12-3 할리데이 11판 솔루션 일반물리학

$$\begin{cases} m_A&=85\ut{kg}\\ m_B&=10\ut{kg}\\ M&=\Sigma m= 95\ut{kg}\\ \end{cases}$$ $$\begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau_{D}&=0\\ \end{cases}$$ $$\begin{cases} 0&=N_C-f_D\\ 0&=N_D-Mg\\ 0&=\br{x_{AD}}m_Ag+\br{x_{BD}}m_Bg-\br{y_{CD}}N_C\\ \end{cases}$$ $$\begin{cases} x_{AD}&={L_{AD}\over L_{CD}}x_{OD}\\ x_{BD}&={1\over 2}x_{OD}\\ y_{CD}&=\sqrt{{L_{CD}}^2-{x_{OD}}^2}\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} N_C &={x_{OD}\over y_{CD}}\cdot\br{{L_{AD}\over L_{CD}}m_A+{m_B\over 2}}g\\ &={x_{OD}\over \sqrt{{L_{CD}}^2-{x_{OD}}^2}}\cdot\br{{L_{AD}\over L_{CD}}m_A+{m_B\over 2}}g\\ &={1475\over8\sqrt{231}}g\\ &\approx 118.96431792914245\ut{N}\\ &\approx 1.2\times10^2\ut{N}\\ \end{aligned}$$ $$\ab{b,c}$$ $$\begin{cases} f_D&=N_C\\ N_D&=Mg\\ \end{cases}$$ $$\begin{aligned} \vec F_D &=f_D\i+N_D\j\\ &=N_C\i+Mg\j\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} F_D &=\sqrt{{N_C}^2+\br{Mg}^2}\\ &=g \sqrt{M^2+\br{x_{OD}\over y_{CD}}^2\br{{L_{AD}\over L_{CD}}m_A+{m_B\over2}}^2}\\ &=\frac{5 }{8}\sqrt{\frac{5424049}{231}}g\\ &\approx 939.1965856775718\ut{N}\\ &\approx 9.4\times10^2\ut{N}\\ \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} \theta &=\tan^{-1}{Mg\over N_C}\\ &=\tan^{-1}{M\over {x_{OD}\over y_{CD}}\cdot\br{{L_{AD}\over L_{CD}}m_A+{m_B\over 2}}}\\ &=\tan^{-1}{152\sqrt{231}\over295}\\ &\approx 1.4437891006369301\ut{rad}\\ &\approx 1.4\ut{rad}\\ \end{aligned}$$