Answer
$T_{1}=1380 N$
Work Step by Step
$m = 300$ $kg$
$A=2.00x10^{-6}$ $m^{2}$
$L = L_{1}=L_{3}=2.0000$ $m$
$L_{2}=L+0.006$ $m$
Young's Modulus for Steel: $E=200x10^{9}$ $\frac{N}{m^{2}}$
Sum of forces in the Y direction: (1) $$T_{1}+T_{2}+T_{3}-mg=0$$
Since $L_{1} = L_{3}$, $\Delta L_{1} = \Delta L_{3}=\Delta L$ and (2) $\Delta L=\Delta L_{2} + 0.006$
Using the tension formula: $$\frac{F}{A}=E\frac{\Delta L}{L} $$
In this problem, F = T so solve for $\Delta L$ we get: (3) $$\Delta L = \frac{TL}{AE} $$
Note that $T_{1}$ and $T_{3}$ are equal as the lengths are equal. For$\Delta L_{2}$: (4) $$\Delta L_{2} = \frac{T_{2}(L+0.006)}{AE}$$
Sub (3) and (4) into (2): (5) $$\frac{TL}{AE}=\frac{T_{2}(L+0.006)}{AE}+0.006 $$
Solve for T2: (6) $$T_{2}=\frac{TL-AE(0.006)}{L+0.006} $$
Let $T = T_{1} = T_{3}$ and combine (1) and (6): $$2T+\frac{TL-AE(0.006)}{L+0.006}=mg$$
Solving for $T$: $$T=\frac{mg(L+0.006)+AE(0.006)}{3L+2(0.006)}$$
Therefore, $$T= \frac{(300)(9.8)(2+0.006)+(2x10^{-6})(200x10^{9})(0.006)}{3(2)+2(0.006)}=1380 N$$
