12-9 할리데이 11판 솔루션 일반물리학

$$ \put \begin{cases} A : \text{Rod}\\ B : \text{Box}\\ \end{cases} $$ $$ \begin{cases} L&=1.82\ut{m}\\ W_{A}&=200\ut{N}\\ W_{B}&=300\ut{N}\\ \theta&=30.0\degree\\ T_{\max}&=412\ut{N}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N_x-T\cos\theta\\ 0&=N_y-W_A-W_B+T\sin\theta\\ 0&=-xW_B-{L\over2}W_A+TL\sin\theta\\ \end{cases} $$ $$\ab{a}$$ $$ T=\frac{1}{2\sin\theta}\left(\frac{2x }{L}W_B +W_A\right)\\ $$ $$ T\le T_{\max}\\ $$ $$ x\le\frac{2 T_{\max} \sin \theta - W_A}{2 W_B}L\\ $$ $$ \begin{aligned} x&\le{4823\over7500}\ut{m}\\ \end{aligned} $$ $$ \begin{aligned} x_{\max}&={4823\over7500}\ut{m}\\ &\approx 0.6430666666666667\ut{m}\\ &\approx 0.643\ut{m}\\ &\approx 64.3\ut{cm}\\ \end{aligned} $$ $$\ab{b}$$ $$N_x=T\cos\theta$$ $$ \begin{aligned} \max(N_x)&=T_{\max}\cos\theta\\ &=206\sqrt3\ut{N}\\ &\approx 356.8024663591887\ut{N}\\ &\approx 357\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$N_y=W_A+W_B-T\sin\theta$$ $$ \begin{aligned} \max(N_y)&=W_A+W_B-T\sin\theta\\ &=294\ut{N}\\ \end{aligned} $$