12-44 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} L&=2.50\ut{m}\\ mg&=500\ut{N}\\ d&=1.50\ut{m}\\ \end{cases} $$ $$ x=\sqrt{L^2-d^2},$$ $$\ab{a}$$ $$ \begin{aligned} \Sigma \tau&=0\\ &=-\frac{x}{2}mg+LP\\ \end{aligned} $$ $$ \begin{aligned} P&=\frac{mgx}{2L}\\ &=\frac{mg}{2L}\sqrt{L^2-d^2}\\ &=200\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$\put x=\sqrt{L^2-d^2}$$ $$\sin\theta=\frac{d}{L},$$ $$ \begin{aligned} \vec P &=-P\sin\theta\i+P\cos\theta\j\\ &=-\frac{dP}{L}\i+\frac{xP}{L}j\taag1\\ \end{aligned} $$ $$ \begin{aligned} \Sigma \vec F&=0\\ &=m\vec g+\vec P+\vec N\\ \end{aligned} $$ $$ \begin{aligned} N&=\abs{\vec N}\\ &=\abs{-m\vec g-\vec P}\\ &=\abs{-m\br{-g\j}-\br{-\frac{dP}{L}\i+\frac{xP}{L}j}}\\ &=\abs{\frac{dP}{L}\i+\br{mg-\frac{xP}{L}}\j}\\ &=\sqrt{\br{\frac{dP}{L}}^2+\br{mg-\frac{xP}{L}}^2}\\ &=100\sqrt{13}\ut{N}\\ &\approx 360.5551275463989\ut{N}\\ &\approx 361\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} f&=\mu N\\ N_x&=\mu N_y\\ \end{aligned} $$ $$ \begin{aligned} \mu&= \frac{N_x}{N_y}\\ &= \frac{\frac{dP}{L}}{mg-\frac{xP}{L}}\\ &=\frac{d P}{L mg-xP}\\ &=\frac{6}{17}\\ &\approx 0.35294117647058826\\ &\approx 0.353\\ \end{aligned} $$