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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?
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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.
".... 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?
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.
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...
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?
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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....
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.
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.
-Mark
The curse of having precise measuring tools is being able to actually see how imperfect everything is.
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...
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