6-48 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m_Ag&=44\ut{N}\\ m_Bg&=30\ut{N}\\ \mu_s&=0.20\\ \mu_k&=0.15\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$\ab{a}$$ $$\Sigma \vec F = 0 \Harr \Delta \vec v=0,$$ $$ \begin{cases} \Sigma F_{Ax} &= 0=T-\mu_s N_A\\ \Sigma F_{Ay} &= 0=N_A-N_C-m_Ag\\ \Sigma F_{Bx} &= 0=m_Bg-T\\ \Sigma F_{Cy} &= 0=N_C-m_Cg\\ \end{cases} $$ $$ \begin{aligned} m_Cg&=\frac{m_Bg}{\mu }-m_Ag\\ &=106\ut{N}\\ &\approx 1.0\times10^2\ut{N} \end{aligned} $$ $$\ab{b}$$ $$\Sigma \vec F = m\vec a,$$ $$ \begin{cases} \Sigma F_{Ax} &= m_Aa=T-\mu_k N_A\\ \Sigma F_{Ay} &= 0=N_A-m_Ag\\ \Sigma F_{Bx} &= m_Ba=m_Bg-T\\ \end{cases} $$ $$ \begin{aligned} a&=\frac{m_Bg-\mu m_Ag}{m_Ag+m_Bg}g\\ &=\frac{117}{370}g\\ &\approx 3.1010217567567566\ut{m/s^2}\\ &\approx 3.1\ut{m/s^2}\\ \end{aligned} $$