12-7 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} E_\text{Iron}&=200\times10^9\ut{N/m^2}\\ m&=210\ut{kg}\\ R&=1.20\ut{mm}\\ p&=2.00\ut{mm}\\ L_{A}&=2.50\ut{m}\\ L_{B}&=L_{A}+p\\ \Delta L_B&=\Delta L_A-p\\ \end{cases} $$ $${F\over A}=E\cdot{\Delta L\over L},$$ $$ \begin{cases} 0&=\Sigma F=T_A+T_B-mg\\ T_A&=E (\pi R^2)\cdot{\Delta L_A\over L_A}\\ T_B&=E (\pi R^2)\cdot{\Delta L_B\over L_B}\\ \end{cases} $$ $$ \begin{aligned} \Delta L_A&=\frac{{p E\pi R^2+mg(L_A+p)}}{ E \pi R^2(2 L_A+p)}L_A,\\ &=\br{\frac{2919 g}{3201280 \pi }+\frac{5}{5002}}\ut{m}\\ &\approx 3.845903854320749\times10^{-3}\ut{m}\\ &\approx 3.85\times10^{-3}\ut{m}\\ &\approx 3.85\ut{mm}\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} T_A&=E (\pi R^2)\cdot{\Delta L_A\over L_A}\\ &=\frac{90 (2919 g+3200 \pi )}{2501}\ut{N}\\ &\approx 1.391876731600904\times10^3\ut{N}\\ &\approx 1.39\times10^3\ut{N}\\ &\approx 1.39\ut{kN}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} T_B &=E (\pi R^2)\cdot{\Delta L_B\over L_B}\\ &=E (\pi R^2)\cdot{\Delta L_A-p\over L_{A}+p}\\ &=\frac{1500 (175 g-192 \pi )}{2501}\ut{N}\\ &\approx 667.5197683990962\ut{N}\\ &\approx 668\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} 0&=\Sigma \tau\\ &=d_A T_A-d_BT_B, \end{aligned} $$ $$ \begin{aligned} {d_A\over d_B} &={T_B\over T_A}\\ &=\frac{\frac{1500 (175 g-192 \pi )}{2501}}{\frac{90 (2919 g+3200 \pi )}{2501}}\\ &=\frac{8750 g-9600 \pi }{8757 g+9600 \pi }\\ &\approx 0.4795825328808613\\ &\approx 0.480\\ \end{aligned} $$