# Stopping force on wire rope from a falling object attached to it

Okay, bit of a weird one. Imagine an object of known mass in free fall, tethered by a wire rope to an infinitely stiff sky hook. When that wire rope eventually tightens, what will be the stopping force on said object, i.e. what is the dynamic tension experienced by the wire?

Using linear motion equations we can determine the final velocity of the object. If the stopping time or distance were known then we could calculate the object's deceleration and therefore the force.

However, how does one determine the deceleration? I thought about using Hooke's law or the wire's strain as a way to determine it's elongation - but one cannot work these out without the force! Also I can't find any reliable info on a wire's stiffness. Is there a rule of thumb for something like this?

Any help or pointers are gratefully received.

• This sounds like a harmonic motion equation? Commented Jan 22, 2020 at 14:41
• 'wire' in this case can be replaced by 'spring', with force exerted being directly proportional to displacement/stretch distance (times Young modulus). If you know the two, you'll know the maximum force (and maximum deceleration) at the end of the travel.
– SF.
Commented Jan 22, 2020 at 14:59
• It depends on the acceleration ( negative) rate ; as in F = M a . Commented Mar 18, 2021 at 14:00
• F * t = M * v, the variable force in the line times the duration of application counters the objects mass times its velcity profile. It will result in simple harmonic motion. Also keep in mind the constant weight, which will be the only load at stasis. (There will be a distribution of load in the wire due to its own weight, but I guess you'll consider that negligible.) Commented Mar 18, 2021 at 14:27

## 2 Answers

Assuming the following properties for the rope:

• $$L$$: undeformed length of the rope
• $$E$$: young's modulus of the rope
• $$A$$: cross-section
• weight of the rope is zero (it greatly simplifies the calculation)

Then defining the x-axis as the vertical axis and

• setting 0 the location that the wire/rope is completely free,
• setting downward as positive axis

then at a distance x (again x is positive below 0), the tension on the rope will be equal to:

$$T(x) = \frac{E A}{L} x$$

At that location on the object there are only two forces:

• tension on the rope T(x)
• Gravity $$m*g$$

Therefore the equation is: $$\sum F_x = m\cdot a_x$$ $$mg - T(x) = m\cdot a_x$$ with some rearranging: $$m\cdot a_x + T(x) = mg$$

Because $$a_x = \ddot{x}$$ $$m\cdot \ddot{x} + T(x) = mg$$

$$m\cdot \ddot{x} + \frac{E A}{L} x = mg$$

You can substitute $$\frac{E A}{L}$$ with $$k$$ and you get the very basic problem of the undamped harmonic oscillator (there are some caveats -e.g. there is only force when x is positive- but I won't go into that):

$$m\cdot \ddot{x} + k x = mg$$

This equation along with the boundary conditions

• $$x(t)=0$$
• $$\dot{x} (t)=v_0$$ : the initial velocity when the rope is at tension for the first time.

is a very basic Initial Value Problem (IVP). You can easily calculate the function for $$x(t)$$

$$x{\left(t \right)} = \frac{g m}{k} + \left(- \frac{g m}{2 k} - \frac{v_{0}}{2 \sqrt{- \frac{k}{m}}}\right) e^{- t \sqrt{- \frac{k}{m}}} + \left(- \frac{g m}{2 k} + \frac{v_{0}}{2 \sqrt{- \frac{k}{m}}}\right) e^{t \sqrt{- \frac{k}{m}}}$$

and with that you can calculate the $$T(x)$$

• +1 though "you can easily calculate" and square-root exponents seems like a bit of an oxymoron :p.
– Wasabi
Commented Mar 19, 2021 at 12:37
• touché.. Although you forgot to mention negative square root exponents. Initially, I didnt' bother to put the equation (because you can literally find in dynamics and vibrations books), and then I thought "what the hell", and I spend literally 3 minutes to fire up a wolphramalpha page to find the solution. I didn't bother to clean up the $e^{\pm\omega t}$. I wasn't even sure if anybody was going to read this so... :-)
– NMech
Commented Mar 19, 2021 at 14:17

Let's say you have a wire of steel with the cross-sectional area $$A$$ and length $$L$$ (for instance, say 20 km). Then, the strain energy of the stretched wire is equal to the kinetic energy of the falling mass once it has reached a depth beyong $$L$$ from the sky hook. Therefore, $$\frac{1}{2}mv^2+mgh=\frac{1}{2}A E_{steel}\epsilon_{steel}^2 \$$ since the work done by strain is a triangular area. We can then calculate for $$\epsilon = \epsilon_{steel}$$, $$\Delta x=\epsilon \cdot 20km.$$ And as has been mentioned in the comment by @Jonathon R swift, the mass will bounce back up.

• The mass has an amount of energy, which is transferred to the 'spring'. You have included Kinetic, but ignored the GPE that is lost by the mass as it continues to move downwards while the wire is stretching. If the mass is travelling fast, and the wire is stiff, this is a fair assumption to simplify the maths. Otherwise, you need to add in an $M*g*\Delta x$ term to the strain energy... And how to find the value of $\Delta x$? Looks like it's integration time... Commented Jan 22, 2020 at 16:05
• thank you, Jonathan. that makes sense in some cases where the wire is very flexible but we usually ignore that such as we ignore the mass of the wire and the changing of its CG. Commented Jan 22, 2020 at 17:37
• That sounds good - but does a steel cable have an equivalent axial stiffness to a round steel bar of the same diameter? Commented Jan 23, 2020 at 13:52
• It depends, at hi strains the cable is stronger, but has to be restrained against unwinding, or else it could spin. Commented Jan 23, 2020 at 19:16
• I think SF has the better explanation. The exact solution will not be found until gaining the understanding on the property of the rope. Here is information on wire ropes, and characteristics of elongation under force. randers-reb.com/fishing-rope/rope-technology/….
– r13
Commented Mar 17, 2021 at 21:20