6-44 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m_A&=5.0\ut{kg}\\ m_B&=3.0\ut{kg}\\ \mu&=0.50\\ \theta&=30\degree\\ \end{cases} $$ $$\Sigma \vec F = m\vec a,$$ $$ \begin{cases} \Sigma F_{Ax} &= m_A a\\ \Sigma F_{Ay} &= 0\\ \Sigma F_{Bx} &= m_B a\\ \Sigma F_{By} &= 0\\ \end{cases} $$ $$ \begin{cases} m_A a &= m_Ag\sin\theta-T\\ 0 &= N_A-m_Ag\cos\theta\\ m_B a &= T-\mu N_B\\ 0 &= N_B-m_Bg\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} T&=\frac{ m_A m_B (\mu+\sin\theta)}{m_A+m_B}g\\ &=\frac{15}{8}g\\ &=18.38746875\ut{N}\\ &\approx 18\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} a&= \bra{\sin\theta-\frac{m_B (\sin\theta+\mu )}{m_A+m_B}}g\\ &=\frac{g}{8}\\ &=1.22583125\ut{m/s^2}\\ &\approx 1.2\ut{m/s^2}\\ \end{aligned} $$