12-29 할리데이 11판 솔루션 일반물리학

$$\begin{cases} L&=6.10\ut{m}\\ mg&=510\ut{N}\\ h&=4.00\ut{m}\\ \theta_0&=70\degree\\ \end{cases}$$ $$\begin{cases} r_f&=h\\ r_{\text{floor}}\tan\theta&=h\\ \br{r_{mg}+\frac{L}{2}\cos\theta}\tan\theta&=h \end{cases}$$ $$\begin{cases} N_{\text{edge }x}&=N_{\text{edge}}\sin\theta\\ N_{\text{edge }y}&=N_{\text{edge}}\cos\theta\\ \end{cases}$$ $$\begin{aligned} f&=\mu N_{\text{floor}}, \end{aligned}$$ $$\begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases}$$ $$\begin{cases} 0&=f-N_{\text{edge }x}\\ 0&=N_{\text{edge }y}+N_{\text{floor}}-mg\\ 0&=r_ff-r_{\text{floor}}N_{\text{floor}}+r_{mg}mg\\ \end{cases}$$ $$\begin{cases} 0&=\mu N_{\text{floor}}-N_{\text{edge}}\sin\theta\\ 0&=N_{\text{edge}}\cos\theta+N_{\text{floor}}-mg\\ 0&=h\mu N_{\text{floor}}-{h\over\tan\theta}N_{\text{floor}}+\br{{h\over \tan\theta}-{L\over2}\cos\theta}mg\\ \end{cases}$$ $$\begin{aligned} \therefore \mu&=\frac{2 L \sin\theta \sin2 \theta}{8 h-L \sin\theta -L \sin3 \theta}\\ &=\frac{244 \sin40 \degree \cos 20 \degree}{701-122 \cos 20 \degree}\\ &\approx 0.2513510216566625\\ &\approx 0.251\\ \end{aligned}$$