Engineering Asked by WestyTea on September 1, 2021
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.
Assuming the following properties for the rope:
Then defining the x-axis as the vertical axis and
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:
Therefore the equation is: $$sum F_x = mcdot a_x$$ $$mg - T(x) = mcdot a_x$$ with some rearranging: $$ mcdot a_x + T(x) = mg $$
Because $a_x = ddot{x}$ $$ mcdot ddot{x} + T(x) = mg $$
$$ mcdot ddot{x} + frac{E A}{L} x = mg $$
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):
$$ mcdot ddot{x} + k x = mg $$
This equation along with the boundary conditions
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 $T(x)$
Answered by NMech on September 1, 2021
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.
Answered by kamran on September 1, 2021
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