솔루션피아

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

$$ \begin{cases} a&=2.5\ut{m}\\b&=3.0\ut{m}\\c&=1.0\ut{m}\\F_1&=20\ut{N}\\F_2&=10\ut{N}\\F_3&=8.0\ut{N}\\\end{cases} $$$$ \begin{cases} \Sigma F_{x}&=0\\\Sigma F_{y}&=0\\\Sigma \tau&=0\\\end{cases} $$$$ \begin{cases} 0&=F_h-F_3\\0&=F_v-F_1-F_2\\0&=dF_v-bF_2-aF_3\\\end{cases} $$$$\ab{a}$$$$ \begin{aligned}F_h&=F_3=8.0\ut{N}\\\end{aligned} $$$$\ab{b}$$$$ \begin{aligned}F_v&=F_1+F_2\\&=30\ut{N}\\\e..

$$ \begin{cases} F_h&=13.4\ut{N}\\F_v&=162.4\ut{N}\\\end{cases} $$$$ \begin{aligned} \Sigma \tau&=-{d\over3}F_t\sin45\degree+d F_v \sin10\degree+dF_h\sin80\degree\\ &=0\end{aligned} $$$$ \begin{aligned}F_t&=3\sqrt2\br{F_h\cos10\degree+F_v\sin10\degree}\\&={3\sqrt2\over5}\br{67\cos10\degree+812\sin10\degree}\\&\approx 175.63212110909666\ut{N}\\&\approx 176\ut{N}\\\end{aligned} $$

$$ \begin{cases} h&=3.00\ut{cm}\\r&=6.00\ut{cm}\\m&=0.415\ut{kg}\\g&=9.80665\ut{m/s^2}\\\end{cases} $$$$ \begin{cases} x&=\sqrt{r^2-y^2}\\y&=r-h\\\end{cases} $$$$ \begin{aligned}\Sigma \tau&=-yF+xmg=0\\F&={x\over y} mg\\&=\frac{ \sqrt{h (2r-h)}}{r-h}mg\\&={83\sqrt3\over200}g\\&=\frac{16279039 \sqrt{3}}{4000000}\ut{N}\\&\approx 7.049030661598812\ut{N}\\&\approx 7.05\ut{N}\\\end{aligned} $$

$$ \begin{cases} \theta_1&=40\degree\\\end{cases} $$$$ \begin{cases} \Sigma F_{x}&=0\\\Sigma F_{y}&=0\\\end{cases} $$$$ \begin{cases} 0&=-T_1\cos\theta_1+T_2\cos\theta_2\\0&=T_1\sin\theta_1+T_2\sin\theta_2-mg\\\end{cases} $$$$ T_2={mg\cos\theta_1\over\sin\br{\theta_1+\theta_2}}$$$$\ab{a}$$$$ \begin{aligned}\min (T_2),\\ {\sin\br{\theta_1+\theta_2}}&=1\\\theta_1+\theta_2&=90\degree\\\theta_2&=50\..

$$ \begin{cases} y&=2.1\ut{m}\\ x&=0.91\ut{m}\\ h&=0.25\ut{m}\\ m&=27\ut{kg}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{cases} p&=y-h\\ q&=h\\ \end{cases} $$ $$ \begin{cases} A&=(0,0)\\ B&=(x,0)\\ C&=(x,y)\\ D&=(0,y)\\ P&=(0,p)\\ Q&=(0,q)\\ \end{cases} $$ $$ \begin{cases} N_{Py}&={mg\over2}\\ N_{Qy}&={mg\over2}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma \tau&=0\\ \end{cases}..

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