솔루션피아

$$\begin{cases} L&=12.0\ut{m}\\\theta&=50.0\degree\\T&=400\ut{N}\\\end{cases}$$$$\begin{cases} \Sigma F_{x}&=0\\\Sigma F_{y}&=0\\\Sigma \tau&=0\\\end{cases}$$$$\begin{cases} 0&=-T+N_x\\0&=-mg+N_y\\0&=TL\cos\theta-mg{L\over2}\sin\theta\\\end{cases}$$$$\ab{a}$$$$ \begin{aligned}mg&={2T\over\tan\theta}\\&=800\tan40\degree\ut{N}\\&\approx 671.279704941824\ut{N}\\&\approx 671\ut{N}\\\end{aligne..

$$\begin{cases} 2x&=3.44\ut{m}\\y&=25.0\ut{cm}\\mg&=3160\ut{N}\end{cases}$$$$\begin{aligned}\tan\theta&={y\over x},\\\end{aligned}$$$$\Sigma F_y=0,$$$$\begin{aligned}0&=2T_y-mg\\T_y&={mg\over2}\end{aligned}$$$$ \begin{aligned}T&={T_y\over \sin\theta}\\&=m g\cdot\frac{ x }{2 y}\sqrt{\frac{y^2}{x^2}+1}\\&=\frac{316 }{5}\sqrt{30209}\ut{N}\\&\approx 1.098462544468404\times 10^4\ut{N}\\&\approx..

$$\begin{cases} H&=59.1\ut{m}\\2R&=7.44\ut{m}\\\Delta x_{\text{top}0} &= 4.01\ut{m}\\\end{cases}$$$$\ab{a}$$$$\Delta x_\com = R,$$$$\begin{aligned}\Delta x_{\text{top}}&=2\Delta x_\com\\&=2R\\\end{aligned}$$$$\begin{aligned}\Ans&=2R-\Delta x_{\text{top}0}\\&=3.43\ut{m}\end{aligned}$$$$\ab{b}$$$$ \begin{aligned}\theta&=\sin^{-1}{2R\over H}\\&=\sin^{-1}{124\over 985}\\&\approx 0.126223229105..

$$T=900\cdot {y\over L} \taag 1$$$$\abs{F_h}=300-300\cdot{y\over L} \taag 2$$$$\begin{cases} \Sigma F_{x}&=0\\\Sigma F_{y}&=0\\\Sigma \tau&=0\\\end{cases}$$$$\begin{cases} 0&=F_a-T\cos\theta+F_h\\0&=N-T\sin\theta-mg\\0&=LT\cos\theta-yF_a\\\end{cases}$$$$\ab{a}$$$$\begin{aligned}F_a&=300\ut{N}\end{aligned}$$$$\ab{b}$$$$ \begin{aligned}\theta&=2\tan^{-1}{1\over\sqrt2}\\&\approx 1.2309594..

$$\begin{cases} \theta&=26\degree\\L&=43\ut{m}\\T&=2.5\ut{m}\\W&=12\ut{m}\\\rho&=3.2\ut{g/cm^3}=3.2\times10^{3}\ut{kg/m^2}\\\mu_s&=0.39\\\end{cases}$$$$\begin{aligned}m&=\rho V\\&=\rho LTW\\&=4.128\times10^6\ut{kg}\end{aligned}$$$$\ab{a}$$$$ \begin{aligned}\Ans&=mg\sin\theta\\&=4.128g\sin26\degree \times10^6\ut{N}\\&\approx 17.74607553468879\times10^6\ut{N}\\&\approx 18\times10^6\ut{N}\\&\ap..

$$\begin{cases} m_A&=430\ut{kg}\\m_B&=45.0\ut{kg}\\\phi&=30.0\degree\\\theta&=45.0\degree\\\end{cases}$$$$\begin{cases} T_x&=T\cos\phi\\T_y&=T\sin\phi\\\end{cases}$$$$\begin{cases} \Sigma F_{x}&=0\\\Sigma F_{y}&=0\\\Sigma \tau&=0\\\end{cases}$$$$\begin{cases} 0&=N_x-T_x\\0&=N_y-T_y-m_Ag\\0&=T_xL\sin\theta-\br{T_y+m_Ag}L\cos\theta-m_Bg\br{L\over2}\cos\theta\\\end{cases}$$$$\ab{a}$$$$ \beg..

$$\begin{cases} m_A&=10\ut{kg}\\m_B&=5.0\ut{kg}\\\theta&=30\degree\end{cases}$$$$0=\vec F_A+\vec F_B+\vec T,$$$$\begin{aligned}\tan\theta&={F_A\over F_B}\\&={\mu m_Ag\over m_Bg}\\\end{aligned}$$$$\begin{aligned}\mu&={m_B\over m_A}\tan\theta\\&={1\over2\sqrt3}\\&\approx 0.2886751345948129\\&\approx 0.29\\\end{aligned}$$

$$\begin{cases} M&=m_b+m_p=89.00\ut{kg}\\T_a&=500\ut{N}\\T_b&=700\ut{N}\\g&=9.80665\ut{m/s^2}\end{cases}$$$$\put {x\over L}=k,$$$$\Sigma \tau=0,$$$$\begin{aligned}0&=-{L\over2}m_bg-\br{x\over L}Lm_pg+LT_y \\&={m_bg}+2km_pg-2T\sin\theta \\\end{aligned}$$$$\begin{cases} 0&={m_bg}+2\br{0}m_pg-2\br{500}\sin\theta \\0&={m_bg}+2\br{1}m_pg-2\br{700}\sin\theta \\89&=m_b+m_p\end{cases}$$$$\ab{a,b,..

$$\begin{cases} \vec F_1&=8.40\i-5.70\j\ut{N}\\\vec F_2&=16.0\i+4.10\j\ut{N}\\\end{cases}$$$$\ab{a,b}$$$$\begin{aligned}\vec F_3&=-\br{\vec F_1+\vec F_2}\\&=-24.4\i+1.6\j\ut{N}\\\end{aligned}$$$$\begin{cases} F_{3x}&=-24.4\ut{N}\\F_{3y}&=1.6\ut{N}\\\end{cases}$$$$\ab{c}$$$$ \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..