6-49 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} F_x&=90\ut{N}\\ m&=35\ut{kg}\\ \mu_s&=0.37\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} f_{s\max}&=\mu_s F_N=\mu_s mg\\ &=126.9961175\ut{N}\\ &\approx 1.3\times10^2\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$\text{Stop}$$ $$\ab{c}$$ $$f=F_x=90\ut{N}$$ $$\ab{d}$$ $$ \begin{cases} F_x=\mu_s F_N\\ F_N+F_y-mg=0\\ \end{cases} $$ $$ \begin{aligned} F_y&=mg-\frac{F_x}{\mu_s}\\ &\approx 99.98950675675675\ut{N}\\ &\approx 1.0\times10^2\ut{N}\\ \end{aligned} $$ $$\ab{e}$$ $$ \begin{cases} F_x+F_2=\mu_s F_N\\ F_N-mg=0\\ \end{cases} $$ $$ \begin{aligned} F_2&=\mu_s mg-F_x\\ &=\frac{259}{20}g-F_x\\ &=36.9961175\ut{N}\\ &\approx 3.7\times10\ut{N}\\ \end{aligned} $$