Length change for a rod of two sections

[Myr73]

A rod is made of two sections joined end to end. The sections are identical, except that one is steel and the other is brass. While one end is held fixed, the other is pulled to result in a change in length of 1.20mm. By how much does the length of each section increase ?
I understand that E (brass) = 1/2 E steel.
However first of all. when it change in length of 1.20 mm, is it talking about the complete change in length or just the one of the steel?
After that, I am unsure how to put the equation together. I understand that Change in L = 1/E (F/A) L0
But considering we don't have F and A. I know a previous explanation just took F/A out, but I can't see how that would work.
Thank you

 

[TSny]

Hello, Myr73. Welcome to PF!

I think they mean the total change in length is 1.20 mm.

Can you find the ratio ΔL_steel/ΔL_brass?

 

[Myr73]

Thank you,
Umm so like {(1/Esteel)(F/A)L0steel }/ {(1/Ebrass)(F/A) L0brass} ??
do F and A cancel there--> 1/Esteel L0Steel / 1/Ebrass Lobrass ?? Idk

 

[TSny]

Thank you,
Umm so like {(1/Esteel)(F/A)L0steel }/ {(1/Ebrass)(F/A) L0brass} ??
do F and A cancel there--> 1/Esteel L0Steel / 1/Ebrass Lobrass ?? Idk

Yes, F and A cancel. Anything else cancel? Note that they said each rod is identical except for the material.

 

[Myr73]

So Lo cancels as well lol.. Then ΔLsteel/ΔLbrass= 1/ESteel / 1 / Ebrass.. Would I cancel the 1's as well? = ESteel/Ebrass?
Does ΔLsteel/ΔLbrass=2 ?

 

[TSny]

$$\frac{\Delta L_{steel}}{\Delta L_{brass}} = \frac{ \frac{1}{E_{steel}}}{\frac{1}{E_{brass}}}$$

Be careful when you simplify this. Do you remember rules for dividing by a fraction?

 

[Myr73]

umm.. ΔLsteel/ΔLbrass= Ebrass/Esteel.= 0.5 ?

 

[TSny]

Yes.

 

[Myr73]

hugh..nice thanks.. however I don't know what to do next.

 

[TSny]

You have two unknowns. What are they?

To solve for two unknowns you need two equations that relate the unknowns. Can you write down two equations that you could use to solve for the unknowns?

 

[Myr73]

The two unknowns are Delta L (brass) and (steel). Or do you mean the F and A.. I am sorry I honestly don't know. on-line classes sometimes leaves me with a shortage of information.

 

[TSny]

Yes, your two unknowns are ΔL_steel and ΔL_brass.

You have found the ratio of these and that gives you one equation to work with.

You will need another relation between ΔL_steel and ΔL_brass. Can you write down a second relation (or equation) based on what's given in the problem?

 

[Myr73]

ok,
Let's see I have --> L0(brass)= L0(Steel) , L = 2L0(b)=2L0(s) or = L0(b) + L0(s)
Delta L = Lfb-L0b + Lfs-L0s ( although I can't see how I can use that)
Delta L= 1.2mm
ok umm --> 1.2mm = Delta L (brass) + Delta L (S)
and 0.5 Delta L Brass= Delta L Steel
--> 1.2=1.5 Delta L (Brass)
--> 1.2/1.5=0.8mm = Delta L (Brass)
1.2-0.8= 0.4mm= Delta L( Steel)
Is that right, or am I way off ? lol

 

[TSny]

ok umm --> 1.2mm = Delta L (brass) + Delta L (S)
and 0.5 Delta L Brass= Delta L Steel
--> 1.2=1.5 Delta L (Brass)
--> 1.2/1.5=0.8mm = Delta L (Brass)
1.2-0.8= 0.4mm= Delta L( Steel)
Is that right, or am I way off ? lol

That looks correct. Good work!

 

[Myr73]

Thank you :).. You help me a lot , I appreciate your kindness