Vertical Cable, Torque, Force applied to Wheels

AI Thread Summary
The discussion revolves around a university project focused on developing a robotic climber that ascends a cable. The team is exploring the effects of cable curvature on the resistance force experienced by the motor, specifically how increased curvature leads to greater friction and makes it harder for the motor to operate. They are seeking methods to calculate the additional torque required for the motor to overcome this resistance, particularly when the cable is under tension. The conversation also touches on the influence of bearing friction and the properties of the cable material, Technora, which may affect the climber's performance. Testing different cable flexibilities and adjusting the design to minimize bending may be necessary for improved functionality.
breadvsrice
Messages
3
Reaction score
0
I am working on a project at my University. It's extra-curricular and not homework related. At the end we will give a report on what we learned during the development. Our team is working on some ideas for a basic robotic climber, that climbs a cable (similar to a seatbelt).

What we decided to try, was to "offset" the wheels on either side of the cable, and adjust their height for proper grip on the cable.

I was hoping someone would know how to calculate forces like as shown in the pic below?
The best I could find so far, was the capstan equation for calculating friction based on the force applied to the cable, and the number of times it is wrapped around the wheel. Ex: π/6 radians.

What I really want to know, is the "resistance force," or how much harder the motor has to work to turn the wheels based on how "curved" the belt is...
As you can see in the picture, the cable "bends" through the tires. So, when a pulling force is applied to the cable (balloon and anchor) it creates a force that wants to "straighten" the cable. When the cable tries to straighten itself, it applies a force to wheel 1 and 2.

Logic tells me that this force will increase friction, and make it harder for the wheels to rotate. (Also pushes the shaft against the walls of the bearings, etc. slight, but still added friction. I will ignore this for now, and focus only on the wheel-cable part)

1) Am I right in assuming the "curvier" the cable is, the higher the straightening force, and the harder for the motor to work?

2) How can I calculate how much harder the motor has to work? (Ex. No stress = 1Nm to start climbing. With stress = 2Nm needed to start climbing. How do I calculate this?)

Here are the pictures:
https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-frc1/t1/400596_10202953978913296_150499901_n.jpg
 
Physics news on Phys.org
There is no frictional drag from the wheels at all if the belt is not slipping relative to the wheels - all the resistance comes from whatever torque is required to keep the wheel rotating about its axis (This is why we use belt drives in so many applications). You will, however, get friction in both guides if the moving belt is rubbing against them (which is why when we use belt drives, we prefer to use tensioner and idler pulleys to change the direction).
 
Increasing the tension will increase the load on the motor but only because it increases the load on the bearings. This should be a modest effect if the bearings are any good. You would need to measure how bearing friction varies with bearing load. I don't believe you can calculate it.
 
Thanks for the quick response! This was our original thought, that the rotation was independent of friction as it isn't slipping, but we noticed that when you held the cable loosely, the climber would go up and down easily. But, if you pulled on both ends of the cable with an decent bit of human-only force, you could keep the climber from sliding down do to gravity. It also made it harder for the climber to go up. Our bearings are ball bearings, 8mm diameter, 4mm center hole, 3mm thick. They are very smooth. Each one is about $3 per bearing, so of decent quality. We didn't think the "only" force making it harder to climb was just because of the extra stress on the shaft/bearing...But maybe that is what the problem was.

So, if we decrease the force at points F1 and F2 in the pic above, it should decrease stress on the bearings, reducing friction there...maybe this is all we need to do. Guess we will just have to test a little and see how it goes. So, "rolling friction" doesn't play apart here?

EDIT: Also, how would you go about calculating the F1 and F2 values anyway? By pulling both ends of the cable, you are definitely causing a force to be applied at those points. Is there an equation to use here?
 
Last edited:
What's the cable made of? If it's compressible but not very elastic then rolling resistance could be a factor.
 
  • Like
Likes 1 person
CWatters said:
What's the cable made of? If it's compressible but not very elastic then rolling resistance could be a factor.

The cable is made of something called "Technora" made by Teijin.
http://www.teijinaramid.com/aramids/technora/

It is moderately flexible, but also kind of stiff. Now that I think about it, the climber might not be able to climb because the cable itself can't "bend" fast enough to make that "S" shape with the wheels...if this is the case, maybe a more flexible cable would work better, or designing the climber so it doesn't need the cable to "bend" or "flex" as much?

Thanks for the insight! Interesting thoughts...
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top