10-27 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} r_A&=8.0\ut{m}\\ F_A&=16\ut{N}\\ \theta_A&=135\degree\\ \end{cases} $$ $$ \begin{cases} r_B&=4.0\ut{m}\\ F_B&=16\ut{N}\\ \theta_B&=160\degree\\ \end{cases} $$ $$ \begin{cases} r_C&=3.0\ut{m}\\ F_C&=24\ut{N}\\ \theta_C&=90\degree\\ \end{cases} $$ $$ \begin{cases} \tau_A&=+r_AF_A\sin\theta_A\\ \tau_B&=+r_BF_B\sin\theta_B\\ \tau_C&=-r_CF_C\sin\theta_C\\ \end{cases} $$ $$ \begin{cases} \tau_A&=+64\sqrt2\ut{N\cdot m}\\ \tau_B&=+64\sin20\degree\ut{N\cdot m}\\ \tau_C&=-72\ut{N\cdot m}\\ \end{cases} $$ $$ \begin{aligned} \Sigma \tau&=\tau_A+\tau_B+\tau_C\\ &\approx 40.39895716472088\ut{N\cdot m}\\ &\approx 40\ut{N\cdot m}\\ \end{aligned} $$