# Mass falling while attached to wire

1. Aug 11, 2013

### mbigras

1. The problem statement, all variables and given/known data
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 x 1011 N/m2). 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 x 109 N/m2. What is the largest possible value of h if the wire is not to break?

2. Relevant equations
$$stress = \delta = F/A$$
$$strain = \epsilon = \Delta l/l_{0}$$
$$\delta / \epsilon = Y$$
Average impact force * distance traveled = $$\Delta KE$$
ultimate strength = $$\delta_{UTS}$$

3. The attempt at a solution
Initially I thought that I couldn't relate ultimate strength to distance traveled once the mass starts pulling on the wire using Young's modulus because the relation between stress and strain at that point is no longer linear. Then I saw a rephrasing of this exact question that included the statement "Assume stress remains proportional to strain throughout the motion". However, in my textbook this statement isn't mentioned, which could be why our answers aren't matching.

But assuming that stress does remain proportional to strain, I calculate the distance the mass traveled using Young's modulus. Using the ultimate strength I found the max force. Relating the two with the work-energy principle I was able to find the change in kinetic energy which is the same as the change in the original gravitational potential energy. Knowing how much poential energy it needs tells me h.
$$\delta_{UTS} = F/A = 1.1*10^{9} N/m^{2}$$
$$\Delta l = \delta_{UTS}*l_{0} / Y = .011 m$$
$$F = \delta_{UTS}*A = 215.985 N$$
$$F*\Delta l = mgh$$
$$h = .48 m$$
This is not the answer the book has which is h = .23 m. Using the work-energy princple seems to me where I might be making some assumptions that aren't realistic. Also the wording "sudden jerk" is throwing me for a loop, maybe assuming the wire stretches all the way isn't realistic. It feels like I'm missing something here.

Last edited: Aug 11, 2013
2. Aug 11, 2013

### voko

Is the diameter 0.5 meters, or millimeters?

Regarding your solution, the work-energy principle is good here, but you made a couple of mistakes. First, in $F \Delta l = mgh$ you assumed that tha force is constant. This is not so, as the force changes with the extension linearly, so $F$ is not constant. Perhaps you know the formula for potential energy in a spring?

Secondly, in that same formula you assumed that the potential energy due to gravity depends only in $h$. But as wire extends, the object gets lower, which decreases its potential energy further.

3. Aug 11, 2013

### rude man

If the wire were a spring with spring constant k so that F = kx, how would you solve for max. F? This problem is entirely analogous so just figure out what k is in terms of the given parameters.

4. Aug 11, 2013

### mbigras

Potential energy of a spring! The energy transfer is from gravitational potential energy to spring potential energy.
$$x = -\frac{l_{0}*\delta_{UTS}}{Y} = -0.011 m$$
$$k = \frac{A*Y}{l_{0}}$$
$$mgh = \frac{1}{2} kx^{2}$$
$$h≈.24m$$
thank you guys.

5. Aug 11, 2013

### rude man

Good shot!
EDIT: Voko is right, you need to rethink what h really is. It's a small correction. Otherwise you're spot on.
(Hint: how far does the mass really fall?).

Last edited: Aug 11, 2013
6. Aug 11, 2013

### voko

You still have a 1 cm difference with the correct result. Pay attention to the second part of my comment in #2.

7. Aug 11, 2013

### mbigras

Ahh yes. I thought it was just a rounding error but now I see. I forgot to include the negative gravitational potential energy that the mass has once it is stretched all the way. So:
$$mgh = \frac{1}{2}kx^{2} -mgx$$
$$h≈.23m$$
with x = 0 at the bottom of the unstretched wire.

Last edited: Aug 11, 2013