Rope strength necessary to stop a fall

Question

Just a random thought experiment, that got me confused.

So when you buy a rope you get info about Tensile Strength(lbs) or kg let's say 1000kg.

That means it can hold an object of that mass.

But the object is stationary and I would like to calculate it for a falling object.

So let's say we have a 10kg object that falls, it was falling for 1s without drag so the speed is 1g *1s = 9.8m/s.

What's the Tensile Strength necessary how to calculate it?

First I was thinking convert it into force F =am but if a = g that would mean that the force is not proportional to speed. And that doesn't feel ok.

We could probably do something with Kinetic energy as it is E=(v^2m)/2 so it scales with v and m. But I don't know what.

My physics is a bit rusty I'm years out of college.

Answer 1

your problem is the bungee jumping problem

you jump from a clip with a rope of a length $~L~$ and stiffens $~k~$ .

what is the maximum of the rope tension ?

you have two cases

I)

free falling case until your height $~y_I~$ is less the the rope length $~L~$

$$y_I=\frac g2\,t^2\quad, v=g\,t$$

from here with

$$y_I=L\quad\Rightarrow~ t_L=\frac{\sqrt{2gL}}{g}\quad, v_L=g\,t_L=\sqrt{2gL}$$

II)

now the elasticity ($~k~$) of the rope is active , you obtain this equation:

$$ m\ddot y(t)=F=m\,g-k\, y(t)$$

You can solve this differential equation with the initial conditions $~y(0)=0~,\dot y(0)=v_L~$ analytically and obtain the force $~F(t)~$. From here the maximal force $~F_m~$ is

$$F_m = mg \sqrt{1 + \frac {2kL}{mg}}\\ F_m(k=0)=m\,g $$

---

with $~k=\frac{E\,A}{L}~$ and $~A=\pi\left(\frac d2\right)^2~$

$$F_m = mg\, \sqrt{1 + \frac {E\,\pi\,d^2}{2\,mg}}\\ \tag 2$$

where $~E~$ is the (tensile) elastic modulus.

your question

If you buy a rope you know the maximal allowable weight ($\;[kg]~$) that it can carry out, you can measure the diameter $~d~$ and you can find out the elastic modulus $~E~$ . with these data applied equation (2) and obtain $~\frac{F_m}{g}~[kg]$ . if the result (with margin ) is less then the maximal allowable weight you can use this rope ,otherwise you can't.

Answer 2

I think that in the equation $F=ma$ the acceleration is due to the stopped fall, not what the body was going through before the fall. Say the rope stopped a body falling with velocity $v$ in $t$ time (stopping times may differ due to elasticity of the rope, and maybe other factors that I don't know of) then the average acceleration $a=\frac{v}{t}$.
But then again average acceleration is of limited use in this case. You'd want to know the highest instantaneous acceleration and compare it to the limit of the rope.
Another way would be to find the maximum extension in the rope. The Young's modulus is defined as $$Y=\frac{Fl}{A\Delta l}$$
Now for a rope $A$ and $l$ are constant (roughly). So we can rewrite this as $$F=k\Delta l$$ Add a negative to account for the direction, and you have a spring! With the maximum mass the rope can withstand $F$ and the constant $k$ which can be experimentally found, we can calculate $\Delta L$ the maxmimum extension in the rope before it snaps.
Then we equate the kinetic energy of the falling body with the potential energy in the rope at maximum extension. $$\frac{1}{2}mv^2=\frac{1}{2}k(\Delta L)^2\\ \implies v=\sqrt{\frac{k}{m}}\Delta L$$

I may have overlooked something, so feel free to correct me. Also don't go buying safety ropes just based off of this answer.