Calculating Young's Modulus for Iron Wire Experiment

[rit]

Homework Statement

I had to do an experiment and create a graph with the data.
We attached one end of the Iron wire to a clamp and then tied on weights of 100g at a time to the other end so that a force can be applied to the Iron wire and so it stretches. (as pizza1512 did but in copper)

Homework Equations

Diameter of wire: 0.31mm (therefore area is 0.302mm²)
Length of wire: 2m70cm (270cm)

The Attempt at a Solution

i worked out that if u apply 100g (1Newton force)
1/2700= 3.7x10^-4 (not sure if right)

And stress is
1/0.302= 3.3

I worked out the Youngs Modulus
3.3/(3.7x10^-4) and i got 8918.92

is that right?
and how do i plot it on a graph

 

[LowlyPion]

Homework Statement

I had to do an experiment and create a graph with the data.
We attached one end of the Iron wire to a clamp and then tied on weights of 100g at a time to the other end so that a force can be applied to the Iron wire and so it stretches. (as pizza1512 did but in copper)

Homework Equations

Diameter of wire: 0.31mm (therefore area is 0.302mm²)
Length of wire: 2m70cm (270cm)

The Attempt at a Solution

i worked out that if u apply 100g (1Newton force)
1/2700= 3.7x10^-4 (not sure if right)

And stress is
1/0.302= 3.3

I worked out the Youngs Modulus
3.3/(3.7x10^-4) and i got 8918.92

is that right?
and how do i plot it on a graph

How did you work that out? From what? What are the units of that expression?

 

[rit]

It is for Strain.
it extended by 1mm
so,
1mm/2700mm (length of wire) =3.7^-4

 

[LowlyPion]

It is for Strain.
it extended by 1mm
so,
1mm/2700mm (length of wire) =3.7^-4

$$E = \frac{\sigma}{\epsilon} = \frac{F*L}{A_o*\Delta L} = \frac{1N * 2.7m}{.302*10^{-6}m^2*.001m}$$

Straighten out your units.

All you have provided is one data point so I'm not sure what kind of graph you're talking about.