Mechanical advantage of a moving pulley vs an anchored pulley

[Rolacycle]

Homework Statement

I understand how to calculate mechanical advantage ratio, I think. The question I have does the MA change when a moving pulley (p6) is the anchor point of another pulley (p3) and does that then make (p3) a moving pulley thus effecting MA ratio?

Relevant Equations

I’m not sure

This is deer feeder pulley system I plan to make unless there is more efficient way to use the same 6 pulleys and pull load centered between the two poles

 

[hutchphd]

What is it you are trying to maximize with your pulleys? This certainly doesn't lift max weight. Do you need side to side freedom etc etc. I have no experience here.
Also pulleys will equilibrate as required by the various Forces (Tensions). Your drawing is not accurate...the tension is the same everywhere in the rope (~30lb in this rig) because that is what a rope does. The top rope, for instance, will not be that straight.

 

[Rolacycle]

It’s about lifting and remaining centered in between the 2 poles holding all this up, not necessary the most effective MA system. And no you’d only raise load to point where p6 stays centered

 

[hutchphd]

Do you want the pull rope to be at the pole? And how high do you need to lift...this will get harder with altitude

 

[Rolacycle]

Do you want the pull rope to be at the pole? And how high do you need to lift...this will get harder with altitude

I certainly don’t want to stand there and hold it up. And it’s got to be where it can be tied off and secured.

 

[Cutter Ketch]

I do not think this will work quite the way you want. It would require high tension to keep the rope as flat as it is at p6, but it clearly isn’t that tight as can be seen from the angles at p4 and p2. What will happen is the angles at p6 and p4 and p2 will flatten out more or less together. This isn’t necessarily a bad thing, and may constitute some extra mechanical advantage, but it seems overly complicated and unnecessary. Also, if you feel strongly about getting the rope relatively flat at p6 you would need several hundred pounds of tension in the rope at which point the lower rope would also be flat.

It would be easy to make it work the way you want. Why not get rid of p6 and p5. Make the top rope a separate fixed rope tied off at both poles. You will have to get a lot of tension (several hundred pounds) to keep the rope from sagging too much, so use a strong cable and a turnbuckle or similar. The lower rope would then also be tied off at 5.

Me personally, I would probably just go p1 to p3 to p2 to p3 to p2 to p3 etc until I had the mechanical advantage I want. Those ropes pulling at angles don’t add as much to the mechanical advantage. However, I see that your way is stable against swinging in one dimension, so that may be important to you.

 

[Cutter Ketch]

P.S.: as it will take a lot of tension to keep the top rope flat you will need strong well anchored posts. You could alleviate that by bracing the tops of the posts apart from each other, but if you are going to do that, why have the top rope at all? Why not make a rigid structure on top?

 

[Lnewqban]

Please, see:
https://www.unistrutohio.com/search?q=Wheel

 

[haruspex]

Not sure what other constraints you have, but I would either lean the side posts into meet at the top (that will,probably require taller posts) or replace the rope section joining them with another post.
You only need four pulleys.
Fix P3 to the top centre of the post structure.
Fix one end of the rope directly above the left side of P4. Take it down and under P4, up over pulley P3 and down to P2, as you have it, then almost vertically up to P1, and finally down to the bracket where you will tie it off.
You can offset P1 a bit to the right. Ideally, when the feeder is raised, the rope should make about the same angle to the vertical on each side of P1.

 

[hutchphd]

P.S.: as it will take a lot of tension to keep the top rope flat you will need strong well anchored posts. You could alleviate that by bracing the tops of the posts apart from each other, but if you are going to do that, why have the top rope at all? Why not make a rigid structure on top?

Its all one rope so the top rope cannot be independently tensioned. This will be flat only if the other ropes are flat...hence the question "how high do you want to lift the weight" which I ask again.

 

[Rolacycle]

Not sure what other constraints you have, but I would either lean the side posts into meet at the top (that will,probably require taller posts) or replace the rope section joining them with another post.
You only need four pulleys.
Fix P3 to the top centre of the post structure.
Fix one end of the rope directly above the left side of P4. Take it down and under P4, up over pulley P3 and down to P2, as you have it, then almost vertically up to P1, and finally down to the bracket where you will tie it off.
You can offset P1 a bit to the right. Ideally, when the feeder is raised, the rope should make about the same angle to the vertical on each side of P1.

Yeah I actually was thinking just before I read your reply about using separate rope not connected to load/tension then anchor a pulley to independent rope, thus not requiring extra pulleys to center the load. This is variation with extra pulley and rope ending at post not pulley.

 

[Rolacycle]

Its all one rope so the top rope cannot be independently tensioned. This will be flat only if the other ropes are flat...hence the question "how high do you want to lift the weight" which I ask again.

The higher you pull load the more tension is applied to p6 thus flattening the leg it’s on. And as that gets closer to 180° p2 and p4 get closer to p3. So yes all pulleys will get closer to 180° in relation to pulleys on either side. The maximum height before system passes the center of the load is 15’. So p6 with load fully raised will find center at any length below 14’11”. And I’ve attached a more accurate drawing of pulleys when load is raised. The first pic was more to show where the pulleys need to be to raise load in center, not to accurately show rope lengths and angles created by pulleys.

 

[Rolacycle]

Its all one rope so the top rope cannot be independently tensioned. This will be flat only if the other ropes are flat...hence the question "how high do you want to lift the weight" which I ask again.

Like this with 2nd rope, just to set fixed anchor point for top pulley’s. I believe

 

[hutchphd]

If that is an accurate depiction of the geometry the tension in the rope might need to be hundreds of pounds and the force pulling sideways on the posts will be nearly twice the tension. Can the posts be higher? Or can you put a beam across? This is the same issue you get putting up a hammock with very little "bow"allowed...the tension goes to huge.

 

[jack action]

unless there is more efficient way to use the same 6 pulleys and pull load centered between the two poles

This would be my way to use more efficiently 6 pulleys:

You need two cables (anchored at the yellow points). The 2 top pulleys are connected at the center of the top cable (thus centering the load). The looser the top cable is, the less tension will be in it. The 5 pulleys over the load gives you 5 nearly vertical load carrier (which is what you want) and a sixth one going to the pulley on the pole ($\approx$ 45°). This gives you a MA of about 5.7, i.e. you need 18 lb to lift the load. This is about twice as efficient as your arrangement in your OP.

You could also connect the pulled cable to the other pole instead of the top pulley (or top cable). This would reduce the MA ($\approx$ 5.4), but might give you more stability as the lower cable will equally pull sideways on either side of the load.

 

[Rolacycle]

If that is an accurate depiction of the geometry the tension in the rope might need to be hundreds of pounds and the force pulling sideways on the posts will be nearly twice the tension. Can the posts be higher? Or can you put a beam across? This is the same issue you get putting up a hammock with very little "bow"allowed...the tension goes to huge.

The 6:1 mechanical advantage system only requires 16.6 lb force to move 100lb load. And it doesn’t have to get to maximum height/ “little bow” to get +8’ off ground, yes I do understand that closer to maximum height it will get exponentially heavier especially the last few degrees as the rope nears 180°

And I probably could add beam in between but was just going for quick and easy not 100% everything it could possibly be. Just old telephone poles few pulleys and couple screws. But the poles are buried 8’ deep so any horizontal force on them is likely moot, I could probably hit it with my truck and not be enough force to change angle of pole, but point taken

 

[Rolacycle]

This would be my way to use more efficiently 6 pulleys:

View attachment 258063​

You need two cables (anchored at the yellow points). The 2 top pulleys are connected at the center of the top cable (thus centering the load). The looser the top cable is, the less tension will be in it. The 5 pulleys over the load gives you 5 nearly vertical load carrier (which is what you want) and a sixth one going to the pulley on the pole ($\approx$ 45°). This gives you a MA of about 5.7, i.e. you need 18 lb to lift the load. This is about twice as efficient as your arrangement in your OP.

You could also connect the pulled cable to the other pole instead of the top pulley (or top cable). This would reduce the MA ($\approx$ 5.4), but might give you more stability as the lower cable will equally pull sideways on either side of the load.

Yeah I see the flaw of my original concept was single rope, instead of independent top pulley anchor line/ or beam.

 

[Cutter Ketch]

Its all one rope so the top rope cannot be independently tensioned. This will be flat only if the other ropes are flat...hence the question "how high do you want to lift the weight" which I ask again.

My P.S. went with my immediately prior post.

 

[Rolacycle]

I do not think this will work quite the way you want. It would require high tension to keep the rope as flat as it is at p6, but it clearly isn’t that tight as can be seen from the angles at p4 and p2. What will happen is the angles at p6 and p4 and p2 will flatten out more or less together. This isn’t necessarily a bad thing, and may constitute some extra mechanical advantage, but it seems overly complicated and unnecessary. Also, if you feel strongly about getting the rope relatively flat at p6 you would need several hundred pounds of tension in the rope at which point the lower rope would also be flat.

It would be easy to make it work the way you want. Why not get rid of p6 and p5. Make the top rope a separate fixed rope tied off at both poles. You will have to get a lot of tension (several hundred pounds) to keep the rope from sagging too much, so use a strong cable and a turnbuckle or similar. The lower rope would then also be tied off at 5.

Me personally, I would probably just go p1 to p3 to p2 to p3 to p2 to p3 etc until I had the mechanical advantage I want. Those ropes pulling at angles don’t add as much to the mechanical advantage. However, I see that your way is stable against swinging in one dimension, so that may be important to you.

The drawings pulley positions were more so I knew it how/where to put pulleys so it’s pull centered. Fully raised 180° rope at p6 would be 15’, and honestly anything over 6’ should be plenty clearance for deer/pigs/cattle. With load raised 7~13’ and weight of load I figured sagging at p6 would add stability

#3 windiest city in USA ...a 30mph wind is a like a Tuesday.

 

[Cutter Ketch]

By the way, you can buy a block and tackle on amazon for twenty bucks

 

[hutchphd]

The 6:1 mechanical advantage system only requires 16.6 lb force to move 100lb load.

Actually the mechanical advantage would be 4:1 (for your original) if all the ropes were vertical (you just count the ropes pulling on the load...I know there are six pulleys...but you need to be careful here.) Because of the angles the tension required is worse=~30lbs as I mentioned previously. As you go up it gets even worse because of angles.
For stability your design makes sense but you must figure how much force you wish to pull and then you can figure how tall the towers need to be worst case. Probably you don't need to worry about side loads (if the posts are rooted as you say.!)
The "two rope" solution may be the way to go...

 

[Rolacycle]

Actually the mechanical advantage would be 4:1 (for your original) if all the ropes were vertical (you just count the ropes pulling on the load...I know there are six pulleys...but you need to be careful here.) Because of the angles the tension required is worse=~30lbs as I mentioned previously. As you go up it gets even worse because of angles.
For stability your design makes sense but you must figure how much force you wish to pull and then you can figure how tall the towers need to be worst case. Probably you don't need to worry about side loads (if the posts are rooted as you say.!)
The "two rope" solution may be the way to go...

4:1 ??
Both sides of rope @ P2/p4/p6 pull upwards from fully lowered position so how’s that 4:1

 

[jack action]

4:1 ??
Both sides of rope @ P2/p4/p6 pull upwards from fully lowered position so how’s that 4:1

If you isolate your 100 lb load, you'll notice that only 4 cables are pulling it upward, that is, the load is divided equally between 4 cables (assuming they have the same tension and are all vertical).

If you isolate your p3-p6 assembly, you'll notice that 2 cables are pulling it upward and 2 are pulling it downward. Similarly, you can also say that there are 2 cables pulling it to the left and 2 cables pulling it to the right. That means that they must be pulling equally, both up-and-down and sideways. So the angle between any two cables will be the same as the angle between the cables on the opposite side (i.e all cables will form an 'X' shape).

Assuming all your cables are vertical, they will all carry 25 lb each, the ones on the p3-p6 assembly going one against each other (hence adding nothing the MA).

 

[FactChecker]

Here is a thought process that I think will convince you of the 4:1 MA.
1) The MA is the ratio between the amount you pull the rope end down and the amount that the height of the weight changes.
2) If the P3 pully was hard-mounted rather than being suspended from P6, then clearly, the ratio of 1) would be 4:1.
3) Since the P3 pully is suspended from P6, the combined length-change of the P6/P3 rope system totals the same as if P3 was hard-mounted. That is, MA = 4:1.

 

[Rolacycle]

If you isolate your 100 lb load, you'll notice that only 4 cables are pulling it upward, that is, the load is divided equally between 4 cables (assuming they have the same tension and are all vertical).

If you isolate your p3-p6 assembly, you'll notice that 2 cables are pulling it upward and 2 are pulling it downward. Similarly, you can also say that there are 2 cables pulling it to the left and 2 cables pulling it to the right. That means that they must be pulling equally, both up-and-down and sideways. So the angle between any two cables will be the same as the angle between the cables on the opposite side (i.e all cables will form an 'X' shape).

Assuming all your cables are vertical, they will all carry 25 lb each, the ones on the p3-p6 assembly going one against each other (hence adding nothing the MA).

ok I do see that p6& p3 offset each other in the entire system, but unfortunately when pulling rope you can’t isolate parts of the system.
And no I don’t believe these will pull evenly. Say the load, rope slack and p6/p3 all laying on ground and I start pulling, 4:1 system but it’s using 6:1 rope. Once slack is gone it will raise p6/p3 until slack between p5—->p4 —>p3—>etc pulls load 10’ off ground. That would need 60’ at 6:1 to lift 10’ even tho u pointed out it is 4:1 MA it’s 6:1 in relation to rope length right? Here’s another diagram with more information and fully raised one please don’t judge it too much lol
Anchor 15’
P5 15’
P1 15’

raise p2/p4 10” need 60’ rope. I do believe once slack is out (which will raise) p6/p3 (to to some height, don’t know that it would matter once slack is gone) they will pull evenly, but not before.

 

[Lnewqban]

I have found this:
https://roperescuetraining.com/physics_mechanical_advantage.php

https://roperescuetraining.com/physics_calcma_tsystem.php

https://roperescuetraining.com/physics_calcma_counting.php

 

[Rolacycle]

I have found this:
https://roperescuetraining.com/physics_mechanical_advantage.php

https://roperescuetraining.com/physics_calcma_tsystem.php

https://roperescuetraining.com/physics_calcma_counting.php

I saw that page and it’s very informational and even found 32:1 MA that’s super easy to make but not sure even if I make overhead solid anchor point if it has room to pull slack out. Poles are already set so that’s only thing I’m not willing to undo/ redo

 

[FactChecker]

but not sure even if I make overhead solid anchor point if it has room to pull slack out.

It's not clear what you mean by "pull slack out".
The separation of the anchors at the top is not essential. As long as you can keep the system from getting tangled up, all the upper anchor points can be together at the central "sagging" point of a cable across the top.

EDIT: I think I understand the slack. This takes some height for the cascade of pulleys.

 

[Rolacycle]

It's not clear what you mean by "pull slack out".
The separation of the anchors at the top is not essential. As long as you can keep the system from getting tangled up, all the upper anchor points can be together at the central "sagging" point of a cable across the top.

EDIT: I think I understand the slack. This takes some height for the cascade of pulleys.

Slack like unused rope used to raise load. Well it’s a deer feeder and it doesn’t refill itself so load has to be dropped and doing so at 6:1 rope length the amount of rope to lift load makes a lot of slack when is laying on ground. And when u re raise load slack with be pulled out of pulleys in 6,5,4,3,2,1 order thus lifting the p6/3 (1st load) combo before raising the p4/2 (2nd and primary load)

and yes the top doesn’t have to be at perfect 180° plain. But the separated anchor is to keep wind at bay and maintain centeral vertical lifting point. MA is nice but honestly 100lb I could life with no advantage but 4:1 my 12yr old daughter can refill it

 

[Lnewqban]

I believe that this will be the limit to the effective MA of P6:
When the angle of the ropes running through a pulley are at 0 degrees, the force will be doubled.
When the angle of the running lines is not 0 degrees, the effectiveness of the pulleys is diminished.
When the angle of the running lines reaches 120 degrees, there is no more mechanical advantage in that pulley.

Look for "120-degree rule" here:
https://www.frostburg.edu/faculty/r...s_preppers_chapters/Ch15-PulleySystems_v2.pdf