12-10 할리데이 11판 솔루션 일반물리학

$$\begin{cases} S_{\text{wood}}&=50\times10^6\ut{N/m^2}\\ \rho_{\text{wood}}&=525\ut{kg/m^3}\\ A_{\text{wood}}&=d^2\\ M&=430\ut{kg}\\ a&=1.9\ut{m}\\ b&=2.5\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases}$$ $$\begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases}$$ $$\begin{cases} 0&=N_x-T_x\\ 0&=N_y-Mg-mg\\ 0&=aT_x-bMg-{b\over2}mg\\ \end{cases}$$ $$\begin{cases} N_x&={b\over2a}\br{m+2M}g\\ N_y&=\br{m+M}g\\ \end{cases}$$ $$m=\rho AL=\rho A(a^2+b^2),$$ $$\begin{cases} N_x&=\frac{25}{76} \left(10353 A+1720\right) g\\ N_y&=\left(\frac{10353 }{2}A+430\right) g\\ \end{cases}$$ $$\begin{aligned} N&\lt{1\over6}N_{\max}\\ \sqrt{{N_x}^2+{N_y}^2}&\lt{1\over6}AS_{\max}\\ \end{aligned}$$ $$A \gtrsim 0.0008420800400491018\ut{m^2}$$ $$\begin{aligned} d &\gtrsim 0.029018615405444516\ut{m}\\ &\gtrsim 0.029\ut{m}\\ &\gtrsim 2.9\ut{cm}\\ \end{aligned}$$