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Evan
08-08-2006, 12:46 AM
While working on the counterbalance of my mill tonight this little problem occured to me.

The question is: How much tension is there in the blue line?

http://vts.bc.ca/pics/pulleypuz.gif

A.K. Boomer
08-08-2006, 01:08 AM
50lbs throughout...

JRouche
08-08-2006, 01:32 AM
50lbs

Neat idea. You might think because there is a total of 100lbs being suspended in air the line would have to be supporting the entire 100lbs.

But, the pulley actually separates the two loads at the centerline of the wheel. JRouche

Evan
08-08-2006, 01:42 AM
Disregarding any friction of the line against the pulley I think we can agree that at the exact top dead center of the pulley we have a portion of line on one side with 50 lbs of tension and the same on the other side with 50 lbs of tension in the opposite direction. So, why doesn't it add to 100 lbs at the point on the exact top? There is 50 lbs of pull in opposite directions.

Suppose the line is elastic like a rubber band. How much will it stretch in total? 50 lbs worth or 100 lbs worth?

gmatov
08-08-2006, 02:06 AM
Fortunately for mankind, there is 50 pounds per line, or NO block and tackle lifting device, such as an EOT or a jackup crane, or a strip mining crane would work.

Would have to go to the books to see the added resistance to pull per wire. They are not 100% efficient.

Were the load per wire to be the actual load, you would never devise a 500 ton crane, the wires would be so thick they would need a drum 20 feet in diameter just to bend around.

Repaired EOTs the last 15 years. 2 to 16 falls, 3/8 to 1 1/8 wire. Formerly had a 500 ton with 1 3/8 wire, 24 falls. Hot end, the melt plant, has 350 ton cranes, I think they, too, are 1 3/8 wire. Didn't count, but probably 24 falls, too. Safety factor is supposed to be 7 to 1. That DOES come in handy when the operator shock loads the crane with a running start.

Cheers,

George

lynnl
08-08-2006, 10:13 AM
".... So, why doesn't it add to 100 lbs at the point on the exact top? There is 50 lbs of pull in opposite directions. "

Because the 100lb force vector at top dead center is acting straight downward, normal to the line (cable, etc.) at that point. ...this is the total 100 lbs at the center of the pulley that JRouche cited.

In fact, at the EXACT TDC in the cable, it would seem to me that there's NO longitudenal tension in the cable. ...that can't be right - can it? ...or is it?

bdarin
08-08-2006, 10:19 AM
I'm with Evan on this one. If the rope were horizontal and being pulled with 50 lbs at each end there'd be 100 lbs of tension on the line. Just cuz it's bent over a pulley doesn't change anything. If it were tied to the post at the top of the pulley then I see 50 lbs per line. Since it's free to move then it seems logical that there would be a total of 100 lbs tension on the line. This assumes the pulley bearing is zero friction. As the friction of the pulley bearing increases, the load would become more divided between the lines. If the pulley bearing became totally locked up then the weight would divide equally to 50 lbs per side. Just a theory on my part, you understand, but it seems logical.

A.K. Boomer
08-08-2006, 10:33 AM
I'm with Evan on this one. If the rope were horizontal and being pulled with 50 lbs at each end there'd be 100 lbs of tension on the line. Just cuz it's bent over a pulley doesn't change anything. If it were tied to the post at the top of the pulley then I see 50 lbs per line. Since it's free to move then it seems logical that there would be a total of 100 lbs tension on the line. This assumes the pulley bearing is zero friction. As the friction of the pulley bearing increases, the load would become more divided between the lines. If the pulley bearing became totally locked up then the weight would divide equally to 50 lbs per side. Just a theory on my part, you understand, but it seems logical.

You put 50lbs of pull in one direction when you lift a 50lb weight off the ground with a rope, in one direction your pulling up 50lbs, in the other the weight is 50lbs --- this does not put 100 lbs on the rope, using the horizontal method just uses one of the 50 lbs to conteract the other just like gravity would have with one weight and lifting that one weight...

Evan
08-08-2006, 10:46 AM
:D

Evan
08-08-2006, 11:02 AM
I'm playing devil's advocate here. We have without question a total suspended weight of 100 lbs. Also, each side is supporting 50 lbs. The pulley is supporting 100 lbs (not counting the pulley etc). The single piece of cable is also supporting a total of 100 lbs.

None of that is in question. But, we don't have two separate lines to support the 100 lbs, just one continous piece.

Cables have elasticity and we will assume it is equal throughout the length of the cable. A length of cable under a single 50 lb load will have the load evenly distributed over the entire length of the cable resulting in a certain amount of elastic deformation over that length.

In this case however we have each 50 lb load distributed over only half the cable, not the entire length. How does this affect the cable and is it different from a single 50 lb load over the entire length?

A.K. Boomer
08-08-2006, 11:43 AM
This is the way i look at it,,, forget about the 50lb wieght on the left side for a second, now providing the pulley has a friction free bearing (just for this example because in the other it really doesnt matter because we are not getting off balance) --------So --- forget about the left side having weight and just anchor the rope to the ground, leave the right side hanging and what do you get with force on the pulley,,, the same as with both weights,,, there is 100 lbs of weight on the pulley system, why? because there is a 2 to 1 ratio of the weight moving down (the right side 50 lbs) to how much the pulley block will move --- if it was mobile,,,, The rope on the right side has 50lbs of pull --- but so does the rope on the left side that is just anchored, now forget about having the anchor ----- zing the rope has no resistance so the 50lb weight drops and the pulley block has no pull on it, how much weight would it take to duplicate the right side 50 lbs? 50 on the left because thats how much pull the left is recieving --- think of the left as a perfectly balanced "anchor"

By the way, this is how we get pinned boats (and bodies) off of rocks in the river, we set up a Z drag and increase our ratio, its not that the rope gets super loaded, its that we have increased ratio,,, the pulley example that Evan has givin is one of the most simplest ways of doing this, the boat would be attached to the pulley block, the force would be the 50lb on the right side and the left side would run back to the bank and be attached to a strong anchor (tree or rock) I also introduced allot of people to vector forces on the river,,,,, the longer the rope the more leverage you have ---- if people are pulling on a rope and you have 500lbs of force between the boat and the people you then anchor it off and then relocate your work force directly between the boat and the anchor and push 90 degree's on the rope between the two, the force is incredible and iv also single handedly pulled trucks out of ditches doing this....

Wirecutter
08-08-2006, 12:27 PM
50lbs

Neat idea. You might think because there is a total of 100lbs being suspended in air the line would have to be supporting the entire 100lbs.

But, the pulley actually separates the two loads at the centerline of the wheel. JRouche

Wow, this is a switch! I would have expected Evan to be on "the other side" of a question like this, spewing supporting facts, formulae, and citations. :D

Another way to look at this problem: Each weight can only exert a downward force of 50 lbs. That accounts for the tension everywhere on the blue line except possibly along the pulley. The explaination for the tension along the pulley involves just a wee bit of trig and physics, which are left as an exercise for the student. :D

-Mark

Wirecutter
08-08-2006, 12:29 PM
Ah yes - I like AK's answer better. Consider the example of a block and tackle, which magnifies force quite nicely, thank you.

-Mark

BobWarfield
08-08-2006, 12:33 PM
I thought the correct answer was always, "Use a bigger wire?"

<G>

BW

A.K. Boomer
08-08-2006, 12:45 PM
By the way, when you use the vector method you are now putting way over the amount of "stretch" on your rope because you are increasing the pull, 50lbs hanging by gravity by a rope strung horizontally between point A and Point B is way over 50lbs stretch on the rope,,, in fact you can exceed any rope ever built with just 50 lbs,,, three things are needed, 1; you have to hang the weight directly between point A and B 2; The stronger the rope the further apart point A and B have to be and 3; the straighter the rope is strung (without sag) the more effective the results will be...

dm1try
08-08-2006, 01:13 PM
The system is static, and if you isolate one weight and a piece of rope attached to it you'll get the following:

http://mmcd.meditprofi.ru/machining/hookpuz.gif

Forces to that piece of rope will be the same...

/dmitry

Evan
08-08-2006, 01:14 PM
Wow, this is a switch! I would have expected Evan to be on "the other side" of a question like this, spewing supporting facts, formulae, and citations.

As I said, I'm playing devil's advocate on this. I want to see somebody else provide the ironclad reason that there is only 50 lbs of tension and the equivalent elastic deformation in the cable.

There isn't any force magnification here. It doesn't even need to be a pully, a frictionless support will do fine.

RPM
08-08-2006, 01:21 PM
Dear Evan,
When I had the exact same problem counterbalancing the weight of my drillpress table, being weak on math, I decided to fit a cable that could take 1501bs anyway.
Hasn't broken yet...
Richard in Los Angeles

Evan
08-08-2006, 01:21 PM
dmitry

At the risk of restating the obvious what we have is this:

http://vts.bc.ca/pics/pulley2.gif

Magic9r
08-08-2006, 01:44 PM
Weight and thus tension is not distributed over the length of the cable unless the load IS the cable.

Scatterplot
08-08-2006, 02:36 PM
http://img.photobucket.com/albums/v240/bertmcmahan/Pulley-1.jpg

The top of the pulley System, the roof, is supporting 100, but the cable is just doing 50. Imagine a 50 lb weight with a cable on it sitting on the floor. Pick it up. What's the tension? 50, duh! Now walk over to a pulley. Put the rope over the pulley. Now, pulling down, it's the same force as it is to hold it up, just in the opposite direction, right? Now back to sans-a-pulley. How much force do you need to hold it up? 50 lbs. In the cable you have 50 lbs pulling each direction, for a tension of 50 lbs, not 100. Going on to the pulley, you laid the cord over the pulley. Now you are pulling Down with 50 lbs, correct? There is still just one weight. Just the 50 lb weight. So now pretend you need something else to hold that weight. Put another 50 lbs on there. Nothing changes. The pulley simply allows you to redirect the force of each weight to cancel each other out (in the rope, excluding it's mounts).

cam m
08-08-2006, 03:47 PM
Iron Clad? This is as close as I can come.

1. The system is at rest - therefore all forces must be balanced.
2. The tension through the length of cable is constant (ignoring the weight of the cable and friction between the cable and the pulley).

If the tensile force is constant through the length of the cable, it must all bear 50 lbs tensile load or there would be an imbalance and the system would no longer be at rest.

winchman
08-08-2006, 04:18 PM
So, how much tension is there in the horizontal section of cable between the two pulleys?
http://img.photobucket.com/albums/1003/winchman/000_13592.jpg

Roger

cam m
08-08-2006, 04:26 PM
See point 2.. Ignoring friction and the incremental weight of the cable itself, the tension in the cable is constant throughout - 0.00001 mm above each weight - between the two pulleys - and... and...

Mcgyver
08-08-2006, 04:49 PM
regarding the stretch Q, assuming loads are well under that required for deformation, one 100lb load over the whole length would stretch it 2x as much as 2x50lbs over 50% of the length - with the same modulus of elasticity, stretch is a function of length and and force and is linear. 100lbs x 1length = 2x 2*50lb*.5length

rantbot
08-08-2006, 05:40 PM
Not so good on Free Body Diagrams around here, are we?

If the tension in the string is anything other than 50 lbs, the weights would accelerate in one direction or another. A bit of "f=ma" is involved.

Evan
08-08-2006, 05:54 PM
Rantbot has it. Mcgyver has it on the stretch. I was waiting for sombody to simply say to tie one side to the ground but holding it is the same.

Now, is this balanced? What is the tension? Is it different at all?

http://vts.bc.ca/pics/pully3.gif

http://vts.bc.ca/pics/pully3.jpg

nheng
08-08-2006, 06:12 PM
The tension is still the same but now you've picked up a mechanical advantage of 2:1 at the 50 lb loads should you chose to pull down with 50 lbs force to lift the 100 lbs. Distance lifted, of course, is half the downward pull distance.

A.K. Boomer
08-08-2006, 08:54 PM
I was waiting for sombody to simply say to tie one side to the ground.

]

Ummm,,, already did that back in post 11:D

Evan
08-08-2006, 09:20 PM
Sorry Boomer, I somehow missed that. I was a bit busy today improving the security at my shop. We have had a couple of breakins on either side of me so I built a couple of "bait" computers that contain entirely junk dead parts but look nice and are easily identifiable. Hopefully if somebody breaks in they run off with those. I also have a camera that is motion activated as well as bars on the windows.

cam m
08-09-2006, 12:34 AM
Evan

As far as more sophisticated configurations go.. without complicating things with Free body diagrams or three dimensional vector solutions... the tension is the same throughout the cable if the system is at rest. If not, there would be motion induced. That is a layman's short cut to the free body diagram of the middle 100 lb wieght supported by the two cables each exerting 50 lbs upward pull. The system is in equilibrium (at rest). For those who are wanting to go further, F=ma=0 at each center of rotation for each pulley and each mass. No acceleration - all forces sum to zero in three dimensions - the system is at rest.

A.K. Boomer
08-09-2006, 12:36 AM
Nice job, even if their "bump keying" at least you will have them on film leaving with the crap merchandise...

Evan
08-09-2006, 12:52 AM
Cam,

I know. Net forces must be zero. I posed this because it is generally counterintuitive. This is especially so when you keep adding weight in a new configuration but the tension doesn't change. It is also directly relevant to some designs.

I asked this question today of an old friend of mine who was also a machinist before he retired and he at first assumed the wrong answer. It's funny how some of the simplest looking problems can be so easy to mistake.

Millman
08-09-2006, 12:55 AM
{{ machinist before he retired and he at first assumed the wrong answer.}} Actually, you have never met or talked to anyone in the REAL world. Have to feel sorry for you.

Evan
08-09-2006, 01:03 AM
We all live in a world of our own making. Perception is 9/10ths of reality. Some of those worlds are a bit stranger than others, apparently.

Millman
08-09-2006, 01:05 AM
{{stranger than others, apparently.}}. Yeah, then there's just people with no common sense.

cam m
08-09-2006, 01:06 AM
Sometimes the simplest solution is the best. The trick is to identify which part doesn't change between rig ups.

Millman

Common sense, like common courtesy, isn't as common as it used to be or should be....

Evan
08-09-2006, 01:10 AM
A definition of "common sense"

A form of evidence that is based on conventional wisdom, tradition, or someone’s personal philosophy or perspective. It is hard to judge the validity and reliability of common sense because little supporting evidence is involved. Most people judge the validity and reliability of common sense by the person citing common sense as the basis for a decision. However, common sense can be a very biased approach to decision making and means nothing more than “what is common to me makes sense.”

Millman
08-09-2006, 01:15 AM
Like I said...just repeating what someone else has already said or done.

Evan
08-09-2006, 02:14 AM
ike I said...just repeating what someone else has already said or done.

So then, how do you learn? Do just make it up as you go along? Or do you study the work that others have said and done? Have you ever opened a textbook?

Millman
08-09-2006, 02:17 AM
I try to quote from personal experience, text books are only a starting experience for the uneducated. After a while, you do not need the text or pictures to qualify.

Evan
08-09-2006, 02:25 AM
The quality of a man's education depends the wisdom of his teachers. This is especially so for the self taught.

Millman
08-09-2006, 02:30 AM
As usual, you are partially correct. A man's wisdom comes from experience...which there is no substitute. All the books in the world only makes you "well read", however that does not make you smart.

John Stevenson
08-09-2006, 02:42 AM
Things must be slow if we get three pages of drawings blocks, strings and pullies instead of making the bloody thing..................

.

Elninio
08-09-2006, 04:03 AM
the force of tension in the wire is 100 lbs, picture it as a tug of war between the wire , 50 lbs in each direction, the pully in this case is just directing the motion. Another way to think of this is as if instead of 50/50, you then had 70/30 and then 90/10 etc.. until you came really close to 100/0, in that case it would be the simple case of a weight on a straight wire of 100 lbs.

kap pullen
08-09-2006, 07:58 AM
the force of tension in the wire is 100 lbs, picture it as a tug of war between the wire , 50 lbs in each direction, the pully in this case is just directing the motion. Another way to think of this is as if instead of 50/50, you then had 70/30 and then 90/10 etc.. until you came really close to 100/0, in that case it would be the simple case of a weight on a straight wire of 100 lbs.

Elninio,

Based on your assumption, if you hook one end to the 50# weight, and the other end to the ground;

The tension is the weight of planet earth, plus 50#?

Just wondering.

Kap

Evan
08-09-2006, 08:40 AM
Things must be slow if we get three pages of drawings blocks, strings and pullies instead of making the bloody thing..................