6-50 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} mg&=45.0\ut{N}\\ \theta&=15.0\degree\\ \mu_s&=0.500\\ \mu_k&=0.340\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{aligned} F_N&=mg\cos\theta\\ &=\frac{45 \(1+\sqrt{3}\)}{2\sqrt{2}}\ut{N}\\ \end{aligned} $$ $$ \begin{aligned} f_{s\max}&=\mu_s F_N\\ &=\frac{45\(1+\sqrt{3}\)}{4\sqrt{2}}\ut{N}\\ &\approx 21.733331091504034\ut{N}\\ \end{aligned} $$ $$ \begin{aligned} f_k&=\mu_k F_N\\ &=\frac{153\(1+\sqrt{3}\)}{20\sqrt{2}}\ut{N}\\ &\approx 14.778665142222744\ut{N}\\ &\approx 14.8\ut{N}\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} \Sigma F&=mg\sin\theta+P_a\\ &=\frac{45\(\sqrt{3}-1\)}{2\sqrt{2}}+3\ut{N}\\ &\approx 14.646857029613432\ut{N}\\ &\approx 14.6\ut{N} \end{aligned} $$ $$\Sigma F\lt f_{s\max},$$ $$ \begin{aligned} f&=\Sigma F\\ &\approx 14.6\ut{N} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Sigma F&=mg\sin\theta+P_b\\ &=\frac{45\(\sqrt{3}-1\)}{2\sqrt{2}}+9\ut{N}\\ &\approx 20.6468570296134322\ut{N}\\ &\approx 20.6\ut{N} \end{aligned} $$ $$\Sigma F\lt f_{s\max},$$ $$ \begin{aligned} f&=\Sigma F\\ &\approx 20.6\ut{N} \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \Sigma F&=mg\sin\theta+P_c\\ &=\frac{45\(\sqrt{3}-1\)}{2\sqrt{2}}+13\ut{N}\\ &\approx 24.646857029613432\ut{N}\\ \end{aligned} $$ $$\Sigma F\gt f_{s\max},$$ $$ \begin{aligned} f&=f_k\\ &\approx 14.8\ut{N}\\ \end{aligned} $$