6-43 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m_1&=2.0\ut{kg}\\ m_2&=1.0\ut{kg}\\ F&=28\ut{N},\theta=35\degree\\ \mu&=0.28 \end{cases} $$ $$\Sigma \vec F = m\vec a,$$ $$ \begin{cases} \Sigma F_{1x} &= m_1 a\\ \Sigma F_{1y} &= 0\\ \Sigma F_{2x} &= m_2 a\\ \Sigma F_{2y} &= 0\\ \end{cases} $$ $$ \begin{cases} m_1 a &= T-\mu N_1\\ 0 &= N_1-m_1g\\ m_2 a &= F\cos\theta-T-\mu N_2\\ 0 &= N_2-F\sin\theta-m_2g\\ \end{cases} $$ $$ \begin{aligned} T&=\frac{m_1 (\cos\theta-\mu\sin\theta)}{m_1+m_2}F\\ &=\frac{56}{3} \(\cos35\degree-\frac{7}{25} \sin35\degree\)\ut{N}\\ &\approx 12.292945319399713\ut{N}\\ &\approx 12\ut{N}\\ \end{aligned} $$