12-35 할리데이 11판 솔루션 일반물리학

$$\begin{cases} m_A&=430\ut{kg}\\ m_B&=45.0\ut{kg}\\ \phi&=30.0\degree\\ \theta&=45.0\degree\\ \end{cases}$$ $$\begin{cases} T_x&=T\cos\phi\\ T_y&=T\sin\phi\\ \end{cases}$$ $$\begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases}$$ $$\begin{cases} 0&=N_x-T_x\\ 0&=N_y-T_y-m_Ag\\ 0&=T_xL\sin\theta-\br{T_y+m_Ag}L\cos\theta-m_Bg\br{L\over2}\cos\theta\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} T&={\cos \theta \over2\sin \br{\theta -\phi}}\br{2 m_Ag+m_Bg}\\ &={1+\sqrt3\over2}905g\\ &\approx 12.12350038855051\times10^3\ut{N}\\ &\approx 12.1\times10^3\ut{N}\\ &\approx 12.1\ut{kN}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} N_x&=T\cos\phi\\ &={3+\sqrt3\over4}905g\\ &\approx 10.499259319275254\times10^3\ut{N}\\ &\approx 10.5\times10^3\ut{N}\\ &\approx 10.5\ut{kN}\\ \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} N_y&=m_Ag+T\sin\phi\\ &={525+181\sqrt3\over4}5g\\ &\approx 10.278609694275256\times10^3\ut{N}\\ &\approx 10.3\times10^3\ut{N}\\ &\approx 10.3\ut{kN}\\ \end{aligned}$$