12-30 할리데이 11판 솔루션 일반물리학

$$\begin{cases} m_1&=40.0\ut{kg}=4m\\ m_2&=10.0\ut{kg}=m\\ L&=0.800\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases}$$ $$\begin{cases} \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases}$$ $$\begin{cases} 0&=N_A+N_B-m_1g-m_2g\\ 0&=-r_1m_1g-r_2m_2g+r_BN_B\\ \end{cases}$$ $$\begin{cases} N_A&=\cfrac{r_B-r_1}{r_B}m_1g+\cfrac{r_B-r_2}{r_B}m_2g\\ N_B&=\cfrac{r_1}{r_B}m_1g+\cfrac{r_2}{r_B}m_2g\\ \end{cases}$$ $$\begin{cases} r_B&={L\over2}\\ r_1&=r_2-{L\over4} \end{cases}$$ $$\begin{cases} N_A&=\br{7-10\cdot\cfrac{r_2}{L}}mg\\ N_B&=\br{-2+10\cdot\cfrac{r_2}{L}}mg\\ \end{cases}$$ $$\ab{a,b}$$ $$r_2={L\over2},$$ $$\begin{cases} N_A&=2mg=196.133\ut{N}\approx 196\ut{N}\\ N_B&=3mg=294.1995\ut{N}\approx 294\ut{N}\\ \end{cases}$$ $$\ab{c,d}$$ $$r_2={L\over4},$$ $$\begin{cases} N_A&={9\over2}mg=441.29925\ut{N}\approx 441\ut{N}\\ N_B&={1\over2}mg=49.03325\ut{N}\approx 49.0\ut{N}\\ \end{cases}$$ $$\ab{e}$$ $$\begin{aligned} N_A&=\br{7-10\cdot\cfrac{r_2}{L}}mg=0,\\ \end{aligned}$$ $$\begin{aligned} r_2&={7\over10}L \end{aligned}$$ $$\begin{aligned} \therefore \overline{ m_2 N_B}&=r_2-r_B\\ &={7\over10}L-{1\over2}L\\ &={L\over5}\\ &={4\over25}\ut{m}\\ &=0.160\ut{m}\\ &=16.0\ut{cm}\\ \end{aligned}$$