok but haruspex claims that C of G strictly below S is necessary for stable horizontal equilibrium. I do not see how that is necessary and I stated why in my previous post. Could you explain haruspex?

 

[mfb]

"but"? There is no contradiction. My post applies to the case of a CoG below S.

Does that sketch help?

 

[Shoelace Thm.]

In your last picture, why did the C of G move from beneath point S? Are you saying that the tilting causes the net torque to longer be 0 about point S?

 

[mfb]

I rotated the setup around the anchor point, and the center of gravity rotated together with the mesh.