12-6 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m_A&=11.0\ut{kg}\\ \theta&=30.0\degree\\ m_B&=7.00\ut{kg}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$T_A=m_Ag\sin\theta,$$ $$ \begin{aligned} \vec T_A &=-m_Ag\sin\theta\cos\theta\i-m_Bg\sin^2\theta\j\\ &=-{11\sqrt3g\over4}\i-{11g\over4}\j\taag1\\ \end{aligned} $$ $$T_B=m_Bg,$$ $$ \begin{aligned} \vec T_B &=-m_Bg\j\\ &=-7g\j\taag2\\ \end{aligned} $$ $$ \begin{aligned} \vec T_S &=-\br{\vec T_A+\vec T_B}\\ &={{11\sqrt3g\over4}\i+{39g\over4}\j}\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} T_S &=\abs{{11\sqrt3g\over4}\i+{39g\over4}\j}\\ &={g\over2}\sqrt{471}\\ &\approx 106.4145795565597\ut{N}\\ &\approx 106\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \phi&=\tan^{-1}{{39g\over4}\over{11\sqrt3g\over4}}\\ &=\tan^{-1}{{13\sqrt3}\over11}\\ &\approx 1.1163690504087669\ut{rad}\\ &\approx 1.12\ut{rad}\\ \end{aligned} $$