Pulling Strategy

[Tim Gardner]

Originally posted by TheTreeSpyder


With no opposite pull the limb will fall? With no friction, the pull must come from hitch?

There is no way you can exert 200# anywhere on a rope with a 100# limb tied and hanging like in your illustration. Where is the extra 100# coming from? The limb only weighs 100#. If it is still attached I can see you being able to PULL the rope and get 200# but not with the limb just hanging from an anchor point.

 

[Tim Gardner]

Originally posted by TheTreeSpyder
Very good on both counts; the extra leg of pull down on pulley on same spar is neutralized by the pull up along the major axis of the spar. But the pressure at the redirect position, still follows the pattern of loading topping out at 2/1 here, as outside constraint. Now, what happens the second that spar starts to move; and seeing as that is the same time frame as Forcing Hinge Strength is there usable force operating un-neutralized?

I can not see how applying tension to the stick along its axis ABOVE the hinge will in any way apply any force at the hinge.

I can see however, by applying tension along the axis of the stick above the hinge would use some of the force exerted on the rope that would otherwise wind up at the hinge. That means that you would get less pull at the hinge.

 

[Nickrosis]

I spent waaay too much time making this, but it demonstrates the principle I believe Spidey is using.

 

[Tim Gardner]

Originally posted by Nickrosis
I spent waaay too much time making this, but it demonstrates the principle I believe Spidey is using.

What I got out of Ken’s illustration is that there is 200 # at the pulley when the pulley is attached to the 100 # weight. In your illustration Nick there would be 200 # at the pulley.

 

[glens]

Originally posted by TheTreeSpyder
i think that in non friction, non leveraged circumstance, you cannot exert 200# against 100#.

Sure you can.  The result will be that the 100# load will accelerate, allowing marginally greater weight to be present at the pulley than that of the static load.

Any time you use a pulley you have a leveraged circumstance of some type.

Here's the answer to your previous question.

Glen

 

[TheTreeSpyder]

Well, i guess where even Glens; i don't know where ya git the 17#, 77# and 200# values with zero friction, no angle factor in line. Just looking at the 3 positions of Load, Pulley , Anchor; i think the diagrams are equivalent.

To place the 200# pull on a line to a friction free pulley to lift 100# i think you would have to move pretty fast with a hard snatch; for in general with only 100# resistance, you can't pick up 200# on it, any more than you can walk over to a canary feather fluttering across the ground lightly, and exert 5# of effort picking it up un tethered. Now, perhaps you can exert 200 foot pounds, to lift 100# 2'.

Tim, i think that there is the same pull at the hinge before the spar starts to move. At the precise moment of movement, i think an upset happens; it is the only way i can de-scribe with these standard numbers, what i think i have observed for years, and tried to explain here in the last few years here.

Nice drawing Nick; i say Spyder with a "Y" for the quiet watcher, the rope thang we do, and the (inter)net. Made it up for email to 'civilians'; started using it on ISA board outta frustration when it wouldn't let me be just "KC"; in a time of Tom's fireside chats there.

 

[Nickrosis]

Originally posted by Tim Gardner
What I got out of Ken’s illustration is that there is 200 # at the pulley when the pulley is attached to the 100 # weight. In your illustration Nick there would be 200 # at the pulley.

Which is what's illustrated.

 

[TheTreeSpyder]

i thought my foam example would answer the query here.

Did you visualize it like this?

 

[TheTreeSpyder]

Applied directly to puzzle

 

[Tim Gardner]

Ken are you trying to illustrate a reversed single whip where the pull is coming from the weight itself?

 

[TheTreeSpyder]

i guess that is what puzzle would be, if i said it better!

 

[Tim Gardner]

Then I believe the power would be 1/2x at the foam or crotch. If you look at a reversed single whip the leg that is pulled on only requires 1/2x the weight of the load to lift. If the load were applying 2x the weight on the pulley (foam or crotch) then it would require 2x the power to lift the load.

 

[TheTreeSpyder]

is not the line tensioned by the weight just as if you pulled it?

If the upside down slingshot was nailed to the ground, and the line laced in the suggested "J" pattern, and the line tension set to 100# on long end, no friction, would not the foam have 2x pull on it as line crushed it down?

Would not the pull leg and the anchor leg both have to shorten 1', for a total of 2' of line slack taken out, to compress foam 1'?

Take out nails, float the 100# slingshot, so that the line tension ont he long leg of line was once again 100#. Does not anchor leg and pull leg have to shorten 1', for a total of 2' of line reeled in,inorder to compress foam 1'?

In putting up 2' of distance in effort, to be funneled into 1' of work in compressing foam would always increase the power beyond the line tension especially in friction free etc.

edit: just caught pic; Tim i think that anchor marked 1/2couldn't be to the load itself, it would have to be to a seperate anchor, then each seperate anchor would hold each 1/2 load; to match each other through pulley. Here, on a single leg, the line tension is 100/full load, for the load is borne on a single support. The weight of the load is the power for the system, like in rigging etc. i think.

 

[glens]

TheTreeSpyder:
Well, i guess where even Glens; i don't know where ya git the 17#, 77# and 200# values with zero friction, no angle factor in line. Just looking at the 3 positions of Load, Pulley , Anchor; i think the diagrams are equivalent.

By "no angle factor in line" I take it to mean the "pulls" are purely vertical, even though shown differently.  The sideways components arise from the offset from center of stem to the point the pulley is anchored.  If you didn't intend such, why the distinction between the first two as a group and the third?

To place the 200# pull on a line to a friction free pulley to lift 100# i think you would have to move pretty fast with a hard snatch; for in general with only 100# resistance, you can't pick up 200# on it...

If you place 100# of tension you will not move the load, you will merely counterbalance it.&nbsp; You must exert <i>more</i> or <i>less</i> force to move it.&nbsp; You could make it be only an ounce different and move it slowly or, for a brief moment, double or triple the force (or halve or third it) and get the thing moving rapidly.&nbsp; Certainly you will not be able to indefinitely keep more than the counterbalance force applied; that's not what I suggested.

Why do you think there will be any load on the stationary line when it's anchored on the same immovable object as the redirect?&nbsp; Don't you think that the "load" would rather be either zero or infinite?&nbsp; When it's a 100_lb weight, there is no load on the line until slack is removed and then it will ever be 100_lb (though inertia will effectively add or subtract load momentarily when direction of travel is changed).&nbsp; When the line is affixed to the same member as is supporting the redirect, the assumed "load" is only present whenever the free end is being pulled and immediately goes to zero when pull ceases.&nbsp; Think of the bald kid in the matrix having said, instead of "there is no spoon", "there is no load".

In my answer to your "riddle", the weight of the entire assembly is given as 200_lb, 100_lb, and 100_lb.&nbsp; That is correct as stipulated.

You may crush your "foam" the same amount between the first two, where one has an actual load hanging, and the other has no load, but the first will weigh 100_lb more at the base of the support than will the second because the first is supporting a load at 2:1 and the second has no load to support.

Really, Ken, have a look at my recent post in off-topic, <a href="http://www.arboristsite.com/showthread.php?postid=164333#post164333">29=30</a>.

You need some scales or load cells to see that although you might crush the foam the same as if there were 200_lb of total force, there is in fact only 100_lb of total downward force in your examples which have the one end tied to the redirect support itself.&nbsp; There is no equal and opposite weight present.

Glen

 

[glens]

TheTreeSpyder:
Would not the pull leg and the anchor leg both have to shorten 1', for a total of 2' of line slack taken out, to compress foam 1'?

Sure, you've got 2:1 against the foam, but you do not have 2:1 against the slingshot.&nbsp; The pull leg will shorten 2' to compress the foam 1', with the result being the anchor leg shortening 1'.&nbsp; The compressing foam causes the anchor leg to shorten, not the other way around as you suggest.&nbsp; If the opposite leg from the pull leg had an actual load on it of 100#, then both legs would shorten by 1' causing the foam to compress (as the result) 1'.&nbsp; I guess I have to ask, just what the hell is your point?

Take out nails, float the 100# slingshot, so that the line tension ont he long leg of line was once again 100#. Does not anchor leg and pull leg have to shorten 1', for a total of 2' of line reeled in,inorder to compress foam 1'?

Again, the anchor end shortening is strictly a <i>result</i> of the foam compressing, not a cause.

In putting up 2' of distance in effort, to be funneled into 1' of work in compressing foam would always increase the power beyond the line tension especially in friction free etc.

NO!&nbsp; The 2' of rope pulled will compress the foam, resulting, <i>finally</i>, in a total load of 100# on the rope.

edit: just caught pic; Tim i think that anchor marked 1/2couldn't be to the load itself, it would have to be to a seperate anchor, then each seperate anchor would hold each 1/2 load; to match each other through pulley. Here, on a single leg, the line tension is 100/full load, for the load is borne on a single support. The weight of the load is the power for the system, like in rigging etc. i think.

I'm still waiting for the image to arrive...

Okay, the image is correct, but does not apply to the upside-down slingshot example, where the anchored end is to the load itself and the figure in the image as "x/2" would be simply "x".

Glen

Instead of 634,014 bytes as BMP format, try 672 bytes as PNG...

 

[Stumper]

Tim, Technically a single whip is an anchored pulley to redirect force. One pulley floating in the system is called a runner-and it does provide a 2/1.

 

[Tim Gardner]

Originally posted by Stumper
Tim, Technically a single whip is an anchored pulley to redirect force. One pulley floating in the system is called a runner-and it does provide a 2/1.

Stumper, yes I know a reversed single whip is called a runner. And yes I know that it gives a 2:1 MA. In my illustration it shows that it is 2:1 just in a different way. I just assumed everyone knew that. If you look at #3209 in The Ashley Book of Knots you will see my source: “A runner is the same as a single whip, reversed. The figure 2 means that the lift is double the effort spent.”

My reason for labeling the illustration they way I did was to show that the pull (the leg that you actually pull with your hands) is 1/2x that of the load being lifted.

 

[Tim Gardner]

Originally posted by glens


Instead of 634,014 bytes as BMP format, try 672 bytes as PNG...

Thanks Glen. I realized that I had posted a bmp and converted it to jpg for my post in Ken’s other thread relating to this subject.

 

[Tim Gardner]

Originally posted by glens


Okay, the image is correct, but does not apply to the upside-down slingshot example, where the anchored end is to the load itself and the figure in the image as "x/2" would be simply "x".

Glen

Glen, I think my image does apply to Ken’s. If you look at the image I posted of the reversed single whip, the working end (the part of the rope coming out of the pulley that is not anchored but pulled on) only requires a pull 1/2 of the weight of the load to lift the load. If you tied the working end of that reversed single whip to the pulley above the shive you would have the same set up as Ken’s “slingshot”. Ken’s “slingshot being the load and pulley in one. I believe that even though the working end is tied off to the pulley it still would only bear 1/2 of the load. The opposite leg leading to the anchor would carry 100% of the load. I could be wrong believing this but until I see proof otherwise I am sticking with it. Maybe Ken will work the details out on how to measure the loads and settle this.

I have included a revised image of the reversed single whip with the parts labeled to hopefully illustrate how it applies.

 

[glens]

Hey Tim.

How is it that you expect the pulley arrangement to work when one end of the rope is attached to the pulley itself?&nbsp; That's like taking the fulcrum away from a lever system and expecting it to still work, isn't it?

Rather like grabbing your ankles and pulling up as hard as you can, I think.&nbsp; Try as you might, it just ain't gonna work.

See my upcoming reply to the "Condensed Mayhem" thread.&nbsp; I've put together an image you might be able to follow.

Glen