Hi there I have a practical physics question. I'm an industrial abseiler, and on the course today we had this question which not even the instructor can answer. If we have a rope connected to a horizontal beam. Say the rope can hold 100kg before it breaks. Now, if we take the top end of that rope and trace it over the beam and back down to the mass so now we have it doubled over, apparently we can now hold 200kg mass before it breaks. This makes sense if each strand of the rope was independent, but they aren't because its actually one rope doubled over the beam. So I'm thinking, wouldn't there be over 100kg acting on that point in opposing directions? I have attached a diagram. Thanks. David.
In both situation the tension of the rope is 1000N(100kg) Even in the second situation you will be able to hold only 100oN tension.You won't be able to double the maximum tension it can hold by bending the rope.But you can hold more masses by dividing the force created by the mass on rope,just by bending it into two or there..etc
Hello. I'm only in high school (don't go too harsh on me, guys...), but I think I see the root of your problem. You are British, right? Often, people from Britain and other parts of the world that have scales in terms of kilograms get confused as to what they are measuring. The kilogram is a unit of mass. What you are looking for when talking about how much a rope can hold is in terms of "weight," which is a force. Weight is the force on an object due to gravity. The force is nearly equally distributed between the two ends of the rope. Thus, each end of the rope is holding a weight of around 980 N(ewtons), because [itex]\sum[/itex][itex]\vec{F}[/itex] = m[itex]\vec{a}[/itex]. Mass (100 kg) times acceleration (9.8 m/s^{2} is typically acceptable as the gravitational acceleration felt on Earth's surface) equals the net force due to gravity. Because force is a vector, it has direction. If you want to find the tension in part of a rope, assume the system is in equilibrium. The net force has to equal zero in the case of equilibrium, so a rope holding 980 N must also have a tension of 980 N going in the opposite direction (we call this direction up, because physics is always creative). Don't take my word for it, though. I have a feeling someone on here has a better, and probably more technical, reason for this.
hey..I'm not a British.. :)...Not even near it I've not confused this weight and mass.... You may have confused this... I have use the brackets to show that 1000N has came from 100kg mass. and also I've taken the gravitational acceleration as 10ms^{-2}. The whole system can hold a mass of 200kg(weighing 2000N) in the second situation...But even in that situation the rope can hold only 100N
Shalikadam: I didn't say we have increased the maximum tension it can hold. Basically what you have done is explain the situation so others can understand what I'm asking. Thankyou for that. So I want to understand the physics behind this phenomenon. Most people who do this course ask this question. If we look at the top centre of the rope in the second diagram of Shalikadam's one would think that T2a and T2b are antagonistic Tensions and thus adding to give Tmax only with 100kg. But in fact as described in this picture we can hold 200kg before we reach Tmax. Is this something to do with breaking the vectors down into artesian coordinates, and that right up the top where I thouogth the rope would have broken? Explanation in vectors and forces and tension would be appreciated. PS: I'm Australian, but generally, Australians and British, (In particular ones who visit forums) are aware of the difference between mass and weight. Thanks for your assistance Mandelbroth. David.
Shalikadam I think Mandelbroth was referring to me. But what you have failed to do is really answer my question. All you did is restate what I was explaining. So please, if you do know "why", instead of "what" that would be appreciated. You just have told me what I already know. Now I want to know why. Since you are so good at diagrams, maybe you can explain in terms of vectors, force and tension maybe. thanks Eagerly awaiting your explanation. David.
All I can think of is that, right at the apex of this rope, the rope is actually horizontal, and so if we break the vector into x and y components, there are no x components up the top and so way up the top at the apex where the rope is horizontal momentarily, there is not tension. That sounds odd though, to me. So does Tension decrease from when we get close to the apex as we approach the apex? David.
It really depends upon friction. If your rope is (relatively) free to slide over the top of the beam then the beam acts as a pulley and the tension is the same in both drops and acts as shalikadm has presented. If however you take several turns around the beam so that the rope cannot slip the forces in each drop will be independent and the capacity doubled. In the indeterminate situation where there is some friction the tension can be different in each drop. Does this help?
This is a quite common confusion about tension. But think about it, the same can be said not only for the point at the top, but for every point on the rope. And you can do the same "analysis" for the situation with only one piece of rope (first drawing). If you take an arbitrary section of the rope, there is a tension "up" and one "down". If the weight is 1000N, what will the tension in the rope be ? Will the rope hold if the breaking tension is 1000 N? Or you only can hold 500 N? How come nobody raised the question about the first case? In the second case, the beam is the one supporting double load.
Can you specify what do you mean when you say that the capacity is doubled? With a rope that breaks at 1000N tension, what will be the maximum load in the two cases (without and with friction)?
The attachment shows a bar weighing 1500 kg suspended by two independent ropes of breaking force 1000kg. This is a stable situation since each rope carries 750 kg force. If for some reason one rope breaks the other will be unable to sustain the load and also break. Involving friction allows this support situation to occur with a single rope.
This is not quite the same problem discussed above. It was a single rope going around a beam. You seem to imply that there is difference in the maximum load, depending of existence of friction between rope and beam. Sorry of I misunderstood your description.Here is the quotation: "It really depends upon friction. If your rope is (relatively) free to slide over the top of the beam then the beam acts as a pulley and the tension is the same in both drops and acts as shalikadm has presented. If however you take several turns around the beam so that the rope cannot slip the forces in each drop will be independent and the capacity doubled. In the indeterminate situation where there is some friction the tension can be different in each drop." None of these cases seem to describe the situation in your last post. Even though the last case (with friction) is somewhat equivalent to having two ropes, as in your sketch. Are you saying that without friction, the maximum load is less than with friction? For the same rope over the same beam?
Bear in mind that I was responding to this question, specifically. Sorry if that was not clear but the rope is separately attached to the body at two places. This is why a diagram is best.
Why are you bringing friction into this? If the rope was a single rope, which passed over a frictionless pulley at the top, it would still be able to support the load just fine. There is no need to worry about friction in this problem. To answer the original question, assuming there is no significant friction, you could indeed model the forces in the rope at the top, where it passes over the pulley (I'll assume there's a pulley here) as being 1000 N in each direction. However, that's not specific to the top of the rope - everywhere in the rope, there are balanced forces of 1 kN in each direction. That's basically the definition of tension. When the rope is single, not doubled over, the support point at the top is pulling up with a force of 1000 N, so you still have 1000 N pulling on each end of the rope in this case (but one end is pulling upwards). With the pulley, you are effectively just redirecting the rope - the tension is unchanged by a pulley, so now, instead of the support point pulling on the other end with 1000N, the weight itself is supplying this force. Since the weight is supplying the force on both ends of the rope though, each force will only be half of the weight, so the tension is cut in half for a given weight (compared to the single rope case). Interestingly enough, friction could actually make the load bearing capacity worse, not better. With no friction, the tension in the entire rope is constant - neither side can support any more than the other. With friction however, the tension in one side could be higher, and the friction over the top would prevent the higher load from being transfered to the entire rope. This means that one side could reach its breaking strength well before the total load reaches twice the rope's breaking strength.
Why did you not read the whole thread? If you suspend a 1000 kg weight from a single rope and take three turns round a capstan and hold the loose end via a spring balance what will the spring balance read?
I think I understand your point of confusion, and it is something that has puzzled many of us when we began studying physics. Suppose you have a rope with tension T that is oriented along the x-axis. The question is, at an arbitrary point x along the rope, what is the force exerted by the part of the rope at x^{+} on the portion of the rope at x^{-}, where x^{+ } is on the positive side of x, and x^{-} is on the negative side of x? The answer to this question is +Ti_{x}, where i_{x} is the a unit vector in the positive x direction. The next question is, at an arbitrary point x along the rope, what is the force exerted by the part of the rope at x^{-} on the portion of the rope at x^{+}? The answer to this question is -Ti_{x}. So there are two forces acting on the point x on the rope, +Ti_{x} and -Ti_{x}, and the (zero mass) point x is in equilibrium. Now suppose that at x = L, the rope is attached to an object. What is the force exerted by the rope on the object? It is -Ti_{x}. What is the force exerted by the object on the rope? It is +Ti_{x}. The point x = L is in equilibrium. The tensile force exerted by a rope on an object is oriented in the direction from the object to the rope. The tensile force exerted by an object on a rope is oriented in the direction from the rope to the object. This can all be explained much more easily using the stress tensor, but that is beyond introductory physics.
I don't know why we are making this so theoretical, the original question was a practical one from a practical man. I posed the question about the capstan to demonstrate the power of friction and winding the rope around the beam. The difference in tension between the two ends of a rope wound round a beam or capstan etc is T_{1} = T_{2}exp(2n∏μ) Where T_{1} & T_{2} are the tensions at each end, n is the number of turns, μ is the coefficient of friction. To illustrate the power of this say the coefficient of friction is as low as 0.2. For six turns the tension at one end will be nearly 2000 times the tension at the other. (Yes two thousand) Consider a cowboy hitching his horse to a beam or rail. If he takes 6 turns of the reins around the beam and hangs a loose end down weighing just 4 ounces or 100 grams the horse will need to pull 1/5 ton or nearly 450 lbs or 2000N to free himself.
Possibly because the original question described a rope attached to the supported body with both ends. There was no free end of the rope. At least this what can be seen in both drawings posted on the previous page. When you mentioned friction, you did not say that you are thinking about someone pulling the free end of a rope, in which case indeed friction is very important (as applied for example to attaching a boat to the poles on the pier). As you said, an image may help When you first mentioned friction, you somehow omitted to say that you changed the setup. I hope my (crude) figure will help to clarify the discussion.
Despite many rumours to the contrary I did no such thing. I specifically outlined and discussed separately two cases, one with a simple passing over the beam, which I likened to a (frictionless) pulley and a second with the case of taking some securing turns around the beam.