12-45 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=300\ut{kg}\\ A&=2.0\times10^{-6}\ut{m^2}\\ L_1=L_3&=2.0000\ut{m}\\ L_2&=L_1+d\\ d&=6.00\ut{mm}\\ g&=9.80665\ut{m/s^2}\\ E_{\text{steel}}&=200\times10^9\ut{N/m^2} \end{cases} $$ $${F\over A}=E{\Delta L\over L},$$ $$ \begin{cases} F_1&=AE\cfrac{\Delta L_1}{L_1}\\ F_2&=AE\cfrac{\Delta L_2}{L_2}\\ \end{cases} $$ $$ \begin{cases} F_1&=AE\cfrac{\Delta L_1}{L_1}\\ F_2&=AE\cfrac{\Delta L_1-d}{L_1+d}\\ \end{cases} $$ $$ \begin{cases} F_1&=2 \Delta L_1\times10^5\\ F_2&=\cfrac{4 (500 \Delta L_1-3)}{1003}\times10^5\\ \end{cases} \taag1$$ $$ \begin{aligned} \Sigma F_{y}&=0\\ &=2F_1+F_2-mg\\ &=2\br{AE\cfrac{\Delta L_1}{L_1}}+\br{AE\cfrac{\Delta L_1-d}{L_1+d}}-mg\\ \end{aligned} $$ $$ \begin{aligned} \Delta L_1&=\frac{L_1 }{A E }\cdot\frac{A d E+ mg (d+L_1)}{ 2 d+3 L_1}\\ &=\frac{276721399}{40080000}\times10^{-3}\ut{m}\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} F_1&=AE\cfrac{\Delta L_1-d}{L_1+d}\\ &=\frac{276721399}{200400}\ut{N}\\ &\approx 1380.8453043912175\ut{N}\\ &\approx 1.4\times10^3\ut{N}\\ &\approx 1.4\ut{kN}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} F_2&=AE\cfrac{\Delta L_1}{L_1}\\ &=\frac{180665}{1002}\ut{N}\\ &\approx 180.30439121756487\ut{N}\\ &\approx 1.8\times10^2\ut{N}\\ \end{aligned} $$