- #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!

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!