솔루션피아

$$ \begin{cases} 2R&=4.8\ut{cm}\\L&=5.3\ut{cm}\\m&=1500\ut{kg}\\G&=3.0\times10^{10}\ut{N/m^2}\\g&=9.80665\ut{m/s^2}\\\end{cases} $$$$\ab{a}$$$$ \begin{aligned}F&=mg\\&=14709.975\ut{N}\\&\approx 1.5\times10^4\ut{N}\\&\approx 15\ut{kN}\\\end{aligned} $$$$\ab{b}$$$${F\over A}=G{\Delta x\over L},$$$$ \begin{aligned}\Delta x&={FL\over AG}\\&={Lmg\over \pi R^2 G}\\&=\frac{53 g}{11520000 \pi }\ut{m}\\&..

$$ \begin{cases} \theta_1&=51.0\degree\\\theta_2&=66.0\degree\\m&=704\ut{kg}\\g&=9.80665\ut{m/s^2}\end{cases} $$$$\ab{a}$$$$ \begin{aligned}T_A&=mg\\&=6.9038816\times10^3\ut{N}\\&\approx 6.90\times10^3\ut{N}\\&\approx 6.90\ut{kN}\\\end{aligned} $$$$\ab{b,c}$$$$ \begin{cases} \theta&=\theta_1+\theta_2-90\degree&=27.0\degree\\\phi&=90\degree-\theta_2&=24.0\degree\\\end{cases} $$$$ \begin{aligned}0..

$$ \begin{cases} r_A&=7.0\ut{cm}\\r_B&=4.0\ut{cm}\\R&=4.0\ut{cm}\\F&=220\ut{N}\\\end{cases} $$$$ \begin{cases} 0&=F-N_A+N_B\\0&=RF-r_AN_A-r_BN_B\end{cases} $$$$ \begin{cases} N_A&=\cfrac{R+r_B}{2r_B}F\\N_B&=\cfrac{R-r_B}{2r_B}F\\\end{cases} $$$$\ab{a,b}$$$$ \begin{cases} N_A&=220\ut{N}\\N_B&=0\\\end{cases} $$

$$ \put \begin{cases} A : \text{Top Ball}\\B : \text{Bottom Ball}\\L : \text{Left Wall}\\R : \text{Right Wall}\\G : \text{Ground}\\\end{cases} $$$$ \begin{cases} \Sigma F_{Ax}&=0\\\Sigma F_{Ay}&=0\\\Sigma F_{Bx}&=0\\\Sigma F_{By}&=0\\\end{cases} $$$$ \begin{cases} 0&=N_{ABx}-N_R\\0&=N_{ABy}-mg\\0&=N_L-N_{ABx}\\0&=N_G-N_{ABy}-mg\\\end{cases} $$$$\theta=45\degree\Rarr N_{ABx}=N_{ABy}$$$$\ab{a,b,c,..

$$ \begin{cases} L&=2.00\ut{m}\\m&=78.0\ut{kg}\\d_h&=3.00\ut{m}\\d_v&=4.00\ut{m}\\g&=9.80665\ut{m/s^2}\end{cases} $$$$x_1=d_h-{L\over2},$$$$\theta=\tan^{-1}{d_h\over d_v}$$$$ \begin{cases} \Sigma F_{x}&=0\\\Sigma F_{y}&=0\\\Sigma \tau&=0\\\end{cases} $$$$ \begin{cases} 0&=N_x-T\sin\theta\\0&=N_y+T\cos\theta-mg\\0&=-x_1 mg+Td_h\cos\theta\\\end{cases} $$$$\ab{a}$$$$ \begin{aligned}T&= {x_1 mg\over..

$$ \begin{cases} \Sigma F_{y}&=0\\\Sigma \tau&=0\\\end{cases} $$$$ \begin{cases} 0&=3F-mg\\0&=x\cdot 2F+L F-{L\over2}mg\\\end{cases} $$$$ \begin{aligned}x&={L\over4}\end{aligned} $$

$$ \begin{cases} m&=1.05\ut{kg}\\ r&=4.2\ut{cm}\\ L&=6.0\ut{cm}\\g&=9.80665\ut{m/s^2} \end{cases} $$ $$\theta=\tan^{-1}{r\over L}$$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N-T\sin\theta\\ 0&=T\cos\theta-mg\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} T &={mg\over \cos\theta}\\ &={mg\over {L\over \sqrt{L^2+r^2}}}\\ &=\frac{21 \sqrt{149} g}{200}\\ ..

$$ \begin{cases} m&=13\ut{kg}\\ \theta&=55\degree\\ \phi&=25\degree\\g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \end{cases} $$ $$ \begin{cases} 0&=T\cos\phi-mg\sin\theta\\ 0&=-T\sin\phi-mg\cos\theta+N\\ \end{cases} $$ $$ \begin{aligned} T&={\sin\theta\over\cos\phi}mg \\ &\approx 1.1522662348306987\times10^2\ut{N}\\ &\approx 1.2\times10^2\ut{N}\\ \end..

$$ \begin{cases} d&=60\ut{m}\\L&=150\ut{m}\\H&=7.2\ut{m}\\z&=5.8\ut{m}\\A&=885\ut{cm^2}\\\rho&=2.8\ut{g/cm^3}\\S_{\text{steel}}&=400\times10^6\ut{N/m^2}\end{cases} $$$$\ab{a}$$$$ \begin{aligned}\Ans&=mg\\&=\rho V g\\&=\rho L d z g\\&=1.4616\times10^8\ut{N}\\&\approx 1.5\times10^8\ut{N}\\&\approx 150\ut{MN}\\\end{aligned} $$$$\ab{b}$$$$ \begin{aligned}{F \over NA}&={1\over2}S_{\text{steel}}\\\end..

(풀이자 주:문제의 맥락이 경첩과 잠금쇠를 잇는 1차원 막대문제로 풀라는 의도로 보입니다. 즉, 정사각형이라는 조건은 무시합니다. 만일 아니라면 답은 달라집니다.)$$ \put \begin{cases} A : \text{Hinge}\\B : \text{Lock}\\\end{cases} $$$$ \begin{cases} m&=15\ut{kg}\\L&=0.91\ut{m}\\g&=9.80665\ut{m/s^2}\end{cases} $$$$x_\com={L\over2}-0.1=0.355\ut{m}$$$$ \begin{cases} \Sigma F_{x}&=0\\\Sigma \tau&=0\\\end{cases} $$$$ \begin{cases} 0&=N_A+N_B-mg\\0&=-x_\com mg+LN_B\\\end..