6-57 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m_1&=30\ut{kg}\\ m_2&=15\ut{kg}\\ \mu_s&=0.60\\ \mu_k&=0.40\\ F&=100\ut{N}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\Sigma \vec F = m\vec a,$$ $$ \begin{cases} \Sigma F_{1x} &= m_1 a_1\\ \Sigma F_{1y} &= 0\\ \Sigma F_{2x} &= m_2 a_2\\ \Sigma F_{2y} &= 0\\ \end{cases} $$ $$ \begin{cases} m_1 a_1 &= -f\\ 0 &= N_1-N_2-m_1g\\ m_2 a_2 &= -F+f\\ 0 &= N_2-m_2g\\ \end{cases} $$ $$ \begin{cases} a_1&=\cfrac{-f}{m_1}\\ a_2&=\cfrac{-F+f}{m_2} \end{cases} $$ $$\title{Static Friction ?}$$ $$ \begin{aligned} \text{if }a_1&=a_2,\\ \frac{-f}{m_1}&=\frac{-F+f}{m_2}\\ \end{aligned} $$ $$ \begin{aligned} f&=\frac{m_1}{m_1+m_2}F\\ &=\frac{200}{3}\ut{N}\\ &\approx 66.66666666666667\ut{N} \end{aligned} $$ $$ \begin{aligned} f_{s\max}&=\mu_s N_2=\mu_s m_2g\\ &=88.25985\ut{N}\\ \end{aligned} $$ $$ \therefore f \lt f_{s\max}$$ $$\title{Static Friction Possible}$$ $$\ab{a,b}$$ $$v_{2\larr1}=0 \Harr \text{Static Friction}$$ $$ \begin{aligned} \therefore \vec a&=\vec a_1=\vec a_2=-\frac{F}{m_1+m_2}\i\\ &=-\frac{20}{9}\i\ut{m/s^2}\\ &\approx -2.2222222222222223\i\ut{m/s^2}\\ &\approx -2.2\i\ut{m/s^2}\\ \end{aligned} $$ $$\title{Additional Answer - Kinetic Friction}$$ $$ \begin{cases} a_1&=\cfrac{-\mu_km_2g}{m_1}\\ a_2&=\cfrac{-F+\mu_km_2g}{m_2} \end{cases} $$ $$ \begin{cases} a_1&=-\cfrac{g}{5}\\ a_2&=\cfrac{2g}{5}-\cfrac{20}{3}\\ \end{cases} $$ $$ \begin{cases} a_1&= -1.96133\ut{m/s^2}\\ a_2&\approx -2.74400676\ut{m/s^2}\\ \end{cases} $$ $$\text{if } v_{2\larr1}\ne0 \Harr \text{Kinetic Friction}$$