Force on Rollers: 100N Pulled Down, 200N On Cylinders?

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In summary, if the surface roughness of the cylinders is smaller than the maximum allowable surface roughness, the force on the rollers will be greater than 0N and less than 200N.
  • #1
Kalgoolie
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Hey,

Got some rope running through four stationary cylinders (pic below)! Now if the rope on the left hand side is pulled down with let's say 100N, I would just like to find out the force on each cylinder.

common sense tells me that the force on all four cylinders will be 100N in alternating directions of course. Am I correct in assuming this?

Also the tension in the rope were the blue and red lines are, would they be equal to 100N as well?

Thanks for any help!

I should specify that the rollers cannot actually spin or turn, so I don't know if they can be treated as pulleys
Edit - Or is it something like this; http://en.wikipedia.org/wiki/Image:Pulley0.svg
Where if I pull down 100N the tension in the rope at the red part is 100N and the force on the cylinder is 200N?
 

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  • #2
thats a toughie!
If you assume that rollers are stationary & rope move(skid) relative to the cylinders, tension on both sides of disk are related as T2/T1 = e^(μ*θ), μ is kinetic friction coefficient & θ is angle of overlap, google a bit, you ll understand the physics behind it.

The tension as you move towards right ll continue to decrease.

If they don't turn, they are not pulleys.

Or is it something like this; http://en.wikipedia.org/wiki/Image:Pulley0.svg
Oh! do you mean downward force on disks?? I thought you meant torque, anyways, downward force is the sum of tensions on both sides. In your case, tension on both sides are not equal, but related by the above formula.
 
  • #3
Is this a statics or dynamics problem? From the statement that they don't turn I would think static, but you're not clear. If it is static, then you still are dealing with the sum of the forces in each direction = 0.
 
  • #4
Thanks a million for the replies.

Sorry should have specified! The rollers stay still and the rope is moved back and forwards through the rollers.

The rollers are held in place by pins on either side I just need to find the shearing force in the pin which I assume is releated to the tension?Below is a front on view, the image in the first post is a side view
 

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  • #5
if they are called rollers, how can they stay still?? Or you mean, the frame for rollers stay still(which is quite obviously is the practical situation), that is why i called it a toughie, problem is still not very clear
 
  • #6
ank_gl said:
if they are called rollers, how can they stay still?? Or you mean, the frame for rollers stay still(which is quite obviously is the practical situation), that is why i called it a toughie, problem is still not very clear

Sorry I keep calling them rollers. Pretty much these cylinders are still and they have a surfance roughness, the rope moves backwards and forwards through the cylinders hence abrazing the rope!

I need to find the forces in the pins holding the cylinders still and the power of the motor required to pull the rope and I believe that this is all related to the tension in the rope!
 
  • #7
Please ignore the post above!

ank_gl from what I have read your approach seems to be correct! Page 7 of this seems to derive the equation as well; http://ocw.mit.edu/NR/rdonlyres/Physics/8-01TFall-2004/84DD1138-93A4-47E0-8F54-CB45EBE8351D/0/exp05b.pdf"
Thanks for leading me in the right direction!

I have run into a problem, I have been given the average surface roughness and the maximum allowable surface roughness of the cylinder (both have units of micro-meters). These problems need the coefficient of friction so I need some way of incorporating the surface roughness into that the formula! All I know is what speed the rope will be moving at and the dimensions of the (stationary cylinders)
 
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  • #8
hey sorry for being a bit late, i was a bit busy.

I don't know of any way to incorporate surface roughness in friction calculations, essentially because friction laws are not based on physical principles, but on practically obtained data, so it is an empirical relation. You should conduct some experiments on those rollers & find out friction.
Also, you need not worry about the speed, as you can assume that kinetic friction remains constant.
 
  • #9
The problem cannot be worked with the information you've given, except to say that the force on the rollers due to the rope will be greater than 0N and less than 200N (I'm assuming these are essentially rappelling brakes). This sort of braking design works perfectly well with smooth surfaces; unless you want to abrade the rope, why make them rough?
 

1. What is the force on the rollers when a 100N weight is pulled down?

The force on the rollers will be 100N. The weight is being pulled down, so the force will be directed downwards.

2. How does the force on the rollers change when the weight is increased to 200N?

The force on the rollers will increase to 200N. The weight is being pulled down with a greater force, so the force on the rollers will also increase.

3. Are the cylinders affected by the weight on the rollers?

Yes, the cylinders will also experience a force of 200N. As the weight is pulling down on the rollers, the cylinders are connected to the rollers and will also experience the same force.

4. What happens to the force if the weight is pulled in a different direction?

If the weight is pulled in a different direction, the force on the rollers will also change in that direction. For example, if the weight is pulled to the left, the force on the rollers will also be directed to the left.

5. Is there a limit to how much weight can be pulled down before the rollers break?

Yes, there is a limit to the weight that can be pulled down before the rollers break. The limit depends on the strength and durability of the rollers and the material they are made of. If the weight exceeds this limit, the rollers may break or become damaged.

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