# Impact force in a fall

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?

• This doesn't actually look like homework, so I'm removing the homework tag. – user4552 Sep 29 '14 at 1:04
• Can you add a diagram please? – BMS Sep 29 '14 at 2:53
• 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. – Standaa - Monica side Sep 29 '14 at 9:16
• 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. – Qmechanic Sep 29 '14 at 9:18
• @Qmechanic: Thanks for pointing me to that info. After reading it, I still don't think this should have the homework tag. – user4552 Sep 29 '14 at 20:34

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.

• 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. – kenik Sep 29 '14 at 1:00
• @kenik $E$ has to be in $\text{Pa}$, not $\text{GPa}$, to get Newtons. – JiK Sep 29 '14 at 7:06
• 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. – DarioP Sep 29 '14 at 7:45
• @JiK If I use Pa I get 115297,46 which doesn't make much sense either. – kenik Sep 29 '14 at 9:20
• @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. – Sanchises Sep 29 '14 at 10:23