12-25 할리데이 11판 솔루션 일반물리학

$$\begin{cases} 2R&=4.8\ut{cm}\\ L&=5.3\ut{cm}\\ m&=1500\ut{kg}\\ G&=3.0\times10^{10}\ut{N/m^2}\\ g&=9.80665\ut{m/s^2}\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} F&=mg\\ &=14709.975\ut{N}\\ &\approx 1.5\times10^4\ut{N}\\ &\approx 15\ut{kN}\\ \end{aligned}$$ $$\ab{b}$$ $${F\over A}=G{\Delta x\over L},$$ $$\begin{aligned} \Delta x&={FL\over AG}\\ &={Lmg\over \pi R^2 G}\\ &=\frac{53 g}{11520000 \pi }\ut{m}\\ &\approx 1.436131451417069\times10^{-5}\ut{m}\\ &\approx 1.4\times10^{-5}\ut{m}\\ &\approx 14\ut{\mu m}\\ \end{aligned}$$