Solving an AS Physics Problem: Steel & Brass Extension

[chaz699]

hello, i am currently going over past exam questions to further my knowledge in as physics when i came across this old question which i don't quite understand,

a length of steel and a length of brass are joined together. This combination is suspended from a fixed support and a force of 80n is applied at the bottom end

each wire has a cross scetional area of 2.4 x 10(-6)m(squared)

length of steel wire = 0.80m
length of brass wire = 1.40m
youngs modulas for steel = 2.0 x 10(11)pa
youngs modulas for brass = 1.0 x 10(11)pa

calculate the total extension produced when the force 80n is applied


basically so far I've substituted stress to be 80n/2.4 x 10(-6)

but I am not sure about the change in length on the strain side of things

someone help?

 

[Chi Meson]

If this is not a calculus-based question, then the answer is simple: the total extension is the sum of the individual extension with 80 N applied.

This ignores the fact that the upper portions of wire will have to include the weight of the lower portions (in addition to the 80 N). This means you'd need to integrate the strain over the length of the wire.

Two reasons why I don't think that's required: on first glance this additional weight is insignificant (very thin wires), and they don't specify which wire is above the other.

So I'm thinking it's the simpler solution. Find the strain of each and apply it to the original length of each to find the extension of each, then add them.

 

[chaz699]

but how do i find the strain on each with the information given

 

[asleight]

but how do i find the strain on each with the information given

$$E=\sigma/\epsilon$$,

where $$E$$ is YM, $$\sigma$$ is stress, and $$\epsilon$$ is strain.

 

[Chi Meson]

and strain is the ratio of change in length over initial length, ∆L/L