5-40 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=8.0\ut{kg}\\ \vec F_1 &= 60\i\ut{N} \end{cases} $$ $$a_{\min}=\frac{\Sigma F_{\min}}{m},$$ $$ \begin{aligned} \abs{\vec a + \vec b}\le a+b \end{aligned} $$ $$\ab{a}$$ $$ \begin{cases} F_2 &= 40\ut{N}\\ F_3 &= 20\ut{N}\\ \end{cases} $$ $$ \begin{aligned} \abs{\vec F_2+\vec F_3} &\le 60\ut{N}\\ \therefore \vec F_2+\vec F_3&=\vec F_1 \end{aligned} $$ $$ \begin{aligned} a_{\min}&=\frac{\vec F_1+(-\vec F_1)}{m}\\ &=0\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{cases} F_2 &= 30\ut{N}\\ F_3 &= 10\ut{N}\\ \end{cases} $$ $$ \begin{aligned} \abs{\vec F_2+\vec F_3} &\le 40\ut{N}\\ \therefore \vec F_2+\vec F_3&=-40\i\ut{N} \end{aligned} $$ $$ \begin{aligned} a_{\min}&=\abs{\frac{60\i-40\i}{m}}\\ &=\abs{\frac{20\i}{8}}\\ &\approx 2.5\ut{m/s^2}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{cases} F_2 &= 40\ut{N}\\ F_3 &= 40\ut{N}\\ \end{cases} $$ $$ \begin{aligned} \abs{\vec F_2+\vec F_3} &\le 80\ut{N}\\ \therefore \vec F_2+\vec F_3&=-\vec F_1 \end{aligned} $$ $$ \begin{aligned} a_{\min}&=\frac{\vec F_1+(-\vec F_1)}{m}\\ &=0\\ \end{aligned} $$