솔루션피아

$$ \begin{cases} mg&=413\ut{N}\\\end{cases} $$$$ \begin{cases} \Sigma F_x&=0\\\Sigma F_y&=0\\\Sigma \tau&=0\\\end{cases} $$$$ \begin{cases} 0&=N_x-T\sin\theta\\0&=N_y+T\cos\theta-mg\\0&=mgL\sin\br{2\theta}-TL\sin\theta\\\end{cases} $$$$\ab{a}$$$$ \begin{aligned}T&=2 mg \cos\theta\\&=413\sqrt3\ut{N}\\&\approx 715.3369835259463\ut{N}\\&\approx 715\ut{N}\\\end{aligned} $$$$\ab{b}$$$$ \begin{aligned..

(모든 블록의 y축 두께는 문제의 결론에 영향을 주지 않으므로 무시합니다.) $$\ab{a}$$ $$ \put \begin{cases} T : \text{Top}\\ L : \text{Middle Left}\\ R : \text{Middle Right}\\ B : \text{Bottom}\\ l : \text{length} \end{cases} $$ $$ \begin{cases} \Sigma F_T&=0=N_T-mg\\ \Sigma F_{L}&=0=N_{L}-N_T-mg\\ \Sigma F_{R}&=0=N_{R}-mg\\ \Sigma F_B&=0=N_B-N_L-N_R-mg\\ \end{cases} $$ $$ \begin{cases} N_T&=mg\\ N_{L}&=2mg\\ N_{R}&=mg\\ N_B&=..

$$ \begin{cases} m_A&=85\ut{kg}\\m_B&=10\ut{kg}\\M&=\Sigma m= 95\ut{kg}\\\end{cases} $$$$ \begin{cases} \Sigma F_{x}&=0\\\Sigma F_{y}&=0\\\Sigma \tau_{D}&=0\\\end{cases} $$$$ \begin{cases} 0&=N_C-f_D\\0&=N_D-Mg\\0&=\br{x_{AD}}m_Ag+\br{x_{BD}}m_Bg-\br{y_{CD}}N_C\\\end{cases} $$$$ \begin{cases} x_{AD}&={L_{AD}\over L_{CD}}x_{OD}\\x_{BD}&={1\over 2}x_{OD}\\y_{CD}&=\sqrt{{L_{CD}}^2-{x_{OD}}^2}\\\end..

$$ \begin{cases} m&=6.40\ut{kg}\\g&=9.80665\ut{m/s^2}\\\end{cases} $$$$ \begin{cases} F&=T_1\\T_2&=2T_1\\T_3&=2T_2\\0&=T_1+T_2+T_3-mg\\T&=2T_3\\\end{cases} $$$$ \begin{aligned}T&={8\over7}mg\\&=71.72864\ut{N}\\&\approx 71.7\ut{N}\\\end{aligned} $$

$$ \begin{cases} m_Ag&=32\ut{N}\\m_Bg&=45\ut{N}\\\phi&=35\degree\\\end{cases} $$$$ \begin{cases} \Sigma \vec F_A&=0\\\Sigma \vec F_B&=0\\\end{cases} $$$$ \begin{cases} \Sigma F_{Ax}&=0\\\Sigma F_{Bx}&=0\\\Sigma F_{Ay}&=0\\\Sigma F_{By}&=0\\\end{cases} $$$$ \begin{cases} 0&=T_2-T_1\sin\phi\\0&=-T_2+T_3\sin\theta\\0&=T_1\cos\phi-m_Ag\\0&=T_3\cos\theta-m_Bg\\\end{cases} $$$$\ab{a}$$$$ \begin{aligne..