How Do You Calculate Wire Extension Under Sudden Loads?

  • #1
giodude
30
1
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!
 
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  • #2
What type of motion will occur when the wire becomes taut?
 
  • #3
Simple harmonic motion?
 
  • #4
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?
 
  • #5
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.
 
  • #6
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?
 
  • #7
It'd occur just before the sudden jerk? So, at PE = 0 and KE = mgh?
 
  • #8
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}##
 
  • #9
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##
 
  • #10
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?
 
  • #11
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##.
 
  • #12
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.
 
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Likes giodude
  • #13
Whats the inexactitude?

Also how does this reconcile with ##v_{f} = \sqrt{2gh}## if ##v = 0## at time ##t##?
 
  • #14
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.
 
  • #15
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!
 
  • #16
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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