12-28 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} 2R&=2.4\ut{cm}\\ m&=670\ut{kg}\\ E_{\text{steel}}&=200\times10^9\ut{N/m^2}\\ g&= 9.80665\ut{m/s^2} \end{cases} $$ $$ {F\over A}= E{\Delta L\over L}, $$ $$ \begin{aligned} \Delta L &={ FL \over A E} \\ &={ mgL \over \pi R^2 E} \\ &={ 67g \over 2880000\pi } L\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} \Delta L &={ 67g \over 2880000\pi } L\\ &=\frac{13140911}{4800000000 \pi }\ut{m}\\ &\approx 8.714337259919424\times10^{-4}\ut{m}\\ &\approx 8.7\times10^{-4}\ut{m}\\ &\approx 0.87\ut{mm}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Delta L &={ 67g \over 2880000\pi } L\\ &=\frac{2378504891}{28800000000 \pi }\ut{m}\\ &\approx 0.02628825073409026\ut{m}\\ &\approx 0.026\ut{m}\\ &\approx 2.6\ut{cm}\\ \end{aligned} $$