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?

  • $\begingroup$ This doesn't actually look like homework, so I'm removing the homework tag. $\endgroup$
    – user4552
    Sep 29, 2014 at 1:04
  • $\begingroup$ Can you add a diagram please? $\endgroup$
    – BMS
    Sep 29, 2014 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. $\endgroup$ Sep 29, 2014 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, 2014 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. $\endgroup$
    – user4552
    Sep 29, 2014 at 20:34

1 Answer 1


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.

  • $\begingroup$ 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, 2014 at 1:00
  • 1
    $\begingroup$ @kenik $E$ has to be in $\text{Pa}$, not $\text{GPa}$, to get Newtons. $\endgroup$
    – JiK
    Sep 29, 2014 at 7:06
  • 1
    $\begingroup$ 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, 2014 at 7:45
  • 1
    $\begingroup$ @JiK If I use Pa I get 115297,46 which doesn't make much sense either. $\endgroup$
    – kenik
    Sep 29, 2014 at 9:20
  • 1
    $\begingroup$ @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, 2014 at 10:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.