11판/12. 평형과 탄성

12-5 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 6. 25. 18:25
$$ \begin{cases} mg&=413\ut{N}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_x&=0\\ \Sigma F_y&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N_x-T\sin\theta\\ 0&=N_y+T\cos\theta-mg\\ 0&=mgL\sin\br{2\theta}-TL\sin\theta\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} T &=2 mg \cos\theta\\ &=413\sqrt3\ut{N}\\ &\approx 715.3369835259463\ut{N}\\ &\approx 715\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} N_x &=mg \sin(2\theta)\\ &={413\over2}\sqrt3\ut{N}\\ &\approx 357.66849176297313\ut{N}\\ &\approx 358\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} N_y &=-mg \cos(2\theta)\\ &=-{413\over2}\ut{N}\\ &= -206.5\ut{N}\\ &\approx -207\ut{N}\\ \end{aligned} $$