# How could a cord withstand a force greater than its breaking strength?

How could a 100 N object be lowered from a roof using a cord with a breaking strength of 80 N without breaking the cord?

My attempt to answer this question is that we could use a counter weight. But I don't really understand the concept behind counterweights so I hope someone can clear that up for me and if there is a better answer I'll love to know it.

• It is useful to ask yourself how you would solve the problem in real life. You have a length of lightweight cord and a heavy, but valuable package to be lowered from a roof. What are your options? – dmckee --- ex-moderator kitten Feb 15 '15 at 16:06
• Either double up the rope or lower the object fast enough that the force on it never exceeds 80N. – Hot Licks Feb 16 '15 at 12:14
• It occurs to me that it might work to run the rope over a pulley and put the 100 N object on one end and an 80 N counterweight on the other end. But it makes my head hurt to think about whether the acceleration of the 80 N counterweight could somehow put more than 80 N force on the rope. – Hot Licks Feb 17 '15 at 23:48
• @HotLicks - and hope they don't meet in the middle... – Floris Mar 14 '15 at 13:40
• Note that the advertised breaking strength also has a safety factor involved. – John Alexiou Nov 15 '16 at 12:59

Breaking strength refers to the maximum tension in the cord. Now, from the sounds of this problem, you've probably been doing force diagrams involving cords. What happens when you attach two cords to a single 100N object (and keep it stationary)? Is the tension in both of those cords 100N? Or is the combined force 100N, so that each just has 50N?

Put another way, most ropes you see will be made of many individual little threads. Each one of them is much weaker than the whole rope. See what I'm getting at?

• If you set this problem up on paper like this, it is indeterminate. We know the force division must be equal but getting here is too difficult for a homework problem. – Joshua Feb 15 '15 at 17:10
• I don't understand why you think it's indeterminate, or why it's too difficult for a homework problem. This sort of setup is totally standard for the usual force diagram students see early in first-year physics. Certainly there are simplifications, but that's standard practice in intro physics courses. – Mike Feb 15 '15 at 18:06
• There is only one vertical equation for statics. F1 + F2 = 100N. To get the second equation requires removing the normal assumption of rope length cannot change, yielding F1 + (k * d1) + F2 - (k * d2) = 100N and d1 = d2. The problem is no longer first semester and you now have a hooks law assumption known to be not true (ropes are not hooks-law springs). Unless you have the insight to know that lim(d->0) f(k, d) -> k * d where f is any spring function you can't solve the problem. – Joshua Feb 15 '15 at 21:41
• @Joshua: If one of the cords were a lot longer, you could tell by the slack which one is pulling all the weight, and if you've ever played on a swing set, you might know that the ropes get a little longer when you sit down, so I think the mechanism by which the weight is distributed is at least somewhat intuitive. With that in mind, you can make a decent case for F1 = F2 due to symmetry. In fact, I think it's much more difficult to conclude from the equations that the system is indeterminate. – Marcks Thomas Feb 15 '15 at 23:00
• How on earth are people getting "indeterminate", "too difficult", "can't be done with statics" and "one of the cords" and "he said there's a rope and a weight, so clearly there's also three pulleys, a counterweight, an elephant, and a banana involved"? You double the rope over to make a roughly 160N rope. It's not a hard question at all, it's a kindergarten level question. – Dewi Morgan Feb 16 '15 at 1:47

You have to accelerate the object towards the ground. (Let it fall a bit.) This creates a bit of "slack" in the cord so that it doesn't break. Figuring out how much acceleration it should have is a good exercise.

EDIT: I figure I might as well work it out since this question has so many views. Note that personally I view doubling the rope as cheating. From the free body diagram, we get, in an obvious notation $$W-T=ma$$ We want to find the minimum acceleration needed, so we plug in the max tension, $80\,\text{N}$: $$100\,\text{N}-80\,\text{N}=20\,\text{N}=ma$$ Assuming we are on Earth, the mass of the package is $$m=\frac{W}{g}=\frac{100\,\text{N}}{9.8\,\text{m/s}^2}=10.2\,\text{kg}$$ From above, we get $$a=\frac{20\,\text{N}}{m}=\frac{20\,\text{N}}{10.2\,\text{kg}}=1.96\,\text{m/s}^2$$

• Despite the down-vote this answer is correct, at least in physics land when you can magic the proper initial conditions into being. Think Atwood's machine. It may not be what is intended in the problem however. – dmckee --- ex-moderator kitten Feb 15 '15 at 16:04
• @dmckee: Is there a more practical solution? (This was a question on an AP Physics quiz a while back and I remember thinking that controlling the acceleration is ridiculous.) – Ryan Unger Feb 15 '15 at 16:08
• Atwood's machine is not the solution. The rope still has to hold the full 100N at the pulley. – Joshua Feb 15 '15 at 17:13
• @Joshua What makes you say that. That the tension is constant across the cord is the core fact that makes the system solvable. You get $a = [(m_h - m_l)/(m_h + m_l)]g < g$ and $T = m_h(g-a) < m_hg$. This is adjusted when you allow either a missive pulley or friction at the pulley, but it but the force acting on the part of the cord supporting the heavy load is still less than $m_hg$. – dmckee --- ex-moderator kitten Feb 15 '15 at 20:43
• Since you can get the weight to the ground without a rope (just let it fall), I assume the purpose of the rope is to land the weight gently. In this case, I'm worried more about the deceleration at the end. "I'm not worried about falling, I'm worried about landing!" – DGM Feb 16 '15 at 18:13

As far as I see, if you want the object to lower at constant speed the only way is actually fixing at least both the extremities of the cord to the object (can't describe this well, let's say your "grip" is on the middle of the cord).

Whatever happens somewhere else, constant speed means that the balance of the forces on the object must be 0, so if you tie it in two points you have $F_g + T + T =0$ (being $F_g$ the gravitational pull on the object and $T$ the tension on the cord, with appropriate signs). With this configuration you get $| T | = | F_g | /2 = 50N$, less than the breaking strength.

• +1 This is the clearest answer (though one might still quibble over signs). – Mike Feb 15 '15 at 22:59
• Well there's no quibbling with absolute values. :) – Mike Feb 16 '15 at 14:23

I would not be surprised if a cord with a given breaking strength of 80N held 100N once. It should be regarded as trash after that though.

You see, they set the breaking strength as guaranteed to hold under the worst-case setup. Select a low-reduction knot (the rating not is the double figure-eight which is common but not the optimal choice) and apply the load smoothly, and you will find the real breaking strength is higher than rated.

It would be unwise to depend on this though.

EDIT: The double figure eight's coefficient is .75 which means the straight line breaking strength is 106N. If we assume 0 margin (as we do in homework problems) this solution requires a coefficient of .943. This means long splice is the only "knot" with a good enough coefficient (the coefficient of the long splice is not known but estimated at 1). While we can make a loop with a long splice, this is really suboptimal.

• If this were a real-world scenario, this is the right answer. But if it's a physics homework problem, they probably mean the actual breaking strength of the cord, not the specified minimum breaking strength. – The Photon Feb 15 '15 at 17:43
• +1 for an entirely accurate practical answer. But I agree with The Photon, in that the original post sounds like a physics homework problem where simplifications and idealizations are expected. If I were grading a physics homework paper that gave this as an answer, I would actually mark it wrong because students need to know what questions are getting at. Plus, even in a real-world situation, a little more imagination would mean you could potentially stay safely within the rating. – Mike Feb 15 '15 at 18:10

Use a ramp, an incline of 53° will work.

Otherwise You need to double up the cord.

The third option is to just carry the 10 kg object down the stairs.

The simplest approach (and what the person asking this question probably was getting at) is to use a pulley like so:

The weight of the object is now shared between the two sides, with each carrying a 50 N load. You end up using twice as much cord. The other advantage is that you now have a "mechanical advantage" and you only need to use a force of 50 N to lower the object. You do need a point where you fix the other end, unless you hold both sides in your hands. In that case the load is evenly balanced between the two halves of the cord by the pulley.

Alternatively you could simply double up the cord - but the tricky thing there is to ensure that they share the load evenly. This is done in practice by twisting the ropes together (yes, that is one reason why ropes since time immemorial are twisted): if one strand carries a larger fraction of the load it tends to straighten out - which makes the other strand "take a longer path" (it becomes more twisty) until it starts carrying more of the load. In this way, twisting ensures sharing of the load. Twisting of the strands also makes a rope more flexible (since strands spend "equal amounts of time" on the inside of the bend and the outside - I put that expression in quotes since it is only approximately true but you get the idea).