6-47 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=28.0\ut{kg}\\ F&=110\ut{N}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{aligned} F_h&=F\cos\theta\\ \end{aligned} $$ $$ \begin{aligned} 0&=F_N+F\sin\theta-mg\\ F_N&=mg-F\sin\theta\\ \end{aligned} $$ $$\ab{a}$$ $$\theta=0\degree$$ $$ \begin{aligned} F_h&=F\cos\theta\\ &=110\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$\theta=0\degree$$ $$ \begin{aligned} F_N&=mg-F\sin\theta\\ &=28g\\ &=274.5862\ut{N}\\ &\approx 275\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$\theta=30.0\degree$$ $$ \begin{aligned} F_h&=F\cos\theta\\ &=55\sqrt{3}\ut{N}\\ &\approx 95.26279441628824\ut{N}\\ &\approx 95.3\ut{N}\\ \end{aligned} $$ $$\ab{d}$$ $$\theta=30.0\degree$$ $$ \begin{aligned} F_N&=mg-F\sin\theta\\ &=28g-55\\ &=219.5862\ut{N}\\ &\approx 220\ut{N}\\ \end{aligned} $$ $$\ab{e}$$ $$\theta=60.0\degree$$ $$ \begin{aligned} F_h&=F\cos\theta\\ &=55\ut{N}\\ \end{aligned} $$ $$\ab{f}$$ $$\theta=60.0\degree$$ $$ \begin{aligned} F_N&=mg-F\sin\theta\\ &=28g-55\sqrt{3}\\ &\approx 179.32340558371178\ut{N}\\ &\approx 179\ut{N}\\ \end{aligned} $$ $$\ab{g,h,i}$$ $$\mu=0.420$$ $$\ab{g}$$ $$\theta=0\degree$$ $$ \begin{aligned} F_h&=110\ut{N}\\ f_s&=\mu F_N\\ &=115.326204\ut{N}\\ \end{aligned} $$ $$\text{Stop}$$ $$\ab{h}$$ $$\theta=30.0\degree$$ $$ \begin{aligned} F_h&=55\sqrt{3}\ut{N}\\ &\approx 95.3\ut{N}\\ f_s&=\mu F_N\\ &=92.226204\ut{N}\\ \end{aligned} $$ $$\text{Move}$$ $$\ab{i}$$ $$\theta=60.0\degree$$ $$ \begin{aligned} F_h&=55\ut{N}\\ f_s&=\mu F_N\\ &=\frac{21}{50} \(28 g+55 \sqrt{3}\)\ut{N}\\ &\approx 155.33657765484108\ut{N}\\ \end{aligned} $$ $$\text{Stop}$$