11-31 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=2.0\ut{kg}\\ r&=2.0\ut{m},\theta_r=45\degree\\ v&=4.0\ut{m/s},\theta_2=30\degree\\ F&=2.0\ut{N},\theta_3=30\degree\\ \end{cases} $$ $$ \begin{aligned} \vec r&=r\cos\theta_r\i+r\sin\theta_r\j\\ &=\sqrt2\i+\sqrt2\j\taag1\\ \end{aligned} $$ $$\theta_v=\theta_r-180\degree-\theta_2,$$ $$ \begin{aligned} \vec v&=v\cos\theta_v\i+r\sin\theta_v\j\\ &=-(\sqrt6+\sqrt2)\i-(\sqrt6-\sqrt2)\j\taag2\\ \end{aligned} $$ $$\theta_F=\theta_r+\theta_3,$$ $$ \begin{aligned} \vec F&=F\cos\theta_F\i+F\sin\theta_F\j\\ &={\sqrt6-\sqrt2\over2}\i+{\sqrt6+\sqrt2\over2}\j\taag3\\ \end{aligned} $$ $$\ab{a,b}$$ $$ \vec L=m(\vec r\times \vec v)=8\k\ut{kg \cdot m^2/s} $$ $$\ab{a}$$ $$L=8.0\ut{kg \cdot m^2/s}$$ $$\ab{b}$$ $$\vec L\gt0\Rarr \text{CounterClockwise}$$ $$\ab{c,d}$$ $$ \vec \tau=\vec r\times \vec F=2\k\ut{N\cdot m}$$ $$\ab{c}$$ $$\tau=8.0\ut{N\cdot m}$$ $$\ab{d}$$ $$\vec \tau\gt0\Rarr \text{CounterClockwise}$$