6-21 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=62\ut{kg}\\ \mu_f&=1.2\\ \mu_b&=0.80\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$\ab{b}$$ $$ \begin{cases} \Sigma F_x&=F_{Nf}-F_{Nb}=0\\ \Sigma F_y&=\mu_fF_{Nf}+\mu_bF_{Nb}-mg=0\\ \end{cases} $$ $$F_{Nf}=F_{Nb}=F_N$$ $$ \begin{aligned} F_N&=\frac{mg}{\mu_f+\mu_b}\\ &=31g\\ &=304.00615\ut{N}\\ &\approx 3.0\times10^2\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \frac{\mu_fF_N}{mg}&=\frac{\mu_f\frac{mg}{\mu_f+\mu_b}}{mg}\\ &=\frac{\mu_f}{\mu_f+\mu_b}\\ &=\frac{3}{5}\\ \end{aligned} $$