12-19 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=1.05\ut{kg}\\ r&=4.2\ut{cm}\\ L&=6.0\ut{cm}\\g&=9.80665\ut{m/s^2} \end{cases} $$ $$\theta=\tan^{-1}{r\over L}$$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N-T\sin\theta\\ 0&=T\cos\theta-mg\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} T &={mg\over \cos\theta}\\ &={mg\over {L\over \sqrt{L^2+r^2}}}\\ &=\frac{21 \sqrt{149} g}{200}\\ &\approx 12.569068956048666\ut{N}\\ &\approx 13\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} N &=mg\tan\theta\\ &=mg{r\over L}\\ &=\frac{147 g}{200}\\ &=7.20788775\ut{N}\\ &\approx 7.2\ut{N}\\ \end{aligned} $$