솔루션피아

12-46 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=95\ut{kg}\\L&=15\ut{m}\\2R&=9.6\ut{mm}\\\Delta L &= 1.3\ut{cm}\\\end{cases} $$$${F\over A}=E{\Delta L\over L},$$$$ \begin{aligned}E&={F\over A}\cdot{L\over\Delta L}\\&={mgL\over \pi R^2\Delta L}\\&=\frac{185546875000 g}{39 \pi }\\&\approx 1.4851141642012506\times10^{10}\ut{N/m^2}\\&\approx 1.5\times10^{10}\ut{N/m^2}\\&\approx 15\ut{GN/m^2}\\\end{aligned} $$

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..

12-44 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} L&=2.50\ut{m}\\mg&=500\ut{N}\\d&=1.50\ut{m}\\\end{cases} $$$$ x=\sqrt{L^2-d^2},$$$$\ab{a}$$$$ \begin{aligned}\Sigma \tau&=0\\&=-\frac{x}{2}mg+LP\\\end{aligned} $$$$ \begin{aligned}P&=\frac{mgx}{2L}\\&=\frac{mg}{2L}\sqrt{L^2-d^2}\\&=200\ut{N}\\\end{aligned} $$$$\ab{b}$$$$\put x=\sqrt{L^2-d^2}$$$$\sin\theta=\frac{d}{L},$$$$ \begin{aligned}\vec P &=-P\sin\theta\i+P\cos\theta\j\\&=-..