12-39 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} 2x&=3.44\ut{m}\\ y&=25.0\ut{cm}\\ mg&=3160\ut{N} \end{cases} $$ $$ \begin{aligned} \tan\theta&={y\over x},\\ \end{aligned} $$ $$\Sigma F_y=0,$$ $$ \begin{aligned} 0&=2T_y-mg\\ T_y&={mg\over2} \end{aligned} $$ $$ \begin{aligned} T&={T_y\over \sin\theta}\\ &=m g\cdot\frac{ x }{2 y}\sqrt{\frac{y^2}{x^2}+1}\\ &=\frac{316 }{5}\sqrt{30209}\ut{N}\\ &\approx 1.098462544468404\times 10^4\ut{N}\\ &\approx 1.10\times 10^4\ut{N}\\ &\approx 11.0\ut{kN}\\ \end{aligned} $$