12-36 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \theta&=26\degree\\ L&=43\ut{m}\\ T&=2.5\ut{m}\\ W&=12\ut{m}\\ \rho&=3.2\ut{g/cm^3}=3.2\times10^{3}\ut{kg/m^2}\\ \mu_s&=0.39\\ \end{cases} $$ $$ \begin{aligned} m&=\rho V\\ &=\rho LTW\\ &=4.128\times10^6\ut{kg} \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} \Ans&=mg\sin\theta\\ &=4.128g\sin26\degree \times10^6\ut{N}\\ &\approx 17.74607553468879\times10^6\ut{N}\\ &\approx 18\times10^6\ut{N}\\ &\approx 18\ut{MN}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Ans&=\mu_smg\cos\theta\\ &=1.609920g\cos26\degree\times10^6\ut{N}\\ &\approx 14.190090268274229 \times10^6\ut{N}\\ &\approx 14 \times10^6\ut{N}\\ &\approx 14\ut{MN}\\ \end{aligned} $$ $$\ab{c}$$ $$p_{\max}=3.6\times 10^8 \ut{N/m^2},$$ $$A=6.4\ut{cm^2},$$ $${F\over A} \lt p_{\max},$$ $$ \begin{aligned} F_{\max} &= p_{\max}\cdot A \end{aligned} $$ $$ \begin{aligned} {-mg\sin\theta+\mu_smg\cos\theta}+N\cdot F_{\max} &\gt 0\\ \end{aligned} $$ $$ \begin{aligned} N&\gt{mg\sin\theta-\mu_smg\cos\theta\over {F_{\max}}}\\ &~~~~~=\frac{43}{240} g (100 \sin 26 \degree-39 \cos 26 \degree)\\ &~~~~~\approx 15.433963829924307\\ \end{aligned} $$ $$ \begin{aligned} N_{\min}&=16 \end{aligned} $$