New Years Diversion

Ermm ... W/2?

Ok ... W/3

W/4 he says with boundless confidence

My guess was one of the above. Then I built one to see and was surprised. This may be why my mom always referred to my fourth grade education as "The lost years".

I think zero.

Drew

That means if you let go of the string, it won't move.

No, I think the mechanical advantage is zero, you will have to pull with a force equal to the weight of the object and extra tackle.

Drew

Zero (or undefined W/0), depending on which way you look at it) while the weight and left pulley free-fall; then the left pulley two-blocks into the weight pulley; and then the mechanical advantage goes to 1, with correction for friction of the rope being pinched/snarled between the pulleys. At that point, it is just one rope up to the top pulley and back down.

At that point, it is just one rope up to the top pulley and back down.

Actually, it's not. It is then a loop around the top and bottom pulleys, from which you are pulling a length of rope. For every inch of rope pulled out, the loop shortens by 1/2 inch. So the theoretical mechanical advantage is 2, but in practice, it seems to be about 1/2, because of the various frictions adding up.

Please check again. In the original picture, it appears that you would be lifting the weight by the rope on the right. However, when the left and lower pulleys collide, they and the weight are a unit. You are now lifting by the left rope, which just goes up over the top pulley and back down through the collided pair. Lift up = pull down = unity mechanical advantage, except for hard-to quantify friction. (Which could be as low as essantially zero if the pulley housings mate well without the sheaves or ropes coming in contact.)

Kinda like I said in #22, but I still think I should get some credit for being the firs to say it wouldn't work in #5 when I said MA = 0...well I didn't say it wouldn't work, I just said you wouldn't get any mechanical advantage.

Drew

Well... OK, I have awarded you the coveted GA, which is accepted as legal tender in many places. Although, I suppose that if the advantage is the usual "distance of rope pull/distance of load movement" then your answer would have to be "infinite" or "divide by zero error."

Right you are!

I was befuddled by laying the tackle on the floor in the double-blocked condition. Then, pulling on the tail 1" shortens the tackle (from pulley 1 to pulley 3) by 1/2", leading me to my erroneous conclusion. With the tackle hanging as drawn, then it works just as you described.

Gadzooks! Playing with ropes has left me daft. "(from pulley 1 to pulley 3)" should read "(from pulley 1 to 2)" or perhaps better, "(between pulleys 1 and 2)".

I get P= W/1 if the moving pulleys contributes no appreciable additional mass to be lift. At the pulley with the weight and W tension on its axis W/2 must be the tension for each string going up. At the fixed pulley this W/2 tension gets translated to the other moving pulley. The thread traveling over the moving pulley with a W/2 axis load adds the other W/2 so P=W/2+W/2=W/1

IT'S A FREE ENERGY DEVICE!!!!!

But, more seriously, I believe that if you pull on the string nothing happens because it is a "paradox" - but if you let the string go the weight will fall.

Take the middle (left) pulley and draw a free body.

1 P pulling down on left (tension in rope)

1 P pulling down on right (tension in rope)

which leads to the "upwards" force of 2P ........... BUT ............

This is where you divide by zero and the world explodes

It cannot be that the upwards force is equal to 2P, because that is the tension in the rope, which we already said has to be 2P..................

my head hurts.

CORRECTION (at the end)

................ which we already said has to be P ...............

The P=1/2W I reason it out the following way

OK first take the rope and tie it to the floor, now you need only to consider the tension in the rope pulling up on the floor

if W=100 lbs then the tension in the rope is 50 lbs acting on both directions

50 lbs <----- left WEIGHT right -----> 50 lbs

both ropes head upwards

Now trace the weight through the rope

On the right side the 50 lbs goes up through the pully at the top and pulls UP on the free floating pully on the left of the diagram.

The rope on the left side, it passes over the free floating pully (on the left side of diagram) and pulls down on the pully with 50 lbs (which pulls up on the floor = and oppossite with 50 lbs)

OK so now the pully on the left is in equlibrium with 50 lbs pulling up in it and 50 lbs pulling down on it.

P=50 lbs counter balancing W=100 lbs

So the mechanical advantage is 2 to 1

The weird thing when you build it I suspect is that the only thing that moves is the pully with the weight goes up and the pully on the left remains still

Guest 11 here

Laying awake in bed last night I was reconcidering this, and yea actually did deduce that as soon as you put more than P=1/2w force on the pully the pully on the left would move down. This is a cool problem because it is so finiky. Mechanical advantage is typicly thought of in terms of work, work typically being technically defined in physics as the displacement of an object one point to another.

I get it if you put the weight on the floor you can't lift it. What I wonder is what if you have a weight hung in the air, what really happens? Ok so the pully on the left moves down but what happens next. Do the pullys stop when the at the bottom is horizontal? Does the pully on the left go below the pully on the right, and allow the weight to go even lower?

This is a great premis for an engineering problem ( a mean one but a great thinking problem ) but to further develop it I'd like a bit more information.

Is there any way to get the riging to work, by pulling on the far right rope? Can you rearange the pullys and rope to create mechanical advantage? <-- same materials but make it work.

And what hapens to this if you put it into a computer simulator, does the computer model actually work or does the computer automaticly load the rope with tension equal to 1/2w + 1 = tension, and maintain equlibrium. I wonder if the 'phun' physics simulator would handle this type of thing.

Is equlibrium able to be concidered mehanical advantage?

And what is this other problem ?

First, yes, you can make it work by adding friction to the pulleys.

The other problem was described but not illustrated. It's a Spanish Burton hung on two ships at once?? I'm working on it, but don't have enough pulleys yet. Supposedly, unless you got absolutely everything perfect, it wouldn't work at all.

Post 7 describes what happens, but to add a bit more:

When the left and lower pulleys clash together, as you keep on pulling, the left pulley goes lowest. At this point, the rope goes up from the weight, over the top pulley, and then kind of zigzags through the touching lower pulleys. Apart from the friction there, it is just a rope up and a rope down, which requires W of pull (plus friction) to lift weight W. Thus mechanical advantage = W/W = 1.

Pretty useless except for entertainment--great puzzle, though!

Seems to me that when the pulleys collide it's analogous to a ratchet cable tie or hose clamp outcome - (but no friction) - hence diametrical force is Pi x P

You would need pulleys with 1/2 rope d deep grooves though, so they did not jamb by differential action of flange velocity v.s. rope velocity.

Mind you all this is predicated on the 'floating pulley' moving, so also with respect to the P pull point.

Given that "dynamic adjustment" to the diagram, the vector of P can be regarded as moveable toward horizontal, so the down force of P is no longer subtracting 100% from the lift, and the resultant force at the floating pulley axis no longer being P + P, and the floating pulley moving outward toward P pull point, which adds in a catenary or 'sling' leverage, somewhat reduced by movement of weight W to a lesser extent in the direction of pull P.

As Packy might say; Jus' pictur'in

It's W/2. The pulley in the left won't move.

Since the cable would have equal tension, the maximum tension woulld be under the top fixed pulley which is W/2. Therefore, the cable pull would be W/2.

Old age aint for sissies.

W/4 because the rope is shortening in 4 places. So ignoring weight of rope and pully it will balance when pulling force is w/4. To lift it means w/4 + an infinitessimal extra force.

w/3 using the same analysis as my past offering but now counting properly. By the way the relative sizes of the pulleys doesn`t matter does it?

To all who've answered. I loved this puzzle because it completely fooled me. I had to build it to see what was happening. I got it from a Royal Sea Scout site on tackles, but the next question is: What is a Royal Sea Scout? Is that like a US Boy Scout? Cause they have another one on there that also baffles me and I haven't been able to get one built well enough to see what they're referring to.

It's a slip knot! A pull on the end will cause the two lower pulleys to come together, the top one rotating anticlockwise. Unless the two wheels align perfectly, the system jams.

At equilibrium, the tension in the rope is constant, otherwise the pulleys would begin to rotate (couple=0). Also the sum of forces applied in opposite directions must be equal. For weight W not to move, the force applied by the ropes must be W/2. Therefore, P=W/2.

Well, not to let it run on too long. You can't lift the weight by pulling on the rope. That just pulls the left hand pulley down. So, the mechanical advantage is either 0 or minus something (I don't even know how to calculate that). When I saw this, I was absolutely sure the pull had to be W/2. Then I built it and pulled, and...oh, crap. Those doggoned Sea Scouts had gotten me good.

Guest (post) #9 pats himself on the back.

Oh so the weight is resting on a surface originally. I assumed that is was initially suspended.

Oh, you're right! Darn! I drew it without the floor. Now I have to put it back together again. Well, that's a lot more important than looking for a job! Thanks for pointing it out.

Drew, Post #5 pats self on back.

Drew

This arrangement reminded me of a Spanish Burton rig (upside-down maybe). My reflexive guess was 3:1 (or 1:3, whichever way you want to say it). But then it looked like the left pulley was free to move downward. My earlier posting, however, was not just a guess. That is EXACTLY what happens. By coincidence, I had 3 pulleys on a shelf for an upcoming project, so I just rigged it up with a 25-lb test weight.

In the US, the Sea Explorers is one of the Boy Scouts senior divisions. I guess the Sea Scouts would be equivalent.

Yep, my next project, when I steal find some more pulleys is to rig a Spanish Burton. Supposedly you could do it on two ships at once?? Why, you might well ask, do I waste my time with 17th century Royal Navy rigging instead of working for a living. Say, have I told you about my plans for a forge?

TYP.

If no friction, why isn't the answer ∏ P ?

Kz

Meaning ∏:1 ratio (when said in the normal math way, as opposed to first blab off the scatty brain)

p=w×2 for more accuracy add gravitational force

actually we are lifting both pulley and weight we need a power to lift weight and pulley weight at a time i think is hint for those person who can calculate it

The length of the rope from the left pulley, over the top pulley, around the bottom pulley, and back to the first pulley cannot change without the rope slipping. The result is that none of the pulleys can turn and it makes no difference what force P is applied, weight W will not move.

Well, I went back into my lair and rebuilt this last night.

First, I sometimes blast guests for being anonymous, but Guest 9 really gets a "tip of me hat" for a terriffic answer!

Anyway, there's no way you can lift weight W until (are you ready) you add friction to the sheaves. Is that wild or what? If you wrap string around the sheave axles and get enough friction, it turns into a reasonable pulley system. My estimate would be that it takes a minimum of about 0.3 - 0.4 friction on each.

I was limited in my research by my wife reminding me of a long list of chores I'm neglecting and how I'm not making a penny for all this work (while neglecting some work I actually do get paid for). God knows what trouble I'm in when she finds out I'm gonna try to make a longbow.

Anyway, it gave me a new perspective on how friction (damping) can actually be desirable.

Guest #9 here again.

Thanks TVP

I should have caught this earlier (thanks hangover)

My world crushing singularity was due to the fact that I was thinking in statics - that the left/middle pulley would not move.

But if you look at the freebody another way:

1 P (tension in rope) pulling down on left side of pulley

1 P (tension in rope) pulling down on right side of pulley

1 P (tension in rope) pulling UP on pulley

Leads to unbalanced forces on the pulley, in which you can then say that the pulley will accelerate downwards.

So much clearer now

Hi TVP45 and a happy new year.

(Sorry for my late reply but I had some days off and now I'm back to work again... ...)

I think that the answer is the following (I made a quick draw):

As you can see, the force gain is 2, i.e. you need W/2 in order to rise a weight W. (Of course, the height will be the half in order to produce the same work.)

Another way to find this out (as far as I can remember) is that the force gain has to do with the supporting points of the rope. There are two such supporting points: A1 and A2 (which are shown in the following figure).

Hence, once again, the result is that F=W/2.

The system is unstable. The pulley on the left would drop down, level with the one on the right, and become an idler. Then both pulleys, as the joining rope becomes horizontal, would move together and the two pulleys would lock together and act as one. Mechanical Advantage would then be 2.

Imagine that the weight is sitting on a table, there is no tension, add weights to a hook at "P". As soon as there is tension in the rope, the left hand pulley has twice as much downward force on it as upward force. The only effect would be that the left pulley lowers and moves to the right hand pulley.

To me it seems that no force applied at all leaves the weight stationary! Lets explain it quickly, as I have to leave internet soon:

Consider the system in equilibrium - by applying the appropriate force to the end of the rope, so there is no acceleration in any part of the system.

First of all, the force that the lowermost pulley experiences in its both sides, is W/2. Consequently, the force that is applied to the topmost pulley on its right side is also W/2 (due to the 1st Newton's Law). Then, the force applied to the left side of the same pulley must also be W/2. Already, at this point we see that the total force that is applied to the topmost pulley from the ceiling is W (to keep it in place), that is as much as the load! So this is enough to keep the system in equilibrium without any extra force applied.

I will sleep on it tonight, and hope I don't see nightmares!

Come on people. Do you realy think that this system will be stationary or unstable or whatever??? Let's simplify the system considering the following figures:

Figure A shows the whole system. The overall force on the pulley 3 is 0 (as two forces F are applied in opposite directions). As this pulley is steady, we can firm this pulley and the associated edge of the rope, as shown in figure B (i.e. this is equivalent with figure A). Furthermore, we can remove the pulley 1 (it simply redirects the rope), as shown in figure C. Now we see that the pulley 3 has a "gain force=1" (it simply redirects the rope) so we can, also, remove this pulley, as shown in figure D. The result is that only the pulley 2 remains and this pulley gives us a "gain force=2".

And this was my initial statement.

"And this was my initial statement."

Which was, and still is, wrong. See Blink's link in #44.

[I'm not suggesting that my analysis was any better].

No problem with steps B, C and D, but A has a fault I guess. As the rope is stretched, it applies an equal force on both sides of pulley No3. So middle pulley experiences a downward force of 2F and an upward of F, therefore it cannot be stationary!

Yeah, the A was wrong. We have not egual and opposite forces applied on pulley 3. Taking a more careful look at the system I see the following:

The weight W is transfered on the two parts of the rope which supports the pulley 2, i.e. we have W/2 on each part. The "W/2" on the right part of the rope is transfered -through the pulley 1- on the pulley 3 (i.e. on its connection with the rope). So the force F that we apply -in order to overcome the weight W- must be: F=W/2+W/2=W.

Hence it is F=W and we get no force gain. I think this is right, don't you think???

(I haven't seen the video yet. But I'll see it.)

As I see it, if you apply no force at all, then the rope on the right of pulley 3 will not have the W/2 tension. Or if it already does have it from before, it will move so that the tension is released. In such a case the weight will not be supported.

Now, if you pin the rope to pulley 3, then the forces will be as depicted, and apparently this pulley, as well as all other ones will be in an equilibrium. Nothing will have to move.

If on the other hand, we apply a F=W/2 force to the rope to sustain the tension W/2 on the right side of pulley 3, then we observe that this pulley will have a net downward force applied to it (equal to W/2). Therefore it will move. The most interesting is that the other forces will not have to change! The picture will be as shown in the figure. That is, there is no force transfer to the other pulleys and they will consequently keep happy by being in equilibrium.

Hi Tkot. I have the sense that you complicate things without reaching a conclusion. E.g. "if you apply no force at all" the weight, of course, will fall down. Or "if you pin the rope to pulley 3" the whole system, of course, will be steady. All these tell us nothing.

"If on the other hand, we apply a F=W/2 force to the rope to sustain the tension W/2 on the right side of pulley 3, then we observe that this pulley will have a net downward force applied to it (equal to W/2). Therefore it will move..." No, it seems that (if you apply F=W/2) pulley 3 will experience an upward (total) force (the other "W/2" applied on the top of it by the attached rope). Of course, pulley 3 will move upwards and pulley 2 will move downwards, as the F=W/2 is not enough to support (overcome) the weight.

Anyway, the issue is to see what force is needed in order to support (overcome) the weight. I have the sense that the figure and the result of my previous post is right (although I was wrong about my estimations on my other posts). I haven't seen the video yet, but let me make a guess: If I'm right, I'll expect to see that if you pull the rope at distance S, then the weight will rise, also, at distance S (and, of course, pulley 3 will move upwards at distance S too). And this will be so because we have force gain=1 (i.e. F=W).

(I'll see the video and I'll estimate whether I'm right or not.)

Well, I saw the video... Amazing... ...

Oumf... All my efforts were a total failure... ... Hmm... At least I tried... ...

Now, if you pin the rope to pulley 3, then the forces will be as depicted, and apparently this pulley, as well as all other ones will be in an equilibrium. Nothing will have to move.

Yes. If you put a knot in the rope so the tail cannot pass though pulley 3, you can move pulley 3 up and down easily, making no change in the length of the loop around pulleys 1 and 2. With the knot, pulling on the tail just causes the loop to circulate, until the blocks collide. Without pulling on the tail, pulley 3 will maintain its position, with no change due to increasing load (such as pulling down on the weight).

Without the knot, the tackle can be held as drawn by George, with an unsupported load, and if a force of about 1/2 w is applied to the tail, the system is stationary. (This is in real life, using reasonably low-friction pulleys, and relatively soft appropriately-sized rope -- but a relatively light object to lift -- about .5kg.) If you then pull downward on the tail, the weight does not move, and pulley 3 moves downward (as described above, and as seen on the video). However, this effect is dependent upon friction. With the pulleys and rope I am using, in this situation the rope does not cause pulley 3 to rotate, (and the tail length out from pulley 3 does not change) but pulleys 1 and 2 rotate as expected.

However, if you then reset the tackle so that pulley 3 is again free to move up and down, and then ease off on the tail, the weight lowers, and pulley 1 does not rotate (in my real world system). You might expect that simply pulling on the tail would then restore the weight position, but it does not (instead the situation above occurs). The system behaves as if is has a (somewhat unpredictable) ratchet.

As TVP has said, little changes in friction change the way the system works.

With low friction pulleys and very light line (which has low internal friction from bending) the weight of pulley 3 prevents the system from being set up as drawn: pulley 3 simply falls until it collides with pulley 2. (You can see this in the video -- the student is applying a very low force to the rope, and pulley 3 almost free-falls.) So real life and idealized physics are quite different in this case -- which is why the explanation with the video describes this as a "trick". (The pulleys are not weightless, the rope has significant bending friction, and the pulleys have a small amount of friction.)

It's worth playing around with real ropes and pulleys, because some of the behavior seems paradoxical (such as pulling in not having the opposite effect of easing out).

I agree generally with what you say. Only that I don't find the following correct:

"Without the knot, the tackle can be held as drawn by George, with an unsupported load, and if a force of about 1/2 w is applied to the tail, the system is stationary"

Maybe I don't get what you really mean, but if you apply a force W/2 to the rope tail, then the pulley 3 will not stay stationary. There will be a net downward force. Alternatively, you can see it if you consider that the rope does indeed form a loop around pulleys 1 & 2, so by pulling the string the loop will rotate, i.e. the pulley 3 will move down.

(By the way, I haven't seen the video yet due to company restrictive policies... maybe because it's a trick and may set my PC in danger!)

The video is part of a physics lecture series from the University of Maryland. Some of the demos are animated, like this one; others are not. No detected computer dangers so far. Try this at home if the company doesn't allow.

As you may know by now, the weight W must be supported, or else it just falls, as does the left pulley. It takes about 3 hands to set this up according to the original drawing. Good thing I once learned how to juggle (a basketball and two ping-pong balls being my biggest challenge. But I ain't the Karamazov Brothers--I don't do chainsaws!) For juggling, search out the Chris Bliss video "Fugly."

Hope this isn't too much of a spoiler ...

Maybe I don't get what you really mean, but if you apply a force W/2 to the rope tail, then the pulley 3 will not stay stationary.

As the device is prepared in the video (which is a little tricky -- see Tornado's most recent post) pulley 3 does indeed drop, and with very little pull on the tail.

However, for my statement to be true, the conditions in parentheses after it are necessary. There is a surprisingly large difference in the way the tackle works in the real world vs non-real world, and with variation of rope internal friction and load. (Without any friction, but with pulley mass, then the whole tackle just falls apart -- there is nothing to support pulley 3.) TVP referred to the effect of friction changing the way it works, and it really does.

The internal friction of the rope is a large contributor to allowing the setup to be staged as it is in the video, which you really must try to view. Better yet, though, is to actually build up the tackle and play with it. Without too much trouble (finding the right load and rope stiffness) you can have it release the load at a very apparent 2:1 ratio, (in which case pulley 1 does not rotate) but then it changes to its "sliding useless" mode when you pull back in (with pulley 3 not rotating and both 1 and 2 rotating). Lay the whole thing on the floor, and it acts, generally, as an ordinary tackle would if it is on carpet, but pretty unpredictably on tile.

I wish this site had video abilities -- I'd take a video and post it.

But definitely rig up some pulleys and try different sizes of rope. It is pretty entertaining. Probably best not to do this at work -- they will think you've gone mad.

Hi Tkot. I understood what you meant by saying that "there will be a net downward force". My previous drawing was (in a way) incomplete. The completed drawing is as follow:

We need to apply a force W/2 at the end of the rope (green arrow) in order just to keep the rope tight around the pulley 3, i.e. just to overcome the W/2 that exists on the rope (at the right part of pulley 3). Just by applying this W/2, the total force on the pulley 3 is a "downward force W/2" (i.e. there are two "downward W/2" and one "upward W/2" applied on this pulley). The result is that -obviously- the pulley 3 will move downward. In other words, the system is unstable and it will collapse.

A is not equivalent to B.

In actual operation, pulley three does not rotate as one pulls on the free end of the rope, as you can see in the video: the hand and pulley 3 move in synch. The rope around pulleys 1 and 2 simply circulates. The system acts like a continuous (endless) loop around pulleys 1 and two to which you have tied a pull cord at the location of pulley three.

Not until pulleys 2 and 3 collide does any meaningful force get transferred.

"Not until pulleys 2 and 3 collide does any meaningful force get transferred." Or, so far, only 'taking up the slack' has been considered by most.

If however you consider that the falling of the floating pulley brings the line substantially horizontal - then you have a "Superman lift" - which is quite an interesting "new view" of what W and the floating pulley then do.

On the other hand she could drill a hole in the table. Depends on how wedded to "well, the rope was hanging vertical at the start" she is.

Oh, thanks a lot. Now I'll have to diversify and pull my string from several new directions. Please don't expect any publications on it, though.... God knows what the peer review process would look like!

writedit.wordpress.com/.../

Good Guess!

I guessed that this would have to be called a Fool's Tackle, and searched accordingly. Sure 'nuf. If you click on the photo on this page, you can see a video of what happens, provided the weight is sitting on a table when the pulleys are positioned more or less as drawn.

Fun stuff, TVP.

Ah yes Blink - but the video ends too soon - continue and you get what I described - new P vector and all.

Blink, nice research to capture precisely what was intended in this challenge. I have a request for anybody who has access to a set of pulleys. Can this Fool's Tackle configuration statically support the weight. If it can will the tension P be equivalent to the weight W? I do realize that it cannot lift W so the weight will have to be supported by something first and then the support removed. I'm curious if my wrong assumption that the weight was initially supported by this tackle can happen. I would expect that if it can, any increase in the downward force applied at P will only cause the free pulley to lower. But what will happen if P is decreased, will the free pulley be lowered or will the weight and pulley be lowered?

I sail, so I have a few pulleys and some rope around. I'll fool around tomorrow.

I played around with some pulleys.

I am now on my way to the loony bin, but will report as best I can while they have me strapped down.

The results are remarkably sensitive to pulley friction, rope internal friction (bending resistance, hysteresis) and pulley weight. With free-running pulleys and relative small rope size (I used pulleys about 50 mm diameter and 5mm rope) it is hard to set up the demonstration as in the video: the weight of pulley 3 (using George Kokatos's numbering) causes it to drop, pulling all the line through.

With thicker line (about 8mm), I could arrange the pulleys so that they could hang in air from my hand, with the weight of pulley 2 more or less balancing the weight of pulley 3 and the tail on the rope: this looked just like TVP's drawing. To get this to work, I used the smaller sheave on a double block (25mm od).

In this configuration, if I tensioned the system, the pull on the tail was about 1/4 the load force, I'd estimate, to maintain a static condition. (This reminds me of the winches used in trimming lines when sailing -- a couple turns around the winch provides so much friction that only a small amount of tension needs to be maintained on the tail to hold a large tension on the line to the sail.) The pulleys I used were low quality (by sailing standards) with plastic sheaves riding on a stainless steel pin. They develop noticeable friction pretty quickly when loaded as does the rope as it bends around small radii. The actual mechanical advantage is not exactly calculable in the usual sense, because with low tension on the loop between pulleys 1 and 3, pulley 2 just tends to fall down when the tail is pulled down.

Once pulley 2 has fallen down, if you continue to pull down on the tail then the weight would be lifted with a 2:1 advantage, but the friction in the colliding pulleys is so high that the actual advantage is nearly reversed: maybe 1:2. This of course depends on the nature of the pulleys and how they are colliding.

If you lay the system out on the floor, the situation appears to be completely different. If you hold the tail of the rope against pulley 2 where it enters, you can move that block up and down (side to side on the floor) and the rest of the rope just circulates, making no change in the distance between pulleys 1 and 3. But if you simply pull the rope tail "down" the tackle seems to work pretty much as you might think it would -- in other words that tends to pull pulleys 1 and 3 together. So with the pulleys laid out on the floor, you'd swear you have a different system altogether.

All this may be clear as mud -- you really have to get some pulleys and play around -- it is pretty interesting.

But the short answer is that the system cannot support or lift any weight to speak of until pulleys 2 and 3 collide -- and just getting it to stand still (as in the video) is a little tricky.

Thanks for the effort.

I just realized that I am switching "pulley 3" with "pulley 2" in some cases in this post. (I could while away the hours, conferrin' with the flowers...)

It depends also on whether you continue to pull P straight downward, in which case you get what I already described. The table limits the girl's hand movement, but if the rig were at the edge of the table and she just kept going straight past the table's edge....