Question

The length of a rod under longitudinal tension $T_1$ is $L_1$ and that under longitudinal tension $T_2$ is $L_2$. What is the actual length of the rod, in the absence of tensions?

• A. $\frac{L_1T_1-L_2T_2}{T_2-T_1}$
• B. $\frac{L_1T_2-L_2T_1}{T_2+T_1}$
• C. $\frac{L_1T_1-L_2T_2}{T_2+T_1}$
• D. $\frac{L_1T_2-L_2T_1}{T_2-T_1}$

Answer 1

Answer: (D)

$\frac{L_1T_2-L_2T_1}{T_2-T_1}$

Answer 2

The length of a rod under longitudinal tension $T$ is given by $L = L_0 \left(1 + \frac{T}{AY}\right)$, where $L_0$ is the original length. This can be written as $L = L_0 + \left(\frac{L_0}{AY}\right)T$. This is a linear relationship between $L$ and $T$. We have two points $(T_1, L_1)$ and $(T_2, L_2)$. The equation of the line passing through these points is $\frac{L - L_1}{T - T_1} = \frac{L_2 - L_1}{T_2 - T_1}$. The actual length of the rod in the absence of tension is $L_0$, which is the value of $L$ when $T=0$. Substituting $T=0$ and $L=L_0$: $\frac{L_0 - L_1}{0 - T_1} = \frac{L_2 - L_1}{T_2 - T_1}$. Solving for $L_0$: $L_0 = L_1 - T_1 \left(\frac{L_2 - L_1}{T_2 - T_1}\right) = \frac{L_1(T_2 - T_1) - T_1(L_2 - L_1)}{T_2 - T_1} = \frac{L_1T_2 - L_1T_1 - T_1L_2 + T_1L_1}{T_2 - T_1} = \frac{L_1T_2 - L_2T_1}{T_2 - T_1}$.