6-13 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=15\ut{kg}\\ \mu_s&=0.60\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{cases} \Sigma F_x&=0\\ \Sigma F_y&=0\\ \end{cases} $$ $$ \begin{cases} F_x-\mu_sF_N&=0 \\ F_N-mg+F_y&=0\\ \end{cases} $$ $$ \begin{aligned} F_x&=\mu_s(mg-F_y)\\ F\cos\theta&=\mu_s(mg-F\sin\theta)\\ F(\theta)&=\frac{\mu_smg}{\cos\theta+\mu_s\sin\theta} \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} F(0)&=\frac{\mu_smg}{\cos0+\mu_s\sin0}\\ &=9g\\ &=88.25985\ut{N}\\ &\approx 88\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} F(50\degree)&=\frac{\mu_smg}{\cos50\degree+\mu_s\sin50\degree}\\ &\approx 80.06051078695634\ut{N}\\ &\approx 80\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} F(-50\degree)&=\frac{\mu_smg}{\cos(-50\degree)+\mu_s\sin(-50\degree)}\\ &\approx 481.87046955311916\ut{N}\\ &\approx 4.8\times10^2\ut{N}\\ \end{aligned} $$