12-26 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} mg&=60\ut{N}\\ L&=3.2\ut{m}\\ F&=50\ut{N}\\ \theta&=25\degree\\ h&=2.0\ut{m}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma F_{y}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=F+N_x-T\cos\theta\\ 0&=N_y-mg-T\sin\theta\\ 0&=hT\cos\theta-LF\\ \end{cases} $$ $$ \begin{cases} T&={F L\over h\cos\theta}\\ N_x&={F(L-h)\over h}\\ N_y&={FL\over h}\tan\theta+mg\\ \end{cases} $$ $$\ab{a}$$ $$T={F L\over h\cos\theta},$$ $$ \begin{aligned} \vec T&={F L\over h\cos\theta}\cos\theta\i+{F L\over h\cos\theta}\sin\theta\j\\ &={F L\over h}\i+{F L\over h}\tan\theta\j\\ &=80\i+80\tan25\degree\j\ut{N}\\ &\approx 80\i+37.30461265239989\j\ut{N}\\ &\approx 80\i+37\j\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{cases} N_x&={F(L-h)\over h}\\ N_y&={FL\over h}\tan\theta+mg\\ \end{cases} $$ $$ \begin{cases} N_x&=30\ut{N}\\ N_y&=\br{80\tan25\degree+60}\ut{N}\\ \end{cases} $$ $$ \begin{aligned} \vec N&={F(L-h)\over h}\i+\br{{FL\over h}\tan\theta+mg}\j\\ &=30\i+\br{80\tan25\degree+60}\j\ut{N}\\ &\approx 30\i+97.3046126523999\j\ut{N}\\ &\approx 30\i+97\j\ut{N}\\ \end{aligned} $$