12-41 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} L_A&=2.40\ut{m}\\ m_A&=54.0\ut{kg}\\ m_B&=68.0\ut{kg}\\ d&=1.60\ut{m}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$ \begin{cases} \Sigma \vec F_A&=0\\ \Sigma \vec F_B&=0\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{Ax}&=0\\ \Sigma F_{Bx}&=0\\ \Sigma F_{Ay}&=0\\ \Sigma F_{By}&=0\\ \Sigma \tau_A&=0\\ \Sigma \tau_B&=0\\ \end{cases} $$ $$\put \vec F=\text{Force }A \lrarr B$$ $$ \begin{cases} 0&=N_{Ax}+F_x\\ 0&=N_{Bx}-F_x\\ 0&=N_{Ay}-m_Ag+F_y\\ 0&=N_{By}-m_Bg-F_y\\ 0&=-{L_A\over2}m_Ag+L_AF_y\\ 0&=-{L_A\over2}m_Bg+dF_x \end{cases} $$ $$ \begin{cases} N_{Ax}&=-\cfrac{L_A}{2d}m_Bg\\ N_{Ay}&=\cfrac{1}{2}m_Ag\\ N_{Bx}&=\cfrac{L_A}{2d}m_Bg\\ N_{By}&=\cfrac{m_A+2m_B}{2}g\\ F_x&=\cfrac{L_A}{2d}m_Bg\\ F_y&=\cfrac{1}{2}m_Ag\\ \end{cases} $$ $$ \begin{cases} \vec N_A&=-\cfrac{L_A}{2d}m_Bg\i+\cfrac{1}{2}m_Ag\j\\ \vec N_B&=\cfrac{L_A}{2d}m_Bg\i+\cfrac{m_A+2m_B}{2}g\j\\ \vec F&=\cfrac{L_A}{2d}m_Bg\i+\cfrac{1}{2}m_A\j \\ \end{cases} $$ $$ \begin{cases} \vec N_A&=-51g\i+27g\j\\ \vec N_B&=51g\i+95g\j\\ \vec F&=51g\i +27g\j \\ \end{cases} $$ $$ \begin{cases} \vec N_A&=-500.13915\i+264.77955\j\ut{N}\\ \vec N_B&=500.13915\i+931.63175\j\ut{N}\\ \vec F&=500.13915\i +264.77955\j\ut{N}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \vec N_A&=-500.13915\i+264.77955\j\ut{N}\\ &\approx \br{-5.00\i+2.65\j}\times10^2\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \vec F&=500.13915\i +264.77955\j\ut{N}\\ &\approx \br{5.00\i +2.65\j}\times10^2\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \vec N_B&=500.13915\i+931.63175\j\ut{N}\\ &\approx \br{5.00\i+9.32\j}\times10^2\ut{N}\\ \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} -\vec F&=-500.13915\i -264.77955\j\ut{N}\\ &\approx \br{-5.00\i -2.65\j}\times10^2\ut{N}\\ \end{aligned} $$