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Suppose I weigh about $75~\text{kg}$ and I am attached to a (non-elastic) rope which is attached to a carabiner which can take at least $20~\text{kN}$. Suppose furthermore that I reach terminal velocity before the rope is tight, a bit far fetched maybe, but pretend that I reach $530~\text{km/h}$.

What force is being applied to the carabiner once the rope finally is tight?

Since $530~\text{km/h}$ is slightly exaggerated, what speed should I expect to reach when free-falling $5$ meters while rock climbing?

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It depends how springy you are! For real, just attach a 75kg weight, in the form of a rigid block of metal. If the rope is not elastic, the force on the carabiner is unlimited, and something will break - carabiner, rope, something. Elasticity is everything. For the speed, just use $x=1/2gt^2$, let x = 5, solve tor t, and multiply by g. –  Mike Dunlavey Jul 31 '13 at 21:00
    
@Gugg I would like to just point out that this isn't really my homework, I was just curious as I'm a beginner rock climber. But perhaps it should go to the homework category nonetheless. –  Pål GD Jul 31 '13 at 21:35
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@PålGD I judged it should. "Applies to questions of primarily educational value - not only questions that arise from actual homework assignments, but any question where it is preferable to guide the asker to the answer rather than giving it away outright." The $530~\text{km/h}$ indicated to me that you are not only interested in practical aspects. –  Glen The Udderboat Jul 31 '13 at 21:44
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This problem cannot be solved without considering the elasticity of the rope and the carabiner together. Also what is attahed to the other side of the rope. This is important too. –  ja72 Aug 1 '13 at 13:17
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closed as off-topic by Dilaton, Qmechanic Aug 7 '13 at 13:00

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2 Answers

As @Kaz suggests, the force being applied on you will be a function of time and will vary, so all you'll get from $m\Delta v=F\Delta t$ is the average force. But if you observe the force-time graph, it will look somewhat like this.

enter image description here

In fact, it will be a lot steeper than this graph.
We can use that to approximate this graph to a triangle having its vertex at $F_{max}$ and its base of length $\Delta t$ on the time-axis.

The impulse, or the area under the graph equals $$J=\int Fdt=\frac 12F_{max}\Delta t$$ We also know that the impulse equals the change in linear momentum given by $$J=m\Delta v$$

Equating the two we get, $$F_{max}=2m\frac{\Delta v}{\Delta t}$$

This $F_{max}$ will be the tensile force felt by the carabiner if we assume it to be massless, or almost massless compared to the person, in which case the string on the other end will also exert the same force $F_{max}$ on it. So if this $F_{max}$ exceeds $20kN$, the carabiner will break.

Also note that $\Delta t$ will have to be calculated experimentally, because I don't think there is a way to derive it using the data at hand.

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This is impossible to answer without a lot of additional information and assumptions. The force applied to the carabiner is, of course:

$$F = ma$$

Your mass is known, but the acceleration isn't. It depends on flexibility in the carabiner itself and in your body, and the linkage between the carabiner and your body. (Of course, we would normally include the rope also, but it is a given that this is inelastic.)

If we know how many seconds is required for you to stop, we can use the momentum-equals-impulse formula to get $f$, namely: $\Delta mv = F\Delta t$. However, this $F$ will just be an average $F$ over the stopping time $\Delta t$, not an instantaneous $F$, and certainly not the maximum $F$, which is what we would like to know.

Over the course of the event, force is a function of time; it is not constant. The carabiner fails if the function, at any instant, that force exceeds its load handling.

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The $F$ above is the total force. The total force can be zero and the carabiner still break if two equal and opposite forces are applied. –  ja72 Aug 1 '13 at 13:41
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