12-5 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} mg&=413\ut{N}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_x&=0\\ \Sigma F_y&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N_x-T\sin\theta\\ 0&=N_y+T\cos\theta-mg\\ 0&=mg\br{L\over2}\sin\br{2\theta}-TL\sin\theta\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} T &=mg \cos\theta\\ &={413\sqrt2\over2}\ut{N}\\ &\approx 357.6684917629731\ut{N}\\ &\approx 358\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} N_x &= {1\over2} mg \sin(2\theta)\\ &={413\over4}\sqrt3\ut{N}\\ &\approx 178.8342458814866\ut{N}\\ &\approx 179\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} N_y &=mg \sin^2\theta\\ &=-{413\over4}\ut{N}\\ &= 103.25\ut{N}\\ &\approx 103\ut{N}\\ \end{aligned} $$