6-1 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \theta&=30.0\degree\\ mg&=60.0\ut{N}\\ \mu_s&=0.250\\ \mu_k&=0.150\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{cases} \Sigma F_x &= 0\\ \Sigma F_y &= 0\\ \end{cases} $$ $$ \begin{cases} F_a+\mu_s F_N-mg\sin\theta &= 0\\ F_N-mg\cos\theta &= 0\\ \end{cases} $$ $$ \begin{aligned} F_a&=\frac{15}{2} \(4-\sqrt{3}\)\ut{N}\\ &\approx 17.00961894323342\ut{N}\\ &\approx 17.0\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{cases} \Sigma F_x &= 0\\ \Sigma F_y &= 0\\ \end{cases} $$ $$ \begin{cases} F_a-\mu_s F_N-mg\sin\theta &= 0\\ F_N-mg\cos\theta &= 0\\ \end{cases} $$ $$ \begin{aligned} F_a&=\frac{15}{2} \(4+\sqrt{3}\)\ut{N}\\ &\approx 42.99038105676657\ut{N}\\ &\approx 43.0\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{cases} \Sigma F_x &= 0\\ \Sigma F_y &= 0\\ \end{cases} $$ $$ \begin{cases} F_a-\mu_k F_N-mg\sin\theta &= 0\\ F_N-mg\cos\theta &= 0\\ \end{cases} $$ $$ \begin{aligned} F_a&=\(30+\frac{9 \sqrt{3}}{2}\)\ut{N}\\ &\approx 37.79422863405995\ut{N}\\ &\approx 37.8\ut{N}\\ \end{aligned} $$