12-12 할리데이 11판 솔루션 일반물리학

$$\begin{cases} \theta_1&=40\degree\\ \end{cases}$$ $$\begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \end{cases}$$ $$\begin{cases} 0&=-T_1\cos\theta_1+T_2\cos\theta_2\\ 0&=T_1\sin\theta_1+T_2\sin\theta_2-mg\\ \end{cases}$$ $$T_2={mg\cos\theta_1\over\sin\br{\theta_1+\theta_2}}$$ $$\ab{a}$$ $$\begin{aligned} \min (T_2),\\ {\sin\br{\theta_1+\theta_2}}&=1\\ \theta_1+\theta_2&=90\degree\\ \theta_2&=50\degree\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} T_2 &={\cos\theta_1\over\sin\br{\theta_1+\theta_2}}mg\\ &=\cos40\degree mg\\ &\approx 0.766044443118978 mg \\ &\approx 0.77 mg \\ \end{aligned}$$