# Breaking two ropes. What can be said about the tensions?

This is a theoretical problem. Suppose we have two ropes of the same material and diameter, but differing in length. We excert force on both ropes until they break. Let rope A be the longer one. What can be said about the stress in rope A compared to the stress in rope B at the moment of breaking, taking in account the deformation required to keep a constant volume?

You would expect them to be the same as the stress-strain diagram is derived for a specific material. But at the same strain of both ropes, the stress in rope A would be higher as it crosssection decreases more.

Stress-strain diagram from http://cdn.transtutors.com/

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• The tension in the two ropes is always the same. If the tension wasn't the same there would be a net force on the junction between the ropes and it would move. Feb 19, 2014 at 10:40
• The ropes are not connected. Imaging them hung parallel to each other. Feb 19, 2014 at 10:46
• Ah, OK, but that doesn't change the answer as the tensions would still be equal. However rope A would have stretched farther before it snapped. Let me know if you want me to elaborate on this and I'll put it in as an answer. Feb 19, 2014 at 10:49
• The longer rope will probably break first as, being longer, it has a bigger chance of flaws that will let it break. Hence the tension in the shorter rope will probably be higher when it breaks. Note the use of the word "probably", twice. Feb 19, 2014 at 10:51
• @ John Rennie. Please eleborate. Did you consider the change in diameter? @hdhondt. As it is a theoretical question, assume no flaws in the rope. Feb 19, 2014 at 10:51

In an ideal scenario, the tension should be the same -as the two ropes have equal breaking strain.

In reality, the longer rope has more chance of failure simply down to random weaknesses being more likely to occur over a greater length.

The correct way to think about the internal forces in the ropes is in terms of stress and strain (eg http://en.wikipedia.org/wiki/Stress%E2%80%93strain_curve).

The stress, $\sigma$, is the force per unit area in the rope, for this system we expect the stress to be constant across the cross section of either of the given ropes so, for a given rope, $\sigma = \frac{F}{A}$ where $F$ is the force in that rope and $A$ is that rope's cross sectional area.

The strain, $\epsilon$, is the ratio of extension ($x$) to natural length ($l$) (ie the length of the rope when no force is applied), that is $\epsilon = \frac{x}{l}$.

In general the stress is a function of strain, $\sigma = f(\epsilon)$; because both our ropes are made of the same material they have the same relation between stress and strain that is: (where subscripts label the rope to which the given property belongs) $$\sigma_A = \frac{F_A}{A} = f(\epsilon_A) = f\left(\frac{x_A}{l_A}\right)$$ $$\sigma_B = \frac{F_B}{A} = f(\epsilon_B) = f\left(\frac{x_B}{l_B}\right)$$ A special case of this is called Hookean elasticity, where the stress and strain are proportional, their ratio being a constant, $E$, called the Young's Modulus: $$\sigma = E \epsilon$$ A rope will break when it's stress reaches a point known as the ultimate strength of that material, $\sigma_{ult}$

From now on I will assume that both ropes are attached to some kind of a rig such that their attachment points are always parallel, the system cannot twist or shear, and the only way that the system can move is by the bars, to which the ropes are attached, being separated by a force, $F_{app}$, being applied.

In this problem, there are two regions whose behaviour is qualitatively different.

The first region is where the total separation of the bars is less than the natural length of the longer rope, in this case the longer rope is slack, ie it has no tension, and the applied force is entirely balanced by the force from the shorter rope: $$F_{app} = F_B = Af\left(\frac{x_B}{l_B}\right)$$ This region continues until the separation of the bars, $x_A+l_A$, is equal to the natural length of the longer rope. If the ultimate strength of the ropes is reached before that, the shorter rope will break and (at least in the instant of the breakage) the tension in the longer rope will be zero.

When the separation of the bars is equal to the natural length of the longer rope: $$x_B+l_B = l_A$$ $$\implies \frac{x_b}{l_A} = \frac{l_A-l_B}{l_B}$$ $$\implies f\left(\frac{x_b}{l_A}\right) = f\left(\frac{l_A-l_B}{l_B}\right)$$

So the tension in the longer rope will be zero when the shorter rope breaks if the ultimate stress is less than the stress at this separation, ie: $$\sigma_{ult} < f\left(\frac{l_A-l_B}{l_B}\right)$$

Otherwise we move into the second region, where the external force is balanced by (possibly different) tensions in each of the ropes. For some given separation of the bars, $d$, we have: $$F_{app} = F_A + F_B = Af\left(\frac{d-l_A}{l_A}\right) + Af\left(\frac{d-l_B}{l_B}\right)$$ Now, the shorter rope will still break first, but the tension in the longer rope when this happens will be non-zero, to find this tension: set the stress in the shorter rope to be the material's ultimate strength, find the bar separation for which this occurs, and substitute this into the the equation for the stress in the longer rope.

To do this you need to know the exact form of the stress-strain relationship, for Hookean behaviour: $$\sigma_{ult} = E \frac{d-l_B}{l_B}$$ $$\implies d = \frac{l_B\sigma_{ult}}{E} + l_B$$ $$\implies F_A = A\sigma_A(d) = A\sigma_{ult}\frac{l_B}{l_A}+AE\left(\frac{l_B}{l_A}-1\right)$$

EDIT:

In general the cross sectional area of the ropes will also be a function of strain, ie $A=g(\epsilon)$. Hence: $$\sigma = \frac{F}{A} = f(\epsilon) \implies F = g(\epsilon)f(\epsilon)$$ A common form for the dependence of the area on strain is: $$A = A_0\epsilon^{-\nu}$$ where $\nu$ is the Poisson's ratio. http://en.wikipedia.org/wiki/Poisson's_ratio

If the volume is conserved during deformation, Poisson's ratio is equal to exactly one half, but there is no fundamental reason that volume should be conserved.

The results above are easy to generalise using this new formula, although the algebra will be complicated. Start with:

$$F_{app} = F_A + F_B = g\left(\frac{d-l_A}{l_A}\right)f\left(\frac{d-l_A}{l_A}\right) + g\left(\frac{d-l_B}{l_B}\right)f\left(\frac{d-l_B}{l_B}\right)$$

then substitute whatever functional dependence you want the area and stress to have for $f$ and $g$, eg $f(\epsilon)=E\epsilon$, $g(\epsilon)=A_0\epsilon^{-\nu}$, find the value of $d$ for which the shorter rope breaks using $\sigma_{ult} = f\left(\frac{d-l_B}{l_B}\right)$, then substitute this into the equation for the tension in rope A. The algebra will be complicated but it's not difficult.

• Thanks but the question is what can be said about the he tension in the ropes the moment they break, not which breaks first. I'm wondering if the elongation of the longer rope deforms the rope that much that the tension (strain) is less. Feb 19, 2014 at 22:34
• When one of the ropes breaks the tension in that rope depends entirely on the properties of that rope's material ie the ultimate tensile stress of that material multiplied by the area of the rope's cross section, what this tension will be depends on the material and can't be found from the information you've given. The tension in the rope which doesn't break at the moment that one of the ropes breaks can be found using my reasoning above.
– JPB
Feb 20, 2014 at 2:22
• When you say: "I'm wondering if the elongation of the longer rope deforms the rope that much that the tension (strain) is less." I don't know what you mean: The elongation in the longer rope deforms which rope? The tension is less than what?
– JPB
Feb 20, 2014 at 2:22
• English not being my native language I made some translation errors. I will edit the question Feb 22, 2014 at 16:27

The question leaves a bit open to interpretation, but generally tension in a stationary rope is simply the applied force. Which will break first is harder to tell, because it depends on other factors, such as whether they are horizontal and subject to gravity and slight drooping, or vertical and subject to gravity, or some other setup. Assuming that they are perfectly ideally identical (excepting length), and defect-free, and the experiment were done in a way that gravity plays no role, then I'd say they would break at the same applied force.