12-35 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m_A&=430\ut{kg}\\ m_B&=45.0\ut{kg}\\ \phi&=30.0\degree\\ \theta&=45.0\degree\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} T&=m_Ag\\ &=430g\\ &=4.2168595\times10^3\ut{N}\\ &\approx 4.22\times10^3\ut{N}\\ &\approx 4.22\ut{kN}\\ \end{aligned} $$ $$\ab{b}$$ $$\Sigma F_{x}=0,$$ $$ \begin{aligned} N_x-T_x&=0 \end{aligned} $$ $$ \begin{aligned} N_x&=T\cos\phi\\ &=430g\cos\phi\\ &=215 \sqrt{3} g\\ &\approx 3.651907451189746\times10^3\ut{N}\\ &\approx 3.65\times10^3\ut{N}\\ &\approx 3.65\ut{kN}\\ \end{aligned} $$ $$\ab{c}$$ $$\Sigma F_{y}=0,$$ $$ \begin{aligned} 0&=N_y-T_{Ly}-T_{Ry}-m_Bg\\ \end{aligned} $$ $$ \begin{aligned} N_y &=T_{Ly}+T_{Ry}+m_Bg\\ &=T\sin\phi+T+m_Bg\\ &=430g\sin\phi+430g+45g\\ &=690g\\ &=6.7665885\times10^3\ut{N}\\ &\approx 6.77\times10^3\ut{N}\\ &\approx 6.77\ut{kN}\\ \end{aligned} $$