6-39 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m_1&=2.0\ut{kg}\\ m_2&=4.0\ut{kg}\\ \theta&=40\degree\\ \mu&=0.25\\g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \Sigma \vec F = m\vec a, $$ $$ \begin{cases} \Sigma F_{1x} &= m_1 a_x\\ \Sigma F_{1y} &= 0\\ \Sigma F_{2x} &= m_2 a_x\\ \Sigma F_{2y} &= 0\\ \end{cases} $$ $$ \begin{cases} m_1 a&=T-\mu N_1\\ 0&=N_1-m_1g\\ m_2 a&=m_2g\sin\theta-T\\ 0&=N_2-m_2g\cos\theta\\ \end{cases} $$ $$ \begin{aligned} T&=(\mu+\sin\theta)\frac{m_1m_2 }{m_1+m_2}g\\ &=\frac{1+4\sin40\degree}{3}g\\ &\approx 11.673674150043333\ut{N}\\ &\approx 12\ut{N}\\ \end{aligned} $$