5-34 할리데이 11판 솔루션 일반물리학

$$\begin{cases} m&=1650\ut{kg}\\ g&=9.80665\ut{m/s^2}\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} F_1&=\frac{mg}{N}\\ &=550g\\ &=5393.6575\ut{N}\\ &\approx 5.39\ut{kN}\\ \end{aligned}$$ $$\ab{b}$$ $$\Sigma \vec F = m\vec a,$$ $$\begin{aligned} 3F_2-mg&= m a\\ F&=\frac{m}{3}(g+a)\\ &=55 (10 g+19)\\ &=6438.6575\ut{N}\\ &\approx6.44\ut{kN}\\ \end{aligned}$$