솔루션피아

$$\begin{cases} \vec F_1&=8.40\i-5.70\j\ut{N}\\\vec F_2&=16.0\i+4.10\j\ut{N}\\\end{cases}$$$$\ab{a,b}$$$$\begin{aligned}\vec F_3&=-\br{\vec F_1+\vec F_2}\\&=-24.4\i+1.6\j\ut{N}\\\end{aligned}$$$$\begin{cases} F_{3x}&=-24.4\ut{N}\\F_{3y}&=1.6\ut{N}\\\end{cases}$$$$\ab{c}$$$$ \begin{aligned}\theta&=\tan^{-1}{1.6\over-24.4}\\&\approx 3.0761126286663285\ut{rad}\\&\approx 3.08\ut{rad}\\\end{ali..

$$\begin{cases} m_1&=85.0\ut{kg}\\m_2&=30.0\ut{kg}\\L_2&=2.00\ut{m}\\m_3&=20.0\ut{kg}\\d&=0.500\ut{m}\\L_1&=L_2+2d\\\end{cases}$$$$\begin{cases} \Sigma F_{1y}&=0\\\Sigma \tau_1&=0\\\Sigma F_{2y}&=0\\\Sigma \tau_2&=0\\\end{cases}$$$$ \begin{cases} 0&=T_{1L}+T_{1R}-T_{2L}-T_{2R}-m_1g\\0&=-dT_{2L}-{L_1\over2}m_1g-\br{L_1-d}T_{2R}+L_1T_{1R} \\0&=T_{2L}+T_{2R}-m_2g-m_3g\\0&=-dm_3g-{L_2\over2}m_2g..

$$\begin{cases} m_1&=40.0\ut{kg}=4m\\m_2&=10.0\ut{kg}=m\\L&=0.800\ut{m}\\g&=9.80665\ut{m/s^2}\end{cases}$$$$\begin{cases} \Sigma F_{y}&=0\\\Sigma \tau&=0\\\end{cases}$$$$\begin{cases} 0&=N_A+N_B-m_1g-m_2g\\0&=-r_1m_1g-r_2m_2g+r_BN_B\\\end{cases}$$$$\begin{cases} N_A&=\cfrac{r_B-r_1}{r_B}m_1g+\cfrac{r_B-r_2}{r_B}m_2g\\N_B&=\cfrac{r_1}{r_B}m_1g+\cfrac{r_2}{r_B}m_2g\\\end{cases}$$$$ \begin{..

$$\begin{cases} L&=6.10\ut{m}\\mg&=510\ut{N}\\h&=4.00\ut{m}\\\theta_0&=70\degree\\\end{cases}$$$$\begin{cases} r_f&=h\\r_{\text{floor}}\tan\theta&=h\\\br{r_{mg}+\frac{L}{2}\cos\theta}\tan\theta&=h\end{cases}$$$$\begin{cases} N_{\text{edge }x}&=N_{\text{edge}}\sin\theta\\N_{\text{edge }y}&=N_{\text{edge}}\cos\theta\\\end{cases}$$$$\begin{aligned}f&=\mu N_{\text{floor}},\end{aligned}$$$$ \..