youngs modulus....

mercury

there are two rods rod1 , made of brass - length L1=2m
area of cross section A1 2x10^-4 m^2 and young's modulus y1= 10^11
rod 2 , made of steel , lenght L2 , cross sectional area A2=10^-4 m^2 and youngs modulus y2 = 2x10^11 .

the question is if the two rods are fixed end to end , and two equal and opposite forces are applied at each end (the ends not fixed to each other-free ends )pulling the combination of both the rods outward,
and the value of the forces applied are 5x10^4 on both sides.
then find the length of rod 2 such that both have equal increase in length dl (read delta L ) as a result of the forces.

here's what i did ,
i assumed that because both the rods are fixed at one end , any extension that takes place will occur in the direction of the force

i.e F = (y1.A1.dl)/L1
i know F,A1,y1,L1 theredore i can find dl
now i substituted this dl in the corresponding equation for rod 2 and then found L2 ( F=( y2.A2.dl)/L2 )
and i got it to be equal to 2m but the answer in the solutions given out was 1.8m

i want to know what is wrong with my assumption because i am sure something is wrong with it...

or if take the lenght of the total rod to be L1+L2 i.e L2+2and take the midpt. of the rod as the origin for calculating ..
oh well i'm very confused..
please help me if u can!
thanks

arcnets

Well I can't see any flaw in your calculation. The steel rod has only half the cross section, but double strength, so the effect should be the same.

The only correction I can think of, is: If a rod gets longer, it will also get thinner. So maybe you have to use different A's.

On the other hand, if you plug in values, you find dl = 5mm. Or .25% of the original length. Not enough to explain the 10% difference. Plus, I doubt the rods can take such a strong force and still be in the elastic domain. Imagine you suspend a 5-ton truck by only a 2m x 9mm brass rod...