# Forces exerted on an anchor point from a falling object

In summary, the conversation discusses the forces exerted on an anchor point when a secured object is dropped. The formula Fmax = mg + sqrt(2mg x E x A x fall factor +(mg)2 ) is given, with explanations of the variables. The formula may not fully account for the "whip" effect of the rope becoming taut and straightening. The speaker is unsure if this formula is the correct method for estimating the load on the anchor point.

Hi,

I'm looking at the forces exerted onto an anchor point when a n object is dropped but is secured to that anchor point witha steel wire rope lanyard.

i can find this but not sure if it is relevant or not to my investigation:
Fmax = mg + sqrt(2mg x E x A x fall factor +(mg)2 )

The mass is 69kg, E is 193GPa, A = 15.6x10-6, fall factor 2 (4m rope 8m total drop)

I know the KE and final velocity but I'm not sure of the stopping distance.

any guidance helps!

Thanks!

You've given a formula. Can you motivate that formula?

At a guess, that is for a massless wire rope which becomes taut and then stretches under the force the falling object. As you've explained, the "fall factor" is the ratio of drop distance to rope length.

I think the formula you mean to write is:$$F_\text{max} = mg + \sqrt{2mg \times E \times A \times \frac{h}{l} + 2mg}$$
At a guess, E is Young's modulus for the rope and A is its cross-sectional area. I've used h for the fall distance and l for the rope length. m is obviously the object's mass and g is the acceleration of gravity.

In reality, I suspect that you will get also get some cushioning from the "whip" effect as a rope that was not straight rapidly becomes straight when it tightens. It will be hard to quantify that and will require at least the rope's linear density as a parameter.

Thanks, this what i meant

I've got the theoretical value, what I'm trying to ascertain is if this is the correct method to estimate the load on the anchor point.

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