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I'm a climber and I constructed myself an anchor that I fixed to a rock wall. To test it, I hooked to it a 12mm in section steel cable with a length of 2,8m and a concrete block of 30kg to the other tip. I then dropped it from anchor level and it held. I am now wondering what kind of impact force was developed in this test. Can you help me please?

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  • $\begingroup$ This doesn't actually look like homework, so I'm removing the homework tag. $\endgroup$ – Ben Crowell Sep 29 '14 at 1:04
  • $\begingroup$ Can you add a diagram please? $\endgroup$ – BMS Sep 29 '14 at 2:53
  • $\begingroup$ A side note, as I happen to pass here and am a climber myself. I'd like to warn you about your anchor. Apparently, you repeatedly applied a factor two of fall on the anchor, with a static rope. By doing so, chances are great that you ruined its mechanical properties. As for all the gear implied in your test, I wouldn't use it anymore for climbing. $\endgroup$ – Stanislasdrg Sep 29 '14 at 9:16
  • $\begingroup$ Hi @Ben Crowell. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Sep 29 '14 at 9:18
  • $\begingroup$ @Qmechanic: Thanks for pointing me to that info. After reading it, I still don't think this should have the homework tag. $\endgroup$ – Ben Crowell Sep 29 '14 at 20:34
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When the mass reaches its lowest point, the steel wire will have increased in length from $L$ to $L+x$. So equating the strain energy of the wire with the initial gravitational potential energy of the ball:

$$\frac{1}{2}kx^2 = mg(L+x) \approx mgL $$

which rearranges to

$$ x = \sqrt{\frac{2mgL}{k}} $$

Note that

$$ k = \frac{EA}{L} $$

where $E$ is Young's Modulus (about 200 GPa for steel), $A$ is the cross-sectional area of the wire, and $L$ is the length.

The largest force that acts is then

$$ f=kx=\sqrt{2mgEA} $$

I'll let you plug the numbers in.

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  • $\begingroup$ Hi, thank you vm for your prompt answer. My numbers add up to 3646. Are those in newtons? It's been a long time since i had physics. $\endgroup$ – kenik Sep 29 '14 at 1:00
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    $\begingroup$ @kenik $E$ has to be in $\text{Pa}$, not $\text{GPa}$, to get Newtons. $\endgroup$ – JiK Sep 29 '14 at 7:06
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    $\begingroup$ Nice to note that the force does not depends on the falling height! This is in principle because a longer rope has a smaller spring constant, so it extends more distributing the impact acceleration on a longer time. It would be a nice high-school-lab experiment to verify this. It should be possible to see that as this formula assumes $L\approx L+x$ this is not completely true. $\endgroup$ – DarioP Sep 29 '14 at 7:45
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    $\begingroup$ @JiK If I use Pa I get 115297,46 which doesn't make much sense either. $\endgroup$ – kenik Sep 29 '14 at 9:20
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    $\begingroup$ @kenik You should realize that steel is extremely stiff - your block will stop with a 'shock' and thus has an extremely large deceleration. Using newton's F=m*a, it's easy to see that the force must be extremely large. Also note that the actual force may even be larger, since this description assumes quasi-static conditions in the steel wire. $\endgroup$ – Sanchises Sep 29 '14 at 10:23

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