Tensile strain produced on two thin rods of different lengths

Question

Suppose there are two thin rods $Y$ and $Z$ with length $L_1$ and $L_2$ respectively. $L_2$ has larger magnitude than $L_1$. Both rods have same density $p$, cross sectional area $A$, Young's Modulus $E$ and force applied perpendicularly to the cross sectional area $F$.

For simple compression or tension we have equation: $${F\over A}= E {\Delta L\over L} $$ that relates tensile stress and strain.

We can also write this equation as: $${FL_1\over AE}={\Delta L_1}$$ and $${FL_2\over AE}= {\Delta L_2}$$ for rod with length $L_1$ and $L_2$ respectively.

We can notice from above equations that same force $F$ causes more change in length ($\Delta L_2$) of rod ($L_2$) (since magnitude of $L_2$ is higher, $\Delta L_2$ would be higher) but my book says more force must be applied to rod with larger length ($L_2$) if we want to keep the ratios ${\Delta L_1\over L_1}$ and ${\Delta L_2\over L_2}$ same for both rods. So I think my concept isn't clear and I am wrong somewhere. Please correct me.

Answer 1

You are correct that—all else being equal—a longer rod will contract proportionally more under an axial load. In this case, however, it sounds like a greater-than-proportional contraction is required, as all rods are required to contract to the same final length. In other words,

$$L_4-\Delta L_4=L_3-\Delta L_3$$

$$\frac{\Delta L_4}{L_4}> \frac{\Delta L_3}{L_3}$$

To avoid confusion, it would be better if the book said something like "The single longer leg must be compressed by a larger relative amount (or larger strain) $\Delta L_4/L_4$ and thus by a force with a larger magnitude $F_4$."

Does this resolve the issue?