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. 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.

  • $\begingroup$ This sounds like a harmonic motion equation? $\endgroup$ – Jonathan R Swift Jan 22 '20 at 14:41
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    $\begingroup$ '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. $\endgroup$ – SF. Jan 22 '20 at 14:59

Let's say you have a wire of steel with the area of A and length L(could be 20 km, say).

Then the strain energy of the stretched wire is equal to the kinetic energy of the falling mass.

$1/2mv^2+mgh=1/2A*E_{steel}\epsilon_{steel}^2 \ $ Because the work done by strain is a triangular area. From here we can calculate

$\epsilon_{steel}\ and\\ \Delta x=\epsilon*20km .$

And as has been mentioned in @Jonathon R swift comment the mass will bounce back up.

  • $\begingroup$ 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... $\endgroup$ – Jonathan R Swift Jan 22 '20 at 16:05
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    $\begingroup$ 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. $\endgroup$ – kamran Jan 22 '20 at 17:37
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    $\begingroup$ That sounds good - but does a steel cable have an equivalent axial stiffness to a round steel bar of the same diameter? $\endgroup$ – WestyTea Jan 23 '20 at 13:52
  • $\begingroup$ It depends, at hi strains the cable is stronger, but has to be restrained against unwinding, or else it could spin. $\endgroup$ – kamran Jan 23 '20 at 19:16
  • $\begingroup$ Sorry, but why are we adding the kinetic and potential energy together? Surely it would be one or the other? $\endgroup$ – WestyTea Feb 5 '20 at 11:58

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