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I was trying to cook up a good exam problem on Newtonian mechanics, and I came up with one that I didn't fully understand myself. Always fun when you can outsmart yourself.

enter image description here

In the figure a cylinder rests in a 90-degree corner that has been tilted so that the sides are at 45-degree angles. We're given the weight of the cylinder. If the surfaces are frictionless, then determining the magnitude of the two normal forces is pretty trivial.

But if the surfaces are not frictionless, then the problem is underdetermined. Because of symmetry, Newton's first law is automatically satisfied in the $x$ direction, and therefore Newton's first law in the $x$ direction gives no useful information, and similarly we get nothing useful from setting the total torque equal to zero. The symmetry of the problem means that there are only two unknowns: the magnitude $N$ of the normal forces, and the magnitude $F$ of the frictional forces. Newton's first law in the $y$ direction gives a single constraint, which is not enough to determine both unknowns. The problem appears to be underdetermined.

What does this mean physically? If I set this up and measured $F$ and $N$, would I really get nonreproducible results? Would the results depend on how hard I jammed the cylinder into the corner? The system would then have a memory of how it was prepared, which seems surprising to me, since this information would have to be stored somehow in its physical state, and that would presumably be because there was some microscopic displacement of one surface relative to the other. But in theory I can make the surfaces arbitrarily smooth and hard, while still maintaining a nonzero coefficient of static friction. After all, we know that static friction can exist, and even be quite strong, between surfaces such as optically flat glass plates.

Or does the system actually not have any memory, meaning that if I try to jam the cylinder down harder, it will just spring back into a standard state when I let go? Would this be a state with $F=0$?

[EDIT] Ja72 and dmckee made some helpful comments that pointed me toward some interesting material. I think this is probably not analogous to the Painleve paradox, since it's static rather than dynamic; if I'm understanding correctly, the resolution of the Painleve paradox involves forces that act like delta functions of time. It seems more similar to the problem dmckee discussed, with a plate supported at four points; the solution seems to be that it's not consistent to describe the plate as perfectly rigid: http://mathtrifles.wordpress.com/2010/09/19/rigid-plate-on-four-supports/ Possibly the solution to my problem is that it's not consistent to assume that the box is perfectly rigid.

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    $\begingroup$ For a different class of underdetermined problems (a object supported at four points in a rectangular grid) you can convince yourself that when you stop neglecting elastic deformation you can "fix" the system (at the cost of making it a larger system). It's not obvious to me if that is the case here. $\endgroup$ – dmckee Oct 19 '14 at 18:17
  • $\begingroup$ I do think that you would have to be very careful in setting up an actual experiment to get the assumption of symmetry to hold. Even a small deviation will break the symmetry and make the system determined again. $\endgroup$ – dmckee Oct 19 '14 at 18:18
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    $\begingroup$ Interesting question that relates to the painleve problem. $\endgroup$ – ja72 Oct 19 '14 at 20:00
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    $\begingroup$ BenCrowell, Isn't the answer just the same as in the example of @BMS ? Even without friction the cylinder will remain static, like the block on a string. Friction is a `response force' i.e. its magnitude is equal to the force applied to it (up to some maximum), so in your case giving the surface a friction coefficient doesn't result in a friction force, because the system was already balanced $\endgroup$ – Michiel Oct 20 '14 at 6:59
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    $\begingroup$ @BenCrowell He's right for static friction with a non-deformation assumption (i.e. the sort we teach in PHYS 101). In that case it is always "just enough to prevent relative motion" up to the point where it fails and motion begins. The difficulty here is that examined at high resolution there can be deformation both normal to and across the surface so both the normal force on one side and the static friction on the other can be supporting the cylinder against motion along one direction (i.e. both can have non-zero values in a quiescent state), and ambiguity arises. Time to break out the FEA. $\endgroup$ – dmckee Oct 21 '14 at 2:52
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This might be one of those problems that looks really simple on the surface but it's surprisingly complicated to prove. There might be a simpler proof, but I couldn't come up with one.

Let's stick to the "ideally rigid" scenario. At an instinctive level, I think everybody will agree that the friction "ought to be zero". It took me several rounds to find a semi-formal way to prove it (I invite anybody to improve this response with a more formal proof).

I'm going to sketch a proof relying on the principle of Virtual Work. For a system to be in equilibrium, the virtual work from all the forces (due to any infinitesimal motion) must be exactly zero.

The cylinder is unconstrained to rotate in place. The normal forces don't produce any virtual work because they are normal to the virtual movement. The weight doesn't produce virtual work because the center of mass doesn't move under a rotation. The only virtual work comes from the friction forces. However, that means that the virtual work from the friction on the left wall has to exactly cancel the virtual work form the friction from the right side, and that can't happen because both of them act in the opposite direction of the virtual rotation (both virtual works have to be negative). The only explanation is that that the friction forces must be exactly zero.

Now, let's assume for the sake of argument that you know for a fact there is a frictional force. Can't that happen? Yes, but only if the apparatus is no longer ideally rigid. It can happen if the cylinder is not just resting on top of the wedge, but it's actually... wedged in.

Wedging requires that the wedge deforms to make space for the cylinder (and/or the cylinder has to squeeze, which is a harder problem to work with).

In a basic view of the problem, the wedge acts like a loaded spring. The load on the spring (wedge) will be due to the weight of the cylinder. But if we push the cylinder down to wedge it in, then there would be an additional load due to excess potential energy transferred to the wedge.

In that scenario, there will be friction, and the friction is countering the additional restorative force from the wedge trying to return the additional potential energy.

A much simpler version of the same problem is a box on a horizontal surface with friction, pushed against a horizontal spring. The friction between the box and the surface can be calculated as $F=k\times x$ where $x$ is the distance the box moved while being pushed against the spring and $k$ is the stiffness of the spring.


Bonus content:

Hey, look, that's an underdeterminated (unidimensional) static problem but we still managed to solve it!

Being statically underdeterminated doesn't mean that you "cannot know the answer", I mean, the Universe obviously knows the answer somehow... It just means that the laws of static are not enough to give the answer to you. You need more sophisticated tools, and more importantly - you need more information about the system. In the simplified example, the additional information is: "one of the components is an ideal elastic object of stiffness $k$ and the "more sophisticated tool" is "Hooke's law for springs".

How about the original problem?

That's harder, because the wedge is not necessarily a simple spring. How it reacts to the wedging depends on a lot of details about the cylinder, the wedge and the environment. In the real world - even assuming a rigid cylinder - the answer might depend on such things as the exact material used for the wedge, its construction, temperature gradients present at the time, the scale of the forces involved and thickness of the walls of the wedge.

A typical (academic perhaps) version of the problem might be:

The wedge is made of a perfectly isotropic elastic material [the material is expected to follow a generalized version of Hooke's law within the expected range of forces it will be subjected to. Also, it deforms the same way in all directions] of elastic modulus $E$ [a generalized version of $k$ for springs]. The cylinder has a weight of $F$; you may consider the cylinder rigid (because it's rigid enough that its own deformations won't contribute significantly to the value of the force). The cylinder was placed on top of the wedge and then pushed down an additional vertical distance $n$. You also need more details on the geometry of the problem. For example: The walls of the wedge are $h$ thick and the cylinder is $d$ in diameter.

Now we're closer to something manageable.

I'll throw in one more simplification for now: "the walls of the wedge are relatively thin compared to their height" (if the problem is numerical, the measurements bear that). This is commonly called "slenderness" and it's an important simplification that allows us to use comparatively simpler tools to analyze the problem. We also want the wedging distance to be relatively small compared to the dimensions of the apparatus. In these conditions, the two sides of the wedge can be analyzed exactly like cantilever beams.

With that information, an undergrad engineering student can determine the force responding to the deformation of the wedge walls due to the wedging ($F = \epsilon\times 3EI/L^3$. $\epsilon$ is the deformation normal to the wall due to the wedging; $L$ and $I$ can be calculated from the geometry of the problem). After that, if you want, you can then determine the frictional force required by using conventional static methods.

If you can't make the slenderness simplification, you will have to resort to other methods. I would have to look it up, but there should be some approximate formulas available in the literature. Or you might have to use numerical simulations (finite element analysis) to determine the deformation suffered by the wedge. If the wedge is not elastic (or the forces exceed the elastic limit for the material), that throws more complications into the problem. You might get to the point where you have to "build it and check it out in a lab".

I liked your thought processes on these parts of your question; let me try to answer them:

What does this mean physically? If I set this up and measured F and N, would I really get nonreproducible results?

Assuming the wedged-in scenario: maybe. As mentioned, the answer depends a lot on tiny details of the apparatus. If you use the same apparatus every time and you wedge it in the same distance, under the same conditions, and you don't exceed the elastic limit, you should get the same results. As an opposite example, if the cylinder is wedged in and you use untreated wood parts, even if everything else stays the same the results might change by the hour depending on the temperature and humidity fluctuations throughout the day.

Would the results depend on how hard I jammed the cylinder into the corner?

Yes.

The system would then have a memory of how it was prepared, which seems surprising to me, since this information would have to be stored somehow in its physical state, and that would presumably be because there was some microscopic displacement of one surface relative to the other.

The information is stored mostly on the deformation of the wedge walls due to the total force applied. It will be small, but it probably won't be microscopical.

But in theory I can make the surfaces arbitrarily smooth and hard, while still maintaining a nonzero coefficient of static friction. After all, we know that static friction can exist, and even be quite strong, between surfaces such as optically flat glass plates.

The harder you make the materials, the less deformation it will take to store the additional energy. But the amount of friction in our wedged-in example depends only indirectly on that and it's affected by other factors. However, if all the materials are perfectly (ideally) rigid, then jamming is not possible and there won't be friction.

Or does the system actually not have any memory, meaning that if I try to jam the cylinder down harder, it will just spring back into a standard state when I let go?

That's exactly what would happen with an elastic wedge and a null or negligible friction coefficient.

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  • $\begingroup$ What is virtual work? $\endgroup$ – Unique Aug 19 at 0:59
  • $\begingroup$ @unique you should post that as a separate question or search for existing questions on the topic. The full definition is a bit tricky and “gets mathematical”. I don’t feel qualified to explain it properly without first refreshing on a lot of material. The gist is that in a static system (it turns out) that any infinitesimal “wiggle” must result in net Work (energy transfer) of zero . You set up work equations assuming the system does move, equate the total Work to zero, and solve the equations for what you seek. For certain types of problems this is more convenient than using Newton’s laws. $\endgroup$ – Euro Micelli Aug 23 at 11:26
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I don't agree that the problem would be under-determined. If the materials are not frictionless, then the friction coefficients must be given as well as the coefficient of elasticity of the materials. This additional information will allow the calculation of the forces!

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There will be no friction if both the cylinder and the walls are perfectly rigid.

Imagine the cylinder is on wheels that have the same friction with the planes as the cylinder would, but no internal resistance as in this diagram

diagram

If there is friction, it would produce torque on the wheels causing them to turn. The wheels will not turn when in equilibrium, so there is no friction acting on them then. Therefore there should also be no friction if the cylinder rests directly on the planes.

However, if the cylinder and/or walls are not perfectly rigid, then if we very slowly lower the cylinder into position, deformations will start to occur after the wheels make first contact with the walls and load is gradually applied. If the lowering is done precisely the first point of contact will be at the equilibrium position of the rigid system, but deformations may allow the cylinder to settle into a lower position, causing the wheels to roll.

If the wheels are prevented from rolling, they will be held in place by friction and the system will come to equilibrium with friction present. Similarly the cylinder without wheels should experience friction in equilibrium if there is deformation as it is lowered. The final size of the friction component will depend on on how the cylinder is lowered and likely differ between the original system and the system with wheels.

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  • $\begingroup$ It seems reasonable to imagine that the amount of friction depends on deformation. However, I don't see any reason to infer that when the box is perfectly rigid, the friction must vanish, and I don't buy the argument in the 2nd paragraph about wheels. I think the assumption of perfect rigidity simply makes the problem underdetermined -- it doesn't make F=0. Solutions with $F\ne 0$ exist for any finite rigidity of the box, so I don't see how one can argue otherwise in the limit of infinite rigidity. $\endgroup$ – Ben Crowell Oct 21 '14 at 19:13
  • $\begingroup$ Can you elaborate the objection to the wheels argument. $\endgroup$ – Daniel Mahler Oct 21 '14 at 19:16
  • $\begingroup$ The wheels argument would simply seem to be an argument that any time there is equilibrium, there can be no static friction; you haven't used anything specific to the present problem. But we know that static friction can exist when there is equilibrium. Changing a solid surface to a wheeled surface simply changes the behavior of the surface. $\endgroup$ – Ben Crowell Oct 21 '14 at 19:20
  • $\begingroup$ The wheels argument would simply seem to be an argument that any time there is equilibrium - This is not true the last two paragraphs use the wheels argument to show that there is friction in equilibrium in the presence of deformation. $\endgroup$ – Daniel Mahler Oct 21 '14 at 19:30
  • $\begingroup$ Solutions with F≠0 exist for any finite rigidity of the box, so I don't see how one can argue otherwise in the limit of infinite rigidity This is consistent with F being inversely proportional to rigidity. The wheels analysis of the non rigid case makes this seem plausible. $\endgroup$ – Daniel Mahler Oct 21 '14 at 19:34

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