Great Pulley Problem

[Parfait's Provider]

John, in your example you have two lines to support 20#, so they only need to have a working load of 10# each if they are both fixed at the apex of the pulley and not free to move. However, if the pulley is free to rotate, then the whole load is supported by one side and the pulley is merely changing the direction of the pull of 20#. In your example, the pulley is not free to rotate, thus, it has no affect and each leg of the line acts independently even though they are the same line. The pulley supports two 10# loads, one on each side, and doesn't have a rotational force on it. If the sheave is free to turn as a result of one leg being untied from the load, then the tension is 20# and your 10# line will break. Further, the load on the pulley axle will increase from 20# to 40#.

Think about that.

[Oswego John]

Carl,

In that last post of yours, you conveyed the thought that I was trying to explain as a solution. Your reasoning is a lot clearer than my attempt at an answer.

Your answer as to how the weight is transferred to the upper surfaces of the block in a variety of horizontal degrees is what I was trying to say.

As the weight of the 20# load is borne by the block, I was alluding to the thought of the weight resting on the wagon drawn by the horse. Re: a horse can pull more than it can carry.

In John's problem, there are two cords sharing the load. Two horses are pulling the wagon that supports the total load of 20#. Each horse is pulling half of the weight resting on the wagon.

This is still not coming out the way I had hoped it would, but it is as explicit as I can make it tonight. Maybe tomorrow I'll receive a new revelation, a flash of brilliance and the problem might be solved to everyone's satisfaction.

Have a good one,
O J

[Didereaux]

Can anybody help me with this? (Darmoose, where are you when I need you?) I’ve been struggling for years to solve the Great Pulley Problem.

It’s not easy to describe, but I’ll write this v-e-r-y s-l-o-w-l-y so that members of the Freewheeling Sect, who are slow readers, will be able to understand, even if they have to move their lips.

Imagine a single block suspended from a boom. Reeve a line through the block and let the two parts hang down either side. Fasten the two free ends together just above deck level, and attach a weight of 20 pounds at the bottom of the loop just formed.

Now, each side of the loop will be supporting a weight of 10 pounds. That’s because there are two lines supporting the total weight. But at the very top, over the top of the pulley, there is just one line. So this line must also withstand a pull of 20 pounds.

How is it that turning a line over a block enables that line to bear twice the weight it would normally be able to bear?

I mean, two lines with a breaking strain of 10 pounds each can support a weight of 20 pounds-—but where those lines turn over a pulley they become a single line, and therefore the loop should break after 10 pounds, but it doesn’t. It breaks after 20 pounds. This has never made sense to me.

The closest I’ve ever got to an answer is that as it goes over the pulley, the forces acting on the line change from torsion to compression, and then back again. But I can only comprehend that very vaguely.

Any brilliant ideas?

John V.

Oh for crying outt loud! AGAIN your off 180 degrees.
1. a block SUSPENDED with a line rove throuh and hanging downward with a 20# wt requires a 20# force pulling DOWNward on the other part. No splitting of load takes place' however the boom must now support a total of 40#...the 20 from the weight plus your 20 pulling on the weight. The fulcrum is the boom
(all that in a frictionless world: in truth you have to apply more than 20# to start lifting the weight and when it is then held in a static condition just 20# are needed to maintain the equilibrium.

2, IF however you tie one end of the line solidly to that boom reeve the line, and shakle the 20# to the becket on the block sitting on the deck and then pull UPWARDS on the other end of the line you need only apply a 10# force to lift the weight!!!!! The block is now splitting the load between you and the boom. the fulcrum is now the load(block)

THINK teeter-totter/seesaw! In your situation BOTH kids are on the same seat, in the second situation each is on a seat opposite each other. ( a terrible analogy, but it's late and thats the best I can do right now)

Why because THEN the block splits the load with you, the boom does 10# of work and you do 10# of work. (this holds more or less until you start pulling at an angle nearing or less than 45 degrees...then you gotta do the trig thing to determine the actual force vectors


But being one of those 'slow' types you refer to I may be wrong. And you have been sailing how long?

If memory serves my grandad explained that to me when I was 6 or 7 years old....and I understood perfectly well. But then he may have spoken veeerrry slowly.

Keep them cheap shots commin JV. heheh

[Ed Haley]

John:
Nice thought exercise.

I think you're using the word "line" in two distinct applications here and that is causing you grief.

There is the physical line composed of fibers and sheathing that has a static breaking strength of 10 pounds. This is simply the strength that the material will bear in any situation.

There are the supporting lines (2) that hold your 20 pound weight. These lines are "pulling" upward (strength of fibers) against gravity with a force of 10 pounds each, since each supporting line "shares" the load (and we are neglecting the weight of the line here). If you had 4 lines supporting the weight, each would have a tension of 5 pounds. In other words, if you divide the weight being suspended by the number of supporting lines from the pulley you'll get the tension in each line (again, neglecting the weight of the lines).

Above the pulley, whatever single attachment point you use to support the system is bearing the total weight of everything.

[John Vigor]

Perhaps I'm laboring under a misapprehension here, so let me restate the problem that's bugging me.

In strictly non-technical terms, we have a rope with a breaking strain of 10 pounds looped over a pulley block. The loose ends hang down, and at each end there is a 10-pound weight.

Now I can see that there is no problem about the rope breaking in the pieces that are hanging down from the pulley. Each rope is supporting only 10 pounds.

But the pulley is supporting 20 pounds. Now, where the rope goes across the top of the pulley, it is a single rope. It is not doubled. Why does this piece of the rope, with a breaking strain of 10 pounds, not break when it has to support a load of 20 pounds from the pulley?

This is a rope with a 10-pound breaking strain supporting a load of 20 pounds--and not because it's sharing the load with another length of rope. This piece of rope is single where it goes over the top of the pulley. The two lengths of rope that are sharing the load turn into one piece at the top. Why doesn't it break?

Or will it break? Am I laboring under a misapprehension? Have I actually got this all wrong?

John V.

[Ed Haley]

In strictly non-technical terms, we have a rope with a breaking strain of 10 pounds looped over a pulley block. The loose ends hang down, and at each end there is a 10-pound weight.

As I understand your setup in your hypothetical example, the pulley is not being used as a machine if the single rope ends hang loose off the sheeve of the pulley on both sides. The pulley is merely a hook. The tension on the loop on the pulley would have a tension of 20 pounds (tug of war, so to speak, of 10 + 10 pounds). The rope would break.

[Didereaux]

Perhaps I'm laboring under a misapprehension here, so let me restate the problem that's bugging me.

In strictly non-technical terms, we have a rope with a breaking strain of 10 pounds looped over a pulley block. The loose ends hang down, and at each end there is a 10-pound weight.

Now I can see that there is no problem about the rope breaking in the pieces that are hanging down from the pulley. Each rope is supporting only 10 pounds.

But the pulley is supporting 20 pounds. Now, where the rope goes across the top of the pulley, it is a single rope. It is not doubled. Why does this piece of the rope, with a breaking strain of 10 pounds, not break when it has to support a load of 20 pounds from the pulley?

This is a rope with a 10-pound breaking strain supporting a load of 20 pounds--and not because it's sharing the load with another length of rope. This piece of rope is single where it goes over the top of the pulley. The two lengths of rope that are sharing the load turn into one piece at the top. Why doesn't it break?

Or will it break? Am I laboring under a misapprehension? Have I actually got this all wrong?

John V.

Read my post previous to this. It explains it FULLY....or did I write to fast for you?

[Oswego John]

To W I M Concern,

One question:

Is the block that the 10# cord passes over free wheeling or locked?

O J

[Didereaux]

To W I M Concern,

One question:

Is the block that the 10# cord passes over free wheeling or locked?

O J

1. when the line is looped through a block which is FIXED it does NOT divide the forces, it acts solely as a turning point (when the two parts of the line are more or less parallel.)

Sol long as the line is a continuous piece so that the force(s) are transferred along its length all the block furnishes is a lack of resistance. However, if there were two lines tied separately say to a becket then once one line or the other is pulled hard enough to lift the weight the second plays no part(unless you pulled the first one at an angle.

On the other hand when you fix one of the ends of the lines and attach the pully to the weight THEN the block divides the forces making the thing attached to one end share equally the load from the end you are attached to.

What I have just described expalins why some get into trouble and pull pieces of deck hardware out of decks and such. Just remember the load divides if it moves, it does nothing if it is fixed: it simply redirects the DIRECTION of the force(s).

[Cathy Monaghan]

CLICK HERE to see how a block and tackle system works. This link is provided by Howstuffworks.com.

And CLICK HERE to see how the Amateur Radio Station K7NV describes the mechanical advantage of a block and tackle.

<center>
Image from Wikipedia


Image from Wikipedia</center>

Wikipedia's description and pictures are pretty good too.

And here's a military article on "BLOCK AND TACKLE, WIRE ROPE,
AND MARLINESPIKE SEAMANSHIP" if you want to get a bit more technical.

[Neil Gordon]

when the line is looped through a block which is FIXED it does NOT divide the forces, it acts solely as a turning point

What he said!

Okay... so if you tie a line to a 10 lb weight, you need to pull up with 10 lbs of force to lift the weight. 10 lbs pulling up + 10 lbs pulling down = 10 lbs of force on the line. (Think about it... if there's no upward force, there's zero force on the line, which lies on the floor in a puddle of rope.)

Okay, as Diderdaux said, a fixed block merely turns the lifting end of the line. Ignore friction and the forces are the same as before, yes? So pass the line through the block so that both ends pull down.

Next, instead of pulling down on the "lifting" end of the line, just tie a 10 lb weight to that end, too. The contraption is in balance... no friction and you still have 10 lbs of force on the line itself, right? Must be if all we did was turn the force, as Didereaux said!

Okay... last thing we do is weld the two weights together. No longer two 10 lb weights, we now have a single 20 lb weight suspended from two ends of a line/pulley system.

Can welding two weights together change the force on the line?

[John Vigor]

Thanks Cathy, I understand the theory and practice of mechanical advantage. That's not what puzzles me.

What I want to know is if the line I have described in my last post will break or not.

Are we all agreed that the line will break? I am battling a stubborn demon in the back of my mind that says it will not break-- but I can't understand why, or justify that position.

Perhaps if I just looped the line over a small peg, instead of running it over a block, it would clear up some of the confusion. If I hung two 10-pound weights from the ends of the lines, which can bear a tension of 10 pounds and no more, will the line break at the peg? Or will the line bear the full 20-pounds because it's being compressed against the peg, so that the force acting on it is no longer tension, but compression, down against the peg?

Can any of you mechanical engineers please help me deal with my demon?

Cheers,

John V.

[darmoose]

john, i may have misunderstood your initial premis, but i think i got it. let me pose it another way.

if we have a rope with a 10lb breaking strength, and we tie it to the boom and simply hang a 20lb weight on it, we can all agree, it will break, no?

if, we take our same rope, but instead of tying it to the boom, we simply reeve it through a block (whether it is free to turn or not), and then tie both ends to our 20lb weight, it will support the weight.

my first response was that the line would still break, but you indicated that it would not, and i presumed you have witnessed this. but, in your last post (you devil) you revealed that you dont know whether it will break.

i think it will break, unless you have actually done this, then i am with you.

hows that
darrell

[darmoose]

john,

i am going way out on a limb here (now there must be a puzzle there somewhere), but i think the line will break, as i said in my first post on this subject. especially since you have admitted that you dont know.

but, look at all the tortured explanations trying to show otherwise, this is amazing.

darrell

[John D.]

John V.

You seem focused on the rope at 12:00 on the pulley. At that point, it has only horizontal forces in it. By the time the rope gets to the top of the pulley, it is no longer supporting the 10-lb weight. By that point, all of the downward force has already been transferred to the quarter of the pulley with which it is in contact. There is no part of the rope that is supporting 20 pounds.

Draw a picture, and break the rope into segments like a stone arch. Do a finite element analysis for each segment, being sure to include the forces exerted by and on the pulley.

It also does not matter whether the pulley is free or fixed. Either way, all of the downward vectors are transferred to the pulley (or tree branch, or kingpin, or whatever)