# Homework Help: Three Difficult Materials Questions (check my answers!)

1. Dec 5, 2007

### DizzyDoo

[SOLVED] Three Difficult Materials Questions (check my answers!)

Hi there. As you can see below, there are three questions on materials, that are quite difficult in my opinion. I've given my best attempt at them, checked them, and I think they are correct. However, as I am prone to making stupid mistakes, and I really want to get all three 100% right, I'd really appreciate it if one or two people could see if they get the same answers as me. Thanks!

1. The problem statement, all variables and given/known data
Question 1: A steel wire of diameter 0.5mm and length 1.84 is suspended vertically from a fixed point and used to support a 60N weight hung from the lower end of the wire. The Young Modulus for steel is 2.0x10^11 Nm^-2. Calculate the extension of the wire.

Question 2: A 3500kg statue is placed on top of a cylindrical concrete (Young Modulus = 2.3x10^10 Nm^-2) stand. The stand has a cross-sectional area of 7.3x10^-2 m^2 and a height of 1.8m. By how much does the statue compress the stand?

Question 3: A 72kg mountain climber hangs on a nylon rope (radius = 6.5mm). If the rope stretches by 5.0x10^-2m what is the un-stretched length of the rope? (Young Modulus of nylon = 1x10^9 NM^-2

2. Relevant equations

YM = Stress / Strain
Stress = Force / Area
Strain = eXtension / Length
Area of circle = (pie)r^2

3. The attempt at a solution

1)

Firstly I changed the diameter of 0.5mm into metres:

0.5 / 1000 = 0.0005m

I then worked out the area of the wire:

(pie)x(0.0005/2)^2 = 1.9x10^-7

I then stuck every value into the Young Modulus equation:

2.0x10^11 = (60 / 1.9x10^-7) / (x / 1.84)

2.0x10^11 = 305577490.7 / (x / 1.84)

Stuck with the extension on the wrong side, so a quick bit of rearranging and...

x = (305577490.7 / 2.0 x 10^11) x 1.8
x = 0.001527887 x 1.8
x = 0.00275m
x = 2.7x10^-3m

2) First off, I get the force from the mass (weight):

3500 x 9.81 = 34335

Stick my information into the young modulus equation:

2.3x10^10 = (34335 / 7.3x10^-2) / (x / 1.8)

Solve and rearrange:

x = (470342 / 2.3x10^10) x 1.8
x = 0.000036
x = 3.6 x 10^-5 However, the question asks for the compression, so is the answer correct?

3) Again, firstly I work out the force with the mass...

72 x 9.81 = 706.32

And change the radius into metres, so I can use it:

6.5mm / 1000 = 0.0065m

Work out the area of the rope:

(pie)x 0.0065^2

Now, stick all that I have into the young modulus equation:

1x10^9 = (706.32 / 0.000132732) / (5.10^-2 / L)

1x10^9 = 5321387.9 / (5x10^-2 / L)

Time to rearrange and solve: (Check my rearranging!)

L = (5x10^-2 x 1x10^9) / 5321387.9

L = 7.2m (Which sounds about right in my common sense check)

I understand that's a lot to check, but hopefully I've made it clear enough for anyone wanting to lend a hand. Thank you muchly for your time!

2. Dec 5, 2007

### vector3

Axial extensions or compression D = (P*L)/(A*YM) --> P = applied force
P/A is the applied stress, L is the length and A is the Area

The answers to 1 and 2 look correct to me.

The answer to 3 something seems to be missing something. D*A*YM / P = L
I get 9.39m but you should check my math.

When an applied force extends an object it is called a "tension" or "tensile" force.
When an applied force compresses or shortens an object it is called "compression" or "compressive" force.