11판/6. 힘과 운동 II

6-25 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 3. 1. 19:14
$$\ab{a}$$ $$\Sigma \vec F = 0 \Harr \Delta \vec v=0,$$ $$ \begin{cases} \Sigma F_x &= 0\\ \Sigma F_y &= 0\\ \end{cases} $$ $$ \begin{cases} 0&=F\sin\theta-f_k\\ 0&=F_N-mg-F\cos\theta \\ f_k&=\mu_kF_N \end{cases} $$ $$ \begin{aligned} F&=\frac{\mu_k \text{mg}}{\sin\theta-\mu_k \cos\theta} \end{aligned} $$ $$\ab{b}$$ $$\Sigma \vec F = 0 \Harr \Delta \vec v=0,$$ $$ \begin{cases} \Sigma F_x &= 0\\ \Sigma F_y &= 0\\ \end{cases} $$ $$ \begin{cases} 0&=F\sin\theta-f_s\\ 0&=F_N-mg-F\cos\theta \\ f_s&\lt\mu_sF_N \end{cases} $$ $$F\sin\theta\lt \mu_s(mg+F\cos\theta),$$ $$\title{Case 1}$$ $$\tan\theta\gt\mu_s,$$ $$ F\lt\frac{\mu_s mg}{\sin\theta-\mu_s\cos\theta} $$ $$\title{Case 2}$$ $$\tan\theta\le\mu_s,$$ $$ F\gt0 $$ $$\title{To Stop Case}$$ $$\text{if } \theta\le\tan^{-1}(\mu_s),$$ $$ \text{Always Stop.} $$ $$ \begin{aligned} \therefore \theta_0&=\tan^{-1}(\mu_s) \end{aligned} $$