Young's modulus problem -- need a hint

[redshift]

There are two wires, one brass the other copper, both 50 cm long and 1.0 mm diameter. They are somehow connected to form a 1m length. A force is applied to both ends, resulting in a total length change of 0.5 mm. Given the respective young's moduluses of 1.3 x 10^11 and 1.0 x 10^11, I'm supposed to find the amount of length change in each section.

Apparently a variation of Hooke's law should be used here, such as F/A=Y(change in length/original length)

I'm stuck on how can I solve this with 2 unknowns (force and change in length)?
Regards

[Clausius2]

There are two wires, one brass the other copper, both 50 cm long and 1.0 mm diameter. They are somehow connected to form a 1m length. A force is applied to both ends, resulting in a total length change of 0.5 mm. Given the respective young's moduluses of 1.3 x 10^11 and 1.0 x 10^11, I'm supposed to find the amount of length change in each section.

Apparently a variation of Hooke's law should be used here, such as F/A=Y(change in length/original length)

I'm stuck on how can I solve this with 2 unknowns (force and change in length)?
Regards

Hello redshift! I'm going to rewritte your problem in terms of stress \sigma (Pa) and unitary deformation \epsilon=\frac{L-L_o}{L_o} where Lo is the original lenght. So that, the stress exerted is the same in each section of the wire:

Hooke's law: \sigma=E_t \epsilon_t=E_1 \epsilon_1=E_2 \epsilon_2 where "Et" (N/m^2) is the apparent Young modulus of the complete wire.

Compatibility of deformations: \bigtriangleup L=\bigtriangleup L_1 + \bigtriangleup L_2


Then, you have three equations for three unknowns: Et, epsilon1 and epsilon2.

Hope this help you a bit.




You've got two unknowns for

[redshift]

I think I get it. Based on your equations, the ratio of the young's moduluses should equal the ratio of the individual increases, that is, 13/10 = L1/L2
Therefore, 10L1 = 13L2
Since, L1 = L2 = 0.5, then L1 = 0.5 - L2. Plugging this into the above gives 10(0.5 - L2) = 13L2, so that the increase of L2 (brass wire) is 0.21 mm.

Many thanks!