솔루션피아

$$\begin{cases} S_{\text{wood}}&=50\times10^6\ut{N/m^2}\\\rho_{\text{wood}}&=525\ut{kg/m^3}\\A_{\text{wood}}&=d^2\\M&=430\ut{kg}\\a&=1.9\ut{m}\\b&=2.5\ut{m}\\g&=9.80665\ut{m/s^2}\end{cases}$$$$\begin{cases} \Sigma F_{x}&=0\\\Sigma F_{y}&=0\\\Sigma \tau&=0\\\end{cases}$$$$\begin{cases} 0&=N_x-T_x\\0&=N_y-Mg-mg\\0&=aT_x-bMg-{b\over2}mg\\\end{cases}$$$$ \begin{cases} N_x&={b\over2a}\br{m+2M}g..

$$\put \begin{cases} A : \text{Rod}\\B : \text{Box}\\\end{cases}$$$$\begin{cases} L&=1.82\ut{m}\\W_{A}&=200\ut{N}\\W_{B}&=300\ut{N}\\\theta&=30.0\degree\\T_{\max}&=412\ut{N}\\\end{cases}$$$$\begin{cases} \Sigma F_{x}&=0\\\Sigma F_{y}&=0\\\Sigma \tau&=0\\\end{cases}$$$$\begin{cases} 0&=N_x-T\cos\theta\\0&=N_y-W_A-W_B+T\sin\theta\\0&=-xW_B-{L\over2}W_A+TL\sin\theta\\\end{cases}$$$$\ab{a}$$..

$$\ab{a}$$$$\com(1)={L\over2}\le L-a_1,$$$$\begin{aligned}a_1\le{L\over2}\taag1\end{aligned}$$$$\therefore \max(a_1)={L\over2}$$$$\ab{b}$$$$\com(12)={L+a_1\over2}\le L-a_2,$$$$\begin{aligned}a_2\le{L-a_1\over2}\taag2\end{aligned}$$$$\therefore \max(a_2)={L\over2}\\(a_1=0)$$$$\ab{c}$$$$\com(123)={L+a_1+a_2\over2}\le L-a_3,$$$$\begin{aligned}a_3\le{L-a_1-a_2\over2}\taag3\end{aligned}$$$$\the..

$$\begin{cases} E_\text{Iron}&=200\times10^9\ut{N/m^2}\\m&=210\ut{kg}\\R&=1.20\ut{mm}\\p&=2.00\ut{mm}\\L_{A}&=2.50\ut{m}\\L_{B}&=L_{A}+p\\\Delta L_B&=\Delta L_A-p\\\end{cases}$$$${F\over A}=E\cdot{\Delta L\over L},$$$$\begin{cases} 0&=\Sigma F=T_A+T_B-mg\\T_A&=E (\pi R^2)\cdot{\Delta L_A\over L_A}\\T_B&=E (\pi R^2)\cdot{\Delta L_B\over L_B}\\\end{cases}$$$$ \begin{aligned}\Delta L_A&=\frac{{..

$$\begin{cases} m_A&=11.0\ut{kg}\\\theta&=30.0\degree\\m_B&=7.00\ut{kg}\\g&=9.80665\ut{m/s^2}\end{cases}$$$$T_A=m_Ag\sin\theta,$$$$\begin{aligned}\vec T_A&=-m_Ag\sin\theta\cos\theta\i-m_Bg\sin^2\theta\j\\&=-{11\sqrt3g\over4}\i-{11g\over4}\j\taag1\\\end{aligned}$$$$T_B=m_Bg,$$$$\begin{aligned}\vec T_B&=-m_Bg\j\\&=-7g\j\taag2\\\end{aligned}$$$$ \begin{aligned}\vec T_S&=-\br{\vec T_A+\vec T_B..

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