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I heard this quote a lot (especially from my teachers) and I always wanted to know whether or not it was actually true. I think that its partially true; a chain with one weak link is weaker than a chain with a lot of links that are slightly weak. So does the force just spread out or what actually happens?

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It is true, although the details depend on the chain.

One simple model is as follows: the chain is composed of links that are solid until the force on them exceeds a threshold, $T_\text{breaking}$, the breaking tension. Once this happens, the link comes apart completely and the chain is broken. This threshold can be different for the different links.

When you apply a tension force $T$ to the whole chain, then as long as it doesn't break, the force does indeed get "spread out" - each link will experience a tension of exactly $T$, the same as all the other links. (This is just how tension forces work.) This means that the chain will break as soon as $T$ exceeds the breaking tension of the weakest link. Therefore under this mode the chain is exactly as strong as its weakest link. I think this model is probably pretty good for most actual chains.

One can certainly imagine other models, which might be more appropriate depending on the chain. For example, each link might have a probability of breaking per unit time, with this probability being proportional to the force exerted, with the constant of proportionality being the strength of the link. In this case the strength of the chain is proportional to the products of the strengths of all the links, so a long chain with lots of slightly weak links can be weaker than a short chain with a single weak link.

But the real point of the expression is that a chain can never be stronger than its weakest link. Because all the links experience the same force, if the weakest link breaks then so does the whole chain. There's nothing you can add in another part of the chain to reduce the force on the weak link and therefore prevent it from breaking.

Of course all of this is assuming you're applying the force steadily to both ends of the chain, with the chain itself either not being very heavy, or resting on a horizontal surface. In the case of a heavy chain hanging downwards, or when a force is applied very suddenly to one end, you can end up with more force being applied to some links than to others. In such a situation the weakest link might not experience very much force, and therefore a stronger link might be likely to break first. I suppose this opens up the question of how you define the strength of a chain. But if it's defined in terms of a uniform steady force being applied to both ends, with no other forces acting, then the old saying is correct.

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    $\begingroup$ Re This is just how tension forces work. That's only true in freshman physics, where one uses massless chains. This is not how tension works in a massive chain. $\endgroup$ Sep 27 '14 at 7:27
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    $\begingroup$ I've edited my answer - see the last paragraph. $\endgroup$
    – Nathaniel
    Sep 27 '14 at 8:01
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No. Not necessarily. A common demonstration done in the physics lab is suspending a heavy object by a string, then below that a string of slightly larger diameter (and strength). If the lower string is pulled down slowly, the thinner string above will break. But if the string is pulled quickly, that string will break (the stronger link) Why? Because the inertia of the heavy object resists being accelerated. Newtons 2nd law.

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    $\begingroup$ This isn't really a case of the thicker string being weaker, though, it's just a case of more force being applied to the stronger string than the weaker one. $\endgroup$
    – Nathaniel
    Sep 27 '14 at 7:52

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