Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Sorry when this sounds like a first year physics homework question, but it's actually not. It arose during a discussion regarding safe practices for rock climbing anchors...

So here we go: Imagine you have a closed loop (a sling) around an object such as a tree. Let's for the moment assume the tree to be frictionless and a cylinder of radius $R$.

You attach a load (via a carabiner, but it doesn't matter here) to that sling. It's pulling away from your tree with force $\vec{F}_l$. Now my question would be: What's the tension in the rope given the angle in the rope at the attachment point.

Well... I know the carabiner in equilibrium isn't moving, so the force $\vec{F}_l$ must be balanced by forces coming from the sling. I can draw the parallelogram of forces. Let's assume I already know the angle that the sling forms at that point. My confusion stems from this: Because I have a closed loop, should there be some sort of "back action" from the tension in one leg of the loop on that on the other leg?

The situation is clear when instead of a closed loop I have piece of rope attached to two points of attachment: Then I can work out the tension in each leg and know that this tension will be transferred to the points of attachment...

Also, it should be obvious that the sling exerts force on the tree, but if I only look at tension, which runs in the direction of the rope, I can't really tell where that force is coming from.

I must confess, back in my first year in physics I'm not even sure we properly did discuss more advanced situations regarding rope tension. It was always the good old frictionless pulleys introducing $90$ or $180$ degree turns...

EDIT: Here's an attached image. The left one shows the setup I'm interested in, the right one shows a situation that I'd know how to solve...

forces in slings

share|improve this question
add comment

1 Answer

Because I have a closed loop, should there be some sort of "back action" from the tension in one leg of the loop on that on the other leg?

Cool question. About the "back action." Imagine, instead of a smooth cylinder, a cylinder with a thin, rectangular bump directly opposite the point that the load hangs beneath (on the top part of the cylinder.) Also, instead of a circular rope which stretches all the way around, picture a regular, two-ended rope which stretches around the underside of the cylinder (that is, the load side), but where each end of it is joined to one side of the bump, on the top.

In this case, there cannot be any back action. But this is the same as the case you describe. Let us illustrate. First, we can cut the bump from the cylinder -- nothing will change, as there is no net force upwards or downwards acting on the bump. Second, we can replace the bump with a piece of rope, and voila -- this is just a rope stretched all the way around.

Since tension only comes from forces acting parallel to the direction that the rope is stretched, and the normal force from the cylinder is always perpendicular, there can be no contribution to the tension from the cylinder.

This means the tension in the circular rope, due to the load, is the same as the tension in two straight, fixed ropes carrying that load.

Also, it should be obvious that the sling exerts force on the tree, but if I only look at tension, which runs in the direction of the rope, I can't really tell where that force is coming from.

The rope now pulls down on the limb a weight equal to the weight of the load (if we take the mass of the rope to be negligible.) The pressure on the cylinder does not come from the tension in the rope.

share|improve this answer
    
Ah, I see it a bit clearer now, thanks for your explanation. So depending on the length of the sling (suppose it's only very slightly longer than $2\pi R$, the circumference of the tree, then the tension should be pretty high, right? I also still wonder how that downwards force is transmitted to the tree. I imagine that there will be some normal force wherever the rope touches the tree, pointing towards the tree, because the rope will put "pressure" on the tree in an attempt to contract... –  Lagerbaer May 14 '13 at 5:37
    
@Lagerbaer Mostly correct but just a technical thing -- the rope doesn't contract, strictly speaking -- it pulls because it has a mass (which we take to be negligible) that pushes against the cylinder. This is the cause of the normal force of the cylinder on the rope. –  santa claus May 14 '13 at 6:29
    
But what about the mass of the load? Imagine replacing the tree with a poor unsuspecting person. That person would feel squished, wouldn't they? –  Lagerbaer May 14 '13 at 6:30
    
@Lagerbaer my bad, I was speaking loosely, sort of referring to my answer. In order for there to be a force on the cylinder there has to be some mass doing the pushing -- this is the mass of the rope. But the combined mass of the rope and load provide the weight that the cylinder feels. –  santa claus May 14 '13 at 6:37
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.