# How Do You Calculate Wire Extension Under Sudden Loads?

• giodude
giodude
Homework Statement
(a) An object of mass 0.5 kg is hung from the end of a steel wire of length 2 m and of diameter 0.5 mm. (Young's modulus = ##2 \times 10^{11} \frac{N}{m^{2}}##). What is the extension of the wire?

(b) The object is lifted through a distance ##h## (thus allowing the wire to become slack) and is then dropped so that the wire receives a sudden jerk. The ultimate strength of steel is ##1.1 \times 10^{9} \frac{N}{m^{2}}##. What is the largest possible value of ##h## if the wire is not to break?
Relevant Equations
##A = \frac{\pi \times d^{2}}{4}##
##k = \frac{AY}{l_{0}}##
##F = mg = kx##
##stress = \frac{\Delta P}{A}##
##strain = \frac{\Delta l}{l_{0}}##
##Y = \frac{stress}{strain}##
Hi, I solved part (a) and will provide my solution below. However, I've been working on part (b) for quite a bit and reviewed the provided, relevant text a few times now but haven't been able to find what I'm missing:

Solution (a):
Using ##A = \frac{\pi \times d^{2}}{4}##, ##k = \frac{AY}{l_{0}}##, and ##F = mg = kx## we have:

##x = \frac{m \times g \times l_{0} \times 4}{\pi \times d^{2} \times Y} = 0.25 mm##

Solution (b):
We can use the cross sectional area and the given ultimate strength of steel to solve for the maximum allowed force:
##1.1 \times 10^{9} = \frac{F}{A}##
##F = 216 N##

I'm also aware that the potential energy at the peak just before the rod is dropped is ##PE = mgh## and that this potential energy is converted entirely to kinetic energy during the sudden jerk such that:

##PE = mgh = F \times \Delta l = KE##

However, I get stuck dealing with the ##\Delta l## term which I'm using to denote the wire extension that occurs during the sudden jerk.

Any suggestions on where I'm going wrong would be much appreciated!

What type of motion will occur when the wire becomes taut?

Simple harmonic motion?

giodude said:
Simple harmonic motion?
Right, so you cannot use ##F\times\Delta l## since F is not constant.
Instead of trying to find the velocity when it becomes taut, think about the conservation of energy from start to finish. What is the change in GPE?

The change in GPE will be mgh from start to finish. Which means KE will have changed from 0 to mgh at the moment of maximum tension.

giodude said:
The change in GPE will be mgh from start to finish. Which means KE will have changed from 0 to mgh at the moment of maximum tension.
Where do you think the maximum tension occurs in the motion?

It'd occur just before the sudden jerk? So, at PE = 0 and KE = mgh?

I think I'm starting to get somewhere. We know, due to the starting state of lifting and dropping that the initial speed, ##v_{0} = 0##. Additionally, we know that ##KE = mgh## at the point of peak tension. Therefore,

## \Delta KE = \frac{1}{2} \times m \times v_{f}^{2} - \frac{1}{2} \times m \times v_{0}^{2} = m \times g \times h##

Plugging in ##v_{0} = 0##, we have:
##\frac{1}{2} \times m \times v_{f}^{2} = m \times g \times h##
##v_{f} = \sqrt{2 \times g \times h}##

We also know that in SHM ##x = A cos(\omega t + \alpha)##, where ##\omega^{2} = \frac{k}{m}##. So,

##v = \frac{dx}{dt} = - \omega A sin(\omega t + \alpha)##
##a = \frac{d^{2}x}{dt^{2}} = - \omega^{2} A cos(\omega t + \alpha)##

KE is maximum where ##a = 0## at ##cos(\omega t + \alpha) = 0## and ##sin(\omega t + \alpha) = 1##. Therefore, ##\sqrt{2gh} = \omega A##

giodude said:
We also know that in SHM ##x = A cos(\omega t + \alpha)##, where ##\omega^{2} = \frac{k}{m}##. So,

##v = \frac{dx}{dt} = - \omega A sin(\omega t + \alpha)##
##a = \frac{d^{2}x}{dt^{2}} = - \omega^{2} A cos(\omega t + \alpha)##

KE is maximum where ##a = 0## at ##cos(\omega t + \alpha) = 0## and ##sin(\omega t + \alpha) = 1##. Therefore, ##\sqrt{2gh} = \omega A##
But you don't want max velocity. What motion variable is maximised when the tension is maximum?

Oh, I see. If ##T = kx - mg = ma## is at its maximum then acceleration needs to be maximum. As a result, we want ##cos(\omega t + \alpha) = 1## and ##sin(\omega t + \alpha) = 0##.

This makes sense because this sets ##x = Acos(\omega t + \alpha)## to its max ##x = A##.

giodude said:
Oh, I see. If ##T = kx - mg = ma## is at its maximum then acceleration needs to be maximum. As a result, we want ##cos(\omega t + \alpha) = 1## and ##sin(\omega t + \alpha) = 0##.

This makes sense because this sets ##x = Acos(\omega t + \alpha)## to its max ##x = A##.
Right.
You still have one small inexactitude, but it's probably insignificant. The loss in GPE will be a little more than mgh.

giodude
Whats the inexactitude?

Also how does this reconcile with ##v_{f} = \sqrt{2gh}## if ##v = 0## at time ##t##?

We also know ##T = 216 N## from the original post, given the ultimate strength of the wire. So,

##T = kx - mg##
##216 N = kx - (0.5 kg)(9.8 \frac{m}{s^{2}})##
##kx = 220.9 N##
##x = \frac{220.9 N}{k} = 0.0113 m##

Dimensionally this resolves properly. However, the solution in the textbook is ##h = 0.23m##. Hmmm.

Oh, think I figured it out. This might be where the inexactitude you mentioned is coming in:

The ##x## in the previous message is the amplitude. Now we can plug it into ##mgh = \frac{1}{2}kA^{2}## which we derive from the conservation of energy equation:
##h = \frac{kA^{2}}{2mg} = 0.2536m##

Hopefully this is correct! Either way, have learned a bunch here. Thank you for your help!

giodude said:
##T = kx - mg##
The total extension is x+(mg)/k.
The small inexactitude is that the lost GPE is mg(h+x), but x is tiny compared to h.

Last edited:

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