0
$\begingroup$

enter image description hereSo, if I take a rope, 'fold' it over something to redirect it and then load both strands in the same direction with a certain force, each Strand of the rope should See 50% of the tension, right?

But what happens at the 'top' of the bend, where both strands 'meet'? Do the forces there add up, so that the rope still sees 100% of the tension at one particular point?

I'm wondering because logically thinking (and in practice) a doubled rope holds double the strength of a the single Strand before breaking. I just dont quite get why, because there seems to be a point where both Split up forces seem to meet.

I'm probably over complicating this Problem, but it just doesnt go in my head somehow.

Any help is welcome!

enter image description here

$\endgroup$
3
  • $\begingroup$ The tension in a rope around pullies is the same at all points of the rope, if you do not take into account the frictions and masses of the pullies and rope which is usually very small. $\endgroup$ Mar 1 at 20:46
  • $\begingroup$ Where you ask "what happens here". Yes, the rope is pulled left with force (F/2). it is also pulled right with force (F/2). Therefore the total tension in the rope there is..... (F/2). The same as in any other part of the grey rope. Remember that a rope in static tension is always pulled to both sides equally. The original error in your diagram is not showing the counter-force pointing upwards at the anchor point, balancing the force acting on the end of the rope. $\endgroup$
    – PcMan
    Mar 2 at 6:50
  • $\begingroup$ @PcMan Thanks. I‘ve added another illustration of my problem. See my reply to silverhaul s answer ... :) $\endgroup$ Mar 2 at 16:11
1
$\begingroup$

So, if I take a rope, 'fold' it over something to redirect it and then load both strands in the same direction with a certain force, each Strand of the rope should See 50% of the tension, right?

Each strand would see 50% of the tension that would have been if a single strand was used. So, tension in each strand would be F/2 where F is the weight / force pulling down on the entire thing.

But what happens at the 'top' of the bend, where both strands 'meet'? Do the forces there add up, so that the rope still sees 100% of the tension at one particular point?

At the point, where you have circled, if you draw a free body diagram, then there will be a force ( i.e. rope tension ) puling to the right of F/2 and a force ( i.e. rope tension ) pulling to the left of F/2 .

That is what we mean when we say , there is some tension in the rope. We mean, that each point in the rope is being pulled left and right by the same force and hence is in equilibrium

$\endgroup$
5
  • $\begingroup$ Thanks for the reply. I’m still a bit confused. I‚be added another illustration of where I have trouble getting it. At the point of the top, the forces are still in equilibrium, yes. But is on that top point not double the force than on every other point (F/2 plus T/2 = F) pulling the rope ‚apart‘? $\endgroup$ Mar 2 at 16:10
  • $\begingroup$ First of all, in your illustration , at the bottom end of the figure, why is T = Fmax ? If T is the tension in the rope ? Then it should be 2T = Fmax . And at the top, there is a force of T towards left and a force of T towards right. There is also a force of 2T downwards due to the 2 halves of rope pulling downwards with T each and there will be a reaction force from the nail = 2T $\endgroup$ Mar 3 at 5:42
  • $\begingroup$ Alright. But wouldnt that mean, that at the top the rope is being 'pulled apart' by a force equal to 2T - T to the left and T to the right? And if Fmax =2T, there would be Fmax on a single point of the rope at the top, the same thats "splitted up" to two strands at the bottom... $\endgroup$ Mar 7 at 15:56
  • $\begingroup$ Alright. But wouldnt that mean, that at the top the rope is being 'pulled apart' by a force equal to 2T - T to the left and T to the right? And if Fmax =2T, there would be Fmax on a single point of the rope at the top, the same thats "splitted up" to two strands at the bottom... $\endgroup$ Mar 7 at 15:56
  • $\begingroup$ Why would there be 2T-T to left ? There would be T to the left and T to the right. $\endgroup$ Mar 11 at 16:59
0
$\begingroup$

Assuming both strands hold a common load in equilibrium (no sliding of the rope over the top "peg"), and assuming a massless rope, the tension in the rope is the same at every cross section of the rope, equal to half the tension for the single strand case. There is a constraining force from the "peg" upward on the rope looped around it equal to $2T$ where $T$ is the tension in each strand.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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