12-21 할리데이 11판 솔루션 일반물리학

$$\begin{cases} L&=2.00\ut{m}\\ m&=78.0\ut{kg}\\ d_h&=3.00\ut{m}\\ d_v&=4.00\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases}$$ $$x_1=d_h-{L\over2},$$ $$\theta=\tan^{-1}{d_h\over d_v}$$ $$\begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases}$$ $$\begin{cases} 0&=N_x-T\sin\theta\\ 0&=N_y+T\cos\theta-mg\\ 0&=-x_1 mg+Td_h\cos\theta\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} T &= {x_1 mg\over d_h\cos\theta}\\ &={5\over6 }mg\\ &=65g\\ &=637.43225\ut{N}\\ &\approx 637\ut{N}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} N_x &=T\sin\theta\\ &= {x_1 mg\over d_h}\tan\theta\\ &={1\over2}mg\\ &=39g\\ &=382.45935\ut{N}\\ &\approx 382\ut{N}\\ \end{aligned}$$ $$\ab{c}$$ $$N_x\gt0 \Rarr\text{Right}$$ $$\ab{d}$$ $$\begin{aligned} N_y &=mg-T\cos\theta\\ &=mg\br{1-{x_1\over d_h}}\\ &={1\over3}mg\\ &=26g\\ &=254.9729\ut{N}\\ &\approx 255\ut{N}\\ \end{aligned}$$ $$\ab{c}$$ $$N_y\gt0 \Rarr\text{Up}$$