How to calculate the characteristics of this compression spring ?

[123nob]

Hello;

I have as shown in the picture, a rope (in green), this rope is located between a support A and a support B. The Support A doesn't move and attached to the ground. The support B apply a force F1 against the rope, this force is made by a helical compression spring.
Because there is a friction between the rope and the two supports A and B, and because the spring push B toward A so we conclude that much the force F1 increase much it is harder to overcome the friction and move the rope.

The question is: How can I calculate the force F1 required to achieve a minimum overcome force T1 of 200 Newtons ? what I mean by this, is what characteristics the spring should have in order to generate a pushing force F1 that require at minimum 200 Newtons applied to the rope in order to move it ?
If the problem is difficult, may someone please give me a link to a similar problem to seek a solution.

Some data:
1. Coefficient of friction Rope/Support : u
2. Rope diameter : d
3. Weight of support B : Neglected

Thank You

 

[Compositepro]

That is a poor way to control tension because the frictional characteristics of rope will change due to a wide variety of factors. A spring or dead weight force on a dancer arm with a free-wheeling pulley on one end can provide for very uniform rope tension. As the rope going over the pulley moves the dancer arm, the arm will turn-off a brake and allow a spool of rope to unwind until the arm comes back to position and turn-on the brake. The brake can be on-off or modulating. You can even pinch the rope as you show in your sketch, except the pinch is controlled by the dancer arm. This way friction is not a a factor in the resulting rope tension.

 

[123nob]

Hi;
I know my approach is poor and dirt. I just want an easy yet simple method with minimal components. When I google that tension control device I get results such as: magneto rheological brake or nonlinear adaptive backstepping control scheme. But is their any easy and clean way to simplify that so the mechanism fix the rope tension into 200N (not adaptive)? Thank you.

 

[desertfox]

Hi 123Nob

A real dirty method would be to assume a friction coefficient between the rope and stop, therefore

T1 = mu*F1 ( mu coefficient of friction)

rearrange to find F1 i.e. T1/mu = F1

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[Compositepro]

Google "ratchet and pawl tension control". This was one of the results:

https://ahrenslooms.com/technical-information/tension-and-brakes/

One of the reasons that it is difficult to find information on simple mechanisms is that there is no money to be made from providing it. If you look for off-the shelf solutions you will mainly find commercial solutions that are very expensive. Every business has overhead and must make a profit. There is nothing wrong with that, but if you don't want to pay someone else to solve your problems you need to learn to solve problems yourself. You are doing that here, but you are getting the hand-holding for free.