12-47 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m_A&=46.0\ut{kg}\\ L&=5.00\ut{m}\\ m_B&=250\ut{kg}\\ d&=0.540\ut{m}\\ \end{cases} $$ $$ \begin{cases} x_1&=d\\ x_2&=L-d\\ x_A&=L\\ x_B&={L\over2}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N_1+N_2-m_Ag-m_Bg\\ 0&=x_1N_1+x_2N_2-x_Am_Ag-x_Bm_Bg\\ \end{cases} $$ $$\ab{a,b}$$ $$ \begin{cases} N_1&=\cfrac{ m_A (x_A-x_2)+m_B (x_B-x_2)}{x_1-x_2}g\\ N_2&=\cfrac{ m_A (x_1-x_A)+m_B (x_1-x_B)}{x_1-x_2}g \end{cases} $$ $$ \begin{cases} N_1&=\cfrac{ 2 d (m_A+m_B)-L m_B}{4 d-2 L}g\\ N_2&=\cfrac{ 2 d (m_A+m_B)-L (2 m_A+m_B)}{4 d-2 L}g \end{cases} $$ $$ \begin{cases} \vec N_1&=\frac{11629}{98}g\j\\ \vec N_2&=\frac{17379}{98} g\j \end{cases} $$ $$ \begin{cases} \vec N_1&\approx 1.163689110714286\j\times10^3\ut{N}\\ \vec N_2&\approx 1.739079289285714\j\times10^3\ut{N} \end{cases} $$ $$ \begin{cases} \vec N_1&\approx 1.16\j\times10^3\ut{N}\\ \vec N_2&\approx 1.74\j\times10^3\ut{N} \end{cases} $$ $$ \begin{cases} \vec N_1&\approx 1.16\j\ut{kN}\\ \vec N_2&\approx 1.74\j\ut{kN} \end{cases} $$