Suppose I had a rope and idler pulley set up. The idler pulley was set at 5 feet tall on a post. The rope hung straight down on the pulling side and on the object being pulled it would be at 90 degrees. Now inbetween the idler pulley and the object I install another pulley level with the main idler pulley. It has no pressure on it and does no work it is just there. I always want the object being pulled to come back at 90 degrees to the post of the main pulley so I make the installed center pulley with an adjustable height. How could I calculate the downward pressure on the center pulley if for example I raised the object to 7 feet high and left the post at 5 feet. The center pulley would always be raised to match the height of the object being pulled.
Think force vectors. Draw a right angle triangle with the horizontal base being the horizontal distance between your two pulleys. Draw the diagonal with its angle being that of the angle between the low and high pulleys. The diagonal represents the pulling tension in the rope and the vertical 3rd side of your triangle is proportional to the vertical force on your added 2nd pulley.
Note these magnitudes are ***proportional: ***you’ll have to do some scaling and know the pull force to calculate actual vertical load.
Would it mak a difference how the middle pulley was centered between the loads? Nevrmind this, just doing as you suggested would handle that answer.
Thinking about this a little more, it might be better to make the length of the diagonal directly proportional to your pull force. Get the angle correct. The lenght of the horizontal and vertical lines represent horizontal and vertical force components and are then whatever length they lay out to be and are proportional to the pull force vector (length).
Nitpick
To calculate the pressures you are going to need the diameter of the pulleys and the diameters and lengths of the axels the pulleys are on.
If you are calculating the force then use force vectors.
Thats surely not the complete list of things to know … the point is made that pressure is force per unit area, force/area (typically sensibly averaged over a non-zero area, because the pressure at a point can’t be measured directly.)
Your system does not work.How can the rope 90 degrees pull anything up ?
Anyway, pulleys must be in a block and tackle arrangement to reduce the required force at the “pulled” section.
Also the rope appears to be pulling at a perpendicular, or almost, to the middle pulleys travel, which is unlikely to behave in a sensible fashion, and in practice friction becomes significant. To that end, we don’t even know what is keeping the middle pulley at 7 foot, or is it put there but left to drop ? Are you expecting friction to keep it there ?
The middle pulley is hung and only has the job of keeping the pulling motion at 90 degrees, not looking for any pulley advantage. The object being pulled would not be weight it would be spring tension.Some of the statements seem to be contradictory. If the first pulley is 5 ft up, and the 2nd pulley is raised to match the height of the object being pulled, how can the 2nd pulley be level with the 1st pulley? Or do the 2nd pulley and object start at 5 ft height, but you are varying this height? (So the rope coming off the 1st pulley is no longer at 90 degrees)?
___
_______|obj| __7 ft from ground
/o |___|
/
|o <-- 5 ft from ground
|
|
|pull down
By the way, you’re using the word “pressure” incorrectly. Pressure is force per unit area. There is no area involved here; you mean “force.”
p.s. If my understanding above is correct, calculating the force on the 2nd pulley is a simple matter of adding the 2 vectors. The tension on each rope is the same, because they are the same rope and the pulley is just an idler (free to move).
Your drawing is correct. The middle pulley would be hung and floating so I realize the force would not be pushing straight down.
I will tell you where I am going with this. It was basicaly just a passing thought for a different way to rig up a catapult. The center pulley would hang from a pivot and could rotate 360" from its anchor point, there would be a throwing arm that would rotate from the anchor point on the center pulley. I would lift the pulley to load it. It would allow me several tuning points I could work from and easily adjust.
It may be a horrible design but I wasn’t able to do the math on it to get some kind of estimates.
So have we answered your question?
Yes as usual you were all very helpful. I have kind of been teaching myself some new math by putting little problems on here and getting answers that I can look up on how to figure out. All the pulleys would be adjustable but I believe the formula or methods I have been given will work for any position.