12-40 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} L&=12.0\ut{m}\\ \theta&=50.0\degree\\ T&=400\ut{N}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=-T+N_x\\ 0&=-mg+N_y\\ 0&=TL\cos\theta-mg{L\over2}\sin\theta\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} m\vec g &={2T\over\tan\theta}\j\\ &=800\tan40\degree\j\ut{N}\\ &\approx 671.279704941824\j\ut{N}\\ &\approx 671\j\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{cases} N_x&=T\\ N_y&=mg\\ \end{cases} $$ $$ \begin{aligned} \vec N&=T\i+mg\j\\ &=400\i+800\tan40\degree\j\ut{N}\\ &\approx 400\i+671\degree\j\ut{N}\\ \end{aligned} $$