Unstable Picture: CR4 Challenge (01/19/10)

Longer cord = more "wrap" on the nail = higher static friction = more stable.

Minimum length of cord = 0 (use two nails with hanging brackets attached to the picture frame).

I am not an engineer but it seems to me that the reason for the instability when using a short cord has something to do with the center of gravity. short cord raises the center thus promoting instability, longer cord lowers the center of gravity of the picture to the nail thus promoting stability.

All easy problems have already been solved.

Ahh - yeah, the center of gravity will be higher with a longer cord, increasing stability.

Mike

Keep in mind that, since this is being suspended rather than supported, a higher center of gravity gives greater stability; a higher center of gravity in a supported item gives lower stability.

"Keep in mind that, since this is being suspended rather than supported, a higher center of gravity gives greater stability; a higher center of gravity in a supported item gives lower stability."

What picture do you have in mind, I see a pendulum and the reverse of your statement.

if you mean the frame with v- shapped cord, It' s related to horizontal forces on the both side of the cord. If you use a long cord, the horizantal force on the cord is less amount than the contrary condition.

If we assume a frictionless wall/picture interface, a frictionless cord with finite elasticity so that it will always form two lines with an included angle intersecting at the nail and the possibility that intersection can move along the length of the string... we still have to evaluate the possible different effects of a wide rectangle frame, square, tall rectangle, circle or oval. Do we assume the cord attaches at the upper corners of a rectangular frame? or somewhat in from the corners (like real-world objects)? What rules apply to the placement of the cord on circular and various oval frames?

Maybe none of that matters... this requires more thought.

I think we should start by defining the position of the Centre of Gravity relative to the two hanging "point on the picture". The general solution will cover all such triangles, and, if we're lucky will only need to take into account the length of the base of this isosceles triangle.

While I have not had this particular problem, and simply have attached picture hanging wire basically intuitively across the frame back so it will hang right I can suggest that it is not really necessary to find studs and use nails for the hang point.

I have often used what they call at the hardware store zipits, which are basically plastic or metal screws with large threads that grab well in common sheetrock.

In their center, after entered into the wall is the hole that accepts a screw to which you may either attach a catch hook, or leave the screw head out some so as to catch the hang wire.

They are great little bits of hardware allowing you to put your picture pretty much wherever you might want, with minimal damage to the wall.

I often used them to hang lights off what are called pigeon plates on locations.

Is there a minimum length of the cord that will make the picture stable?

Yes. The cord should be twice as long as the distance from the centre of the cord to one of its two ends. No more, no less.

As for an actual value, I have a feeling Trigonometry is involved.

Yes. The cord should be twice as long as the distance from the centre of the cord to one of its two ends. No more, no less.

So, the cord should be twice as long as half it's length? Wow! GA for clever doublespeak!

Perhaps we may append... .... twice the length... as half the distance.... with a (nominal) length for wrapping back... no more, no less....

I would like to add one more detail.

Position the nail in the middle of the cord.

Assuming you attach a longer string lower on the frame there is no difference between the center of gravity of the mass and the point of suspension.

It isn't clear to me why, but I'm going to conjecture that it has something to do with the tension on the wire. A straight cord between two points has an incredible amount of mechanical advantage when pulled in the center and so for a given weight the tension on a straight cord is much greater than the tension on even a slightly longer cord.

A 30⁰ sling geometry I'd imagine - i.e. the angle between strings at the nail is 120⁰

However, in hanging one attaches the string to the frame such that the nail is concealed by the work. (as in art work)

This means the nail is about 1" below the frame top. So in the finish, that point relative to the CofG of the work, is the 'stability factor'.

That being the case, the anchor point can be higher and string could be 'straight' (though would bend slightly due to elasticity of string and frame)

Therefore minimum length is frame anchor point distance + tie off allowances.

Sensible length minimum (for frame forces) is 111.8 % of anchor point distance, plus tie off allowance.

I question the hypothesis. If the nail is directly above the CG of the picture, why would it move? therefor, we must assume a beginning eccentricity.

If the nail is not quite directly above the CG and the chord is very taught, it could slide all the way until the nail blocks against the higher end of the chord, think of using a rod instead of a chord.

But we have a chord.

If there is a beginning eccentricity, the chord on one side is longer than the other, for the CG to move under the nail, the picture will be rotated.

Without more information, I think it is a non(sense) challenge because if we use a longer chord, and again start with a beginning eccentricity, it will still move, less obviously because it will be in smaller proportion, but it will move.

We're looking for the length of chord where the picture "moves" from unstable to stable equilibrium.

Surely you both mean "cord" as in "piece of string" rather than "chord" which is a trigonometric term?

Yes there is a minum length of cord that will give best stability. It could also be very long and the picture hung from a hook attached to a picture rail. These are found in old houses that had high ceilings. The picture rail was about 30cm ( 1 foot ) down and went all round the room.

Intuitively i would guess that the shortest, and most stable would be to attach the cord to the top corners of the picture at a length that subtended an angle of 90 deg at the nail.

For this configuration the length of the cord would be d x √2 ; where d = distance between the attachment points of the cord.

If the picture were to hang free of the wall (using a long nail, say) it would form a pendulum, and any disturbance would make it swing. If it hangs against the wall, the longer the pendulum, the more likely is the friction on the wall to prevent it swinging. Next is the effect of friction between the nail and the string. if there is little or no friction (nylon string) then any disturbance of the picture may actually rotate it and again, the longer the string, the better. The question might better be framed: is there a maximum length of string following which no further improvement in stability is achieved in practice? Peter Harris

The key is the locus of the centre of gravity as deviation from the desirable hanging position occurs. If the locus is downwards the length is too short and is unstable. With greater lengths the locus is upwards. The minimum length for stability is when the locus is virtually flat.

There is a difference between a portrait or landscape hanging. For portraits, I believe they are always stable.

I think it's fair to assume that the Centre of Gravity (CoG) and hanging points on the picture frame are fixed. And therefore the general solution should cover all isosceles triangles with base W (the nominal width of the frame) and height h (the distance from the CoG to the line between the fixing points).

It is clear that the nail traces an ellipse with the two fixing points as foci.

We just need to work out how the length of the line between the nail and the CoG changes for small values of alpha.

Note that the centre of the ellipse is at centre of the base of the isosceles triangle. The ellipse does not go through the CoG.

This is a great question. I wish I had more time.

The curvature of the ellipse at the nail has to exactly match the curvature of a circle with its centre at the CoG.

I think I'll have a thrash at this one! Imagine the cord is taught on the back of the picture. When the picture is hung, if the centre of gravity of the picture is not directly below the mounting nail, the unbalanced weight of the picture will tilt the picture causing it to slide to the side with most weight! This sliding will continue until the picture falls on the floor or by some miracle, catches on the 'D' ring and hangs at an impressive angle!! Therfore....the longer the cord, the more margin of error between mounting nail and COG! (I'm starting to sweat now!!) This 'tolerance' also depends on the dimensions of the picture, ie a portrait will hang easier that a landscape! (I'm starting to dig a hole, I can feel it!!) So, if you place your pictures COG outside of the tolerance zone so that it's unbalanced weight causes one of the the mounting rings (on the picture) to exceed the height of the wall nail, it will slide and end up on the floor or at an impressive angle! (Are you following me!!) So, how long is the optimum length, simple.....long enough to hang the picture without seeing the nail, there's nothing worse that seeing the nail!! I would try to come up with a mathematical formula that includes the weight of the picture and the 'max out of tolerance' and the angle the picture moves through before one of the picture mounting points is on the same horizontal plane as the mounting nail, but I don't know where to start so I won't even try!

Anyway Happy 2010

I am assuming that the attach points are at or above the center of gravity of the picture.

There are only two situations. Either the center of gravity is lowest when it is centered, or it is highest when it is centered. So, you only need to consider two cases: when it is centered and the extreme case when it is suspended at one end.

Suspended at one end, the attach point on the picture and the nail will be the same point. The center of gravity will be the distance from that attach point to the center of the picture.

Therefore, for the picture to be stable when it is centered, the distance from the CG of the picture to the nail must be greater than the distance from the CG of the picture to one of the attach points.

NO.

It's all about where you mount the cord in relation to the frame.

Too low and the picture will flip upside down.

Higher up.. Viola!

...f you could remove all possible slack removed from the cord. then it might slip side to side, but only a frame and cord of infinite strength will allow for that.

Where I come from... any small disturbance that is moving framed artwork is a major disturbance

It seems to me that where on the frame the hanging points are place is as important to the stability of the frame as is the length of cord used. So, if the hanging points are near the horizontal axis, there will be less stability because of the relative position to the frame's center of gravity. If we treat the whole as if the hanging points were placed at the horizontal axis, then we can consider the frame as a bar and treat the problem from that perspective. In this case, the minimum length of cord used will equal 1.414 times the length of the horizontal axis to achieve stability (or, 1 1/2 times, more or less, the length of the horizontal axis). In this way, the cord won't show above the frame. The problem with this is that hanging from the horizontal axis will allow the the frame to tilt forward. This works in our favor as moving the hanging points upward will add more stability simply by taking advantage of the lower (longer ratio) center of gravity until we reach a point where the cord starts to show above the frame. Just a quick guess, though.

You guys are making this way too difficult. As a victim of picture obsessed mate/repressed interior decorator I have engaged this problem in a practical manner. I simply loop the wire over the nail and you have it. In fact it becomes an effort to move it if you don't get it level at the first attempt.

It's pretty straightforward. The distance from the center of gravity to the nail has to be greater than the distance from the COG to the hook eye on either side.

Now, there are some "rules" that framers work by. One is that the wire (cord) or hanger never be visible. The other is that the point of attachment of the hook eye be above the COG to avoid flipping and to have the proper tilt out from the wall; they usually use one-third down for this.

This means that "portrait" style pictures can be hung stably, but not "landscape" style. They deal with this by using two hangers, by adding "bumpers" on the two lower corners, or by adding a weight to the bottom rail.

Coming from a large family myself, adding to that my wife's large family and, not forgetting about my 3 step-children and 7 grandchildren... I hang A LOT of pictures.

I don't know the answer to this question and I probably never will because, from experience, wire is the hardest to use. Especially when hanging multiple pictures in an arrangement where location, in both planes, is critical. I use bracket cleats instead.

I found this article to be a pretty nice resource when figuring out how to hang the 20+ framed family photos I have.

But, I do have to admit that the proposed question is intriguing from a math/physics perspective... I may just have to start laying out a couple free body diagrams of some typical frame dimensions and see what I can come up with... although I can't do it now... need to actually start focusing on the tasks that pay the bills.

JavaHead

The hook of the picture is right at the back center of the picture, thats why when you used a short cord or rope it will be unstable

Rgds

Bai

For almost all pictures the center of gravity will be the exact center of the picture. On the back of the picture draw a circle from the center of gravity that is large enough to just go beyond the sides of the picture frame but not the top. Put your attachment points as high above the center of gravity as is possible but still inside the circle. Now attach your string so the hanging point is just outside your circle and the top angle is greater than 90 degrees. You may have to play with the circle size to do this. Now inorder to move off the hanging point you must raise the center of gravity so the picture will remain stable. As stated by others, landscape type frames can't be made stable and still hide the string.

GA to Duckinthepond. I have used the two hangers (hangers, nails, screws, etc.) in the wall technique to hang large pictures an mirrors with great success. Once positioned, they do not move. You can slide it left or right to adjust the tilt and the tilt never changes over time.

Although you answer solves the problem, it does not answer the question. You must be an engineer.

To answer the question, I offer the following: Attach the hanging wire to the back of the frame at about 1/4 to 1/3 down. The wire should be loose enough to nearly reach to the edge. With the wire pulled taught, measure the distance from the top edge of the frame to the apes of the hang wire. Measure the wall accordingly and sink the attachment into the wall.

If we are only looking for ways to stabilize, do what my Mum used to do, she would loop around the nail. From the top, twist a downward loop, hook it on the nail, center the picture and then let the wire take the weight.

i don't know whether this should be off topic or not; another with a fix, rather than an answer, has a GA vote!

The picture will be most stable when the string forms a 120 degrees included angle. The reason is that, in this situation, the components of the weight that act like tension stresses in the string, will be equal to the weight itself.

Have built many frames and found this an interesting question. I have always installed the eyelet screws on the frame about 1/3 of the way down the frame, then attach the wire with little to no slack. Too much slack in the wire lets the top of the frame droop down from the wall (which I find un-attractive). The no-slack wire method holds the frame to the wall at the points where the eye-screws touch and at the bottom of the frame. This gives a better appearance on the wall, and gives it greater stability.

Having hung many pictures, like everyone else, I know what I do (with 1 nail) but it took some thought to figure out why. Here is my shot at it - hope it makes sense.

A properly hung picture does not show the nail or cord and does not have the top of the frame lean out far enough to be obvious. Unless the cord can hang on the end of the nail at a point recessed inside the frame equal too or past the depth COG, the picture top will lean slightly away from the wall (this is most often the case).

The shortest cord is unstable for two reasons.

1) If the two picture support points are closer to the top of the picture frame, there is little "leaning" outward at the top and little or no friction between the wall and the bottom of the picture frame. Unless the picture is hung with the nail exactly vertically above the COG it is unstable because any resultant rotational forces on the picture is greater than the friction drag of the picture against the wall. As you lengthen the cord, you increase the "lean" and also the frictional drag of the frame against the wall. When the cord length is long enough for the support points to be at or lower than the COG, you start to experience undesired visual "lean" of the picture top away from the wall, but it does not tip (rotate) as easily.

2) The lateral component of the forces from the cord to the frame supports are the greatest and the vertical supportive forces are proportionally the least (and the cord contact with the nail is at it's minimum), so that any slight force or vibration along with any misalignment of the nail and the picture COG will cause the picture to easily rotate until the COG is aligned with the nail. As you lengthen the cord, you decrease the lateral component forces with respect to the vertical components. The relatively greater vertical components stabilize the picture because it will require a greater rotating force on the frame from a misaligned nail & COG to make it tilt from the desired "top being horizontal" position. As the cord lengthens, it's area of contact with the nail increases also, but cord slippage on the nail is not usually the cause of instability and not practically calculable.

I could be wrong, but because of these two contributions with the lengthening of the cord, you could not calculate a minimum length without knowing more about the geometry of the picture & frame itself. A simplistic approach would be a minimum stability threshold where the the vertical and horizontal components of the force at the frame supports are equal (cord is 45° from vertical), but depending on the picture size (W x H), and the other aforementioned constraints, this may not be achievable.

My method for this primitive 1 nail & 2 support hanging is pretty simple, but exact. As others have stated, this mounting will not work well with a picture that is more than twice as wide as it is high. Hold the picture and have your better half tell you when it is properly located (your opinion does not matter). Make a small mark with light, erasable pencil at the top and one side of the frame. Measure and make a vertical mark 1/2 the picture width in from your side mark. Measure from the top of the frame to the bottom of the top frame edge and make a mark at the center, just greater than this dimension, below the top of the picture you marked previously (put your nail in here). If they are not already present on the frame, install the 2 supports on the right and left frame (centered on each style) between 1/3 and 1/2 the picture height from the top of the frame. Attach the cord to one support and run it over your finger located at the underside of the top frame and attach it to the other support (simulating the actual cord placement). Do not cut off the excess cord until the top matches your mark and the "boss" has approved the result. Hang the picture cord and slide it on the nail until it lines up with your edge mark (hopefully placing the nail directly over the COG - checking for tilt stability in case the COG is not the center of the picture frame). If this is not stable enough (live near the railroad, etc.) stick a dab of tack putty at the lower back corners of the frame so that they touch the wall when released.

It' s related to horizontal forces on the both side of the cord. If you use a long cord, the horizantal force on the cord is less amount than the contrary condition.

Is not this a leverage problem?

Solving this practically seems relatively simple: either loop the cord round the nail or put in two nails, so, I thought it was better to look for the theoretical solution.

As I pointed out before it is clear that the nail traces out an ellipse relative to the two support pegs on the picture at the foci.

But it seems easier to plot the ellipse relative to a fixed CoG (Centre of Gravity)

I've done a spread sheet of the positive quadrant:-

I've made W (half the width of the picture) =1

Then for various values of h (the height of the fixing pegs above the CoG) I've done a goal seek to adjust a (the half cord length) until the ellipse approximates to a circle (centre CoG) at the top

Cell A15 is just h+b
Cell A16 is A15 + 0.00001 (the increment is increased gradually to stop the sheet using too many rows)
Cell F16 is =(D16/D15-1)*10trillion
column C is just a copy of column A so that the x,y plot comes out conventionally.

I did the goal seek to make cell F16 = 0 by changing a (F3)

Optimum values of a for various values of h came out as follows:-

Clearly when the cord is longer than these optimum lengths the picture is in stable equilibrium, and, when it is shorter the picture would be in unstable equilibrium.

Does this make sense?

Where's Fyz when you need him?

Hummmm the official answer has no explanation. But I added a small section to my spread sheet, and it looks as though my answer and the official answer agree:-

There must be some simple explanation for this, but I can't see what it is at the moment.

Just to confirm the relationship between the official answer and my laborious analysis of the ellipse.

Fyz has now supplied the formula for the radius of curvature of the ellipse, in particular at the end of the minor axis. R=a²/b

Using my original drawing, but now we can ignore d.

Effectively we are given W and h.

Now the ellipse curvature at the top has to "match" the curvature of a circle at the CoG.

R=b+h but R=a²/b
so a²/b= b+h
i.e a²=b²+bh
but a²=W²+b²
so W²=bh and re-arranging this
W/b = h/W

Those are the tangents of ½A and ½B (from the official solution)

I assume the drawing was distorted to allow you to fit everything in. As the distortion may add to the confusion, I'll take the liberty of commenting that, at equilibrium the quadrilateral becomes cyclic, which places the CofG very much lower than shown in your drawing.

Thanks again Fyz: the drawing is distorted because it's the one I drew originally, before I knew what the answer was. For completeness here's a 16 by 9 and a 9 by 16 picture with the ellipse; the circle with centre at the CoG which has the same curvature as the ellipse at the top, and, the circle through the CoG highlighting the cyclic quadrilateral.

I think you've just about covered everything of significance for the horizontally symmetrical case.
I'm thinking how to analyse the situation when the mountings are not equidistant from the CofG. I suspect that the cyclic solution may still apply* - but of course we have to add a constraint - that the two sections of cord must be at equal angles to the horizontal when the CofG is directly below the nail.

*This is just a guess, so I probably shouldn't have even written it. Whatever the result is, it (and its development) are probably well-known - just "not to me".

Fyz

To my way of thinking the "answer" is a hard way to make d a constant for n pictures.

(d being distance from nail to C of G, that has now acquired a circle of 2d)

But all it seems to achieve is 'equal pendulum' - on an arbitrary value based in artwork format ratio.

Seems far harder than a stick d long over which you could loop the string and tie it off.

As for 'optimum string' - is that the tensile for the land scape quadrupled for the tall portrait? so like 4 times L? (assuming equal mass)

Seems far more difficult than uniform sling geometry.

I await - still confused as to what, if anything, is being 'optimized' in any of this.

Aside from; "Yes Madam, all of our exhibits keep the same time"

Assuming that the "sling" is constructed from friction-free webbing, this should be simpler than picking up an oblong load inside the webbing - (2D instead of 3D).

I know nothing about the background, but I can only speculate that the difference is that with a hoist you can ensure that there is no slippage across the hook?

(At least the maths for mountings equidistant from the CofG is relatively straightforward)

Hi Fyz,

Hoisting "D's" may be;

2D (tie off to lifting eyes)

2D in effect; loops around the load, coming to a single strand, the hook engages (imagine a rope with a lasso at each end) This is used for instance in lifting bundles of 'sticks', you need to compress, or lifting a smooth object the load loops 'grip', to maintain sling geometry (not slide together), bottom lifting an object to keep it upright, or in fact to adjust/turn an object to upright.

3D; full loops to the hook - figure 8 sling - but only used if the loops cannot slide together along the load. Never used for a thin load, such as a plate on edge, or like this picture, as it can topple sideways in the sling (our picture can/will fall forward in "non gripping/containing" loops)

But, the sling geometry is overall, is the same rule of thumb. 45⁰ to 60⁰ sling angle (90 - 120⁰ at nail) sling tension is similar to load mass. 30⁰ sling tension is twice load mass _{(in principal for riggers and dog-men, not U lot right?)}

It is ill advised to go below 30⁰ as sling tension rises dramatically.

So if faced with lifting the portrait in this picture puzzle, (maintaining arbitrary d) you would use the bottom of the frame as the lift interface and either "load loop" or corner loop and place guiding eyes on the top rail to maintain the load upright.

For simplicity, using the bottom corners as defined tie off points (and adding frictionless perfectly aligned sling thru-eyes on the top rail) you would have a totally different sling 'nail angle' - yet identical 'pendulum' (assuming d = k).

So, given the top rail is structurally adequate those thu-eyes could become the tie off points.

This would then be the "shortest string" and equally stable as any other configuration (where d is constant) (and fair tension - whereas the 'official answer' is way not)

Or being bounded by the tie off points in the puzzle is "lineal thinking pre-cursive that makes the pendulum math consistent"

Hook slippage (read force to displace the nail along Randalls' ellipse - thus reciprocally lateral force to displace the load to cause slippage at the nail)

As demonstrated above, sling filament tension should be in the range of equal to x2 load mass.

There by, for a "frictionless hook" to slide on the sling it must change the tension in one or other sling filaments at grater rate than C of G change x mass provides (as you and Randal have demonstrated).

In the real world, with a properly slung load, you cannot move a hook once the mass is engaged. To balance a load level requires a good eye for symmetry of set up and load C of G assessment (and n trial lift/look/relax/adjust/lift/s)

The load may be balanced and level and the sling asymmetric (to an extent)

Hence my comment that the whole is a 'body in unison'

An interesting qualification to "you cannot move a hook" is; as the sling angle becomes greater than 60 (less than 120 at nail) moving the hook at greater 'taken weight' is progressively easier.

E.g. if you used 80⁰ (20 at nail) sliding the hook effort is lower, because sling tension is lower.

Or the lower the angle at the nail the easier to displace the nail along the ellipse as you are overcoming a friction component rising less against sling vector geometry change forces.

Similarly, at the other extreme, a virtually flat sling, this is also the case (due to reduced geometry induced forces).

Both meaning; with a frictionless hook and a squatter landscape, or taller portrait, this "rule" is totally ARSE ABOUT! in lateral disturbance venerability of the load.

Similarly in the landscape application, irrespective of width; the load will be stable using loops (or eyes) IF the sling geometry is between 30 and 60 degrees and symmetrical with C of G.

With such a long load you would aim for the loops (or eyes) at 25% in from the ends (you know, about where that guy said)

So;

The answer applied "stability" considered as an impact to the picture laterally and in line with the C of G.

Frictionless nail case

Original fig 1, the picture 'pendulums' at d set Hz and the string displaces at the nail. Eventually the sling tension re-equalises and the whole then settles to original position.

Portrait fig. the same displacements occur - but sling angle is far less re-centring, so string displacement will be greater. But we are frictionless - so it settles to original.

Landscape fig, as above - (frictionless)

Conclusions;

1. Anything frictionless will settle to original position – Math Proof Pointless

2. Any object 'hung' where d is some distance above C of G is 'stable'

3. Any group of objects of equal mass hung by equal d are equally stable.

Extrapolation

A, Change any or all d's in all as above - (frictionless) – and all that changes is the pendulum Hz

B. The only way to not have the picture return to equilibrium is to make the sling so flat that the nail can be 'further outside the circle about the C of G, intersecting the (effective) tie off points - than it was in the original d.

I.e. make the sling so flat that d can grow with nail displacement.

In the friction case and nominating a force such that;

Original fig will just pendulum and not displace on the string.

Portrait is highly likely to displace along the string, so not regain level

Landscape; displacing force is nearer the tie points, so is greater in effect in de-weighting the nearer sling filament, producing slip at the nail.

Therefore if 'stability' of fig 1 is unity, both portrait and landscape (hung to the rule) are less resistant to a lateral force and less likely to return to level.

But if all were hung employing a 30⁰ sling (top corner tie off - per puzzle) all would be equally resistant to displacement.

Just wouldn't keep the same time.

(as it happens this "rule" is not new to me – it was as practically flawed then as it is theoretically now. What it's use is; is hanging from a fixed rail at a uniform 'center of art' eye line)

Found these in another thread - stole them.

"stability"?

(Regard the end if the container as the picture fitted with "guiding eyes" on top rail)

note; the 'limiting' load shares (yellow) at "nail"

note; were there 'lock ins' at the braces near the 'scuff pads', the chain would be "more optimal" - and 'limiting' on load shares is no longer required ^{_{(other than for chain breakage, lock-in failure etc.)}}

On a slope

The guiding eyes make it very stable. If you draw a circle through the load-share and the two guiding eyes you will see that the upper arrangement would be stable with the CofG anywhere below the top of the text. And this is when the load sharing is perfect (no constraint or friction). The stability range will be even greater if the load-share on these hoists is designed to allow a mismatch but limit it to some proportion of the load (I have no knowledge of this area...)?

PS. I assumed the "features" at the top edge of the container were the guiding eyes. Rereading what Kyzine wrote I may have misunderstood (i.e. were these just the scuff pads?)

There are no guiding eyes on the container - the things on the chain are rubber anti scuff pads (it's a new unit, I'd say)

I was just using those for reference points where you would put the guiding eyes on our picture - or like the truck on the slope - the picture would float away from the sling - (fall forward)

The load share is just a bit of bar, in effect, with 3 holes, center being pivot, ends to chains. It is a limited seesaw and mainly to account for bent or load distorted containers. So you don't get all the weight on 2 chains (diagonals) and the thing flopping back and forth 'shocking' the slack chains (and whole caboodle).

But as you can see the container has to tip a long way to get its C of G outside the chain triangle. Though they are meant to be evenly loaded - many are not - yet very few are dropped by these rigs (and that's usually for other reasons)

So I hope you can see why I'm having difficulty with unquantified "stability" in the question/solution

Thank you, that is much clearer. So the limited load share effectively allows one of the chains to lengthen by some maximum amount while the other shortens.

In that case I'd say the support remains stable so long as the movement of the loadshare is insufficient to tip the CofG outside the triangle defined by the chains.

indeed - "stabilize" the C of G, and it's stable.

Fixing at guiding eyes would prevent this "tip" - and remarkably shorten the string

Or optimum string length is about tensile in sling geometry

The chosen or dictated attachment points, set practical string length for that sling.

(for that "optimized length" see my first post #10)

Silly old man - the requirement of equal angles between each cord and the vertical between the nail to the CofG implies that the four points are concyclic if (and only if) the ends of the cord are equidistant from the CofG.
So pictures with unequal distances between the mountings and the CofG will not have any equilibria (stable or otherwise) with the four points concyclic.

It also means that (for the original problem with cord ends equidistant from the CofG), if the cord is shorter than the minimum for the central ("straight") position of the picture to be stable, the other two possible equilibrium positions will be when the pin, the ends of the cord, and the CofG are concyclic.

It's now easy to show that these two equilibria must almost always be stable, because:
. Equilibria that are adjacent to an unstable equilibrium are either stable equilibria or are zero-slope inflection points - and these latter are extremely rare. (I suspect that these alternate equilibria are always stable, but formal proof eludes me at the moment).

Fyz

Are we sure we have this right?

If I redraw my string to replicate B on the landscape its one hell of a long string.

Redraw B on the portrait and it's remarkably closer to the G of G

Just asking

Yes I think so: not sure about your drawings but I think this probably agrees with what most people have said they've found in practice.

Got to go: talk to you tomorrow.

Hi Randall, I very much admire what you have done above, trying to put a number on this bit of string. I think we all thought it's length was the 'challenge", so be clear this is not in any way critical of you. And thanks for the sketch fixing.

It's the official answer that is confusing me.

So, looking at that;

I would contend that if the picture is adjudicated stable in the original fig. it is the distance to the center of gravity that is 'adequate', against the pendulum self correction properties,(pendulum length d)

Therefore; that d only need be duplicated to afford the same stability in any configuration.

In the landscape fig, the official solution, d increases so extraordinarily, to me, it means a huge excess of string.

In the portrait configuration, the official solution reduces d and produces such a 'flat' - so greater stress in string/frame - and means a less stable picture - or produces too short a string.

"Frame vertical" is another thing all together.

If there was zero friction at the nail, forces and angles in the "sling" will equalize - if the tie-offs are symmetric - the frame will be vertical.

So if the string is not "forced to slide" on the nail - the pendulum is vertical and the frame is vertical.

But given there is friction at the nail; you can cause the picture to not self level.

But, in both cases; the picture and 'sling' operate as a single body. (the shape of the mass is irrelevant - d determines stability)

So; I still fail to see any correlation (per the official answer) between angle B and "optimum string length".

Of course, the official answer and Randall's only apply if there is absolutely no friction. In this case the cord will slide until the centre of gravity is at its lowest possible point. Even the smallest amount of friction between nail and string is enough to make the perfectly-symmetrical position stable, and the only additional condition needed is that the line joining the ends of the cord passes above the centre of gravity.

Agreed; except for "official answer" applying.

What has <A to <B to do with anything?

"If the angle at the nail is bigger than the angle at the diagonals, the picture is unstable." = a totally erroneous hypothesis, as stability is quantified by d.

"For stability, the angle at the diagonals must be equal or bigger than the angle at the nail." - when extrapolated, to 'landscape', is meaningless in terms of;

"Is there a minimum length of the cord that will make the picture stable?"

and ridiculous in terms of lifting.

Imagine the 'landscape' "rule" applied to slinging a shipping container or rail car.

Any sling will make it stable - so long as the hook point is outside a circle centered at the CofG, drawn coincident with the sling attachment points. All 'angles' are about anchorage and stress in the sling.

But I am still open to an explanation as to how the "official answer" answers the question/problem.

Until then it remains to me like;

The platypus lives in the bank

There is a bank in town

Therefore the platypus lives in town.

"Any sling will make it stable - so long as the hook point is outside a circle centered at the CofG, drawn coincident with the sling attachment points. All 'angles' are about anchorage and stress in the sling."

I think you misdescribed circle. I believe that the CofG needs to lie on the circumference of the circle, not at its centre. The description using the two angles is exactly equivalent to this, because "the opposed angles of a cyclic quadrilateral are complementary"

Fyz

Very busy is what. The "answer" is, of course, only correct if the coefficient of friction of the the cord is zero and the picture does not rub against the wall. The CofG would then need to be below the centre of curvature of the top edge of the ellipse. Unfortunately, I haven't time to look for an elegant geometric solution. However, solution using the characteristic equation of the ellipse to define its curvature is relatively straightforward.

In practice, the vertical force on the cord is equal to the mass of the picture, and near the "unstable" equilibrium the offsetting force along the cord is quite small. This is why I am able to hang my pictures with their cords very much shorter than the challenge implies; this is just as well, as I prefer the cord to be so short as to be hidden behind the picture, and given lath-and-plaster walls, I prefer to minimise the number of holes...

P.S. I should have added some geometric details to assist others to compete the picture(?!):

Call the length of the cord 2.a, the distance between the mountings for the cord 2.c, the length of the minor axis of the ellipse 2.b. If you consider the length of cord required to reach circumference along the major axis, it is readily apparent that 2.a is also the length of the major axis.

The radius if curvature of the ellipse is R=(L1.L2)^3/2/(a.b) where L1 and L2 are the lengths of the two sections of the cords. At the desired equilibrium point the cord sections are of equal length a and the expression reduces to R=a²/b.

Clearly, in the absence of friction the neutral position for the centre of gravity is coincident with the centre of curvature, or a distance R below the nail.

Now, we can see that R/a = a/b, also that the line Nail-CofG coincides with the minor axis. This means that the triangle Nail.CordEnd.CofG is right-angled, and therefore that the angle Nail-CordEnd-CofG is equal to the angle CordEnd-CordEnd-Nail, which is equivalent to "the official answer" to the challenge.

Apologies for the lack of drawings and consequent non-clarity - I'm unlikely to get near a sensible drawing package for quite a while. So please feel free to draw, clarify, expand, whatever...

Fyz

Guys,

I guess that this must be classsified "Off Topic"

I hang all my picture with two nails, agreeing with Duckinthepond.

The pictures all have a piece of wire crossing the back of the picture from about 1/3rd down. Attached by screw eyes or whatever else is suitable for the frame materials.

The wall has two "nails" or fixings per picture.

There is no wire showing , and the pixtures remain relatively flat against the wall.

As an additional precaution I attach tiny "felt" pads to the bottom corners of the picture. This provides additional stability;though I do not believe that it is essential, but is very easy to do!

Also I have found that this is a good way of hanging a row of pictures along a wall all at the same height.

Sleepy