12-24 할리데이 11판 솔루션 일반물리학

$$\begin{cases} \theta_1&=51.0\degree\\ \theta_2&=66.0\degree\\ m&=704\ut{kg}\\ g&=9.80665\ut{m/s^2} \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} T_A&=mg\\ &=6.9038816\times10^3\ut{N}\\ &\approx 6.90\times10^3\ut{N}\\ &\approx 6.90\ut{kN}\\ \end{aligned}$$ $$\ab{b,c}$$ $$\begin{cases} \theta&=\theta_1+\theta_2-90\degree&=27.0\degree\\ \phi&=90\degree-\theta_2&=24.0\degree\\ \end{cases}$$ $$\begin{aligned} 0&=\vec T_A+\vec T_B+\vec T_C\\ &=\br{-T_A\cos\phi\i-T_A\sin\phi\j}+\br{T_B\sin\theta\i+T_B\cos\theta\j}+T_C\i\\ \end{aligned}$$ $$\begin{cases} 0&=-T_A\cos\phi+T_B\sin\theta+T_C\\ 0&=-T_A\sin\phi+T_B\cos\theta\\ \end{cases}$$ $$\begin{cases} T_B&=\frac{\sin\phi}{\cos\theta}mg\\ T_C&=\frac{\cos\br{\theta+\phi}}{\cos\theta}mg\\ \end{cases}$$ $$\begin{cases} T_B&\approx 3.1515612399524152\times10^3\ut{N}\\ T_C&\approx 4.876230813496548\times10^3\ut{N}\\ \end{cases}$$ $$\begin{cases} T_B&\approx 3.15\times10^3\ut{N}\\ T_C&\approx 4.88\times10^3\ut{N}\\ \end{cases}$$ $$\begin{cases} T_B&\approx 3.15\ut{kN}\\ T_C&\approx 4.88\ut{kN}\\ \end{cases}$$